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Important Questions for Class 12 Biology Chapter 11- Organisms And Populations Very Short Questions and Answers

Question 1. Win do cattle and goats generally not browse on Caiotropis plants growing in an abandoned field? Give any one reason.
Answer: Caiotropis produces highly poisonous cardiac glycosides so cattle or goats never browse on this plant.

OR
Give two reasons as to why a weed such as Caiotropis flourishes in abandoned fields,
Answer: Dry hairy seeds help in dissemination or are not grazed by animals as they produce poisonous substances.

Important Questions for Class 12 Biology Chapter 11- Organisms And PopulationsShort Question And Answers

Question 1.

1. Observe the schematic representation given above and answer the following questions :

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1. Identify A and B.
2. Calculate the growth rate of bacteria in a curd sample, where I million bacteria increased to two million, within one hour.

Answer: A – Mortality B – Natality

The per cent growth or birth rate per individual

1. A – Mortality B – Natality
2. The per cent growth or birth rate per individual

= \(\frac{\text { Final population }- \text { Initial population }}{\text { Initial population }} \times 100\)

= \( \frac{2 \text { Million }-1 \text { Million }}{1 \text { Million }} \times 100\)

= \(\frac{1}{1} \times 100=100 \%\)

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2. Identify the type of pyramid given above. Write (lie identifying feature based on which of you identified it.

Answer: The given pyramid is expanding. The population of pre-reproductive is higher than the post-reproductive population, which makes the expanding pyramid of population.

Question 2. Explain the pollination mechanism involved in the co-evolution of the two species, namely Ophrys (orchid) and its insect pollinator bees (& bumble bees).
Answer:

• Orchids show a bewildering diversity of floral patterns many of which have evolved to attract the right pollinator insect (bees and bumblebees) and ensure guaranteed pollination by it
• The Mediterranean orchid Ophrys employs ‘sexual deceit’ to get pollination done by a species of bee. One petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings.
• The male bee is attracted to what it perceives as a female, ‘pseudocopulates’ with the flower, and during that process is dusted with pollen from the flower When this same bee pseudocopulates’ with another flower, it transfers pollen to it and thus, pollinates the flower, so we can see how co-evolution operates.
• If the female bee’s colour patterns change even slightly for any reason during evolution, pollination success will be reduced unless the orchid flower co-evolves to maintain the resemblance of its petal to the female bee

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Question 3.Given below an equation describing the growth pattern of a population: dN/dt = rN

1. Mention the type of growth in the model or growth pattern of the population described by the given equation.
2. What does V in the equation signify?
3. Mention the type of growth curve that will be obtained if the population density (N) is plotted against time (t).
4. According to you, will the resource availability be limited or unlimited for (his type of growth in a given population?

Answer:

1. Exponential growth
2. The r in this equation is called the ‘intrinsic rate of natural increase’
3. Logistic Growth Curve
4. The resource availability will be limited for this type of growth in a given population

Question 4. Describe the two basic processes which contribute to an increase in the population density of an area.
Answer:

Natality and Immigration contribute to an increase in population density.

1. Natality refers to the number of births during a given period in the population that are added to the initial density.
2. Immigration is the number of individuals of fee same that have come into the habitat from elsewhere during the period under consideration.

Question 5. Explain commensalism with the help of an example from the animal world.
Answer: Commensalism is the interaction in which one species benefits and the other is neither harmed nor benefited.

• An example of commensalism is the interaction between a sea anemone that has stinging tentacles and the clown fish that lives among them.
• The fish gets protection from predators which stay away from the stinging tentacles. The anemone does not appear to derive any benefit from hosting the clownfish.

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Question 6. Why do some organisms enter into diapause while some others into aestivation? Give one example of each of such organisms.
Answer:

• Diapause: To avoid unfavourable conditions, Example-Zooplankton
• Aestivation: To avoid summer-related problems or to avoid heat/desiccation, for Example, Snail Fish.

Question 7. If in a population of size ‘N’, the birth rate is represented as it and the death rate as ‘d’ the increase or decrease in ‘N’ during a unit period ‘t’ will he :

⇒\(\frac{d N}{d T}=(b-d) \times N\)

The equation given above can also be represented as :

⇒\(\frac{d N}{d T}=r \times N\)

What does V represent? Write any one significance of calculating V for any population.

Answer:

r = intrinsic rate of natural increase, it is an important parameter for assessing the impacts of any biotic or abiotic factor on population growth.

Question 8. Explain (the role played by predators in a community.
Answer:

• Acting as ‘conduits’ for energy transfer across trophic levels, predators play important roles. They keep prey populations under control.
• Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species.

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Question 9. Explain the difference between commensalism and mutualism types of interactions, with the help of a suitable example of each. 
Answer:

1. Mutualism – Both the interacting species are benefitted example Uehen or Mycorrhizae or Fig And Wasp
2. Commensalism – One species benefits and the other does not benefit from loss example orchid gunving on a mango branch or a barnacle on the back of a whale or cattle egret And grazing cattle.

Question 10. The graph given below shows the different types of growth curves of different species.

Answer the questions :

1. Name the type of growth curve ‘as shown in the graph.
2. State one reason the growth equation ‘b’ is said to be logistic.
3. what is ‘K’ representing in the equation ⇒\(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{rN}\left[\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right]\) given along the logistic curve?

Answer:

1. Exponential or geometries or T-shaped
2. Resources for the growth of most animal populations are finite and become limiting sooner or later
3. Carrying capacity.

Question 11. Differentiate between an ’Expanding age pyramid’ and a ’Stable age pyramid’. Substantiate your answer with diagrams.
Answer:

Question 12. Explain with the help of an example each any three ways the ecologists use to measure the population density of different organisms rather than by calculating their absolute number.
Answer:

• By measuring the per cent cover or biomass which may be more meaningful, in cases like in an area where a large number of Parthenium is there but only one banyan tree densities of microorganisms in a culture medium ”
• Measuring relative densities instead of absolute densities of organisms, eg the number of fish caught per trap in a lake is good enough to estimate population size.

By estimating the population size indirectly without actually seeing or counting them, e.g counting tiger population in national parks based on their pug marks or faecal pellets

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Question 13. Explain the logistic growth pattern of a population. Why do population growth patterns of all organisms ultimately follow it?
Answer:

• A population growing in a habitat with limited resources, shows a lag phase, followed by phases of acceleration and deceleration and finally an asymptote when the population density reaches the carrying capacity,.
• A plot of population density about time results in sigmoid curve Since resources for growth of most organisms are finite, and become limiting sooner or later the logistic growth pattern is ultimately followed.

Question 14. Explain with the help of two examples how certain plants have evolved morphological and chemical defences against primary consumers such as cows and goats.
Answer:

Plants have evolved an astonishing variety of morphological and chemical defences against herbivores.

• Thorns (Acacia, Cactus)are the most common morphological means of defence Many plants produce and store chemicals that make the herbivore sick when they are eaten, inhibit feeding or digestion, disrupt its reproduction or even kill it.
• You must have seen the weed Caloiropisgrowing in abandoned fields The plant produces highly poisonous cardiac glycosides and that is why you never see any cattle or goats browsing on this plant.
• A wide variety of chemical substances that we extract from plants on a commercial scale (nicotine, caffeine, quinine, strychnine, opium, etc.) are produced by them actually as defences against grazers and browsers

Important Questions for Class 12 Biology Chapter 11- Organisms and Populations Long Question And Answers

Question 1. Study the age pyramids; A’, ‘B’ and ‘c’ Of the human population given below the questions that follow :

1. Identify pyramids ‘B’ and ‘C’
2. Write the basis on which the above pyramids are plotted.

Answer:

1. B-Stable population
   1. C- Declining population
2. Age Distribution of male and female human population

Question 2.

1. Compare, giving reasons, the (J-shaped and S-shaped) models of population growth of a species.
Answer:

Question 3. Explain “fitness of a species” as mentioned by Darw in.
Answer: When resources are limited. Competition occurs between individuals, the fittest will survive, and reproduce to leave more progeny

Important Questions for Class 12 Biology Chapter 11- Organisms and Populations Case Study-Based Question And Answers

Question 1. Read the following passage and answer any four questions from 1 to 5.

• Acacia plants are particularly common in drier tropical and subtropical environments in the world.
• The swollen thorn acacias, form obligate mutualisms with Pseudomyrmex.
• A species of ants restricted to the New World Swollen thorn acacias show several characteristics related to their obligate association with ants, including enlarged thorns with a soft, easily excavated pith; year-round leaf production, enlarged foliar nectaries; and leaflet tips modified into concentrated food sources called Beltian bodies.
• The thorns provide living space, while the foliar nectaries provide a source of sugar and liquid.
• Beltian bodies are a source of oils and protein. Resident ants vigorously guard these resources against encroachment by nearly all comers, including other plants.

1. The association between the genus of Acaci and species of ants depict population interactions, known as :

1. Competition
2. Amensalism
3. Mutualism
4. Predation

Answer: 1. Mutualism

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2. In exchange for food and shelter, ants protect Acacias from the attacks of:

1. Fungi
2. Bacteria
3. Herbivores
4. Carnivores

Answer: 3. Herbivores

3. The above interaction suggests that the relationship between the two species is an example of:

1. Competitive release
2. Competitive exclusion
3. Co-evolution
4. Resource partitioning

Answer: 3. Co-evolution

4. The removal of resident ants from the Acacias will lead to :

1. Reduced growth of Acacias
2. Increased growth of Acacias
3. Reduced population of ant species
4. Increased population of ant species

Choose the correct alternative from the above statements :

1. Only 1 is true
2. 1 and 3 are true
3. 3 and 4 are true
4. 1 and 4 are true

Answer: 2. 1 and 3 are true

5. Given below is a graphical representation of ants and the Acacia shoots with an abundance of herbivorous insects :

The conclusion drawn from the above data is :

1. Acacia shoots will have higher rates of growth with resident ant species.
2. Acacia shoots will have a neutral effect on growth with or without resident ant species
3. Acacia shoots will have higher rates of growth without resident ant species.
4. Growth of Acacia shoots is independent of resident ant species

Answer: 3. Acacia shoots will have higher rates of growth without resident ant species.

Important Questions For Class 12 Biology Chapter 10 Biotechnology And its Applications Very Short Questions And Answers

Question 1. Nematode Specific Genes were introduced into the tobacco host plant using a vector

1. pBR322
2. Plasmid
3. Bacteriophage
4. Agrobacterium

Answer: 4. Agrobacterium

Question 2. ‘Cry genes’ that code for insecticidal toxins are present in

1. Cotton bollworms
2. Nematodes
3. Corn borer
4. Bacillus

Answer: 4. Bacillus thuringiensis

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Question 3. Mention the chemical change that pro-insulin undergoes, to be able to act as mature insulin.
Answer: Removal of C – peptide (from pro-insulin)

Question 4. What are Cry genes? In which organism are they present H
Answer: The genes that code for lit toxin Cry proteins or toxic proteins are present in Bacillus thingies

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Important Questions For Class 12 Biology Chapter 10 Biotechnology And its Applications

Short Question And Answers

Question 1. Name the Indian crop variety for which in 1997 an American company got a patent right through the IS Patent and Trademark Office. Why did the company claim it to be an invention or a novelty?
Answer:

• An American company got patent rights on Basmati rice through the US Patent and Trademark Office.
• This allowed the company to sell a ‘new’ variety of Basmati in the US and abroad
• This ‘new’ variety of Basmati had been derived from Indian farmer’s varieties. Indian Basmati was crossed with semi-dwarf varieties and claimed as an invention or a novelty.

Question 2. Name one toxin gene isolated from B.thuringiensis and its target pest.
Answer: Toxin gene crvlAc cell Ab targets pest-cotton Bol 1 worm crylAb, controls corn borer

OR
Why does the toxin produced by B.thuringiemis not kill the Bacillus?
Answer: Bt Toxin protein exists as inactive protoxins, the inactive toxin is converted into an active form of toxin only in the presence of the alkaline pH which is not available in the Bacillus.

Question 3. What are cry proteins? With the help of a suitable example, explain how it acts as a biological pesticide.
Answer:

• Cry Protein The insecticidal protein which is produced by the soil bacterium Bacillus thuringiensis is called cry protein
• For example, proteins encoded by the genes cryl.Ac and cryllAb control the cotton bollworms. The club controls corn borer.

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Question 4.

1. One of the potential uses of genetic engineering is in the correction of a gene defect that has been diagnosed in a child or embryo, Explain how gene therapy is of help in ADA deficiency.
Answer:

• Gene therapy is a collection of methods that allows the correction of a gene defect that has been diagnosed in a child or embryo. Here genes are inserted into a person’s cells and tissues to treat a disease. Correction of a genetic defect involves the delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non-functional gene.
• The disorder is caused due to the deletion of the gene for adenosine deaminase. In some children ADA deficiency can be cured by bone marrow transplantation: in others, it can be treated by enzyme replacement therapy, in which functional ADA is given to tire patients by injection. However, the problem with both of these approaches is that they are not completely curative.
• As a first step towards gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient, as these cells are not immortal, the patient requires a periodic infusion of such genetically engineered lymphocytes, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure

2. A schematic diagram of matured human insulin is given below :

How is the process of its formation naturally in the human body different from that of its formation by rDNA technology? Explain,

• Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulfide bridges In mammals, including humans, insulin is synthesized as a pro-hormone which contains an extra stretch called the C peptide.
• This peptide is not present in the mature insulin and is removed during maturation into insulin. The main challenge for the production of insulin using rDNA techniques was getting insulin assembled into a mature form.
• Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin, and introduced them in plasmids to produce insulin chains.
• Chains A and B were produced separately, extracted, and combined by creating disulfide bonds to form human insu1in

Question 5. What are transgenic animals? How are they being used for vaccine safety and chemical safety testing? Explain.
Answer:

Transgenic animals: Animals that have had their DNA manipulated to possess and express an extra or foreign or transgene.

• Transgenic mice are being developed for use in testing the safety of the vaccine before they are used in humans or transgenic mice are being used to test the safety of the polio vaccine, if successful and found reliable they could replace the use of monkeys to test the safety of batches of the vaccine.
• Transgenic animals are made that carry genes that make them more sensitive to toxic substances than non-transgenic animals, they are exposed to the toxic substances, and the effects are studied (which allows to obtain results in less time)

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Question 6. A child is born with ADA-deficiency

1. Suggest an ex-phi in a procedure for a possible life-long (permanent) cure.
2. Name any other possible treatment for this disease.

Answer:

1. In gene therapy, lymphocytes from the blood of a patient are grown in a culture outside the body, functional ADA cDNA is introduced into these lymphocytes, and these cells are returned to the patient’s body at an early embryonic stage.
2. Bone marrow transplantation, enzyme replacement therapy

Question 7. How has the use of agrobacterium as vectors helped in controlling Meloidegyne incognita infestation in tobacco plants? Explain in the correct sequence.
Answer:

• Using the Agrohucteriwti vector nematode specific genes are introduced into the host plant.
• Sense and antisense strands of mRNA are produced.
• ds RNA is formed.
• ds RNA initiates RNAi
• Prevents translation of mRNA or silencing of mRNA of parasite or nematode.
• The parasite will not survive.

Question 8.

1. What are transgenic animals?
2. Name the transgenic animal having the largest number amongst all the existing transgenic animals.
3. Mention any three purposes for which these animals are produced.

Answer:

1. Animals that have had their DNA manipulated to possess and express an extra/foreign gene
2. Mice
3. Normal physiology and development
   1. Study of disease.
   2. Biological products
   3. Vaccine safety
   4. Chemical safety testing

Question 9. Explain the various steps involved in the production of artificial insulin.
Answer: Two DNA sequences corresponding to A and B polypeptide chains of human insulin were prepared, and these were introduced into Kco/I to produce A and B chains separately, these chains were extracted and combined by creating disulfide bonds.

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Question 10. What was the challenge for the production of insulin using rDNA techniques? How did Eli Lilly produce insulin using rDNA technology?
OR
Explain how Eli Lilly, an American company produced insulin by recombinant DNA technology.
Answer:

• In mammals, including humans, insulin is synthesized as a pro-hormone which contains an extra stretch called the C peptide. This C peptide is not present in the mature insulin and is removed during maturation into insulin.
• The main challenge for the production of insulin using rDNA techniques was getting insulin assembled into a mature form. In 1983, Eli Lilly an American company prepared two DN A sequences corresponding to A and B, chains of human insulin, and introduced them in plasmids of Jf.coli to produce insulin chains.
• Chains A and B were produced separately, extracted, and combined by creating disulfide bonds to form human insulin.

Question 11. Name the organism from which the ‘cry’ genes are isolated. Mention with the help of suitable examples why and how bio-technologists have made use of ‘cry’ genes.
Answer:

• Bt toxin genes were isolated from Bacillus t and incorporated into several crop plants such as cotton. The choice of genes depends upon the crop and the targeted pest, as most – Bt toxins are insect-group sped tic.
• The toxin is coded by a gene crylAc named cry. There are a number of them, for example, the proteins encoded by the genes crylAc and cryllAb control the cotton bollworms, and that of crylAb controls corn borer.
• B. thingie misforms protein cn stats during a particular phase of their growth. These crystals contain a toxic insecticidal protein. The Bt toxin protein exists as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilises the crystals.

The activated toxin binds to the surface of midgut epithelial cells and creates pores. that cause cell swelling and lysis and eventually cause the death of the insect.

Important Questions For Class 12 Biology Chapter 10 Biotechnology And its Applications Long Question And Answers

Question 1. Read the following paragraph and answer the questions that follow :

Biotechnology revolves around the “gene of interest”, to open various avenues for human welfare in health, medicine pharma, agriculture etc, using different techniques, tools, and processes. One of the breakthroughs of biotechnology in medicine is the gene therapy.

1. Name the human disease for which gene therapy was used for the first time.
2. Explain the steps of gene therapy carried out to cure (the disease using the lymphocytes of (lie patient. Why is (his therapy not a permanent cure for the disease?
3. Write the possible permanent cure of the disease by the gene therapy that is in progress.

Answer:

1. ADA deficiency
2. As a first step towards gene therapy. lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a “retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes.
3. If the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

Question 2.RINA interference (RNAi) holds great potential as a therapeutic agent for the treatment of human diseases and as a biocontrol agent for controlling pests in the field of agriculture.
Answer:

• The graph given below illustrates the use of RNAi for the potential treatment of disorders of cholesterol metabolism. Some people possess genetic mutations with elevated levels of the ‘ApoB’ gene which predisposes them to coronary artery diseases Lowering the amount of’ApoB’ can reduce the number of lipoproteins and lower the blood cholesterol.
• Tracy Zimmerman and her colleagues used RNAi in 2006 to reduce the level of of’ApoB’ in non-human primates Cynomofgus monkeys One group of monkeys was given RNAi treatment (Small interfering RNAs. siRNAs) (doses mg/kg siRNAs), a second group of monkeys were given RNAi treatment (doses 2.5 mg/kg siRNAs) and a third group of monkeys were injected with saline as control

The results of the experiments are illustrated in the graph given below :

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1. Write your interpretation from the bars X and Z obtained after 264 hours of treatment of monkeys with saline and 2.5 night siRNAs treatment respectively, in comparison to the bars obtained after 24 hours of treatment with saline and 2.5 mg/kg siRNAs.
Answer:

• Lowering the amount of’ApoB’ can reduce the number of lipoproteins and lower the blood cholesterol. According to the graph, we can infer that administering RNAi to the monkeys did indeed reduce the level of the ‘ApoB’ gene which in turn led to lower levels of Relative serum cholesterol
• Also, the reduction in Relative serum cholesterol level was found to be directly proportional to the amount of Si RNA administered, that is, the group of monkeys that were given RNAi treatment (Small interfering RNAs, siRNAs) (doses 1 mg/kg siRNAs) had less Relative serum cholesterol than the group of monkeys were injected saline as control and more Relative serum cholesterol than the group of monkeys was given RNAi treatment (doses 2 5 mg/kg siRNAs).

2. A tobacco plant made transgenic using RNA interference is protected from the parasite Meloidogym incognitia. How is the transgenic tobacco plant able to prevent itself from infestation by the nematode? Explain briefly,
Answer:

1. A nematode Mdoidegyne incognitia infects the roots of tobacco plants and causes a great reduction in yield. A novel strategy was adopted to prevent this infestation which was based on the process of RNA interference (RNAi) RNAi takes place in all eukaryotic organisms as a method of cellular defense.
2. This method involves the silencing of a specific mRN A due to a complementary dsRNA molecule that binds to and prevents translation of the mRN A (silencing).
   The source of this complementary RNA could be infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate. I sing Agrobacterium vectors, nematode-specific genes were introduced into the host plant. The introduction of DNA was such that it produced both sense and anti-sense RNA in the host cells.
3. These two RNAs being complementary to each other formed a double-stranded (dsRNA) that initiated RNAi and thus, silenced the specific mRNA A of the nematode. The consequence was that the parasite could not survive in a transgenic host expressing specific interfering RNA. The transgenic plant therefore got itself protected from the parasite.

Question 3. Explain the three different approaches used in the treatment of a person suffering from Adenosine Deaminase (ADA) Deficiency.
Answer:

1. In some children ADA deficiency can be cured by bone marrow transplantation; in others
2. It can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection.
3. However, the problem with both of these approaches is that they are not completely curative. As a first step towards gene therapy, lymphocytes from the patient’s blood are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are returned to the patient.
4. However, as these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

Question 4. Insulin in the human body is secreted by the pancreas as prohormone or proinsulin. The schematic poly pet ide structure of proinsulin is given below. This proinsulin needs to undergo processing before it becomes functional in the body. Answer the questions that follow

1. State the change the proinsulin undergoes at the time of its processing to become functional.
2. Name the technique the American company Eli Lilly used to commercialize human insulin.

How are the two polypeptides of a functional I insulin chemically held together?

Answer:

1. a ‘C’ Peptide is removed
2. rDNA technology or Recombinant DNA Technology.
3. Disulfide bonds.

Important Questions for Class 12 Biology Chapter 9 – Biotechnology: Principles And Processes Very Short Questions And Answers

Question 1. After the separation of DNA fragments by gel electrophoresis and staining with ethidium bromide, a student placed the gel in the UV chamber under the I-V light. State a reason for doing so.
Answer:

Explanation: After Gel electrophoresis the separated DNA is visualized after staining in ethidium bromide followed by exposure to UV light. The stained DN A fragments appear as orange-contoured bands.

Question 2.Assertion: When DN A from two different sources are cut fey the same restriction enzyme, the resultant DNA fragments have different kinds of sticky ends’.
Reason: These can be joined together end-to-end using DN A ligases.

1. Both Assertion and Reason are true, and Reason is the correct explanation. Assertion,
2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
3. The assertion is true, but Reason is false.
4. Both Assertion and Reason are false.

Answer: 1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

Question 3. In biotechnology experiments, ‘molecular scissors’ used are

1. Plasmid
2. Restriction enzymes
3. Vectors
4. Sigma factor

Answer: 2. Restriction enzymes.

Question 4. Why do DNA fragments move towards the anode during gel electrophoresis?
Answer: DNA fragments are negatively charged.

Question 5. Write the function of a Bioreactor. 
Answer: To produce the recombinant product in large quantities

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Important Questions for Class 12 Biology Chapter 9 – Biotechnology: Principles And Processes Very Short Questions And Answers Short Question And Answers

Question 1. How is the use of “microinjection” different from using the ‘method of biolistics’ in biotechnology? Explain.
Answer:

Recombinant DNA is directly injected into the nucleus of an animal cell in a method called microinjection. In another method, cells are bombarded with high-velocity micro-particles of gold or tungsten coated with DNA in a method known as biolistics or gene gun.

Question 2. How are DN A fragments visualized during gel-elect phoresis? What is elution?
Answer: Separated DNA fragments stained with ethidium bromide, followed by exposure to UV radiations, removal of DNA bands from agarose gel, and its extraction from gel is elution.

Question 3. Explain the principle that helps in the separation of DMA fragments in Gel electrophoresis.
Answer:

Since DNA fragments are actively charged molecules they can be separated by forcing them to run e towards an anode-pole under an electric field through a medium (matrix), DNA fragments separate according to their through-seeing effect picked in agarose gel (matrix)

Question 4. 

1. Given below is the stepwise schematic representation of the process of electrophoresis.

Identify the ‘alphabets’ representing

1. Anode end
2. The smallest or lightest DNA strand in the matrix
3. Agarose gel

2. What is elution? State the importance of elution in this process.
Answer:

1. Anode- S end
2. R
3. T

2. The separated bands of DN A are cut out from the agarose gel and extracted from the gel piece. This step is known as elution.

Importance:- The DNA fragments purified in this way are used in constructing recombinant DN A by joining them with cloning vectors.

Question 5.

1. Simple stirred-tank bioreactors are used to produce large quantities of recombinant proteins, stirring the contents and mixing them with oxygen. Write any four other advantages of using a stirred tank.
Answer:

• A stirred-tank reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents. The stirrer facilitates oxygen availability throughout the bioreactor that is tire bioreactor has an agitator system, an oxygen delivery system, a foam control system, and a temperature control system.
• pH control system, fermentation is contamination-free, and sampling ports so that small volumes of the culture can be withdrawn periodically.

2. After downstream processing, the product of the biosynthetic stage cannot be marketed directly. Why? Give two reasons.
Answer:

• After completion of the biosynthetic stage, the product has to be subjected to a series of processes before it is ready for marketing as a finished product The processes include separation and pu rill a ton. which are collectively referred to as downstream processing
• The product has to be formulated with suitable preservatives Such formulation has to undergo thorough clinical trials as in the case of drugs Strict quality control testing for each product is also required.
• The downstream processing and quality control testing vary from product to product.

Question 6. Explain only with the help of a self-explanatory diagram, the three basic steps of Polymerase Chain Reaction (PCR),
OR
Explain three steps involved in polymerase chain reaction.
OR
Describe the technique that is very effectively used to get a large amount of desired DNA for research and detailed investigation.
Answer:

• By using PCR (polymerase chain reaction) denatured desired DNA,
• Separate into 2 strands white each acting as a template
• For each strand, a separate set of pr mortised (two times).
• With the help of dextrin) nucleotides and I as polymerase DNA polvoiearase isolated from Thermus aquaticus)
• Results in extension of DNA primer

The process of PCR takes place in three steps

• Desaturation Two strands of DNA are separated by heating Each strand acts as a template.
• Annealing Two sets of primers are attached and annealed to the separated DNA strands
• Extension Taq polymerase catalyzes the extension of primers using genomic DNA as a template and nucleotides provided in the reaction.

Question 7. Explain the three steps carried out in the formation of recombinant DNA using the enzyme EcoRI.
OR
Explain the action of EcoRI on DNA technology experiments.
Answer:

EcoRI cuts vector DNA. foreign DNA gene of interest. at palindromic site 3’CTTAAG 5’, 5’ GAATTC 3′ (between bases G and A only), sticky end (overhanging stretch of bases) formed at each strand Joining of sticky ends from DNA fragments by the enzyme DNA Ligase, Recombinant DNA(rDNA) is formed.

Question 8. Name any two natural cloning vectors. Give reasons that make them act as cloning vectors. Write the two characteristics the engineered vectors are made to possess
Answer:

Plasmids, bacteriophages, ability to replicate within bacterial cells. high copy number within the bacterial cells Characteristics of engineered Vectors: easy linking of foreign DNA, Selection of recombinants from non-recombinants or selectable markers.

Question 9. Name the most commonly used bioreactor in biotechnology labs. Mention the most essential components this bioreactor must have to provide the optimum conditions to the culture medium, resulting in the production of a large volume of the desired product.
Answer:

The most commonly used bioreactor is the stirring type.

It has the following components.

1. Stirring type agitator system,
2. O₂ Delivery System
3. Foam control system
4. Temperature control system
5. pH control system

Question 10.

1. How has the development of bioreactors helped in biotechnology?
2. Name the most commonly used bioreactor and describe its working.

Answer:

1. Larger biomass large volume of culture can be processed leading to higher yields of desired specific products (protein or enzymes), under controlled condition
2. Stirring type

• • Mixing of reactor contents evenly (with an agitator system or a stirrer)
  • Facilitates oxygen availability
  • Temperature, PH1 foam control under optimum conditions

Question 11. Explain the roles of the following with the help of an example each in recombinant)NA technology:

1. Restriction Enzymes –
2. Plasmids

Answer:

1. It recognizes a specific sequence of base pairs or palindromes. and cuts the D’NA strand at a specific site example EcoRI Hind 11
2. Act as vectors cloning of desired alien gene or foreign gene eg pBR322 or  plasmid of Salmonella or plasmid of Agrohacterimive Plasmid or Tumour including Plasmid

Question 12.

1. Explain the significance of ‘palindromic nucleotide sequence’ in the formation of recombinant DNA
Answer:

Palindromic nucleotide sequence The significance of the Palindromic nucleotide sequence is that the sequences read the same in both directions. This is important during the process of replication, transcription, and directional repair mechanisms.

2. Write the use of restriction endonuclease in the above process.
Answer:

• The same restriction endonuclease binds to both the vector and the foreign DNA, cutting each of the two strands of the double helix at specific points in their sugar-phosphate backbone of recognition sequence for restriction endonucleases ‘ palindromic sequence of vector and foreign DNA.
• Cut strands a little away from the center of the palindrome sites creates overhanging stretches or ticky ends

Question 13. Describe the roles of heat, primers, and the bacterium Thermus aquatic us in the process of PC R.
Answer:

1. Heat – Denaturation or separation of DN A into two strands
2. Primer– Enzyme DNA Polymerase extends the primers using the nucleotides provided in the reaction and the genomic DNA as a template.
3. Thermus aqimfiats – Source of thermostable DNA polymerase: Taq polymerase

Question 14.Mention the role of (1) selectable marker, (2) Ori, and (3) drop-in Ecoli cloning vector pBR322.
Answer:

Selectable marker in addition to the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

Origin of replication (ori): This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. So, if one wants to recover mam copies of the target DNA it should be cloned in a vector whose origin supports high
copy number

Repressor of primer (Rop): It is a small dimeric protein that participates in the mechanism that controls the copy number of the plasmid It codes for the proteins involved in the replication of the plasmid

Important Questions for Class 12 Biology Chapter 9 – Biotechnology: Principles And Processes Long Question And Answers

Question 1. Read the paragraph given below and answer the questions that follow:

Enzyme Taq polymerase is extracted from a eubacteria microorganism Thermits aquaticus from Yellowstone National Park in Montana. ISA and isolated by Chien et al. (1976). Taq polymerase successfully replaced the DNA polymerase from E.coli that was being used in PCR earlier and this shift revolutionized the PCR technique,

1. Taq polymerase after its discovery replaced E.coli DNA polymerase in the PCR technique. Explain giving reasons why the need felt for the change.
2. What is a primer and its importance in PCR?
3. What is the importance of PC R as a diagnostic tool?

Answer:

• The Lincoln DNA polymerase cannot cany out PCR at high temperatures (as it becomes inactive) whereas thermostable DNA polymerase (is isolated from a bacterium.
• Thermits that remain active during the high temperature-induced denaturation of double-stranded DNA.
• A primer is a small segment of DNA that binds to a complementary strand of DNA.
• Primers are necessary to start the functioning of DNA polymerase enzyme and therefore are necessary in polymerase chain reaction.
• PCR is important because it can generate several copies of a DNA sequence in a very short period.
• It is also important in forensic science as a tool for genetic engineering. Early detection of diseases like cancer AJDS genetic disorders.

Question 2. Restriction endonucleases are a class of bacterial enzymes that recognize a specific short sequence of nucleotides within a double-stranded DNA molecule The natural purpose of these enzymes is to protect bacteria from pathogens, notably bacteriophages. There are different classes of restriction enzymes, but type-11 restriction enzymes are widely used in manipulating DNA as they recognize short sequence nucleotides that are typically palindromes.

1. Name a specific restriction endonuclease and write the palindromic nucleotide sequence in the DNA recognized by this enzyme. Also, indicate the site at which it cuts.
2. A piece of DNA is cut by a restriction enzyme. The fragments are then separated by gel electrophoresis and stained by ethidium bromide. Write the principle on which gel electrophoresis works.

Answer:

1. EcoRl comes from Escherichia Coli RY 13 Recognition Sequence 5-GAATTC 3′

3’-CTTAAG-5’

The site at which it Cut 5’— G AATTC—3′

3’—CTTAA G—5′

2. The cutting of DNA by restriction endonucleases results in the fragments of DNA These fragments can be separated by a technique known as gel electrophoresis. Since DNA fragments are negatively charged molecules they can be. separated by forcing them to move towards the anode under an electric field through a medium/matrix. Nowadays the most commonly used matrix is agarose which is a natural polymer extracted from seaweeds. The DNA fragments separate (resolve) according to their size through the sieving effect provided by the agarose gel.

• Hence, the smaller the fragment size, the farther it moves. The separated DNA fragments can be visualized only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV You can see bright orange colored bands of DNA in an ethidium bromide-stained gel exposed to UV light. The separated bands of DN A are cut out from the agarose gel and extracted from the gel piece

This step is known as elution. The DNA fragments purified in this way are used in constructing recombinant DNA by joining them with cloning vectors.

Question 3. Explain how an antibiotic resistance gene in a cloning vector (plasmid pBR322) helps in selecting the recombinants front the non-recombinants. 
Answer:

• In the closing vector pBR322 the genes encoding for resistance to antibiotics such as ampicillin. tetracycline is considered as a selectable marker:
• These markers are used to identify and eliminate non-transformants and permit the growth of trial formats in media containing antibiotics.

For example, you can ligate a foreign DNA at the BartiH I site of the tetracycline resistance gene in the vector pBR322. The recombinant plasmids will lose tetracycline resistance due to the insertion of foreign DNA but can still be selected from non-recombinant ones by plating the transformants on a tetracycline-containing medium.

• The transformants growing on an ampicillin-containing medium are then transferred to a medium containing tetracycline.
• The recombinants will grow in an ampicillin-containing medium but not on that containing tetracycline. But. recombinants will grow on the medium containing both antibiotics.

Question 4.

1. Write the mechanism that enables Agrobacterium impatiens to develop tumors in their host dicot plant.
2. State how Agrobacterium tumefadmsand some retroviruses have been modified as useful cloning vectors.

Answer:

1. file bacterium Agrobacterium tumefacient delivers a piece of DN A known as T-DNA present in its plasmid to transform the host plant cell into a tumor (and direct tumor cells to produce the chemical required by the pathogen).
2. The Ti plasmid of A.tumifaciens has been disarmed 1 modified so that it is no longer pathogenic to the host plant but is still able to use the mechanism to deliver genes of interest into plants. Retroviruses have also been disarmed or modified now and are used as cloning sectors to transfer desirable genes into animal cells

Question 5.

1. Name the most commonly used bioreactor. Why are these bioreactors used?
2. How is the operation in a bioreactor carried out to achieve the desired end

Answer:

1. Stirred tank bioreactor, to obtain large quantities of desired products from the culture medium containing cloned organisms with genes of interest
2. By providing optimum growth conditions for the living materials such as temperatures of F₁ substrate salts or vitamins oxygen(and four conditions)

(OR) Explain the process of amplification of genes of interest using PCR technique.
Answer: PCR – Technique

1. Denaturation – The two strands of the gene of interest are separated as DNA templates under high-temperature
2. Annealing – The two DNA primers attached to the two separated DNA template strands
3. Extension – Taq polymerase extends the primers (in 5′ → 3′ using deoxynucleotides provided in the medium)

The Cycle is repeated to gel the multiple copies of the gene of interest

Important Questions for Class 12 Biology Chapter 9 – Biotechnology: Principles And Processes Case Study-Based Questions

Read The Following Passage And Answer Any Four Questions From 1 To 5 :

• Experiments involving cloning genes and expressing proteins require the use of host cells to receive the foreign cloned gene In some experiments, prokaryotes such as l’.. coli and Bacillus subtil is, and eukaryotes such as the budding yeast (Saccharomyces are used as host cells for DNA cloning.
• These host cells are relatively easy to grow in the laboratory and have been studied extensively for decades. Their genetics have been well-understood and therefore can be manipulated to make them appropriate hosts. Many types of cells can be converted into biochemical factories using rDNA technology to produce various kinds of biomolecules. coli and B. subtil are both commonly used as host cells for DNA cloning Fortunately, humans have become very experienced at cultivating microbes cheaply and efficiently on large and small production scales.
• Over the centuries, brewers and bakers have learned to employ yeast cells to manufacture beer, bread, and related food products. In terms of impact on human health, probably the most important product made by bacteria is antibiotics.

Question 1. The most commonly used eukaryotic microorganism used in biotechnology is :

1. E.coli
2. Bacillus
3. Saccharomyces cerevisiae
4. Drosophila

Answer: 1. E.coli

Question 2. Over the centuries, brewers and bakers have learned to employ yeast cells to manufacture man’s household products. Select the option with all the correct answers from the given list:

1. Bread. Idli, Roquefort cheese
2. Bread, Toddy, Swiss cheese
3. Dosa, Idli, Bread
4. Lipases, Pectinases, Zymase

Answer: 3. Dosa, Idli, Bread

Question 3. The most common products made by certain bacteria that have a great impact on human health are:

1. Antibiotics.
2. Bioactive molecules
3. Enzymes
4. Fermented drinks

Answer: 1. Antibiotics.

Question 4. The best-known host cells for DNA cloning and producing various kinds of biomolecules are:

1. Agrobacteriumtumefaciens
2. Escherichia
3. Bacteriophage lambda
4. Bacteriophage X174

Answer: 2. Escherichia

Question 5. The enzyme that is not required to manipulate the genetics of the microorganism to convert them into biochemical factories is :

1. Restriction endonuclease
2. DNA polymerase
3. Lactase
4. Ligase

Answer: 3. Lactase

Important Questions for Class 12 Biology Chapter 8 – Microbes In Human Welfare Question And Answers

Question 1. What for are Cyclosporin A anil Streptokinase bioactive molecules prescribed by a doctor
Answer:

• Cyclosporin A – It is used as an immunosuppressive agent.
• Streptokinase – It is used as a clot bluster for removing clots from blood vessels.

Question 2. Assertion: Large holes in ‘Swiss cheese’ are due to the production of a large amount of carbon dioxide by a specific microbe

Reason: The specificity of the characteristic texture, flavour and taste of Swiss cheese is due to the use of the bacterium Propionibacterium Sherman.

1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.
2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
3. The assertion is true, but Reason is false
4. Both Assertion and Reason are false.

Answer: 1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion

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Question 3. Some cyanobacteria in aquatic and terrestrial environments that enrich the soil by fixing atmospheric nitrogen are

1. Rhizobium and Azotobacter
2. Azospirillum and Glomus
3. Anabaena and Nostoc
4. Azospirillumand Azotobacter

Answer: 3. Anabaena and Nostoc

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 4. The microbes commonly used in kitchens are

1. Lactobacillus and Yeast
2. PeniciiHum and Yeast
3. Microspora and E.coli
4. Rhizopus and

Answer: 1. Lactobacillus and

Important Questions for Class 12 Biology Chapter 8 – Microbes In Human Welfare Short Question And Answers:

Question 1. Fanners are often suggested to use the following organisms in their cropland to improve soil fertility. Explain.

1. Rhizobium
2. Anabaena

Answer:

1. Rhizobium is a bacterium found in soil that helps in fixing nitrogen in leguminous plants
2. It attaches to the roots of the leguminous plant and produces nodules. These nodules fix atmospheric nitrogen and convert it into ammonia that can be used by the plant for its growth and development.
3. Anabaena plays a significant role in farming where it is used as a biofertilizer and soil stabilizer.

Question 2. Organic farmers use Trichoderma and liacuas as biological control agents. Explain.
Answer:

1. Trichoderma – species are free-living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens Baculoviruses are pathogens that attack Insects and other arthropods.
2. Baculovirus – The majority of baculovirus ruses used as biological control agents are in the genus Nucleopolyhedrovtrus. These viruses are excellent candidates for species-specific, narrow-spectrum insecticidal applications. They have been shown to have no negative impacts on plants, mammals, birds, fish or even on non-target insects.

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Question 3. A particular cyanobacterium is spread by (lie farmers in their fields while growing paddy). Name the cyanobacteria used and give two advantages of it.
Answer:

Cyanobacteria like Anaebena. Nosiocetc can fix atmospheric No, decompose organic wastes and residues, detoxify heavy metals, pesticides, and other xenobiotics, catalyze nutrient cycling, suppress the growth of pathogenic microorganisms in soil and water, and produce some bioactive compounds that contribute to plant growth.

Question 4. Study the given diagrams of Sewage Treatment Plants (STP) and answer the questions that follow:

1. Which of the two Sewage Treatment Plants? (A) or (B), will be more effective in treating human excreta in municipal waste?
2. How is the primary effluent treated in the aeration tanks till there is a significant reduction in the BOD of (the effluent?

Answer:

1. Sewage Treatment Plants (A) will be more effective in treating human excreta in municipal waste.
2. The primary effluent is taken to aeration tanks, where it is constantly agitated mechanically. Air is pumped into it. periodically A large number of aerobic heterotrophic microbes grow in the aeration tank to form floes.
3. Due to microbial activity, the organic matter gets digested The microbes convert it into microbial biomass and release the minerals Due to the breakdown of organic matter, the BOD (biochemical oxygen demand) of the wastewater reduces to about 10 to 15% of raw sewage.

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Question 5. Write the different components of activated sludge. Explain the different ways it can be used further in the sewage treatment process.
Answer:

• Dining secondary treatment BOD- of sewage or wastewater is reduced significantly the effluent is then passed into a settling tank where the bacterial “floes are allowed to sediment called activated sludge.
• The floes are masses of bacteria associated with fungal filaments to form mesh-like structures.

Question 6.

1. Name the genus of baculovirus that acts as a biological control agent despite being a pathogen. Justify by giving three reasons that make it an excellent candidate for the job,
Answer:

Genus – Nucleopolyhedrovirus, species-specific, Narrow spectrum. No negative impact on (plants or mammals or birds or fish) non-target organisms.

2. ‘’Micro-organisms play an important role in the biological treatment of sewage.” Justify
Answer:

Primary effluent is passed into aeration tanks, constantly agitated and the air is pumped in. This allows the growth of useful aerobic microbes into flow or masses of bacteria and fungal filaments), these microbes consume organic matter and reduce the Biochemical oxygen demand (BOD) of effluent.

Question 7.

1. Mention the difference in the level of BOD before and after the secondary treatment of sewage water.
2. Write the importance of floes during the secondary treatment of sewage.

Answer:

1. Before – High. After- Low

2. Reduces BOP Biochemical Oxygen Demand

Question 8. Expand ‘LAB’. How are LABs beneficial to humans? (Write any two benefits).
Answer:

Lactic Acid Bacteria,

1. They produce acid which partially digests the milk protein or sets milk into curd.
2. They improve nutrition and quality by producing Vitamin B.
3. Check disease-causing microbes in our stomach

Question 9. Your advice is sought to improve the nitrogen content of the soil to be used for the cultivation of a noil-leguminous terrestrial crop.

1. Recommend two microbes that can enrich the soil with nitrogen.
2. Why do leguminous crops not require such enrichment of the soil?

Answer:

1. Azospiri/fum Azotobacter Anabaena Nosioc Oscillaloriu rankia
2. They can fix atmospheric nitrogen, due to the presence of Rhizobium which is a V-fixing bacteria in their root nodules.

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Question 10. Why are microbes like Spirulina being produced on a commercial scale? Mention its two advantages.
Answer:

• As a source of food protein.
• Reduces environmental pollution or solves the problem of hunger and malnutrition or a rich source of protein or low-cost production.

Question 11.Name (the microbes that help the production of the following products commercially:

1. Statin
2. Citric acid
3. Penicillin
4. Butyric acid

Answer:

1. Monascus purpureas
2. Aspergillus
3. PenicilHum notatum
4. Clostridium butylicum

Question 12. Name the first antibiotic discovered and by whom.
Answer: Penicillin by A. Fleming.

Question 13. What is the pathogenic property of baculovirus, used as a biological agent? Name the genus of these organisms.
Answer:

• Baculoviruses are pathogens that attack insects and other arthropods.
• The majority of baculoviruses used as biological control agents are in the genus

Question 14. Explain the changes that mills undergo when a suitable starter or inoculum is added to it. flow does the end product formed prove to be beneficial for human health?
Answer:

Lactobacillus lactic acid bacteria (LAB) present in inoculum, grows in milk at a suitable optimum temperature, multiplies converting milk to curd,-and produces acids that coagulate and partially digest the milk proteins. This improves its nutritional quality by increasing Vitamin B₁₂, and LAB checks diseases causing.

Question 15. Why does an organic farmer intentionally not use toxic chemicals to kill the pests which damage the last crops? Explain giving three reasons. 
Answer:

• Toxic chemicals can have adverse side effects cause biomagnification,
• it kills both useful and harmful life forms indiscriminately,
• It eradicates pests not control pests,
• Beneficiary predatory and parasitic insects which depend upon them as food or hosts would not be able to survive,
• It disturbs the food chain food webs or vibrant ecosystems.

Question 16.

1. Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.
Answer:

• Reduces dependence on toxic chemicals.
• Protects our ecosystem or environment.
• Protects and conserves non-target organisms they are species-specific.
• These chemicals being non-biodegradable may pollute the Permanently.
• These chemicals being non-biodegradable may cause biomagnification.

2. Give an example of a bacterium, a fungus and an insect that are used as biocontrol agents.
Answer:

• Bacteria Bacillus thuringiensis Fungus Trichodcnva.
• Insect Lady bird or Dragonfly or Moth.

Question 17. The three microbes are listed below. Name the product produced by each one of them and mention their use.

1. spergillusniger
2. Trichodermtt polysporum
3. Monascus purpure

Answer:

1. Aspergillus niger – Citric Acid, natural preservative or flavouring agent.
2. Trichodennapolysporum – Cyclosporin A, immunosuppressive agent.
3. Monascus purpureas – Statin, blood cholesterol-lowering agent.

Question 18. Baculoviruses are good examples of biocontrol agents. Justify by giving three reasons.
Answer:

• Species-specific narrow-spectrum insecticidal application
• They have no negative impact on plants mammals or birds or fish or non-target insects
• They are beneficial for 1 PM (Integrated Pest Management) or the Pest Management Programme.

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Question 19. Secondary treatment of the sewage is also called Biological treatment. Justify this statement and explain the process.
Answer:

Involves biological organisms such as aerobic and anaerobic microbes or bacteria and fungi to digest or Consume organic waste Primary effluent is passed into an aeration tank where vigorous growth of aerobic microbes (floes) takes place. BOD is reduced (microbes consume a major part of organic matter), effluent is passed to a settling tank where floe sediment produces activated sludge, and sludge is pumped to an anaerobic sludge digester to digest bacteria and fungi.

Question 20. Identify A, B, C, D, and E and find the following table :
Answer:

1. Streptokinase
2. ‘clot buster’
3. cyclosporin A
4. Trichoderma  polysporrum.
5. Monascus  purpureus
6. Blood-cholesterol-lowering agents.

Important Questions for Class 12 Biology Chapter 8 – Microbes In Human Welfare Long Question And Answers

Question 1. Describe the process of secondary treatment given to municipal wastewater (sewage) before it can be released into fresh waterbodies. Mention another benefit provided by this process.
Answer:

Process of secondary treatment Passing of primary effluent into a large aeration tank which is constantly agitated mechanically and air is pumped into it allowing vigorous growth of useful aerobic microbes into floes
↓
Microbes consume a major part of organic matter in effluent which significantly reduces BOD
↓
Now effluent is passed into a settling tank where floes are allowed to settle or sediment called activated sludge
↓
Digestion of activated sludge by anaerobic microbes and effluents from secondary treatment can be released into river or stream
↓
This resulted in the production of Biogas (CH₄, IBS and CO₂) which can be used as a source of energy.

Class 12 Biology Chapter 7 Human Health And Disease Very Short Question And Answers

Question 1. write the scientific name of the causative Lent of pneumonia in humans and mention one specific symptom of the disease.
Answer:

Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for die disease pneumonia. The symptoms of pneumonia include fever, chills, cough and headache.

Question 2. Expand MALT and mention any one location of it in the human body.
Answer:

The mucosa-associated lymphoid tissue (MALT), also called mucosa-associated lymphatic tissue, is a diffuse system of small concentrations of lymphoid tissue found in various submucosal membrane sites of the body, such as the gastrointestinal tract, nasopharynx, thyroid, breast, lung, salivary glands, eye, and skin.

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Question 3. For early detection of cancer, 3-D images of tissues are essential. Name the technique and the basis on which it can generate a three-dimensional image of changes in the IMtig tissue.
Answer:

Medical imaging is the technique and process of imaging the interior of a body for clinical analysis and medical intervention, as well as visual repair of the juice the organs
Of tissues (physiology). Medical imaging seeks to reveal internal! structures are hidden by the skin and bones, as well as to diagnose and treat disease.

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 4. The main barrier that prevents the entry of micro-organisms into our body is

1. Antibodies
2. Macrophages
3. Monocytes
4. skin

Answer: 4. Skin

Question 5. Colostrum provides passive immunity to human infants as it contains antibody

1. IgA
2. IgM
3. tgE
4. igG

Answer: 1. IgA

Question 6. Name any two physiological barriers that provide innate immunity. 
Answer:

Acid in Stomach, Saliva in mouth

Question 7. Name the pathogen which causes Typhoid. Name the test that confirms the disease.
Answer:

Salmonella typhl,w idal test

Question 8. How does the body respond when harmony in produced hy is released in its blood?
OR
Write the role of interferons.
Answer:

Chill and high fever occur, in regular intervals every 3 to 4 days.
OR
Virus-infected cells secrete proteins called interferons, which protect non-infected cells (from further viral infection )

Question 9. Name two diseases whose spread can be controlled by the eradication of Aedes mosquitoes.
Answer:

Dengue and Chikunguniya.

Question 10. How do cytokine barriers promote innate immunity in humans?
Answer:

Interferon (proteins) which is secreted by virus-infected cells protect non-infected cells from further viral infection

Question 11. Flow do monocytes act as a cellular barrier in humans to provide innate immunity?
Answer:

Phagocytosis of microbes or destroy microbes.

Question 12. Name the condition in vertebrates where the body attacks self-cells.
Answer:

Autoimmunity.

Class 12 Biology Chapter 7 Human Health And Disease Short Questions And Answers

1. A body developed some allergic reactions when he straight entered his air-conditioned room after a game of football outside his house. Write any two symptoms that could be noticed in such a condition. How does our body combat such conditions?
Answer:

Symptoms of allergic reactions include sneezing, watery eyes, runny nose and difficulty in breathing. The immune system overreacts by producing antibodies called Immunoglobulin E (IgE). These antibodies travel to cells that release chemicals, causing an allergic reaction

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Question 2.

1. Write the Scientific name of the plant from where natural cannabinoids are obtained.

2. Mention the parts of the plant that are used for extracting the drug.

3. How does the drug affect the human body?
OR
The epithelial lining of our intestine is considered a secondary lymphoid organ. Justify the statement.
Answer:

1. Condition saliva

2. Cannabinoids are obtained from the inflorescence of the plant.

3. These drugs affect the cardiovascular system of the body. They affect brain areas that influence pleasure, memory, thinking, concentration, movement and coordination.
OR
There is lymphoid tissue also located within the lining of the major tracts (respiratory, digestive and urogenital tracts) called mucosal-associated lymphoid tissue (MAtT) It constitutes about 50 per cent of the lymphoid tissue in the human body

Question 3. The major cause of death among infants is because of Vcute Respirators Infection ( XRI) of the lower respiratory tract, which further affects the alveoli. Name the associated disease, and a causative bacteria and give any two symptoms of the disease.
Answer:

The associated disease is pneumonia and the most common bacteria causing pneumonia is Streptococcus pneumoniae.

Symptoms:-

Fast or Laboured Breathing – Breathing patterns would be rapid but shallow, directed from the stomach instead of the chest, accompanied by wheezing.

Pale Skin – The skin around the lips and face starts turning blue (a sign of decreased oxygen in the bloodstream).

Pain – Depending on the infected part, he will experience pain in the lungs or the abdomen. Especially when coughing or breathing deeply

Question 4.

1. In our country YVC’O and other Non-Government Organisations (NGOs) are doing a lot to educate people about AIDS. Expand ’NACO’. Enlist any three ways by which transmission of HIV- AIDS infection occurs.
Answer:

National Aids Control Organization

Ways of Transmission

• Mother-to-child transmission is the most common way that children get HIV & This is called perinatal transmission It is less common because of advances in HIV prevention and treatment,
• Used needles, syringes, and other injection equipment may have someone raise blood on them, and blood can carry HIV  with someone who has HIV
• The primary way in which HIV’ is transmitted from person to person is when a partner has unprotected vaginal or anal sex with an HIV-positive partner.

2. write the scientific name of the plant and the part from which opioids are extracted. How does it affect our body functions?
Answer:

• Opioids are extracted from the late plant called tie opium poppy, whose scientific name is Papaver souvenir.
• Receptors are present m the central nervous system, cardiomiscible system and gastrointestinal tract.
• Opioids act as a depressant and slow down body functions.

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Question 5. It is often observed that the chances of a person suffering from measles in his or her lifetime are low if he or she has suffered from the disease in their early childhood. Justify the statement.
Answer:

Memory cells develop during measles in early childhood, subsequent encounters with the same pathogen elicit a highly intensified secondary anamnestic response.

Question 6. A student on a field trip suddenly felt breathlessness and started to sneeze very badly. Name this response and explain what it is due to.
Answer:

Allergy Allergens (dust or pollen mites etc.) are due to the release of chemicals like histamine serotonin (from mast cells).

Question 7. Causative organisms of some diseases gain entry into the human body through mosquito bites and make humans suffer from the disease. Name one such: 

1. protozoan disease along with the scientific name of the causative organism.
2. Helminths disease along with the scientific name of the causative organism.

Answer:

1. Malaria → Plasmodium vivax Plasmodium malaria Plasmodium falciparum
2. Elephantiasis/Filariasis → Wuchereria bancrofti Wuchcicria malayi

Question 8. Write the functions of bone marrow as the primary lymphoid organ and lymph nodes as the secondary lymphoid organs.
Answer:

Bone Marrow-lymphocytes are produced here, develop and mature into antigen-sensitive lymphocytes

Lymph nodes – trap the microorganism or antigens from the tissue fluid, the trapped antigens activate the lymphocytes (present in lymph nodes) to cause an immune response.
OR
What is a vaccine? State the type of immunity that it induces.
Answer:

The vaccine is a weakened/inactivated pathogen, or its antigenic protein, Active immunity

Question 9.

1. Name the source plant of heroin drug. Mow is it obtained from the plant?
Answer:

Papaver somtiifermnlPoppy plant.

Extracted from the latex of the plant or acetylation of morphine (obtained from the latex of the plant)

2. Write the effects of heroin on the human body.
Answer:

Depressant slows down body function

Question 10. Why is the structure of an antibody molecule represented as ILL2? Name any two types of antibodies produced in the human body.
Answer:

L ₂ – Two light or small polypeptide chains,

Ha,” two heavy or longer polypeptide chains.

I _gA / I _gM / I _gE / I _gG

Question 11. Mention one application for each of the following :

1. Passive immunization
2. Antihistamine
3. Colostrum
4. Cytokinin-barrier

Answer:

1. Provide preformed antibodies anti-toxins for quick response in ca microbes! tetanus) or snake bite.
2. Reduces symptoms of allergy
3. Provides passive immunity, antibodies or I_gV to newborns.
4. Protection of non-infected cells from further viral infection

Question 12.

1. Write the complete name of the diagnostic test for AIDS. Explain the principle it works on.
Answer:

ELISA Enzyme Linked Immune Sorbent Assay. It is based on the principle of antigen-antibody interaction where a pathogen can be detected by the presence of antibodies (proteins, glycoproteins, etc.) on it

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2. Name the type of genetic material present in the AIDS-causing pathogen.
Answer:

MIX is a retrovirus, which means it carries single-stranded RNA as its genetic material rather than the double-stranded DNA human cells carry.

Question 13. A patient complains of suffering from constipation, stomach ache, stool with blood clots and excess mucous. The physician diagnosed it as amoebiasis, after a stool test.

1. Write the scientific name of the microbe identified in the stool sample.
2. How do you think, the patient must have contracted it?
3. Write your suggestions to the patient to avoid infection in future.

Answer:

1. Entamoeba histolytica
2. I houseflies act as mechanical carriers and serve to transmit the parasite from the faeces of infected persons to food and food products, thereby contaminating them. Drinking water and food contaminated by faecal matter are the main sources of infection.
3. Perform hand hygiene frequently, especially before handling food or eating, and after using the toilet or handling faecal matter Wash hands with liquid soap and water, and rub for at least 20 seconds.

Question 14. Name any two techniques which can be used to generate a three-dimensional image to detect cancers of the internal organs and explain any one of the techniques.
Answer:

Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful in detecting cancers of the internal organs.

Magnetic resonance imaging (MRI): Magnetic resonance imaging (MRI)  is a spectroscopic imaging technique used in medical settings to produce images of the inside of the human body MRI is based on the principles of nuclear magnetic resonance (NMR). which is a spectroscopic technique used to obtain microscopic chemical and physical data about molecules.

Magnetic resonance imaging is accomplished through the absorption and emission of energy of the radio frequency (RF) range of the electromagnetic spectrum

MRI Uses

1. Utilizes non-ionizing radiation (unlike X-rays)
2. Ability to image in any plane, (unlike CT scans).
3. Very low incidents of side effects.
4. Ability to diagnose, visualize, and evaluate various illnesses.

Question 15. Given Below Is A Diagrammatic Representation Of the Immune System Of The Human Body:

1. Identify ‘ X’and ’Y’ in the given diagram.
2. Explain two major functions of the organs that you have identified,

Answer:

1. X – Lymph Nodes

Y – Thymus

2. Lymph Nodes –

1. The removal of excess fluids from body tissues.
2. Absorption of fatty acids and subsequent transport of fat, and chyle, to the circulatory system
3. Production of immune cells (such as lymphocytes, monocytes, and antibody-producing cells called plasma cells).
4. There is lymphoid tissue also located within the lining of the major tracts (respiratory, digestive and urogenital tracts) called mucosa-associated lymphoid tissue (MALT) It constitutes about 50 per cent of the lymphoid tissue in the human body.

Thymus – The thymus gland is a lobular structure located between the lungs behind the sternum on the ventral side of the aorta.

• The thymus plays a major role in the development of the immune system.
• This gland secretes the peptide hormones called thymosins.
• Thymosins play a major role in the differentiation of T – T-lymphocytes, which provide cell-mediated immunity.
• Thymosins also promote the production of antibodies to provide humoral immunity.
• Thymus is degenerated in old individuals resulting in a decreased production of thymosins.
• As a result, the immune responses of old persons become weak.

Question 16. Answer the following questions concerning “opioids”, the commonly abused drug :

1. Where in our body are the specific opioid receptors present?
2. What is heroin chemically known as?
3. Write the scientific name of the plant from which opioids are extracted.

Answer:

1. Opioids are drugs, which bind to specific opioid receptors present in our central nervous system and gastrointestinal tract.
2. Heroin commonly called smack is chemically known as diacetylmorphine.
3. Papaver sonmifentni.

Question 17. Name the two primary lymphoid organs in humans. Explain their functions in providing immunity.
Answer:

Bone Marrow: The main lymphoid organ where all blood cells including lymphocytes are produced

Thymus:

1. A lobed organ located near the heart arid beneath the breastbone
2. Quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained it reduces to a very small size

Question 18.

1. Name the causative agents of pneumonia and the common cold.
2. How do these differ in their symptoms?
3. Mention two symptoms common to both.

Answer:

1. Streptococcus pneumoniae Haemophilus influenza, rhinoviruses

2. Different symptoms

3. Common symptoms

Or

1. Write the scientific names of the causative agent and vector of malaria, and write its symptoms.
2. Name any two diseases spread by Aedes sp.

Answer:

1. Plasmodium vivax P. falciparum P.malariae,vector-female Anopheles mosquito Symptoms chills, high fever
2. Dengue, Chikungunya

Question 19. Name is a human disease, its causal organism, symptoms (any three) and sector, spread by intake of water and food contaminated by human faecal- matter. 
Answer:

Amoebiasis (Amoebic dysentery). Entamoeba histolytica, constipation abdominal pain cramps or stools with excess mucus or blood clots & causing agent is Housefly.
OR
Ascariasis. A scar is, internal bleeding muscular pain fever anaemia or blockage of intestinal passage and The causing agent is Housefly.
OR
Typhoid, Salmonella n phi, high fever weakness stomach pain or constipation headache or loss of appetite and the causing agent is Housefly.
OR

1. Why is there a fear among the guardians that their adolescent wards may get trapped in drug or alcohol abuse?
2. Explain ‘addiction” and ‘dependence’ in respect of drug/alcohol abuse in youth.

Answer:

1. Adolescents are easily affected by this. (vulnerable to) peer pressure adventure or curiosity excitement or experimentation or media
2. Addiction- Psychological attachment to certain effects such as Euphoria or temporary feeling of well-being Dependence- Tendency of the body to show withdrawal syndrome or symptoms if regular doses of drug/alcohol are abruptly discontinued.

Question 20.

1. What is an “allergic reaction”?
Answer:

The exaggerated response of the immune system to certain antigens present in the environment is called allergy The substances to which such an immune response is produced are called allergens.

2. Name any two drugs used to quickly reduce the symptoms of allergy.
Answer:

Anti-histamine, adrenalin and steroids

3. Why do more and more children in metro cities of India suffer from allergies and asthma?
Answer:

More and more children in metro cities of India suffer from allergies and asthma due to sensitivity to the environment. This could be because of the protected environment provided early in life.

Class 12 Biology Chapter 7 Human Health And Disease Long Question And Answer

Question 1. It is commonly observed that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction. The result of Ibis parental inhibition is that the children go astray sometimes. 

1. Explain the reasons that you feel are behind such embarrassment amongst some parents to freely discuss such issues with their growing children.
2. By taking one example of a local plant and animal, how would you help these parents overcome such inhibitions about reproduction and sexuality?

Answer:

1. Illiteracy or conservative attitudes or misconceptions social myths etc. are the main reasons.
2. If a student gives the clarity of the concept of reproduction and sexuality by taking an example of a plant and an animal to reproductive organs, gamete formation fertilization, sexual behaviour etc.

Question 2.

1. Differentiate between active and passive immunity.
Answer:

2. Comment on the role of vaccination and immunization in keeping the human population healthy.
Answer:

Role of vaccination or immunization:

• Antibodies produced in the body against antigens neutralize pathogenic agents.
• Vaccines also generate memory cells (B and T cells) that recognize quickly subsequent exposure and control the growth of pathogens with massive production of antibodies.
• Preformed antibodies or antitoxin protect our body from deadly microbes like tetanus and against snake venom.

Class 12th Biology Important Questions Chapter 6 Evolution Very Short  Question And Answers

Question 1. Mention one example each from plants and animals exhibiting divergent evolution.
Answer:

Thorn of Bougaimilira and tendrils of Cucurbita, forelimbs of whales bats, cheetahs,s and humans (all mammals or vertebrate hearts or vertebrate drains

Question 2. Write the names of the following:

1. A 15 ape-like mya primate
2. A 2 mya primate that In ed in East African grasslands

Answer:

1. Dryopithecus
2. Australopithecines or Australopithecus or Homo habilis

Question 3. State two postulates of Oparin and Haldane concerning the origin of life.
Answer:

1. The first form of life could have come from existing non-living organic molecules RNA And Protein
2. Formation of life was preceded by chemical evolution or formation of diverse organic, molecules from inorganic coast entrants

“molecular basis of inheritance class 12 notes bank of biology “

Class 12th Biology Important Questions Chapter 6 Evolution Short Question And Answers

Question 1. Wings of birds and wings of butterflies contribute to locomotion, and the evolution of such organs is a result.
Answer:

They are not anatomically similar in structure but perform the same function, hence these are analogous structures type of evolution is convergent evolution similar habitats of birds and butterflies have resulted in. the selection of similar adaptive features (wings.) in different groups of organisms, but towards the same function convergent evolution.

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 2. According to the Darwinian theory of natural selection, the rate of appearance of new forms is linked to the life cycle or the life span of an organism. Explain with the help of an example.
Answer:

A colony of bacteria (say A) growing in a given medium has built-in variation in terms of the ability to utilize a feed component, a change in the medium composition would bring out only that part of the population B) that can survive under the new conditions In due course of time this variant population outgrows the others and appears as new species thus organisms with shorter life-cycle or life-span will undergo evolution faster or for the same thing to happen in fish or fowl would take millions of years as life spans of these animals are in years.

Question 3. Explain the Ilardy-Weinberg principle with the help of an algebraic equation.
Or
With the help of an algebraic equation, how did Ilardy-Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations?
Answer:

The Principle says that allele frequency in a population is stable and is constant from generation to generation, the gene pool remains constant and expressed as p² +2pq+ q²/ ( p+q)²

• Where p² = frequency of individuals with A A genotype
• Where q²= frequency of individuals with a genotype
• Where 2pq = frequency of individuals with Aa genotype

“biology class 12 ch 6 “

Question 4. What is adaptive radiation? How did Darwin explain it?
Answer:

The process of the evolution of different species in a given geographical area starting from a point and radiating to other areas of geography (habitats) is called adaptive radiation He observed that there were many different varieties of finches (Darwin’s finches) in the same island, and all those varieties evolved ou the island itself, from the original seed-eating features many other forms with altered beaks arose to become insectivorous or vegetarian finches.

Question 5.

1. Differentiate between analogous and homologous structures.
Answer:

1. Aitalogous-Anatomically embryonic development, origin, and structure are not similar though perform similar functions and are a result of convergent evolution.
2. Honiologus – Anatomically embryonic development, origin, and structure are similar (but perform different functions) and are a result of divergent evolution.

2. Select and write analogous structures from the list given below :

1. Wings of butterflies and birds
2. Vertebrate hearts
3. Tendrils of bougainvillea and cucurbita
4. Tubers of sweet potato and potato

Answer: 1 Or 4

Question 6. How can the Hardy- Weinberg Equilibrium be affected? Explain giving three reasons.
Answer:

1. Gene Migration Or Gene Flow: When the migration of a section of the population occurs to another place and gene frequencies change in the original as well as in the new population.
2. Genetic drift: If the same change occurs by chance or new genes or alleles are added to the new population and these are lost from the old population.
3. Mutation: Pro existing advantageous mutations when selected will result in new phenotypes.
4. Genetic recombination: Variation in characteristics will be there because of genetic recombination, during meiosis and also due to random fusion of gametes.
5. Natural selection: Heritable variations enabling better survival enabled organisms to reproduce and leave a greater number of progeny.

Question 7. Write the characteristics of Raniapithecus, Drvopitheeiis, and Neanderthal man.
Answer:

• Raniapithecus: I fairy Or walked like gorillas and chimpanzees, more man-like.
• Drvopitheeiis: Hairy or walked like gorillas and chimpanzees, more aped ike.
• Neanderthal man: brain size is 1400cc, used hides to protect their body or buried their dead,

Question 8. Excessive and continuous use of pesticides has resulted in the evolution of some new species of pests. Explain what must have led to this. What is this type of evolution called?
Answer:

• Excess use of herbicides, pesticides, etc. only resulted in the selection of resistant varieties in a much lesser time scale.
• Hence, resistant organisms seel Is appear in a time scale of months or years and not centuries.
• These are examples of evolution by anthropogenic action.
• This also tells us that evolution is not a directed process in die sense of determinism.
• It is a stochastic process based on chance events in nature and chance mutation in the organisms.

“class 12th biology molecular basis of inheritance “

Question 9. What type of organs eye of an Octopus and that of a human called? Give another example from the animal group and one from the plants of such organs. Name and explain the evolutionary process they exhibit.
Answer:

• The eyes of the octopus and mammals are an example of analogous organs. Sweet potato (root modification) and potato.
• Wings of butterflies and birds.
• The similar habitat has resulted in the selection of similar adaptive features in different groups of organisms but toward the same function, analogous structures are a result of convergent evolution – different structures evolving for the same function and hence having similarity.

Class 12th Biology Important Questions Chapter 6 Evolution Long Question And Answers

Question 1.

1. Describe the observations made on the collection of white-winged moths and dark-winged moths in England between the years 1850 and 1920. What did these observations lead to?
Answer:

Before industrialization tree bark was covered with light-colored lichens, In this background white-winged moth survived but dark colored moth was picked out. by predators, post-industrialization tree trunks became- dark due to industrial smoke and soot, under this condition, and the white-winged moth did not survive due to predators, while the dark-winged/melanized moth survived, this showed that organism that is better adapted to survive are selected by Nature/Natural selection.

2. How is the use of herbicides, pesticides, and antibiotics by humans for various purposes, comparable with the observations made on moths in the above question? What is this type of phenomenon called?
Answer:

Excessive use of these chemicals has resulted in the selection of resistant varieties, in u much lesser time (scale).

These are examples Of evolution by anthropogenic action.

Question 2. Describe S.E. Miller’s experiment. Comment on the observations he made and his contribution towards the origin of life on Earth.
Answer:

High temperature (800″C), high energy radiation, reducing atmosphere created, by electric discharge in a closed flask, containing CH4, IE. NIC. and water vapors in the experimental setup.

bank of biology class 12 molecular basis of inheritance

Observation and Contribution

• Formation of amino acids
• The first form of life arose slowly through evolutionary forces in front of living molecules or abiogenesis.

Question 3. Taking an example of white-winged moths and dark-winged moths of England in pre and post-industrialised eras, explain evolution by natural selection. 
Answer:

An interesting observation supporting evolution by natural selection comes from England In a collection of moths made in the 1850s, therefore before industrialization set in, it was observed that there were more white-winged moths on trees than dark-winged or melanic moths.

• However, in the collection carried out from the same area, but after industrialization, in 1920, there were more dark-winged moths in the same area, therefore the proportion was reversed.
• The explanation put forth for this observation was that ‘predators will spot a moth against a contrasting background’.
• During the post-industrialization period, the tree trunks became dark due to industrial smoke and soot. Under this condition, the white-winged moth did not survive due to predators, dark winged or melanised moth survived.
• Before industrialization set in, thick growth of almost white-colored lichen covered the trees – in that background, the white-winged moth survived but the Clark color. red moth was picked out by predators.
• Lichen will not grow in areas that are polluted. Hence, moths that were able to camouflage themselves, therefore, hide in the background, survived.
• This understanding is supported by the fact that in areas where industrialization did not occur example in rural areas, the count of matric moths was low.
• This showed that in a mixed population, those that can better adapt, survive and increase in population size. Remember that no variant is completely wiped out.

Important Questions for Class 12 Biology Chapter 5 – Molecular Basis of Inheritance Very Short Question And Answers

Question 1. Given the low are the observations drawn in HGP. Select the options that show the correct observations.

1. The human genome contains 3164.7 billion base pairs.
2. The average gene consists of 3000 bases.
3. Less than 2% of the genome codes for proteins.
4. Chromosome one has the most genes ( 2698 )
   1. 1 and 2
   2. 2 and 3
   3. 3 and 4
   4. 1 and 3

Answer: 2. 2 and 3

Question 2. The phosphodiester linkage in the formation of a nucleotide involves the bonding between

1. Phosphate group and OH of 3’C of a nucleoside
2. Phosphate group and OH of 5’C of a nucleoside
3. Phosphate group and H of 3’C of a nucleoside
4. Phosphate group and H of 5’C of a nucleoside

Answer: 2. Phosphate group and OH of 5’C of a nucleoside.

” principles of inheritance and variation”

Question 3. The switching ‘on’ and ‘off of the lac operon in prokaryotes is regulated by

1. Glucose
2. Galactose
3. Lactose
4. Fructose

Answer: 3. Lactose.

Question 4.For ‘in-vitro’ DNA replication, which one of the following substrates needs to be added along with the necessary enzymes the DNA template, and specific conditions?

1. Ribonucleotide triphosphate
2. Deoxyribonucleoside triphosphate
3. Deoxyribonucleotide triphosphate
4. Ribonucleoside triphosphate

Answer: 2. Deoxy r i bo nu Pelosi triphosphate

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 5. Which one of the following factors will associate transiently with RNA polymerase to terminate transcription in prokaryotes?

1. Sigma factor
2. Rho factor
3. Delta factor
4. Theta factor

Answer: 2. Rho factor

Question 6.Choose the correct pair of codons with their corresponding amino acids from the following list:

1. UAG: Glycine
2. AUG: Arginine
3. UUU: Phenylalanine
4. UGA: Methionine

Answer: 3. UUU: Phenylalanine

Question 7. During the elongation process of translation, the peptide bond formation between amino acids is catalyzed

1. Ribosomal RNA
2. Protein in the small subunit of the ribosome
3. Protein to the large subunit of the ribosome
4. Transfer RNA

Answer: 1 . Ribosomal RNA

Question 8. A region of the coding strand of DNA has the following nucleotide sequence: 5-TGCGCCA – 3’ The sequence of bases on niRNA transcribed by this DNA strand would be:

1. 3′ – ACGCGGT – 5′
2. 5′ – ACGCGGT – 3′
3. 5′ – UGCGCCA – 3’
4. 3′ – UGCGCCA – 5′

Answer: 3.  5′ – UGCGCCA – 3′

Question 9. A DNA molecule is 160 base pairs long. It has 20# adenine. How many cytosine bases are present in this DNA molecule?

1. 192
2. 96
3. 64
4. 42

Answer: 2. 96

” principles of inheritance and variation class 12″

Question 10. A template strand in a bacterial DNA has the following base sequence:
5′ – TTTAACGAGG – 3′

1. 5′ – AAATTGCTCC- 3′
2. 3′ -AATTGCTCC-5′
3. 3′ – AAAUUGCUCC – 3’
4. 5’ – CCUCGUUAAA – 3′

Answer: 3. 5′ – CCUCGUUAAA – 3′

Question 11. tRNA has an that has bases complementary to the codon. Its actual structure is a compact molecule which looks like. Select the option that has the correct choices for the two ‘blanks’

1. Amino acid acceptor end, clover-leaf
2. Anticodon loop, clover-leaf
3. Amino acid acceptor end, inverted L
4. Anticodon loop, inverted L

Answer: 3. Anticodon loop, inverted L

Question 12. Which type of RNA is correctly paired with its function? 

1. Small nuclear RNA – Processes rRNA
2. Transfer RNA: attaches to amino acid
3. Ribosomal RNA: involved in transcription
4. Micro RNA: involved in translation

Answer: 2. Transfer RNA: attaches to an amino acid.

Question 13. The figure given below has labeling (1), (2), and (3), which two labeling in the given figure are components of a nucleosome? Select the correct option.

1. 1 – H1 histone, 2 – DNA
2. 1 – DNA, 2 – Histone Detainer
3. 2 – DNA, 3 – HI Histone
4. 2 – Histone octamer, 3 – DNA

Answer: 4. 2 – Histone octamer, 3 – DNA

“inheritance class 12 “

Question 14. Which one of the following diagrams is a correct depiction of a Polynucleotide chain To DNA?

Answer:  2

Question 15. In molecular biology who proposed that genetic information flows in one direction?

1. Hargobind Khoraim
2. Francis Crick
3. Watson mid-Crick
4. Marshall Nirenhcrg

Answer: 2. Francis Crick

Watson and Crick proposed that genetic information flows in one direction.

Question 16. Meseison and Stahl carried out centrifugation in CsC12 density gradient to separate :

1. DNA from RN A
2. DNA from protein
3. The normal DNA from rN-DNA
4. DNA from tRNA

Answer: 3. The normal DNA from rN-DNA

Question 17. Write the dual purpose served by Deoxyribonudeoside triphosphates in polymerization.
Answer:

Acts as a substrate towards the reaction and provides energy to the poh memorization ie energy source (from the initial two phosphates’)

Question 18. Name one amino add, which is coded by only one codon.
Answer:

Methionine Tryptophan

Question 19. Write the conclusion Griffith arrived at the end of his experiment with Streptococcus pneumoniae.
Answer:

He concluded that 11 ml of the R .sbaiu bacteria had somehow been transformed h\ the local ‘killed S strain bacteria.

Important Questions for Class 12 Biology Chapter 5 – Molecular Basis of Inheritance Short Question And Answers

Question 1. Draw a labeled schematic representation of the Central Dogma of Molecular Biology as proposed by Francis Crick.
Answer:

Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA → RNA → Protein

Question 2. State four salient observations drawn from the Human Genome Project.
Answer:

1. Some of the salient observations drawn from the human genome project are as follows :
2. The human genome contains 3164.7 million bp.
3. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
4. The total number of genes is estimated at 30,000-much lower than previous estimates of 80.000 to 1.40.000 genes. Almost: all (99.9 percent) nucleotide bases are the same in all people.

The functions are unknown for over 50 percent of the discovered genes.

“genetics questions and answers pdf “

Question 3. Given below is one of the strands of a DNA segment :

1. Write its complementary strand.
Answer:

(Polarity, Nucleotide sequence)

2. Write a possible RNA strand that can be transcribed from the above DNA molecule formed.
Answer:

(Polarity, Nucleotide sequence)

Question 4. Draw a schematic diagram of a transcription unit with the polarity of the DNA strands and label the coding strand, template strand, and terminator.
Answer:

Question 5. A segment of a DNA molecule comprises 546 nucleotides. How many cytosine nucleotides would be present in it if the number of adenine nucleotides is 96?
Answer:

⇒\(\begin{aligned}
& \text { Given total Nucleotides }=546 \\
& \mathrm{G}+\mathrm{C}=546-192=354 \text { because } \mathrm{G}=\mathrm{C} \text { so } \mathrm{C}=354 / 2, \\
& \text { Cytosine }=177
\end{aligned}\)

6. Although a prokaryotic cell has no defined nucleus, DNA is not scattered throughout the cell ‘Explain.
Answer:

DNA is negatively charged & positively charged proteins and can hold it in places and large loops (in a region termed as nucleoid)

Question 7. Differentiate between the genetic codes given below :

1. Unambiguous and Universal
2. Degenerate and Initiator

Answer:

Question 8.

1. Draw a polynucleotide chain (four nucleotides long) of DNA having four variable
nitrogenous bases.
Answer:

“principles of inheritance and variation short notes “

2. Draw a neat labeled diagram of the nucleosome. Name the basic amino acid residues present mainly in the nucleosome.
Answer:

Amino Acid: Lysine and Arginine

Question 9.

1. Explain the events occurring in a ‘Replicating Fork” during the replication of DNA.
Answer:

DNA-dependent DNA polymerase catalyzes the polymerization, of deoxynucleotides Or deoxyribonucleotide triphosphates.

• The polymerization takes only in one direction 5’→3‘, on one strand (the template with polarity 3′ .5) the replication is continuous,  while on the other (the template with polarity 5″ →3′) it is discontinuous, – the discontinuous)’ synthesized fragments are later joined by the enzyme DNA Haase.

2. Name the different types of RNA polymerases in a eukaryotic cell. Write their roles in transcription.
Or
3. Name the three RNA polymerases found in eukaryotic cells and mention their functions.
Answer:

RNA polymerase 1 →transcribes rRNAs(28S or 18S or 5.8S).

RNA polymerase 2→ transcribes precursor of mRNA or hnRNA or heterogenous nuclear RNA

RNA polymerase 3 → transcribes tRNA or 5srRNA or snRNAs or small nuclear RNAs

Question 10. Explain the mechanism of translation that occurs in the ribosomes in a prokaryote.
Answer:

1. Charging of tRNA Or aminoacylation of tRNA. the small subunit of ribosome binds to RNA
   (5′ end), for initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognized only by initiator tRNA
2. In the elongation phase, an amino acid with tRNA sequentially binds to the appropriate codon on mRNA(forming complimentary base pairs with tRNA anticodon),
3. Ribosome moves from codoil to codon along the mRNA and amino acids are added one by one in the two sites of the large subunit joined by a peptide bond.
4. Termination occurs when a release factor binds to the stop codon and releases the complete po 1 ypepti de.

Question 11. Explain the role of regulatory genes in a lac operon. Why is the regulation of lac operon called negative regulation?
Answer:

• Regulatory gene-gene codes for the repressor of the lac operon, the repressor protein (synthesized by the I gene, binds to the operator site of the operon, and prevents the RNA polymerase from transcribing the operon
• The repressor of the lac operon is synthesized constitutively all 1 the time, and thus the operon is in a switched-off position generally, it is switched on only when lactose is present in the culture medium of the E.coli when the operon gets ‘switched on’

Question 12.

1. Expand VNTR and describe its role in DNA fingerprinting.
Answer:

VNTR -Variable Number of Tandem Repeat(s) -used as a probe (because of its high degree of polymorphism)

2. List any two applications of DNA fingerprinting techniques.
Answer:

Forensic science or criminal investigation (any point related to forensic science (determining population and genetic diversity paternity testing or maternity testing or study of evolutionary biology.

13. Why is the DNA molecule a better hereditary material than the RNA molecule?
Answer:

The DNA molecule is a better hereditary material.

1. It is more stable (due to the presence of thymine and not uracil as in RNA).
2. Less reactive than RNA (as RNA has 2″ – Oil making it more reactive).
3. Being less reactive, DNA is not easily degradable (RNA being more reactive is easily degradable).
4. The rate of mutation is slow (The rate of mutation in RNA is faster)

Question 14. Explain the post-transcriptional modifications the hn-RNA undergoes in eukaryotic cells.
Answer:

1. Splicing → lustrous are removed and exons are joined
2. Capping → Methyl guanosine triphosphate Or map is added to the 5’ end of hnRNA
3. Tailing → Polyadenylate residues are added to 3’etui in a template-independent manner

Question 15.

1. List the two methodologies involved in the human genome project. Mention how they were used.
Answer:

Expressed Sequence Tags, Identifying all the genes that are expressed as RNA Sequence Annotation, sequencing the whole set of genome coding or noncoding sequences, and later assigning different regions with functions.

2. Expand ’YAC and mention what was it used for.
Answer:

Yeast Artificial Chromosomes used as cloning vectors (cloning/amplification)

Question 16. Identify, A, B, C, D, E, and F In the table given below :
Answer:

1. Nitrogenous base
2. OH of PC pentose sugar
3. N-glycosidic linkage
4. phosphate group
5. phosphodiester linkage
6. 3′-5’ phosphodiester linkage

Important Questions for Class 12 Biology Chapter 5 – Molecular Basis of Inheritance Long Question And Answers

Question 1.

1. How did Matthew Meselson and Franklin Stall? experimentally prove that DNA replication
Answer:

Matthew Meselson and Franklin Stahl performed the following experiment in 1958 : (They grew E.coli in a medium containing NiliCl ( ‘.NHs the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that is N was incorporated into newly synthesized DNA (as well as other nitrogen-containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (Please note that 15N is not a radioactive isotope, and it can be separated from 14N only based on densities).

2. The use of heavy isotope of nitrogen by Meselson and Stahl demonstrated a semi-conservative mode of replication of a DNA molecule.” Explain how they arrived at this conclusion.
Answer:

Then they transferred the cells into a medium with normal “14NH4Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently by oil CsCl gradients to measure the densities of DNA.

Question 2.

1. Name and describe the technique which is an important tool of forensic science.
Answer:

The technique is DMA fingerprinting DNA profiling DNA typing Or DNA test.

1. It is a technique to identify a person based on his/her DNA specificity.
2. This technique was invented by Sir Alec Jeffery (1984).
3. In India, DNA Fingerprinting has been started by Dr. V.K. Kashyap & Dr. Lai Ji Singh.
4. DNA of humans is almost (99.9%) the same but a very small amount (0.1 that differs from person to person,
5. The human genome has 3.3 x 10 bp
6. 0.1 % different = 3.3 x 1 bp
7. These differences are mainly due to – Repetitive DNA sequences.
8. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
9. Differences in DNA sequences.
10. Arises due to mutations.
11. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in a human population with a frequency greater than 0.01.
12. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.

2. Mention any two applications of this technique other than its use in forensic studies.
Answer:

To determine the population and genetic diversities of the population.

Question 3. Describe the experiment carried out by Mershey and Chase. Write the conclusion they arrived at.
Or
1. Hershey and Chase carried out their experiment in three steps: infection, blending, and centrifugation. Explain each step.
2. Write the conclusion and interpretation of the result they obtained.

Answer:

They grew viruses on a medium containing radioactive phosphorus r P) and some on radioactive sulfur ( “Si radioactive D.NA contained in viruses grown on radioactive phosphorus, the radioactive protein contained in virus grown in radioactive sulfur. Radioactive phages were allowed to attach to E.rolibacteria, the virus coats were removed from bacteria by agitating in a blender, and the virus particles were separated from the bacteria on centrifugation. Bacteria infected with viruses containing radioactive DNA were radioactive, whereas bacterial cells infected with viruses containing radioactive protein did not show radioactivity.

Conclusion – DNA is therefore the genetic material that is passed from virus to bacteria.

Question 4. List the different components of a Lac Operon. Explain the role of these components, when the operon is in an ‘open state’.
Answer:

Components: Regulatory gene/the T gene (inhibitor), and three structural genes i.e. z, y, and a Repressor, which is synthesized (all-the-time -constitutively from the I gene, is inactivated by interaction with the inducer (lactose or also lactose), this allows RNA polymerase access to die promoter, turning on the transcription of these three genes in the lac operon, which in turn produces enzymes responsible for digestion of lactose (ß – galactosidase break lactose to glucose and galactose)

Question 5. Write the five important goals of the Human Genome Project (HGP).
Answer:

Goals of HGP

1. Identify all the approximately 20.000-25.000 genes in human DNA.
2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA:
3. Store this information in databases
4. Improve tools for data analysis
5. Transfer related technologies to other sectors, such as industries:
6. Address the ethical, legal, and social issues (El. SI) that may arise front the project.

Question 6. Compare the processes of DN A replication and transcription in prokaryotes.
Answer:

Similarities:

Both the processes involve –

1. Unwinding of the helix and separating the two DNA strands
2. Breaking the hydrogen bonds between the bases/pairs
3. Follow the complementary base pair rule
4. Polymerization occurs in a 5′ → 3′ direction
5. linking Polvmeriaction of nucleotides

Dissimilarities:

1. DNA replication Transcription
2. DNA nucleotides added are RNA nucleotides added are ATP, GTP, CTP, TTP ATP, GTP, CTP. UTP
3. Deoxyribose sugar is the part of Ribose sugar is the part of nucleotide Adenine pairs with Thymine Adenine with Uracil
4. Both strands copied Only one strand copied.
5. Resulting in two DNA molecules resulting in the formation of an RNA molecule (Any other correct dissimilarity)

Question 7.

1. Explain Griffith’s ‘transforming principle’ experiment.
Answer:

When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate produce smooth shiny colonies (S) because the S strain bacteria have a mucus (polysaccharide) coat, Mice infected with the S strain(x indent) die from pneumonia infection while others produce rough colonies (R), but mice infected with the R strain do not develop pneumonia. Griffith observed that heat-killed S strain bacteria when injected into mice did not kill them, When he injected a mixture of heat-killed S and live R bacteria, the mice died. Moreover, he recovered living S bacteria from the dead mice.

2. In the above experiment, ‘’heat which killed one type of bacteria, did not destroy the properties of genetic material.’” Justify
Answer:

The two DN A strands get separated by heating and come together, when appropriate conditions are provided heat does not destroy the genetic properties.

Question 8.

1. State the ‘Central dogma’ as proposed by Francis (“rick. Are there any exceptions to it? Support your answer with a reason and an example.
Answer:

Yes, in some viruses flow of information is in reverse direction/reverse transcription e.g. Retrovirus/HIV.

2. Explain how the biochemical characterization (nature) of the ‘Transforming Principle’ was determined, which was not defined in Griffith’s experiments.
Answer:

Protein DNA and RNA were purified from heat-killed S strain/smooth Streptococci! s/Diploco( cits pneumoniae

1. Protein + Protease → transformation secured (R cell to S type)
2. RNA + RNA use → transformation occurred (R cell to S type)
3. DNA + DNA use → transformation inhibited

Hence, DNA alone is the transforming material

Question 9.

1. Why does DNA replication occur in small replication forks and not in its entire
length?
Answer:

DNA being very long, requires high energy for opening along its entire length

2. Why is DNA replication continuous and discontinuous in a replication fork?
Answer:

DNA-dependent DNA polymerase catalyzes polymerization only in one direction, i.e.Two strands of DNA are anti-parallel and have opposite polarity

3. State the importance of the origin of replication in a replication fork.
Answer:

The site where replication originates

4. What is an operon? Explain the functioning of the lac option when in an open state.
Answer:

The arrangement where a (Polycistronic) structural gene is regulated by a common promoter and regulatory genes.

Lactose acts as an inducer and binds with repressor protein, RNA polymerase freely moves over the structural genes, and transcribes lac miRNA, which in turn produces enzymes-transacetylase. permease, ß-galactosidase (by lac z), is responsible for the digestion of lactose.

Question 10.

1. Describe the structure and function of a t-RNA molecule. Why is it referred to as an adapter molecule?
Answer:

Clover-leaf shaped/inverted L-shaped molecules have an anti-codon loop with bases complementary to specific codons. has an amino acid acceptor end. It reads the code on one hand and binds with the specific amino acid on the other hand.

2. Explain the process of splicing of hn-RNA in a eukaryotic cell.
Answer:

Introns are removed, and exons are joined in a definite order called splicing.

Important Questions Class 12 Biology Chapter 4 Principles Of Inheritance And Variation Very Short Questions And Answers

Question 1. The case of Down’s syndrome in humans is;

1. Extra copy of an autosome
2. Extra copy of a sex chromosome
3. Absence of an autosome
4. Absence of a sex chromosome

Answer: 1. Extra copy of an autosome

Question 2. Which of the following features shows the mechanism of sex determination in honeybees’

1. An offspring formed from the union of a sperm and egg develops as a male
2. Males have half the number of chromosomes than that of female
3. The females are diploid having 32 chromosomes ”
4. Males have fathers and can produce sons-.

questions on principles of inheritance and variation

Answer: 2. Males have half the number of chromosomes than females.

Question 3. Select the incorrect pair:

1. Sickle-cell anaemia: Autosomes like recessive
2. Haemophilia Autosome linked recessive trait,
3. Colourblindness Sex sex-linked recessive trait
4. Thalassemia: Autosome-linked recessive trait

Answer: 2. Haemophilia: Autosome-linked recessive trait

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 4. An example of a human trait where a single gene can exhibit multiple phenotypic expression is

1. Phenylketonuria
2. Cystic fibrosis
3. Thalassemia
4. Haemophilia

Answer: 1. Phenylketonuria

Question 5. The life cycle of Drosophila melanogaster is completed in

1. 7 days
2. 14 days
3. 21 days
4. 28 days

Answer: 2. 14 days

Question 6. How many types of gametes would developed by an organism with genotype AaBBCcDD?

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

Answer:  4. 4

Question 7.Assertion (A): There is an expression of only one gene of the parental character in a Mendelian Monohybrid cross in Ft generation

Reason (R): In a dissimilar pair of factors one member of the pair dominates the other

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true, but Reason (R) is false.
4. Assertion (A) is false, but Reason (R) is true.

Answer: 1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

” class 12 biology chapter 5 questions and answers”

Question 8. In Pisum sativum, the flower colour may be Violet (V) or White (v). What proportion of the offspring in a cross of VV vv would be expected to be violet 

1. 25%
2. 50%
3. 75%
4. 100%

Answer: 4. 100%

Question 9. Which one of the gene pairs is expected to give a ratio of 1 1: I I in the progeny of a Mendelian Dihybrid cross

1. AaBb×AbBb
2. A ABB × AaBb
3. AaBb × aabb
4. A ABB × aabb

Answer: 3. AaBh x aabb

Question 10. The progeny of a cross between two snap-dragon plants heterozygous for flower colour, bearing different coloured flowers would be:

1. 25% red, 50%
2. pink, 25% white
3. 50% red, 50% white
4. 75% red, 25% white

Answer: 2. 25% red. 50% pink, 25% white.

Question 11. Study the given pedigree of a family and select the trait that shows this pattern of inheritance

1. Autosomal recessive, Phenylketonuria
2. Sex-linked recessive. Colour-blindness
3. Autosomal dominant. Myotonic dystrophy
4. Sex-linked dominant. Vitamin I) resistant rickets

Answer: 1. Autosomal recessive. Phenylketonuria

Question 12. A child with blood group A has a father with blood group B and a mother with blood group AB What would be the possible genotypes of the parents and the child Choose the correct option:

Answer: 3. Father – I^Bi, Mother – I^Ai^B, Child – I ^A i

Question 13. In a hybrid Mendelian cross, garden pea plants heterozygous for violet flowers and round seeds are crossed with homozygous white flowers and wrinkled seeds. The genotypic and phenotypic ratio of F, progeny would be:

1. 9:3:3: 1
2. 1:2:2: 1
3. 1:1.1.1
4. 3: 1

Answer: 3. 1: 1: 1: 1

Question 14. Colour blindness is a sex-linked recessive trait in humans. A man with normal colour vision marries a woman who is colourblind. What would be the possible genotypes of the parents, the son and the daughter of this couple?

Answer: 1. Mother- XX, Father – X^CY, Daughter- X^CX, Son – XY

Question 15. Given below are the pairs of contrasting traits in Pisum sativum as studied by Mendel Select the incorrectly mentioned option from the table given below:

Answer: 1 Or 4

Question 16. How many types of gametes can be produced in a diploid organism which is heterozygous for 4 loci?

1. 4
2. 8
3. 16
4. 32

Answer: 3.  16

principles of inheritance and variation pyq neet

Question 17. Given below is a Karyotype obtained after analysis of foetal cells for a probable genetic disorder.

Based on the above Karyotype, the chromosomal disorder detected in the unborn foetus and the consequent symptoms the child may suffer from are

1. Down’s syndrome: Gynecomastia, overall masculine development
2. Down’s syndrome: Furrowed tongue, short stature
3. Klinefelter’s syndrome Gynecomastia, Masculine development.
4. Klinefelter’s syndrome: Rudimentary ovaries, short stature

Answer: 3. Klinefelter’s syndrome: Gynecomastia, Masculine development.

Question 18 The recombinant Frequency between the four linked genes is as follows:

1. Between X and Y is 40%.
2. Between Y and Z is 30%.
3. Between Z and W is 10%.
4. Between W and X. is 20%.

Select the option that shows the correct order of the position of W, X. Y and Z genes on the chromosome:

1. Y – X -Z -W
2. Y – W – Z – X
3. X- Y -Z -W
4. Z – X – Y – W

Answer: 2. Y – W – Z – X

Question 19. Write the possible genotypes of a person with blood group ‘B
Answer:

The genotype for blood group B is H1 in homozygous condition and H is in heterozygous condition

Question 20. Write the dominant traits in pea plants observed by Mendel concerning:

1. colour of a pea pod.
2. flower position.

Answer:

1. Green pod colour was the dominant
2. The terminal flower position was recessive to the axial position

Question 21. Write the symbolic representation used in a pedigree chart showing

1. a carrier mother and
2. a sufferer son, concerning haemophilia.

Answer:

A carrier mother→

A sufferer’s son →

Question 22. Assert ion: The progenies of a test cross can be easily analysed to predict the genotype of the test organism.

Reason: In a typical test cross, an organism showing a recessive phenotype is crossed with a recessive parent instead of self-crossing 

1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion
2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion
3. The assertion is true, but Reason is false
4. Both Assertion and Reason are false.

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

Question 23.

1. Name a human genetic disorder due to the following :

1. An additional X-chromosome in a male
2. Deletion of one X-chromosome in a female

Answer:

1. Klinefelter’s Syndrome
2. Turner’s Syndrome

2. State what does aneuploidy lead to.
Answer:

Abnormal number of chromosomes in a cell. Down’s Syndrome or Turner’s Syndrome or Klinefelter’s Syndrome

Question 24. State Mendel’s Law of Independent Assortment.
Answer:

When two pairs of traits (characters) are combined in a hybrid segregation of one pair of characters is independent of the other pair of characters.

Question 25. Write one example of each of the organisms exhibiting

1. Male heterogamety, and
2. Female heterogamety.

Answer:

1. Human or Drosophila or Grasshopper
2. Birds Chicken

Question 26. Write the sex of a human having XXY chromosomes with 22 pairs of autosomes. Name the disorder this human suffers from.
Answer:

Mate, Klinefelter’s syndrome

Question 27. Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Answer:

Test cross

Question 28. A colour-blind bos is born to a couple with normal colour vision. Write the genotype of the parents.
Answer:

Mother-Xc
X Father-XY

Important Questions Class 12 Biology Chapter 4 Principles Of Inheritance And Variation Short Question And Answers

Question 1.

1. Write two closely linked genes that control a-Thalassemia.
Answer:

Thalassemia is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16 of each parent and it is observed due to mutation or deletion of one or more of the four genes

2. Differentiate between Thalassemia and Sickle cell anaemia based on their effect on the globin molecule of haemoglobin.
Answer:

Thalassemia differs from sickle-cell anaemia in. that the former is a quantitative problem of synthesising too few globin molecules while the latter is a qualitative problem of synthesising an incorrectly functioning globin

Question 2.

1. Mendel did not explain the expression of incomplete dominance in plants, (give an example of a flower exhibiting incomplete dominance. Name and state the Law of Mendel that the genes which exhibit incomplete dominance follow.
Answer:

• Antirrhinum Snapdragon or Dog flower or Four o’clock plant or A Law of segregation.
• Allele or factors of a pair segregate from each other such that a gamete receives only one of the two factors.

2. Your teacher gave you a tall pea plant and asked you to find out whether the plant is homozygous or heterozygous. Mow will you proceed to find the genotype of the given plant?
Answer:

The Genotype was found by test cross. The Crossing of Unknown plant with recessive parent.

Question 3. Why is the frequency of red-green colour blindness more in human males than in females? Explain.
Answer:

The gene for colour blindness is located on the X chromosome in humans, it is a recessive gene, since human males have a single X chromosome the recessive gene always expresses when present, whereas in human females as they have two X chromosomes (the trait is expressed only if both the sex chromosomes have this reason e gene

Question 4. A lipophilic father can never pass the gene for haemophilia to his son. Explain.
Answer:

It is a sex-linked recessive disorder in which the X chromosome has the haemophilic gene, the Son inherits an ‘S’ chromosome from the father and the gene for haemophilia is not present on the S chromosome

Question 5.

1. What happens when chromatids fail to segregate during cell division cycle? Explain your answer with an example.
Answer:

Failure of segregation of chromatids during the cell division cycle results in the gain or loss of a chromosome (s), called aneuploidy. For example, Down’s syndrome results in the gain of an extra copy of chromosome 21

2. ABO blood groups are a good example of co-dominance. Justify.
Answer:

In blood groups I ‘ and I1’ are present together they both express their types of sugars this is because of co-dominance, lienee red blood cells have both A and B types of sugar polymers

Question 6.

1. Generally it is observed that human males suffer more than human females, who rarely suffer from it. Explain giving reasons.
Answer:

This is a sex-linked X chromosomes recessive disease, the heterozygous female carrier for haemophilia may transmit the disease to sons (mate progeny), and the possibility of a female becoming a haemophimophilia is extremely rare haemophilia because the mother of such a female has to be at least carrier and the father should be haemophilic.

2. F₁ progeny of pea plant bearing violet flowers and snapdragon plant bearing red flowers were soiled to produce their respective F₂ progeny. Compare the phenotypes, the genotypes and the pattern of inheritance of their respective F₁ progeny.
Answer:

Question 7.

1. Differentiate between pleiotropy and polygenic inheritance by taking one example of each.

2. How is polygenic inheritance different from pleiotropic? Give one example of each.

Answer:

Question 8.

1. How does mutation occur?
Answer:

Eosstdeletion or gain (insertion duplication or addition) or change in position of DNA segments/chromosome

2. Differentiate between point mutation and frameshift mutation.
Answer:

Mutation due to a change in a single base pair of DNA is point mutation. Insertion or deletion of one or two bases changes the reading frame from the point, of insertion or deletion.

Question 9. Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings?
Answer:

In birds, sex determination takes place by the ZW -method;

• In Birds, female heterogamety female produces (Z) type and (W) type of gametes in the case of bird males in hornogamety (ZZ) and female is heterogamety.
• In 11 humans being male heterogamety male produces (X) type and (Y) type of gametes.

Question 10. Both Haemophilia and Thalassemia are blood-related disorders in humans. Write their causes and the difference between the two. Name the category of genetic disorder they both come under.
Answer:

Mendelian disorder

Important Questions Class 12 Biology Chapter 4 Principles Of Inheritance And Variation Long Question And Answer

Question 1. The cytological observations made in several insects led to the development of the concept of the genetic or chromosomal basis of the sex-determination mechanism. The honey bee is an interesting example to study the mechanism of sex determination. Study the schematic cross between the male and the female honey bees given below and answer the questions that follow :

1. Identify the cell divisions V and ‘IT’ that lead to gamete formation in female and male honey bees respectively.
Answer:

• ‘A’-Meiosis
• ‘B’-Mitosis

2. Name the process ( that leads to the development of the male honey bee (drone).
Answer:

‘C’-Parthenogenesis

Question 2. T.H. Morgan carried out a cross on Drosophila involving genes for body colour (y+/Y) and genes for eye colour (w +/w). Study the schematic representation of the cross-opto I I generation and answer the questions that follow :

1. Name the kind of cross it represents.
2. Identify and write the dominant phenotype concerning eye colour.
3. What are these genes located on the chromosome shown referred to as?

Answer:

1. Dihybrid cross
2. The red-eye phenotype of w
3. Linked genes or sex-linked genes

Question 3. Mendel crossed a homozygous pea plant having yellow and round seeds with another pea plant hearing green and wrinkled seeds, lie found that in some of the IS populations new combination of parental characters was observed. How will you explain the appearance of a new combination of parental characters in IS offspring? Support your answer with the help of Punned Square.

Answer: When two pairs of traits are combined in a hybrid segregation of one pair of characters is independent of the other pair of characters.

 Phenotypic ratio

Genotypic ratio

Question 4.

1. Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination concerning genes as studied by them.
Answer:

Trosophikttnehmogasier
They observed that two genes (located closely on a chromosome) did not segregate independently of each other (Fe ratio deviated very significantly from 9:3:3: 1) lightly linked genes tend to show very fewer (lesser) recombinant frequency of parental traits/show higher (more) frequency of parental type loosely linked genes show higher percentage (more) of recombinant frequency of parental traits Or lower frequency percentage of parental type genes present on the same chromosome are said to be linked and the recombinant frequency depends on their relative distance on the chromosome.

2. How did Sturtevant explain gene mapping while working with Morgan?
Answer:

He used the frequency of recombination between gene pairs on the same chromosome, as a measure of the distance between genes and ‘mapped’ their position on the chromosome.

Important Questions For Class 12 Biology Chapter 3 – Reproductive Health Very Short Questions And Answers

Question 1. Assertion (A): Through Reproductive and Child Health (RCI) programs in India; we could bring down the population growth rate.

Reason (R): A rapid increase in MMR and IMR were the reason, along with other reasons for this.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true, but Reason) is not the correct explanation of Assertion (A).
3. Assertion (A) is true, but Reason (R) is false.
4. Assertion t A) is false, but Reason (R) is true.

Answer: Assertion (A) is true, but Reason (R) is false.

Question 2. Assertion (A): Sterilisation methods are generally advised for male or female partners as a terminal method to prevent any more pregnancies.

Reason (R): These techniques are less effective and have high reversibility.

1. Both Assertion (A) and Reason (R) are true and Reason (R)is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true, but Reason (R ) is not the correct explanation of Assertion (A).
3. Assertion (A) is true, but Reason (R) is false.
4. Assertion (A) is false, but Reason (R) is true.

Answer: 3. Assertion (A) is true, but Reason (R) is false.

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 3. A specialized procedure to form an embryo in the laboratory in which sperm is directly, injected into the ovum is:

1. HIT
2. IUI
3. ICSI
4. ZIFT

Answer: 3. ICSI

biology class 12 chapter 4

Question 4. Listed below are all reproductive tract infections except

1. Genital herpes
2.  Filariasis
3. Trichomoniasis
4. Syphilis

Answer: 2. Filariasis

Question 5. Write the full name of the technique used for the transfer of early embryos in the uterus of the mother for further development. Write the minimum number of blastomeres the embryo must have before being transferred.
Answer:

The zygote or early embryo with up to 8 blastomeres is transferred into the Fallopian tube and the process is called Zygote Intra Fallopian Transfer or ZIFT.

Question 6.

1. Assertion: A statutory ban on amniocentesis for sex determination is to legally check increasing female foeticide.

Reason: In amniocentesis, some of the amniotic fluid that has the developing fetus is taken to analyze the chromosomes in the fetal cells.

1. Both Assertion and Reason are true, and Reason is the caned explanation of the Assertion.
2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
3. The assertion is true, but Reason is false.
4. Both Assertion and Reason are false.

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

 2. Assertion: Our laws permit legal adoption and it is as yet, one of the test methods for couples looking for parenthood.

Reason: Emotional, religious, and social factors are also no deterrents to the legal adoption of orphaned and destitute children in India.

1. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
2. Both Assertion and Reason are true, but Reason is not the correct explanation of the Assertion.
3. The assertion is true, but Reason is false.
4. Both Assertion and Reason are false.

Answer: 1. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

Question 7. Give one reason to justify a statutory ban on amniocentesis.
Answer:

Prevent female foeticide.

Important Questions For Class 12 Biology Chapter 3 – Reproductive Health Short Question And Answers

Question 1. Why do doctors suggest some monied couples go for ‘IVF’? Explain the steps carried out in the process of ‘IVF’.
Answer:

Childless couples could be assisted in having children through IVF

Ova from the wife/donor (female) and sperm from the husband or donor(male) are collected and are induced to form a zygote under simulated conditions in the laboratory, the zygote or early embryos (with up to 8 blastomeres) could then be transferred into the fallopian tube(ZIFT), and embryos with more than 8 blastomeres transferred into the uterus(lUT)

” class 12 biology chapter 4 questions and answers”

Question 2. A doctor after conducting certain tests on a pregnant woman advised her to undergo M.T.P., as the fetus she was earning showed trisomy of the 21^st chromosome.

1. State the cause of trisomy of the 21^st chromosome.
Answer:

Cause non-disjunction or failure of segregation of chromatids of the 21^st chromosome during gamete formation, leading to a gain of a chromosome.

2. Why was the pregnant woman advised to undergo M.T.P. and not to complete the full term of her pregnancy?
Answer:

Mother was advised to undergo MTP because Trisomy of the 21^st Chromosome would lead to s, Down’s syndrome or an individual is short-statured with furrowed tongue or broad palm with characteristic palm crease or retarded physical or mental and psychomotor development.

Question 10. List any two types of IEDs that are available for human females and state their mode of action.
Answer:

1. Copper releasing fUDs (CuT, Cu7, Multiload 375)
2. Cu ions released suppress sperm motility and. the fertilising capacity of sperms,
3. The hormone-releasing TUDs (Progcstascrt, LNG-20)
4. The hormone-releasing JUDs. in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperm.

Important Questions For Class 12 Biology Chapter 3 – Reproductive Health Long Question And Answer

Question 1. IVF is a very popular method this day that is helping childless couples to bear a child. Describe The different steps that are carried out in this technique. Would you consider Gamete Intrafallopian Transfer (GIFT) as an IN F₁ Give a reason in support of your answer.
Answer:

1. Ova from the wife or donor (female) and sperms from the husband or donor (male) are collected and induced to form a zygote, in simulated conditions in the laboratory, the zygote or early embryos up to 8 blastomeres) then transferred into the fallopian tube(ZIFT), and embryos with more than 8 blastomeres. into the uterus (JUT) to complete its further development.
2. No, GIFT cannot be considered an IVF technique because fertilization takes place in the female body,

Question 2.

1. Explain the steps involved in in vitro fertilization popularly known as the test tube baby program.
Answer:

In vitro fertilization followed by embryo transfer (ET) is one such method. In this method, popularly known as the test tube baby program, ova from the wife or donor (female) and sperms from the husband or donor (male) are collected and induced to form a zygote under simulated conditions in the laboratory. The zygote or early embryos (with up to 8 blastomeres) could then be transferred into the fallopian tube (ZIFF-zygote intrafallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT intra uterine transfer), to complete its further development.

class 12 reproductive health ncert solutions

2. State the importance of this program.
Answer:

• A large number of couples all over the world including India are infertile, i.e., they are unable to produce children despite unprotected sexual cohabitation.
• The reasons for this could be many physical, congenital, diseases, drugs, immunological or even psychological,
• In India, often the female is blamed for the couple being childless, but more often than not, the problem lies in the male partner.
• Specialized health care units (infertility clinics, etc.) could help in the diagnosis and corrective treatment of some of these disorders and enable these couples to have children.

Sexual Reproduction In Flowering Plants Very Short Question And Answers

Question 1. The hilum in a typical angiosperm ovule represents the junction between –

1. Integuments and the embryo sac.
2. Embryo sac and its ocellus
3. The body of the ovule and the funicle
4. Nocellus and the funicle

Answer: 3. Body of the ovule and the funicle.

Question 2. In the given diagram of a transverse section of a young amber Choose the labelling showing the correct placement of the wall layers from the table given below.
Answer:

class 12 biology chapter 2 important questions and answers

1. Tapetum
2. Endotliecium
3. Epidermis
4. Middle layers

Read and Learn More Important Questions for Class 12 Biology Chapter Wise

Question 3. The term used for the embryo entering into the state of inactivity as the seed matures is

1. Quiescent
2. Parthenogenesis
3. Parthenocarpy
4. Donnancy

Answer: 4. Dormancy

Question 4. The ploidy of the apomictic embryo developed from the integument cells and megaspore mother a cell without reduction division respectively will be

1. 2n and 2n
2. n and n
3. 2n and n
4. 3n and 2n

Answer: 1. 2n and 2n

Question 5. Below is a diagrammatic representation of a mature embryo sac of a typical angiosperm plant. Choose the option showing the correct labelling for the parts W, X, Y and Z from the table below.
Answer:

reproduction in flowering plants class 12 questions and answers

Answer: 3. Y – Micropylar end. X – Antipodals, Y-Synergids, Z – Central cell

Question 6. Given below is a figure of an angiosperm plant showing two different types of flowers ‘X’ and the possible type of pollination in them :

Select the correct option for the flower (X) and flower (Y) and the possible type of pollination from the given table:

Answer: 3. Flower X Chasmogamous self Or cross-pollination Flower Or Cleistogamous, self-pollination

Question 7. An undifferentiated sheath covering the root cup of a monocotyledonous embryo is:

1. Scutellum
2. Coleorhiza
3. Coleoptile
4. Epi blast

Answer: 2. Coleorhiza

Question 8. Residual persistent nucellus in black pepper is known as :

1. Perisperm
2. Pericarp
3. Putvinus
4. Perianth

Answer: 1. Perisperm

Question 9. Amongst tile insects, the dominant biotic pollinating agents are –

1. Ants
2. Wasps
3. Beetles
4. Bees

Answer: 4. Bees

Question 10. A genetic mechanism which prevents inbreeding depression in the majority of atigiospenu plants is

1. Parthenogenesis
2. Parthenoeurpy
3. Mutation
4. Self-incompatibility

Answer: 3. Self-incompatibility

” reproduction in flowering plants questions and answers pdf class 12″

Question 11. Self-pollination is fully ensured if

1. The flower is bisexual.
2. The style is longer than the filament.
3. The flower is cleistogamous.
4. The time of pistil and another maturity is different.

Answer: 3. The flower is cleistogamous.

Question 12. Some flowers, selected for artificial hybridization, do not require emasculation but bagging is essential for them. Give a reason.
Answer:

As some Bowers are unisexual, to prevent contamination of their stigma with unwanted pollen grains.

Question 13. Write one advantage and one disadvantage of cleistogamy to flowering plants.
Answer:

Cleistogamy Advantage:

Assured seed set maintains pure lines.

Cleistogamy Disadvantage:

No variation or only parental characters are preserved or it can lead to inbreeding depression.

Sexual Reproduction In Flowering Plants Short Question And Answers

Question 1:

1. Write the two crucial changes the seeds undergo while reaching maturity that enable them to be in a viable state until the onset of favourable conditions.
Answer:

Seeds remain in a state of dormancy-induced desiccation and the hormone abscisic acid until conditions for growth become favourable.

2. Name the oldest viable seed excavated from the Arctic Tundra as per the records.
Answer:

The oldest is that of a lupine, Lupinus an excavated front Arctic Tundra.

Question 2. Explain the mechanism of pollination in marine seagrasses like Zostcra.
Answer:

In marine Seagrasses, female flowers remain submerged in water and the pollen grains are released inside the water. Pollen grains in many such species are long, ribbon and they are carried passively inside the water; some of them reach the stigma and achieve pollination.

Question 3. What is ‘bagging’? State its importance in the artificial hybridization of flowering plants.
Answer:

Bagging – Process of covering emasculated or female or artificially pollinated flowers with a bag of suitable size generally made up of butter paper (to prevent contamination of stigma with unwanted pollen).

Importance – The desired pollen grains are used for pollination or the stigma is protected from contamination from unwanted pollen. It combines desirable characteristics to produce commercially superior varieties.

Question 4:

1. You are given castor and bean seeds. Which one of the two would you select to observe the endosperm?
Answer:

Castor

2. The development of endosperm precedes that of embryos in plants. Justify.
Answer:

Endosperm stores reserve food materials or provide nutrition to the developing embryo.

Question 5. Write any two ways by which apomictic seeds may be developed in angiosperms.
Answer:

• Develops from a diploid egg cell (formed without reduction division) which grows into an embryo without fertilization.
• Develops from the nuclear cell which divides and protrudes into the embryo sac and develops into an embryo

Question 6. A pollen grain in angiosperm at the time of dehiscence from an anther could be 2-celled or 3-celled. Explain. Mow arc the cells placed within the pollen grain when shed at a 2-celled stage?
Answer:

• In the 2-celled stage, the mature pollen grain contains a generative and vegetative cell, whereas in the 3- 3-celled stage one vegetative cell and two male gametes are present.
• The generative cell floats in the cytoplasm of the vegetative cell.

Question 7:

1. When a seed of orange is squeezed, many embryos, instead of one are observed. Explain how it is possible.
Answer:

Polyemhryony and nucellar cells surrounding the embryo sac start dividing, protrude into the embryo sac and develop into many embryos.

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2. Are these embryos genetically similar or different? Comment. 
Answer:

These embryos are genetically similar, as produced from nucellar cells by mitotic division/formed without fertilisation (but different from the embryo formed by fertilization)

Question 8. Draw a schematic transverse section of a mature anther of an angiosperm. Label its epidermis, middle layers, tapetum, eudothecium, sporogenous tissue and the connective.
Answer:

Question 9. Differentiate between wind-pollinated and insect-pollinated flowers.
Answer:

Wind pollinated: Pollen grains are light sticky or well-exposed stamens or Large feathery stigmas or single ovule in each flower or Small flowers packed in inflorescence.

Insect pollinated: Pollen grains sticky or Moral rewards or Fragraut or Neetar rich or Large Flower.

Question 10:

1. When and where do tapetum and synergids develop in (lowering plants? Mention their functions.
Answer:

1. Tapetum – Mierosporogenesis. Microsporang (Anther), nourishes the developing pollen grains.
2. Synergids – Megusporogenesis. Megasporangiumtovule), synergids have filiform apparatus to guide the pollen tube into it.

2. Where are the following structures present in a male gametophyte of an angiosperm? Mention the function of each one of them.

1. Germ pore
2. Sporopollenin
3. Generative cell

Answer:

1. Germ pore- Pollen grain exiting, site from where pollen tube emerges.
2. Sporopollenin – Exine of the pollen grain, protects the pollen grains from high temperature and strong acids And alkalis or enzymes or adverse conditions.
3. Generative cells – Pollen grains give rise to two male gametes.

Question 11. Draw L.S. of an embryo of grass and label its parts.
Answer:

Question 12. Draw a labelled diagram of a mature male gametophyte of an angiosperm.
Answer:

Question 13. Do you think apomixis can be compared with asexual reproduction? Support your answer, by giving one reason. Flow is apomixis beneficial to farmers? Explain.

Answer:

Yes, seeds are produced without fertilisation.

Production of hybrid seeds is costly, if hybrids with desirable characteristics can be made into apomicts, there is no segregation of characters in the hybrid progeny, and farmers can continue using hybrid seeds year after year and do not have to buy hybrid seeds.

Question 14:

1. Can a plant flowering in Mumbai be pollinated In pollen grains of the same species growing in New Delhi? Provide explanations for your answer.
Answer:

Yes. Artificial hybridisation in pollen grains of the flower is introduced artificially on the stigma of another flower.

2. Draw the diagram of a pistil where pollination has successfully occurred. Label the parts involved in reaching the male gametes to their desired destination.
Answer:

Diagram with following labelling Stigma, Pollen tube, Synergid or Filiform Apparatus. Micropyle

Question 15:

1. Draw a diagram of Pistil showing pollen tube growth in angiosperm and label

1. Stigma
2. male gametes
3. micropyle end
4. Ovule.

Answer: 1.

2. Write the function of micropyle.
Answer:

The pollen tube, after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus.

Sexual Reproduction In Flowering Plants Long Question And Answers

Question 1. Study the figures given below of the development of megaspores in an angiosperm and answer the questions that billow :

1. Describe the developmental events in the nucellus of the ovule. What is this type of development of megaspore referred to as?
Answer:

• Enclosed within the integuments is a mass of cells called the nucellus.
• Cells of the nucellus have abundant reserve food materials.
• Located in the nucleus is the embryo sac or female gametophyte.
• This method of embryo sac formation from a single megaspore is termed monosporic development.

 2. How many free nuclear mitotic divisions will the functional megaspore undergo to form a mature embryo sac?
Answer:

3 mitotic division

3. Describe the structure of a typical female gametophyte of a flowering plant.
Answer:

• A characteristic distribution of the cells within the embryo sac.
• Three cells are grouped at the micropylar end and constitute the egg apparatus.
• The egg apparatus, in turn, consists of two synergids and one egg cell.
• The synergids have special cellular thickenings at the micropylar tip called filiform, apparatus, which plays an important role in guiding the pollen tubes into the syttergid.
• Three cells are at the chalazal end and are called the antipodal.
• The large central cell, as mentioned earlier, has two polar nuclei.
• Thus, a typical angiosperm embryo sac. at maturity, though 8-nucleate is 7-celled.

Question 2:

1. Name the specific part of the anther and the process responsible for the development of a male gametophyte in an angiosperm.
Answer:

porogenous tissue or Microsporungium Microspore mother cell Pollen mother cell or PMC Microsporogenesis.

2. Draw a labelled diagram of a mature male gametophyte (3-celled) of an angiosperm. Write the functions of each labelled part
Answer:

• Vegetative cell – It has an abundant food reserve
• Male gametes – Participate in double fertilisation or one male gamete fuses with the egg and the other fuses with two polar nuclei or secondary nucleus (any two)
• Exilic – Made up of the most resistant organic material sporopollenin or can withstand high temperatures or strong acids or alkalis or no enzyme can degrade it.
• In tine – Contributes pollen tube formation.
• Germ pore – Region from where pollen tubes arise.

Question 3. Explain the post-pollination events up to double fertilisation, that occur in an angiosperm.
Answer:

The Pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores.

• The contents of the pollen grain or vegetative cell generative cell or two male gametes move into the pollen tube.
• The pollen tube grows through the tissues of the stigma and pollen tube to reach the ovary.
• The pollen tube enters (through micropyle) the synergids through the filiform apparatus, pollen tube releases two male gametes in the cytoplasm of the synergids.
• One of the male gametes fuses with the egg cell or female gamete completing syngamy. to form (diploid) zygote.
• The other male gamete fuses with two polar nuclei in the (central cell) to produce (a triploid) primary endospermic cell, three haploid cells fuse called triple fusion, and two types of fusion syngamy and triple fusion are called double fertilisation.

Question 4:

1. Describe any two devices in a flowering plant which pretend both autogamy and geitonogamy.
Answer:

Dioecy/production of unisexual flowers (in different plants) Self sterility or self-incompatibility or intraspecific incompatibility lists is a genetic mechanism and prevents self pollen (front the same flower or another flower of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

2. Explain the events up to double fertilisation after the pollen tube enters one of the synergids in an ovule of an angiosperm.
Answer:

Pollen tube releases 2 male gametes in the cytoplasm of synergid

• One male gamete fuses with egg cells or syngamy. resulting in a diploid zygote
• Other male gamete fuses with polar nuclei or triple fusion, to form diploid PEN (Primary Endosperm Nucleus) or PEC (Primary Endosperm Cell)

Question 5:

1. Draw a diagrammatic sketch of a transverse section of an anther of an angiosperm. Label its different walls and the tissue forming microspore mother cells.
Answer:

2. Describe the process of microsporogenesis up to the formation of a microspore.
Answer:

Sporogenous tissue → Microspore mother cell → Microspore tetrad

3. Write the function of germ pore1 in a pollen grain of an angiosperm.
Answer:

Germ pores allow the germinating or growing pollen tube with contents of the pollen grain or male gametes + vegetative cell to come out of the pollen grains

Question: 6

1. State one difference and one similarity between geitonogamy and xenogamy.
Answer:

GeitOnogamy – Transfer of pollen grains from the anther to the stigma of another flower of the same plant. Although is functionally cross-pollination involves 1 a pollination. agent, genet it ally it m similar to autogamy since the pollen grains come Item the sunn plant

Xenogamy – transfer of pollen grains from the anther to the stigma of a different plant This is the only tv[x of pollination which during pollination brings genetically different types of pollen grains to the stigma.

2. Explain any three devices developed in flowering plants to discourage self-pollination and encourage cross-pollination.
Answer:

Flowing plants have Haw developed many devices to discourage sell-pollination and to encourage cross-pollination.

• At some speed. pollen and base and stigma negativity are not synchronised.
• Lit her the pollen is released before the stima becomes receptive or stigma becomes receptive much before the release of pollen.
• In some other species, the anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.