Haritha

MCQs on States of Matter for NEET

by

NEET Chemistry For States Of Matter Multiple Choice Question

Question 1. Intermolecular forces are forces of attraction and repulsion between interacting particles that will include

  1. Dipole-dipole forces
  2. Dipole-induced dipole forces
  3. Hydrogen bonding
  4. Covalent bonding
  5. Dispersion forces.

Choose the most appropriate answer from the options given below:

  1. 1, 2, 3, 5 are correct.
  2. 1, 3, 4, 5 are correct.
  3. 2, 3, 4, and 5 are correct.
  4. 1, 2, 3, 4 are correct.

Answer: 1. 1, 2, 3, 5 are correct.

Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms and molecules).

This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together, i.e., covalent bonds.

Question 2. Dipole-induced dipole interactions are present in which of the following pairs?

  1. HCl and He atoms
  2. SiF4 and He atoms
  3. H2O and alcohol
  4. Cl2 and CCl4

Answer: 1. HCl and He atoms

HCI is polar (μ ≠ 0) and He is non-polar (μ = 0) giving dipole-induced dipole interactions.

Question 3. Which one of the following is the correct order of interactions?

  1. Covalent < hydrogen bonding < van der Waals < dipole-dipole
  2. van der Waals < hydrogen bonding < dipole- dipole < covalent
  3. van der Waals < dipole-dipole < hydrogen bonding < covalent
  4. Dipole-dipole < van der Waals’ < hydrogen bonding < covalent.

Answer: 2. van der Waals < hydrogen bonding < dipole- dipole < covalent

The strength of interaction follows the order: van der Waals < hydrogen-bonding < dipole-dipole < covalent. It is so because the bond length of the H-bond is larger than that of a covalent bond.

Also covaient bond is strongest because, tire greater the extent of overlapping, the stronger the bond formed.

Read and Learn More NEET MCQs with Answers

Question 4. Which of the following statements is wrong for gases?

  1. Confined gas exerts uniform pressure on the walls of its container in all directions.
  2. The volume of the gas is equal to the volume of the container confining the gas.
  3. Gases do not have a definite shape and volume.
  4. The mass of a gas cannot be determined by weighing a container in which it is enclosed.

Answer: 4. Mass of a gas cannot be determined by weighing a container in which it is enclosed.

Mass of the gas = Mass of the cylinder including gas – Mass of empty cylinder.

So, the mass of a gas can be determined by weighing the container in which it is enclosed.

Thus, the statement (4) is wrong or gases

MCQs on States of Matter for NEET

Question 5. Which of the following options is the correct graphical representation of Boyle Law?

States Of Matter Graphical rtepresentation of Boyles Law

Answer: 4

The graph in option (4) correctly the Boyle’s law.

Question 6. Choose the correct option for a graphical representation of Boyle’s law, which shows a graph of pressure vs volume of a gas at different temperatures.

States Of Matter Pressure vs Volume Of A Gas At Different Temperatures

Answer: 1

According to Boyle’s law, \(P \propto \frac{1}{V}\) (at constan T)

The graph between P and V is a hyperbola

Question 7. At 25°C and 730 mm pressure, 380 mL of dry oxygen was collected. If the temperature is constant, what volume will the oxygen occupy at 760 mm pressure?

  1. 569 mL
  2. 365 mL
  3. 265 mL
  4. 621 mL

Answer: 2. 365 mL

At 25°C and 730 mm pressure, 380 mL of dry oxygen was collected. If the temperature is constant

V1 = 380 mL, P1 =730 mm, V2 = ? P2= 760 mm.

From Boyle’s Iaw, P1V1 = P2V2

⇒ \(V_2=\frac{P_1 V_1}{P_2}=\frac{730 \times 380}{760}=365 \mathrm{~mL}\)

Question 8. Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by a definite fraction of its volume at

  1. 0°C
  2. Its critical temperature
  3. Absolute zero
  4. It’s Boyle temperature.

Answer: 1. 0°C

According to Charles’ law which states that the volume of the given mass of a gas increases or decreases by 1/27.3 of its volume at 0°C for a degree rise or fall of temperature at constant pressure.

∴ \(V_t=V_0\left(1+\frac{t}{273}\right) \text { at constant } P \text { and } n \text {. }\)

Question 9. Which one is not the correct mathematical equation for Daltons’ Law of partial pressure? Here p = total pressure of the gaseous mixture

  1. \(p=p_1+p_2+p_3\)
  2. \(p=n_1 \frac{R T}{V}+n_2 \frac{R T}{V}+n_3 \frac{R T}{V}\)
  3. \(p_i=x_i p\), where \(p_i=\) partial pressure of \(i^{\text {th }}\) gas \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture
  4. \(p_i=x_i p_i^{\circ}\), where \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture, \(p_i^{\circ}=\) pressure of \(i^{\text {th }}\) gas in pure state

Answer: 4. \(p_i=x_i p_i^{\circ}\), where \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture, \(p_i^{\circ}=\) pressure of \(i^{\text {th }}\) gas in pure state

Question 10. A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O1 gas is behaving ideally). The pressure inside the flask in the bar is (Given R = 0.0831 L bar K-1 mol-1)

  1. 2.5
  2. 498.6
  3. 49.8
  4. 4.9

Answer: 4. 4.9

A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O1 gas is behaving ideally).

V = 10 L, Mass of O1 = 64 g, T = 300 K

According to the ideal gas equation, PV = nRT

P = \(\frac{n R T}{V}=\frac{64}{32} \times \frac{0.0831 \times 300}{10} \Rightarrow P=4.986 \mathrm{bar}\)

Question 11. Choose the correct option for the total pressure (in atm) in a mixture of 4 g O2 and 2 g H2 confined in a total volume of one litre at 0° C is [Given R = 0.082 L atm mol-1 K-1, T = 273 K]

  1. 26.02
  2. 2.518
  3. 2.602
  4. 25.18

Answer: 4. 25.18

Applying PV = \(\frac{w}{M} R T\)

∴ \(P_{\mathrm{O}_2}=\frac{4}{32} \times \frac{0.0821 \times 273}{1}=2.80 \mathrm{~atm}\)

∴ \(P_{\mathrm{H}_2}=\frac{2}{2} \times \frac{0.0821 \times 273}{1}=22.4 \mathrm{~atm}\)

Now, according to Dalton’s law, \(P_{\text {total }}=P_{\mathrm{O}_2}+P_{\mathrm{H}_2}=2.80+22.4=25.2 \mathrm{~atm}\)

Question 12. A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2 is [Use atomic masses (in g mol-1): N2= 14, Ar = 40]

  1. 9 bar
  2. 12 bar
  3. 15 bar
  4. 18 bar.

Answer: 3. 15 bar

A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar

Number of moles of \(\mathrm{N}_2=\frac{7}{28}\) = 0.25 mol

Number of moles of Ar = \(\frac{8}{40}\) = 0.2 mol

Mole fraction of \(\mathrm{N}_2=\frac{0.25}{0.25+0.2}=\frac{0.25}{0.45}=0.55\)

Partial pressure of N2 gas = mole fraction x total pressure = 0.55 x 27 = 14.85 ≈ 15 bar

Question 13. The volume occupied by 1.8 g of water vapour at 374°C and 1 bar pressure will be [Use R = 0.083 bar L K-1 mol-1]

  1. 96.66 L
  2. 55.87 L
  3. 3.10 L
  4. 5.37 L

Answer: 4. 5.37 L

m = \(1.8 \mathrm{~g} \Rightarrow n=\frac{m}{M}=\frac{1.8}{18}=0.1 \mathrm{~mol}\)

T = \(374^{\circ} \mathrm{C}=647 \mathrm{~K}, P=1\) bar, R=\(0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)

V = \(\frac{n R T}{P}=\frac{0.1 \times 0.083 \times 647}{1}=5.37 \mathrm{~L}\)

Question 14. Equal moles of hydrogen and oxygen gases are placed in a container with a pinhole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

  1. 3/8
  2. 1/2
  3. 1/8
  4. 1/4

Answer: 3. 1/8

Equal moles of hydrogen and oxygen gases are placed in a container with a pinhole through which both can escape.

Let the number of moles of each gas = x

Fraction of hydrogen escaped \(=\frac{1}{2} x\)

⇒ \(\frac{r_{\mathrm{O}_2}}{r_{\mathrm{H}_2}}=\sqrt{\frac{M_{\mathrm{H}_2}}{M_{\mathrm{O}_2}}} \Rightarrow \frac{n_{\mathrm{O}_2} / t}{\frac{x}{2} / t}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}\)

⇒ \(\frac{n_{\mathrm{O}_2} / t}{\frac{x}{2} / t}=\frac{1}{4} \Rightarrow n_{\mathrm{O}_2}=\frac{1}{8} x\)

Hence, the fraction of oxygen escaped = 1/8

Question 15. What is the density of N2 gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K-1 mol-1)

  1. 1.40 g/mL
  2. 2.81 g/mL
  3. 3.41 g/mL
  4. 0.29 g/mL

Answer: 3. 3.41 g/mL

PV=nRT

PV = \(\frac{w}{M} R T\left[n=\frac{\text { Weight of the gas taken }(W)}{\text { Mol. mass of gas }(M)}\right]\)

P = \(\frac{w}{M} \times \frac{R T}{V} ; P=\frac{d R T}{M} \quad\left[\text { Density }=\frac{\text { Mass }}{\text { Volume }}\right]\)

d = \(\frac{P M}{R T}=\frac{5 \times 28}{0.0821 \times 500}=3.41 \mathrm{~g} / \mathrm{mL}\)

Question 16. 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively to effuse through a pinhole under similar conditions. If the molecular mass of gas B is 36, the molecular mass of gas A will be

  1. 96
  2. 128
  3. 32
  4. 64

Answer: None

50 mL of each gas A and of gas B takes 150 and 200 seconds respectively to effuse through a pinhole under similar conditions.

According to Graham’s law of diffusion, \(\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{M_2}{M_1}}, r_A=\frac{V_A}{T_A}, \quad r_B=\frac{V_B}{T_B}\)

⇒ \(\frac{V_A / T_A}{V_B / T_B}=\sqrt{\frac{M_B}{M_A}}\)

⇒ \(V_A=V_B, T_A=150 \mathrm{sec}, T_B=200 \mathrm{sec}, M_B=36, M_A=?\)

⇒ \(\frac{T_B}{T_A}=\sqrt{\frac{M_B}{M_A}} \Rightarrow \frac{200}{150}=\sqrt{\frac{36}{M_A}}\)

⇒ \(\frac{4}{3}=\sqrt{\frac{36}{M_A}} \text { or } \frac{4 \times 4}{3 \times 3}=\frac{36}{M_A} \text { or } M_A=\frac{36}{4 \times 4} \times 3 \times 3=20.25\)

Question 17. A certain gas takes three times as long to effuse out as helium. Its molecular mass will be

  1. 27 u
  2. 36 u
  3. 64 u
  4. 9u

Answer: 2. 36 u

According to Graham’s law of diffusion, \(r \propto \frac{1}{\sqrt{d}} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)

Rate of diffusion = \(\frac{\text { Volume of gas diffused }(V)}{\text { Time taken }(t)}\)

∴ \(\frac{V_1 / t_1}{V_2 / t_2}=\sqrt{\frac{M_2}{M_1}}\)

If the same volume of two gases diffuses, then \(V_1=V_2\)

∴ \(\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}\)

Here \(t_2=3 t_1, M_1=4 \mathrm{u}, M_2=\)?

∴ \(\frac{3 t_1}{t_1}=\sqrt{\frac{M_2}{4}} \Rightarrow 3=\sqrt{\frac{M_2}{4}} \Rightarrow 9=\frac{M_2}{4} \Rightarrow M_2=36 \mathrm{u}\)

Question 18. Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. The molecular mass of B will be

  1. 50.00 u
  2. 12.25 u
  3. 6.50 u
  4. 25.00 u

Answer: 2. 12.25 u

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively.

We know that \(\frac{r_A}{r_B}=\frac{V / t_A}{V / t_B}=\sqrt{\frac{M_B}{M_A}}\)

⇒ \(\frac{t_B}{t_A}=\sqrt{\frac{M_B}{M_A}} \Rightarrow \frac{10}{20}=\sqrt{\frac{M_B}{49}}\)

⇒ \(\left(\frac{10}{20}\right)^2=\frac{M_B}{49} \Rightarrow \frac{100}{400}=\frac{M_B}{49}\)

⇒ \(M_B=\frac{49 \times 100}{400}=12.25 \mathrm{u}\)

Question 19. A gaseous mixture was prepared by taking equal moles of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is

  1. 0.5 atm
  2. 0.8 atm
  3. 0.9 atm
  4. 1 atm

Answer: 1. 0.5 atm

A gaseous mixture was prepared by taking equal moles of CO and N2. If the total pressure of the mixture was found 1 atmosphere

⇒ \(p_{\mathrm{CO}}+p_{\mathrm{N}_2}=1 \mathrm{~atm}\)

⇒ \(2 p_{\mathrm{N}_2}=1\)

⇒ \(p_{\mathrm{N}_2}=\frac{1}{2}=0.5 \mathrm{~atm}\)

Question 20. A bubble of air is underwater at a temperature 15°C and a pressure of 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

  1. The volume will become greater by a factor of 1.6.
  2. The volume will become greater by a factor of 1.1.
  3. The volume will become smaller by a factor of 0.70.
  4. The volume will become greater by a factor of 2.5

Answer: 1. Volume will become greater by a factor of 1.6.

A bubble of air is underwater at a temperature 15°C and a pressure of 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar,

From ideal gas equation, \(V \propto \frac{T}{P}\)

Given \(T_1=15+273=288 \mathrm{~K}, P_1=1.5\) bar \(T_2=25+273=298 \mathrm{~K}, P_2=1\) bar \(V_1 \propto \frac{288}{1.5}\)

i.e., \(V_1 \propto 192\) and \(V_2 \propto \frac{298}{1}\)

∴ \(\frac{V_2}{V_1}=\frac{298}{192}=1.55 \approx 1.6\)

Question 21. The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J K-1 mol-1)

  1. 275216 Pa
  2. 13409 Pa
  3. 41648 Pa
  4. 31684 Pa

Answer: 3. 41648 Pa

Given, the mass of CH4, w = 6g

Volume of CH4, V= 0.03 m³

T = 129°C = 129 +273 = 402 K, R = 8.314 J K-1 mol-1

Molecular mass of CHr, M = 12.01 + 4 x 1.01 = 16.05

PV = nRT = \(\frac{w}{M} R T\)

∴ P = \(\frac{w}{M} \frac{R T}{V}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}\)

= \(41647.7 \mathrm{~Pa} \approx 41648 \mathrm{~Pa}\)

Question 22. Which of the following mixtures of gases does not obey the Daltorx law of partial pressure?

  1. Cl2 and SO2
  2. CO2 and He
  3. O2 and CO2
  4. N2 and O2

Asnwer: 1. Cl2 and SO2

∴ \(\mathrm{Cl}_2+\mathrm{SO}_2\) \(\underrightarrow{\text { Sunlight }}\) \(\underset{\mathrm{Sulphuryl chloride}}{\mathrm{SO}_2 \mathrm{Cl}_2}\)

Dalton’s law of partial pressure is applicable only in those cases where gases are non-reacting. As Cl2 and SO2 react to form SO2Cl2 this law is not obeyed in a given case.

Question 23. At what temperature, the rate of effusion of N2 would be 1.625 times the rate of SO2 at 50°C?

  1. 373°C
  2. 620°C
  3. 100°C
  4. 173°C

Answer: 3. 10°C

∴ \(r_1=1.625 r_2\) and \(T_2=50^{\circ} \mathrm{C}=323 \mathrm{~K}\)

We know that \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1} \times \frac{T_1}{T_2}}\)

or \(1.625=\sqrt{\frac{64}{28} \times \frac{T_1}{323}}\) or \(T_1=\frac{(1.625)^2 \times 28 \times 323}{64}=373.15 \mathrm{~K}=100.15^{\circ} \mathrm{C}\)

Question 24. 50 mL of hydrogen diffuses out through a small hole of a vessel, in 20 minutes. The time taken by 40 mL of oxygen to diffuse out is

  1. 32 minutes
  2. 64 minutes
  3. 8 minutes
  4. 12 minutes

Answer: 2. 64 minutes

Volume of hydrogen = 50 mI; Time for diffusion (t) = 20 min and volume of oxygen = 40 mI.

Rate of diffusion of hydrogen (r1) = 50/20 =2.5 mL/min

Rate of diffusion of oxygen (r2) = 40/t mL/min

Since the molecular mass of hydrogen (M1) = 2 and that of oxygen (M2) = 32

Therefore \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \Rightarrow \frac{2.5}{40 / t}=\sqrt{\frac{32}{2}} \Rightarrow \frac{t}{16}=4 \Rightarrow t=64 \text { minutes }\)

Question 25. Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of I atm but also a concentration of 1-mole litre-1? (R = 0.082 litre atm mol-1 deg-1)

  1. At STP
  2. When V = 22.4 litres
  3. When T=12K
  4. Impossible under any conditions

Answer: 3. When T= 12K

PV = nRT or \(P=\frac{n}{V} R T=C R T\)

Hence, \(1=1 \times 0.082 \times T \Rightarrow T=\frac{1}{0.082}=12 \mathrm{~K}\)

Question 26. The correct value of the gas constant ‘R’ is close to

  1. 0.082 litre-atmosphere K
  2. 0.082 litre-atmosphere K-1 mol-1
  3. 0.082 litre-atmosphere-1 K mol-1
  4. 0.082 litre-1 atmosphere-1 K mol.

Answer: 2. 0.082 litre-atmosphere K-1 mol-1

Question 27. Select one correct statement. In the gas equation, PV = nRT

  1. n is the number of molecules of a gas
  2. V denotes the volume of one mole of the gas
  3. n moles of the gas have a volume of V
  4. P is the pressure of the gas when only one mole of gas is present.

Answer: 3. n moles of the gas have a volume of V

In the ideal gas equation, PV=nRT

n moles of the gas have volume V

Question 28. At constant temperature, in a given mass of an ideal gas

  1. The ratio of pressure and volume always remains constant
  2. Volume always remains constant
  3. Pressure always remains constant
  4. The product of pressure and volume always remains constant.

Answer: 4. The product of pressure and volume always remains constant.

According to Boyle’s law at a constant temperature, \(P \propto \frac{1}{V} \text { or } P V=\text { constant }\)

Question 29. If P, V M, T and R are pressure, volume, molar mass, temperature and gas constant respectively, then for an ideal gas, the density is given by

  1. \(\frac{R T}{P M}\)
  2. \(\frac{P}{R T}\)
  3. \(\frac{M}{V}\)
  4. \(\frac{P M}{R T}\)

Answer: 4. \(\frac{P M}{R T}\)

Ideal gas equation is, \(P V=n R T=\frac{m}{M} R T\)

or \(P M=\frac{m}{V} R T=d R T\) [here d= density]

⇒ d = \(\frac{P M}{R T}\)

Question 30. The correct gas equation is

  1. \(\frac{V_1 T_2}{P_1}=\frac{V_2 T_1}{P_2}\)
  2. \(\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)
  3. \(\frac{P_1 T_1}{V_1}=\frac{P_2 V_2}{T_2}\)
  4. \(\frac{V_1 V_2}{T_1 T_2}=P_1 P_2\)

Answer: 2. \(\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)

∴ \(\frac{P V}{T}=\text { constant or } \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \Rightarrow \frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)

Question 31. By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?

  1. 2.0
  2. 2.8
  3. 4.0
  4. 1.4

Answer: 4. 1.4

Average velocity = \(\sqrt{\frac{8 R T}{\pi M}}\)

When T becomes 2T then average velocity = \(\sqrt{\frac{8 R(2 T)}{\pi M}}\)

i.e., \(\sqrt{2}\) or 1.41 times increase.

Question 32. The temperature of a gas is raised from 27°C to 927°C. The root mean square speed of the gas

  1. Remains same
  2. \(\text { gets } \sqrt{\frac{927}{27}} \text { times }\)
  3. Gets halved
  4. Gets doubled.

Answer: 4. Gets doubled.

T1 =27°C = 300 K and T2 = 927°C= 1200 K

We know that root means square speed (v) ∝ √T.

Therefore root mean square speed of the gas, when its temperature is raised = \(\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=2 \text { times }\)

Question 33. The ratio among most probable velocity, mean velocity and root mean square velocity is given by

  1. \(1: 2: 3\)
  2. \(1: \sqrt{2}: \sqrt{3}\)
  3. \(\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}\)
  4. \(\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}\)

Answer: 4. \(\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}\)

Most probable velocity, \(\left(u_{m p}\right)=\sqrt{\frac{2 R T}{M}}\)

Mean velocity, \((\bar{v})=\sqrt{\frac{8 R T}{\pi M}}\)

Root mean square velocity, \(\left(u_{\text {r.m.s }}\right)=\sqrt{\frac{3 R T}{M}}\)

∴ \(u_{m p}: \bar{v}: u_{r, m, s}=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}}=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}\)

Question 34. The root mean square velocity at STP for the gases H2, N2, O2 and HBr are in the order

  1. \(\mathrm{H}_2<\mathrm{N}_2<\mathrm{O}_2<\mathrm{HBr}\)
  2. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{N}_2<\mathrm{H}_2\)
  3. \(\mathrm{H}_2<\mathrm{N}_2=\mathrm{O}_2<\mathrm{HBr}\)
  4. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{H}_2<\mathrm{N}_2\)

Answer: 2. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{N}_2<\mathrm{H}_2\)

PV = \(\frac{1}{3} m n u^2=\frac{1}{3} M u^2\) or \(u=\sqrt{3 P V / M}\),

At STP, \(u \propto \sqrt{\frac{1}{M}}\) and molecular masses of \(\mathrm{H}_2, \mathrm{~N}_2, \mathrm{O}_2\) and \(\mathrm{HBr}\) are 2, 28, 32 and 81 .

Question 35. The root mean square velocity of a gas molecule is proportional to

  1. \(m^{1 / 2}\)
  2. \(m^0\)
  3. \(m^{-1 / 2}\)
  4. m

Answer: 3. \(m^{-1 / 2}\)

PV = \(\frac{1}{3} m N u^2\), here u= root mean square velocity.

Now \(u^2=\frac{3 P V}{m N}\) or \(u \propto \frac{1}{\sqrt{m}}\)

Question 36. The energy absorbed by each molecule (A2) of a substance is 4.4 x 10-19 J and bond energy per molecule is 4.0 x 10-19 J. The kinetic energy of the molecule per atom will be

  1. \(2.2 \times 10^{-19} \mathrm{~J}\)
  2. \(2.0 \times 10^{-19} \mathrm{~J}\)
  3. \(4.0 \times 10^{-20} \mathrm{~J}\)
  4. \(2.0 \times 10^{-20} \mathrm{~J}\)

Answer: 4. \(2.0 \times 10^{-20} \mathrm{~J}\)

The energy absorbed by each molecule (A2) of a substance is 4.4 x 10-19 J and bond energy per molecule is 4.0 x 10-19 J.

Energy absorbed by each molecule = 4.4 x 10-19 J

The energy required to break the bond = 4.0 x 10-19 J

The remaining energy is converted to kinetic energy

= (4.4 x 10-19 – 4.0 x 10-19 )J = 0.4 x 10-19 J  per molecule

∴ Kinetic energy per atom = 0.2 x 10-19 J or 2 x 10-20 J

Question 37. If a gas expands at a constant temperature, it indicates that

  1. The kinetic energy of molecules remains the same
  2. Number of the molecules of gas increases
  3. The kinetic energy of molecules decreases
  4. The pressure of the gas increases.

Answer: 1. Kinetic energy of molecules remains the same

The average translationalK’E. of one molecule of an ideal gas will be given by \(E_t=\frac{K. E .}{N_A}=\frac{3 / 2 R T}{N_A}=\frac{3}{2} k T\)

where \(R / N_A=\) Boltzmann constant

i.e. \(E_t \propto T\)

Question 38. The average molar kinetic energy of CO and N2 at the same temperature is

  1. KE1 = KE2
  2. KE1 > KE2
  3. KE1 < KE2
  4. Can’t say anything. Both volumes are not given.

Answer: 1. KE1 = KE2

K.E = \(\frac{3}{2}\)RT (for one mole of a gas)

As temperatures are the same and KE is independent of molecular mass KE1 = KE2.

Question 39. The average kinetic energy of an ideal gas, per molecule in S.I. units, at 25°C will be

  1. \(6.17 \times 10^{-20} \mathrm{~J}\)
  2. \(7.16 \times 10^{-20} \mathrm{~J}\)
  3. \(61.7 \times 10^{-20} \mathrm{~J}\)
  4. \(6.17 \times 10^{-21} \mathrm{~J}\)

Answer: 4. \(6.17 \times 10^{-21} \mathrm{~J}\)

Temperature (T) = 25°C=298K.

Therefore, K.E. per molecule = \(\frac{3 R T}{2 N_A}=\frac{3 \times 8.314 \times 298}{2 \times\left(6.02 \times 10^{23}\right)}=6.17 \times 10^{-21} \mathrm{~J}\)

Question 40. At STP, 0.50 mol H2 gas and 1.0 mol He gas

  1. Have equal average kinetic energies
  2. Have equal molecular speeds
  3. Occupy equal volumes
  4. Have equal effusion rates.

Answer: 1. Have equal average kinetic energies

Because average kinetic energy depends only on temperature K.E = \(\frac{3}{2}\) K T

Question 41. Internal energy and pressure of a gas per unit volume are related as

  1. \(P=\frac{2}{3} E\)
  2. \(P=\frac{3}{2} E\)
  3. \(P=\frac{1}{2} E\)
  4. \(P=2 E\)

Answer: 1. \(P=\frac{2}{3} E\)

PV = \(\frac{1}{3} m n u^2=\frac{1}{3} M u^2\)

= \(\frac{2}{3} \cdot \frac{1}{2} M u^2=\frac{2}{3} E\) (because \(\frac{1}{2} M u^2=E\))

or P = \(\frac{2}{2} E\) per unit volume.

Question 42. A closed flask contains water in all its three states solid, liquid and vapour at 0°C. In this situation, the average kinetic energy of water molecules will be

  1. The greatest in all the three states
  2. The greatest in vapour state
  3. The greatest in the liquid state
  4. The greatest in the solid state.

Answer: 2. The greatest vapour state

Velocity and hence average KE. of water molecules is maximum in the gaseous state.

Question 43. Which is not true in the case of an ideal gas?

  1. It cannot be converted into a liquid.
  2. There is no interaction between the molecules.
  3. All molecules of the gas move at the same speed.
  4. At a given temperature, PV is proportional to the amount of the gas.

Answer: 3. All molecules of the gas move at the same speed.

Molecules in an ideal gas range with different speeds. Due to collision between the particles their speed changes.

Question 44. A gas at 350 K and 15 bar has a molar volume 20 per cent smaller than that of an ideal gas under the same conditions. Tire correct option about the gas and its compressibility factor (Z) is

  1. Z < 1 and repulsive forces are dominant
  2. Z > 1 and attractive forces are dominant
  3. Z > 1 and repulsive forces are dominant
  4. Z < 1 and attractive forces are dominant.

Answer: 4. Z < 1 and attractive forces are dominant.

A gas at 350 K and 15 bar has a molar volume 20 per cent smaller than that of an ideal gas under the same conditions.

∴ \(V_{\text {ideal }}=V, V_{\text {real }}=V-0.2 \mathrm{~V}=0.8 \mathrm{~V}\)

Z = \(\frac{V_{\text {real }}}{V_{\text {ideal }}}=0.8\)

Question 45. A gas such as carbon monoxide would be most likely to obey the ideal gas law at

  1. Low temperatures and high pressures
  2. High temperatures and high pressures
  3. Low temperatures and low pressures
  4. High temperatures and low pressures.

Answer: 4. High temperatures and low pressures

Real gases show ideal gas behaviour at high temperatures and low pressures.

Question 46. Maximum deviation from ideal gas is expected from

  1. \(\mathrm{CH}_{4(g)}\)
  2. \(\mathrm{NH}_{3(g)}\)
  3. \(\mathrm{H}_{2(g)}\)
  4. \(\mathrm{N}_{2(g)}\)

Answer: 2. \(\mathrm{NH}_{3(g)}\)

NH3 is a polar molecule, thus more attractive forces between NH3 molecules.

Question 47. For real gases, van der Waals’ equation is written as \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T \) where a and b are van der Waals’ constants. Two sets of gases are

  1. \(\mathrm{O}_2, \mathrm{CO}_2, \mathrm{H}_2\) and \(\mathrm{He}\)
  2. \(\mathrm{CH}_4, \mathrm{O}_2\) and \(\mathrm{H}_2\)

The gases given in set-1 in increasing order of 2 and gases given in set-2 in decreasing order of 1 are arranged below. Select the correct order from the following:

    1. \(\mathrm{He}<\mathrm{H}_2<\mathrm{CO}_2<\mathrm{O}_2\)
    2. \(\mathrm{CH}_4>\mathrm{H}_2>\mathrm{O}_2\)
    1.  \(\mathrm{O}_2<\mathrm{He}<\mathrm{H}_2<\mathrm{CO}_2\)
    2. \(\mathrm{H}_2>\mathrm{O}_2>\mathrm{CH}_4\)
    1. \(\mathrm{H}_2<\mathrm{He}<\mathrm{O}_2<\mathrm{CO}_2\)
    2. \(\mathrm{CH}_4>\mathrm{O}_2>\mathrm{H}_2\)
    1. \(\mathrm{H}_2<\mathrm{O}_2<\mathrm{He}<\mathrm{CO}_2\)
    2. \(\mathrm{O}_2>\mathrm{CH}_4>\mathrm{H}_2\)

Answer: 3. \(\mathrm{H}_2<\mathrm{He}<\mathrm{O}_2<\mathrm{CO}_2\) and \(\mathrm{CH}_4>\mathrm{O}_2>\mathrm{H}_2\)

Question 48. van der Waals’ real gas, acts as an ideal gas, at which conditions?

  1. High temperature, low pressure
  2. Low temperature, high pressure
  3. High temperature, high pressure
  4. Low temperature, low pressure

Answer: 1. High temperature, low pressure

At low-pressure and high-temperature van der Waals, real gas acts as an ideal gas and is observed to obey PV = nRT relation.

At very low pressure when the gas volume is quite large the space occupied by the molecules themselves becomes negligible comparatively and because the molecules are then far apart, the force of mutual attraction becomes too feeble, the real gas would satisfy the postulates of kinetic theory.

As the temperature is raised, the volume of the gas increases and

we can consider \(\left(P+\frac{n^2 a}{V^2}\right)\) term as P and at low pressure (V – nb) term as V. \(\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T\) (van der Waals equation)

This equation becomes PV = nRT

This is an ideal gas equation

Question 49. When is deviation more in the behaviour of a gas from the ideal gas equation PV = nRT.

  1. At high temperatures and low pressure
  2. At low temperatures and high pressure
  3. At high temperatures and high pressure
  4. At low temperatures and low pressure

Answer: 2. At low temperatures and high pressure

At low temperatures and high pressure, there is a deviation from the ideal behaviour in gases.

Question 50. A gas is said to behave like an ideal gas when the relation PV/T = constant. When do you expect a real gas to behave like an ideal gas?

  1. When the temperature is low.
  2. When both the temperature and pressure are low.
  3. When both the temperature and pressure are high
  4. When the temperature is high and pressure is low.

Answer: 4. When the temperature is high and pressure is low.

A gas is said to behave like an ideal gas when the relation PV/T = constant.

At high temperatures and low pressure, the effect of a/V² and b is negligible.

As we know, PV = nRT (Ideal gas equation)

PV = \(R T \text { or } \frac{P V}{R T}=1\)

∴ Z=1 [Z is compressibility factor]

Hence gas shows ideal behaviour.

Question 51. In van der Waals’ equation of state for a non-ideal gas, the term that accounts for intermolecular forces is

  1. \((V-b)\)
  2. \((R T)^{-1}\)
  3. \(\left(P+\frac{a}{V^2}\right)\)
  4. R T

Answer: 3. \(\left(P+\frac{a}{V^2}\right)\)

Vander Waals’ equation for 1 mole is \(\left(P+\frac{a}{V^2}\right)\)(V-b)=RT

Here, \(\left(P+\frac{a}{V^2}\right)\) represents the intermolecular forces and (V-b) is the correct volume.

Question 52. Given van der Waals’ constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquified?

  1. NH3
  2. H2
  3. O2
  4. HO2

Answer: 1. NH3

Given van der Waals’ constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59

van der Waals constant signifies the intermolecular forces of attraction between the particles of gas.

So, the higher the value of ‘a’, the easier will be the liquefaction of gas.

Question 53. An ideal gas, obeying the kinetic theory of gases cannot be liquefied, because

  1. It solidifies before becoming a liquid
  2. Forces acting between its molecules are negligible
  3. Its critical temperature is above 0°c
  4. Its molecules are relatively small in size.

Answer: 2. Forces acting between its molecules are negligible

A gas can only be liquefied, if some forces of attraction are acting in its molecules. According to kinetic theory, an ideal gas is devoid of the force of attraction in its molecules, therefore it cannot be liquefied.

Question 54. The beans are cooked earlier in a pressure cooker because

  1. The boiling point increases with increasing pressure
  2. The boiling point decreases with increasing pressure
  3. The extra pressure of the cooker softens the beans
  4. Internal energy is not lost while cooking in a pressure cooker

Answer: 1. Boiling point increases with increasing pressure

The more the pressure, the greater the boiling point.

 

NEET Chemistry MCQs On Classification Of Elements And Periodicity In Properties

by

NEET Chemistry For Classification Of Elements And Periodicity In Properties Multiple Choice Questions

Question 1. The IUPAC name of an element with atomic number 119 is

  1. Ununennium
  2. Unnilennium
  3. Unununnium
  4. Ununoctium.

Answer: 1. Ununennium

The IUPAC name of an element with atomic number 119 is ununennium.

Question 2. Identify the incorrect match.

Classification Of Elements And Periodicity In Properties Periodic Trends Properties

  1. 1-A
  2. 2-B
  3. 3-C
  4. 4-D

Answer: 4. 4-D

Unnilunium – Mendelevium ⇒ (1)-(A)

Unniltrium – Lawrencium ⇒ (2)-(B)

Unnilhexium – Seaborgium ⇒ (3)-(C)

Unununnium – Roentgenium ⇒ (4)-(D)

Question 3. The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?

  1. Carbon family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^2\)
  2. Oxygen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^4\)
  3. Nitrogen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^6\)
  4. Halogen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^5\)

Answer: 1. Carbon family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^2\)

The electronic configuration of the element with Z = 114 (Flerovium) is \([\mathrm{Rn}] 5 f^4 6 d^{10} 7 s^2 7 p^2\)

Hence, it belongs to the carbon family which has the same outer electronic configuration.

Read and Learn More NEET MCQs with Answers

Question 4. An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d3 4s2 you will place it in

  1. Fifth group
  2. Fifteenth group
  3. Second group
  4. Third group.

Answer: 1. Fifth Group

The electronic configuration of an atom: 1s2 2s2 2p6 3s2 3p6 3d3 4s2

In the configuration, the last electron of the atom is filled in d-subshell as 3d3. Thus, this element belongs to the d-block of the periodic table with group number VB or 5.

NEET Chemistry MCQs On Classification Of Elements And Periodicity In Properties

Question 5. The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p3 What is the atomic number of the element, which is just below the above element in the periodic table?

  1. 36
  2. 49
  3. 33
  4. 34

Answer: 3. 33

The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p3 What is the atomic number of the element

The atomic number of the given element is 15 and it belongs to group 15. Therefore the atomic number of the element below the above element = 15 + 18 = 33

Question 6. If the atomic number of an element is 33, it will be placed in the periodic table in the

  1. First group
  2. Third group
  3. Fifth group
  4. Seventh group.

Answer: 3. Fifth Group

The electronic configuration of an element with Z = 33 is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3

Hence, it lies in the VA or the 15th group

Question 7. The electronic configuration of four elements is given below. Which elements do not belong to the same family as others?

  1. \([\mathrm{Xe}] 4 f^{14} 5 d^{10} 4 s^2\)
  2. \([\mathrm{Kr}] 4 d^{10} 5 s^2\)
  3. \([\mathrm{Ne}] 3 s^2 3 p^5\)
  4. \([\mathrm{Ar}] 3 d^{10} 4 s^2\)

Answer: 3. \([\mathrm{Ne}] 3 s^2 3 p^5\)

Elements (1), (2) and (4) belong to the same group since each one of them has two electrons in the valence shell. In contrast, element (3) has seven electrons in the valence shell, and hence it lies in another group.

Question 8. The element expected to form the largest ion to achieve the nearest noble gas configuration is

  1. N
  2. Na
  3. O
  4. F

Answer: 1. N

N3-, O2-, F– and Nat have 10 electrons each, hence these are isoelectronic. For isoelectronic species, the size of the species decreases as the nuclear charge increases. Hence, the size decreases as N3- > O2- > F– > Na+

Hence, among the given elements, nitrogen is expected to form the largest ion to achieve the nearest noble gas configuration.

Question 9. For the second-period elements, the correct increasing order of first ionization enthalpy is

  1. Li < Be < B < C < O < N < F < Ne
  2. Li < Be < B < C < N < O < F < Ne
  3. Li < B < Be < C < O < N < F < Ne
  4. Li < B < Be < C < N < O < F < Ne

Answer: 3. Li < B < Be < C < O < N < F < Ne

As we move across a period, ionisation enthalpy increases, because of increased nuclear charge and decrease in atomic radii. However, abnormal values are observed for Be, N and Ne due to the extra stability of half-filled and fully-filled orbitals.

Thus, the actual order is, Li<B<Be<C<O<N<F<Ne.

Question 10. Match the oxide given in column A with its property given in column B

Classification Of Elements And Periodicity In Properties Nomenclature Elements

Which of the following options has all the correct pairs?

  1. (1)-B, (2)-A, (3)-D, (4)-C
  2. (1)-C, (2)-B, (3)-A, (4)-D
  3. (1)-A, (2)-D, (3)-B, (4)-C
  4. (1)-B, (2)-D, (3)-A, (4)-C

Answer: 4. (1)-B, (2)-D, (3)-A, (4)-C

Na2O – Basic oxide, AI2O3 – Amphoteric oxide, N2O – Neutrai oxide, Cl2O7 – Acidic oxide.

Question 11. Which of the following oxides is most acidic in nature?

  1. MgO
  2. BeO
  3. BaO
  4. CaO

Answer: 2. BeO

In metals, on moving down the group, metallic character increases, so basic nature increases hence most acidic will be BeO

Question 12. In which of the following options the order of arrangement does not agree with the variation of the property indicated against it?

  1. I < Br < Cl < F (increasing electron gain enthalpy)
  2. Li < Na < K < Rb (increasing metallic radius)
  3. Al3+ < Mg2+ < Na+ < F– (increasing ionic size)
  4. B < C < N < O (increasing first ionisation enthalpy)

Answer: 1. I < Br < Cl < F (increasing electron gain enthalpy) and 4. B < C < N < O (increasing first ionisation enthalpy)

The correct order of increasing negative electron gain enthalpy is: I < Br < F < CI due to electron-electron repulsion in small-sized F atom and the correct order of increasing first ionisation enthalpy is B < C < O < N due to extra stability of half-filled orbitals in N-atom.

Question 13. The formation of the oxide ion, O2-(g) from the oxygen atom requires first an exothermic and then an endothermic step as shown below:

  • \(\mathrm{O}_{(g)}+e^{-} \rightarrow \mathrm{O}_{(g)}^{-} ; \Delta_f H^{\circ}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
  • \(\mathrm{O}_{(g)}^{-}+e^{-} \rightarrow \mathrm{O}_{(g)}^{2-} ; \Delta_f H^{\circ}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Thus, the process of formation of O2- in the gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that,

  1. O– ion has a comparatively smaller size than the oxygen atom
  2. Oxygen is more electronegative
  3. The addition of an electron in oxygen results in a larger size of ion
  4. Electron repulsion outweighs the stability gained by achieving noble gas configuration.

Answer: 4. Electron repulsion outweighs the stability gained by achieving noble gas configuration.

Question 14. Which of the following orders of ionic radii is correctly represented?

  1. \(\mathrm{H}^{-}>\mathrm{H}^{+}>\mathrm{H}\)
  2. \(\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}\)
  3. \(\mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+}\)
  4. \(\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{N}^{3-}\)

Answer: None

Cations lose electrons and are smaller in size than the parent atom, whereas anions add electrons and are larger in size than the parent atom. Hence, the order is H–>H>H+.

For isoelectronic species, the ionic radii decreases with increase in atomic number i.e., orclear charge. Hence, the correct orders are \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+} \text {and } \mathrm{N}^{3-}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\)

Question 15. Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O?

  1. Al < Ca < O < C <F
  2. Al < O < C < Ca < F
  3. C < F < O < Al < Ca
  4. Ca < Al < C < O < F

Answer: 4. Ca < Al < C < O < F

Electron gain enthalpy becomes less negative from top to bottom in a group while it becomes more negative from left to right within a period.

Question 16. In which of the following arrangements the given sequence is not strictly according to the property indicated against it?

  1. HF < HCl < HBr < HI: increasing acidic strength
  2. H2O < H2S < H2Se < H2Te: increasing pKa values
  3. NH3 < PH3 < AsH3 < SbH3: increasing acidic character
  4. CO2 < SiO2 < SnO2 < PbO2: increasing oxidising power

Answer: 2. H2O < H2S < H2Se < H2Te: increasing pKa values

The acidic strength of hydrides increases with an increase in molecular mass.

Thus, the order of acidic strength is

HF<HCl<HBr<HI

H2O < H2S < H2Se < H2Te

NH3 < PH3 < AsH3 < SbH3

and as acidic strength increases, pKa decreases. Thus order of \(\mathrm{p} K_p, \mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{Te}\)

Question 17. Identify the wrong statement in the following.

  1. Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.
  2. Amongst isoelectronic species, the greater the negative charge on the anion, the larger the ionic radius.
  3. The atomic radius of the elements increases as one moves down the first group of the periodic table.
  4. Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.

Answer: 1. Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.

As positive charge on the cation increases, effective nuclear charge increases. Thus, atomic size decreases.

Question 18. What is the value of electron gain enthalpy of Na+ if IE1 of Na = 5.1 eV?

  1. -5.1 eV
  2. -10.2 eV
  3. +2.55 eV
  4. +10.2 eV

Answer: 1. -5.1 eV

⇒ \(\mathrm{Na} \rightarrow \mathrm{Na}^{+}+e^{-} ; \Delta H=5.1 \mathrm{eV}\)

⇒ \(\mathrm{Na}^{+}+e^{-} \rightarrow \mathrm{Na} ; \Delta H=-5.1 \mathrm{eV}\)

Question 19. Which of the following oxides is amphoteric?

  1. SnO2
  2. CaO
  3. SiO2
  4. CO2

Answer: 1. SnO2

SnO2 reacts with acid as well as base. So, SnO2 is an amphoteric oxide.

⇒ \(\mathrm{SnO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{SnO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_3+\mathrm{H}_2 \mathrm{O}\)

Question 20. The correct order of the decreasing ionic radii among the following isoelectronic species is

  1. \(\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}\)
  2. \(\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}\)
  3. \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)
  4. \(\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Cl}^{-}>\mathrm{S}^{2-}\)

Answer: 3 \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)

S2- > Cl– > K+ > Ca2+

Among isoelectronic species, ionic radii increase with the increase in negative charge. This happens because the effective nuclear charge (Zeff) decreases.

Similarly, ionic radii decrease with an increase in positive charge as Zeff increases.

Question 21. Which of the following represents the correct order of increasing electron gain enthalpy with a negative sign for the elements O, S, F and Cl?

  1. Cl < F < O < S
  2. O < S < F < Cl
  3. F < S < O < Cl
  4. S < O < Cl < F

Answer: 2. O < S < F < Cl

Cl atom has the highest electron affinity in the periodic table. F being a member of group 17 has higher electron gain enthalpy than S which belongs to group 16. This in turn is higher than the electron affinity of O atom. Thus, Cl >F>S>O

It is worth noting that the electron gain enthalpy of oxygen and fluorine, the members of the second period, have less negative values of electron gain enthalpy than the corresponding elements sulphur and chlorine of the third period.

This is due to small size of the atoms of oxygen and fluorine. As a result, there is a strong inter-electronic repulsion when an extra electron is added to these atoms, i.e., the electron density is high and the addition of an extra electron is not easy.

Question 22. Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is

  1. Mg < Ca < Cl < P
  2. Cl < P < Mg < Ca
  3. P < Cl < Ca < Mg
  4. Ca < Mg < P < Cl

Answer: 2. Cl < P < Mg < Ca

The atomic radii decrease on moving from left to right in a period, the order of sizes for Cl, P and Mg is Cl < P < Mg. Down the group size increases. thus, overall order is Cl<P<Mg<Ca.

Question 23. Among the following which one has the highest cation to anion size ratio?

  1. CsI
  2. CsF
  3. LiF
  4. NaF

Answer: 2. CsF

The cation-to-anion size ratio will be maximum when the cation is of the largest size and the anion is of the smallest size.

Among the given species, Cs+ has the maximum size among given cations and F– has the smallest size among given anions, thus CsF has the highest rc/ra ratio.

Question 24. Amongst the elements with the following electronic configurations, which one of them may have the highest ionisation energy?

  1. \(\mathrm{Ne}\left[3 s^2 3 p^2\right]\)
  2. \({Ar}\left[3 d^{10} 4 s^2 4 p^3\right]\)
  3. \(\mathrm{Ne}\left[3 s^2 3 p^1\right]\)
  4. \(\mathrm{Ne}\left[3 s^2 3 p^3\right]\)

Answer: 4. \(\mathrm{Ne}\left[3 s^2 3 p^3\right]\)

Among options (1), (3) and (4), option (4) has the highest ionisation energy because of extra stability associated with, a half-filled 3p-orbital.

In option (2), the presence of 3d10 electrons offers a shielding effect, as a result, the 4p3 electrons do not experience much nuclear charge and hence, the electrons can be removed easily.

Question 25. Identify the correct order of the size of the following.

  1. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
  2. \(\mathrm{Ar}<\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
  3. \(\mathrm{Ca}^{2+}<\mathrm{Ar}<\mathrm{K}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
  4. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{S}^{2-}<\mathrm{Cl}^{-}\)

Answer: 1. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)

Among isoelectronic ions, ionic radii of anions is more than those of cations. Further size of the anion increases with an increase in negative charge and size of the cation decreases with an increase in positive charge.

Question 26. With which of the following electronic configuration an atom has the lowest ionisation enthalpy?

  1. \(1 s^2 2 s^2 2 p^3\)
  2. \(1 s^2 2 s^2 2 p^5 3 s^1\)
  3. \(1 s^2 2 s^2 2 p^6\)
  4. \(1 s^2 2 s^2 2 p^5\)

Answer: 2. \(1 s^2 2 s^2 2 p^5 3 s^1\)

The larger the atomic size, the smaller the value of the ionisation enthalpy. Again higher the screening effect, the lesser the value of ionisation potential. Hence, option (2) has the lowest ionisation enthalpy.

Question 27. Which one of the following ionic species has the greatest proton affinity to form a stable compound?

  1. \(\mathrm{NH}_2^{-}\)
  2. \(\mathrm{F}^{-}\)
  3. \(\mathrm{I}^{-}\)
  4. \(\mathrm{HS}^{-}\)

Answer: 1. \(\mathrm{NH}_2^{-}\)

In going from Ieft to right across a period in the periodic table, the basicity (l.e., proton affinity) decreases as the electronegativity of the atom possessing the lone pair of electrons increases.

Hence, the basicity of NH2– is higher than F–. On moving down a group, as the atomic size increases, basicity decreases.

Hence, F is more basic than I– and HO– is more basic than HS–. Hence, among the given ionic species, NH2– has maximum proton affinity.

Question 28. Which of the following is the most basic oxide?

  1. SeO2
  2. Al2O3
  3. Sb2O3
  4. Bi2O3

Answer: 4. Bi2O3

⇒ \(\mathrm{SeO}_2 \longrightarrow\) acidic oxide,

⇒ \(\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Sb}_2 \mathrm{O}_3 \longrightarrow\) amphoteric, \(\mathrm{Bi}_2 \mathrm{O}_3 \longrightarrow\) basic oxide.

Question 29. What is the correct relationship between the pH of isomolar solutions of sodium oxide, Na2O (pH1), sodium sulphide, Na2S (pH2), sodium selenide, Na2Se (pH3) and sodium telluride Na2Te (pH4)?

  1. \(\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4\)
  2. \(\mathrm{pH}_1>\mathrm{pH}_2=\mathrm{pH}_3>\mathrm{pH}_4\)
  3. \(\mathrm{pH}_1<\mathrm{pH}_2<\mathrm{pH}_3<\mathrm{pH}_4\)
  4. \(\mathrm{pH}_1<\mathrm{pH}_2<\mathrm{pH}_3=\mathrm{pH}_4\)

Answer: 1. \(\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4\)

⇒ \(\begin{array}{l|l}
\mathrm{Na}_2 \mathrm{O} & \text { Basic character } \\
\mathrm{Na}_2 \mathrm{~S} & \text { decreases down the group } \\
\mathrm{Na}_2 \mathrm{Se} & \\
\mathrm{Na}_2 \mathrm{Te} &
\end{array}\)

pH ∝ basic character

Hence, pH1 > pH2 > PH3 > pH4

Question 30. Ionic radii are

  1. Inversely proportional to the effective nuclear charge
  2. Inversely proportional to the square of the effective nuclear charge
  3. Directly proportional to the effective nuclear charge
  4. Directly proportional to the square of effective nuclear charge.

Answer: 2. Inversely proportional to the square of effective nuclear charge

Question 31. The ions \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) are isoelectronic. Their ionic radii show

  1. A significant increase from O2- to Al3+
  2. A significant decrease from O2- to Al3+
  3. An increase from O2- to F– and then decrease from Na+ to Al3+
  4. A decrease from O2- to F– and then increase from Na+ to Al3+

Answer: 2. A significant decrease from O2- to Al3+

The ions \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) are isoelectronic.

Amongst isoelectronic ions, the ionic radii of anions are more than that of cations. Further size of the anion increases with an increase in the +ve charge and the size of the cation decreases with an increase in the +ve charge.

Hence, correct order is \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} .\)

Question 32. Which of the following orders is wrong?

  1. \(\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3\)-acidic
  2. \(\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}-1^{\text {st }}\) IP
  3. \(\mathrm{Al}_2 \mathrm{O}_3<\mathrm{MgO}<\mathrm{Na}_2 \mathrm{O}<\mathrm{K}_2 \mathrm{O}-\) basic
  4. \(\mathrm{Li}^{+}<\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Cs}^{+}-\) ionic radius.

Answer: 2. \(\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}-1^{\text {st }}\) IP

Li, Be, B, C – these elements belong to the same period. Generaliy the value of 1’t ionisation potential increases on moving from left to right in a period, since the nuclear charge of the elements also increase in the same direction.

But the ionisation potential of boron (B → 2s² 2p¹) is lower than that of beryllium (Be → 2s²) since in the case of boron, 2p¹ electron has to be removed to get B+ while in the case of Be (2s²), s-electron has to be removed to get Be+ (2s¹). p-electron can be removed more easily than s-electron so the energy required to remove electron will be less in case of boron.

The order will be Li < B < Be < C.

Question 33. The correct order of 1st ionisation potential following elements Be, B, C, N, O is

  1. B<Be<C<O<N
  2. B<Be<C<N<O
  3. Be<B<C<N<O
  4. Be<B<C<O<N

Answer: 1. B<Be<C<O<N

The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionisation energy. The ionisation potential increases as the size of the atom decreases. Atoms with fully or partly filled orbitals have high ionisation potential.

Question 34. Which of the following elements has the maximum electron affinity?

  1. I
  2. Br
  3. Cl
  4. F

Answer: 3. Cl

Among the halogens the electron affinity value of ‘F’ should be maximum. But due to small size, there is interelectronic repulsion thus, there is difficulty in the entry of new electrons. Thus, the E.A. value is slightly lower than chlorine andthe order is I < Br < F < Cl.

Question 35. The first ionization potentials (eV) of Be and B respectively are

  1. 8.29,8.29
  2. 9.32,9.32
  3. 8.29,9.32
  4. 9.32,8.29

Answer: 4. 9.32,8.29

⇒ \({ }_4 \mathrm{Be} \rightarrow 1 s^2 2 s^2,{ }_5 \mathrm{~B} \rightarrow 1 s^2 2 s^2 2 p^1\)

Due to the stable fully-fiIled ‘s’-orbital arrangement of electrons in the ‘Be’ atom, more energy is required to remove an electron from the valence shell than the ‘ B’-atom. Therefore ‘Be’ has a higher ionisation potential than ‘B’.

Question 36. Which one of the following is the correct order of the size of iodine species?

  1. \(I^{+}>I^{-}>\mathrm{I}\)
  2. \(I^{-}>I^{\text {I }^{+}}\)
  3. \(I>I^{-}>I^{+}\)
  4. \(I^{\text {I }^{+}}>I^{-}\)

Answer: 2. \(I^{-}>I^{\text {I }^{+}}\)

Positive ion is always smaller and negative ion is always larger than the parent atom.

Question 37. Which of the following ions is the largest in size?

  1. K+
  2. Ca2+
  3. Cl–
  4. S2-

Answer: 4. S2-

Since all of these ions contain 18 electrons each, these are isoelectronic. For isoelectronic ions, the anion having a large negative charge is the largest in size i.e., S2-.

Question 38. Which of the following has the smallest size?

  1. \(\mathrm{Al}^{3+}\)
  2. \(\mathrm{F}\)
  3. \(\mathrm{Na}^{+}\)
  4. \(\mathrm{Mg}^{2+}\)

Answer: 1. \(\mathrm{Al}^{3+}\)

These are isoelectronic ions (ions with the same number of electrons) and for isoelectronic ions, the greater the positive charge, the greater the force of attraction on the electrons by the nucleus and the smaller the size of the ion. Thus, Al3+ has the smallest size.

Question 39. Among the following oxides, the one which is most basic is

  1. ZnO
  2. MgO
  3. Al2O3
  4. N2O5

Answer: 2. MgO

Al2O3 and ZnO are amphoteric. N2O5 is strongly acidic. MgO is the rnost basic.

Question 40. Which of the following has the largest size?

  1. Na
  2. Na+
  3. Na–
  4. Can’t be predicted.

Answer: 3. Na–

The cations are always smaller than the neutral atom and anions are always larger in size, Na– > Na > Na+

Question 41. \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+} \text { and } \mathrm{Si}^{4+}\) are isoelectronic. The order of their ionic size is

  1. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}<\mathrm{Si}^{4+}\)
  2. \(\mathrm{Na}^{+}<\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)
  3. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)
  4. \(\mathrm{Na}^{+}<\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}<\mathrm{Si}^{4+}\)

Answer: 3. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)

In isoelectronic ions, the size of the cation decreases as the magnitude of the positive charge increases

Question 42. In the periodic table from left to right in a period, the atomic volume

  1. Decreases
  2. Increases
  3. Remains same
  4. First decreases then increases.

Answer: 4. First decreases then increases.

Within a period from left to right, atomic volume first decreases and then increases.

Question 43. Which electronic configuration of an element has an abnormally high difference between second and third ionization energy?

  1. \(1 s^2, 2 s^2, 2 p^6, 3 s^1\)
  2. \(1 s^2, 2 s^2, 2 p^6, 3 s^1, 3 p^1\)
  3. \(1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^2\)
  4. \(1 s^2, 2 s^2, 2 p^6, 3 s^2\)

Answer: 4. \(1 s^2, 2 s^2, 2 p^6, 3 s^2\)

An abnormally high difference between 2nd and 3rd ionisation energy means that the element has two valence electrons, which is the case in configuration (4)

Question 44. One of the characteristic properties of non-metals is that they

  1. Are reducing agents
  2. Form basic oxides
  3. Form cations by electron gain
  4. Are electronegative.

Answer: 4. Are electronegative.

Question 45. Which one of the following has the minimum value of the cation/anion ratio?

  1. NaCl
  2. KCl
  3. MgCl2
  4. CaF2

Answer: 3. MgCl2

The order of ionic size for given ions will be K2+ > Ca2+ > Mg2+ and that of CI– > F–. Therefore, MgCl2 has a minimum value of cation/anion (Mg2+/Cl–) ratio.

Question 46. Which of the following sets has the strongest tendency to form anions?

  1. Ga, Ni, Tl
  2. Na, Mg, Al
  3. N, O, F
  4. V, Cr, Mn

Answer: 3. N, O, F

N, O and F are highly electronegative non-metals and will have the strongest tendency to form anions by gaining electrons from metal atoms.

Question 47. Elements of which of the following groups will form anions most readily?

  1. Oxygen family
  2. Nitrogen family
  3. Halogens
  4. Alkali metals

Answer: 3. Halogens

As halogens have seven electrons (ns²np5) in the valence shell, they have a strong tendency to acquire the nearest inert gas configuration by gaining an electron from the metallic atom and forming halide ions easily.

Question 48. In the periodic table, with the increase in atomic number, the metallic character of an element

  1. Decreases in a period and increases in a group
  2. Increases in a period and decreases in a group
  3. Increases both in a period and the group
  4. Decreases in a period and the group.

Answer: 1. Decreases in a period and increases in a group

Metallic character decreases in a period and increases in a group

Question 49. Which of the following atoms will have the smallest size?

  1. Mg
  2. Na
  3. Be
  4. Li

Answer: 3. Be

The atomic size decreases within a period from left to right, therefore Li > Be and Na > Mg. The size increases in a group from top to bottom. Hence, the size of Na is greater than Li. Overall order Na > Mg > Li > Be. Thus, Be has the smallest size.

 

MCQs on Structure of Atom for NEET

by

NEET Chemistry For Structure Of Atom Multiple Choice Questions

Question 1. Select the correct statements from the following

  1. Atoms of all elements are composed of two fundamental particles.
  2. The mass of an electron is 9.10939 x 10-31 kg.
  3. All the isotopes of a given element show the same chemical properties.
  4. Protons and electrons are collectively known as nucleons.
  5. Dalton’s atomic theory regarded the atom as an ultimate particle of matter.

Choose the correct answer from the options given below:

  1. 1 and 5 only
  2. 2, 3 and 5 only
  3. 1, 2 and 3 only
  4. 3, 4 and 5 only

Answer: 2. 2, 3 and 5 only

Atoms of all elements are composed of three fundamental particles: electrons, Protons, and neutrons.

Protons and neutrons are collectively known as nucleons

Hence, 2, 3, and 5 are correct statements

Question 2. From the following pairs of ions which one is not an iso-electronic pair?

  1. \(\mathrm{Fe}^{2-}, \mathrm{Mn}^{2-}\)
  2. \(\mathrm{O}^{2-}, \mathrm{F}\)
  3. \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}\)
  4. \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{3-}\)

Answer: 1. \(\mathrm{Fe}^{2-}, \mathrm{Mn}^{2-}\)

Among the given pairs of ions, Fe2+ and Mn2+ is not an iso-electronic pair.

Question 3. The number of protons, neutrons and electrons in \({ }_{71}^{175} \mathrm{Lu}_3\) respectively, are:

  1. 71, 104 and 71
  2. 104, 71 and 71
  3. 71, 71 and 104
  4. 175,104 and 71

Answer: 1. 71, 104 and 71

⇒ \({ }_{71}^{175} \mathrm{Lu}_3\) Number of protons = Number of electrons =

Atomic number = 71

Number of neutrons = Mass number – Atomic number

= 175 – 71 = 104

Read and Learn More NEET MCQs with Answers

Question 4. Be2+ is isoelectronic with which of the following ions?

  1. \(\mathrm{H}^{+}\)
  2. \(\mathrm{Li}^{+}\)
  3. \(\mathrm{Na}^{+}\)
  4. \(\mathrm{Mg}^{2+}\)

Answer: 2. \(\mathrm{Li}^{+}\)

⇒ \(\begin{array}{cc}
\text { Species } & \text { No. of electrons } \\
\mathrm{Be}^{2+} & 2 \\
\mathrm{H}^{+} & 0 \\
\mathrm{Li}^{+} & 2 \\
\mathrm{Na}^{+} & 10 \\
\mathrm{Mg}^{2+} & 10
\end{array}\)

 

Question 5. Isoelectronic species are

  1. \(\mathrm{CO}, \mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{C}_2^{2-}\)
  2. \(\mathrm{CO}^{-}, \mathrm{CN}^{-} \mathrm{NO}, \mathrm{C}_2^{-}\)
  3. \(\mathrm{CO}^{+}, \mathrm{CN}^{+}, \mathrm{NO}^{-}, \mathrm{C}_2\)
  4. \(\mathrm{CO}, \mathrm{CN}, \mathrm{NO}, \mathrm{C}_2\)

Answer: 1. \(\mathrm{CO}, \mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{C}_2^{2-}\)

Species having the same number of electrons are called isoelectronic species.

The no. of electrons in \(\mathrm{CO}=\mathrm{CN}^{-}=\mathrm{NO}^{+}=\mathrm{C}_2^{2-}=14\). So, these are isoelectronic species

Question 6. The ion that is isoelectronic with CO is

  1. \(\mathrm{CN}^{-}\)
  2. \(\mathrm{N}_2^{+}\)
  3. \(\mathrm{O}^{2-}\)
  4. \(\mathrm{N}_2^{-}\)

Answer: 1. \(\mathrm{CN}^{-}\)

Since both CO and CN– have 14 electrons, therefore these are isoelectronic species (i.e. having the same number of electrons)

Question 7. Which one of the following is not isoelectronic with O2-?

  1. \(\mathrm{Tl}^{+}\)
  2. \(\mathrm{Na}^{+}\)
  3. \(\mathrm{N}^{3-}\)
  4. \(\mathrm{F}^{-}\)

Answer: 1. \(\mathrm{Tl}^{+}\)

The number of electrons in O2- N+2, F-, and Nan is 10 each, but a number of electrons in Tl+ is 80.

Question 8. A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1, 368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is [speed of light. c = 3.0 x 108 ms-1]

  1. 21.92 cm
  2. 219.3 m
  3. 219.2 m
  4. 2192 m

Answer: 2. 219.3 m

A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1, 368 kHz (kilohertz).

v = \(\frac{c}{\lambda} ; \lambda=\frac{c}{v}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{1368 \times 10^3 \mathrm{~s}^{-1}}=219.3 \mathrm{~m}\)

Question 9. Which of the following series of transitions in the spectrum of hydrogen atom tails in the visible region?

  1. Brackett series
  2. Lyman series
  3. Balmer series
  4. Paschen series

Answer: 3. Balmer series

Lyman series: UV region

Balmer series: Visible region

Paschen series: IR region

Brackett series: IR region

Question 10. Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck’s constant, h = 6.63 x 10-34J s, speed of light, c = 3 x 108 m s-1)

MCQs on Structure of Atom for NEET

Question 11. The value of Planck’s constant is 6.63 x 10-34. The speed of light is 3 x 1017 nm s-1. Which value is closest to the wavelength in the nanometer of a quantum of light with a frequency of 6 x 1015 s-1?

  1. 50
  2. 75
  3. 10
  4. 25

Answer: 1. 50

The value of Planck’s constant is 6.63 x 10-34. The speed of light is 3 x 1017 nm s-1.

c = vλ

λ = \(\frac{c}{v}=\frac{3 \times 10^{17}}{6 \times 10^{15}}=50 \mathrm{~nm}\)

Question 12. According to the law of photochemical equivalence, the energy absorbed (in ergs/mole) is given as (h = 6.62 x 10-27 ergs, c = 3 x 1010 cm s-1 NA = 6.02 x 1023 mol-1)

  1. \(\frac{1.196 \times 10^8}{\lambda}\)
  2. \(\frac{2.859 \times 10^{16}}{\lambda}\)
  3. \(\frac{2.859 \times 10^5}{\lambda}\)
  4. \(\frac{1.196 \times 10^{16}}{\lambda}\)

Answer: 1. \(\frac{1.196 \times 10^8}{\lambda}\)

We know that, E = \(\frac{h c N_A}{\lambda}\)

= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda}\)

= \(\frac{1.1955 \times 10^8}{\lambda}=\frac{1.196 \times 10^8}{\lambda} \mathrm{ergs} \mathrm{mol} \mathrm{m}^{-1}\)

Question 13. The energies E1 and E2 of the two radiations are 25 eY and 50 eV respectively. The relation between their wavelengths i.e., λ1 and λ2 will be

  1. \(\lambda_1=\lambda_2\)
  2. \(\lambda_1=2 \lambda_2\)
  3. \(\lambda_1=4 \lambda_2\)
  4. \(\lambda_1=\frac{1}{2} \lambda_2\)

Answer: 2. \(\lambda_1=2 \lambda_2\)

The energies E1 and E2 of the two radiations are 25 eY and 50 eV respectively.

∴ \(E_1=\frac{h c}{\lambda_1}\) and \(E_2=\frac{h c}{\lambda_2} ; \frac{E_1}{E_2}=\frac{h c}{\lambda_1} \times \frac{\lambda_2}{h c}=\frac{\lambda_2}{\lambda_1}\)

or \(\frac{25}{50}=\frac{\lambda_2}{\lambda_1}\) or \(\frac{1}{2}=\frac{\lambda_2}{\lambda_1} \Rightarrow \lambda_1=2 \lambda_2\)

Question 14. The value of Planck’s constant is 6.63 x 10-34 s. The velocity of light is 3.0 x 108 m s-1. Which value is closest to the wavelength in nanometers of a quantum of light with a frequency of 8 x 1015 s-1?

  1. \(2 \times 10^{-25}\)
  2. \(4 \times 10^1\)
  3. \(5 \times 10^{-18}\)
  4. \(3 \times 10^7\)

Answer: 3. \(5 \times 10^{-18}\)

The value of Planck’s constant is 6.63 x 10-34 s. The velocity of light is 3.0 x 108 m s-1.

Applying v = c/λ

λ = \(\frac{c}{v}=\frac{3 \times 10^8}{8 \times 10^{15}}=37.5 \times 10^{-9} \mathrm{~m}=37.5 \mathrm{~nm}=4 \times 10^1 \mathrm{~nm}\)

Question 15. For given energy, E = 3.03 x 10-19 joules corresponding wavelength is (h = 6.626 x 10-34 J sec, c = 3 x 108 m/sec)

  1. 65.6 nm
  2. 6.56 nm
  3. 3.4 nm
  4. 656 nm

Answer: 4. 656 nm

E = \(\frac{h c}{\lambda} \Rightarrow \lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.03 \times 10^{-19}}=656 \mathrm{~nm}\)

Question 16. What will be the longest wavelength line in the Balmer series of spectra?

  1. 546 nm
  2. 656 nm
  3. 566 nm
  4. 556 nm

Answer: 2. 656 nm

The longest wavelength means the lowest energy.

We know that relation for wavelength \(\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

(\(R_{\mathrm{H}}\), Rydberg constant = \(109677 \mathrm{~cm}^{-1}\))

For \(n_1=2, n_2=3\)

∴ \(\frac{1}{\lambda}=109677\left(\frac{1}{(2)^2}-\frac{1}{(3)^2}\right)=15233\)

or, \(\lambda=\frac{1}{15233}=6.56 \times 10^{-5} \mathrm{~cm}\)

= \(6.56 \times 10^{-7} \mathrm{~m}\)

= \(656 \mathrm{~nm}\)

Question 17. If the radius of the second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of the third Bohr orbit of the Li2+ ion?

  1. 158.7 pm
  2. 1587 pm
  3. 1.587 pm
  4. 158.7 Å

Answer: 1. 158.7 pm

The radius of the second Bohr orbit of the He+ ion is 105.8 pm,

Radius \(=r_0 \times \frac{n^2}{Z}\)

For \(\mathrm{He}^{+}, n=2 ; Z=2\)

∴ \(\mathrm{r}_{\mathrm{He}^{+}}=r_0 \times \frac{2 \times 2}{2} ; 105.8=r_0 \times 2\)

∴ \(r_0=\frac{105.8}{2} \mathrm{pm}\)

For \(\mathrm{Li}^{2+} ; n=3 ; Z=3\)

∴ \(r_{\mathrm{Li}^{2+}}=r_0 \times \frac{(3)^2}{3}=\frac{105.8}{2} \times 3=158.7 \mathrm{pm}\) or 1.587Å

Question 18. Based on equation E = \(-2.178 \times 10^{-18} \mathrm{~J}\left(\frac{Z^2}{n^2}\right),\) certain conclusions are written. Which of them is not correct?

  1. The equation can be used to calculate the change in energy when the electron changes orbit.
  2. For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
  3. The negative sign in the equation simply means that the energy of the electron bound to the nucleus is lower than it would be if the electrons were at an infinite distance from the nucleus.
  4. The larger the value of n, the larger is the orbit radius.

Answer: 2. For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.

The electron is more tightly bound in the smallest allowed orbit.

Question 19. According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

  1. n = 6 to n = 1
  2. n = 5 to n = 4
  3. n = 6 to n = 5
  4. n = 5 to n = 3

Answer: 3. n = 6 to n = 5

We know that \(\Delta E \propto\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\) where \(n_2>n_1\)

∴ n = 6 to n = 5 will give the least energetic photon.

Question 20. The energy of the second Bohr orbit of the hydrogen atom is -328 kJ mol-1; hence the energy of the fourth Bohr orbit would be

  1. \(41 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  2. \(-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  3. \(-164 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  4. \(-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 2. \(-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(E_n=-K\left(\frac{Z}{n}\right)^2\)

Z=1 for hydrogen; n=2

∴ \(E_2=\frac{-K \times 1}{4} \Rightarrow E_2=-328 \mathrm{~kJ} \mathrm{~mol}^{-1} ; K=4 \times 328\)

∴ \(E_4=\frac{-K \times 1}{16} \Rightarrow E_4=-4 \times 328 \times \frac{1}{16}=-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 21. The frequency of radiation emitted when the electron falls from n = 4ton=lina hydrogen atom will be (Given ionization energy of H = 2.18 x 10-18 J atom-1 and h = 6.626 x 10-34 J s)

  1. \(1.54 \times 10^{15} \mathrm{~s}^{-1}\)
  2. \(1.03 \times 10^{15} \mathrm{~s}^{-1}\)
  3. \(3.08 \times 10^{15} \mathrm{~s}^{-1}\)
  4. \(2.00 \times 10^{15} \mathrm{~s}^{-1}\)

Answer: 3. \(3.08 \times 10^{15} \mathrm{~s}^{-1}\)

E = hv or v = E/h

For \(\mathrm{H}\) atom, \(E=\frac{-21.8 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{atom}^{-1}\)

ΔE= \(-21.8 \times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)=20.44 \times 10^{-19} \mathrm{Jatom}^{-1}\)

v = \(\frac{20.44 \times 10^{-19}}{6.626 \times 10^{-34}}=3.08 \times 10^{15} \mathrm{~s}^{-1}\)

Question 22. In a hydrogen atom, the energy of the first excited state is -3.4 eV. Then find out K.E. of the same orbit of the hydrogen atom.

  1. +3.4 eV
  2. +6.8 eV
  3. -13.6 eV
  4. +13.6 eV

Answer: 1. +3.4 eV

Kinetic energy = \(\frac{1}{2} m v^2\) = \(\left(\frac{\pi e^2}{n h}\right)^2 \times 2 m\)

(because \(v=\frac{2 \pi e^2}{\pi h}\))

Total energy, \(E_n=-\frac{2 \pi^2 m e^4}{n^2 h^2}=-\left(\frac{\pi e^2}{n h}\right)^2 \times 2 m=-K . E\).

∴ Kinetic energy = -En

Energy’s first excited state is -3.4 eV.

∴ The kinetic energy of the same orbit (n = 2) will be +3.4 eV

Question 23. Who modified Bohr’s theory by introducing elliptical orbits for electron path?

  1. Rutherford
  2. Thomson
  3. Hund
  4. Sommerfeld

Answer: 4. Sommerfeld

Sommerfeld modified Bohr’s theory considering that in addition to circular orbits electrons also move in elliptical orbits.

Question 24. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)

  1. 4.77
  2. 1.06
  3. 0.13
  4. 2.12

Answer: 4. 2.12

For nth orbit of ‘H’ atom, rn = n² x r1

⇒ the radius of 2nd Bohr’s orbit.

r2 = 4 x r1 = 4 x 0.53 = 2.12Å

Question 25. In Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed?

  1. \(2.389 \times 10^{-12} \text { ergs }\)
  2. \(0.239 \times 10^{-10} \text { ergs }\)
  3. \(2.15 \times 10^{-11} \mathrm{ergs}\)
  4. \(0.1936 \times 10^{-10} \mathrm{ergs}\)

Answer: 4. \(0.1936 \times 10^{-10} \mathrm{ergs}\)

The energy of an atom when n = 1

∴ \(E_1=-\frac{1312}{(1)^2}=-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Similarly energy when n = 3, \((E_3)=-\frac{1312}{(3)^2}=-145.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The energy absorbed when an electron jumps from n = 1 to n = 3, \(E_3-E_1=-145.7-(-1312)=1166.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

= \(\frac{1166.3}{6.023 \times 10^{23}}=193.6 \times 10^{-23} \mathrm{~kJ}\)

= \(193.6 \times 10^{-20} \mathrm{~J} \quad\left[1 \text { joule }=10^7 \text { ergs }\right]\)

∴ \(193.6 \times 10^{-13} \text { ergs }=0.1936 \times 10^{-10} \text { ergs }\)

Question 26. The radius of a hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar state is

  1. 0.53 Å
  2. 1.06 Å
  3. 0.17 Å
  4. 0.265 Å

Answer: 3. 0.17 Å

Due to the ground state, the state of hydrogen atom (n) = 1;

The radius of hydrogen atom (r) = 0.53 Å

The atomic number of Li (Z) = 3

Now, radius of \(\mathrm{Li}^{2+} \text { ion }=r \times \frac{n^2}{Z}=0.53 \times \frac{(1)^2}{3}\) =0.17Å

Question 27. The energy of an electron in the nth Bohr orbit of a hydrogen atom is

  1. \(\frac{13.6}{n^4} \mathrm{eV}\)
  2. \(\frac{13.6}{n^3} \mathrm{eV}\)
  3. \(\frac{13.6}{n^2} \mathrm{eV}\)
  4. \(\frac{13.6}{n} \mathrm{eV}\)

Answer: 3. \(\frac{13.6}{n^2} \mathrm{eV}\)

Energy of an electron in nth Bohr orbit of hydrogen atom = \(\frac{-13.6}{n^2} \mathrm{eV}\)

Question 28. The spectrum of He is expected to be similar to that

  1. H
  2. Li+
  3. Na
  4. He+

Answer: 2. Li+

Both He+ and Li+ contain 2 electrons each.

Question 29. If r is the radius of the first orbit, the radius of the nth orbit of H-atom is given by

  1. rn²
  2. rn
  3. r/n
  4. r2n2

Answer: 1. rn2

The radius of nth orbit of H-atom = r0n2

where r0 = radius of the first orbit.

Question 30. In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, a0 = 52.9 pm]

  1. 211.6 pm
  2. 211.6 π pm
  3. 52.9 π pm
  4. 105.8 pm

Answer: 2. 211.6 π pm

Bohr radius, ao = 52.9 pm

n = 2, rn = n²a0= (2)2a0= 4x 52.9 pm = 211.6 pm

The angular momentum of an electron in a given stationary state can be expressed as in the equation,

mvr = \(n \cdot \frac{h}{2 \pi}=2 \times \frac{h}{2 \pi}=\frac{h}{\pi} \Rightarrow m v r \pi=h\)…..(1)

de-Broglie equation, \(\lambda=\frac{h}{m v} ; \lambda m v=h\)…..(2)

From equations (1) and (2), we get λ = πr

Putting the value of r, λ = 211.6 π pm

Question 31. A 0.66 kg ball is moving at a speed of 100 m/s. The associated wavelength will be (h = 6.6 x 10-34 J s)

  1. \(6.6 \times 10^{-32} \mathrm{~m}\)
  2. \(6.6 \times 10^{-34} \mathrm{~m}\)
  3. \(1.0 \times 10^{-35} \mathrm{~m}\)
  4. \(1.0 \times 10^{-32} \mathrm{~m}\)

Answer: 3. \(1.0 \times 10^{-35} \mathrm{~m}\)

According to de-Broglie equation, \(\lambda=\frac{h}{m v}\)

Given, \(h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; m=0.66 \mathrm{~kg} ; v=100 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\lambda=\frac{6.6 \times 10^{-34}}{0.66 \times 100}=1 \times 10^{-35} \mathrm{~m}\)

Question 32. If uncertainty in position and momentum are equal, then uncertainty in velocity is

  1. \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)
  2. \(\sqrt{\frac{h}{\pi}}\)
  3. \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
  4. \(\sqrt{\frac{h}{2 \pi}}\)

Answer: 3. \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

From the Heisenberg uncertainty principle, \(\Delta p \cdot \Delta x \geq \frac{h}{4 \pi}\) or \(m \Delta v \times \Delta x \geq \frac{h}{4 \pi}\)

or \((m \Delta v)^2 \geq \frac{h}{4 \pi}\) (because \(\Delta x=\Delta p\)) or \(\Delta v \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 33. The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 x 10-18 g cm s-1. The uncertainty in electron velocity is (mass of an electron is 9 x 10-28 g)

  1. \(1 \times 10^5 \mathrm{~cm} \mathrm{~s}^{-1}\)
  2. \(1 \times 10^{11} \mathrm{~cm} \mathrm{~s}^{-1}\)
  3. \(1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)
  4. \(1 \times 10^6 \mathrm{~cm} \mathrm{~s}^{-1}\)

Answer: 3. \(1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)

The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 x 10-18 g cm s-1.

Uncertainty in momentum (mΔv) = 1 x 10-18 g cm s-1

Uncertainty in velocity (Δv) = \(\frac{1 \times 10^{-18}}{9 \times 10^{-28}}=1.1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)

Question 34. Given: The mass of an electron is 9.11 x 10-31 kg, the Planck constant is 6.626 x 10-34 J s, and the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is

  1. \(5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}\)
  2. \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)
  3. \(5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\)
  4. \(5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 2. \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\Delta x \cdot m \Delta v=h / 4 \pi\)

0.1 x \(10^{-10} \times 9.11 \times 10^{-31} \times \Delta v=\frac{6.626 \times 10^{-34}}{4 \times 3.143}\)

∴ \(\Delta v=\frac{6.626 \times 10^{-34}}{0.1 \times 10^{-10} \times 9.11 \times 10^{-31} \times 4 \times 3.143}\)

= \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 35. The uncertainty in the momentum of an electron is 1 x 10-5 kg m/s. The uncertainty in its position will be (h = 6.62 x 10-34 kg m²/s)

  1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  2. \(80 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  3. \(50 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\Delta x \times \Delta p=\frac{h}{4 \pi}\) (Heisenberg uncertainty principle)

⇒ \(\Delta x=\frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 10^{-5}}=5.27 \times 10^{-30} \mathrm{~m}\)

Question 36. The de Broglie wavelength of a particle with a mass 1 g and velocity of 100 m/s is

  1. \(6.63 \times 10^{-35} \mathrm{~m}\)
  2. \(6.63 \times 10^{-34} \mathrm{~m}\)
  3. \(6.63 \times 10^{-33} \mathrm{~m}\)
  4. \(6.65 \times 10^{-35} \mathrm{~m}\)

Answer: 3. \(6.63 \times 10^{-33} \mathrm{~m}\)

λ = \(\frac{h}{m \nu}=\frac{6.63 \times 10^{-27} \mathrm{erg} \mathrm{sec}}{1 \mathrm{~g} \times 10^4 \mathrm{~cm} / \mathrm{s}}\)

= \(6.63 \times 10^{-31} \mathrm{~cm}=6.63 \times 10^{-33} \mathrm{~m}\)

Question 37. The position of both, an electron and a helium atom is known within 1.0 nm. Further, the momentum of the electron is known within 5.0 x 10-26 kg m s-1. The minimum uncertainty in the measurement of the momentum of the helium atom is

  1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  2. \(80 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  3. \(50 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
  4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

The position of both, an electron and a helium atom is known within 1.0 nm. Further, the momentum of the electron is known within 5.0 x 10-26 kg m s-1.

According to the uncertainty principle, the product of uncertainty in position and uncertainty in momentum is constant for a particle.

i.e., \(\Delta x \times \Delta p=\frac{h}{4 \pi}\)

As, Δx =1.0 nm for both electron and helium atom, so Δp is also the same for both the particles.

Thus, the uncertainty in the momentum of the helium atom is also 5.0 x 10-26 kg m s-1.

Question 38. Uncertainty in position of an electron (Mass = 9.1 x 10-28 g) moving with a velocity of 3 x 104 cm/s accurate upto 0.001% will be (Use h/(4π) in uncertainty expression where h = 6.626 x 10-27 erg second)

  1. 5.76 cm
  2. 7.68 cm
  3. 1.93 cm
  4. 3.84 cm

Answer: 3. 1.93 cm

Mass of an electron (m) = 9.1 x 10-28 g

Velocity of electron (v) = 3 x 104 cm/s

Accuracy = 0.001% = \(\frac{0.001}{100}\) and

Planck constant (h)= 6.626 x 10-27 erg-second.

We know that actual velocity of the electron \((\Delta v)=3 \times 10^4 \times \frac{0.001}{100}=0.3 \mathrm{~cm} / \mathrm{s}\)

Therefore, uncertainty in the position of the electron, \((\Delta x)=\frac{h}{4 \pi m \Delta v}=\frac{6.626 \times 10^{-27}}{4 \pi \times\left(9.1 \times 10^{-28}\right) \times 0.3}=1.93 \mathrm{~cm}\)

Question 39. Which of the following statements do not form a part of Bohr’s model of the hydrogen atom?

  1. The energy of the electrons in the orbits is quantized.
  2. The electron in the orbit nearest the nucleus has the lowest energy.
  3. Electrons revolve in different orbits around the nucleus.
  4. The position and velocity of the electrons in the orbit cannot be determined simultaneously.

Answer: 4. The position and velocity of the electrons in the orbit cannot be determined simultaneously.

It is Heisenberg’s uncertainty principle and not Bohr’s postulate.

Question 40. The relation between nm (nm = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number

  1. \(n_m=2 l^2+1\)
  2. \(n_m=l+2\)
  3. \(l=\frac{n_m-1}{2}\)
  4. \(l=2 n_m+1\)

Answer: 3. \(l=\frac{n_m-1}{2}\)

∴ \(n_m=2 l+1 \quad \text { or } \quad l=\frac{n_m-1}{2}\)

Question 41. Identify the incorrect statement from the following.

  1. All the five 5d orbitals are different in size when compared to the respective 4d orbitals.
  2. The five 4d orbitals have shapes similar to the respective 3d orbitals.
  3. In an atom, all five 3d orbitals are equal in energy in a free state.
  4. The shapes of dxy, dyz, and dzx orbitals are similar to each other; and \(d_{x^2-y^2} \text { and } d_{z^2}\) are similar to each other

Answer: 4. The shapes of dxy, dyz, and dzx orbitals are similar to each other; and \(d_{x^2-y^2} \text { and } d_{z^2}\) are similar to each other

The shapes of dxy, dxz,, dyz, and \(d_{x^2-y^2}\) are similar to each other, whereas that of d; is different from others. AIl five 3d orbitals are equivalent in energy. The d-orbitals for which n is greater than 3 (4d, 5d, …) also have shapes similar to 3d-orbital but differ in energy and size.

Question 42. 4d, 5p, 5/and 6p orbitals are arranged in the order of decreasing energy. The correct option is

  1. 5f> 6p > 4d > 5p
  2. 5f> 6p> 5p> 4d
  3. 6p > 5f> 5p > 4d
  4. 6p > 5f> 4d > 5p

Answer: 2. 5f> 6p> 5p> 4d

The higher the value of (n + 1) for an orbital, the higher its energy. However, if two different types of orbitals have the same value of (n + 1), the orbital with the lower value of n has lower energy. Therefore, the decreasing order of energy of the given orbitals is 5f > 6p > 5p > 4d.

Question 43. Orbital having 3 angular nodes and 3 total nodes is

  1. 5p
  2. 3d
  3. 4f
  4. 6d

Answer: 3. 4f

Number of spherical/radial nodes in any orbital =n-l-1

Number of planar/angular nodes in orbital = l = 3

.’. Totalnumberofnodesinanyorbital= n -1=3

∴ n = 4

Thus, the orbital is 4f

Question 44. Which one is a wrong statement?

  1. The total orbital angular momentum of an electron in the s-orbital is equal to zero.
  2. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
  3. The electronic configuration of the N atom is Structure Of Atom Hunds Rule
  4. The value of m for dz2 is zero.

Answer: 3. The electronic configuration of the N atom is

Structure Of Atom Hunds Rule

According to Hund’s rule of maximum multiplicity, the correct configuration of ‘N’ is

Question 45. Which one is the wrong statement?

  1. The uncertainty principle is \(\Delta E \times \Delta t \geq \frac{h}{4 \pi}\)
  2. Half-filled and fully-filled orbitals have greater stability due to greater exchange energy, greater symmetry, and a more balanced arrangement.
  3. The energy of 2s-orbital is less than the energy of 2p-orbital in the case of hydrogen-like atoms.
  4. de-Broglie’s wavelength is given by \(\lambda=\frac{h}{m v}\) where m = mass of the particle, v = group velocity of the particle.

Answer: 3. The energy of 2s-orbital is less than the energy of 2p-orbital in the case of hydrogen-like atoms

In the case of hydrogen-like atoms, energy depends on the principal quantum number only. Hence, 2s-orbital will have energy equal to 2p-orbital.

Question 46. Consider the following sets of quantum numbers:

Structure Of Atom Set Of Quantum Numbers

Which of the following sets of quantum numbers is not possible?

  1. (1), (2), (3) and (4)
  2. (2), (4), and (5)
  3. (1) and (3)
  4. (2), (3), and (4)

Answer: 2. (2), (4) and (5)

  1. Represents an electron in 3s orbital.
  2. This is not possible as the value varies from 0, 1, … (n-1).
  3. Represents an electron in 4/orbital.
  4. Is not possible as the value of varies from -l… +l.
  5. This is not possible as the value of z varies from -l …+l, It can never be greater than l.

Question 46. The following quantum numbers are possible for how many orbitals? n = 3, l = 2, m = +2

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 1. 1

n = 3, l= 2, m = +2

It symbolizes one of the five d-orbitals (3d).

Structure Of Atom Symbols Of Five d Orbitals

Question 47. For which of the following sets of four quantum numbers, an electron will have the highest energy?

Structure Of Atom Set Of Four Quantum Numbers

Answer: 2.

The energy of an electron depends on the value of (n + 1). The subshell are 3d, 4d, 4p and 5s, out of which 4d has highest energy

Question 48. The electronic configuration of calcium atoms can be written as

  1. [Ne]4p²
  2. [Ar]4s²
  3. [Ne]4s²
  4. [Kr]4p²

Answer: 2. [Ar]4s²

The atomic number of Ca = 20

∴ Electronic configuration of Ca = [Ar]4s²

Question 49. For azimuthal quantum number l = 3, the maximum number of electrons will be

  1. 2
  2. 6
  3. 0
  4. 14

Answer: 4. 14

l = 3 means f subshell

Maximum no. of electrons in f subshell = 14

Structure Of Atom Azimuthal Quantum Number

NEET Biology Notes – Plant Growth And Development

by

NEET Biology Notes For Plant Growth And Development Growth

Growth is an irreversible increase in size, volume, and weight of a part or whole of an organism. An irreversible increase in size, volume, or weight is called apparent growth as it is the external manifestation of growth. Formation at cellular materials or protoplasm is the real growth. Growth is a measurable or quantitative phenomenon that can be measured in relation to time. The growth of living beings is internal or intrinsic.

Plant growth is diffused like that of animals only during the early embryonic stages. Later on, plants develop specific areas, called meristem, for growth. On account of meristems, plant growth is localized.

Characteristics Of Plant Growth

  1. Growth is localized.
  2. Growth continues throughout life.
  3. There is an increase in the number of parts.
  4. It is open-ended.
  5. The younger one or seedling can be quite different from an adult.
  6. The juvenile stage may have different traits.

Plant Growth Differentiation: Growth is invariably associated with differentiation. Differentiation is a permanent localized qualitative change in size, biochemistry, structure, and function of cells, tissues, or organs, for example, fiber, vessel, tracheid, sieve tube, mesophyll, leaf, etc. Some examples of differentiation are as follows

  1. Enlargement, lignocellulosic wall thickening, and emptying in case of tracheids
  2. Loss of end wall in case of vessel elements
  3. Loss of nucleus and perforation of end wall in sieve tube members
  4. Deposition of suberin and tannins in cells
  5. Differential wall thickening (in guard cells)
  6. Secretion of mucilage in root cap

Development is the sequence of changes that occur in the structure and functioning of an organism, organ, tissue, or cell involving its formation, growth differentiation, maturation, reproduction, senescence, and death.

Read and Learn More NEET Biology Notes

Characteristics Of Growth

  • Primary Growth: It is the formation of primary permanent tissues and organs. It is caused by the activity of apical and intercalary meristems.
  • Secondary Growth: It is an increase in girth. It occurs by two types of lateral meristems, vascular cambium and cork cambium.
  • Efficiency Index: It is the rate of growth. It is measured by calculating the increase in size, diameter, or area per unit time.

plant growth and development

NEET Biology Notes For Plant Growth And Development Growth Rate

An increase in growth per unit of time is called growth rate. Growth rate may result in arithmetic or geometric growth.

Arithmetic Growth: Arithmetic growth is a type of growth in which the rate of growth is constant and an increase in growth occurs in arithmetic progression 2, 4, 6, 8, 10, and 12. Arithmetic growth is found in root or shoot elongating at a constant rate.

The meristematic cells at the growing point divide in such a fashion that one daughter cell remains meristematic while the other cells grow and differentiate. The process continues. Mathematically, arithmetic growth is expressed as:

Lt = L0 + rt

where Lt is the length after time t, L0 is the length at the beginning, and r is the growth rate or elongation per unit time.

NEET Biology Plant Growth And Development Arithmetic Growth Curve

Geometric Growth: Geometric growth is quite common in unicellular organisms when grown in a nutrient-rich medium. Here, every cell divides. The daughter cells grow and divide. The granddaughters repeat the process and so on. The number of cells is initially small so the initial growth is slow. Later on, there is rapid growth at an exponential rate.

Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes a slowing down of growth. It leads to the stationary phase. (There may be actually a decline.) Plotting the growth against time will give a typical sigmoid or S-curve.

NEET Biology Plant Growth And Development Geometric Growth Curve

The S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues, and organs of plants.

Exponential Growth (Law Of Compound Interest): Growth is dependent on three factors: initial size (W0), rate of growth (r), and time interval (t) for which the rate of growth can be retained.

W1 = W0ert

Here W1 is the final size, W0 is the initial size, r is the growth rate, 1 is the time of growth, while e is the base on natural logarithms. The magnitude of r or the rate of growth has been called the efficiency index by Blackman (1919).

Absolute And Relative Growth Rate: Quantitative comparisons between the growths of various systems can be made by measuring their absolute and relative growth rates.

Absolute Growth Rate: The absolute growth curve is the actual growth curve obtained by plotting growth against time. It is commonly S-shaped. The absolute growth rate is the total growth per unit time. A graph plotted for absolute growth rates for various times of the grand period of growth appears bell-shaped.

” plant growth and development short notes”

The peak is formed when the growth rate is the fastest. The period of increasing growth is depicted by the first part of the curve while the period of decreasing growth rate is shown by the second part of the curve.

NEET Biology Plant Growth And Development Absolute Or Actual Growth Curve And Absolute Growth Rate Curve

Relative Growth Rate: It is the growth per unit time per unit initial growth.

Relative growth rate = \(\frac{\text{Growth in given time period}}{\text{Measurement at start of time period}}\)

Suppose two leaves have grown by 5 cm² in one day. The initial size of leaf A was 10 cm² while that of leaf B was 15 cm². Though their absolute growth is the same, the relative rate of growth is faster in leaf A.

NEET Biology Notes For Plant Growth And Development Measurement Of Growth

To measure growth, various methods/instruments are used.

  1. Direct method
  2. Horizontal microscope
  3. Auxanometer
    • Arc auxanometer
    • Pfeffer’s auxanometer
  4. Crescograph (Developed By J.C. Bose): A highly sensitive growth-measuring instrument that can magnify growth by 10000 times.

NEET Biology Notes For Plant Growth And Development Growth Hormones

In all plants, minute quantities of certain substances are found (plant growth regulators or phytohormones), which regulate growth and differentiation.

Five major types of growth substances are recognized: aux-ins, gibberellins, cytokinins, abscisic acid, and ethylene.

Auxins: Auxins are weak organic acids having unsaturated ring structures.

  • Charles Darwin conducted his experiments concerning growth on canary grass (Phalaris canariensis) and found that the bending movement of coleoptiles in uni-lateral light was due to a chemical.
  • Went is credited with the discovery of auxin.
  • Auxins are synthesized mainly in apices and exhibit polar transport through parenchyma.
  • KogI and Haagen Smith (1931) found that human urine contained a growth substance, which was isolated and given the name auxin-a (auxentriolic acid). In 1934, Kogl and coworkers isolated another compound, auxin-b (auxenolinic acid) from com germ oil and heteroauxin (now known as IAA or indole-3-acetic acid, C10H9O2N), from human urine. It is the only natural auxin.
  • The precursor of auxin is the tryptophan amino acid.
  • The optimum concentration in the stem apex is 10 ppm while in the root apex, it is 0.0001 ppm.
  • Auxin is active in a free state and can be easily extracted. Bound auxin is inactive and meant for storage. For example, IAA-aspartic acid, IAA-inositol, and IBA-alanine.

Functions Of Auxins

  • Auxins promote cell elongation by loosening cell wall microfibrils, solubilization of carbohydrate reserve, and increased respiration.
  • Responsible for phototropism and geotropism.
  • Promote apical dominance (in the presence of an apical bud, the growth of lateral buds are inhibited due to auxin secreted by the apical bud).
  • Promote root initiation in cuttings (by NAA, IBA).
  • Delay of abscission of leaves by preventing the formation of an abscission layer.
  • Prevention of lodging in cereals.
  • Induce parthenocarpy (production of seedless fruits).
  • Selective weedicide.
  • Have feminizing effect (increase number of female flowers in plants, for example, Cannabis).
  • Seasonal activity of cambium is promoted by auxin.
  • Healing of injury is effected through auxin-induced division in cells around the injured area.
  • Auxin induces negative potential in cell membranes.
  • In legumes, IAA stimulates nodule formation.

Antiauxins: They inhibit auxin activity. For example, triiodobenzoic acid (TIBA), PCIB (p-chlorophenoxy isobutyric acid)

Bioassay Of Auxins

  1. Avena curvature test
  2. Split pea test
  3. Root growth inhibition test

Applications Of Synthetic Auxins

  • Rooting: IBA, IBA-alanine, and NAA are used.
  • Parthenocarpy: IAA, IBA.
  • Weedicide: 2,4-D, 2,4-5-T are used for killing broad-leaved weeds (generally dicot).
  • Flowering: NAA and 2,4-D for litchi and pineapple.
  • Storage: Methyl ester of NAA for the storage of potato.
  • Pre-harvest Fruit Drop: 2,4-D for citrus fruits; NAA for tomato.
  • Prevention Of Lodging: Naphthalene acetamide (NAAM).
  • Vegetable Crops: Chlorophenoxypropionic acid is used to improve the quality of vegetable crops by inhibiting flower formation. For example, lettuce.

Dwarf Shoots: NAA is used for increasing dwarf shoots and a number of fruits in apples.

Gibberellins: In the early twentieth century, Japanese farmers noticed balance or foolish seedling disease of lice. As a result of the disease, certain rice seedlings grew excessively tall, the disease was caused by the fungus Gibherella fuikuroi (perfect state of Fusarium moniliform).

ch 15 biology class 11 notes

  • Yabuta and Suxniki (1930) isolated the growth-inducing hormone and called it gibberellin.
  • Chemically, all gibberellins are terpenes, a complex group of plant chemicals related to lipids. All are weak acids and have gibbane ring skeletons. GA3 is the commonest. GA24 and GA25 are found only in fungi. The precursor is acetyl CoA.
  • Gibberellins are synthesized in the apices of young leaves, embryos, buds, and roots and are transported through the xylem.

Applications Of Gibberellins

  • Internodal Elongation: Like auxins, the main effect of gibberellins is on stem elongation. Gibberellins stimulate stem elongation and leaf expansion but do not affect roots. Thus, gibberellins restore normal-size arid growth to genetically dwarf varieties of pear and maize.
  • Bolting: In many plants, leaf development is profuse, while internodal growth is retarded. This term of growth is called rosette, for example, cabbage, radish, and henbane. Just before the reproductive phase, internodes elongate enormously, causing a marked increase in height.

The stem sometimes elongates five to six times the original height of the plant. This is called bolting. Bolting requires either long days or cold nights and gibberellins treatment.

  • Germination Of Seeds: Gibberellins promote seed germination (especially in cereals).
  • Control Of Flowering: Gibberellins promote flowering in long-day plants and inhibit it in short-day plants. These also control sex expression in certain species. In general, the application of gibberellins promotes the production of male flowers in female plants of Cannabis.
  • Control Of Fruit Growth: Along with gibberellins, auxins control fruit growth and development. Gibberellins cause parthenocarpy in pome fruits (apple, pear, etc.).
  • Vernalization: Gibberellins can substitute vernalization.
  • Dormancy: Gibberellins overcome the natural dormancy of buds, tubers, seeds, etc.

Commercial Application Of Gibberellins

  • Fruit Growth: Increase the number and size of grapes, tomatoes, etc. Pomalin (a mixture of GA3 and BAP) is used for such purpose.
  • Malt: Increase the yield of malt from barley.
  • Overcoming Dormancy: In photoelastic seeds of tobacco and lettuce
  • Delay Ripening: in citrus.
  • Induce Flowering: In long-day plants, in non-inductive periods.

Antigibberellins: Certain chemicals are antagonistic to gibberellins. For example, Phosphan-D, Amo-1618, CCC, and maleic hydrazide.

Bioassay Of Gibberellins

  1. Induction of α-amylase in barley endosperm test
  2. Dwarf maize test
  3. Dwarf pea test

Cytokinins: They are basic hormones and are purine (adenine) derivatives. Cytokinins are substances that act primarily on cell division and have little or no effect on extension growth. In 1955, Miller et al. separated it from herring sperm DNA and yeast DNA and called it kinetin (because of its involvement in cell division, i.e., cytokinesis). Later on, the substance was identified as 6-furfuryl aminopurine. Subsequently, the term cytokinin was adopted by Letham.

  • The first naturally occurring cytokinin to be chemically identified was from young maize (Zea mays) grains in 1963 and was called zeatin, which is benzyl amino purine (BAP). Cytokinins are a part of t-RNA.
  • Cytokinins are mostly synthesized in roots, seeds, and developing fruits. Coconut milk and apple fruit extracts are rich in cytokinins. Some other cytokinins are dihydrozeatin, IPA (isopentanyl adenine).

Applications Of Cytokinins

  • Cell-division: Cytokinins are quite abundant wherever rapid cell division occurs, especially in growing tissues.
  • Morphogenesis: Cytokinins promote cell division. In the presence of auxins, cytokinins promote cell division even in non-meristematic tissues. In tissue cultures, mitotic divisions are accelerated when both auxin and cytokinin are present. The ratio of cytokinins to auxins also controls cell differentiation and morphogenesis.
  • Apical Dominance: Cytokinins and auxins act antagonistically in the control of apical dominance.
  • Delay In Senescence: Cytokinins delay the senescence of plant organs by controlling protein synthesis and mobilization of resources. This is called the Richmond-Lang effect. Cytokinins are also called anti-aging hormones.
  • Flowering: Cytokinins also induce flowering in certain species of plants such as Lemna and Wolffia and are also responsible for breaking the dormancy of seeds of some plants.
  • Favors Transport: Phloem transport is promoted.
  • Favors Salt Accumulation: Accumulation of salts in the cells is promoted.
  • Favors Sex-Expression: Promote femaleness.
  • Temperature/disease Resistance: Increase resistance to low and high temperatures and diseases.

Commercial Applications of Cytokinins

  1. Tissue culture
  2. The shelf life of vegetables and cut flowers is increased
  3. Overcoming senescence

Bioassay Of Cytokinins

  1. Chlorophyll preservation test.
  2. Cell division test

“plant growth “

Ethylene: Ethylene is the only natural plant growth regulator in gaseous form and is effective in the concentration of 0.01-10 ppm.

Ethylene is produced by most or all plant organs but maximum production occurs in ripening fruits and during senescence. High concentrations of auxin induce the formation of ethylene. Though it is a gas, it does not generally move through air spaces in plants. Rather, it escapes from the plant surface. The precursor of ethylene is methionine.

Applications Of Ethylene

  • Growth: Ethylene inhibits stem elongation and stimulates its transverse expansion. As a result, the stem looks swollen.
  • Abscission: It accelerates the abscission of leaves, flowers, and fruits.
  • Fruit Ripening: Its chief effects are on the ripening of fruits accompanied by a rise in the rate of respiration (climacteric). It causes the dehiscence of dry fruits.
  • Flowering: The application of ethylene induces flowering in pineapple.
  • A commercial compound “ethephon” breaks down to release ethylene in plants. It is particularly applied to rubber plants for the flow of latex.
  • It decreases the sensitivity to gravity. Roots become apogeotropic; seedling develops a tight epicotyl hook.
  • It has a feminizing effect.
  • Bioassay: Triple response test.

Growth Inhibitors

  • For a long time, it has been suspected that dormancy is caused by inhibitors. A group of scientists led by Wareing initiated studies to find them. In 1964, pure crystals of a substance were isolated called dormin. It was found to be similar to another compound isolated from young cotton fruits in 1963 by Addicott.
  • This substance accelerated abscission and was called abscission 2. In 1967, it was decided to call it abscissic acid (ABA). Since then, it has been found in all groups of plants (from mosses to higher plants). In liverworts and algae, a compound lunular acid has been found to have activities similar to ABA.
  • Chemically, ABA is a dextrorotatory cA-sesquiterpene. ABA is synthesized in leaves and transported through the xylem and phloem. It is a major inhibitor of growth in plants and is antagonistic to all three growth promoters. Its precursor is violaxanthin (in chloroplast).

Applications Of ABA

  • Stoppage Of Cambial Activity: It inhibits mitosis in vascular cambium.
  • Bud Dormancy: It induces axillary buds to become dormant as the winter approaches.
  • It plays a role in seed development, maturation, and dormancy.
  • Transpiration: It is a “stress hormone” and helps the plant to cope with adverse environmental conditions by closing stomata (antitranspirant).
  • It may be sprayed on tree crops to regulate fruit drop at the end of the season.
  • Application of ABA to green oranges turns them yellow by inducing the synthesis of carotenoids.
  • It induces flowering in some short-day plants such as strawberries and blackcurrant.

NEET Biology Notes For Plant Growth And Development Dormancy

In favorable conditions, if a viable seed fails to germinate, this condition is called dormancy and if the viable seed fails to germinate due to unfavorable conditions, it is called “quiescence.” Dormancy may be due to:

  1. Seed coat impermeable to gases, for example, apple, or water, for example, Trigonella; or seed coat mechanically resistant, for example, Capsella, Amaranthus.
  2. Immaturity of the embryo, for example, G. biloba.
  3. Specific light requirement: Some seeds require light for germination and are called positive photoblastic seeds (for example, Lactuca sativa, Nicotiana tobaccum, Lythrum, etc.). Lettuce (Lactuca) is induced by red light and inhibited by far red light. Some seeds show inhibition in germination due to light exposure and are called negative photoblastic, for example, onions, lily, phlox, etc.
  4. Dormancy due to chilling temperature requirement, for example, Polygonum.
  5. After-ripening: Some seeds have a mature embryo but do not germinate immediately due to the absence of growth hormone. They require a period of after-ripening during which they attain the power to germinate. For example, oats, barley, wheat, etc.
  6. Due To Germination Inhibitors: Some chemicals such as organic acids, phenolics, tannins, alkaloids, lactones, mustard oil, etc., inhibit germination (for example, ferulic acid in tomato pulp).

“plant growth and development neetprep “

Methods To Break Dormancy

  1. Scarification: It is a method of softening and weakening of seed coat by acids, alcohol, or knife.
  2. Stratification: After ripening treatment, low temperature (0-10°C) with O2.
  3. Light exposure.
  4. Low temperature + Gibbcrellin + O2 treatment, etc.

NEET Biology Notes For Plant Growth And Development Seed Germination

Seed germination is of two types:

  1. Epigeal: Hypocotyl grows first, cotyledons come out of the soil as in cucurbits, mustard, castor, onion, tamarind, etc.
  2. Hypodeal: Epicotyl grows first, cotyledons remain underground as in rice, maize, mango, Fabaceae, etc. Whenever a seed germinates inside the fruit, it is vivipary as in Rhizophora, Sonneralia, and Heritiera (mangrove plants).

NEET Biology Notes For Plant Growth And Development Cytochrome

Borthwick and Hendrick, in the 1950s, plotted the action spectrum of wavelengths showing their relative effectiveness in stimu¬lating seed germination. The wavelength most effective for promoting germination was 660 nm (red) and for inhibition of germination about 730 nm (far red). They also demonstrated that only brief exposures of light were necessary and that the effects of red light were reversed by far red fight and vice versa.

  • The pigment responsible for this was isolated in 1960 and was called phytochrome by Butler. It is a blue-green pigment existing in two interconvertible forms: PFR or P730 (absorbs far-red lights) and PR or P660 (absorbs red light).
  • By absorbing red light, PR is converted to PFR rapidly. PFR absorbing far-red light is converted to PR rapidly. PFR is the physiologically active form; PR is inactive. The table describes the effects of red light and far red light on plant growth.

NEET Biology Plant Growth And Development Cytochrome Formula

Effects Of Red Light And Far Red Light On Plant Growth

Effects Of Red Light And Far Red Light On Plant Growth

Phytochrome is responsible for various photomorphogenic processes in plants such as the growth and development of plant organs; germination of seed, pollen, and spores; flowering; differentiation of stomata; epinasty and abscission; etc.

NEET Biology Notes For Plant Growth And Development Photoperiiqdiibm

The response of plants to changes in the relative lengths of day and night is called photoperiodism.

Photoperiod: The relative lengths of dark and light periods in a day vary from place to place and from season to season. The length of the light period is called the photoperiod. At equators, the day length is of 12 hours duration throughout the year.

Types Of Plants According To Photoperiodic Requirements For Flowering

NEET Biology Plant Growth And Development Types Of Plants According To Photoperiodic Requirments For Flowering

For SDP, PR/PFR > 1 while for LDP, PFR/PR > 1 is critical for flowering.

Photoperiodic stimulus is perceived by the leaf. When proper photoperiod is perceived, a flowering hormone called florigen is synthesized in the leaf and is transported to the bud where flowering occurs. Florigen, a hypothetical hormone, is chemically similar to gibberellins.

Difference Between Long-Day And Short-Day Plants

NEET Biology Plant Growth And Development Differences Between Long Day And Short Day Plants

NEET Biology Notes For Plant Growth And Development Vernalization (Yarovization)

The term vernalization was coined by the Russian agronomist Lysenko to refer to the method of accelerating the flowering ability of biennials or winter annuals, by exposing their soaked seeds to low temperatures for a few weeks.

  • However, presently the term is used in a wider sense to include the promotion of flowering in plants by exposing them to low temperatures at any stage in their life cycle. It has been found that some plants especially biennials and perennials are stimulated to flower by exposure to low temperatures.
  • This promotive effect of temperature on flowering is called vernalization. The vernalization was first studied in Europe on the winter varieties of cereals such as wheat, barley, oats, and rye by Klippart.

Site Of Vernalization: The sites of vernalization are the shoot tip, embryo tip, and root apex. As a result of vernalization, a hormone called vemalin (by Melcher) is synthesized.

Requirements Of Vernalization

  1. Low temperature: 0°C-5°C
  2. Period of low temperature: A few hours to a few days
  3. Actively dividing cells (meristematic cells)
  4. Water
  5. Aerobic condition
  6. Proper nourishment

As a result of vernalization, the vegetative phase of the plant is shortened and flowering is initiated. Therefore, the duration of crops is reduced.

NEET Biology Notes For Plant Growth And Development Senescence

Senescence is the study of aging in plants. It is of the following types:

Sequential Senescence: In many perennial plants, the apical meristems continue to produce new buds and leaves, while the older leaves and lateral organs undergo senescence and die. For example, Eucalyptus, mango.

Shoot Senescence: In some perennials such as banana and Gladiolus, the aboveground part of the shoot dies every year after producing flowers and fruits. However, the underground parts survive and give rise to new shoots again in the following year.

Simultaneous Or Synchronous Senescence: In some trees such as elm, Dalbergia, and maples, all the leaves are shed in late autumn (October).

Whole Plant Senescence: In monocarpic plants that flower and produce fruits only once in life, for example, wheat, rice, mustard, etc.

NEET Biology Notes For Plant Growth And Development Abscission

It is a natural separation or shedding of leaves, foliage branches, fruits, floral parts, etc., from plants. It is generally seasonal. A special narrow zone develops in the area of abscission called as abscission zone. Two distinct layers develop in the abscission zone:

  1. Separation layer (upper layer) and
  2. Protective layer (lower suberized and lignitized layer).

NEET Biology Notes For Plant Growth And Development Points To Remember

Strasburger studied growth in roots by marking it at equal intervals with Indian ink.

  1. Ethylene causes respiratory climacteric.
  2. The most widely occurring cytokinin in plants is isopentenyl adenine (IPA).
  3. Sleep disease (enrolling of petals in opened flowers) is caused by ethylene. Even 1 ppm of ethylene prevents the opening of flower buds.
  4. Zinc is important for auxin synthesis.
  5. Clinostat is an instrument used for eliminating the effect of gravity.
  6. Phytochrome is a regulatory pigment. It regulates several light-dependent developmental processes in it called photomorphogenetic processes.
  7. Ethylene is the most widely used plant growth regulator in agriculture.

 

NEET Biology Notes For Plant Growth And Development Assertion Reasoning Questions And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: If a plant is kept horizontally, auxin accumulates on the lower surface.

Reason: The displacement of statoliths and other cell organelles to lower surfaces modifies the translocation pattern of auxins.

Answer: 1. If both Assertion and Reason are true and the Rea¬son is the correct explanation of the Assertion.

Question 2. Assertion: Only bud and embryo can be vernalized.

Reason: Vernalization requires dividing cells.

Answer: 3. If Assertion is true, but Reason is false.

Question 3. Assertion: Phytochrome, a protein, has regulatory functions.

Reason: Various morphogenetic processes are regulated by it.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 4. Assertion: Auxin treatment causes acidification of the cell wall and helps in cell elongation.

Reason: Loosening of cell wall microfibrils occurs.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 5. Assertion: Cytokinins are anti-aging hormones.

Reason: They cause changes in osmotic potential by increasing the volume of mature cells.

Answer: 4. If both Assertion and Reason are false.

 

 

NEET Biology Notes – Photosynthesis In Higher Plants

by

NEET Biology Short Notes PDF Photosynthesis In Higher Plants

Photosynthesis: Photosynthesis (photo = light, synthesis = putting together) is the process of the formation of simple sugars by green plants, some bacteria, and some protistans from water, soil, or from carbon dioxide in the air in the presence of sunlight and chlorophyll.

  • By the process of photosynthesis, solar energy is trapped by autotrophic organisms and stored in the form of chemical energy. Annually, about 75 x 1012 kg of carbon (in the form of CO2) is fixed through photosynthesis, producing about 1700 million tonnes of dry matter.
  • About 90% of it is carried out in the oceans (largely by phytoplanktons and algae). Only 0.2% of the light energy falling on earth is utilized by photosynthetic organisms. Photosynthesis first appeared in cyanobacteria.

” photosynthesis in higher plant”

NEET Biology Short Notes Contributions By Some Scientists

Joseph Priestley: Vegetation purifies (dephlogiston) air.

Jan Ingen-Housz: Discovered photosynthesis.

Lavoisier: Phlogiston is CO2 and dephlogiston is O2.

de Saussure: H2O is involved in photosynthesis, CO2 is used and O2 is evolved.

Dutrochet: The established connection between chlorophyll and photosynthesis.

Ruben, Kamen: Using O18 (H2O), he found that O2 involved in the photosynthesis comes from H2O (in Chlorella).

Read and Learn More NEET Biology Notes

NEET Biology Important Topics Photosynthetic Pigments

The two types of photosynthetic pigments present in higher plants are chlorophyll (discovered by Pelletier and Caventau) and carotenoids.

Chlorophyll: Chlorophylls are specialized lipid molecules embedded in the thylakoid membrane. They are the main pigment concerned with the harvesting or trapping of solar energy. Amoff and Allen (1966) recognized nine types of chlorophyll:

  1. Chlorophyll-a,
  2. Chlorophyll-b,
  3. Chlorophyll-c,
  4. Chlorophyll-d,
  5. Chlorophyll-e,
  6. Bacteriochiorophyll-a,
  7. Bacteriochlorophyll-b,
  8. Chlorobium chlorophyll-650, and
  9. Chlorobium chlorophyll-666.

Out of these, the two important types of chlorophyll found in green plants are chlorophyll-a and chlorophyll-b.

Chlorophyll-a: universal pigment found in all green plants (C55H72O5N4Mg).

Chlorophyll-b: found in green algae and higher plants (C55H70O6N4Mg).

Chlorophyll-c: found in diatoms, dinoflagellates, and brown algae (C55H32O5N4Mg).

Chlorophyll-d: found in red algae.

The chlorophyll molecule is asymmetrical and consists of two parts:

  1. Porphyrin head (15 x 15 Å; hydrophilic)
  2. Phylol fail (20 Å; hydrophobic)

The porphyrin ring is a Mat. square structure containing four smaller pyrrole rings (tetrapyrrole) with a magnesium atom at the center. The porphyrin ring has several side groups which alter the properties of the pigment. In chlorophyll-a, there is a -CH3 (methyl) group, while in chlorophyll-b, it is replaced by – CHO (aldehyde) group.

” photosynthesis in higher plants class 11 notes pdf download”

An isocyclic cyclopentanone ring is also attached to the third pyrrole group. The head is joined to a long hydrocarbon tail (phylol). Phylol esterification is absent in chlorophyll-c.

NEET Biology Photosynthesis In Higher Plants Structure Of Chlorophyll-a

Chloiophyll-a is the most abundant photosynthetic pigment It is the only pigment found in all photosynthetic plants Chlorophylls absorb light near both ends of the visible spectrum, the blue and red light, and transmit or reflect green light, Therefore, these appear to be green in color.

Chlorophylls are synthesized from the precursor protochlorophyll. Its synthesis starts from glycine and succinyl CoA.

Fluorescence: Isolated chlorophyll-a when dissolved in organic solvents and exposed to light emits red color immediately. It is called fluorescence. It is, in fact, a manifestation of loss of light energy absorbed in excess. Actually, when the absorbed energy is not utilized for a purpose (phosphorylation), it is emitted as radiations of longer wavelength in green plants. Of course, this is utilized to synthesize chemical energy (ATP).

Carotenoids: Carotenoids arc yellow to orange pigments and are specialized lipids that absorb light strongly in the blue-violet range. These are called shield pigments as they protect chlorophyll from photo-oxidation (photobleaching) by the light of high intensity and also from oxygen produced during photosynthesis. These absorb light and transfer it to chlorophyll for use in pho¬tosynthesis, and therefore, function as a light-harvesting complex. Carotenoids are of two types:

  1. Carotenes (C40H56) are orange in color and xanthophylls are yellow in color. Carotenes are hydrocarbons, most of them being tetraterpenes, for example, lycopene
  2. Xanthophylls (C40H56O2) are very similar to carotenes but contain oxygen, for example, violaxanthin, and fucoxanthin.

Phycobilins: Phycobilins are proteinaceous pigments found in phycobilisomes. They are soluble in hot water and do not contain magnesium and tail. They are generally of two types: phycocyanin (for example, blue-green algae) and phycoerythrin (for example, red algae). These pigments are accessory pigments and also help in chromatic adaptation (Gaidukov phenomenon).

photosynthesis in higher plants short notes

Quantasome: Park and Biggins (1964) loaned the term quantasome for a group of pigment molecules required for carrying out a photochemical reaction. These are situated as small units on the membranes of thylakoids. Each quantasome consists of about 230 chlorophyll molecules, carotenoids, quinone compounds, sulpholipids, phospholipids, proteins, etc.

NEET Biology Important Topics Absorption And Action Spectra

Different plant pigments absorb only certain wavelengths of light and these wavelengths are not absorbed at the same rate. A curve obtained by plotting the amount of absorption to different wavelengths of light by a particular pigment is called an absorption spectrum.

An action spectrum is a curve showing the effectiveness of different wavelengths of light in stimulating the process being investigated. The effectiveness of different wavelengths of light on photosynthesis is measured and plotted by quantum yield or amount of action (expressed through CO2 reduction, O2 evolution, etc.)

NEET Biology Photosynthesis In Higher Plants Graph Of Absorption Spectra Of Chlorophyll-a And Chlorophyll-b And Carotenes

Hill Reaction: Hill working on Stellana media demonstrated that the evolution of O2 occurs from illuminated chloroplast in the presence of hydrogen or electron acceptors but in the absence of CO2, it produces assimilatory power for use in CO2 assimilation. Hill reaction is now considered to be equivalent to light reaction. Hill oxidants are hydrogen acceptors. The common ones are ferricyanide, benzoquinone, dichlorophenol indophenol, NADP+ (natural H+ acceptor in photosynthesis).

Quantum Yield: It is the number of O2 produced per quantum of light absorbed. It represents the rate of photosynthesis.

Quantum Requirement: It is the number of light quanta needed for the production of one O2 molecule or reduction of one O2 molecule.

Red Drop: Emerson measured quantum yield in Chlorella by exposing it to different monochromatic light. The yield was almost constant in all wavelengths but dropped suddenly in red light. It is called red drop.

Enhancement Effect: Emerson worked on chlorella and gave the concept of two photosystems. When Emerson and other scientists provided monochromatic light, then quantum yield suddenly dropped down (called red drop). When Emerson gave combined light, then photosynthesis increased. This is called the Emerson enhancement effect.

  1. 680 nm ↑: Red drop effect
  2. 680 nm ↓ + 680 nm ↑: Emerson effect

Thus, both lights had a synergistic effect, i.e., one helped the other. It leads to the conclusion that two groups of pigments, one absorbing in the lower range and the other absorbing in the higher range of wavelengths, are involved in photosynthesis and are called pigment systems.

NEET Biology Important Topics Pigment Systems

It is a group of 250-100 pigment molecules having chlorophyll- a, chlorophyll-b, carotenoids, etc. One molecule of chlorophyll-a functions as a reaction center or trapping center and the other acts as a light-harvesting complex (LHC) or antenna molecule. These antenna molecules collect light energy and transfer this energy (by inductive resonance process) to the reaction center where the primary photochemical act occurs (quantum conversion), i.e., absorption of a photon and release of an electron.

NEET Biology Photosynthesis In Higher Plants Pigment System

Differences Between Pigment Systems 1 And 2

NEET Biology Photosynthesis In Higher Plants Differences Between Pigment System 1 And Pigment System 2

NEET Biology Photosynthesis In Higher Plants Pigment System 1 And Pigment System 2

NEET Biology Important Topics Photolysis Of H2O

The photolysis of water occurs in the lumen of the thylakoid and requires minerals Mn+, CI–, Ca+2, and an oxygen-evolving complex (OEC).

2H2O → 4H+ + 4e– + O2

Released electrons are attracted toward reduced reaction center P680 of PS-2, while H+ accumulates in the lumen and finally reduces NADP.

NEET Biology Important Topics Mechanism Of Photosynthesis

Photosynthesis occurs In Two Phases:

  1. Light Reaction: Solar energy is trapped by chlorophyll and is stored in the form of chemical energy (ATP) and as reducing power (NADPH).
  2. Bark Reaction: Reducing capacity or NADPH and the energy of ATP are utilized in the conversion of CO2 to carbohydrates.

Warburg and others, using reaction inhibitors, proved that light and dark reactions occur independently.

1. Light Reaction: Emerson and Arnold (1932) subjected unicellular algae, Chlorella, to brief flashes of light or continued light at high intensities and observed O2 evolution The extent of photosynthesis (per unit of light energy provided to plant) was higher when light and dark periods were alternated (photosynthetic efficiency increases in intermittent light).

  • It showed that the process of photosynthesis consisted of distinct light and dark reactions (the products of the light phase are not utilized immediately in the dark phase, so photosynthetic efficiency decreases in continuous saturating light).
  • During Photosynthesis, 264 g of CO2 and 216 g of water are utilized to produce 108 g of water and 192 g of O2.

Redox Potential: It is the tendency of an atom/molecule having low redox potential to lose electrons (electron donors) while those with high redox potential to accept electrons (gain electrons). Hence, electrons move from substances having low redox potentials to those having high redox potentials.

Assimilatory Power: NADPH + H+ and ATP produced in light reaction constitute assimilatory power.

Photophosphorylation: The generation of ATP in light reaction is called photophosphorylation. It was first explained by P. Mitchell who gave chemiosmotic theory. As a result of photolysis and quinone pump, H+ concentration in the thylakoid lumen increases by 1000-2000 times, and as a result, a proton motive force develops. This force is responsible for ATP synthesis on the head of CF0-CF1 a particle plugged into the thylakoid membrane.

Difference Between Photophosphorylation

NEET Biology Photosynthesis In Higher Plants Difference Between Cyclic And Non Cyclic Photophosphorylation

NEET Biology Photosynthesis In Higher Plants Cyclic And Non Cyclic Photophosphorylation In Thylakoid Membrane

2. Dark Reaction (C3 Cycle, Calvin Cycle): The biosynthetic phase (dark reaction) is so-called because it is independent of light. The ATP and NADPH produced by the light reactions are utilized in the dark reaction to reduce carbon dioxide to carbohydrates by a process called carbon fixation. It occurs in the stroma. The process comprises a series of reactions controlled by enzymes. The sequence of these reactions was determined in Chlorella and Scenedesmus by Calvin, Benson, and

Bassham uses radioactive carbon 14C and techniques such as chromatography and autoradiography. Therefore, it is also known as the Calvin cycle or Calvin Benson cycle. The enzyme for CO2 fixation is RuBisCo. It is a large enzyme found in stroma. It is the most abundant protein on earth and constitutes 16% of the chloroplastic protein. It shows a bifunctional nature having both carboxylase and oxygenase activity. The dark reaction is also known as the black reaction. The whole reaction can be studied in three parts

photosynthesis notes class 11

NEET Biology Photosynthesis In Higher Plants Dark Reaction C3 Cycle And Melvin Calvin Cycle

  1. Carboxylation (acceptance of CO2 by RuBP—CO2 acceptor)
  2. Glycolytic reversal
  3. Regeneration of RuBP

The formation of a molecule of glucose requires 18 ATPs and 12 NADPH + H+ (or to reduce 60O2).

NEET Biology Important Topics Photorespiration

Since 1920, it has been known that higher concentrations of oxygen inhibit photosynthesis (Warburg effect). However, the reason was discovered in 1971. It was shown that RuBP carboxylase (RuBisCo) accepts not only CO2 but also oxygen as a substrate. If oxygen is accepted, the following reaction occurs:

NEET Biology Photosynthesis In Higher Plants Photorespiration Formula

Oxygen is a competitive inhibitor of CO2 fixation. Any increase in O2 concentration would favor the uptake of oxygen rather than CO2, and thus inhibit photosynthesis.

  • The phosphoglycolate is immediately converted to glycolate. The peroxisomes present in the cell metabolize the glycolate into glycine and glycine into seine and carbon dioxide without the production of ATP or NADPH. This process is called photorespiration. Photorespiration is defined as a light-dependent uptake of oxygen and the output of carbon dioxide.
  • Photorespiration has no relation with normal respiration called dark respiration. Both of these resemble only in one point, i.e., O2 is used and CO2 is released. Photorespiration depends upon light for the supply of RuBP. RuBP is available only when photosynthesis is operating, as RuBP is regenerated in the Calvin cycle.
  • The function of photorespiration is to recover some of the carbon from the excess glycolate. However, there is a wasteful loss of carbon as CO2 (when glycine is oxidized to serine) and energy. Although ATP is produced when glycine is oxidized to serine, the overall process is energy-consuming. Up to one-fourth of the photosynthetically fixed CO2 may be lost by photorespiration.

With the increase in temperature, light intensity, and oxygen concentration, the affinity of RuBisCo for CO2 decreases, and its affinity for oxygen increases. Thus, a rise in the temperature means more loss of photosynthetically fixed carbon by photorespiration. Photorespiration reduces the potential yield of plants growing in the tropics by 30-40%.

NEET Biology Photosynthesis In Higher Plants Photo Respiration Process

Differences Between Normal Respiration And Photorespiration

NEET Biology Photosynthesis In Higher Plants Differences Between Normal Respiration And Photorespiration

NEET Biology Important Topics Dicarboxylic Acid Cycle (Cooperative photosynthesis)

H.P. Kortschak and C.E. Hartt (1965) found that in sugarcane (a tropical plant), leaves removed CO2 more efficiently from the atmosphere, and the first products of photosynthesis were acids containing four-carbon acids (for example, malic, oxaloacetate, and aspartic), rather than the 3C acid PGA.

Since then, the same has been found true for many important tropical plants including monocots (such as maize, Sorghum, and Eleusine) as well as dicots (such as Amaranthus and Euphorbia sp.). These plants are called C4 plants. On the other hand, the plants in which the first product for photosynthesis is O3 acid PGA are called C3 plants. Calvin cycle, in fact, is a C3 pathway.

NEET Biology Photosynthesis In Higher Plants Photosynthetic Reactions Taking Place In The Mesophyll And Bundle Sheath Cells

In 1966, two Australian scientists, Hatch and Slack, showed that C4 plants were much more efficient in CO2 utilization than C3 plants. Such plants do not show photorespiration. In 1967, Hatch and Slack explained the manner of CO2 fixation and reduction in such plants. The new carbon pathway in C4 plants is called the Hatch-Slack pathway.

  • The C4 plants possess a characteristic leaf anatomy. Their vascular bundles are surrounded by two rings of cells. The inner ring, called bundle sheath cells, contains starch-rich chloroplasts lacking grana which differ from those in the mesophyll cells that make the outer ring.
  • The chloroplasts in these plants are, therefore, called dimorphic. This peculiar anatomy is called kranz anatomy because kranz means crown or halo, which refers to two distinct rings of cells. Correlation between Kranz anatomy and C4 photosynthesis was established by Dowton and Treguna.

Since every CO2 molecule has to be fixed twice, hence C4 pathway is more energy-consuming than the C3 pathway. The C pathway requires 18 ATP and 12 NADPH2, for the synthesis of one molecule of glucose.

On the other hand, the C4 pathway requires 30 ATP and 12 NADPH2. However since otherwise tropical plants lose more than one-fourth of the photosynthetic carbon in photorespiration, the C4 pathway is an adaptive mechanism for minimizing the loss.

NEET Biology Important Topics Cam Cycle (In Succulent Plants)

The leaves of plants belonging to the Crassulaceae family. for example. certain cacti, orchids, pineapple, Kahuwhoe, and Scdum undergo crassulacean acid metabolism and such plants are called Crassulacean acid metabolism (CAM) plants.

  1. All CAM plants are succulent inhabit. As such, stomata remain open in night and closed during the day (scotoactive stomata).
  2. CO2 is fixed during the night (dark) to malic acid via PEP carboxylase. This CO2 comes from respiration (breakdown of starch) as well as from the atmosphere. Malic acid gets stored in vacuoles.
  3. The CAM plants also contain the enzyme of the Calvin cycle. Thus, during the daytime, malic acid breaks into pyruvate and CO2. While CO2 enters the Calvin cycle, pyruvate is used up to regenerate PEP.
  4. The succulents, therefore, synthesize plenty of organic acids from CO2 during the night (when stomata are open) and plenty of carbohydrates during the day (when stomata are closed).
  5. Like the Calvin cycle, the CAM cycle also operates in the mesophyll cell. None of these have shown chloroplast dimorphism as is found in C4 plants.
  6. It should be remembered that the slow-growing desert succulents exhibiting the CAM cycle have the slowest photosynthetic rate, while the species possessing C4 pathway possess the highest rates. Thus, CAM plants are although not as efficient as C4 plants, they are definitely better suited to adverse conditions (i.e., conditions of extreme dissipation).

NEET Biology Photosynthesis In Higher Plants Schematic Representation Of Acidification In Dark And Deacidification In Light

Blackman (1905) proposed the law of limiting factor which states that “when a biological process is conditioned as to its rapidity by a number of separate factors, the role of the process is limited by the pace of the slowest factor.”

NEET Biology Important Topics Factors Affecting Photosynthesis

Photosynthesis is regulated by many factors.

Light: Light is the most important factor for photosynthesis because it is used as a source of energy. Normally, plants utilize sunlight, but marine algae also use moonlight. Photosynthesis even occurs in electric light.

  1. Quality Of Light: Photosynthesis occurs in the visible part of the spectrum. Photosynthesis is maximum in polychromatic light or white light.
  2. Intensity Of Light: Only 1-4% of light is utilized in photosynthesis. In general, the rate of photosynthesis is higher in intense light than in diffused light. (Upto 10% light is utilized in sugar cane, i.e., the most efficient converter). For a complete plant, the rate of photosynthesis increases with an increase in light intensity, except for very high light intensity where the “solarization” phenomenon occurs, i.e., photo-oxidation of different cellular components including chlorophyll occurs.

CO2: The normal concentration of CO2 in the atmosphere is 360 ppm. By increasing CO2 concentration 15-20 times, the rate of photosynthesis increases, but after that it decreases. C4 plants show saturation at about 360 ppm.

Temperature: Generally photosynthesis is inhibited at 0°C, but some conifers even photosynthesize up to 35°C. Opuntia photosynthesizes up to 55°C. The Q10 value for photosynthesis is 2 up to about 40°C, and above it, the denaturation of enzymes begins, and hence, the rate of photosynthesis decreases.

H2O: As less than 1% of the total water absorbed by the plant is utilized in photosynthesis, so water rarely acts as a limiting factor. In water-deficient conditions, photosynthesis is found to be decreased.

Chlorophyll Content: I fall other factors are favorable, an increase in chlorophyll leads to an increase in photosynthesis.

Accumulation Of End Products: It results in a decrease in the rate of photosynthesis.

Photosynthesis In Higher Plants Points To Remember

Photosynthesis is an anabolic endergonic, oxidation-reduction process.

  • PAR (photosynthetically active radiation) is the wavelength included in the range 400-700 nm.
  • Destruction of chlorophyll due to high light intensity is called solarization.
  • The most common limiting factor for photosynthesis is CO2.

Light Compensation Point: The light intensity at which the rate of photosynthesis is equal to the rate of respiration (in the morning and evening).

  • DCMU (dichlorophenyl dimethyl urea)—a herbicide—inhibits PS-2 and oxygen release in the light phase.
  • In C3 plants, more CO2 is released in light than in dark due to photorespiration.
  • The yield of C3 plants is increased by increasing CO2 in the atmosphere but not in C4 plants.
  • Chlorophyll-b absorbs more of a red wavelength than chlorophyll-a.
  • Chloride ions stimulate the quick release of electrons from water.
  • Phaeophytin is chlorophyll-a without Mg++.
  • The size of quant some is about 180 x 155 x 100 Å
  • In C4 plants, the bundle sheath cells possess chloroplasts without grana.
  • Atriplex hastata show’s Calvin cycle whereas A triplex roseus shows the Hatch-Slack cycle.
  • The reaction center in bacteria is B-890 and the reducing agent is NADH2.
  • In bacterial photosynthesis, the type of photosynthesis is anoxygenic and cyclic.

CO2 Compensation Point: The CO2 concentration at which the rate of photosynthesis is equal to the rate of respiration.

For C3 plants → 50-100 ppm

For C4 plants → 0-10 ppm

NEET Biology Photosynthesis In Higher Plants Question And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: In C4 plants, the chloroplasts of bundle sheath cells are granular.

Reason: PS-2 is mostly found in the appressed part of granum.

Answer: 4. If both Assertion and Reason are false.

Question 2. Assertion: Dark reactions of photosynthesis are temperature-controlled processes.

Reason: Most of the reactions are enzymatic in nature.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 3. Assertion: Dark acidification of cytoplasm occurs in CAM plants.

Reason: Organic acids are decarboxylated during the night.

Answer: 3. If Assertion is true, but Reason is false.

Question 4. Assertion: Assimilatory power in photosynthesis is generated in ETS occurring in 23 thylakoid membranes.

Reason: They are needed for CO reduction.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 5. Assertion: Light-harvesting complexes (LHC) on thylakoid membranes broaden the range of light absorption.

Reason: They transfer electrons to the reaction center.

Answer: 3. If Assertion is true, but Reason is false.

NEET Biology Notes – Transport In Plant

by

NEET Biology Notes Transport In Plant Importance Of Water

Water is one of the most important constituents of the protoplasm. It makes up 80-90% of the fresh weight of most herbaceous plants and over 50% of the fresh weight of woody plants. It acts as a solvent in which the oases, minerals, and other solutes enter the plant cells and move from cell to cell and from organ to organ.

  • Water is a reactant or reagent in many important processes such as the hydrolysis of starch to sugar. Water maintains turgidity which is essential for cell enlargement, growth, and maintaining the form of herbaceous plants.
  • Turgor is also important in the opening and closing of stomata, movement of leaves, flower petals, and various specialized plant structures. Water is also needed for the hydration of protoplasm, in the mobility of gametes, dehiscence, etc.

Diffusion: Diffusion is the movement of particles (or molecules or ions) of a substance from a region of its higher concentration to the region of lower concentration down the concentration gradient until the molecules are evenly distributed throughout the available space.

The rate and direction of a diffusing substance depend upon the concentration of that substance at different spots and is independent of the presence of other diffusing substances. The rate of diffusion is affected by many factors, for example,

transport plants

Rate ∝ Temperature

NEET Biology Transport In Plant Diffusion Formula

Diffusion Pressure (DP): It is the pressure exerted by a substance due to the tendency of its particles to diffuse. It is also defined as the potential ability of a molecule or ion to diffuse. It is directly proportional to the diffusing particles.

DP ∝ Concentration

DP ∝ Temperature

The DP of pure water is maximum.

Facilitated Diffusion: Particles that are lipid soluble can easily pass directly through the cell membrane as it is mainly made of it. The hydrophilic solutes find it difficult to pass through the membrane. Their movement has to be facilitated. For this, the membranes possess aquaporins and ion channels.

NEET Biology Transport In Plant Facilitated Diffusion

Some carrier proteins allow transport only if two types of molecules move together. This is called cotransport. It is of two types. In the symport method of cotransport, both molecules cross the membrane in the same direction at the same time. In the antiport method of cotransport, both molecules move in opposite directions. When a molecule moves across a membrane independent of another molecule, the process is called uniport.

Read and Learn More NEET Biology Notes

NEET Biology Transport In Plant Types Of Transport Of Carrier Proteins

NEET Biology Notes Transport In Plant Osmosis (Diffusion Through Membrane)

Osmosis may be defined as “the passage of solvent molecules from a region of their higher concentration to a region of their lower concentration through a semipermeable membrane.” In all biological systems, the solvent is water.

Types Of Membranes

  • Permeable Membranes: Such membranes allow diffusion of both solvent and solute molecules or ions through them, for example, the cellulose wall of cells. Lignified cell walls are also quite freely permeable to water.
  • Impermeable Membranes: Such membranes prohibit the diffusion of both solvent and solute particles through them, for example, heavily cutinized or suberized cell walls in plants.
  • Semipermeable Membranes: Such membranes allow the diffusion of solvent molecules but do not allow the passage of solute molecules. Such membranes form a perfect partition between two osmometers, for example, membranes of collodion, parchment paper, and copper ferrocyanide membranes.

NEET Biology Transport In Plant Osmosis Types Of Membrances

Differentially Permeable Membranes: However, all membranes found in plants allow some solutes to pass through them along with the solvent molecules. Such membranes are called differentially permeable membranes, for example, all biological membranes.

NEET Biology Notes Transport In Plant Osmotic Pressure

Osmotic pressure (OP) is the “maximum pressure which develops in a solution when it is separated from pure water by a semipermeable membrane.” It is also defined as “the pressure needed to prevent the passage of pure water into an aqueous solution through a semipermeable membrane thereby preventing an increase in the volume of the solution”.

NEET Biology Transport In Plant Osmotic Pressure

The osmotic pressure depends upon

  1. The concentration of solute particles,
  2. Ionization of solute particles,
  3. Hydration of solute particles, and
  4. Temperature.

An increase in the concentration of solutes in the solution increases the osmotic pressure. If the solute ionizes in the solution, the number of particles increases, thus raising the osmotic pressure. If the solute molecules are hydrated, the water molecules bound with the solute will be unable to diffuse and hence increase the osmotic pressure.

” transportation in plants”

  • An increase in the temperature raises the osmotic pressure of the solution.
  • Plant cells exhibit a considerable range of variations in osmotic pressure. In land plants, it varies from 5 to 30 atm.
  • In aquatic plants, it varies from 1 to 3 atm. Plants of arid regions possess high osmotic pressure. The highest osmotic pressure is recorded in the halophytic plant, Atriplex confertifolia, i.e., 202.5 atm.
  • The osmotic pressure of an electrolyte is two to three times greater than a non-electrolyte.
  • Osmotic pressure can be calculated by OP = miRT where m is the molar concentration, i is the ionization constant, R is the gas constant, and T is the temperature (273 °C) (π). Osmotic pressure is numerically equal to osmotic/solute potential (ψs) but has a positive value ψs = π.

Measurement Of Osmotic Pressure: The methods for measuring osmotic pressure are

  1. Plasmolytic method by de Vries
  2. Pleffer osmometer
  3. Berkeley/Hartley osmometer
  4. Cryoscopic osmometer

NEET Biology Notes Transport In Plant Turgidity And Turgor Pressure

If a plant cell is placed in a hypotonic solution or pure water (i.e., a solution of higher water potential), the water starts moving into the cell by osmosis. As the volume of the protoplast increases, it begins to exert pressure against the cell wall and thus stretches it.

  • The pressure exerted by the protoplast against the cell wall is called turgor pressure (TP). The cell wall, being rigid, exerts an equal and opposite pressure on the protoplast, which is called wall pressure (WP).
  • The two pressures are equal and opposite in direction. As the turgor pressure of the cell increases, the cell becomes turgid. However, a stage is reached when osmotic pressure is exactly balanced by turgor pressure.
  • At this point, the amount of water leaving the cell equals that entering the cell. Hence, there is no further net movement of water and the cell is said to be in equilibrium with the exterior solution.

NEET Biology Notes Transport In Plant Diffusion Pressure Deficit (Suction Pressure)

The diffusion pressure of a pure solvent is maximum. The addition of solutes results in a decrease in the diffusion pressure. This deficit is termed diffusion pressure deficit (DPD). The greater the concentration of a solution, the greater is its DPD.

  • Such a solution will tend to take up water and become normal; therefore, water moves from low DPD to high DPD. The value of DPD for a cell is always positive. DPD for a cell is always positive and is calculated as (QP – TP).
  • DPD is an index of the water-absorbing capacity of cell. In a flaccid cell, TP = 0. Therefore, DPD = OP.
  • In a turgid cell, the entry of water into the cell causes the development of TP. At a point, TP = OP. Therefore, DPD = 0. Therefore, in a fully turgid cell, there is no net movement of water into the cell.

NEET Biology Notes Transport In Plant Chemical Potential

Chemical potential is the free energy of one mole of a particular chemical in a multi-component system. The lamer the chemical potential of a substance, the greater its tendency to undergo a chemical reaction. The chemical potential of water is called water potential and is given the value of zero at prevailing temperature and pressure.

NEET Biology Notes Transport In Plant Water Potential

Currently, the term water potential is used by biologists to describe the tendency of water molecules to move from one place to another. It is denoted by the symbol “ψ” (the Greek letter, psi).

Components Of Water Potential: Water potential is the difference in the free energy per unit molal volume of water in the pure state and in a system. For a solution, the value of water potential is always less than zero or negative. Water potential is determined by the following relation

Water Potential (ψ) = Solute potential (ψs) + Pressure potential (ψp)

Solute Potential (ψs): Also called osmotic potential, it is defined as the amount by which Tyr is reduced as a result of the addition of solute. It is always negative.

Pressure Potential (ψp): It is equal to TP and is positive except in plasmolyzed cell and in xylem vessels where it is negative. In terms of water potential, osmosis can be redefined as “the movement of water molecules from a region of higher water potential to the region of lower water potential through a differentially permeable membrane.”

Water potential may be regarded as the tendency of water to leave a system. The more the water potential, the greater is the tendency to leave. If two systems are in contact, water moves from the system with higher water potential to that with lower water potential or less negative ψw to more negative ψw.

Importance Of Osmosis: Osmosis is involved in

  1. Water absorption by roots
  2. Maintenance of turgidity
  3. Turgor movements in Mimosa and Desmodium.
  4. Stomatal movement
  5. Self-dispersal of fruits (autochory)
  6. Drought and frost resistance

NEET Biology Notes Transport In Plant Plasmolysis

When a cell is placed in a hypertonic solution, exosmosis occurs. Due to the loss of water to the external solution, TP of the cell decreases, hence the cell shape slightly changes. This stage is called limiting plasmolysis.

  • If the cell continues to be in a hypertonic solution, more and more water is lost, therefore protoplast starts shrinking. It first leaves the comers of the cell and this stage is called as incipient plasmolysis.
  • If more water is lost, the protoplast shrinks but remains in contact with the wall in one or two places. The space between the cell wall and the protoplast is occupied by the outer solution. This stage is called evident plasmolysis. The turgor pressure of a cell at limiting plasmolysis is equal to zero, whereas at incipient plasmolysis and evident plasmolysis, TP of the cell is negative.

NEET Biology Notes Transport In Plant Imbibition

The adsorption as well as absorption of a liquid by a solid without forming a solution is called as imbibition. The solid sub¬stance is called imbibing, whereas liquid is called as imbibe. The liquid particles are held in between the particles of solid by adsorption and capillarity.

The phenomenon depends upon the affinity of imbibant to imbibe. As a result of imbibition, the following events occur

Volume Change: The volume of the system increases, i.e., swelling. The total volume of water imbibed, plus the imbibing material is less after imbibition than before.

Heat Release: This is due to the compact arrangement of water molecules which liberate some of their kinetic energy.

Development Of Pressure: If the imbibing is confined, pressure may be developed. For example, in dry pea seeds, it is 1000 bars.

“transport in plants “

Imbibants present in plants are generally hydrophilic in nature. Among plant imbibing, hydrocolloids (for example, agar-agar, etc.) are the best imbibing followed by proteins, starch, and cellulose. Imbibition is affected by a number of factors such as temperature, pressure, etc., against the imbibing surface, the texture of imbibing, electrolytes, and pH.

NEET Biology Notes Transport In Plant Hypotonic Hypertonic And Isotonic Solutions

A solution with a higher concentration of solutes is said to be the hypertonic solution. A lower concentration of solutes in a solution is called a hypotonic solution, and two solutions having the same concentration of solutes are called isotonic solutions.

NEET Biology Transport In Plant Hypertonic And Hypotonic Solutions

If a cell is placed in pure water or hypotonic solution, there is a net movement of water into the cell (endosmosis); if it is placed in a hypertonic solution, there is a net movement of water from the cell outwards (exosmosis); if the cell is placed in an isotonic solution, the amount of water leaving the cell equals that entering the cell and therefore, there is no net movement of water.

NEET Biology Transport In Plant Effect Of Isotonic Solution Hypertonic Solution And Hypotonic Solution

NEET Biology Notes Transport In Plant Classification Of Soil Moisture

The amount of water that can be held by soil depends upon the total pore space in the SCA. Water is present in the spaces between the soil. Various types of groundwater are

Run-off Water: Water that flows along the surface of the soil and reaches to the nearest water body constitutes run-off water. This water is not available to the plants.

Gravitational Water: Water that percolates down through the soil macropores (50 piM diameter) under the influence of gravity and reaches to the water table is called gravitational water. This type of water is also not available to the plants.

Capillary Water: Water retained by soil micropores (20 pm diameter) in the form of fine capillaries constitutes capillary water. This type of water is available to the plants.

Hygroscopic Water: Water held in the form of a very thin film around the soil particles due to adsorption constitutes hygroscopic water. This type of water is not available to the plant.

Chemically Combined Water: Water that combines with inorganic salts of the soil in the form of water of hydration constitutes chemically combined water. This type of water is also not available to the plants.

Water Vapors: They are present in soil air spaces. Normally, they are not available to the plants. Under certain conditions, they are useful in the phenomenon of night recovery.

NEET Biology Notes Transport In Plant Absorption Of Water And Pathway Of Water Across The Root

Water is mainly absorbed by the root hair zone. Root hairs are tubular elongations of the external wall of epiblema cells. They range in length from less than a millimeter to about a centimeter and are usually 10 pm in diameter. They have a vacuole filled with salt, sugars, and organic acid.

Hence, the osmotic pressure of the root hair cell is high (approx. 3-8 atm) as compared to that of soil sap (approx, less than 1 atm). The movement of water from the root hair cell to the xylem may occur by two methods.

  1. Apoplast Pathway: In this method, water passes from the root hair cell to the xylem through the walls of intervening cells without crossing any membrane or cytoplasm. The apoplastic movement of water beyond the cortex is blocked due to the presence of Casparian strips in the endodermal cells. The major movement of water through cortical cells occurs by this method, as cortical cells offer the least resistance.
  2. Symplast Pathway: In this method, water passes from cell to cell by crossing the plasma membrane. Therefore, it is also known as a transmembrane pathway. This may occur by two methods
    • Non-Vacuolar Symplast Pathway: In this method, water passes between adjacent cells through plasmodesmata. It does not enter into the vacuoles.
    • Vacuolar Symplast Pathway: In this method, water crosses the tonoplast surrounding the vacuole. This pathway offers a lot of resistance. Beyond the cortex through the endodermis and pericycle water is forced to move through the symplast pathway.

NEET Biology Transport In Plant Apoplastic And Symplastic Pathways

NEET Biology Transport In Plant Three Pathways Of Water Movement

Mechanism Of Water Absorption: Water is mainly absorbed by two different mechanisms

  1. Passive (water is absorbed through roots)
  2. Active (water is absorbed by roots)

Water absorption by rapidly transpiring plants is called passive absorption because forces responsible for water uptake develop in shoots and is transmitted to roots through which water enters. Active absorption depends on forces developing in roots and is found in low-transpiring plants. The table describes the differences between passive and active absorption

Difference Between Passive Absorption And Active Absorption

NEET Biology Transport In Plant Differences Between Passive Absorption And Active Absorption

Factors Affecting Water Absorption: The rate of water absorption decreases with

  1. Increased salt concentration in soil
  2. Decreased soil temperature
  3. Decreased soil aeration (O2)
  4. Decreased soil moisture

Ascent Of Sap: The water absorbed by roots has to be conducted upwards so as to meet the needs of tissues there. This vertical conduction of water from the root to aerial parts of the plant is called the ascent of water or ascent of sap. Various theories explaining the ascent of sap have been put under three categories

  1. Vital Force Theories: These theories explain the ascent of sap by the force developing in living cells.
    • Relay Pump Theory Or Clambering Theory (By Godlewski): According to it, water rises in the stem by the rhythmic changes in osmotic pressure of the xylem parenchyma and medullary rays. The water is relayed in a staircase-like manner to the next higher cell. Now it has been established that water moves through tracheids and vessels. Hence, this theory has been rejected.
    • Pulsation Theory (By J.C. Bose): According to this theory, water rises in the stem due to pulsatory activity in the innermost layer of the cortex. This theory is not accepted nowadays, as neither these pulsations are universal in occurrence nor they are able to explain the magnitude of the ascent of sap.
    • Root Pressure Theory (By Priestley): Due to the movement of water from the soil into the root hairs and from there to xylem cells, hydrostatic pressure develops. This is called root pressure (the term given by Stephan Hales).
      • This pressure pushes the water up in the xylem vessels. The root pressure can be easily observed and measured when a freshly cut stump continues to exude water (bleeding) from its xylem vessels. The development of root pressure is an active process.
      • Stocking has defined root pressure as a pressure developing in tracheary elements of the xylem as a result of the metabolic activity of roots. Root pressure depends upon the active secretion of salts or other solutes into the xylem sap, thereby lowering its osmotic potential.
      • This involves the utilization of metabolically produced energy and is inhibited by respiratory inhibitors such as cyanide, lack of oxygen, and low temperatures. The positive hydrostatic pressure generated by root pressure (maximum 2 atm) is not sufficient to push up water to more than a few meters.
      • It cannot account for water movement up the xylem in tall trees. Also, actively transpiring plants and tall trees, especially conifers, do not generate root pressure. But it is a contributing factor in many plants. It may be sufficient for the transport of water in slowly transpiring herbaceous plants.
  2. Physical Force Theories: According to these theories, some physical force or dead cells are responsible for the ascent of sap.
    • Capillarity Theory (By Boehm): This theory explains the rise of the water column due to the joint force of capillarity and atmospheric pressure, as the xylem vessels have a narrow diameter and act as capillaries. This theory has been rejected as it does not explain the ascent of sap in tall trees because capillary force developed by a narrow vessel of 0.03 mm diameter would support a water column of about 1 m only.
    • Imbibition Theory (By Such): This theory explains the rise of the water column due to the force of imbibition. If it were true, water must rise through the walls of xylem vessels and not through their lumen as it always occurs. Hence, the concept was rejected.
  3. Cohesion-tension Or Transpiration Pull Theory: This is the most widely accepted theory proposed by Dixon and Joley (1894). The main features of the theory are as follows
    • Right from the root hairs to the leaves on top, water forms a continuous column in the plants.
    • Water molecules have high cohesive forces between them, i.e., these tend to stick to each other, because, being polar, they are electrically attracted to each other by hydrogen bonds. The high cohesive force means that a relatively large tension is required to break the water column.
    • In other words, the water column has a great tensile strength. The magnitude is generally 10-30 MPa.
    • The lignin cellulosic cell walls of xylem vessels have a strong affinity (adhesion) for water molecules. Therefore, a strong adhesive force exists between the walls of the xylem vessels and water, i.e., water tends to “stick” to the vessel wall.
    • As the water is lost from the leaf surface by transpiration, the DPD of the leaf cells increases.
    • As a result, the cells develop low water potential and water from the leaf veins (xylem) moves into leaf cells. The xylem vessels, in turn, draw water from the xylem of the main stem. A negative (pulling) pressure is thus exerted by all the leaves on the stem. The combined pressure, called transpiration pun, is strong enough to pull up the column of water to great heights.
    • The whole column of water moves. Water potential as low as -3 MPa or -30 bars has been measured in the leaves borne on tree tops.
    • Such a low water potential is sufficient to create pulling pres-sure which can overcome the gravitational pull and resistance offered by the capillaries of xylem vessels.
  4. Criticism Of Transpiration Pull Theory: The theory assumes tracheids to be more efficient than vessels. Dixon believed that partition walls of the tracheids confer stability on tensile stressed transpiration stream.

NEET Biology Notes Transport In Plant Transpiration

Less than 5% of absorbed water is used by plants and the rest is lost in transpiration.

Types Of Transpiration

  • Cuticular: Loss of water vapor from the general surface through the cuticle. Commonly, it is 3-10% of the total but in herbs and ferns it may be up to 50%.
  • Lenticular: Loss of water vapor from lenticels or aerating pores in the bark of trees or fruits, etc. It is hardly 0.1% of the total transpiration.
  • Bark: It is approximately 1% of the total transpiration.
  • Stomatal: Major form of transpiration constituting 50-97% of the total transpiration.

Number And Distribution Of Stomata: In most plants, there are 1000-60000 stomata per square centimeter of leaf surface. The total pore area is approximately 1-2% of the total leaf area. On the basis of their distribution, stomata are divided into the following categories

  • Apple Type (Mulberry Type): Stomata are present only on the lower surface of the leaf, for example, apple.
  • Potato Type: Stomata are present on both surfaces of the leaf but more on the lower surface, for example, potato, pea, tomato, and many other dicot plants.
  • Oat Type: The number of stomata is equal on both surfaces of the leaf, for example, oat and many other monocots.
  • Water Lily Type: The stomata are present only on the upper surface of the leaf, for example, water lily and most floating plants.
  • Potamogeton Type: Stomata arc cither absent or vestigial, for example, Potamogeton.

Structure Of Stomata: Each stoma is surrounded by two specialized cells called guard cells. They have chloroplasts and arc green and are surrounded by two or more specialized epidermal cells called accessory or subsidiary cells. Guard cells are of two types: kidney-shaped and dumbbell-shaped. Dumbbell-shaped guard cells are found in the family Gramineae and Cyperaceae. Kidney-shaped guard cells have an inner thick wall and an outer thin and elastic wall.

Dumbbell-shaped guard cells have thin-walled ends and thick-walled middle regions. The cell wall bordering the stomatal pore is thicker than that of next to the surrounding cells. Lotfield classified stomata as follows on the basis of their daily opening and closing movement

  • Alfalfa Type: Open the whole day, closed in night. For example, pea, radish, mustard, turnip, apple, grapes.
  • Potato Type: Open the whole day and night except for a few hours in the evening. For example, onion, potato, banana.
  • Barley Type: Open only for a few hours in day. For example, maize, wheat, and other cereals.
  • Equisetum Type: Always open.

Mechanism Of Opening And Closing Of Stomata: Stomata function as turgoroperated valves. When the osmotic concentration of guard cells increases, water comes in guard cells becomes turgid and stomata are open. Whenever the osmotic concentration of the guard cell decreases, water moves out, and guard cells become flaccid, and hence become closed. This increase and decrease in osmotic concentration is explained by a number of theories mentioned below.

NEET Biology Transport In Plant Structure Of Stomata

Photosynthetic Theory (von Mohl And Schwendener): This theory proposes that in the morning, as soon as light is available, the chloroplasts of guard cells start photosynthesis. As a result, sugars are produced, which increase the osmotic concentration of guard cells. Water comes in, guard cells become turgid, and stomata open. This theory was not accepted because the photosynthetic activity of guard cell chloroplasts seems to be negligible and sugar does not occur in detectable quantity in guard cells.

Starch ⇔ Sugar Interconversion Hypothesis (Classical Theory): The theory was proposed by Sayre and Scarth and was modified by Steward on the basis of the activity of the phosphorylase enzyme.

NEET Biology Transport In Plant Strach Sugar Interconversion Hypothesis Formula

According to this theory, guard cells contain starch and phosphorylase enzyme, which causes the conversion of starch into glucose at higher pH which is developed by the utilization of CO2 in photosynthesis.

Glucose increases the osmotic concentration of guard cells, raising the turgor pressure and causing the opening of stomata. At night, CO2 accumulates in cell and intercellular spaces, thus lowering the pH at which phosphorylase causes the conversion of glucose to starch. The osmotic concentration of guard cells decreases. Hence, water moves out and, therefore, stomata are closed.

Objections: In some families, starch is absent, for example, onion. Glu¬cose does not appear in detectable quantity in the guard cells of open stomata, rather malic acid accumulates.

Active K+ Ion Uptake Theory (Levitt 1974): According to this theory, the pH of guard cells rises in the day due to the assimilation of CO2 in the photosynthesis and the uptake of H+ ions by chloroplast and mitochondria from cytoplasm. At this higher pH, starch is converted into phosphoenolpyruvate (PEP). PEP combines with CO2 with the help of PEPCase (phosphoenol pyruvate carboxylase) and forms oxaloacetic acid which gets changed into malic acid. Malic acid dissociates into malate and H+ ions.

  • There occurs an efflux of H+ ions and the influx of K+ ions, which forms potassium malate after reacting with malate. Potassium malate is a highly osmotically active substance which is stored in vacuoles. This raises the osmotic concentration of guard cells, water moves in, cells become turgid, and stomata are open. In the night, the process is reversed.
  • The potassium, chloride, and malate ions have a prominent role in stomatal opening. The ions accumulate in the vacuole of guard cells, lowering the water potential and thereby increasing water uptake and subsequently opening the stomata.

NEET Biology Transport In Plant Active K+ Ion Uptake Mechanism

Factors Affecting Transpiration: Blue light induces maximum opening of stomata. Red and blue lights constitute the action spectrum of transpiration.

A rise in the temperature of the leaf increases the vaporization of water within the leaves resulting in a greater DP gradient of vapors, which diffuses out into the air rapidly. The rate of transpiration is generally doubled with about every 10° = C rise in temperature. High CO2 closes stomata and cytokinins open stomata.

  • Wind Speed: If the wind velocity is high, the rate of transpiration is also high.
  • Relative Humidity: If relative humidity is low, there would be a high rate of transpiration and vice versa.
  • Root/shoot Ratio: An increase in the root/shoot ratio causes an increase in the rate of transpiration. Suolt plants have a high root/shoot ratio.
  • Plant Factors: Number and distribution of stomata, number of open stomata, water status of the plants, canopy structure, etc.

Significance Of Transpiration: The following roles have been given in relation to transpiration by different biologists

  • Since plants absorb far more amounts of water than is actually used by plants, transpiration helps in the removal of excess water.
  • Removal of water in the form of vapors has a cooling effect on the leaves. Transpiration, thus, does not allow the leaf temperature to rise to detrimental levels.
  • The transpiration pull created by transpiration is responsible for the ascent of water in tall trees.
  • Passive absorption of water through the roots takes place due to the negative pressure developing in the shoots as a result of transpiration.
  • Development of mechanical tissues, growth of root system, increasing ash and sugar content of fruits, and development of resistance are other beneficial effects of transpiration.
  • Many chemicals (antitranspirants) have been found to reduce the rate of transpiration without affecting CO2 uptake, for example, phenylmercuric acetate (PMA) abscisic acid (ABA), and CO2. Silicon emulsion and low viscosity wax cover stomata as a film, allowing CO2 and O2 but resisting diffusion of water.

NEET Biology Notes Transport In Plant Guttation

Plants growing under humid conditions in moist warm soil often exhibit droplets of water along the margins of their leaves. This phenomenon is commonly seen in oats, tomatoes, cucumbers, garden nasturtium, saxifrage, etc. The loss of water in the form of liquid is called guttation.

In moist and humid conditions, the rate of absorption of water greatly exceeds transpiration. The root pressure is built up which pushes the water up in the xylem ducts, from where it comes out on the leaf surface through special structures called hydathodes.

NEET Biology Transport In Plant Guttation Process

Hydathodes are present at the tips of veins in the leaves. A hydathode consists of a pore in the epidermis followed by large intercellular spaces and loosely arranged parenchyma called epithem and blindly ending xylem elements. Guttated water contains inorganic and organic salts and is not pure.

NEET Biology Notes Transport In Plant Translocation Of Organic Solutes

Organic solutes such as glucose, and sucrose produced during photosynthesis are translocated through phloem tissue. The transport of photosynthates from the production centers (leaves) to the consumption centers (apices, roots, fruits, tubers) through the phloem is called translocation of organic solutes or long-distance transport. Translocation through phloem occurs in upward, downward, and radial directions from the source (leaves) to the sink (apices, roots, fruits, tubers, etc.)

NEET Biology Transport In Plant Translocation Of Organic Solutions

Chemical analysis of the phloem sap revealed the presence of sugars up to 90%. Sucrose constitutes 5-15% of the total sugars. Other sugars such as raffinose (triose). staehyose (tetrose), and verbascose (pentose) are also present in small quantities.

  • This analysis strongly suggests that the phloem is the tissue concerned with the translocation of organic solutes. Phloem sap may be collected by using sap-sucking aphids.
  • The tracer technique (Rabideau and Bun, 1945) supplied 14CO2 to a leaf during photosynthesis. Sugars synthesized in this leaf were labeled with 14C (tracer). The presence of labeled sugars (radioactivity) in the phloem showed that solutes are translocated through the phloem.

Theories Of Translocation Of Organic Solutes: Protoplasmic Streaming Hypothesis: This theory was proposed by a Dutch botanist, Hugo de Vries, in 18S5 and was supported by C.F. Curtis (1935). According to this theory, the protoplasm of the sieve tubes shows continuous streaming from one end to the other.

  • Organic solutes (sugars) that enter the sieve tube are passively carried by the streaming protoplasm from one end of the sieve tube to the other. Solutes move from one sieve tube to the next sieve tube by diffusion through the pores of the sieve plate.
  • Thus, the streaming protoplasm acts as a conveyor belt or two-way escalator. Different substances move in different directions at the same time in the same sieve tube.

Pressure Flow Or Mass Flow Hypothesis: According to this theory proposed by Munch (1929) and elaborated by Crafts (1938), organic solutes are translocated “en masse’’ through the sieve tubes from the supplying end or source leaves to the consumption end or sink (roots, fruits, tubers).

  • Mesophyll cells synthesize sugars during photosynthesis. As these get dissolved in the cell sap, the osmotic concentration and DPD of mesophyll cells increase (ψw decreases). Water enters the mesophyll cells from the xylem. The turgor pressure or pressure potential (ψp) of the mesophyll cells increases.
  • Sugars dissolved in water move from mesophyll cells into the symplast system of sieve tubes. Solutes are carried “en masse” through the symplast to finally reach the consumption centers.
  • At the consumption end, food materials (solutes) are either used up (as in roots) or are stored in an insoluble form (as in fruits, or tubers). Hence, the osmotic concentration and. consequently, the turgor pressure in these cells will be low. Thus, a continuous turgor pressure gradient gets established across the symplast between the cells of the source (leaf) and the cells of the sink (root).
  • Water returns to the source (leaf) through the apoplast system.

Object To Mass Flow Hypothesis: The bidirectional transport of solution in the same sieve tube needs explanation. The slime content and other fibrils of the sieve tube reduce the speed of the flow of solutes even under high pressure.

Mass flow is not a purely physical process as described by Munch Phloem cells utilize 0.1 0.5% of sucrose translocated through them. This is evidence to show that phloem translocation is an active process and requires metabolic energy.

Factors Affecting Translocation Of Solutes

  • Temperature: The optimum temperature for translocation is between 20° and 30°C. The file rate of translocation increases with an increase in temperature. The temperature influences the root more than the shoot since it acts as a sink for the sugars.
  • Light: The root/shoot dry weight ratio increases with increased light intensities. This indicates that translocation to root increases as compared to shoot when light intensity is increased.
  • Metabolic Inhibitors: The metabolic inhibitors inhibit carbohydrate translocation. These include dinitrophenol (DNP), arsenite, azide, fluoride, and hydrogen cyanide.
  • Mineral Deficiencies: The absorption and translocation of sucrose by a leaf is facilitated by boron. It helps sucrose to move easily through the cell membranes in the form of the boron-sucrose complex.
  • Hormones: Sucrose is much more efficiently translocated when growth regulators are applied, such as kinetin, IAA, and gibberellic acid.

NEET Biology Notes Transport In Plant Points To Remember

The unit for water potential is bar or Pascal (1 MPa = 10 bar) The osmotic pressure of 1 molar solution of a non-electrolytic would be 22.4 atm at 0°C. The equimolar concentrations of two solutions of non-ionizing substances will have the same osmotic pressure.

  • The value of the osmotic potential of an electrolyte will be greater by the degree of its dissociation into ions at a given temperature.
  • Bacteria do not survive in salted pickles because these get plasmolyzed.
  • Common salt kills weeds by plasmolysis.
  • Plasmolysis can be demonstrated in the epidermal peel of the Rhoeo discolor leaf.
  • A plasmolyzed cell regains normal condition if placed in a hypotonic solution. It is called deplasmolysis.
  • The government of H2O occurs from the high value of ψw to low value of ψw, i.e., from less negative value to more negative value of ψw.
  • The auxin-treated cells show an increase in their metabolism. Respiration in these cells increases and more energy is provided for the absorption of water.
  • At low temperatures, water in the intercellular spaces freezes into ice, thus having higher OP. It causes exosmosis of water from cells causing desiccation.
  • If a freshwater plant is transferred to marine water, it dies due to exosmosis.
  • The approximate water potential of root hairs is -3 to -43 bar.
  • Ringing experiments to prove the ascent of water through the xylem were first conducted by Malpighi (1672), Stephan Hales (1727), and Hartig.
  • Root pressure is measured by a manometer.
  • Stocking (1956) considered root pressure as an active process responsible for guttation and bleeding in plants.
  • Maximum root pressure recorded in plants is 2 bar which is sufficient to raise the water column to a height of 20 m.
  • Root pressure is absent in gymnosperms (some of the tallest trees are gymnosperms).
  • Photometers are used for measuring/comparing the rates of transpiration.

CoCl2 paper method is also used to compare the rates of transpiration. Moisture coming out of stomata turns blue CoCl2 paper to pink. Porometers are used for assessing the total pore area (stoma).

  • Generally, stomata are photoactive (open during the day and close at night). But in succulents such as Bryophyllum, Opimtia, and cacti, stomata close during the day and open at night (scotoactive).
  • Transpiration in old stems and fruits occurs through lenticels.
  • The fresh weight of a plant or leaf would be maximum in the morning and minimum in the afternoon.
  • If half of the total number of stomata on a leaf closes down, the rate of transpiration is not reduced by half.
  • Stomata are absent or non-functional in submerged hydrophytes.
  • Cytokinins help in the opening of stomata while ABA (abscisic acid) and low O2 close stomata.
  • Curtis (1926) considered transpiration “as a necessary evil” in plants.
  • Plants growing at high altitudes show xeromorphy (adaptation to minimize transpiration).

Transpiration Ratio: the amount of water lost per unit of dry matter produced during the growing season of a plant.

  • In Saxifraga, the rate of guttation is high during flowering.
  • In Colocasia antiquarian, the rate of guttation is very high.
  • Mechanical shock causes stomatal closure.

Stomatal index = \(\frac{E}{E + S}\) x 100 where E is the number of epidermal cells and S is the number of stomata.

Transpiration index = Leaf test tuneAVater test time.

A psychrometer is used to measure relative humidity and rate of transpiration.

The diameter of tree decreases during the day. It is due to the narrowing of tracheary elements due to the development of negative pressure. It is measured by dendrograph.

Matric Potential, ψm: It is used for surfaces that bind water. It is also negative. For example, soil particles, cell wall, cytoplasm, etc.

Gravity Potential, ψg: It denotes the effect of gravity on ψw. It depends on the height (h) of water above the reference state of water, the density of water, and acceleration due to gravity. The value of ψm negligible up to a height of 5 m from the reference level and also the value of ψm is ignored.

∴ ψw = ψs + ψp

NEET Biology Notes Transport In Plant Multiple Choice Questions And Answers

Question 1. What determines the diffusion of water from one cell to another?

  1. OP
  2. WP
  3. DPD
  4. TP

Answer: 3. DPD

Question 2. The term “water potential” was used for the first time by

  1. Slatyer and Taylor
  2. Stocking
  3. Sachs
  4. Boehm

Answer: 1. Slatyer and Taylor

Question 3. In a hypertonic solution, the water potential of a cell

  1. Increases
  2. Decreases
  3. First increases and then decreases
  4. No chance occurs

Answer: 2. Decreases

Question 4. As a result of endosmosis, of a cell

  1. Increases
  2. Decreases
  3. Remains same
  4. Become zero

Answer: 1. Increases

Question 5. Which of the following equations is wrong?

  1. ψs = -π
  2. DPD = OP + TP
  3. ψw = ψs + ψp
  4. OP = CRT

Answer: 2. DPD = OP + TP

Question 6. When a cell is fully turgid, which of the following will be zero?

  1. Osmotic pressure
  2. Turgor pressure
  3. Wall pressure
  4. Suction pressure

Answer: 4. Suction pressure

Question 7. Osmotic potential is numerically equal to

  1. TP
  2. DPD
  3. OP
  4. WP

Answer: 3. OP

Question 8. The first sign of shrinkage of cell is detectable at

  1. Limiting plasmolysis
  2. Incipient plasmolysis
  3. Evident plasmolysis
  4. Permanent plasmolysis

Answer: 2. Incipient plasmolysis

Question 9. Osmotic pressure depends upon

  1. Concentration of solutes
  2. Temperature
  3. Ionization
  4. All of these

Answer: 4. All of these

Question 10. The water potential of a plasmolyzed cell will be

  1. ψw = -ψs + ψp
  2. ψs = ψp
  3. ψw = 0
  4. ψw  = -ψs – ψp

Answer: 4. ψw  = -ψs – ψp

Question 11. The water potential of soil at field capacity and wilting point are, respectively,

  1. —1.5 MPa and-0.01 MPa
  2. -0.1 MPa and -0.01 MPa
  3. -0.01 MPa and-1.5 MPa
  4. -0.01 MPa and -0.1 MPa

Answer: 3. -0.01 MPa and-1.5 MPa

Question 12. Which is not a characteristic of imbibition?

  1. It is a reversible phenomenon.
  2. Heat is generated.
  3. Involve capillarity and adsorption.
  4. It is a property of hydrophobic and lyophobic colloids.

Answer: 4. It is a property of hydrophobic and lyophobic colloids.

Question 13. The concept of apoplast and symplast imbibition was given by

  1. Munch
  2. Kramer
  3. Renner
  4. Dixon

Answer: 1. Munch

Question 14. The type of soil water that is available to roots is

  1. Gravitational water
  2. Hygroscopic water
  3. Capillary water
  4. Chemically combined water

Answer: 3. Capillary water

Question 15. Passive absorption is controlled by

  1. Transpiration
  2. Capillarity
  3. Presence of solutes in soil
  4. Temperature of atmosphere

Answer: 1. Transpiration

Question 16. If the texture of soil becomes fine, then the rate of movement of water through it will be

  1. Faster
  2. Slower
  3. Nil
  4. Same

Answer: 2. Slower

Question 17. At low temperatures, the rate of water absorption decreases due to

  1. Decreased viscosity of water
  2. Increased permeability
  3. Reduced rate of diffusion
  4. Increased root growth

Answer: 3. Reduced rate of diffusion

Question 18. Water absorption in plants is enhanced by

  1. Increased transpiration
  2. Decreased transpiration
  3. Decreased salt absorption
  4. Increased photosynthesis

Answer: 1. Increased transpiration

Question 19. The phenomenon of water uptake at the expense of energy by the cell and usually again osmotic phenomenon is known as

  1. Osmosis
  2. Active absorption
  3. Passive absorption
  4. Imbibition

Answer: 2. Active absorption

Question 20. Root pressure is measured by

  1. Osmometer
  2. Manometer
  3. Barometer
  4. Auxanometer

Answer: 2. Manometer

Question 21. When the temperature of soil is 0°C, then

  1. The absorption of water increases
  2. The absorption of water is not affected by temperature
  3. The absorption of water decreases
  4. The soil will lose capillary water

Answer: 3. The absorption of water decreases

Question 22. The osmotic theory for active water absorption was given by

  1. Bennet Clarks
  2. Thimann
  3. Atkin and Priestley
  4. Dixon

Answer: 3. Atkin and Priestley

Question 23. Root pressure is maximum when

  1. Transpiration is high and absorption is very low
  2. Transpiration is very low and absorption is high
  3. Transpiration is very high and absorption is also high
  4. Transpiration and absorption both are low

Answer: 2. Transpiration is very low and absorption is high

Question 24. The pulsation theory of the ascent of sap was given by

  1. Godlewski
  2. Dixon
  3. Tansley
  4. Sir J.C. Bose

Answer: 4. Sir J.C. Bose

Question 25. Which is not true for root pressure?

  1. Positive hydrostatic pressure
  2. Maximum during the day and minimum during night
  3. Magnitude is 1-2 bars
  4. Develops due to the metabolic activity of root.

Answer: 2. Maximum during the day and minimum during night

Question 26. A tension (transpiration pull) of 1 atm can pull water to a height of approx

  1. 10 feet
  2. 10 m
  3. 1 m
  4. 1 feet

Answer: 2. 10 m

Question 27. Amphistomatic leaves are generally found in

  1. Dicots
  2. Monocots
  3. CAM plants
  4. Aquatic plants

Answer: 2. Monocots

Question 28. The cutinized wall of epidermal cells is

  1. Permeable
  2. Semipermeable
  3. Impermeable
  4. Selective permeable

Answer: 3. Impermeable

Question 29. The stomata are widely open in

  1. Red light
  2. Blue light
  3. Greenlight
  4. Yellow light

Answer: 2. Blue light

Question 30. When transpiration is rapid

  1. ψw of epidermal cells decreases
  2. A negative pressure develops in the xylem vessel
  3. Water is absorbed through the root passively
  4. All of these

Answer: 4. All of these

Question 31. Which is not true regarding stomata?

  1. They are turgor-operated valves.
  2. They have differentially thickened walls in guard cells.
  3. They open when OP of the guard cell decreases.
  4. They show photoactive openings in CAM plants.

Answer: 3. They open when OP of the guard cell decreases.

Question 32. Cuticular transpiration is approx

  1. 50% in herbs and ferns
  2. 97% in most of the plants
  3. 1% of the total transpiration
  4. 50% in most of the plants

Answer: 1. 50% in herbs and ferns

Question 33. The rate of transpiration and stomatal movement are measured by, respectively,

  1. Porometer and photometer
  2. Potometer and pyrometer
  3. Potometer and tensiometer
  4. Hygrometer and porometer

Answer: 2. Potometer and pyrometer

Question 34. According to starch ⇔ sugar interconversion theory of stomatal opening, a stomatal opening is preceded by

  1. Increase in H+ concentration
  2. Decrease in the pH of the guard cell
  3. Increase in the pH of guard cell
  4. Inactivation of phosphorylase enzyme

Answer: 3. Increase in the pH of guard cell

Question 35. When stomata open in night only, they are called

  1. Photoactive stomata
  2. Scotoactive stomata
  3. Hydathodes
  4. All of these

Answer: 2. Scotoactive stomata

Question 36. Which of the following does not happen during stomatal opening?

  1. Accumulation of K+ ions in guard cell
  2. Lowering of osmotic pressure of guard cell
  3. Creation of water potential gradient between guard cell and subsidiary cell
  4. Increased thickening of the inner wall of the guard cell

Answer: 4. Increased thickening of the inner wall of the guard cell

Question 37. High amount of malate in guard cell during stomatal opening is due to

  1. Import from subsidiary cell
  2. Hydrolysis of starch
  3. Photosynthesis in guard cell
  4. Hydrolysis of proteins

Answer: 4. Hydrolysis of proteins

Question 38. Active K+ exchange mechanism for the opening and clos¬ing of stomata was given by

  1. Darwin
  2. Levitt
  3. Scarth
  4. Fujino

Answer: 2. Levitt

Question 39. Which of the following is a metabolic antitranspirant?

  1. PMA, CO2
  2. Colorless plastics and waxes
  3. Aspirin, ABA
  4. Both (1) and (3)

Answer: 4. Both (1) and (3)

Question 40. Which is not an advantage of transpiration?

  1. Development of mechanical tissues
  2. Development of root system
  3. Distribution of minerals
  4. Fixation of nitrogen

Answer: 4. Fixation of nitrogen

Question 41. The plant factor which affects the rate of transpiration is

  1. Leaf area
  2. Temperature
  3. Humidity
  4. Wind speed

Answer: 1. Leaf area

Question 42. In guttation, water is lost in the form of

  1. Water vapors
  2. A dilute solution of sugars
  3. Pure liquid water
  4. Dilute solution of salts and organic substances

Answer: 4. Dilute solution of salts and organic substances

Question 43. Which of the following is an effective adaptation for better gaseous exchange in plants?

  1. Presence of multiple epidermis
  2. Presence of hair on the lower epidermis
  3. The presence of a waxy cuticle covering the epidermis of the leaves
  4. Location of stomata on the lower surface of leaves and side turned away from direct sun rays

Answer: 4. Location of stomata on the lower surface of leaves and side turned away from direct sun rays

Question 44. The conditions under which transpiration would be most rapid are

  1. Excess of water in soil
  2. Low humidity, high temperature, turgid guard cells, and moist soil
  3. Low velocity of the wind
  4. High humidity

Answer: 2. Low humidity, high temperature, turgid guard cells, and moist soil

Question 45. The transpiration index is equal to

  1. \(\frac{\text{Leaf test time}}{\text{Water test time}}\)
  2. \(\frac{\text{Water test time}}{\text{Leaf test time}}\)
  3. The amount of water lost per unit of dry matter produced during the growing season
  4. \(\frac{\text{PWP}}{\text{Field capacity}}\)

Answer: 1. \(\frac{\text{Leaf test time}}{\text{Water test time}}\)

Question 46. Translocation of photosynthates occurs in the form of

  1. Starch
  2. Glucose
  3. Sucrose
  4. 3PGA

Answer: 3. Sucrose

Question 47. Although a girdled (up to bast) tree may survive for some time, it will eventually die because

  1. Water will not move upward
  2. Water will not move downward
  3. Sugars and other organic solutes will not move downward
  4. Sugars and other organic solutes will not move upward

Answer: 3. Sugars and other organic solutes will not move downward

Question 48. When stomata open, the pH of guard cells.

  1. Increases
  2. Decreases
  3. Remains same
  4. Both (1) and (2)

Answer: 1. Increases

Question 49. Water lost in pulsation is

  1. Pure water
  2. Impure water
  3. In vapor form
  4. Either (1) and (2)

Answer: 2. Impure water

Question 50. What will happen if plant cells are placed in a hypertonic solution

  1. Turgid
  2. Plasmolyzed
  3. Deplasmolyzed
  4. Lysed

Answer: 2. Plasmolyzed

Question 51. Loss of water from the tips of leaves is called

  1. Bleeding
  2. Gustation
  3. Respiration
  4. Transpiration

Answer: 2. Gustation

Question 52. Root pressure is measured by

  1. Manometer
  2. Potomcter
  3. Auxanomcter
  4. Osmometer

Answer: 1. Manometer

Question 53. Which of the following apparatus is commonly used to measure the rate of transpiration?

  1. Porometer
  2. Altimeter
  3. Potomcter
  4. Luxmeter

Answer: 3. Porometer

Question 54. Leaves of the Nehnnbo plant are

  1. Epistomatic
  2. Hypostomatic
  3. Amphistomatic
  4. None of these

Answer: 1. Epistomatic

Question 55. 0.1 M solution has the water potential of

  1. -2.3 bars
  2. 0 bar
  3. 22.4 bars
  4. +2.3 bars

Answer: 1. -2.3 bars

Question 56. A small mesophytic twig with green leaves is dipped into water in a big beaker under sunlight. It demonstrates

  1. Photosynthesis
  2. Respiration
  3. Transpiration
  4. None of the above

Answer: 3. Transpiration

Question 57. Which one is not related to transpiration?

  1. Regulation of plant body temperature
  2. Absorption and distribution of mineral salt
  3. Circulation of water
  4. Bleeding

Answer: 4. Bleeding

Question 58. Stomata can open at night also in

  1. Xerophyte
  2. Gamelophyte
  3. Hydrophyte
  4. None of these

Answer: 1. Xerophyte

Question 59. Who had said that “transpiration is a necessary evil”?

  1. Curtis
  2. Steward
  3. Andersen
  4. J.C. Bose

Answer: 1. Curtis

Question 60. Stomata open during day because the guard cells have

  1. Thin outer walls
  2. Kidney shape structure
  3. Chlorophyll
  4. Large nuclei

Answer: 1. Thin outer walls

Question 61. Stomata open and close due to

  1. Turgor pressure change
  2. Hormone change
  3. Temperature change
  4. All of the above

Answer: 1. Turgor pressure change

Question 62. In plasmolyzed cell, the space between cell wall and protoplasm is occupied by

  1. Hypotonic solution
  2. Hypertonic solution
  3. Isotonic solution
  4. Distil water

Answer: 2. Hypertonic solution

Question 63. In CAM plants, stomata are

  1. Closed at night and open during the day
  2. Closed during the day and open at night
  3. Never closes
  4. Never opens

Answer: 2. Closed during the day and open at night

Question 64. The real force responsible for the movement of water from cell to cell is

  1. OP
  2. TP
  3. DPD
  4. WP

Answer: 3. DPD

Question 65. Which of the following have sunken stomata?

  1. Nerium
  2. Mangifera
  3. Hydrilla
  4. Zea mays

Answer: 1. Nerium

Question 66. When a plasmolyzed cell is placed in a hypotonic solution then water will move inside the cell. Which force causes this?

  1. DPD
  2. OP
  3. WP
  4. None of these

Answer: 1. DPD

Question 67. Rate of transpiration is measured by

  1. Manometer
  2. Auxanometer
  3. Potometer
  4. Barometer

Answer: 3. Potometer

Question 68. If a cell shrinks when placed in a solution, the solution becomes

  1. Hypotonic
  2. Hypertonic
  3. Isotonic
  4. Pure solvent

Answer: 2. Hypertonic

Question 69. If cell A with DPD 4 bar is connected to cell B, C, and D whose osmotic pressure and turgor pressure are, respectively, 4 and 4, 10 and 5, 7 and 3 bar, the flow of water will be

  1. B to A, C, and D
  2. A to D, B, and C
  3. C to A, B, and D
  4. A to B, C, and D

Answer: 1. B to A, C, and D

Question 70. Guard cell controls

  1. The intensity of light entering
  2. Photosynthesis
  3. Closing and opening of stomata
  4. Change in green color

Answer: 3. Closing and opening of stomata

Question 71. Active transport

  1. Releases energy
  2. Requires energy
  3. Produces ATP
  4. Produces a toxic substance

Answer: 2. Requires energy

Question 72. Velamen tissues are associated within

  1. Haustorial function
  2. Assimilation
  3. Absorption of moisture
  4. Nutrition

Answer: 3. Absorption of moisture

Question 73. Cohesion tension theory regarding the ascent of sap was given by

  1. Dixon and Jolly
  2. J.C. Bose
  3. Cristian Wolf
  4. Godlewski

Answer: 1. Dixon and Jolly

Question 74. Velamen tissue is found in

  1. Mesophytes
  2. Epiphytes
  3. Hydrophytes
  4. Xerophytes

Answer: 2. Epiphytes

Question 75. In a fully turgid plant cell, which one is zero?

  1. Turgor pressure
  2. Wall pressure
  3. Suction pressure
  4. None of these

Answer: 3. Suction pressure

Question 76. Who proposed the cohesion theory of the ascent of sap?

  1. Strasburger
  2. Godlewski
  3. Western
  4. Oixon and Jolly

Answer: 4. Oixon and Jolly

Question 77. The most accepted theory for ascent of sap is

  1. Relay pump theory
  2. Pulsation theory
  3. Root pressure theory
  4. Transpiration pull cohesion theory

Answer: 4. Transpiration pull cohesion theory

Question 78. The transport of water and salt is mediated by

  1. Xylem
  2. Sieve tubes
  3. Sclerenchyma
  4. Phloem

Answer: 1. Xylem

Question 79. The removal of a ring of tissue outside the vascular cambi¬um from the tree trunk kills it because

  1. Water cannot move up
  2. Food does not travel down and roots become starved
  3. Shoots become starved
  4. Annual rings are not produced

Answer: 2. Food does not travel down and roots become starved

Question 80. Wilting of the plant is present in

  1. Moss
  2. Fern
  3. Algae
  4. Angiosperm

Answer: 4. Angiosperm

Question 81. Root hair absorbs water from the soil on account of

  1. Turgor pressure
  2. Osmotic pressure
  3. Suction pressure
  4. Root pressure

Answer: 3. Suction pressure

Question 82. Increased humidity in the atmosphere decreases the rate of

  1. Transpiration
  2. Photosynthesis
  3. Glycolysis
  4. Growth

Answer: 1. Transpiration

Question 83. In osmosis, there is movement of

  1. Solute only
  2. Solvte only
  3. Both (1) and (2)
  4. Neither solute nor solvent

Answer: 2. Solvte only

Question 84. Guttation takes place through

  1. Lenticles
  2. Pneumatophores
  3. Stomata
  4. Hydathodes

Answer: 4. Hydathodes

Question 85. Which of the following statements is correct?

  1. Cell membrane is involved only in exosmosis
  2. Cell membrane is involved only in endosmosis
  3. Cell membrane is involved both in exosmosis and endosmosis
  4. None of the above

Answer: 3. Cell membrane is involved both in exosmosis and endosmosis

Question 86. The root hairs absorb which of the following types of water?

  1. Capillary water
  2. Hygroscopic water
  3. Gravitational water
  4. All of the water

Answer: 1. Capillary water

Question 87. If flowers are cut and dipped in dilute NaCl solution, then

  1. Transpiration will become low
  2. Endoosmosis occurs
  3. No bacterial growth takes place
  4. Absorption of solute inside flower cell takes place

Answer: 1. Transpiration will become low

Question 88. A plant cell is plasmolyzed in a solution that is

  1. Hypotonic
  2. Hypertonic
  3. Isotonic
  4. Concentration no means

Answer: 2. Hypotonic

Question 89. Turgidity in guard cells is controlled by

  1. Chloride
  2. Malic acid
  3. Potassium
  4. Potassium, chloride, and malic acid

Answer: 4. Potassium, chloride, and malic acid

Question 90. Stomata are not found in

  1. Algae
  2. Mosses
  3. Ferns
  4. Liverworts

Answer: 1. Algae

Question 91. In which of the following, the rate of transpiration is high?

  1. CAM plant
  2. C3 plants
  3. C2 and C4 plants
  4. C4 plants

Answer: 1. CAM plant

Question 92. Cell sap is found in which cell organelle?

  1. Nucleolus
  2. Chloroplast
  3. Vacuole
  4. Golgi apparatus

Answer: 3. Vacuole

Question 93. Which one of the following fixes nitrogen?

  1. TMV
  2. Yeast
  3. Nostoc
  4. Denitrifying bacteria

Answer: 3. Nostoc

Question 94. Active transport of ions by the cell requires

  1. High temperature
  2. ATP
  3. Alkaline pH
  4. Salts

Answer: 2. ATP

Question 95. To initiate cell plasmolysis, the salt concentration must be

  1. Isotonic
  2. Hypotonic
  3. Hypertonic
  4. Atonic

Answer: 3. Hypertonic

Question 96. The basis of stomatal opening is

  1. Endosmosis
  2. Plasmolysis of guard cells
  3. Decrease in cell up concentration
  4. Exosmosis

Answer: 1. Endosmosis

Question 97. Plants absorb carbon dioxide from

  1. Vegetative
  2. Heterocyst
  3. Both vegetative and heterocyst
  4. None of these
  1. Millets
  2. Cereals
  3. Carbohydrates present in the soil
  4. Atmosphere

Answer: 4. Atmosphere

Question 98. Transpiration will increase with the increase of

  1. Humidity
  2. Temperature
  3. Carbon dioxide
  4. Sulfur dioxide

Answer: 2. Temperature

Question 99. If it is possible to drop a small particle through the stomata of a leaf, what will you conclude?

  1. It will fall on the earth surface.
  2. It will stop on the lower epidermis.
  3. It will stop on mesophyll cells.
  4. It will stop on vascular tissue.

Answer: 3. It will stop on mesophyll cells.

Question 100. During transpiration, turgidity in guard cells is controlled by

  1. Potassium
  2. Bromine
  3. Sodium
  4. Oxalic acid

Answer: 1. Potassium

Question 101. Apparatus used for measuring the transpiration

  1. Evapometer
  2. Potometer
  3. Osmometer
  4. Tensiometer

Answer: 2. Potometer

Question 102. Which of the following theories gives the latest explanation for the closure of stomata?

  1. ABA theory
  2. Munch theory
  3. Starch-glucose theory
  4. Active KT transport theory

Answer: 1. ABA theory

Question 103. The sugarcane plant has

  1. Dumbbell-shaped guard cells
  2. Pentamerous flowers
  3. Reticulate venation
  4. Capsular fruits

Answer: 1. Dumbbell-shaped guard cells

NEET Biology Notes Transport In Plant Assertion Reasoning Questions And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: Xerophytes have high water-retaining capacity.

Reason: They have high OR.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 2. Assertion: There is an indirect relationship between the rate of respiration and water absorption.

Reason: Increased metabolism increases mineral uptake.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 3. Assertion: Root pressure is a dynamic and always a positive hydrostatic pressure.

Reason: It is a universal phenomenon and develops under absorption lag.

Answer: 3. If Assertion is true, but Reason is false.

Question 4. Assertion: Stomata has delegated the task of providing food while preventing thirst.

Reason: They are made for gaseous exchange.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 5. Assertion: During stomata opening, there is a relative change in TP of the guard cell and subsidiary cell.

Reason: TP of the subsidiary cell decreases during the opening.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

 

 

NEET Biology Notes – Mineral Nutrition

by

NEET Biology Notes For Mineral Nutrition Introduction

The sum total of various processes by which an organism with-draws and utilizes the substances required for the development and sustaining life-related processes is called nutrition, and these substances are called nutrients.

  • The absorption, distribution, and metabolism of various mineral elements by plants is called mineral nutrition. About 60 elements have been reported from plant ash (obtained after heating dry plant matter at 600°C). Out of these, 30 are present in all the plants.
  • Inorganic nutrients are classified as essential and non-essential elements. Seventeen elements have been placed in essential elements. These are the elements without which the reproduction and life cycle of a plant cannot be completed. The essential elements are C, H, O, N, P. K. S, Mg, Ca, Fe, Mo, Mn, Ni, Zn, B, Cl, and Cu.

Amon and Stout proposed the following criteria of the essentiality of an element:

” mineral nutrition in plants”

  • The element must be absolutely necessary for supporting normal growth and reproduction.
  • The requirement of the element must be specific and not replaceable by another element.
  • The element must be directly involved in the nutrition of the plant.
  • A deficiency of minerals causes diseases in plants. Magnesium. for example, is a constituent of the chlorophyll molecule and is essential for photosynthesis. It cannot be replaced by any other element for the same function. It is also required as a cofactor by many enzymes involved in cellular respiration and metabolic pathways. Similarly, iron is a constituent of cytochromes.

Depending Upon The Quantity In Which These Elements Are Present In Cell, They Are Classified As:

Macronutrients: They are the elements that are present in a quantity of 1—10 mg/g dry wt. of the cell. They are C, H, O, N, P, S, K, Mg, and Ca.

Micronutrients: They are the elements that are present in a quantity equal to or less than 0.1 mg/g dry wt. of the cell (often less than 1 ppm). They are Fe, Mn, Mo, Ni, Zn, B, Cl, and Cu.

NEET Biology Notes For Mineral Nutrition Culture Medium

Soils normally contain sufficient quantities of essential minerals. However, three important elements need to be replenished in crop fields as they are depleted by repeated cultivation. These x fertilizer elements called critical elements are nitrogen, phosphorus, and potassium (NPK).

  • The common sources of these elements used in India are nitrate of sodium, ammonium sulfate, ammonium nitrate, ammonium chloride, urea, etc. The NPK fertilizers comprising bags of fertilizers are labeled 17-18-19 or 15-15-15 or other combinations. These numbers refer to the percentage by weight of nitrogen, phosphorus, and water-soluble potassium
  • To determine what elements are essential tor plant growth and deficiency symptoms of an essential element, a well-defined nutrient medium has to be used. Seeds are grown in highly washed pure sand in a glass, glazed porcelain, or plastic container and supplied with a carefully made-up nutrient solution.

Arnon And Hoagland’s Medium: Amon and Hoagland prescribed a medium containing micronutrients. Iron was earlier supplied as ferrous sulfate, but it often precipitated out. This problem has now been solved by dissolving ferrous sulfate along with a chelating agent Na-EDTA (disodium salt of ethylene diamine tetraacetic acid).

Read and Learn More NEET Biology Notes

NEET Biology Notes For Mineral Nutrition Culture

Solution culture is performed in glass jars or polythene bottles. The container is covered with black paper after pouring the solution into them. Black paper has two functions

  1. Prevention of the growth of algae.
  2. Prevention of reaction of roots with light.

Seeds are allowed to germinate over the cork. Cotyledons are removed after seedling formation. The plant is properly sup¬ported with the help of split cork. The solution is aerated at regular intervals and is changed after two to three days.

NEET Biology Notes For Mineral Nutrition– Hydroponics

The commercial technique of soil-less culture is called hydroponics, which was first developed by Goerick (1940). The culture is performed in large tanks of metal or RCC tanks. The tanks are covered with wire mesh. Tanks are provided with aerating and circulating techniques. Seeds are suspended in the solution from the wire mesh with the help of threads. As the plant grows, additional support is provided. it depicts the diagrammatic views of the hydroponic technique.

NEET Biology Mineral Nutrition Hydroponics Technique

NEET Biology Notes For Mineral Nutrition Liebig’s Law Of The Minimum

As per this law, “the yield of a crop plant is determined by the amount of the necessary element which is present in minimum quantity in proportion to the demand of the plant.”

NEET Biology Notes For Mineral Nutrition General Functions Of Mineral Elements

Frame Work Elements: They form carbohydrates that form cell walls, for example, C, H, and O.

Protoplasmic Elements: They form protoplasm, for example, C, H, O, N, S, P.

Catalytic Elements: Fe, Cu, Zn, Mo, Mg, Mn, K (activator of over 40 enzymes)

Balancing Elements: Ca, Mg, and K counteract the toxic effect of other minerals.

Storage Elements: C, N, S, P.

Critical Elements: N, P, K.

Elements Affecting Cell Permeability: Monovalent cations (Na+, K+) increase the permeability of membranes while divalent and trivalent cations decrease it.

Toxic Elements: Al, As, Hg, Pb, Ag.

Non-mineral Elements: C, H, O, N. Nitrogen is a mineral as well as a non-mineral element.

Functional Elements: They are non-essential in most plants but have a definite activity in some species, for example, silicon in grasses.

NEET Biology Notes For Mineral Nutrition Nitrogen Fixation

Atmosphere is the ultimate source of nitrogen. Nitrogen is a highly inert gas. It cannot be used directly but has to be combined with C, H, N, and O to form compounds before being used. Higher plants utilize nitrogen in the oxidized forms such as nitrate (NO3–) and nitrite (NO2–) or in the reduced form (NH4) made available to plants by the nitrogen fixers.

The best-known nitrogen-fixing symbiotic bacterium is Rhizobium. This bacterium lives in the soil to form root nodules in plants belonging to the family Leguminosae such as beans, gram, groundnut, and soya bean. Root nodules are little outgrowths on roots.

  • When a section of the root nodule is examined, it appears pinkish due to the presence of a pigment called leghemoglobin. Like hemoglobin, leghemoglobin is an oxygen scavenger. The enzyme that catalyses the fixation of nitrogen is nitrogenase which functions under anaerobic conditions. Leghemoglobin combines with oxygen and protects nitrogenase.
  • Free-living microorganisms such as the cyanobacteria can also fix nitrogen. Some cyanobacteria also have symbiotic as¬sociation with plants. They are found in lichens. Anthoccros. Azollcu and coralloid roots of Cycas.
  • In the process of biological nitrogen fixation, the dinitrogen molecule is progressively reduced by the addition of pairs of hydrogen atoms so that the three bonds between the two nitrogen atoms are cleaved and ammonia is formed.

These reactions occur only in the presence of a single enzyme nitrogenase. The process of nitrogen fixation is energy-intensive.

NEET Biology Mineral Nutrition Nitrogen Fixation

Requirements Of N2 Fixation

  • Nitrogenase enzyme complex (synthesized by nif genes of bacteria) is the seat of nitrogen reduction and contains Mo, Fe, and S.
  • It is a strong reducing agent, for example, NADPH2, FMNH2, ferredoxin, etc.
  • It takes place under anaerobic conditions.
  • The energy source is ATP.
  • Cofactors included are TPP, CoA, Mg+2.
  • Electron and H+ donor (generally glucose-6-phosphate)

NEET Biology Notes For Mineral Nutrition Process Of Nodule Formation

Nodules are little out growths on the roots. When a root hair of a legume comes in contact with Rhizobium, there occurs an ex-change of plant and bacterial signals. Bacteria secrete nod fac-tors which result in the curling of root hair tips.

The plant responds by forming an infection thread that grows inward and carries the bacteria to the cortical cells of the root. Some of bacteria enlarge to become membrane-bound structures called bacteroids which are the seat of N2 fixation. Plant flavones act as the inducers of nod genes which specify early events of modulation.

NEET Biology Mineral Nutrition Nodule Formation

” mineral nutrients in plants”

Cortical cells are stimulated to divide rapidly. It is due to auxin secreted by plants and cytokinin contributed by bacteria. Nodules are pink in color due to leghemoglobin which is an oxygen scavenger and protects nitrogenase. Its heme comes from bacteria and globin from legumes.

Process Of N2 Fixation Can Be Summarized As: N2 + 8e– + 8H+ + 16 ATP → 2NH3 + H2 + 16ADP + 16Pi

Ammonium ions can be taken up by higher plants but plants are more adapted to absorb nitrate (NO3–) than ammonium ions (NH4+) from the soil. Soil bacteria such as Nitrosomonas and Nitrococcus convert ammonia to nitrite (NO2–) ions. Nitrobacter oxidizes nitrite to nitrate. This process of converting ammonia into nitrate, a form of nitrogen more available to plants, is called nitrification.

NEET Biology Mineral Nutrition Nitrificication Formula

This process is an oxidation process and releases energy which is used by nitrifying microbes.

NEET Biology Notes For Mineral Nutrition Nitrate Assimilation

The process of nitrate reduction to ammonia is called nitrate assimilation and is accomplished in two steps mediated by two specific enzymes. First, the nitrate is reduced to nitrite by an enzyme called nitrate reductase. This enzyme is a flavoprotein and contains molybdenum.

The nitrite ions are then reduced to ammonia by an enzyme called nitrite reductase. Ferredoxin is the most detect source of electrons for nitrite reduction and hence it occurs specifically in leaves. Therefore, nitrite ions formed in other parts of the plant are transported to leaves arid and further reduced to ammonia. Nitrite reductase does not require molybdenum but contains copper and iron.

NEET Biology Notes For Mineral Nutrition Synthesis Of Amino Acids

There are two main processes for amino acid synthesis.

1. Reductive Animation: From glutamic acid all other amino acids can be synthesized by the process of transamination.

NEET Biology Mineral Nutrition Synthesis Of Amino Acids Reductive Amination

2. Transamination: It involves the transfer of amino groups from one amino acid to the keto group of the keto acid. The reactions are catalyzed by the enzyme transaminase.

NEET Biology Notes For Mineral Nutrition Mineral Absorption

Mineral Absorption Occurs By Two Ways:

  1. Passive and
  2. Active.
  1. Passive Mineral Absorption: The main theories of passive mineral absorption. In most cases, the movement of mineral ions into the root occurs by diffusion. Molecules or ions diffuse from a region of their higher concentration to a region of their lower concentration. The movement of mineral ions into root cells as a result of diffusion is called passive absorption.
    • Donnan Equilibrium Theory: This theory was proposed by Donnan (1911). The entry of ions into the cell across the plasma membrane to maintain electrical equilibrium is called Donnan equilibrium. Some anions/cations get firmly attached to the inner surface of the plasma membrane (fixed and non-diffusible ions). To neutralize these, ions of opposite charges gain entrance in the cell passively against the concentration gradient without energy expenditure.
    • Ion Exchange Theory: It was proposed by Jenny and Overstreet (1938). The exchange of anions and cations absorbed on a colloidal fraction of the soil (clay and humus) with the ions adsorbed on the root surface is referred to as ion exchange. The small space in which the cations and anions attached to the surface of roots and particles oscillate is called oscillation volume. It is of two types
    • Contact Exchange: This includes the exchange of cation and anions from the root with similarly charged ions of soil solution.
    • Carbonic Acid Exchange: This includes the exchange of H+ and CHO3– ions from the root with similarly charged ions of the soil solution.
    • Bulk Flow/Mass Flow Theory: It was proposed by Hylmo (and supported by Kramer). According to it, the movement of ions occurs through roots along with the stream of water under the influence of transpiration.
  2. Active Mineral Absorption: The uptake of mineral ions against a concentration gradient is called active absorption. Such movement of minerals requires the expenditure of energy by the absorbing cell. This energy is derived from respiration and is supplied through ATP. Hence, when the roots are deprived of oxygen, they show a sudden drop in the active absorption of minerals. Mostly minerals are absorbed by active mechanisms. Various theories are given for the active uptake of minerals.
    • Carrier Concept (By Van den Honert): Carriers are specific proteins in membranes that form complexes with ions. This complex is capable of moving across the membrane and releasing ions on the inner side. There are separate carriers for cations and anions.
    • Cytochrome Pump (By Lundegardh And Burstrom): Cytochromes are iron-containing proteins in the membrane. They transport electrons from the inner surface to the outer surface of the membrane and also transport anions from the outer surface to the inner surface. Anions are transported actively while cation transport is passive. Anion uptake increases the rate of respiration called salt respiration.
    • Protein Lecithin Carrier (by Bennet and Clark): Carrier is a protein associated with a phosphatide called lecithin. It is an amphoteric earner and transports both cations and anions.

NEET Biology Notes For Mineral Nutrition Mobility

The elements that cannot move freely in the plants (for example, Ca, B, S, Fe, etc.) are called immobile elements. Deficiency symptoms of such elements first appear in young leaves. Those elements that can move from old leaves to young leaves and growing tips (N, P, K, Cl, Mg) are called mobile elements. Their deficiency affects old leaves.

” where are the mineral nutrients mostly used in plants”

NEET Biology Notes For Mineral Nutrition Points To Remember

Sodium (Na) is required by a desert shrub Atriplex and not by most of the other species.

  1. Efforts are being made to develop varieties of plants that can mine metals from the soil so that that soil can be reclaimed for agriculture practice, called phytore¬mediation.
  2. Foliar application of Fe, Mn, and Cu is more efficient than application through the soil.
  3. Aeroponics (by Soifer Hillel and David Durger) is a system where roots are suspended into some plastic vessel and misted with oxygenated, nutrient-en-riched water.
  4. Winogradsky (1891) discovered biological nitrogen fixation.
  5. Cobalt is found mostly in hydathodes and is required in N2 fixation.
  6. Selenium is found in Astragalus.
  7. Excess of manganese may, in fact, induce the deficiency symptom of iron, magnesium, and calcium (toxicity of micronutrients).
  8. Gallium accumulates in Lcinnci and Aspergillus.
  9. Phytotron: When a plant is grown in controlled conditions of temperature, and light. pH, etc.
  10. Asparagine and glutamine are two amides formed from aspartic acid and glutamic acid, respectively, in a plant. Amides are the storage form of nitrogen.
  11. Stem nodules are found in Sesbania.
  12. Leaf nodules are found in Pavetta. Dioscorea.
  13. Associative symbiosis (loose symbiosis) is found in tropical grass (with Azotobacter) and maize and sorghum (with Azospirillum).
  14. True humus plant is Wullschleigelia aphylla.
  15. Single ion channels (discovered by Neher and Sakmann) are transmembrane proteins meant for the entry of specific ions.
  16. Aquaporins are water-filled pores in membranes.

NEET Biology Notes For Mineral Nutrition Multiple Choice Questions And Answers

Question 1. Which is not a criterion for the essentiality of a mineral?

  1. Direct role in metabolism
  2. Requirement is specific
  3. Deficiency causes hunger signs
  4. Dispensable for growth

Answer: 4. Dispensable for growth

Question 2. Essential elements are

  1. Only macronutrients
  2. Only micronutrients
  3. Both macro and micronutrients
  4. C, H, O, and N only

Answer: 3. Both macro and micronutrients

Question 3. Which is not a trace element?

  1. Mn
  2. Cu
  3. Mo
  4. K

Answer: 4. K

Question 4. Which is not a true statement regarding macronutrients?

  1. Macronutrients form plant structure.
  2. Macronutrients become toxic when present in excess.
  3. Macronutrients have no role in electron transfer.
  4. Macronutrients develop osmotic potential.

Answer: 2. Macronutrients become toxic when present in excess.

Question 5. Choose the false statement regarding micronutrients.

  1. Micronutrients become toxic in excess.
  2. Micronutrients do not cause osmotic potential.
  3. Micronutrients have little role in protoplasmic structure.
  4. Micronutrients play a secondary role in enzyme activation.

Answer: 4. Micronutrients play a secondary role in enzyme activation.

Question 6. Deficiency in plant growth and disorders caused by the reduced availability of a critical element is called

  1. Critical deficiency
  2. Secondary deficiency
  3. Primary deficiency
  4. Complete deficiency

Answer: 3. Primary deficiency

Question 7. Who prescribed a medium containing microelements for the first time?

  1. Gericke
  2. Arnon Ploagland
  3. Knop
  4. Stout

Answer: 2. Arnon Ploagland

Question 8. Excess of manganese may induce the deficiencies of

  1. Iron
  2. Calcium
  3. Magnesium
  4. All of these

Answer: 4. All of these

Question 9. A partial mineral element is

  1. N
  2. P
  3. K
  4. S

Answer: 1. N

Question 10.The deficiency of which element causes the deficiency of nitrogen?

  1. Mo
  2. K
  3. Mn
  4. S

Answer: 1. Mo

Question 11. Minerals associated with redox reactions are

  1. N, Cu
  2. Fe, Cu
  3. Fe, K
  4. Mn, Mo

Answer: 2. Fe, Cu

Question 12. Minerals that maintain cation-anion balance in cells are

  1. Cl, K
  2. Fe, Cu
  3. K, P
  4. Ca, Fe

Answer: 1. Cl, K

“mineral nutrients “

Question 13. Interveinal chlorosis is due to the deficiency of

  1. Fe
  2. Mn
  3. N
  4. B

Answer: 1. Fe

Question 14. Match columns A and B correctly.

NEET Biology Mineral Nutrition Match The Column

  1. (1) → (C), (2) → (A), (3) → (D), (4) → (B)
  2. (1) → (C), (2) → (A), (3) → (B), (4) → (D)
  3. (1) → (C), (2) → (B), (3)→(D), (4)→ (A)
  4. (1) → (A), (2) → (B), (3) → (C), (4) → (D)

Answer: 2. (1) → (C), (2) → (A), (3) → (B), (4) → (D)

Question 15. Which of the following groups of elements are mobile?

  1. Fe, Ca, B
  2. N, P, K
  3. B, K, Ca
  4. Ca, Mg, K

Answer: 2. Fe, Ca, B

Question 16. Which of the following elements are required for chlorophyll synthesis?

  1. Fe and Mg
  2. Mo and Ca
  3. Cu and Ca
  4. Ca and K

Answer: 1. Fe and Mg

Question 17. If chloroplast is burnt, then which of the following is left?

  1. Magnesium
  2. Manganese
  3. Iron
  4. Sulfur

Answer: 1. Magnesium

Question 18. Which one of the following is a sulfur-containing amino acid?

  1. Valine
  2. Methionine
  3. Tryptophan
  4. Phenylalanine

Answer: 2. Methionine

Question 19. Copper deficiency leads to

  1. Exanthema
  2. Whiptail of cauliflower
  3. Little leaf condition
  4. Interveinal chlorosis

Answer: 1. Exanthema

Question 20. Phosphorus is found maximum in

  1. Roots
  2. Fruits
  3. Flowers
  4. None of these

Answer: 2. Fruits

Question 21. Which of the following is required for auxin synthesis?

  1. Calcium
  2. Zinc
  3. Sugars
  4. Proteins

Answer: 2. Zinc

Question 22. Reversible binding of cations, a property possessed by clay particles, is known as

  1. Retentive capacity
  2. Cation exchange
  3. Adsorption
  4. Chelation

Answer: 2. Cation exchange

Question 23. A characteristic of ion channels is/are

  1. They are transmembrane proteins functioning a selective pores
  2. They were discovered by Neher and Sakman
  3. They are gated canners
  4. All of these

Answer: 4. All of these

Question 24. The theory of Donnan equilibrium explains the process of some

  1. Fixed diffusible cations on the inner side
  2. Fixed non-diffusible anions on the inner side
  3. Non-fixed diffusible anions on the inner side
  4. Non-fixed diffusible cations on the inner side

Answer: 2. Fixed non-diffusible anions on the inner side

Question 25. Mineral salts which are absorbed by the roots from the soil, are in the from of

  1. Very dilute solution
  2. Dilute solution
  3. Concentrated solution
  4. Very concentrated solution

Answer: 1. Very dilute solution

Question 26. The movement of electrolytes through the roots is a general

  1. Against the electrochemical gradient and requires energy
  2. Along the electrochemical gradient and does not require energy
  3. A passive process
  4. Dependent on aquaporins

Answer: 1. Against the electrochemical gradient and requires energy

Question 27. Ionic uptake against electrochemical gradient without the expenditure of metabolic energy can be explained by

  1. Ion exchange
  2. Donnan equilibrium
  3. Carrier proteins
  4. Both (1) and (2)

Answer: 4. Both (1) and (2)

Question 28. Transpiration pull or water tension in leaf is responsible for which one of the following methods of absorption of minerals by the plants from soil?

  1. Active absorption of minerals
  2. Mass flow
  3. Donnan equilibrium
  4. Ionic exchange

Answer: 2. Mass flow

Question 29. If nitrogen is bubbled in the rooting medium, active absorption of minerals will

  1. Increase
  2. Decrease
  3. Remain same
  4. Stop immediately

Answer: 2. Decrease

Question 30. During ionic flux, the uptake of ions into inner space is

  1. Active
  2. Passive
  3. Energy-dependent
  4. Both (1) and (3)

Answer: 4. Both (1) and (3)

Question 31. Carrier proteins for active salt uptake

  1. Have pores
  2. Form complex with ions
  3. Function under transpiration pull
  4. All of these

Answer: 2. Form complex with ions

Question 32. The translocation of solute is

  1. Equal to the rate of translocation of water
  2. Dependent on transpiration pull
  3. Through xylem vessel
  4. None of these

Answer: 4. None of these

Question 33. Find the odd one (with respect to the critical element).

  1. Nitrogen
  2. Potassium
  3. Nickel
  4. Phosphorus

Answer: 3. Nickel

Question 34. The process of conversion of NH4 → NO2  → NO3 is called

  1. Ammonification
  2. Nitrification
  3. N2 fixation
  4. Denitrification

Answer: 2. Nitrification

Question 35. Which of the following is/are diazotrophs?

  1. Rhizobitnn and Azotobacter
  2. Frankia and Klebsiella
  3. Anabaena and Nos toe
  4. All of these

Answer: 4. All of these

Question 36. Which is not true for nitrogenase enzyme in root nodules in legumes?

  1. Synthesized by nit genes of Rhizobium
  2. Site of reduction of N2 into NH3
  3. It is a Mo-Fe protein
  4. Resistant to O2 concentration

Answer: 4. Resistant to O2 concentration

Question 37. Cell division in root nodules is promoted by se creted by plants and secreted by bacteria.

  1. Auxin, Cytokinin
  2. Cytokinin, Auxin
  3. Auxin, Leghemoglobin
  4. Nitrogenase, Leg hemoglobin

Answer: 1. Auxin, Cytokinin

Question 38. Conversion of NO3 → NO2 → NH4 is called is catalysed by

  1. Nitrate assimilation; nitrate and nitrite reductase
  2. Nitrification; nitrate and nitrite reductase
  3. Ammonification; glutamate dehydrogenase
  4. Denitrification; transaminase

Answer: 1. Nitrate assimilation; nitrate and nitrite reductase

Question 39. Transported and storage forms of nitrogen in plants are

  1. Amides
  2. Polypeptides
  3. Amino acids
  4. α-ketoglutaric acids

Answer: 1. Amides

Question 40. The amino acid which plays a central role in nitrogen metabolism is/are

  1. Glutamic acid
  2. α-ketoglutaric acid
  3. Aspartic acid
  4. Double-aminated keto acids

Answer: 1. Glutamic acid

Question 41. The amino acid which plays a central role in nitrogen metabolism is/are

  1. Anthoceros
  2. Aulosira
  3. Nostoc
  4. Groundnut

Answer: 4. Groundnut

Question 42. Nitrite reductase enzyme is used to convert

  1. Nitrate into nitrite ion
  2. Nitrogen of the atmosphere into ammonia
  3. Ammonia into nitrates
  4. Nitrite to ammonium ion

Answer: 4. Ammonia into nitrates

Question 43. What do hemi parasites absorb from the host?

  1. Water and minerals
  2. Sugar
  3. Both (1) and (3)
  4. Nothing

Answer: 2. Sugar

Question 44. The small rootless aquatic herb in which a portion of the leaf forms a tiny sac or bladder that traps water insects is

  1. Dionaea
  2. Utricularia
  3. Sarracenia
  4. Drosera

Answer: 2. Utricularia

Question 45. The process of conversion of NO2, NO3, NH3 to N2 is called and is done by

  1. Nitrification, Nitrosomouas
  2. Denitrification, Pseudomonas
  3. Nitrate assimilation, Nitrogenase
  4. Ammonification, Bacillus

Answer: 4. Ammonification, Bacillus

Question 46. Which of the following elements are essential for the photolysis of water?

  1. Ca and Cl
  2. Mn and Cl
  3. Zn and I
  4. Cu and Fe

Answer: 2. Mn and Cl

Question 47. Which of the following is related with the transfer of food material?

  1. Xylem
  2. Collenchyma
  3. Phloem
  4. Parenchyma

Answer: 3. Phloem

Question 48. Which of the following elements is most mobile in plant metabolism?

  1. Calcium
  2. Phosphorus
  3. Carbon
  4. Magnesium

Answer: 2. Phosphorus

Question 49. The process of converting ammonia to nitrate by bacteria is known as

  1. Ammonification
  2. Nitrification
  3. Nitrogen fixation
  4. Denitrification

Answer: 2. Nitrification

Question 50. Root nodules that are present in plants are meant for fertilizers and are found in/on

  1. Certain leguminous plants
  2. Casuarina
  3. Ainas
  4. All of the above

Answer: 4. All of the above

Question 51. Agriculturists have reported about 40-50% higher yields of rice by applying

  1. Azolla
  2. Cyanophycean members
  3. Mycorrhizae
  4. Thom forest

Answer: 1. Azolla

Question 52. A nutrient element essential for the formation of microtubules of the mitotic spindle apparatus during cell division is

  1. Phosphorus
  2. Sulfur
  3. Calcium
  4. Zinc

Answer: 3. Calcium

Question 53. The non-symbiotic N2 fixer is

  1. Anabaena
  2. Rhizobium
  3. Azotobactor
  4. Azolla

Answer: 3. Azotobactor

Question 54. The N2 fixing bacterium associated with root nodules of legumes is known as

  1. Azotobacter
  2. Nitrobacter
  3. Lactobacillus
  4. Rhizobium

Answer: 4. Rhizobium

Question 55. The bacteria that converts nitrate into molecular nitrogen is called

  1. Nitrifying bacteria
  2. Methanobactcria
  3. Diazotrophic bacteria
  4. Denitrifying bacteria

Answer: 4. Denitrifying bacteria

Question 56. The bacterium capable of anaerobic N2 fixation is known as

  1. Clostridium
  2. Bacillus
  3. Azotobacter
  4. Rhizobium

Answer: 1. Clostridium

Question 57. Which element is essential for the photolysis of water?

  1. Nitrogen
  2. Manganese
  3. Carbon
  4. Oxygen

Answer: 2. Manganese

Question 58. Which of the following can utilise molecular nitrogen (N2) as a nutrient for growth?

  1. Rhizobium
  2. Spirogyra
  3. Mucor
  4. Methancoccus

Answer: 1. Rhizobium

Question 59. Sinks are related to

  1. Transport of organic solutes
  2. Stomata
  3. Enzymes
  4. Phytochrome

Answer: 1. Transport of organic solutes

Question 60. Supply ends in the transport of solute are

  1. Green leaves and storage organs
  2. Root and stem
  3. Xylem and phloem
  4. Hormones and enzyme

Answer: 1. Green leaves and storage organs

Question 61. Which of the following is a biofertilizer?

  1. Funaria
  2. Fern
  3. Anabaena
  4. Fungus

Answer: 3. Anabaena

Question 62. Mo is related with

  1. N2 fixation
  2. Flower induction
  3. Chromosome contraction
  4. Carbon collection

Answer: 1. N2 fixation

Question 63. Which one of the following elements is present in chlorophylls?

  1. Manganese
  2. Magnesium
  3. Copper
  4. Iron

Answer: 2. Magnesium

Question 64. Which one of the following bacteria has the potential for nitrogen fixation?

  1. Nitrosomonas
  2. Nitrobacter
  3. Nitrosococcus
  4. Rhizobium

Answer: 4. Rhizobium

Question 65. For nitrogen fixation, the pigment useful is

  1. Nitrogenase
  2. Hemoglobin
  3. Myoglobin
  4. Leghemoglobin

Answer: 4. Leghemoglobin

Question 66. Which of the following is a symbiotic bacteria?

  1. Rhizobium
  2. Azotobacter
  3. Clostridium
  4. Streptomyces

Answer: 1. Rhizobium

Question 67. The metal ion involved in stomatal regulation is

  1. Fe
  2. Mg
  3. Zn
  4. K

Answer: 4. K

Question 68. Legume plants are important for crop production because they

  1. Help in NO2 fixation
  2. Do not help in NO2 fixation
  3. Increase soil fertility
  4. All of these

Answer: 3. Increase soil fertility

Question 69. Which of the following is a nitrogen-fixing organism?

  1. Some BGA
  2. Rhizobium
  3. Both (1) and (2)
  4. Aspergillus

Answer: 3. Both (1) and (2)

Question 70. Which of the following bacteria is involved in the two-step conversion of NH3 into nitrate?

  1. Azotobacter and Nitrosomonas
  2. Nitrosomonas and Nitrobacter
  3. Azotobacter and Achromobacter
  4. Pseudomonas and Nitrobacter

Answer: 2. Nitrosomonas and Nitrobacter

Question 71. A metal ion involved in stomatal regulation is

  1. Iron
  2. Potassium
  3. Zinc
  4. Magnesium

Answer: 2. Potassium

Question 72. The plant ash is an indication of

  1. Organic matter of plant
  2. Waste product
  3. Mineral salts absorbed by plants
  4. None of these

Answer: 3. Mineral salts absorbed by plants

Question 73. Plant ash has a maximum content of

  1. Mg
  2. Fe
  3. K
  4. B

Answer: 1. Mg

Question 74. Which of the following is a part of cytochrome?

  1. Mg
  2. Zn
  3. Fe
  4. Ca

Answer: 3. Fe

Question 75. Food in plants is translocated in the form of

  1. Glucose
  2. Starch
  3. Sucrose
  4. Fructose

Answer: 3. Sucrose

Question 76. Which of the following is not related to N2 fixation?

  1. Rhizobium
  2. Anabaena
  3. Pseudomonas
  4. Azotobacter

Answer: 3. Pseudomonas

Question 77. Which of the following is not caused by a deficiency of minerals?

  1. Chlorosis
  2. Etiolation
  3. Shortening of internodes
  4. Necrosis

Answer: 2. Etiolation

Question 78. The mineral present in cell walls is

  1. Na
  2. Ca
  3. K
  4. Mg

Answer: 2. Ca

Question 79. What happens when we inoculate Rhizobium in wheat fields?

  1. No increase in production (nitrogen content of soil remains same)
  2. A lot of increase in production (nitrogen content of soil increase)
  3. Fertility of soil decreases
  4. Fertility of soil increases

Answer: 1. No increase in production (nitrogen content of soil remains same)

Question 80. Nitrifying bacteria are able to

  1. Convert atmospheric nitrogen into soluble form
  2. Convert ammonia to nitrate
  3. Ammonia to nitrogen
  4. Nitrate to nitrogen

Answer: 2. Convert ammonia to nitrate

Question 81. Magnesium is found in

  1. Chlorophyll
  2. Carotenoid
  3. Phycobilin
  4. Cytochrome

Answer: 1. Chlorophyll

Question 82. Which of the following is a trace element?

  1. S
  2. Mg
  3. Cu
  4. P

Answer: 3. Cu

Question 83. Which one of the following organisms may respire in the absence of oxygen?

  1. Azotobacter
  2. Clostridium
  3. Rhizobium
  4. Lactobacillus

Answer: 2. Clostridium

Question 84. Which of the following is not a trace element?

  1. Zn
  2. Mn
  3. Mg
  4. Cu

Answer: 3. Mg

Question 85. Symbiotic microorganism is

  1. Clostridium
  2. Azotobacter
  3. Rhizobium
  4. Chromatium

Answer: 3. Rhizobium

Question 86. Essential mineral nutrients are the elements

  1. In the absence of this plants cannot complete their life cycle
  2. Which cannot be replaced by another element in its function
  3. Which are directly associated with the plant metabolism
  4. All of the above

Answer: 4. All of the above

Question 87. Stomatal movement is controlled by

  1. Na
  2. Mg
  3. K
  4. P

Answer: 3. K

Question 88. Which of the following enzymes fixes nitrogen?

  1. Nitrate reductase
  2. Nitrogenase
  3. PEP carboxylase
  4. RuBisCo

Answer: 2. Nitrogenase

Question 89. The bacterium capable of anaerobic nitrogen fixation is

  1. Azatobacter
  2. Rhizobium
  3. Bacillus
  4. Clostridium

Answer: 4. Clostridium

Question 90. In plant metabolism, phosphorus plays a major role to

  1. Evolve oxygen during photosynthesis
  2. Create aerobic condition
  3. Generate metabolic energy
  4. Evolve carbon dioxide during respiration

Answer: 3. Generate metabolic energy

Question 91. Photosynthetic food material is transported in the form of

  1. Glucose
  2. Sucrose
  3. Starch
  4. Fructose

Answer: 2. Sucrose

Question 92. Chlorosis is caused due to the deficiency of

  1. Mg
  2. Ca
  3. B
  4. Mn

Answer: 1. Mg

Question 93. The major portion of the dry weight of plants comprises

  1. Nitrogen, phosphorus, and potassium
  2. Calcium, magnesium, and sulfur
  3. Carbon, nitrogen, and hydrogen
  4. Carbon, hydrogen, and oxygen

Answer: 4. Carbon, hydrogen, and oxygen

Question 94. Which one of the following mineral elements plays an important role in biological nitrogen fixation?

  1. Copper
  2. Manganese
  3. Zinc
  4. Molybdenum

Answer: 4. Molybdenum

Question 95. Stomata of CAM plants

  1. Are always open
  2. Open during the day and close at night
  3. Open at night and close during the day
  4. Never open

Answer: 3. Open at night and close during the day

Question 96. Stomata of a plant open due to

  1. The influx of potassium ions
  2. Efflux of potassium ions
  3. Influx of hydrogen ions
  4. The influx of calcium ions

Answer: 1. Influx of potassium ions

Question 97. Plants deficient of the element zinc show its effect on the biosynthesis of plant growth hormone

  1. Auxin
  2. Cytokinin
  3. Ethylene
  4. Abscisic acid

Answer: 1. Auxin

Question 98. In which one of the following is nitrogen not a constituent?

  1. Idioblast
  2. Bacteriochlorophyll
  3. Invertase
  4. Pepsin

Answer: 1. Idioblast

Question 99. Gray spots of oats are caused by the deficiency of

  1. Cu
  2. Zn
  3. Mn
  4. Fe

Answer: 3. Mn

Question 100. The most abundant element present in the plants is

  1. Iron
  2. Carbon
  3. Nitrogen
  4. Manganese

Answer: 2. Carbon

Question 101. The ability of Venus flytrap to capture insects is due to

  1. Chemical stimulation by the prey
  2. A passive process requiring no special ability on the part of the plant
  3. Specialized muscle-like cells
  4. Rapid turgor pressure changes

Answer: 4. Rapid turgor pressure changes

Question 102. The deficiencies of micronutrients not only affect the growth of plants but also vital functions such as photosynthetic and mitochondrial electron flow. Among the list given below, which group of three elements shall affect both photosynthetic and mitochondrial electron transport the most?

  1. Cu, Mn, Fe
  2. Co, Ni, Mo
  3. Mn, Co, Ca
  4. Ca, K, Na

Answer: 1. Cu, Mn, Fe

Question 103. Potometer works on the principle of

  1. The amount of water absorbed equals the amount that transpired
  2. Osmotic pressure
  3. Root pressure
  4. The potential difference between the tip of the tube and that of the plant

Answer: 1. Amount of water absorbed equals the amount that transpired

Question 104. Farmers in a particular region were concerned that the premature yellowing of leaves of a pulse crop might cause a decrease in the yield. Which treatment could be the most beneficial to obtain maximum seed yield?

  1. Removal of all yellow leaves and spraying the remaining green leaves with 2,4,5-trichloro phenoxy acetic acid
  2. Application of iron and magnesium to promote the synthesis of chlorophyll
  3. Frequent irrigation of the crop
  4. Treatment of plants with cytokinins along with a small dose of nitrogenous fertilizer

Answer: 2. Application of iron and magnesium to promote the synthesis of chlorophyll

Question 105. Sulfur is an important nutrient for optimum growth and productivity in

  1. Fiber crops
  2. Oil seed crops
  3. Pulse crops
  4. Cereals

Answer: 2. Oil seed crops

Question 106. A plant requires magnesium for

  1. Cell wall development
  2. Holding cells together
  3. Protein synthesis
  4. Chlorophyll synthesis

Answer: 4. Chlorophyll synthesis

Question 107. Which of the following is a flowering plant with nodules containing filamentous nitrogen-fixing microorganisms?

  1. Cicer arietinum
  2. Casuarina equisetifolia
  3. Crotalaria juncaea
  4. Cycas revolute

Answer: 2. Casuarina equisetifolia

Question 108. About 98% of the mass of every living organism is composed of just six elements including carbon, hydrogen, nitrogen, and oxygen.

  1. Calcium and phosphorus
  2. Phosphorus and sulfur
  3. Sulfur and magnesium
  4. Magnesium and sodium

Answer: 2. Phosphorus and sulfur

Question 109. Which one of the following elements is not an essential micronutrient for plant growth?

  1. Ca
  2. Mn
  3. Zn
  4. Cu

Answer: 1. Ca

Question 110. Carbohydrates are commonly found as starch in plant storage organs. Which of the following five properties of starch (1-5) make it useful as a storage material?

  1. Easily translocated
  2. Chemically non-reactive
  3. Easily digested by animals
  4. Osmotically inactive
  5. Synthesized during photosynthesis
  1. (1), (3), and (5)
  2. (1) and (5)
  3. (2) and (3)
  4. (2) and (4)

Answer: 4. (2) and (4)

Question 111. Nitrogen fixation in the root nodules of Anlus is brought about by

  1. Frankia
  2. Azorhizobium
  3. Bradyrhizobium
  4. Clostridium

Answer: 1. Frankia

Question 112. Guard cells help in

  1. Fighting against infection
  2. Protection against grazing
  3. Transpiration
  4. Guttation

Answer: 3. Transpiration

Question 113. Manganese is required in

  1. Chlorophyll synthesis
  2. Nucleic acid synthesis
  3. Plant cell wall formation
  4. Photolysis of water during photosynthesis

Answer: 4. Photolysis of water during photosynthesis

Question 114. An element playing an important role in nitrogen fixation is

  1. Manganese
  2. Zinc
  3. Molybdenum
  4. Copper

Answer: 3. Molybdenum

Question 115. Which one of the following is not a micronutrient?

  1. Zinc
  2. Boron
  3. Molybdenum
  4. Magnesium

Answer: 4. Magnesium

Question 116. The chief water-conducting elements of xylem in gymnosperms are

  1. Transfusion tissue
  2. Tracheids
  3. Vessels
  4. Fibers

Answer: 2. Tracheids

Question 117. Which one of the following structures between two adjacent cells is an effective transport pathway?

  1. Endoplasmic reticulum
  2. Plasmalemma
  3. Plasmodesmata
  4. Plastoquinones

Answer: 3. Plasmodesmata

Question 118. One of the free-living, anaerobic nitrogen fixers is

  1. Rhizobium
  2. Azotobacter
  3. Beijerinckia
  4. Rhodospirilium

Answer: 4. Rhodospirilium

Question 119. The common nitrogen fixer in paddy fields is

  1. Oscillatoria
  2. Frankia
  3. Rhizobium
  4. Azospirillum

Answer: 4. Azospirillum

Question 120. The transport of food material in higher plants takes place through

  1. Transfusion tissue
  2. Tracheids
  3. Sieve elements
  4. Companion cells

Answer: 3. Sieve elements

Question 121. Study the cycle and select the option that gives the correct words for all four blanks A, B, C, and D.

NEET Biology Mineral Nutrition Atmospheric Cycle

NEET Biology Mineral Nutrition Atmospheric Cycle

Answer: 1

Question 122. Leguminous plants are able to fix atmospheric nitrogen through the process of symbiotic nitrogen fixation. Which one of the following statements is not correct during this process of nitrogen fixation?

  1. Nodules act as sites for nitrogen fixation.
  2. The enzyme nitrogenase catalyzes the conversion of atmospheric N2 to NH3.
  3. Nitrogenase is insensitive to oxygen.
  4. Leghemoglobin scavenges oxygen and is pinkish in color.

Answer: 3. Nitrogenase is insensitive to oxygen.

Question 123. Which one of the following is correctly matched?

  1. Apoplast—Plasmodesmata
  2. Potassium—Readily immobilization
  3. Balance of rice seedlings—F. Skoog
  4. Passive transport of nutrients—ATP

Answer: 2. Potassium—Readily immobilization

Question 124. Which one of the following is a wrong statement?

  1. Root nodule-forming nitrogen fixers live as aerobes under free-living conditions.
  2. Phosphorus is a constituent of cell membranes, certain nucleic acids, and all proteins.
  3. Nitrosomonas and Nitrobacter are chemoautotrophs.
  4. Anabaena and Nostoc are capable of fixing nitrogen in a free-living state also.

Answer: 2. Phosphorus is a constituent of cell membranes, certain nucleic acids, and all proteins.

Question 125. The nitrogen-fixing microbe associated with Azolla in rice fields is

  1. Anabaena
  2. Frankia
  3. Tolypothrix
  4. Spimlina

Answer: 1. Anabaena

Question 126. Best defined function of Manganese in green plants is

  1. Calvin cycle
  2. Nitrogen fixation
  3. Water absorption
  4. Photolysis of water

Answer: 4. Photolysis of water

Question 127. For its activity, carboxypeptidase requires

  1. Zinc
  2. Iron
  3. Niacin
  4. Copper

Answer: 1. Zinc

Question 128. For its action, nitrogenase requires

  1. High input of energy
  2. Light
  3. Mn2+
  4. Super oxygen radicals

Answer: 1. High input of energy

NEET Biology Notes For Mineral Nutrition Assertion Reasoning Question And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: Some mineral nutrients are essential.

Reasoning: They can be synthesized by the plants.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 2. Assertion: Ca++ cannot replace H+ adsorbed on clay or humus particles.

Reasoning: The retentive capacity of Ca+2 is more than that of H+.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 3. Assertion: Chelating agents used in improving the availability of some minerals in soil are actually electron acceptors.

Reasoning: They increase the solubility of some minerals in acidic soils.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 4. Assertion: N, P, and K are called critical elements.

Reasoning: They become deficient easily in soil due to leaching and higher requirements.

Answer: 4. If both Assertion and Reason are false.

Question 5. Assertion: When cation uptake exceeds anion uptake, certain changes occur in the ionic composition of the cell.

Reasoning: H2O and organic acids produced within cell dissociate and H+ moves outside the cell.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

 

 

NEET Biology Notes – Cell Cycle And Cell Division

by

NEET Biology Notes For Cell Cycle And Cell Division

Cell Division: The process of formation of a new cell from the pre-existing cell is called cell division.

Why Cell Divides:

  1. A cell divides due to a change in the nuclear-cytoplasmic ratio. This ratio is also known as the karyoplasmic ratio (KPR).
  2. A cell divides due to a change in surface area volume ratio (S/V ratio).

In small cells, the S/V ratio and KPR are high. As a cell increases in size, these two ratios decrease. In order to maintain these two ratios, a cell has to divide. Cells also divide due to the following points:

  1. \(\frac{R N A}{D N A}\) ratio determines the type of division.
  2. If \(\frac{R N A}{D N A}\) > 1 → Mitosis (reported by Hotta)
  3. If \(\frac{R N A}{D N A}\) < 1 → Meiosis (reported by Hotta)

Types of Cell Division: Cell division is further divided into direct and indirect cell divisions. These are explained below.

NEET Biology Cell Cycle And Cell Division Types Of Cell Divison

NEET Biology Notes For Cell Cycle And Cell Division Cell Cycle

Cell cycle was discovered by Howard in Vicia faba. Every dividing cell must pass through the cell cycle. The sequence of events through which a cell duplicates its content and divides into daughter cells is called the cell cycle.

Read and Learn More NEET Biology Notes

Stages Of Cell Cycle: The division of a cell includes the following sequence of stages

NEET Biology Cell Cycle And Cell Division Depiciting The Formation Of Daughter Cell From One Cell

The cell cycle depicts the formation of daughter cells from one cell.

NEET Biology Cell Cycle And Cell Division The Cell Cycle

Interphase: Interphase is a longer phase of the cell cycle than the 1-phase. It is commonly called a resting phase because no visible changes in the nucleus take place. Interphase represents generation time. It is metabolically the most active phase, hence called the energy phase.

Synthesis of DNA, RNA, protein, ATP, enzyme, etc., takes place, hence the size of the nucleus increases. Chromosomes are present in the form of a chromatin network in this phase. This phase covers more than 95% time of the cell cycle.

Steps Of Interphase

G1-phase: G1-phase is also called post-mitotic, pre-synthetic, first gap, first growth phase.

  • Longest phase of interphase.
  • Synthesis of RNA and non-histone protein starts here and continues up to G2 phase.
  • Cell increases in its size to the maximum (double).
  • No synthesis of DNA takes place, hence no change in DNA content.
  • The cell divides under stress conditions, so the synthesis of ATP and amino acids takes place at the last stage of G1-phase, hence called antiphase (end of G1-phase).
  • The decision of cell division takes place in this phase, hence it is called the restriction point or checkpoint.
  • It is the most variable phase in different plant species. It takes different time period in different organisms. G0-phase (quiescent phase) is a temporary arrest of cell division and the cell undergoes differentiation, for example, heart cell, root tip.

cell cycle and cell division pdf notes

S-Phase (Synthetic Phase): The replication of DNA takes place, hence replication of chromosomes takes place. The DNA content becomes doubled, but no change in the number of chromosomes so the ploidy level remains the same in the cell.

NEET Biology Cell Cycle And Cell Division Synthetic Phase

The duplication of centriole and synthesis of histone protein takes place in animal cells in S-phase. It is called the invisible phase because replicated chromosomes are not visible.

G2-phase: This phase is also known as the post-synthetic or premitotic phase second gap or second growth phase.

  • G2 is called the period of cytoplasmic growth.
  • Duplication of cell organelles takes place except cen-triole.
  • Synthesis of RNA and non-histone protein is continued up to this stage.
  • Synthesis of tubulin protein required for spindle formation takes place.
  • Repairing of damaged DNA takes place.
  • Synthesis of cyclin-dependent protein kinase (CDC- kinase) takes place, which controls the cell cycle.
  • Kinase enzyme causes the phosphorylation of the nuclear membrane, hence it disappears in late prophase.
  • G1 to S-transition (major checkpoint) requires G2-cyclin + CDC-2 kinase.
  • G2– to M-transition (minor checkpoint) requires mitotic cyclin + CDC-2 kinase. It is called the maturation-promoting factor (MPF).

NEET Biology Notes For Cell Cycle And Cell Division Mitosis (Equational Division)

The term mitosis was coined by W. Flemming. Mitosis was discovered by Strassburger in plant cell and by W. Flemming in an animal cell. Mitosis is a type of indirect cell division in which one mother cell divides into two daughter cells in which the number of chromosomes is similar to the number of mother cells and both daughter cells are also similar to each other.

Generally, mitosis takes place is somatic cells, but also takes place in reproductive cells to increase the number. In animals, mitosis only takes place in diploid somatic cells, while in plants, it takes place in both haploid and diploid cells. A brief outline of the mechanism of mitosis is given below.

NEET Biology Cell Cycle And Cell Division Mechanism Of Mitosis

Karyokinesis: Karyokinesis takes place in the following phases

1. Prophase

  • The longest phase of M-phase.
  • Condensation and dehydration of the chromatin network take place resulting in the formation of chromosomes.
  • Chromosomes are much elongated and ends are not distinct. Hence, early prophase is also called the ball of wool or spireme stage.
  • Chromosomes appear in the form of double strands.
  • It is composed of two chromatids, and both chromatids are attached to centromere.
  • All chromosomes become arranged at the periphery of the nucleus.
  • During the late prophase, the nuclear membrane and nucleolus disappear.

NEET Biology Cell Cycle And Cell Division Prophase Stage

  • During prophase, the viscosity and refractive index of cytoplasm increases.
  • In animal cells, the already-divided centriole reaches to poles from which astral ray develops. Hence, animal mitosis is called amphiastral.
  • In plant cells, mitosis is called anastral due to the absence of an astral ray.
  • At the end of this phase, the cell does not show a Golgi body, ER, nucleolus, or nuclear envelope.

2. Metaphase

  • During metaphase, maximum dehydration of chromosome takes place. Hence, the chromosome becomes much shorter and thicker.
  • Hence, the size of the chromosome is measured in this stage. The morphology of chromosomes can be most easily studied in this phase.
  • All chromosomes are arranged at the central part of the cell to form an equatorial or metaphase plate.

NEET Biology Cell Cycle And Cell Division Metaphase Stage

  • The formation of a single metaphase plate takes place in mitosis.
  • Appearance of spindle fiber takes place.
  • Spindle fiber is composed of tubulin protein (97%) and RNA (3%).
  • The two types of spindle fiber are (1) continuous fiber (these attach to both poles of the cell) and (2) chromosomal fiber (one end attaches to the pole and the other end attaches to the kinetochore of the chromosome).

The arrangement of chromosomes in metaphase is called congression. Centromeres of chromosomes are aligned toward the equatorial plate and arms are toward the pole.

3. Anaphase

  • It is the shortest phase of the M-phase.
  • During early anaphase, the division of the centromere takes place. Hence, both chromatids of a chromo¬some separate from each other, and the separated chromatids are called daughter chromosomes.
  • During late anaphase, daughter chromosomes move to their respective pole due to
    • Contraction of spindle fiber
    • Relaxation of interzonal fiber which forms between two daughter chromosomes.

NEET Biology Cell Cycle And Cell Division Anaphase Stage

  • The shape of the chromosome is observed in anaphase.
  • The movement of chromosomes toward the pole requires 30 ATPs.
  • Cytokinesis also starts during anaphase.

4. Telophase

  • In telophase, all events are reverse of prophase.
  • Chromosomes reach to their poles.
  • Spindle fibers disappear. The nuclear membrane and nucleolus reappear at both poles.
  • Hydration and de-condensation of chromosomes take place.
  • It results in the formation of two daughter nuclei.
  • At in this stage, the chromosomes lose their individuality.
  • The nucleolus, Golgi body, and ER are reformed.

NEET Biology Cell Cycle And Cell Division Early Telophase And Late Telophase

Cytokinesis: The division of cytoplasm is known as cytokinesis.

1. In Plants: Cytokinesis takes place by cell plate formation. Phragmoplast develops between two daughter nuclei by the deposition of spindle fiber and vesicles of GB. The growth of phragmoplast takes place centrifugally. These phragmoplasts form a cell plate.

NEET Biology Cell Cycle And Cell Division Formation Of Cell Plate In Plants

2. In Animals: Cytokinesis takes place by cell furrow method. Contraction in the cell membrane takes place between two daughter nuclei which move centripctally forming two daughter cells.

NEET Biology Cell Cycle And Cell Division Division Of A Mother Cell Into daughter Cells In Animals

Significance Of Mitosis

  • In unicellular organisms, it increases the number of individuals.
  • In multicellular organisms, it is responsible for the growth of the body and the repairing of body parts.
  • It maintains the number of chromosomes in all cells of organisms.
  • It also maintains S/V ratio and KPR.

NEET Biology Notes For Cell Cycle And Cell Division Mitosis Points To Remember

Number of mitosis required for n number of cells is (n – 1), where n is the required number of cells, A number of cells formed in n generation is 2″, where n is the number of generations.

  1. Chemicals that induce mitosis are auxin, gibberellins, cytokinin, insulin, and lymphokines.
  2. Mitotic poison includes those chemicals that inhibit mitosis, for example, chalones, mustard gas (agglutinates the chromosomes), ribonuclease, colchicine, cyanide (inhibits prophase), azides (inhibits prophase), and X-ray.
  3. Eumitosis: It is extranuclear mitosis, i.e., nuclear membrane degenerates.
  4. Pre-mitosis: It is intranuclear mitosis, i.e., the nuclear membrane does not degenerate but intranuclear spindle formation takes place, for example, fungi, protozoa, some algae, etc.
  5. Free Nuclear Division: Karyokinesis is not followed by cytokinesis and multinucleate cell forms, for example, Rhizopus, Vaucheria, Mucor, slime mold, etc.
  6. Denomitosis: In this type of mitosis, the nuclear membrane does not disappear and it is only found in dinoflagellates.
  7. Endomitosis was discovered by Meyer. The division of chromosomes takes place, but no division of the nucleus. Hence, all divided chromosomes remain present in the same nucleus. That is why the number of chromosomes has doubled.
  8. Endomitosis produces polyploidy, for example, tapetum cells. Endomitosis can be artificially induced by colchicines. Such mitosis is called C-mitosis (colchicines- induced mitosis).
  9. Colchicines dissolve microtubules and hence split or stop spindle fiber formation.
  10. Colchicines are produced from Colichicum autums of Liliaceae.

NEET Biology Notes For Cell Cycle And Cell Division Meiosis

Meiosis takes place in diploid cells called as meiocytes. This division occurs only in reproductive cells. The term meiosis was coined by Fanner and Moore. It was discovered by von Benden and Winiwarter.

  • It is a type of indirect cell division in which one diploid reproductive mother cell divides into four haploid daughter cells.
  • Daughter cells are quite different from each other. The number of chromosomes in a daughter cell reduces to half of the mother cell.
  • Two divisions of the nucleus and only one division of the chromosome take place in meiosis.
  • The structural division of chromosomes takes place in meiosis II and the numerical division of chromosomes takes place in meiosis I.
  • In angiosperms, in the bud stage, the anther and ovule show meiosis.
  • Anther and ovule is the most suitable structures for meiosis study.
  • In algae, the zygote undergoes meiosis.
  • In bryophyta, pteridophyta, and gymnosperms, SMC (spore mother cell) undergoes meiosis.

In Fungi:

NEET Biology Cell Cycle And Cell Division Meiosis

  • Brachymeiosis was discovered by Gynenvaughan in fungi. Double reduction takes place in chromosome number in a tetraploid cell to produce a haploid cell.
  • The S-phase of meiosis is much longer than mitosis. Replication of DNA takes place here.
  • G2-phase is either very short or absent in meiosis.
  • Interkinesis is the period in between meiosis-1 and meiosis-2. It may or may not be present. It is generally present in animal cells. There is no S-phase and no DNA synthesis. Hence, called interkinesis.
  • Only duplication of centrioles takes place during interkinesis.

Types Of Meiosis

  1. Zygotic Or Initial Meiosis: Thallophyta (algae, fungi).
  2. Sporic Or Intermediate Meiosis: Bryophta to angiosperm.
  3. Gametic Or Terminal Meiosis: Animal and some brown algae.

Mechanism Of Meiosis: Meiosis includes the following phases

NEET Biology Cell Cycle And Cell Division Mechanism Of Meiosis

Meiosis 1

1. Prophase 1: It is the longest phase of meiosis, hence divided into five phases.

Leptotene: In this phase, condensation and dehydration of the chromatin network initiate.

  • Chromosomes appear as elongated, single-stranded, as shown. Beaded structures represent the chromomere.
  • Chromosomes always appear in homologous pairs; hence, they are diploid.
  • Each chromosome is composed of two chroma¬tids but these chromatids are not visible.
  • Both ends of the chromosome become attached to the nuclear lamina of the nuclear membrane and form a loop-like structure. Hence, this is also called the bouquet stage.

NEET Biology Cell Cycle And Cell Division Leptotene Phase

Zygotene: Dehydration of chromosomes continues. Hence, it becomes shorter and thicker.

  • Both chromatids of a chromosome are called diad.
  • The formation of recombination nodules takes place at several points in divalent chromosomes.
  • Exchange of part of nonsister chromatid takes place. This process is called as crossing over.
  • Formation of the synaptonemal complex takes place between bivalent chromosomes composed of ribonucleic protein (RNA 5% + Protein 95%) reported by Montrose

NEET Biology Cell Cycle And Cell Division Zygotene Phase

Pachytene: Both chromatids of chromosomes become visible, hence each bivalent is composed of chromatids called tetrad.

  • The formation of recombination nodules takes place at several points in divalent chromosomes.
  • Exchange of part of non-sister chromatid takes place. This process is called as crossing over.

NEET Biology Cell Cycle And Cell Division Pachytene Phase

Diplotene: Crossing over is an enzyme-mediated process and the enzyme involved is called recombinase.

  • In the oocyte of the same vertebrate, the diplotene stage can last for months or years.
  • The breaking of the synaptonemal complex starts. Hence, both chromosomes of bivalent tend to separate from each other called terminalization.
  • Both chromosomes are not separated completely due to the formation of an X-like structure at the point of crossing over called chiasmata.
  • Charismata is the result of crossing over.
  • Terminalization starts here but does not complete here.
  • The process of degeneration of the synaptonemal complex is called desynapsis.

NEET Biology Cell Cycle And Cell Division Diplotene Phase

Diakinesis: Both chromosomes of bivalent become separate, i.e., the process of terminalization is complete here. The nuclear membrane and nucleolus disappear.

NEET Biology Cell Cycle And Cell Division Diakinesis

2. Metaphase 1: The formation of a double metaphase plate takes place at the equator. Spindle fibers appear which attach to the centromeres and pull the chromosomes.

NEET Biology Cell Cycle And Cell Division Metaphase 1

3. Anaphase 1: Both chromosomes of a bivalent move to opposite poles due to contraction of the spindle fiber. This is called disjunction. There is no division of centromere.

NEET Biology Cell Cycle And Cell Division Anaphase 1

4. Telophase 1: Chromosomes continuously move and reach to their respective poles. The nuclear membrane and nucleolus reappear. It results in the formation of two daughter cells containing a haploid number of chromosomes and each chromosome is composed of two chromatids.

NEET Biology Cell Cycle And Cell Division Telophase 1

Meiosis 2: Both daughter nuclei divide by mitosis to form four haploid daughter nuclei. Each daughter nucleus is haploid but the chromosome is composed of a single chromatid due to the division of centromere in anaphase 2.

NEET Biology Cell Cycle And Cell Division Meiosis 2

Different stages of meiosis are collectively shown.

cell cycle and cell division pdf notes

Cytokinesis: Cytokinesis is mainly of two types

  1. Successive Type: Cytokinesis takes place after meiosis 1 and 2 resulting in the formation of the isobilateral tetrad, for example, a monocot. It is of advanced type. In successive types, three cytokines are involved, first after meiosis 1 and two after meiosis 2.
  2. Simultaneous Type: Cytokinesis only takes place after meiosis 2 resulting in the formation of tetrahedral tetrad, for example, dicot.

NEET Biology Cell Cycle And Cell Division Different Stages Of Meiosis

Significance Of Meiosts

  • Mciosis maintains the number of chromosomes in each generation of organisms during sexual reproduction.
  • Crossing over lakes places in meiosis which produces variation, which plays an important role in evolution.
  • It is also required to complete the sexual life cycle of organisms.
  • Number of meiosis required for:
  • Formation of n number of pollen grams = \(\frac{n}{4}\)
  • Formation of n number of eggs = n
  • Formation of n number of seeds = n + \(\frac{n}{4}\)
  • The formation of n number of seeds in Cypraceae is n + n, i.e., 2n.

NEET Biology Notes For Cell Cycle And Cell Division Assertion Reasoning Type Questions And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

cell cycle notes pdf

Question 1. Assertion: Endomitosis does not cause karyokinesis or cytokinesis.

Reason: In endomitosis, mitosis occurs within the nucleus.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 2. Assertion: Synaptonemal complex develops between two synapsed homologous chromosomes.

Reason: Mitosis cannot be completed without the synaptonemal complex.

Answer: 3. If Assertion is true, but Reason is false.

Question 3. Assertion: During anaphase 2, the chromatids of a chromosome separate.

Reason: The Centromere of a mitotic chromosome divides during anaphase.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion

Question 4. Assertion: Dictyotene stage occurs in females only.

Reason: Gametogenesis rests for a long period at the diplotene stage in females.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 5. Assertion: Each chromosome of bivalent attaches with two spindles in metaphase.

Reason: In metaphase, bivalents migrate toward metaphase plate.

Answer: 4. If both Assertion and Reason are false.

NEET Biology Notes – Biomolecules

by

NEET Biology Notes For Biomolecules Introduction

Biomolecules Introduction: Protoplasm is a complex mixture of both organic and inorganic compounds.

  • Molecules found in the protoplasm of cells are called biomolecules.
  • The collection of various types of molecules in a cell is called the cellular pool.
  • The cellular pool consists of various types of biomolecules such as water, inorganic materials, and organic compounds.
  • Small molecules of low molecular weight, simple molecular conformations, and higher solubilities are called macromolecules. These include minerals, water, amino acids, simple sugars, and nucleotides.

A List Of Representative Inorganic Constituents Of Living Tissues

NEET Biology Biomolecules A List Of Representative Inorganic Constituents Of Living Tissues

  • Various minerals found in cells have many uses.
  • Mitochondria are rich in manganese.
  • Molybdenum is necessary for the fixation of nitrogen catalyzed by the enzyme nitrogenase.
  • Copper occurs in cytochrome oxidase.
  • Magnesium is essential for a large number of enzymes, particularly those utilizing ATP.
  • Calcium and magnesium decrease the excitability of nerves and muscles.
  • Sodium and potassium are responsible for the maintenance of extracellular and intracellular fluids through the osmotic effects of their concentrations. These two ions are also responsible for the maintenance of membrane potential and transmission of electrical impulses in the nerve cells. Both in cells and in extracellular fluids, diabasic phosphate (HPO42-) and monobasic phosphate (H2PO42-) act as acid-base buffers to maintain the W ion concentration.
  • The most abundant element in cell/living matter is oxygen (O > C > N > H ).
  • Fe++ and Cu++ are found in cytochromes.
  • The concentration of cations inside the cell is K > Na > Ca.

Read and Learn More NEET Biology Notes

A Comparison Of Elements Present In Non-Living And Living Matter

NEET Biology Biomolecules A Comparison Of Elements Present In Non Living And Living Matter

NEET Biology Notes For Biomolecules Methods To Analyze Chemical Composition

In order to study the various biomolecules found in living tissues (a vegetable, a piece of liver, etc.), the tissue is ground in trichloroacetic acid (Cl3CCOOH) using a pestle and mortar.

  • The resultant slurry is strained through cheese cloth or cotton and we obtain two fractions.
  • The filtrate is called an acid-soluble pool while the retentate is called an acid-insoluble fraction.
  • The acid-soluble pool represents roughly the cytoplasmic composition.
  • The macromolecules from the cytoplasm and organelles become the acid-insoluble fraction.
  • Chemicals present in both fractions are further separated by various analytical techniques and identified.

Average Composition Of Cells

NEET Biology Biomolecules Average Composition Of Cells

The acid-soluble pool contains chemicals called biomicromolecules as they have a small molecular mass of 18-800 daltons approximately.

biomolecules

  • Biomacromolecules are large in size, high molecular weight, and complex molecules that are formed by the condensation of biomacromolecules.
  • Their molecular mass is in the range of 10,000 daltons and above.
  • Biomacromolecules are of three types—proteins, nucleic acids, and polysaccharides.
  • Polymers occur in the form of threads. They are folded variously to form three-dimensional shapes required for their functioning.
  • Depending upon the molecular weight and solubility, biomolecules are divided into two categories.
  • Micromolecules: These are small-sized, have low molecular weight, simple molecular structure, and high solubility in the intracellular fluid matrix. These include water, minerals, gases, carbohydrates, lipids, amino acids, and nucleotides.
  • Macromolecules: These are large-sized, have larger molecular weight, complex conformation, and low solubility in the intracellular fluid matrix. They are generally formed by the polymerization of macromolecules. These include polysaccharides, proteins, and nucleic acids.

NEET Biology Biomolecules Types Of Molecules

Analytical techniques, when applied to the compound, give us an idea of the molecular formula and the probable structure of the compound.

  • All the carbon compounds that we get from living tissue can be called biomolecules. However, living organisms also have inorganic elements and compounds.
  • When the tissue is fully burnt, all carbon compounds are oxidized to gaseous form (CO2, water vapor) and are removed.
  • The residue is called ash. This ash contains inorganic elements such as calcium, magnesium, etc.
  • Inorganic compounds such as sulfate, phosphate, etc., are also seen in the acid-soluble fraction. Therefore, elemental analysis gives the elemental composition of living tissues in the form of H, O, Cl, C, etc., while the analysis of compounds gives an idea of the kind of organic and inorganic constituents present in living tissues.
  • From a chemistry point of view, one can identify functional groups such as aldehydes, ketones, aromatic compounds, etc. But from a biological point of view, we shall classify them into amino acids, nucleotide bases, fatty acids, etc.

NEET Biology Notes For Biomolecules Primary And Secondary Metabolites

The most exciting aspect of chemistry deals with isolating thousands of compounds, small and big, from living organisms, determining their structure and if possible, synthesizing them.

  • If one were to make a list of biomolecules, such a list would have thousands of organic compounds including amino acids, sugars, etc. We can call these biomolecules as metabolites.
  • In animal tissues, one notices the presence of all such categories of compounds. For example, proteins, carbohydrates, fats, amino acids, and nucleic acids.
  • These are called primary metabolites. However, when one analyzes plant, fungal, and microbial cells, one would see thousands of compounds other than these primary metabolites which are called secondary metabolites such as alkaloids, flavonoids, rubber, essential oils, antibiotics, colored pigments, scents, gums, and spices. Secondary metabolites are differentiated from primary metabolites in Table.
  • Primary metabolites have identifiable functions and play known roles in normal physiological processes. Many of the secondary metabolites are useful to human welfare (for example, rubber, drugs, spices, scents, and pigments). Their physiological role is unknown.
  • Some secondary metabolites have ecological importance too, which are mentioned in Table.

Difference Between Primary And Secondary Metabolites

NEET Biology Biomolecules Differences Between Primary And Secondary Metabolites

Some Secondary Metabolites

NEET Biology Biomolecules Some Secondary Metabolites

NEET Biology Notes For Biomolecules Carbohydrates

Carbohydrates are mainly composed of carbon, hydrogen, and oxygen. Carbohydrates are so called because in most of them, the proportion of hydrogen and oxygen is the same as that in water (H2O), i.e., 2:1. These are also known as saccharides (compounds containing sugar).

  • Carbohydrates are produced by green plants during photosynthesis. These constitute about 80% of the dry weight of plants.
  • Carbohydrates Are Divided Into Three Main Classes: monosaccharides, oligosaccharides, and polysaccharides.

NEET Biology Biomolecules Classification Of Carbohydrates And Their Examples

Monosaccharides: These are single saccharide units with CnH2nOn general formula which cannot be hydrolyzed further into smaller carbohydrates. These are composed of three to seven carbon atoms and are classified according to the number of C atoms as trioses (3C), tetroses (4C), pen-toses (5C), hexoses (6C), and heptoses (7C). Of these, pentoses and hexoses are the most common. Monosaccharides are important as energy sources and as building blocks for the synthesis of large molecules.

  • All monosaccharides are either aldoses or ketoses. The simplest monosaccharides include trioses, for example, glyceraldehyde and dihydroxyacetone.
  • Tetroses (for example, erythrose) are rare. Erythrose takes part in the synthesis of lignin and anthocyanin pigments.
  • Ribose, ribulose, xylulose, and arabinose are pentoses. Xyluloses and arabinoses polymerize to form xylans and arabans which are cell wall materials.
  • Glucose, fructose, mannose, and galactose are hexoses. These are white, sweet-tasting, crystalline, and extremely soluble in water.
  • Glucose is called “universal sugar” and is also known as dextrose grape sugar or corn sugar.

Fructose is called fruit sugar and is also known as levulose. It is a naturally occurring sweetest sugar. Honey has two sugars: dextrose and levulose.

Heptoses have seven carbon atoms per molecule of sugar with the general formula C7H14O7, for example, sedoheptulose. It is an intermediate of respiratory and photosynthetic pathways.

  • Pentoses and hexoses of monosaccharides occur in solid forms, i.e., open chain and ring form. There are two types of ring chains, i.e., pyranose ring, which has a hexagonal shape with five carbon atoms and one oxygen atom, and furanose ring, which has a pentagonal shape with four carbon atoms and one oxygen atom.
  • Monosaccharides have a “tree” aldehyde or ketone group which can reduce Cu++ to Cu+ Hence, these are also called reducing sugars.
  • Monosaccharides have two important chemical properties: first, sugars having a free aldehyde or ketone group can reduce Cu++ to Cu+ These are called reducing sugars. This property is the basis of Benedict’s test and Fehling’s test to detect the presence of glucose in urine.
  • Second, the aldehyde or ketone group of monosaccharides can react and bind with an alcoholic group of another organic compound to join the two compounds together. This bond is called the glycosidic bond. This bond can be hydrolyzed to give the original reactants.

NEET Biology Biomolecules Open Chain And Ring Form Of Different Monosaccharides

Differences Between Reducing And Non-Reducing Sugars

NEET Biology Biomolecules Differences Between Reducing And Non Reducing Sugars

NEET Biology Notes For Biomolecules Carbohydrates Points To Remember

Derived Monosaccharides

  • Deoxysugar: Loss of oxygen atom at the second carbon of ribose; yields deoxyribose, a constituent of DNA.
  • Aminosugar: Monosaccharides having an amino group, for example, glucosamine, and galactosamine.
  • Sugaracid: Examples, ascorbic acid, glucuronic acid, galacturonic acid.
  • Sugaralcohol: Examples, glycerol, and mannitol (present in brown algae).

” biological molecules “

Oligosaccharides: They are the condensation product of two to nine monosaccharides. These include trisaccharides, tetrasaccharides, hexasaccharides, heptasaccharides, etc.

Disaccharides

  • Disaccharides are formed by the condensation reactions between two monosaccharides (usually hexoses).
  • The bond formed between two monosaccharides is called a glycosidic bond.
  • It normally forms between C-atoms 1 and 4 of the neighboring units (1,4 bond).
  • Once linked, the monosaccharide units are called residues.
  • A molecule of sucrose is formed from a molecule each of glucose and fructose.

NEET Biology Biomolecules Structural Formulae Of Common Disaccharides

  • Sucrose is the storage product of photosynthesis in sugarcane and sugarbeet.
  • Lactose or milk sugar is found in human milk and cow’s milk.
  • It is formed from one glucose molecule and one galactose molecule.
  • Maltose or malt sugar is formed from two molecules of glucose during the germination of starchy seeds.
  • Maltose and lactose are reducing disaccharides.
  • Sucrose does not reduce Cu++ to Cu+. Hence, sucrose is a non-reducing sugar.

Trisaccharides

  • Sugars composed of three monosaccharide units are called trisaccharides (for example, raffinose).
  • Raffinose is a common trisaccharide found in plants.
  • Upon hydrolysis, it yields one molecule each of glucose, fructose, and galactose.
  • Larger oligosaccharides are attached to the cell membrane and enable cell-cell recognition due to their presence.
  • Trisaccharides also take part in antigen specificity.

Polysaccharides

  • These are polymers of monosaccharides and are branched or unbranched linear molecular chains.
  • These are insoluble carbohydrates and are considered to be non-sugars.
  • Starch, glycogen, cellulose, pectin, hemicellulose, and inulin are examples of polysaccharides.
  • Body cells store carbohydrates as polysaccharides since these are easy to store and can be easily converted back into simple carbohydrates upon hydrolysis. These are in more condensed form and they have high molecular weight. These cannot pass through the plasma membrane.

Based on the types of structural components involved, polysaccharides are of two types:

  • Hoinopolysaccharides: These consist of only one type of monosaccharide monomer for example starch, glycogen and cellulose, fructan, xylan, araban, and gnlactan.
  • Heteropolysaccharides: These consist of more than one type of monosaccharide monomer for example chitin, agar, arabanogalactans, arabanoxylans etc.
  • Based On Their Functions, Polysaccharides Are Of Three Main Types: storage (for example, starch and glycogen), structural (for example, chitin, cellulose), and mucopolysaccharides (for example, keratan sulfate, chondroitin sulfate, hyaluronic acid, agar, alginic acid, carrageenin. and heparin).

Storage Polysaccharides: Starch is found abundantly in rice, wheat, and other cereal grains and legumes, potato, tapioca, and bananas.

  • It is formed during photosynthesis and serves as an energy-storing material.
  • Glycogen found in the liver and muscles stores energy in mammals.
  • Storing carbohydrates in the form of polysaccharides has two advantages: first, during their formation, many molecules of water are removed from monosaccharides. This helps in condensing the bulk to be stored. Second, unlike small carbohydrates, polysaccharides are relatively easy to store.
  • When necessary, polysaccharides are broken down by enzymes for the release of energy.

Starch

  • Starch is a polymer of a-D-glucose. It is the major reserve food in plants.
  • Starch has two components: amylose (unbranched polymer) and amylopectin (branched polymer).
  • Amylopectin: Consists of 2000-200,000 glucose molecules forming a straight chain and shows branching (after 25 glucose units). The branching point has a-1,6 glycosidic linkages.
  • Amylose: Consists of a-1,4 glycosidic linkages between a-d glucose molecules. It is a straight chain of 200-1000 glucose units. Starch forms helical secondary structures; each turn consists of six glucose units.

NEET Biology Biomolecules Amylopection

NEET Biology Biomolecules Structure Of Amylose Showing Alpha 1 4 Linkage

Starch Points To Remember: Starch molecules accumulate in the form of layers (stratifications) around a shifting organic center (hilum) to form starch grains.

  • Hilum is made up of protein. In eccentric starch grains, hilum lies on one side. These are found in potatoes.
  • In concentric starch grains, hilum is present in the center. These are found in wheat, maize, and pea.
  • Dumb-bell-shaped starch grains are found in the latex of Euphorbia.
  • Starch grains with a single hilum are called simple (for example, maize) but those with more than one hilum are called compound (for example, potato, rice).
  • Starch turns blue with iodine as the helices in starch hold I2.

NEET Biology Biomolecules Strach Grains From Different Sources

Glycogen: Glycogen is the animal equivalent of starch; many fungi also stoic it. Glycogen turns red-violet with iodine. It consists of 30,000 glucose units joined by α-1.4 bonds, much more branched than starch. Each branch point has α-1,6 linkages and branching occurs after 10-14 glucose units.

Inulin: It is an unusual polysaccharide and a polymer of fructose. It is stored particularly in roots and tubers of the family Compositae, for example, Dahlia tubers.

Structural Polysaccharides (Cellulose) (Hexosan Polysaccharide)

  • Cellulose is the main structural unbranched homopolysaccharide of plants.
  • One molecule of cellulose has about 6000 β-glucose residues.
  • Cotton fibers contain the largest amount (90%) of cellulose among natural materials.
  • Wood contains between 25 and 50% cellulose, the rest being hemicellulose and lignin.
  • Fibers of cotton, linen, and jute are used for textiles and ropes.
  • The artificial fiber rayon is manufactured by dissolving cellulosic materials in alkali and by extruding and coagulating the filaments.
  • By treatment with other chemicals, cellulose is converted into cellulose acetate (used in fabrics, cellulosic plastics, and shatter-proof glass), cellulose nitrate (used in propellant explosives), and carboxymethyl cellulose (added to ice creams, cosmetics, and medicines to emulsify and give a smooth texture).

“intro to biomolecules “

NEET Biology Biomolecules Diagrammatic Representation Of A Portion Of Glycogen

Cellulose can be hydrolyzed to soluble sugars. Microbes can then convert these sugars to form ethanol, butanol, acetone, methane, and other useful chemicals.

  • Cellulose is an unbranched homopolysaccharide of β-glucose.
  • Cellulose is the most abundant carbohydrate in the biosphere.
  • Cellulose is produced by plants and is used for building cell walls. It is also the most abundant organic compound in the biosphere.
  • Wood and cotton contain large quantities of cellulose.
  • Chitin is a polysaccharide found in the exoskeleton of insects, crabs, and prawns.
  • Chitin is similar to cellulose in many ways except that its basic unit is not glucose, but a similar molecule that contains nitrogen (N-acetylglucosamine).
  • Although chitin is soft and leathery, it becomes hard when impregnated with calcium carbonate or certain proteins.
  • The insolubility of these polysaccharides in water helps to retain the form and strengthens the structure of organisms.
  • Pectin and hemicelluloses are structural polysaccharides.
  • Middle lamella which binds the cells together is composed of calcium pectate.
  • Due to calcium pectinate, the water absorption capacity of the cell wall is increased.
  • Fruit walls contain a high percentage of pectin.
  • During ripening, pectin breaks down into simple sugars resulting in the sweetening and loosening of fruits.
  • Hemicellulose is a mixture of n-xylose linked by β-1,4 glycosidic bond.
  • Xylans, Arabians, and Galatians are hemicelluloses. Food such as dates have hemicellulose as the reserve food.

Mucopolysaccharides

  • The slimy substances produced by plants are called mucilages.
  • When you soak the seeds of isabgol (Phuuago ovate) or cut the fruit of okra (bhindi), you will notice the presence of a slimy substance.
  • Mucilages are polysaccharides formed from galactose and mannose.
  • Many seaweeds yield mucilages of commercial value such as agar, alginic acid, and carrageenin.
  • Mucopolysaccharides are found in the cell walls of bacteria and in the connective tissues of animals as well as in body fluids.
  • These bind proteins in cell walls and connective tissue and water in interstitial spaces thereby providing lubrication in ligaments and tendons.
  • The vitreous humor of the eye and synovial fluid also contain mucopolysaccharides.
  • Hyaluronic acid is found in connective tissue and in cell walls.
  • Keratin sulfate and chondroitin sulfate occur in cartilage, cornea, and skin, and impart strength and flexibility to them.
  • Keratin sulfate consists of acctylglucosaminc, galactose, and sulphuric acid. It provides strength and flexibility to the skin and cornea.
  • Hyaluronic acid consists of d-glucuronic acid or iduronic acid and n-acetyl glucosamine, present in the vitreous humor of the eye, synovial fluid, cerebrospinal fluid, etc.
  • Heparin is a polymer of α-1,4 glucosamine and glucuronic acid.
  • It is an anticoagulant present in human blood.
  • The husk of Plantago ovata and the mucilage of Aloe barbadensis are medicinally used.
  • Agar, alginic acid, and carrageenin are obtained from marine algae.
  • Artificial silk is a polysaccharide prepared from rayon.

“biomolecule definition “

Differences Between Oligosaccharides And Polysaccharides

NEET Biology Biomolecules Differences Between Oligosaccharides And Polysaccharides

NEET Biology Notes For Biomolecules Amino Acids

Amino acids are small molecules made of carbon, hydrogen, oxygen, nitrogen, and in some cases, sulfur.

  • Each amino acid has a free amino group, a free carboxyl group, and R as the side chain as the same substituents on the same carbon atom.
  • The amino group lends basic character while the carboxylic group lends basic character to the molecule
  • Lysine and arginine are basic amino acids because they carry two amino groups and one carboxylic group.
  • Glutamic acid (glutamate) and aspartic acid (aspartate) contain one amino and two carboxyl groups each and are classified as acidic amino acids.

NEET Biology Biomolecules Side Chain Of A Basic And An Acidic Amino Acid

  • Alanine, glycine, and valine are neutral amino acids as these contain one amino and one carboxyl group each.
  • There are 20 different amino acids coded by our DMA that differ in the side chain.
  • Most amino acids are laevorotatory while glycine is optically inactive.
  • There are three important non-protein amino acids.
  • They are ornithine, citrulline (both are involved in the ornithine cycle to synthesize urea), and diaminopimelic acid.

NEET Biology Biomolecules Examples Of Polar And Nonpolar Amino Acids

  • A particular property of amino acids is the in/able nature of -NH2 and -COOH groups. Hence, in the solutions of different pH, the structure of amino acids changes.
  • There are two types of amino acids, viz., essential and non-essential amino acids.
  • There are seven essential amino acids in animals whereas eight essential amino acids in man.
  • Threonine is an additional essential amino acid in human beings.
  • Two amino acids, viz., arginine and histidine, are semi-indispensable amino acids as they can be synthesized by human beings but very slowly.

biomolecules functions

NEET Biology Biomolecules Essential And Non Essential Amino Acids

Differences Between Essential And Non-Essential Amino Acids

NEET Biology Biomolecules Differences Between Essential And Non Essential Amino Acids

NEET Biology Notes For Biomolecules Amino Acids Points To Remembers

Amino acids are classified into the following groups:

  • Neutral Amino Acid: With one -NH2 and one -COOH group, for example, glycine, and alanine (non-polar).
  • Acidic Amino Acid: Have an extra -COOH group (mono-amino dicarboxylic), for example, glutamic and aspartic acid.
  • Basic Amino Acid: Have additional NH2 group (diamino monocarboxylic), for example, arginine, and lysine.
  • Sulfur-Containing Amino Acid: Have sulfur, for example, cysteine, cystine, and methionine.
  • Alcoholic Amino Acid: Have —OH group, for example, serine, and threonine.
  • Aromatic Amino Acid: Has cyclic structure having a side chain with -COOH and -NH2 groups, for example, phenylalanine, tryptophan, and tyrosine.
  • Heterocyclic Amino Acid: N is present in the ring. for example. proline, histidine, hydroxyproline.
  • Semi-Essential Amino Acid: Arginine and histidine are semi-essential amino acids required by children.

Protein amino acids are levorotatory and α-type except glycine. Glycine: simplest amino acid, involved in the formation of heme.

Functions Of Amino Acids

  • Besides their principal function as building blocks lor proteins, specific amino acids are also converted into different types of biologically active compounds.
  • For example, tyrosine is converted into the hormones thyroxin and adrenaline, as well as the skin pigment melanin. Glycine is involved in the formation of heme and tryptophan in the formation of the vitamin nicotinamide as well as the plant hormone indole-3-acetic acid.
  • After the removal of the amino group, the carbon chain of many amino acids is converted into glucose.
  • On losing the carboxyl groups as carbon dioxide, amino acids form biologically active amines such as histamine. Histamine is required for the functioning of muscles, blood capillaries, and gastric juices.
  • Ornithine and citrulline are components of the urea cycle.
  • Antibiotics contain non-protein amino acids.
  • Amino acids form organic acids which form glucose by gluconeogenesis.
  • Lysine is an essential amino acid because it is not formed in the body and has to be provided through diet.

NEET Biology Notes For Biomolecules Proteins

Berzelius coined the term protein. Proteins are heteropolymers of amino acids. Two amino acids can join through the amino group of one and the carboxylic group of the other amino acid forming an anhydrous bond (CO-NH linkage) also known as a peptide bond by the loss of water molecule.

  • A protein is a heteropolymer and not a homopolymer.
  • Collagen is the most abundant protein in the animal world and RuBisCO (ribulose biphosphate. carboxylase oxygenase) is the most abundant protein in the whole biosphere.
  • N and C refer to the two termini of every protein. Single-letter codes and three-letter abbreviations of amino acids are also indicated.

NEET Biology Biomolecules Classification Of Amino Acids

Structure Of Proteins: The four levels of protein structure are

1. Primary Structure: The sequence of amino acids in the polypeptide chain gives the protein its primary structure.

  • The primary structure is important as it determines the specificity of a protein but does not make a protein functional.
  • To be functional, the protein must have a particular three-dimensional structure (conformation).
  • A functional protein contains one or more polypeptide chains.
  • The sequence of amino acids in the chain determines where the chain will bend or fold and where the various lengths will be attracted to each other.

NEET Biology Biomolecules Primary Struture Of A Potion Of A Hypothetical Protein

2. Secondary Structure: Through the formation of hydrogen bonds, peptide chains assume a secondary structure.

  • When a chain is arranged like a coil, it is called an a-Helix.
  • When two or more chains are joined together by intermolecular hydrogen bonds, the structure is called a pleated sheet.
  • Helical structure is found in the keratin of hair and pleated structure is found in silk fibers.
  • Each protein has a specific secondary structure also.
  • It generally takes the form of an extended spiral spring, the a-helix, whose structure is maintained by many hydrogen bonds that are formed between adjacent -CO and -NH groups. The H-atom of the NH group of one amino acid is bonded to the O-atom of the CO group of three amino acids away. A protein which is entirely helical is keratin.
  • The other type of secondary structure is called a β-pleated sheet. Here, two or more chains are joined together by intermolecular hydrogen bonds as in silk fibers.
  • A special secondary structure is observed in collagen or tropocollagen helix which has three strands or polypeptides coiled around one another. The coil is strengthened by the establishment of a hydrogen bond between the -NH group of the glycine residue of each strand with the -CO group of the other two strands. The locking effect is due to proline and hydroxyproline.

“biomolecules carbohydrates “

NEET Biology Biomolecules Secondary Structure Of Proteins

3. Tertiary Structure: Usually, the polypeptide chain bends and folds extensively and forms a compact “globular” shape to obtain functional conformation. This is termed as the tertiary structure.

  • In a large protein such as hemoglobin, or in the case of an enzyme, the molecule undergoes further folding and coiling to attain functional conformation.
  • The coils and folds of the protein molecule are so arranged as to hide non-polar amino acid side chains inside and expose the polar side chains.
  • The three-dimensional conformation of a protein brings distant amino acid side chains closer.
  • The active sites of proteins such as enzymes are thus formed.
  • The conformation of proteins is easily changed by pH, temperature, and chemical substances, and hence the function of proteins is liable and subject to regulation.

NEET Biology Biomolecules Various Types Of Bonds Interactions Found During The Coiling Of Polypeptide Chain

4. Quarternary Structure: Many complex proteins consist of an aggregation of polypeptide chains held together by hydrophobic interactions and hydrogen and ionic bonds.

  • Their precise arrangement constitutes the quaternary structure.
  • In aqueous media, proteins carry both cationic and anionic groups on the same molecule.
  • The ionic state of the protein depends on the pH of the medium.
  • A protein-rich in basic amino acids such as lysine and arginine exists as a cation and behaves as a base at the physiological pH of 7.4 (basic protein), for example, histones of nucleoproteins.
  • Similarly, a protein with acidic amino acids exists as an anion and behaves as an acid, for example, most blood proteins (acidic proteins).

NEET Biology Biomolecules Cartton Showing A Secondary Structure And A Teritary Structure Of Proteins

Types Of Proteins: On the basis of constitution, proteins are classified as simple, conjugated, and derived.

  1. Simple Proteins
    • Simple proteins me composed of amino acids only.
    • Some are small, globular molecules mostly soluble in water and not coagulated by heat (for example, histones).
    • As the size of the protein molecule increases, it becomes less soluble and its heat coagulability increases.
    • For example, larger globular proteins (such as egg albumin, scrum globulins, and glutelins of wheat or rice) are coagulated by heat.
    • Fibrous proteins have long molecules and are insoluble in water (for example, keratin of skin and hair, and collagen of connective tissues).
  2. Conjugated Proteins
    • Conjugated proteins are formed by binding of a simple protein with a non-protein called a prosthetic group, for example, nucleoproteins have nucleic acids as their prosthetic group.
    • The conjugated proteins are of the following types
    • Nucleoproteins: Prosthetic group is a nucleic acid, for example, protamines.
    • Metalloproteins: The prosthetic group is a metal, for example, hemoglobin.
    • Chromoproteins: Prosthetic group is a pigment, for example, cytochromes.
    • Phosphoprotcins: Prosthetic group is a phosphoric acid, for example. casein of milk.
    • Lipoproteins: Prosthetic group is lipids, for example, chylomicrons, HDL. LDL, etc.
    • Glycoproteins: Prosthetic group is carbohydrates, for example, mucins.
    • Glycoproteins (and glycolipids) play an important role in cell recognition.

NEET Biology Biomolecules Classification Of Proteins Based On Different Criteria

The specificity of this recognition depends upon the particular sequence of sugars in carbohydrate portions. Ribulose bisphosphate carboxylase (an enzyme) present in large amounts in chloroplast stroma is the world’s most common protein.

  • Storage proteins include albumin of egg and those that occur in seeds (glutelin of wheat). Prolamines are storage proteins.
  • Protamines are basic proteins associated with the DNA of chromosomes, they are rich in lysine and arginine.
  • Keratin and fibroin form protective structures.
  • Antibodies arc defense proteins.
  • Snake venom, ricin of castor, and bacterial toxins are proteinaceous in nature.
  • Actin and myosin are essential for muscle contraction.
  • Microtubules have tubulin protein.
  • Hemoglobin and myoglobin are transport proteins.

Ovalbumin and glutelin are storage proteins present in cereals. Ferretin is an iron-storing protein of animal tissues. The types of prolamines and glutelins found in wheat are gliadin and glutenin.

  • Insulin and parathormone are proteinaceous hormones.
  • Fibrinogen and thrombin are blood-clotting proteins.
  • Rhodopsin and iodopsin are photoreceptor pigments. These are present in rods and cones of the retina and are proteins.
  • Proteins having all essential amino acids are called first-class proteins.
  • Monellin, a protein, is the sweetest chemical obtained from an African berry.
  • Cheese is a denatured protein.
  • Resilin is a perfectly elastic protein found in the wings of some insects.

Some Proteins And Their Functions

NEET Biology Biomolecules Some Protein And Their Functions

NEET Biology Notes For Biomolecules Lipids

Lipids are made of carbon, hydrogen, and little oxygen. The number of oxygen atoms in a lipid molecule is always less as compared to the number of carbon atoms.

  • Sometimes small amounts of phosphorus, nitrogen, and sulfur are also present.
  • Lipids are insoluble in water, but soluble in non-polar solvents such as chloroform and benzene. Lipids contain fatty acids which may be saturated or unsaturated.
  • Fatty acids are organic acids with a hydrocarbon chain ending in a carboxyl group (COOH).

Fatty acids are called saturated if they do not have any double bonds between the carbons of the molecular chain, for example, palmitic acid (16 C) and stearic acid (18 C).

NEET Biology Biomolecules A Triglyceride Fat

CH3(CH2)14COOH Palmitic acid

CH3(CH2)16COOH Stearic acid

The general formula of saturated fatty acids is CnH2nO2.

  • Their melting point is high.
  • Unsaturated fatty acids have one or more double bonds between the carbons of the chain.
  • The 18-C unsaturated fatty acids oleic, linoleic, and linolenic acids have 1, 2, and 3 double bonds, respectively.

CH3(CH2)7CH = CH(CH2)7COOH Oleic acid

The general formula of unsaturated fatty acids is CnH2n-2xO2 (where,v = number of double bonds). The arachidonic fatty acid has four double bonds.

Differences Between Saturated Fally Acids And Unsaturated Fatty Acids

NEET Biology Biomolecules Differences Between Saturated fatty Acids And Unsaturated Fatty Acids

They have a bend at each double bond which keeps them in liquid form at ordinary temperature. They are called polyunsaturated fatty acids (PUFA) when they have more than one double bond in them.

  • They are also called drying oils because they have a tendency to solidify on exposure.
  • Oils of groundnut, mustard seed, sesame seed, and sunflower are rich in unsaturated fatty acids.
  • The unsaturated fatty acids have lower melting points than those of saturated fatty acids.
  • In lipids, fatty acids are usually in the form of esters.
  • Just as acids and bases react to form salts, similarly organic acids react with alcohol to form esters. Here alcohol is glycerol.
  • Plants can synthesize all fatty acids.
  • Animals cannot synthesize linoleic, linolenic, and arachidonic acid.
  • These are called essential fatty acids. Their deficiency causes sterility, kidney failure, and stunted growth. The classification of lipids is given in detail

NEET Biology Biomolecules Classification Of Lipids

1. Simple Lipids: Simple lipids are esters of fatty acids with alcohol. For example, fats. oils, and waves.

  • The simplest alcohol in fats is glycerol (trihydroxypropane).
  • Triglycerides are common in nature.
  • Fats are esters of fatty acids with glycerol (glycerine).
  • Each molecule of glycerol can react with three molecules of fatty acids.
  • Depending on the number of fatty acids that are attached to the glycerol molecule, the esters are called mono-, di-. or triglycerides.
  • Fats that are generally liquid at room temperature are called oils.
  • Oils are rich in unsaturated fatty acids and consequently have low melting points.
  • On hydrogenation, the unsaturated fatty acids become saturated and the oil becomes a solid fat (vanaspati and margarine).
  • Waxes are another class of simple lipids. They are formed by the combination of a long-chain fatty acid with a long-chain alcohol. Waxes play an important role in protection. They form water-insoluble coatings on the hair and skin of animals and stems, leaves, and fruits of plants.

Beeswax is formed from palmitic acid (C16H32O2) and medical alcohol (C30H61O2). Bee wax is also called as Hexacosyl palmitate. secreted by worker bees. Lanolin (wool fat), forms a waterproof coat around the animal fur.

  • Bacteria that cause tuberculosis and leprosy produce a wax (wax-D) that contributes to their pathogenicity.
  • Cutin is formed by the cross etherification and polymerization of hydroxy fatty acids and other fatty acids without esterification by alcohols other than glycerol. Cuticle has 50-90% cutin.
  • Suberin is the condensation product of glycerol and phellonic acid. It makes the cell wall impermeable to water.

2. Compound Lipids: These lipids contain an additional group along with fatty acids and alcohols, for example, phospholipids, glycolipids, lipoproteins, and chromolipids.

Phospholipids

  • These are straight-chain compounds of glycerol, fatty acids, and phosphoric acid.
  • In these, only two fatty acids are attached to the glycerol molecule, and the third hydroxyl group of glycerol is esterified to phosphoric acid instead of fatty acid.
  • Depending upon the type of phospholipid, this phosphate is also bound to a second alcohol molecule which can be choline, cthanolamine, inositol, or serine.
  • Common phospholipids are lecithin and cephalin.
  • Phospholipids are amphipathic molecules having hydrophilic (water-loving) polar regions and hydrophobic (water-repelling) non-polar regions.
  • They are the basic constituents of biomembranes.
  • Many phospholipids arrange themselves in a double-layered membrane in aqueous media (lipid bilayer).
  • Cephalin is found in the brain and acts as an insulation material for nerves and also participates in blood coagulation.
  • Lecithin takes part in cell permeability, osmotic tension, and surface conditioning of cells.

NEET Biology Biomolecules Struture Of A Phospholipid

  • The hydrocarbon chains of the fatty acids are the non-polar tails of the molecule.
  • The phosphate and nitrogenous/non-nitrogenous groups form the polar head group of the molecule.
  • Many phospholipid molecules may a nan go themselves in a double-layered membrane (lipid bilayer) in aqueous media. These have one or more simple sugars.

NEET Biology Biomolecules Molecular Structure Of A Phospholid Molecule

Glycolipids: They are lipids having sugar residues. Two common glycolipids are cerebrosides and gangliosides.

  • Glycolipids Composition: Glycolipids contain fatty acids, alcohol sphingosine, and sugar such as galactose, glucose, etc.
  • Glycolipids Function: The glycolipids are components of cell membranes, particularly in the myelin sheath of nerve fibers on outer surfaces of nerve cell, and in chloroplast membranes.

Lipoproteins: Lipoproteins contain lipids (mainly phospholipids) and proteins in their molecules.

Function: Membranes are composed of lipoproteins. Lipids are transported in the blood plasma and lymph as lipoproteins. Lipoproteins occur in milk and egg yolk.

Chromolipids: These contain pigments such as carot¬enoids, for example, carotene, and vitamin A.

3. Derived Lipids: These are isoprenoid structures, for example, steroids, terpenes, carotenoids, and prostaglandins.

  • Sterols
    • Sterols belong to a class of lipids that are not straight-chain compounds.
    • These are composed of fused hydrocarbon rings and a long hydrocarbon side chain. One of the examples is cholesterol.
    • Cholesterol is found in animals only.
    • It exists either free or as a cholesterol ester with a fatty acid.
    • Cholesterol is also the precursor of hormones such as progesterone, testosterone, estradiol, and cortisol.
    • Another steroid compound, diosgenin, produced by the yam plant (Oioscorea) is used in the manufacturing of antifertility pills.
  • Prostaglandins
    • It is a group of hormone-like unsaturated fatty acids that function as messenger substances between cells.
    • They are derived from arachidonic acid and related C20 fatty acids.
    • Prostaglandins occur in human seminal fluid, menstrual fluid, amniotic fluid, and a number of tissues.
    • They also circulate in blood.
    • They produce a variety of effects in different organs.
    • Prostaglandins regulate the production of acid in the stomach and stimulate the contraction of smooth muscles.
    • They are used to induce labor because they cause uterine contractions.
    • These can reduce the effects of asthma and gastric acidity. Analgesics such as aspirin inhibit prosta-glandin synthesis.
    • Cholesterol helps in the absorption of fatty acids and the formation of sex hormones, vitamin D, and bile salts. Potato is rich in cholesterol.
    • Terpenes are lipid-like carbohydrates formed of isoprene units (C5H8)n, for example, menthol, camphor, carotenoids, etc.

NEET Biology Notes For Biomolecules Nucleotides

Nucleotide is a group of small complex molecules forming a part of the information transfer system in cells. They are the basic units of nucleic acids. They also participate in energy transfer systems.

  • Nucleotides contain carbon, hydrogen, oxygen, nitrogen, and phosphorus.
  • Each nucleotide is made up of a cyclic nitrogenous base, a pentose, and one to three phosphate groups.
  • The nitrogenous bases occurring in nucleotides are either a purine or a pyrimidine.
  • Major purines are adenine and guanine. Thymine, uracil, and cytosine are pyrimidines.

NEET Biology Biomolecules Strutures Of Purines And Pyrimides

  • The sugar pentose is either ribose or deoxyribose.
  • The nucleotides are, thus, called ribonucleotides or deoxyribonucleotides.
  • Examples of ribonucleotides and deoxyribonucleotides are adenylic acid (AMP) and deoxyadenylic acid (dAMP), respectively.
  • Ribonucleotides arce the basic units of ribonucleic acids (RNA) and deoxyribonucleotides arc the basic units of deoxyribonucleic acids (DNA).
  • Nucleotides are mono-, dk, or triphosphate of nucleosides. For example, adenylic acid or adenosine monophosphate (AMP).
  • Adenosine diphosphate (ADP) and adenosine triphosphate (ATP) arc higher adenine nucleotides.
  • Nucleotides with more than one phosphate group arc are called higher nucleotides, for example, ATP and ADP. Likewise. other purines and pyrimidines can also form higher nucleotides.
  • Higher nucleotides of purines and pyrimidines occur in the free state, for example. ATP and ADP. Their third and second phosphate bonds can release about 8 kcal or more of tree energy per mole on hydrolysis. This far exceeds the energy released on the hydrolysis of most other covalent bonds. Therefore, these phosphate bonds of higher nucleotides are called high-energy bonds.
  • Nucleotides of the vitamins nicotinamide and riboflavin in occur either freely or in combination with specific proteins, thus working as coenzymes. They do not participate in the formation of nucleic acids. Instead, they act along with oxidizing enzymes and participate in oxidation reactions occurring in the cell.

Nicotinamide And Riboflavin Nucleotides

NEET Biology Biomolecules Nicotinamide And Riboflavin Nucleotides

Functions Of Nucleotides

  • Purine and pyrimidine nucleotides polymerize to form nucleic acids.
  • Higher purine and pyrimidine nucleotides, particularly ATP, store energy in their high-energy phosphate bonds.
  • They arc formed during photosynthesis and respiration.
  • Hydrolysis of the phosphate bonds of ATP releases their bond energy for driving energy-dependent reactions and processes.
  • Nicotinamide and riboflavin nucleotides act as coenzymes of oxidizing enzymes.

NEET Biology Biomolecules Molecular Structure Of Ribonucleic Acid

NEET Biology Notes For Biomolecules Nucleic Acids

First discovered by Meischer, nucleic acids are giant molecules having a variety of functions. There are two major types of nucleic acids deoxyribonucleic acid or DNA and ribonucleic acid or RNA.

  • DNA is found mainly in the nucleus but also occurs in chloroplasts and mitochondria.
  • It is the genetic material and contains all the information needed for the development and existence of an organism.
  • RNA occurs as genetic material in some viruses.
  • Nucleic acids are linear polymers of purine and pyrimidine nucleotides.
  • The nucleotides are linked serially by phosphate groups, each linking the C’5(5′ – C) and C’3(3′ – C) of the pentoses of the successive nucleotides.
  • A DNA molecule consists of a double chain of nucleo¬tides, whereas RNA consists of a single chain.
  • The nucleotides of DNA contain the bases adenine (A), thymine (T), guanine (G), and cytosine (C), while RNA contains A, G, C, and uracil (U) instead of T.
  • The backbone of the nucleic acid is uniformly made up of alternating pentose and phosphate groups.
  • The pentose in DNA is deoxyribose (C5H10O4) and that in RNA is ribose (C5H10O3).

Double Helix Structure Of DNA

  • In the double-stranded DNA, the bases of the opposite strands pair in a specific relationship by means of hydrogen bonds.
  • A always pairs with T and G always pairs with C. This complementarity is known as the base pairing rule.
  • Nucleic acids exhibit a wide variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson-Crick model.

NEET Biology Biomolecules A Polynucleotide Strand Of DNA And Watson Crick Model O DNA Double Helix

  • The double helix model says that DNA exists as a double helix.
  • The two strands of polynucleotides arc antiparallel i.e., run in the opposite direction.
  • The backbone is formed by the sugar-phosphate-sugar chain.
  • The nitrogen bases are projected more or less perpendicular to this backbone but face inside.
  • A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand.
  • There are two hydrogen bonds between A and T.
  • There are three hydrogen bonds between G and C.
  • Each strand appears like a helical staircase.
  • Each step of the ascent is represented by a pair of bases.
  • At each step of ascent, the strand turns 36°.
  • One full turn of the helical strand would involve 10 steps or 10 base pairs.
  • Attempt drawing a line diagram.
  • The pitch would be 34 Å.
  • The rise per base pair would be 3.4 Å.
  • This form of DNA with the above-mentioned salient features is called B-DNA.

In 1950, Erwin Chargaff found that in any DNA molecule:

  1. The amount of purines and pyrimidines is equal, i.e., A+ G = T + C.
  2. The amount of adenine is always equal to that of thymine; and the amount of guanine is always equal to that of cytosine (i.e., A = T and G = C).
  3. The base ratio (A + T)/(G + C) may vary from one species to another but is constant for a given species.
  4. The deoxyribose sugar and phosphate components occur in equal proportions.

Structure Of RNA

  1. RNA is usually single-stranded, but sometimes (as in reovirus and rice dwarf virus) it is double-stranded.
  2. RNA does not follow Chargaff’s rules, i.e., 1: a 1 ratio does not exist between purines and pyrimidines bases due to its single-stranded nature and lack of complementarity.

NEET Biology Biomolecules Secondary Struture Of DNA

Types Of RNA: There are three types of non-genetic RNA.

  1. Messenger RNA (mRNA): It is produced in the nucleus and carries the information for the synthesis of proteins; it was discovered by Jacob and Monod (1961).
  2. Ribosomal RNA (rRNA): It is the largest RNA and constitutes about 80% of the total cellular RNA. It is found in the ribosomes where protein synthesis takes place.
  3. Transfer RNA or soluble RNA or adaptive RNA (s-RNA, t-RNA): It is the smallest type of RNA and constitutes about 10-15% of the total cellular RNA. These are found in the cytoplasm and are of different types (as many types as those of amino, acids, usually 20). Their function is to collect amino acids from the cytoplasm for protein synthesis. t-RNA molecule is folded to form a clover leaf-like structure. This structure was given by Holley.

NEET Biology Biomolecules A Polynucleotide Strand Of RNA

NEET Biology Notes For Biomolecules Nucleic Acids Points To Remember

The study of X-ray diffraction patterns of DNAs isolated from various organisms by Wilkins, Franklin, and Astbury revealed that DNA has a right-handed helical structure.

  1. Using all the available chemical and physical information, James Watson and F.C. Crick of Cambridge gave the double helix model of DNA for which they were awarded the Nobel Prize in 1962.
  2. The width between the two backbones is constant and equal to the width of a base pair (i.e., the width of a purine + a pyrimidine).
  3. Along the axis of the molecule, the base pairs are spaced at intervals of 0.34 nm. Therefore, one complete turn of the double helix comprises 3.4 nm (10 base pairs).
  4. There is no restriction on the sequence of bases in one chain. However, due to the rule of base pairing, the sequence of one chain determines the sequence in the other. The two chains are thus said to be complementary.
  5. As a result, the (purine) adenine in either chain is associated with (pyrimidine) thymine in the other. Similarly, the (purine) guanine in either chain is associated with the (pyrimidine) cytosine in the other.
  6. The two chains are held together by hydrogen bonding between the bases (joined together in pairs), a single base from one chain being hydrogen-bonded to the complementary base from the complementary chain.
  7. The adenine-thymine pair has two hydrogen bonds and the guanine-cytosine pair has three hydrogen bonds.
  8. The double helix has a diameter of 20 Å, i.e., the distance between two strands is 19.8 Å (or 20 Å).
  9. DNA with a higher percentage of G = C has more density than those with a higher percentage of A = T.
  10. Upon heating at temperatures above SO-90 C, the two strands of DNA uncoil and separate (denaturation). On cooling, the strands come closer and are rejoined together (denaturation/annealing). The low melting area of DNA is A = T base pairs.
  11. 1 μm of DNA contains about 3000 base pairs.

DNA is mostly right-handed. This type of DNA exists in four forms:

  1. B form: The usual DNA, having 10 base pairs per turn.
  2. A form: Having 11 base pairs (instead of 10 base pairs per turn), the base pairs are not perpendicular to the axis, but are tilted.
  3. C form: Like B form, but have nine base pairs per turn.
  4. D form: Like B form, but have eight base pairs per turn.

DNA with left-handed coiling is called Z-DNA. In this, the repeating unit is dinucleotide. In some cases, as in ∅ x 174 and S-13 viruses, the DNA is single-stranded.

NEET Biology Biomolecules Representation Of Small Molecular Weight Organic Compounds In Living Tissuess

Palindromic And Repetitive DNA: DNA duplex possessing areas of the same sequences of nucleotides is called palindromic DNA. Repetitive DNA has a sequence of nitrogen bases repeated several times in tandem.

NEET Biology Notes For Biomolecules Enzymes

Enzymes are proteinaceous, biocatalysts. The first enzyme was discovered by Buchner. The term enzyme was given by Kuhne.

  • Zymase (from yeast) was the first discovered enzyme (Buchner)
  • The first purified and crystalized enzyme was urease (by J.B. Sumner) from the Canavaliat/jack bean (labia plant).
  • The proteinaceous nature of the enzyme was established by Northrop and Sumner.
  • Enzymes are biocatalysts made up of proteins (except ribozyme), which increase the rate of biochemical reactions by lowering the activation energy.
  • The first discovered ribozyme was L19 RNAase by T. Cech from rRNA of a protozoan Tetrahymena thermophila and RNAase P or Ribonuclease P by Altman in a prokaryotic cell (Nobel prize).

General Properties Of Enzymes

  • Large-sized biomolecules, colloidal nature with high molecular weight.
  • The large size (equal to colloid particles) provides more surface area and, hence possesses a large number of active sites. A large number of substrates get converted into products by one molecule of enzyme at a time.
  • The highest molecular weight is of the enzyme pyruvate dehydrogenase complex (46 lakh), which participates in the link reaction of respiration.

Proteinous Nature Of Enzymes

  • The monomer unit of an enzyme is an amino acid.
  • Amino acids are linked together to form a polypeptide chain.
  • Most of the enzymes are arranged in the tertiary structure of protein or globular proteins except isoenzyme (quaternary structure).
  • The tertiary structure of proteins provides stability and water-soluble nature to enzymes.
  • The synthesis of enzymes occurs on ribosomes under the control of genes.

Enzymes Specificity

  • Enzymes are specific in the case of pH, temperature, and substrate.
  • The common pH range for enzyme activity is 6-8.
  • Every enzyme works on a specific pH, pepsin: 2.5 pH, hydrolase: 4-5.
  • RuBisCO, per case: 8.5 pH, trypsin: 8.5 pH.

NEET Biology Biomolecules Effect Of pH The Rate Of Reaction

Enzymes Temperature

  • The common range of temperature for enzyme activity is 20-40°C.
  • Enzymes work on the body temperature of an organism not on environmental temperature.
  • Plant enzymes are affected by environmental temperature change as plants do not show homeostasis.
  • At low temperatures, enzymes become functionally inactive and at high temperatures, denatured.

Enzymes Substrate

  • Every enzyme works on a specific substrate.
  • The substrate binds at the active site of the enzyme which is made up of a specific sequence of amino acids and recognizes its substrate.
  • An example is succinic dehydrogenase acts on succinic acid while pyruvate dehydrogenase acts on pyruvic acid.
  • Enzymes increase the rate of reaction by decreasing activation energy.
  • Activation Energy: Minimum amount of energy required more than the free energy of reactants to reach the transition state of a chemical reaction or to undergo the chemical reaction.

NEET Biology Biomolecules Activation Energy

Enzymes Turn Over Number (TON): The number of reactant molecules converted into products by one molecule of enzyme in a unit of time. The highest TON is of carbonic anhydrase (360 lakh/minute)

NEET Biology Biomolecules Turn Over Number

Enzymes KM Constant

  • Enzymes follow the Michaelis-Menten reaction kinetics.
  • It represents the substrate concentration at which the rate of enzymatic reaction becomes half of the maximum velocity or rate.
  • If an enzyme passes a high KM constant then its affinity towards the substrate is low and the rate of reaction is also low.
  • The energy required for a chemical reaction to proceed is called activation energy.
  • Enzymes lower the activation energy. (Remember that enzymes cannot start the chemical reaction.)

NEET Biology Biomolecules Substrate Concentration

  • With the increase in concentration of substrate, the enzymatic velocity also increases.
  • At a certain value, all the active sites of enzyme molecules are saturated, and an increase in the substrate concentration does not increase the velocity of the enzymatic reaction. (The concentration of substrate at which the velocity of enzymatic action reaches half of its maximum value is called KM value or Michaelis constant).

The higher is the affinity of an enzyme for a substrate, the lower is its KM value, i.e., KM value ∝ 1/Affinity

Ki Constant (Enzyme Inhibitor Complex Dissociation Constant): The substrate concentration at which the enzyme inhibitor complex dissociates and the reaction becomes normal. It is applicable only for competitive reversible inhibitions.

Structure Of Enzyme

  • Simple Enzymes: They are made up of proteins only, for example, pepsin, trypsin, etc.
  • Conjugated Enzymes: They are made up of protein as well as non-protein parts.
    • Co-enzymes: Co-enzymes are non-protein, organic groups, which are loosely attached to apoenzymes. They are generally made up of vitamins.
    • Prosthetic Group: When a non-protein part is tightly or firmly attached to apoenzymes, a prosthetic group is formed.
    • Metal Activators/Co-Factors/Metallic Factor: Loosely attached inorganic co-factor, for example, Mn, Fe, Co, Zn, Ca, Mg, Cu, etc

NEET Biology Biomolecules Concept Map Enzymes

Enzyme Active Site: The part of the polypeptide chain made up of a specific sequence of amino acids at which a specific substrate is to be bound and catalyzed is known as an active site. The specific sequence of amino acids at the active site is determined by genetic codes.

Enzyme Allosteric Site: Besides the active site, some enzymes possess additional sites, at which chemical other than the substrate (allosteric modulators) are bound. These sites are known as allosteric sites and enzymes with allosteric sites are called as allosteric enzymes, for example, hexokinase, and phosphofructokinase.

NEET Biology Notes For Biomolecules Enzymes Points To Remember

  • Endoenzymes: Enzymes that are functional only inside the cells. For example, enzymes of metabolism.
  • Exoenzymes: Enzymes catalyze the reactions outside the cell. For example, enzymes of digestion, some enzymes of insectivorous plants, zymase complex of fermentation, etc.
  • Proenzyme/zymogen: These are precursors of enzymes or inactive forms of enzymes. For example, pepsinogen, trypsinogen, etc.
  • Isoenzymes: Enzymes having Similar action, but little difference in their molecular configuration are called isoenzymes. Sixteen forms of α-amylase of wheat, five forms of LDH (lactate dehydrogenase), and three forms of per case are known. These isoenzyme forms are synthesized by different genes and are tissue- and organ-specific.
  • Inducible Enzymes: When the formation of the enzyme is induced by substrate availability. For example, lactase, nitrogenase, β-galactosidase, etc.
  • Extremozymes: Enzymes that may also function at extremely adverse conditions (very high temperature), for example, Taq polymerase.
  • Abzynics: When monoclonal antibodies (Mab) are used as enzymes.
  • Biodetergents: Enzymes used in washing powders are known as detergents, for example, amylase, lipase, and proteolytic enzymes.

House-keeping/constitutive Enzymes: These are always present in constant amounts and are also essential to cells. For example, enzymes of cell respiration.

NEET Biology Biomolecules Metallo Enzymes

Classification Of Enzymes

  • Enzyme names such as ptyalin (salivary amylase), pepsin, and trypsin give no indication of their action.
  • Other enzymes such as amylase, sucrase, protease, and lipase were named after the substrates on which they act: amylose (starch), sucrose, protein, and lipids, respectively.
  • Still others were named according to the source from which they were obtained papain from papaya, bromelain from pineapple (belongs to the family Bromeliaceae).
  • Some like DNA polymerase indicate its specific action, polymerization.
  • Duclaux (1883) provided a system for naming enzymes by adding the suffixase at the end of the enzyme name.
  • In this system, each enzyme ends with an -ase and consists of two parts, the first part indicates its substrate and the second the reaction catalyzed.
  • For example, glutamate pyruvate transaminase transfers an amino group from the substrate glutamate to another substrate pyruvate. However, arbitrary names such as ptyalin and trypsin still continue to be used because of their familiarity.

Enzymes Are Grouped Into Six Major Classes:

  1. Class 1 Oxidoreductases: These catalyze oxidation or reduction of their substrates and act by removing or adding electrons (and/or H+) from or to substrates, for example, cytochrome oxidase oxidizes cytochrome.
  2. Class 2 Transferases: These transfer specific groups from one substrate to another. The chemical group transferred in the process is not in a free state, for example, glutamate pyruvate transaminase.
  3. Class 3 Hydrolases: These break down large molecules into smaller ones by the introduction of water (hydrolysis) and the breaking of specific covalent bonds. Most digestive enzymes belong to this category, for example, amylase which hydrolyzes starch, and lipases.
  4. Class 4 Lyases: These catalyze the cleavage of specific covalent bonds and removal of groups without hydrolysis, for example, histidine decarboxylase cleaves the C-C bond in histidine to form carbon dioxide and histamine.
  5. Class 5 Isomerases: These catalyze the rearrangement of molecular structure to form isomers, for example, phosphohexose isomerase changes glucose-6-phosphate to fructose-6-phosphate (both are hexose phosphates).
  6. Class 6 Ligases: These catalyze the covalent bonding of two substrates to form a large molecule. The energy for the reaction is derived from the hydrolysis of ATP. Pyruvate carboxylase combines pyruvate and carbon dioxide to form oxaloacetate at the expense of ATP.

Factors Affecting Enzyme Activity: The activity of an enzyme can be affected by a change in the conditions which can alter the tertiary structure of the protein. These include temperature, pH, change in substrate concentration, or binding of specific chemicals that regulate enzyme activity.

  • Enzymes generally function in a narrow range of temperatures and pH.
  • Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH.
  • Activity declines both below and above the optimum value.
  • Low temperature preserves the enzyme in a temporarily inactive state, whereas high temperature destroys enzymatic activity because proteins are denatured by heat.

NEET Biology Biomolecules Effect Of Change In pH And Temperature And Concentrartion Of Substrate On Enzyme Activity

Enzyme Optimum Temperature

  • Enzymes generally work over a narrow range of temperatures.
  • Usually, it corresponds to the body temperature of the organism. For instance, human enzymes work at the normal body temperature.
  • Each enzyme shows its highest activity at a particular temperature called the optimum temperature
  • Activity declines both above and below the optimum temperature.
  • Every enzyme has a specific optimum temperature.
  • According to the general rule of thumb, Q10 (temperature coefficient) for enzymes is 2-3, i.e., in between minimum and optimum temperature (5-40°C), the rate of reaction increases 2-3 times with rise in 10°C temperature.

If the temperature is reduced to near or below freezing point, the enzymes are inactivated (not denatured).

NEET Biology Biomolecules Effect Of Temperature On The Velocity Of Enzyme Action

  • Most enzymes show maximum activity in a temperature range of 25-40°C.
  • Enzymes are thermolabile, i.e., they get denatured at high temperatures.
  • The loss of catalytic properties begins at 35°C and is almost complete around 60°C. However, dried enzyme extracts can endure temperatures of 100-120°C or even higher. That is why dry seeds can endure higher temperatures than germinating seeds.
  • Thermal stability is thus an important quality of some enzymes isolated from thermophilic organisms.

Enzyme Optimum pH

  • Each enzyme shows its highest activity at a specific pH. This is called optimum pH.
  • Activity declines both above and below the optimum pH.
  • Most intracellular enzymes function best around neutral pH.
  • Some digestive enzymes have their optimum pH in the acidic or alkaline range. For example, the protein-digesting enzyme pepsin found in the stomach has an optimum pH of 2.0.
  • Another protein-digesting enzyme, trypsin, found in the duodenum functions best at an alkaline pH of 8.0.

Enzyme Concentration Of Substrate: With an increase in substrate concentration, the velocity of the enzymatic reaction rises at first.

  • The reaction ultimately reaches a maximum velocity (Vmax) which is not exceeded by any further rise in the concentration of substrate.
  • This is because the enzyme molecules are fewer than the substrate molecules and after the saturation of these molecules, there are no free enzyme molecules to bind with the additional substrate molecules.

How Enzymes Speed Up Reactions: A certain amount of energy is necessary to initiate any chemical reaction. This is called activation energy or free energy of activation.

  • In a population of molecules of each substrate, the majority have average kinetic energy, some have higher and some have lower than the average energy.
  • Under normal temperature, only the molecules having relatively high energy are likely to react to form the product. Therefore, the reaction takes place very slowly.
  • One way to make the reaction go faster is to raise the temperature of the mixture.
  • Heat increases the kinetic energy of the molecules, causing their collisions and reactions.
  • The other method of quickening the reaction is by adding an enzyme.
  • The enzyme lowers the activation energy of the reaction and allows a large number of molecules to react at a time.
  • Exactly how the enzymes lower the activation energy is not clear. However, it is known that the enzymes combine with the substrate molecules and bring them close together which favors their collisions in the most suitable directions and locations for the reaction to occur.
  • The inorganic catalysts work in the same manner.
  • It is now held that conformational changes in the active sites of the enzymes actually “push” the substrate molecules toward an interaction.
  • The hydrolysis of starch into glucose is an organic chemical reaction.
  • The rate of a physical or chemical process refers to the amount of product formed per unit time. The rate can also be called velocity if the direction is specified for a given reaction.
  • Reactants absorb energy from surroundings to climb the hill of activation energy (EA) and reach the unstable, short-lived, transition state Enzyme speeds up the reaction by reducing the uphill climb to the transition state. In the transition state, the reactants are in an unstable condition and a reaction can occur.

Mechanism Of Enzyme Action: Two hypotheses have been put forward to explain the mode of enzyme action.

1. Lock And Key Hypothesis

  • This hypothesis was given by Emil Fischer (1894).
  • According to this hypothesis, both enzyme and substrate molecules have specific geometrical shapes.
  • It is similar to the system of lock and key, which have special geometrical shapes in the region of their activity.
  • The active sites contain special groups having -NH2, -COOH, -SH, etc., for establishing contact with the substrate molecules.

NEET Biology Biomolecules Activation Energy Requirements Of Uncatalyzed And Enzyme Catalyzed Reactions

  • Just as a lock can be opened by its specific key, a substrate molecule can be acted upon by a particular enzyme. This also explains the specificity of enzyme action.
  • After coming in contact with the active site of the enzyme, the substrate molecules or reactants form a complex called enzyme-substrate complex.

NEET Biology Biomolecules Lock And Key Hypothesis To Show The Specificity Of Enzymes

Enzyme + Substrate ⇔ Enzyme -Substrate Complex

Enzyme-Substrate Complex ⇔ Enzyme + End Products

  • In the enzyme-substrate complex, the molecules of the substrate undergo chemical change and form products.
  • The product no longer fits into the active site and escapes in the surrounding medium, leaving the active site free to receive more substrate molecules. This theory explains how a small concentration of an enzyme can act upon a large amount of the substrate. It also explains how the enzyme remains unaffected at the end of a chemical reaction.
  • The lock and key theory explains how a substance having a structure similar to the substrate can work as a competitive inhibitor.

2. Induced Fit Hypothesis

  • This hypothesis was proposed by Koshland (1960).
  • According to this hypothesis, the active site of the enzyme does not initially exist in a shape that is complementary to the substrate but is induced to assume the complementary shape as the substrate becomes bound to the enzyme.
  • According to Koshland, “the active site is induced to assume a complementary shape in much the same way as a hand induces a change in the shape of a glove.”
  • An active site of an enzyme is a crevice or a pocket into which the substrate fits. Thus, enzymes through their active site catalyze reactions at a high rate. Hence, according to this model, the enzyme (or its active site) is flexible.

NEET Biology Biomolecules Induced Fit Theory Of Enzyme Action

The active site of the enzyme contains two groups:

  1. The buttressing group is meant to support the substrate.
  2. The catalytic group is meant to catalyze the reaction.

When a substrate comes in contact with the buttressing group, the active site changes to bring the catalytic group opposite the substrate bonds to be broken.

Isoenzymes: Multiple molecular forms of an enzyme (synthesized by different genes) occurring in the same organism and having a similar substrate activity are called iso-enzymes. Over 100 enzymes are known to have iso-enzymes such as

  1. α-amylase of wheat endosperm has 16-iso-enzymes.
  2. Lactic acid dehydrogenase has five iso-enzymes.
  3. Alcohol dehydrogenase has four iso-enzymes.

Site of Enzyme Action: All enzymes are produced in the living cells. About 2000 enzymes have been recorded. These are of two types with regard to the site where they act: intracellular and extracellular.

  1. Intracellular Enzymes
    • Most of the enzymes remain and function inside the cells. They are called intracellular enzymes or endoenzyines.
    • Some are dissolved in the cytoplasmic matrix.
    • The water extract of ground-up liver cells contains all the 11 enzymes necessary to change glucose to lactic acid.
    • Certain enzymes are bound to particles such as ribosomes, mitochondria, and chloroplast.
    • The respiratory enzymes needed to convert lactic acid to carbon dioxide and water are found in mitochondria.
  2. Extracellular Enzymes
    • Certain enzymes leave the cells and function outside them.
    • They are called extracellular enzymes or exoenzymes.
    • They mainly include the digestive enzymes, for example, salivary amylase, gastric pepsin, and pancreatic lipase; secreted by the cells of the salivary glands, gastric glands, and pancreas, respectively.
    • Lysozyme present in tears and nasal secretion is also an exo-enzyme.
    • Rennet tablets, containing the enzyme rennin from the calf’s stomach, are used to coagulate milk protein carcinogen for cheese (casein) formation.

Inhibition of Enzyme Activity

  • Any substance that can diminish the velocity of an enzyme-catalyzed reaction is called an inhibitor.
  • Reversible inhibitors bind to enzymes through noncovalent bonds.
  • Dilution of the enzyme-inhibitor complex results in the dissociation of the reversibly-bound inhibitor and recovery of enzyme activity.
  • Irreversible inhibition occurs when an inhibited enzyme does not regain activity upon dilution of the en¬zyme-inhibitor complex.
  • Some irreversible inhibitors act by forming covalent bonds with specific groups of enzymes; for example, the neurotoxic effects of certain insecticides are due to their irreversible binding at the catalytic site of the enzyme acetylcholinesterase.
  • The two most commonly encountered types of inhibition are competitive and non-competitive.

NEET Biology Biomolecules Types Of Enzyme Inhibition

Competitive Inhibition: This type of inhibition occurs when the inhibitor binds reversibly to the same site that the substrate would normally occupy and, therefore, competes with the substrate for that site.

  • Effect On Vmax: The effect of a competitive inhibitor is reversed by increasing [S]. At a sufficiently high substrate concentration, the reaction velocity reaches the Vmax observed in the absence of an inhibitor.
  • Effect On Km: A competitive inhibitor increases the apparent for a given substrate. This means that in the presence of a competitive inhibitor, more substrate is needed to achieve (1/2) Vmax, for example, sulpha drugs for folic acid synthesis in bacteria and inhibition of succinic dehydrogenase by malonate.

NEET Biology Biomolecules Effect Of A Competitive Inhibitor On the Reaction Velocity Versus Substrate Plot

Non-competitive inhibition: This type of inhibition is recognized by its characteristic effect on Vmax. Non-competitive inhibition occurs when the inhibitor and substrate bind at different sites on the enzyme. The non-competitive inhibitor can bind cither free enzyme or the ES complex, thereby preventing the reaction from occurring.

NEET Biology Biomolecules A Non Competitive Inhibitor Binding To Free Enzyme Substrate Complex

  • Effect On Vmax: Noncompetitive inhibition cannot be overcome by increasing the concentration of substrate. Thus, non-competitive inhibitors decrease Vmax of the reaction.
  • Effect On Km: Non-competitive inhibitors do not interfere with the binding of substrate to enzyme. Thus, the enzyme shows the same Km in the presence or absence of the non-competitive inhibitor. For example, cyanide kills an animal by inhibiting cytochrome oxidase.

 

NEET Biology Notes For Biomolecules Assertion Reasoning Questions And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: Heparin is a natural anticoagulant inside the blood vessels.

Reason: It is an example of homopolysaccharide.

Answer: 3. If Assertion is true, but Reason is false.

Question 2. Assertion: Hemoglobin is a monomeric protein.

Reason: It is made up of two polypeptide chains.

Answer: 4. If both Assertion and Reason are false.

Question 3. Assertion: Saturated fatty acids are non-essential fatty acids.

Reason: They can be synthesized in an animal body.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 4. Assertion: Lipids provide more energy as compared to carbohydrates on oxidation.

Reason: Lipid is the first respiratory substance.

Answer: 3. If Assertion is true, but Reason is false.

Question 5. Assertion: In protoplasm, protein functions as a buffer.

Reason: The protein molecule is amphoteric.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 6. Assertion: Phospholipids form a bimolecular layer in an aqueous medium.

Reason: Phospholipid molecules are amphipathic.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 7. Assertion: Starch is the storage polysaccharide in plants.

Reason: Starch is a polymer of p-glucose.

Answer: 3. If Assertion is true, but Reason is false.

Question 8. Assertion: Lecithin is important in membranes.

Reason: It has an amphipathic nature.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 9. Assertion: Glucose is dextrose.

Reason: The open chains of glucose have four asymmetrical carbons.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 10. Assertion: Histones are acidic proteins.

Reason: These join all nucleic acids.

Answer: 4. If both Assertion and Reason are false.

Question 11. Assertion: Disaccharides show optical activity.

Reason: Cellobiose is an example of a disaccharide.

Answer: 4. If both Assertion and Reason are false.

Question 12. Assertion: Isabgol is used as a medicine.

Reason: The husk of isabgol contains mucilage.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion

Question 13. Assertion: Monellin is the sweetest chemical.

Reason: Monellin is a protein.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 14. Assertion: Natural silk is made up of protein.

Reason: Artificial silk is a polysaccharide.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 15. Assertion: Specific substrate binds at the active site of the enzyme.

Reason: Enzymes increase the activation energy of the substrate.

Answer: 3. If Assertion is true, but Reason is false.

Question 16. Assertion: Enzymes become denatured at high temperatures.

Reason: The tertiary structure of proteins gets damaged at high temperatures.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 17. Assertion: All enzymes are proteins.

Reason: Ribozymes are enzymes without proteins.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 18. Assertion: Allosteric modulators accelerate or retard the rate of catalysis of an allosteric enzyme.

Reason: Allosteric modulators modulate the configura¬tion of the active site of enzyme.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

NEET Biology Notes – Cell The Unit Of Life

by

NEET Biology Notes Cell The Unit Of Life Introduction

Cell Structure And Functions Introduction: The cell is the fundamental unit of structure and function of all life. It is a unit of biological activity delimited by a differentially permeable membrane and is capable of self-reproduction.

NEET Biology Notes Cell The Unit Of Life What Is Cell

The cell is the structural and functional unit of a living organism. The widely accepted definition of the cell was given by Lowey and Sikewitz.

  • Cell is the unit of biological activity container and nucleus and is able to divide in a medium free from other living organisms.
  • Cytology is the study of the state of different components of a cell.
  • Cell biology is the study of the structure, function, and reproduction of cells.
  • The cell was first observed by Marcello Malpighi and he called it saccade.

NEET Biology Notes Cell The Unit Of Life Historical Details

The father of cytology is Robert Hooke and the father of cell biology is Swanson. The term “cell” was first coined by Robert Hooke in his book Micrographia. Robert Hook observed cell wall in the dead cork cell. The first living cell was observed by Leeuwenhoek.

NEET Biology Cell The Unit Of Life Micrographia

  • Alfonso Corti first observed the living substance of a cell.
  • This living substance was called protoplasm by Purkinje in animal cell and by Von Mohl in plant cell.
  • Hammerling was the first to call the nucleus as the brain or master or controlling center of the cell by using a grafting experiment with Acetabiilarici (unicellular marine green algae) which is the largest unicellular organism among the plants.
  • The largest cell in animals is the egg of an ostrich.
  • The largest cellular component is the nucleus.
  • The smallest cell organelle is ribosomes.
  • The term cytoplasm was coined by Strassburgcr.
  • The term hyaloplasm was coined by Preffer.

” cell unit of life”

NEET Biology Cell The Unit Of Life Historical Views

NEET Biology Notes Cell The Unit Of Life Cell Theory

Cell theory was proposed by Schleiden (botanist) and Schwan (zoologist) in 1839. According to the cell theory, All living organisms are composed of cells.

  • The cell is the unit of structure and function of life form.
  • All cells are basically similar in chemical composition and metabolism.
  • A cell is bound by a membrane and contains a nucleus.
  • The activity of a multicellular organism is the result of the interaction and activities of its component cells.

Read and Learn More NEET Biology Notes

Cell Lineage Theory or Common Ancestry Theory

  • Proposed by Rudolf Virchow.
  • Omnis cellula e cellulcr. All cells originate from pre-existing cells.
  • Virchow and Nageli suggested the importance of the reproduction of cells and suggested that it is responsible for the continuity of life.

Exception of Cell Theory: Viruses/viroids do not follow cell theory but mycoplasma, bacteria, Vaucheria, Mucor, and Rhizopus also do not follow cell theory in certain points.

NEET Biology Notes Cell The Unit Of Life Overview Of Cell

You have earlier observed cells in an onion peel and/or human cheek cells under the microscope. Let us recollect their structure. The onion cell which is a typical plant cell has a distinct cell wall as its outer boundary and just within it is the cell membrane.

  • The cells of the human cheek have an outer membrane as the delimiting structure of the cell. Inside each cell is a dense membrane-bound structure called the nucleus. This nucleus contains chromosomes which in turn contain the genetic material, DNA.
  • Cells that have membrane-bound nuclei are called eukaryotic whereas cells that lack a membrane-bound nucleus are called prokaryotic. In both prokaryotic and eukaryotic cells, a semi-fluid matrix called cytoplasm occupies the volume of the cell.
  • The cytoplasm is the main arena of cellular activities in both plant and animal cells. Various chemical reactions occur in it to keep the cell in the “living state.”

NEET Biology Notes Cell The Unit Of Life Cell Shape And Size

Depicts various forms of cell shapes and sizes.

NEET Biology Cell The Unit Of Life Cell Shape And Size

Type Of Cells

  • Based On Differentiation
    • Undifferentiated Cell: Cells having dividing abilities, i.e., meristem in plants and stem cells in animals.
    • Differentiated Cell: Post-mitotic cell formed by division of undifferentiated cells. For example, parenchyma, collenchyma, tracheids, vessels, etc.
    • Dedifferentiated Cell: If a differentiated cell reverts back into an undifferentiated cell, it is known as a dedifferentiated cell. For example, the formation of cork cambium and cambium in dicot root.
    • Re-differentiated Cell: A cell formed by the division of a differentiated cell, for example, the formation of a secondary cortex and cork, is called a re-differentiated cell.
  • Based On Structure
    • Prokaryotic Cell (Daugherty): Nucleus not well organized and membrane-bound cell organelles absent. For example, monerans.
    • Eukaryotic Cell (Daugherty): Nucleus well organized and membrane-bound cell organelles present.
    • Mesokaryotic Cell (Dodge): The Nucleus well organized, but DNA lacks histone protein.

NEET Biology Notes Cell The Unit Of Life Prokaryotic Cell (Eubacteria)

Historical Detail: Antony Von Leeuwenhoek discovered bacteria from stored rain water and teeth scum and called them wild animalcules. He is known as the discoverer of the microbial world or the wonder world of microbes. He used the term “Dicrkens.”

” cell biology notes”

  • Ehrenberg gave the term “bacteria.”
  • Nageli placed bacteria in Schizomycetes, called “fission fungi.”
  • Louis Pasteur proposed the germ theory of disease. He discovered that bacteria is the causing agent of chicken cholera.
  • Pasteur invented antirabies vaccine. He gave the term “microorganism.” He is known to be the father of modem microbiology and sterilization techniques.
  • Robert Koch gave “Koch’s Postulates.”
  • Joseph Lister developed the technique of aseptic culture.
  • D. A. Bergey gave a classification of bacteria in the Manual of Determinative Bacteriology.
  • Se’dillot used the term “microbe” for animalcules.

NEET Biology Cell The Unit Of Life A Prokaryotic Cell

Habitat: Eubacteria are cosmopolitan in distribution. They are present in water, soil, air, and in plant and animal bodies.

Habitat Size: Bacteria range from 0.1-1.5 μm in diameter and 2-10 μm in length. The smallest bacteria is rod-shaped Dialister pneumosintes (0.15-0.3 μm long) present in the nasopharynx of man during the early stage of influenza.

NEET Biology Cell The Unit Of Life Different Sizes of Eubacteria

Spirillum Laidlaw, Epulopsciumfishelsoni (600 μm x 80 μm), and Thiomargarita ramibiensis (750 μm) are among the largest unicellular bacteria. The filamentous bacterium Beggiatoa mirabilis is the largest bacterium (16-15 μm diameter and up to several centimeters long).

Habitat Shape: Cohn classified bacteria into four types based on their shapes

  1. Coccus (pl. cocci): Spherical or nearly sphericaL small, and always non-flagellated.
    • Micrococcus: Occurs singly, for example, Micrococcus lutens. M. roseus.
    • Diplococci: Found in pairs, for example, Diplococcus pneumonia.
    • Streptococci: Cells remain attached to form a chain. For example, Streptococcus lactis, Leptotricha buccalis.
    • Staphylococci: Irregular bundles of cells or grape-like clusters, for example, Staphylococcus aureus.
    • Sarcina: Three-dimensional geometrical figures like cubes, for example, Sarcina.
  2. Bacillus (pl. bacilli): Rod-shaped or cigarette-like with rounded or blunt ends. Most common shape. Motile or non-motile. They may occur as:
    • Monobacillus: Single bacillus
    • Diplobacillus: Occurs in a group of two bacilli
    • Streptobacilli: Found in a chain, for example, Streptobacillus. When the cells form a chain and have a much larger area of contact with each other, these are said to have formed trichomes.
      • For example, Beggiatoa. If the cells are lined side by side like matchsticks and at angles to one another, the arrangement is said to be palisade-like.
      • For example, Corynebacterium diphtheria. In many bacteria (for example, Streptomyces), cells are arranged to form unicellular long, branched filaments called hyphae.
  3. Vibrio (Singular Vibrion): Bacteria with less than one complete twist or turn. These resemble a comma (,) in appearance, for example, Vibrio cholerae.
  4. Spirilla (Singular Spirillum): Coiled forms of bacteria exhibiting twists with one or more turns giving a spiral appearance, for example, Spirillum minus.

Other Uncommon Shapes

  • Stalked Bacterium: For example, Caulobacter.
  • Budding Bacterium: For example, Rhodomicrobium.
  • Pleomorphic: Occurs in more than one form, for example, Rhizobium, Coiynebacteriurm, Azotobacter, and Mycobacterium.

Flagella Structure: Instead of the “9 + 2” arrangement of tubulin-containing microtubules, there is simply a single filament of a globular protein called flagellin.

NEET Biology Cell The Unit Of Life Structure of Flagellum

Parts Of A Flagellum

Flagellum Basal Body: It is the most complex portion of the flagellum and has four rings (L, P, S, and M) in gram-negative bacteria and two rings (S and M) in gram-positive bacteria.

Flagellum Hook: It is made up of different protein subunits.

Flagellum Filament: The longest and the most obvious portion of the flagellum. The protein molecules are arranged in a spiral manner. The filament is 20 nm wide and 1-70 nm long and consists of eight vertical rows of flagellin.

Flagellar Arrangement

  • Atrichous: Flagella absent, for example, Lactobacillus, and Pasteurella.
  • Monotrichous: Bacteria with single flagella, for example, Vibrio, Thiobacillus.
  • Cephalotrichous: Bacteria with many flagella attached at one end, for example, Pseudomonasfluoresce.
  • Lophotrichous: Bacteria with many flagella attached at both ends, for example, Spirillum volitions.
  • Amphitrichous: Bacteria with a single flagellum at each end, for example, Nitrosomonas
  • Peritrichous: Bacteria with flagella all over the body, for example. E, coil, Clostridium tetani.

“cell the unit of life structure and function “

NEET Biology Cell The Unit Of Life Flagellar Arrangement

Pilus And Fimbriae: Pilus is an elongated, tubular, sharing structure for conjugation between two bacteria. They are. made up of pilin protein and are smaller than flagella. Fimbriae are small brush-hair-like fibers that are supporting structures for attachment of the bacterial cell

Gram Staining Technique: The gram staining technique was introduced by Christian Gram in 1884. The stain works on the cell wall of bacteria. depicts the procedure for Gram staining.

NEET Biology Cell The Unit Of Life Procedure For Gram Staining

Gram-negative bacteria become colorless on treatment with a destaining agent due to a thin cell wall and more lipids in the cell wall. The table shows the basic difference between the structures of Gram-positive and Gram-negative bacteria.

Difference Between Gram Positive And Gram-negative Bacteria

NEET Biology Cell The Unit Of Life Differences Between Gram PositiveAmd Gram Negatives Bacteria

Bacterial Cell Envelope

  • The cell envelope consists of the outermost glycocalyx, middle cell wall, and innermost cell membrane.
  • Glycocalyx protects cells and also helps in adhesion. It is represented by either a slime layer or a capsule.
  • The slime layer is composed of dextrin, dextran, and levan, while the capsule is made of polysaccharides and D-glutamic acid.
  • The slime layer protects the cells from loss of water and nutrients.
  • Capsule provides gummy and sticky characters to the cell.

Bacterial  Cell Wall

  • Cell wall is made up of peptidoglycan, mucin, or mucopeptide.
  • The Glycan portion forms the backbone of peptidoglycan which is composed of alternating units of NAM (N- acetyl muramic acid) and NAG (N-acetyl glucosa¬mine) joined together by β-1,4-linkage.
  • The tetrapeptide chain is attached with NAM.
  • leichoic acid is an acidic polymer consisting of a carbohydrate (glucose), phosphate, and an alcohol. It performs several functions such as binding metals and acting as a receptor site for some viruses.
  • Teichoic acid maintains cells at low pH to prevent the degradation of cell walls by self-produced enzymes.
  • Porins function as channels for the entry and exit of hydrophilic low molecular-weight substances.
  • The outer layer of the cell wall in gram-negative bacteria contains lipopolysaccharides (LPS) that act as the main surface antigens in the cell wall.

“what is the basic unit of life “

Mesosome (Chondroid; Fitz James): The plasma membrane shows enfolding into the cell and this enfolding is called mesosome. It is particularly found in Gram-positive bacteria. The various types of mesosomes are as follows:

  • Central Mesosome: It holds the nucleoid, and helps in the separation of nucleoid and septa formation.
  • Peripheral Mesosome: It helps in storing respiratory enzymes such as succinic dehydrogenase, cytochrome oxidase, etc.

Plasmid (Lederberg And Hayes): In addition to the normal chromosomal DNA, some extrachromosomal genetic elements are often found in bacteria. These elements are called plasmids. In fact, these are circular pieces of DNA that have extra genes. These are capable of autonomous replication in the cytoplasm of the bacterial wall. Plasmid is circular, supercoiled, double-stranded naked DNA. It is also called mini chromosomes.

Types Of Plasmids

  • Sex-plasmid: It carries sex fertility factor responsible for the transfer of genetic material during conjugation.
  • R-plasmid: Confers resistance to antibiotics, having resistance transfer factor (RTF).
  • Col-plasmid: Produces special proteins colicins (bacteriocin) to kill other bacteria.
  • Degradative Plasmid: Decomposes hydrocarbons in petroleum.
  • Ti- and Ri-plasmid: “Ti” refers to tumor-inducing plasmids and “Ri” refers to rhizogene plasmids. These are large plasmids with about 200 kbp.

Plasmids Respiration: Depending upon the mode of respiration and their capability to perform alternate mode of respiration, bacteria arc of the following types

  • Obligate Aerobes: These can perform only aerobic respiration, for example, Bacillus subtilis.
  • Obligate Anaerobes: These can perform only anaerobic respiration, for example, Clostridium botulinum.
  • Facultative Aerobes: These are anaerobic forms but can live in the presence of O2, for example, Chlorobium.
  • Facultative Anaerobes: These are aerobic forms but can live anaerobically also, for example, Pseudomonas.
  • Aerotolcrant Anaerobes: These bacteria continue to perform anaerobic respiration even in the presence of O2, for example, lactic acid bacteria.
  • Anaerotolerant Aerobes: Aerobic bacteria continue to perform aerobic respiration even in the absence of free O2 by using O2 of oxidized salts, for example, denitrifying bacteria.

Plasmids Reproduction

  • By Binary Fission: It is the common method of reproduction under favorable conditions.
  • By Endospore Formation: Discovered by Cohn in hay bacteria (Bacillus subtilis), endospores are thick-wailed, and highly dehydrated. and resistant spores formed under adverse conditions. Their wall is differentiated into three to four layers.
  • Endospores are commonly formed in genera such as Bacillus and Clostridium. One endospore is formed per bacterial cell. So they are more a means of perennation than reproduction. Cortex and cytoplasm contain Ca2+ and an anticoagulant dipicolinic acid, which prevents the protoplasm from coagulating at high temperatures.
  • Inclusion Bodies: Reserve material in prokaryotic cell are stored in cytoplasm in the form of inclusion bodies. These are not bounded by any membrane. For example, phosphate granules, glycogen granules, etc.

NEET Biology Cell The Unit Of Life Structure Of Endospore

NEET Biology Notes Cell The Unit Of Life Eukaryotic Cell

NEET Biology Cell The Unit Of Life Eukaryotic Cell Flowchart

Cell Organelles

  1. Plastids
  2. Mitochondria
  3. ER
  4. Golgi body
  5. Lysosome
  6. Ribosomes
  7. Microbodies
  8. Cytoskeleton

Cell Inclusion

  1. Secretory Product: enzyme nectar
  2. Excretory Product: essential oils, alkaloids. gums, latex
  3. Storage Product: carbohydrates, fats, nitrogenous substances

Hyaloplasm = Cytoplasm – (Cell organelles + Cell inclusion)

Protoplasm: Life is impossible without protoplasm, which is the living substance of a cell. Protoplasm is present in all living cells and performs all vital functions of a cell. Hence, J. Huxley rightly defined it as the “physical basis of life.” Dujardin named it as sarcode. Purkinje renamed it as protoplasm.

Protoplasm Cell Wall

  • Discovered by Robert hook.
  • Present in the kingdom Plantae, Fungi, Monera, and most of the members of kingdom Protista.
  • Absent in Animalia, Mycoplasma, gametes, and zoo-spores.
  • The cell wall is the non-living component and protective layer of cell.
  • Originated during cytokinesis from the vesicle of the Golgi body (phragmoplast).

Protoplasm Components Of Cell Wall

  1. Matrix: It is the non-cellulosic component mainly composed of H2O(30-60%), pectin (2-8%), hemicellulose (5-15%), glycoprotein (1-2%), and lipid (0.53%).
  2. Fiber: Mainly composed of cellulose.
  3. Deposition: Deposition only takes place on the secondary wall in the form of silica, pectin, suberin, cutin, lignin, etc.

NEET Biology Cell The Unit Of Life Components Of Cell Wall

Protoplasm Structure Of Cell Wall

  • Middle Lamella: It is the cementing layer between two cells which is formed first among all the layers during cytokinesis.
    • Chemically composed of calcium pectate, but a small amount of magnesium pectate may be present or absent.
    • Absent in the free surface of the cell.
    • The softening of fruits and retting of fibers is due to the dissolution of middle lamella by the pectolytic enzyme.
  • Primary Wall: First layer of cell, thickness 1-3 pm and elastic.
    • Cellulosic fibers are loosely scattered in the matrix.
    • It is only a permanent layer in the parenchyma and meristem.
    • The process of formation of the primary wall is called into susceptions.
  • Secondary Wall: The second wall is more thick and present inside the primary wall. The thickness is 3-10 pm.
    • Mainly composed of cellulose and arranged in three layers (S1 – S3) and shows a threeply structure.
    • Matrix is present between these layers.
    • Deposition of suberin, and lignin takes place on the secondary wall.
    • The process of the formation of a secondary wall is known as accretion (apposition).
  • Tertiary Wall: The tertiary wall is only present in the tracheids of gymnosperms and composed of xylan and cellulose.

NEET Biology Cell The Unit Of Life Structure Of Cell Wall

Structure Associated with Cell Wall

Plasmodesmata: It is the cytoplasmic connection between two neighboring plant cells. The term “plasmodesmata” was coined by Strassburger and discovered by Tangle.

NEET Biology Cell The Unit Of Life Structure Associated With Cell Wall

Pits: Depression present on cell wall lacking secondary wall.

Components Of Pits

  • Pit Membrane: Made up of primary wall and middle lamella. It is permeable.
  • Pit Chamber: It is the area where the secondary cell is absent.
  • Pit Pore: It is the open passage between two cells.

“the basic unit of life “

Types Of Pit:

  1. Simple pit and
  2. Border pit.

NEET Biology Cell The Unit Of Life Components Of Pits

Cell Membrane (Plasma Membrane, Plasmalemma): The Protoplasm of all types of cell is surrounded by a double-layered, living, selectively permeable membrane called cell membrane.

  • Cell Membrane Origin: Cell membrane originates de-novo (independently synthesized) from its constituent chemical but does not originate from any other structure.
  • Cell Membrane Chemical Composition: Cell membrane is chemically composed of phospholipids proteins and carbohydrates.
    • Protein = 20-70%
    • Lipid = 20-79%
    • Carbohydrate = 1.5%
    • Water = 20%
    • Enzymes
    • Amino acid

Cell Membrane Note:

  • Lipids are found in the form of phospholipids, glycolipids, and sterols.
  • Phospholipid forms the basic component of cell membrane.
  • Upids show flip-flop movement (movement of lipids inside and outside of cell membrane).
  • It provides stability, elatedly, and fluidity of the membrane.
  • The fluidity of the membrane is due to the hydrophobic non-polar tail part of fatty acid.
  • Phospholipids is amphipathic in nature due to the presence of both hydrophilic and hydrophobic nature.

Models Of Cell Membrance

  • Trilamellar Model Or Sandwich Model
    • This model was given by Danielli and Davson.
    • The arrangement of lipid (L) and protein (P) molecules is called PLLP.
    • The bimolecular lipid layer is sandwiched between the single layer of protein on both sides.
    • Protein is globular and α type.
    • This is a hypothetical model which was proposed before the discovery of the electron microscope.
  • Unit Membrance Concept
    • This model was given by Robertson.
    • All membranous systems of a cell are composed of lipoprotein.
    • The molecular organization and chemical composition is similar to that of the lamellar model.
    • This model is different from the trilamellar model, that is, the protein is of β-type.
    • Robertson also proposed measurements of this membrane. The total thickness of the membrane is equal to
    • Both inner and outer proteins are different in nature. The outer protein is muco protein and the inner protein is nonmuco protein.
    • A core of 8Å diameter is also present in the protein layer which helps in transportation.
  • Fluid Mosaic Model
    • This model was given by Singer and Nicholson.
    • Widely accepted model which clearly explains selectively permeable proportions of cell membranes.
    • This model can be explained as a protein iceberg in the sea of lipids.
    • Lipids are present in two layers in which the head of fatty acids faces towards the outside and tail towards the inner side.
    • Protein is of α-type and is present in two forms.
    • Extrinsic Protein: Small size protein present outside the lipid layer and can be easily separated. It forms 30% of the total protein.
    • Intrisic Protein; large-sized protein embedded in lipid molecules and cannot be easily separated. It forms 70% of the total protein.

NEET Biology Cell The Unit Of Life Sandwich Model

NEET Biology Cell The Unit Of Life Unit Membrane Model

NEET Biology Cell The Unit Of Life Fluid Mosaic Model Of Plasma Membrane

Asymmetry Of Plasma Membrane: Two surfaces of the plasma membrane are not similar because

  1. Extrinsic proteins present more towards the inner side.
  2. Lecithin (lipid) is present on the outer side.
  3. Cephalin (lipid) is present on the inner side.
  4. The carbohydrate of oligosaccharide nature is present towards the outer side of the membrane and attaches to both lipid and protein called glycolipid and glycoprotein. respectively, and collectivity is called glycocalyx. Glycocalyx is the eye and ear of the membrane, i.e., the recognition center.

Structures Associated With Plasma Membrane

  • Microvilli: Finger-like structure developing from the cell membrane in the alimentary canal.
  • Mesosome: Enfolding of the cell membrane in bacteria. Lomasome: Enfolding of the cell membrane in fungi.
  • Lamellosome: Enfolding of the cell membrane in blue-green algae (BGA).
  • Desmosome: Two animal membranes are separated by each other by intercellular spaces in which the cell membrane develops thickening due to the deposition of protein from which tonofibrils originate. This constitutes a desmosome.
  • Hemidesmosome: Thickening present only on one membrane.
  • Terminal Bars: Desmosomes without tonofibril.
  • Tight Junction: Two animal membranes are closely attached together. Hence, no intercellular space.
  • Gap Junction: Two animal cells are connected with each other by protein pipes.

NEET Biology Cell The Unit Of Life Desmosomes

Experiment To Show Fluidity Of Cell Membrane: Larry Frye and Michael Edidin, at Johns Hopkins University, labeled the plasma membrane proteins of a mouse cell and a human cell with two different markers and fused the cells. Using a microscope, they observed the markers on the hybrid cell. The mixing of the mouse and human membrane proteins indicates that at least some membrane proteins move sideways within the plane of the plasma membrane.

NEET Biology Cell The Unit Of Life Experiment Of Fluidity Of Cell Membrane

Functions of Plasma Membrane: Transport

  • Passive Transport: The transport of substances through a membrane along a concentration gradient is called passive transport it is downhill transport without the expenditure of energy.
  • Active Transport: The transport of substance through the membrane against a concentration gradient with the expenditure of energy is called as active transport. Active transport is highly selective, and unidirectional.
  • Bulk Transport: The transport of substances through the vesicle of the membrane is called as bulk transport.

Types Of Transport

  • Endocytosis: The intake of foreign substances through vesicles is known as endocytosis, The endocytosis of a solid particle is called phagocytosis (cell eating) and the endocytosis of liquid molecules is called pinocytosis (cell drinking).
  • Phagocytosis results in the formation of phagosomes and pinocytosis results into the formation of pinosomes.
  • Exocytosis: Exocytosis or endocytosis geophagy or cell vomiting is the outward movement of a substance from the cell.

NEET Biology Cell The Unit Of Life Types Of Transport

Endomembrane System (GERL System): The term endomembrane system was coined by Navikoff The of cell is composed of a Golgi body, endomembrane system, and vacuole (called endoplasmic reticulum (ER), Iysosome, GERL system).

NEET Biology Cell The Unit Of Life Endomembrane System

Golgi Body: The common names of the Golgi body are mitochondria, idioms, dictyosome, Dalton complex, traffic police, the middle man of the cell, apparato reticular, and Golgi complex. Golgi complex was discovered by Camillo Golgi in the nerve cell of an owl and cat and the term “Golgi body” was coined by Cajal.

Golgi Body Occurrence: Only found in the eukaryotic cell except for mature RBCs and sperm of mammals, mature sieve tube cells, mature sperm of bryophyte, and Pteridophyta. In the plant cell, the Golgi body is called a dictyosome. In fungi, the Golgi body is unicistemal, because its components are not packed.

Golgi Body Structure: It appears as a flattened sac-like structure bounded by a single membrane without any ribosome on its surface. It shows polarity due to differences in the structure of both ends. The convex end is called cis-face or forming (F-face) and the concave end is called transface or (maturing) M-face.

Golgi Body Is Composed Of Four Components:

  1. Cistenae,
  2. Tubule,
  3. Vesicle, and
  4. Golgian vacuole.
  1. Golgi Body Cistenae
    • It is the most important component of the Golgi body.
    • It is found in the form of stacks.
    • Each stack contains 4-18 cistemae.
    • It shows polarity due to the presence of two faces.
    • Slightly curved and encloses a cavity inside
  2. Golgi Body Tubule: It looks like a branched network-like structure.
  3. Golgi Body Vesicle: Vesicles are formed by budding of the terminal end of tubular smooth ER. They are associated with the convex surface of cistemae.
  4. Golgi Body Golgian Vacuole: It is formed by the budding of the terminal end maturing face of Golgi apparatus or surrounding of cisternae. Golgi body is present in the cytoplasmic area and is free from other cell organelles. Hence, the area is called the zone of cell exclusion.

NEET Biology Cell The Unit Of Life Golgi Complex In Stereoscopic View

Function Of Golgi Body

  • Processing and packaging of material: Vesicle enters into the Golgi body at cis- or F-face and the secretary vesicle comes out from the Golgi body from the trans- or M-face. Hence, called the traffic police.
  • Formation of primary Iysosome
  • Glycosidation is the formation of glycolipids from lipids.
  • Glycosylation is the formation of glycoprotein from protein.
  • Extracellular secretion: It secretes mucilage; hence, root caps are rich in the Golgi body.
  • Intracellular secretion of pectin and other cellular materials.
  • Formation of acrosome of sperm; hence, acrosome is also called modified Golgi body.
  • Transport vesicle formed by endoplasmic reticulum.
  • Lysosomes and secretory vesicles originate from the M-face of the Golgi body.

Endoplasmic Reticulum: Endoplasmic reticulum was discovered by Porter and Thomson and the name was given by Porter and Khallan. ER is present in all eukaryotic cells except mature RBCs, monocytes of WBC, undifferentiated cells eggs, etc. It is completely absent in prokaryotic cells.

  • Endoplasmic Reticulum Structure: The endoplasmic reticulum forms a hollow tubular network such as the structure in cytoplasm and is composed of three components
  • Endoplasmic Reticulum Cisternue: Simple unbranched tube-like structures; 40-50 pm in diameter; generally arranged parallel and interconnected to each other.
  • Endoplasmic ReticulumTubule: Tubules are branched tube-like structures 50-200 pm in diameter.
  • Endoplasmic Reticulum Vesicle: Small vacuole-like structures, 25-500 pm in diameter. Ribosomes are not attached. The endoplasmic reticulum is attached to a nuclear membrane at one end and a cell membrane at the other end.

NEET Biology Cell The Unit Of Life Cisternae

NEET Biology Cell The Unit Of Life Tubules

NEET Biology Cell The Unit Of Life Vesicles

Types Of Endoplasmic Reticulum

Rough ER (RER): The rough surface of the endoplasmic reticulum is due to the presence of ribosomes. Only a large unit (60S) of the ribosome is attached to the surface of RER with the help of two types of glycoprotein called Ribo- phorrin-1 and Ribophorrin-2.

  • RER forms two-thirds of the total ER of cells.
  • RER is only composed of cisternae and tubules.
  • RER bears a pore at the point of attachment of the ribosome.

Smooth ER (SER): Smooth endoplasmic reticulum surface is smooth due to the absence of ribosomes.

  • SER forms only one-third of the total ER of cells.
  • SER is composed of only tubule and vesicle.

Endoplasmic Reticulum Note:

  • The ER of muscle is called the sarcoplasmic reticulum, in the retina of the eye is called the myeloid body, and in the nerve cell is called NissTs granule.
  • RER is found in cells involved in protein synthesis.
  • RER is basophilic in nature due to the high content of RNA. Hence, also called ergastoplasm or basophilic body.
  • RER originates from the nuclear membrane, but SER originates from RER.
  • SER is found in cells that are involved in the synthesis of lipids, glycoproteins, and hormones.
  • The liver cell has both types of RER and SER.

Common Functions Of RER And SER

  • The endoplasmic reticulum gives mechanical support to the cytoplasm since it forms a network-like structure inside the cell.
  • Transport of material within cell.
  • It helps in the formation of the nuclear membrane.
  • Both RER and SER increase surface area for enzymatic action.

Functions Of SER

  • Synthesis of lipid and glycogen
  • Detoxification (protection of cell by the toxic effect of chemicals by using cytochrome P450)
  • Helps in muscle contraction by release and uptake of Ca
  • Formation of organelles such as spherosomc and glyoxysome

Functions Of RER

  • Synthesis and transport of protein
  • Formation of SER
  • Synthesis of lysosomal enzyme

Lysosome

  • Lysosome Common Name: Suicidal bag, disposal unit, recycling center, scavenger of cell, and atom bomb of cell. Lysosomes originate from the Golgi body and they were discovered by Christian de Duve. The term was coined by Navikaff.
  • Lysosome Occurrence: Mainly found in eukaryotic animal cells, but also found in fungi and root tip cell of maize and cotton.
  • Lysosome Structure: They are the smallest single membrane-bound cell organelles and are mainly found in animal cells. Lysosomes are small, oval, or spherical structures filled with about 50 types of hydrolytic enzymes collectively called acid hydrolyses. The marker enzyme of lysosomes is acid phosphatase.

All enzymes work at acidic pH. The acidic condition is maintained by the pumping of proton inside lysosomes. The excess of lipid-soluble vitamins, steroidal sex hormones, bile salts, X-rays, and UV rays are called membrane destabilizers. Lysosome membrane has some stabilizers as cortisone, cholesterol, heparin, etc. Lysosomes show polymorphism found in four different forms

  1. Primary Lysosomes: Newly formed lysosomes containing inactive enzymes. They are produced from the transface of the Golgi body.
  2. Secondary Lysosomes Or Digestive Vacuoles Or Hetrophagosomes: Secondary lysosomes contain primary lysosomes and food vacuoles (pinosomes or phagosomes).
  3. Tertiary Lysosomes Or Residual Bodies: Lysosomes containing undigested food material.
  4. Autophagic Vacuoles: They are formed by the union of many primary lysosomes around old and dead organelles.

Enzymes produced by lysosomes are classified into four groups.

  1. Proteases,
  2. Acid phosphatases,
  3. Sulphatases, and
  4. Ribonucleases, deoxyribonucleases.

Lysosome Function

  • Perform extracellular and intracellular digestion.
  • Removal of dead cell organelles. Hence, act as scavengers.
  • Digestion of harmful substances of cells.
  • Autophagy: If a person is on prolonged last, then lysosomes provide energy by digesting the stored food of cell.
  • Autolysis: Disappearance of the tail of a tadpole larva of a frog during metamorphosis is due to an enzyme cathepsin.
  • In old age, the number of lysosomes increases, and the number of mitochondria decreases.

Mitochondria: Mitochondria are the largest cell organelles and the second largest cellular component of animal cells. They are the second largest cell organelles in plant cells and the third largest cellular component of plant cells.

  • Mitochondria were discovered by Kollikar in the striated flight muscle of insects, and he called it as sarcosome. The term mitochondria was given by Benda. All mitochondria of a cell are called chondrites.
  • A mitochondrion is also known as a sarcosome, bioblast, plasmosome, chondroplasty, plastochondria, and a powerhouse of the cell. It is also considered as a cell within cell.
  • The vital stain for mitochondria is Ganus grcen-B which does not kill the mitochondria. The number of mitochondria is variable in different organisms. The number of mitochondria depends on the metabolic activities of a cell.

Mitochondria Number

  1. 1 in Chlorella, Trypanosoma, and Microasterias
  2. 25 in human sperm cell
  3. 300 -400 in human kidney cell
  4. 500-1000 in lever cell
  5. 50,000 in giant amoeba Chaos chaos
  6. 140,000-150,000 in the egg of sea urchin
  7. 500,000 in-flight muscle cells of insect
  8. Plant cells have less number of mitochondria than animal cells because plant cells have both chloroplast and mitochondria for ATP production

Mitochondria Size: The smallest mitochondria are found in yeast cells and the largest in the oocyte of Rana pipens.

Mitochondria Occurrence: Found in all eukaryotic cells except mature RBC.

Mitochondria Structure: The mitochondrion is a double membrane-bound sac-like structure. Each membrane is 60-70 A thick and composed of lipoprotein. Both membranes arc structurally and functionally quite differently from each other. The outer membrane is continuous, fully permeable, rich in lipids, and contains porin (protein). The outer membrane is poorer in protein.

  • The lipid-protein ratio of the outer membrane is 40:60.
  • The lipid-protein ratio of the inner membrane is 20:80.
  • The inner membrane contains cardiolipin (having four fatty acids).
  • The inner membrane is infolded as a finger-like structure called a crista.
  • Cristae are meant to increase the physiological active area of the inner membrane.
  • Cristae bear stalked particles called exosomes.

The cavity of mitochondria is divided into two chambers.

  1. Outer Chamber Or Perimitochondrial Space: It is the space between the outer and inner membranes. The thickness of the outer chamber is 60-80 A and it is filled with the homogenous fluid of phospholipids.
  2. Inner Chamber: It is the space enclosed by the inner membrane. It is filled with a proteinaceous liquid called matrix. The matrix contains DNA (with high GC ratio), RNA,70S ribosomes, and all enzymes of Kreb’s cycle except succinic dehydrogenase. (SDH). SDH is present in the inner membrane of mitochondria.

NEET Biology Cell The Unit Of Life Structure Of Mitochondria

Mitochondria Chemical Composition

  1. Protein: 65-70%
  2. Lipid: 25-30%
  3. RNA: 5-6%
  4. Ribosome: 70S type

Two States Of Mitochondria

  1. Orthodox: It is the inactive state of mitochondria where ATP production is not involved. Hence, the matrix becomes broad and the perimitochondriai space becomes narrow.
  2. Condensed: The active state of mitochondria is involved in ATP production. Hence, the matrix becomes narrow, but the outer chamber becomes broad.

Mitochondrial DNA

  • Mitochondrial DNA forms only 1% of the total DNA of the cell.
  • It is much smaller than the nuclear DNA. Hence, contains few information.
  • It is double-stranded, circular, and Naked (Histone absent)
  • GC content is high. Hence, Tm (melting temperature) is high.
  • Mitochondria have their own DNA polymerase enzyme. Hence, it replicates its own DNA.
  • Mitochondrial DNA is responsible for cytoplasmic inheritance. For example, male sterility in maize, and petite character in yeast.

Mitochondria contain about 70 types of enzymes which is 70% of the total enzymes of the cell, which indicates the highest concentration of enzymes in cell mitochondria.

Oxysome: Oxysome is also called F0-F1 particle elementary particle or subunit of Fernandez and Moran or ATPase particle.

Structure Of Oxysome

  • Head (F1 Particle): Site of ATP synthesis
  • Stalk
  • Base (F0 Particle): Proton tunnel

The space between two exosomes is 100Å. The number of exosomes is 104-105 per mitochondrion.

NEET Biology Cell The Unit Of Life Structure Of A Crista And An Oxysome

Oxysome Note: Mitochondria are called semi-autonomous cell organelles because they have their own DNA, RNA, and 70S type of ribosome, but they also depend on the nucleus and cytoplasm for most of the metabolic activity. Mitochondria are similar to bacteria cells or prokaryotic cells in the following points

  1. Presence of porin protein in the outer membrane
  2. Mitochondria show binary fission of division
  3. Presence of 70S type of ribosomes
  4. DNA double-stranded, circular, and naked

On these bases, mitochondria may be called as “cell within the cell” “intracellular prokaryotic parasite” or “bacterial endosymbiont.”

Functions Of Mitochondria

  • Mitochondria act as the site of respiration except during glycolysis and anaerobic respiration.
  • They act as the site of oxidative phosphorylation (formation of ATP by the oxidation of reduced co-enzyme).
  • They are also called as the powerhouse of cells.
  • Mitochondria form the sperm tail.
  • The synthesis of many amino acids and fatty acids occurs in the mitochondria.
  • Mitochondria also take part in lipid synthesis.
  • Mitochondria may store and release calcium according to requirements.

Ribosome: In plant cells, ribosome was discovered by Robison and Brown, and in animal cells, by Palade. Ribosomes are the smallest nonmembranous and most abundant cell organelle that are negatively charged. The number of ribosomes is 104-105 per cell. Ribosomes are also called as palate particles or protein factories or engines of cell or ribonucleic acid proteinic particles (r-RNA + Protein).

Ribosome Occurrence: Ribosomes are found in both prokaryotic and eukaryotic cells except mature RBCs. In a prokaryotic cell, ribosomes are found freely in the cytoplasm or form polysomes (ribosomes + mRNA). The function of polysomes is the synthesis of multiple copies of a single protein. In eukaryotic cells, two types of ribosomes are found:

  1. Cytoplasmic Ribosomes: 80S type of ribosomes are found in cytoplasm freely or attached with ER or nuclear membrane.
  2. Organelle Ribosomes: 70S type of ribosomes are found in mitochondria and plastids.

Hence, the ribosome is also called organelles within organelles.

Ribosome Type: On the basis of sedimentation coefficient, ribosomes are two types

  1. 70S Type: It is found in both prokaryotic and eukaryotic cells. The ratio of r-RNA and protein is 60:40. The large subunit is 50S (34-protein + 5S r-RNA + 23S r-RNA)and the small subunit is 30S (23-protein + 16S r-RNA).
  2. 80S Type: It is reported only in eukaryotic cells. The ratio of r-RNA and protein is 40:60. The smaller unit is 40S (33-protein + 18S r-RNA) and the larger unit is 60S (40-protein + 5S r-RNA + 5.8S r-RNA + 28S r-RNA).

Ribosome Origin: In prokaryotes, ribosome synthesis takes place in the cytoplasm, but in eukaryotes, it is synthesized in nucleolus.

Association And Dissociation Of Ribosomes: Both subunits of ribosomes are found separately in the cytoplasm. The association and dissociation of ribosomes depend on the concentration of Mg+2 ions. In the high concentration of Mg+2 ions, both subunits associate to form a complete ribosome called dimmer but in low concentration, it dissociates. Both subunits are assembled at the time of protein synthesis.

Functions Of Ribosomes

  1. Site of protein synthesis, hence called protein factory.
  2. Free ribosomes synthesize non-secretary proteins, while ER-associated ribosomes synthesize secretory proteins.
  3. Proper folding of proteins requires chaperones.

Plastid: Plastids were discovered by Hackel. They are only found in plant cells. They are the largest cell organelle and the second largest cellular component. A plastid originates from a protoplastid (found in meristem). All plastids of a plant are collectively called plastidome, which was coined by Dangered. Schimper discovered different types of plastids. Plastids can be divided into three types on the basis of the presence or absence of the pigment.

NEET Biology Cell The Unit Of Life Plastid

  1. Chloroplast: Green plastids (chlorophyll present); are found in the green parts of plants.
  2. Chromoplast: Colored plastids (chlorophyll absent but other pigments present); provide coloration to flowers and fruits.
  3. Leucoplast: Colorless plastids (pigments and grana absent). These are concerned with the storage of carbohydrate proteins and lipids in the underground parts of the plant. These are the largest plastids. Different types of leucoplasts are as follows
    1. Elaioplast Or Oleoplast: Stores fat
    2. Amylop Last: Stores starch
    3. Aleuroplast: Stores protein

Plastid Note: All three forms of plastid, chloroplast leucoplast, and chromoplast are interconvertible. For example, in tomatoes, chloroplasts change into leucoplasts, and then in chromoplasts. In potatoes, leucoplasts change into chloroplasts. In mango or chili, chloroplasts change into chromoplasts. In carrots, leucoplasts change into chloroplasts.

Chloroplast

  • The number, shape, and size of the chloroplast are variable.
  • In higher plants, the chloroplast is discoid in shape and bounded by a double membrane.
  • Each membrane is 90—100 A thick and composed of lipoprotein.
  • Both membranes are structurally and functionally different.
  • The outer membrane is fully permeable, rich in lipid, and contains porin in protein.
  • The inner membrane is selectively permeable and rich in protein.
  • Chloroplast may be called “cell within the cell” “intra-cellular prokaryotic parasite” or “bacterial endosymbiont.”
  • Chloroplast is also a semi-autonomous structure like mitochondria.
  • The DNA of chloroplast is double-stranded, circular, and naked (without histone).
  • The DNA has high GC content, hence high Tm. (melting point).
  • Chloroplast is also responsible for cytoplasmic inheritance, for example, plastid inheritance in Mirabilis jalapa.

Structure Of Chloroplast: The cavity of chloroplast is divided into two chambers

  1. Outer Or Periplasmic Space: The space bounded by outer and inner membranes. It is 100-200 A thick.
  2. Inner Chamber: The space enclosed by the inner membrane. It is filled with a proteinaceous liquid called stroma. It contains DNA, RNA, 70S type of ribosomes, and all enzymes (RuBisCO) taking part in the Kelvin cycle.

NEET Biology Cell The Unit Of Life Structure Of Chloroplast

RuBisCO: RuBisCO forms 50% part of the stroma and about 16% part of the total chloroplast protein. It is considered as the most abundant enzyme or protein on earth. It is also known as RuBP carboxylase, RuBP oxygenase, and carboxy dismutase.

In the stroma of chloroplast, several double membrane-bound sac-like structures called baggy trousers or thylakoids are reported. The term “thylakoid” was coined by Menke. Thal-alkaloid is considered as the structural unit of the chloroplast.

  • Thallakoid is an analogous structure to the cristae of mitochondria.
  • Several thylakoids become stack-like structures and form granum.
  • The number of thylakoids in a granum is 2-100.
  • The number of grana in one chloroplast is 40-60.
  • Grana are interlinked with each other by fret channel or stroma lamella which may lack pigments.
  • In the membrane of thylakoid, about 200-300 quantasomes are present.

NEET Biology Cell The Unit Of Life Rubisco

Quantasome: Quantasome was discovered by Parks and Biggin, but the term was coined by Park and Pon.

  • It contains about 230 chlorophylls (160 chl-a, 70 chl- b) + 50 carotenes. Quantasome absorbs one quantum of light.
  • Quantasome is called the functional unit of chloroplast, or the unit of photosynthesis.
  • Pigments associated with quantasomc form two types of photosystems (PS). Photosystems are composed of two parts
  • Reaction center and
  • Light-harvesting complex (LHC) or antenna.
  • The reaction center contains one molecule of chl-a and LHC contains other pigments.
  • Each photosystem is composed of 250—400 pigments.
  • LHC is concerned with the light-harvesting mechanism. Different wavelengths of light are absorbed by LHC.
  • The reaction center (chl-a) is concerned with the conversion of light energy into chemical energy.
  • On this basis, chl-a is called the main pigment and others are called necessary pigments.

NEET Biology Cell The Unit Of Life Components

Type Of Photosystems

  1. PS-1: Located at the non-appressed part of grana and stroma thylakoid reaction is P700 (fret channel).
  2. PS-2: Located at the appressed part of grana and the reaction center is P680. The reaction center of bacteria is B810.

Photosystems Note:

  • Chloroplast is agranular in algae and bundle sheath cells C4 of plant
  • If thylakoids are directly scattered in the cytoplasm, then they are called chromatophores.
  • Chloroplast in which thylakoids are present but not stalked is called adrenal chloroplast
  • In higher plants, grana contains pigments: chl-a, chl-b, and carotene.
  • In algae, xanthophyll is the main pigment and the other pigment is phycobilin (phycocyanin and phycoerythrin).

Functions Of Chloroplast: Chloroplast is the site of photosynthesis. The light reaction takes place in grana and the dark reaction in the stroma. Hence, the chloroplast is called the glucose factory. The thylakoid membrane is the site of photophosphorylation (synthesis of ATP by using light)

Structure Of Chlorophyll: Chlorophyll without Mg is called pheophytin which acts as an electron acceptor. Chl-a contains -CH3 group present at the third carbon of the second pyrrole ring. Chl-b contains -CHO group present at the third carbon of the second pyrrole ring.

NEET Biology Cell The Unit Of Life Molecular Strurcure Of Chlorophyll

Microbodies: Microbodies are single membrane-bound cell organelles concerned with oxidation reactions other than respiration. Micro¬bodies include

  1. Peroxisomes,
  2. Glyoxysomes, and
  3. Spherosomes.
  1. Peroxisome
    • Discovered by de Duve in the leaf cell of spinach.
    • Peroxisome is mainly found in the mesophyll cells of C3 plant.
    • It is concerned with photorespiration which involves glycolate metabolism.
    • Catalase is the largest enzyme and peroxidase is the smallest enzyme of peroxisome.
    • It is mainly concerned with the formation and destruction of H2O2. Hence, it protects cell from the toxic effect of H2O2.
    • Peroxisomes originated from ER.
    • Peroxisome converts glycolic acid into glycine by the help of glycolic acid oxidase enzyme.
    • In animals peroxisomes found in lever and kidney cells and are concerned with the synthesis and degradation of lipids.
  2. Glyoxysome: Glyoxysomc was discovered by Beevers and Tolbert from the germinating seed of the castor. It is mainly found in the germinating fatty seeds (caster, ground nut, etc.) and also found in some fungi (Neurospora and yeast). It is concerned with the following activity
    • β-oxidation of fatty acid.
    • Conversion of fat into carbohydrate. This process is called glyoxylate metabolism which requires two enzymes that are found in glyoxysome.
    • These enzymes are
    • Isocitrate lyase,
    • Malate synthase. These two enzymes convert acetyl CoA into glucose.
    • Glyoxysomes originated from smooth ER. They are destroyed in cell if the germination of seeds is completed.
    • In animal cell, β-oxidation of fatty acid takes place in mitochondria.
  3. Spherosome Or Oleosome: Spherosome was discovered by Pemer and Hanstein. Spherosome originates from smooth ER and is filled with several hydrolytic enzymes. The term was coined by Dangeard. Spherosomes are half-membrane-bound cell organelles filled with several hydrolytic enzymes. They are only found in plant cells, hence called plant lysosomes. The enzymes of spherosomes are different from those of lysosomes.
    • Spherosome Or Oleosome Functions
      • Synthesis, storage, and translocation of fat.
      • Synthesis of the intermediate compound of suberin.

Centriole: A Centriole is a cylindrical or rod-like structure that is not bounded by any membrane and is present close to the nucleus, mainly in the eukaryotic animal cell.

  • Centriole is also found is some motile plant cells. For example, motile algae, zoospores of algae, antherozoids of bryophytes, and pteridophytes.
  • Centrioles are absent in prokaryotic cells, amoeba, gymnosperm, and angiosperm.
  • Centrioles are found in pairs situated at right angles to each other called diplosomes.
  • Diplosome is found in the exclusive area of cytoplasm which is free from any other cell organelle. This inclusion area is called cytocentrum centrosphere kinoplasm or zone of cell exclusion.
  • Diplosome and cytocentrum are collectively called Centrosomes.
  • Centrioles are found in pairs situated at right angles to each other called diplosomes.
  • Diplosome is found in the exclusive area of cytoplasm which is free from any other cell organelle. This inclusion area is called cytocentrum centrosphere kinoplasm or zone of cell exclusion.
  • Diplosome and cytocentrum are collectively called Centrosomes.
  • Centriole does not contain RNA or DNA but duplicates directly during the end of the S-phase of the cell cycle.

NEET Biology Cell The Unit Of Life Pair Of Centripoles And TS Of A Centripole

Duplication completes in G2 phase. Centriole is a cart wheel-like structure and shows 9 + 0 organization.

  • It has nine peripheral triplet fibers but no central fiber.
  • Each peripheral triplet fiber is composed of three microtubules which represent A, B, and C from inside to outside. Hence, the total number of microtubules in a centriole is 27 (9 x 3).
  • The central part of centriole is called HUB which has no microtubule.
  • HUB is attached to the peripheral fiber by pretentious spokes.
  • Each peripheral triplet fiber is linked with each other by an A-C or C-A linker.
  • Nine amorphous masses of protoplasm are present on the peripheral part of the centriole. These are called masses (MTG micro tubule generator), and pericentriolar satellite/nucleating center. It is concerned with the formation of new centriole.

Centriole Function:

  • Centrioles polymerize microtubules for the formation of spindle fibers and astral rays during cell division.
  • They form the basal body of cilia and flagella.
  • They also form spindle fibers during cell division.

Cilia And Flagella: Both cilia and flagella are locomotory organs and both show more or less similar structural organization. Cilia are smaller in size, more in number, present on the entire surface of cell, and beat in a coordinated synchronous manner. Flagella are fewer in number, more elongated, generally pre¬sent at the anterior end of the cell, and beat independently.

Eukaryotic Cilia/Flagella: The cilia and flagella have a microtubular composition. Both are bounded by a unit membrane. The bounded space is filled with matrix. In the matrix, there remain embedded nine doublets and two singlets of microtubules. Two singlets are located in the middle and called a central axoneme. Nine doublet fibrils are arranged peripherally around the central fibrils called peripheral axoneme.

Functions Of Cellia And Flagella

  • Both flagella and cilia act as the organ of locomotion. Cilia may help in the aeration and circulation of food.
  • Cilia may beat in a metachronous (beat one after one) or synchronous (beat simultaneously) rhythm.
  • The cilia of the respiratory tract help in retaining the dust.

Cytoskeleton: An elaborate system of network-like structure, which is composed of proteins and filamentous structures present in the cytoplasm of eukaryotic cells to support the cytoplasm, is collectively called as cytoskeleton. It includes three structures

  • Peripheral axonemes are linked to the central axoneme by radial spokes.
  • Peripheral axonemes are linked with each other by A-B linker.
  • A side arm develops from microtubule A of the peripheral axoneme.
  • The central and peripheral axonemes are composed of tubulin protein.
  • The side arm is composed of dynein protein which has ATP are activity.
  • A-B linker is composed annexin type of protein.
  • Energy required for the movement of cilia or flagella is provided by dynein protein. It has ATPase zyme which hydrolyzes ATP into ADP and releases energy.

NEET Biology Cell The Unit Of Life Section Of Cilia Or Flagella As Different Parts

  1. Microtubule
  2. Intermediate filament
  3. Microfilament
  1. Microtubule
    • Discovered by de Robertis and Franchi. Microtubules are unbranched, hollow, non-contractile filaments of tubulin protein that form the basal structure of centriole, basal body of cilia and flagella, spindle fiber, and axoneme of cilia and flagella.
    • The assembly and disassembly of microtubules require GTP and Ca+2 ion.
    • Microtubules are responsible for cell motility and maintenance of the cell shape. They also show polarity.
    • The boundary of the microtubule is composed of 13 parallel protofilaments.
  2. Microfilament
    • Discovered by Paleviz.
    • Microfilament is a branched or unbranched contractile solid filament composed of actin protein and found just below the cell membrane.
    • Responsible for cyclosis, sol-gel interconversion, pseudopodia formation, and formation of cleavage furrow during cell division.
  3. Intermediate Filament: It is a hollow, branched non-contractile filament composed of three types of acidic protein
    • Keratin (acidic in nature)
    • Vimentin (acidic in nature)
    • Desmin (acidic in nature)
    • It forms a nuclear matrix or nuclear lamina present on the surface of the inner nuclear membrane
    • It provides stability to the cell.
    • It is involved in the formation of a scaffold for chromatin.

Nucleus: Nucleus is the largest cellular component in a cell. It is an extracytoplasmic component. The nucleus was discovered by Robert Brown in the root cell of an orchid. A cell may be uninucleate, binucleate, or multinucleate.

NEET Biology Cell The Unit Of Life Nucles

If the nucleus degenerates in a mature cell, then the cell is called enucleate, for example, a mature sieve tube, RBC of the animal. Hammerling called the nucleus as the brain of a cell or controlling the center of a cell by using a grafting experiment with two species of Acetabularia (unicellular largest marine green alga): A. crenulata and A. mediterranea.

Structure Of Nucleus: Nucleus can be easily distinguished into the following four parts nuclear membrane, nucleoplasm, nucleolus, and chromatin.

NEET Biology Cell The Unit Of Life Structure Of Nucleus

NEET Biology Cell The Unit Of Life Ultra Struture Of Interphase Nucleus

Nuclear Membrane

  • The nuclear membrane is also called nucleolemma or karyotheca.
  • The nucleus is also bounded by the double unit membrane of lipoprotein.
  • Each membrane is 90-100 Å thick and composed of lipoprotein.
  • The space, enclosed by two membranes (75 Å) is called perinuclear space and is 100-300 Å thick.
  • The nucleus membrane is punctured by several pores called nuclear pores.
  • Nuclear pores allow the movement of m-RNA.
  • These pores may be circular or octagonal.
  • Pore complex = Pore (1000 Å) + Annulus (a special type of protein that plugs the nucleus pore)
  • The outer membrane is connected with the cell membrane through ER and also bears ribosomes (60S unit) on its surface.
  • A network of intermediate filaments is present on the inner surface of the nucleus membrane called nuclear lamina or nucleus matrix.
  • The nucleus lamina functions as the attachment point of the telomere during cell division (leptotene stage of meio- sis-I).

Nucleoplasm

  • A part of protoplasm present within the nucleus is called nucleoplasm.
  • It is more denser than cytoplasm. Hence, the nucleus can be easily observed under a microscope.
  • Nucleoplasm is mainly made up of nucleic acids, his-tone proteins, enzymes (DNA, R.NA polymerase), li¬pids, minerals (P, K+, Na, Ca, Mg), NHC proteins, and NHC non-histone chromosomal proteins.

Nucleolus

  • Nucleolus was discovered by Fontana and the term was given by Bowman.
  • It is the largest part of the nucleus which occupies about 35% of the nucleus.
  • It is the site of the synthesis of r-RNA (except 5S RNA) and subunit of RNA.
  • It is a non-membranous structure composed of proteins, Ca+2 ions, and r-RNA but lacks DNA.
  • The nucleolus stores RNA and synthesizes both subunits of ribosomes. DNA which synthesizes RNA is known as r-DNA of nucleolus.
  • The nucleolus is composed of four parts:
  • Pars Granulosa: Granular part of nucleolus
  • Pars Amorpha: Amorphous part
  • Pars Fibrosa: Fibrous part
  • Pars Chromosoma: Chromosomal part
  • The nucleolus is found at the secondary constriction (nucleolus-organizing region, NOR) of the chromosome.
  • In haploid cells, generally, it is 1 and in diploid cells, it is 1-4.
  • A maximum number of nucleoli is found in the oocyte of Xenopus (1600 nucleoli).

Chromatin Network

  • During the interphase stage of cell division, nucleoplasm contains a fine network-like structure called a chromatin network.
  • The term “chromatin” was coined by W. Flemming.
  • Chromatin is fine fiber composed of DNA, RNA, and protein that takes differential stains in basic dye (Feulgen, acetocarmine, hematoxylin). This staining technique is called heterozygosis.
  • The chromatin network becomes condensed during cell division and forms rod-like structures called chromosomes.

Types Of Chromatin

  1. Heterochromatin: The dark stain part of chromatin which is genetically inactive and has highly coiled or tightly packed DNA with histone is called heterochromatin.
  2. Euchromatin: The light stain part of chromatin which is genetically active and DNA is loosely packed with histone transcription occurs in euchromatin.

Types Of Heterochromatin

  1. Facultative: It is the inactive and condensed part of one of the two homologous chromosomes that take part in the formation of bar body, Y-spot, and drum stick.
  2. Constitutive: It is rich in repetitive DNA and present in the same region of both chromosomes of a homologous pair.

Chromosome: The word chromosome has two words: chroma and soma. Chroma means color and soma means body.

  • A chromosome is a rod-like or thread-like structure present in the nucleus of the eukaryotic cell and only visible during cell division.
  • The chromosomal shape is observed during anaphase, but the size is measured during metaphase due to maxi¬mum coiling.
  • Chromosomes were discovered by Hofmeister in the pollen mother cell of Tradescantia.
  • It was named as chromosome by Waldeyer for its deeply straining nature with basic dye.
  • The number of chromosomes in a particular species is fixed.
  • Chromosomes are composed of two chromatids which are held together at a point called centromere.
  • Each chromatid is composed of an elongated fiber called chromonemata. Chromonemata bear several beaded structures called chromomere. The outer covering of the chromosome is called a pellicle.

Maximum Number Of Chromosomes: In higher plants, Poa literosa (Poaceac family) has 266 chromosomes.

  • In lower plants, Ophioglossum (pteridophyte; Adder’s tangus fern) has 1262 chromosomes.
  • In animals, Aulacantha (radiolarians) has 1600 chromosomes.

Minimum Number Of Chromosomes

  • In higher plants, Haplopappus gracilis (Asteraceae) has four chromosomes.
  • In lower plants, Mucor has two chromosomes.
  • In animals, Ascaris has two chromosomes.

Centromere: The Centromere is technically the non-chromatid part of the chromosome. It is the attachment point of spindle fiber during cell division. The attachment point of the centromere is called the kinetochore. A chromosome can fold at the centromere, hence this point is called primary constriction.

NEET Biology Cell The Unit Of Life A Chromosomes Bearing Kinetochore

Types Of Chromosomes: On the basis of the position of centromere, chromosomes are of four types

  1. Metacentric: Centromere in the central position (V shape).
  2. Submetacentric: Centromere in the subcentral position (L shape).
  3. Acrocentric: Centromere in the subterminal position (J shape).
  4. Telocentric: Centromere in the terminal position (I shape).
  5. Acrocentric Or Holocentric: Centromere absent (I shape).

NEET Biology Cell The Unit Of Life Types Of Chromosomes Based On The Position Of Centromere

Salient Features Of Chromosomes

  • The part of the chromosome after the secondary constriction is called a satellite.
  • Chromosome-bearing satellite is called SAT-chromo-some (marker chromosome).
  • SAT stands for sine acid thymonucleinico or satellite.
  • Secondary constriction is the site of the formation of nucleolus, hence called nucleolus-organizing region (NOR).
  • The terminal end of the chromosome is called the telomere which is the point of chromosome sealing.
  • The telomere is the sealing end of a chromosome.
  • Nucleosome is the fundamental stage of the packaging of the eukaryotic chromosomal DNA.
  • The term “nucleosome” was coined by Oudet. According to this model, the nucleosome represents the bead-on-string organization.
  • Metaphasic chromosomes are composed of two DNA molecules.
  • Anaphasic chromosomes are composed of one DNA molecule.
  • The chromosomes present in the plastids are called as plastogenes.

The chromosomes present in the mitochondria are called chondriogenes.

  • Q-banding: It is the AT-rich region of the chromosome.
  • C-banding: It is the constitutive heterochromatin part.
  • R-banding: It is the sulfur-deficient region.
  • G-banding: It is the sulfur-rich region and low GC content of chromosomes.

Functions Of Chromosomes

  • Since the number of chromosomes in a particular species is constant, hence it is used to determine species.
  • It plays an important role in the cell division of eukaryotic cell.

Special Types Of Chromosomes: Some special types of chromosomes or supplementary chromosomes are:

  1. Giant chromosomes
    • Lampbrush chromosome
    • Salivary chromosome
  2. Small-sized chromosomes
    • β-chromosome

Lampbrush Chromosome Or Diplotene Chromosome: The Lampbrush chromosome was discovered by Ruckert in the oocyte of a shark, although first observed by W. Flemming in an amphibian oocyte. It was also found in the oocytes of several vertebrates and invertebrates which produce large-sized and yolky eggs.

NEET Biology Cell The Unit Of Life Lampbrush Chromosome

  • It is suggested that the loop represents an operon composed of several cistrons which are composed of spacer DNA.
  • It is the largest and reproductive chromosome which is three times larger than polytene or salivary chromosome.
  • It is a bivalent chromosome found in homologous pair and present permanently in the diplotene stage of prophase 2 of mitosis, hence called diplotene chromo¬some which does not undergo cell cycle.
  • Each chromosome is composed of two parts: the main axis and loop.
  • The main axis is composed of two chromatids that run parallel to each other and each chromatid is composed of DNA coated with RNA and protein.
  • The loop axis is composed of DNA and the matrix is composed of RNA and NH-protein.
  • Each loop is the active site of RNA synthesis (m- RNA), hence also the site of the formation of yolk (protein).
  • If the loop is treated with DNAase enzyme, then the loop axis dissolves. It indicated that the loop is composed of DNA.
  • If the loop is treated with RNAase enzyme, then the matrix gets dissolved. It means the matrix is composed of RNA and protein.

Salivary Or Polytene Chromosome: The salivary chromosome was discovered by Balbini from the salivary gland of the Chironomus larva (insect). It is also found in the salivary gland and mal-pighian tubule of Drosophila as well as the endosperm and antipodal cells of plants.It is a multithreaded chromosome, composed of several chromonemata, hence called polytene chromosome.

NEET Biology Cell The Unit Of Life Polytene Chromosome

  • Number Of strands In Drosophila: 1024
  • Number Of Strands In Chironomus: 4096
  • The term “polytene chromosome” was coined by Roller and Railing.
  • It is the second largest somatic chromosome which is 2000 times larger than the mitotic metaphase chromosome of Drosophila.
  • Each polytene chromosome represents a pair of homologous chromosomes which after repetitive endomitosis form multi-stranded structures, but all strands remain attached to a common centromere called chromocenter.
  • After staining, the polytene chromosome transverse band appears which is composed of dark and light bands alternating each other.
  • It also shows swelling at several intervals.
  • A dark band rich in DNA is considered a chromomere where DNA is supercoiled.
  • The inter or light band represents the inactive part of the chromonemata.
  • Swelling is composed of several loops called puffs or Balbiani rings which represent the active site of RNA synthesis

β-chromosome: γ-chromosome is much smaller than the normal chromosome found in maize

  • It is heterochromatin and genetically inert.
  • It has a slow rate of replication and get destroyed later on.
  • It affects the viability of seeds and the fertility of chromosomes.

Karyotype And Idiogram: A characteristic pattern of a set of chromosomes present in a diploid somatic cell is called karyotype. The diagrammatic representation of a karyotype is called an ideogram. All chromosomes of the cell are present at the mitotic metaphase stage and arranged in the order of decreasing size.

NEET Biology Cell The Unit Of Life Karyotype Of Human Male

NEET Biology Cell The Unit Of Life Karyotype Of Human

  1. SAT (Chromosome): 13, 14, 15, 21, 22
  2. NOR (Chromosome): 13, 14, 15,20, 21
  3. Smallest Autosomal Chromosome: 21st pair
  4. Smallest Sex Chromosome: Y chromosome

Types Of Karyotype

  • Asymmetric: Karyotype having few metacentric and few other telocentric chromosomes, and there is much difference between the smallest and the largest chromosomes. It indicates advanced features of the organism.
  • Symmetric: Karyotypes with several metacentric chromosomes, and there is a gradual change in the size of chromosomes. It indicates a primitive type of karyotype.

Significance Of Karyotype

  • With the help of karyotype, any change in the chromosome and any abnormality can be detected.
  • It indicates the primitiveness and advancement of the organism.

Human Karyotype

  • In humans, WBC is used in the preparation of karyotypes.
  • WBC is first treated with colchicine to arrest cell inducting division.
  • It is further stained with fuelgen and an idiogram is prepared.
  • Tejo and Levan proposed the diploid number of chromosomes in humans for example 23 pairs= 46 chromosomes.
  • Denver classified these chromosomes into seven groups represented by the karyotype formula.

Types Of Chromosomes Based On Karyotype Formula

NEET Biology Cell The Unit Of Life Types Of Chromosomes based On Karyotype Formula

Significance Of Karyotype

  • Evolutionary relationship between different species by karyotype analysis.
  • Various types of abnormalities present in chromosomes can be easily identified.
  • Indicates primitiveness or advancement of the organism.

Genome: A haploid set of chromosomes is called genome.

Plasma: Hereditary factors or genes present in the cytoplasm are called as plasma.

 

NEET Biology Notes Cell The Unit Of Life Assertion Reasoning Type Questions And Answers

In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Mark the correct answer.

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
  2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.
  3. If Assertion is true, but Reason is false.
  4. If both Assertion and Reason are false.

Question 1. Assertion: RBC membrane is highly flexible.

Reason: The amount of external protein in the cytoplasmic face of the membrane is more.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 2. Assertion: Cells of Zona reticularis contain a large number of SER.

Reason: They are present in the adrenal cortex.

Answer: 2. If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Question 3. Assertion: Centriole does not form any compartment in a cell.

Reason: Centriole is a non-membranous cell organelle.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 4. Assertion: Janus green B is a vital stain for locating mitochondria.

Reason: Janus green is oxidized by cytochrome a-, present in mitochondria.

Answer: 1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

Question 5. Assertion: Lysosomes help in the digestion of foreign particles in the animal cells.

Reason: They have respiratory enzymes.

Answer: 3. If Assertion is true, but Reason is false.