Class 12 Biology

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.

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

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

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.

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.

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:

\(\mathrm{A}+\mathrm{T}=\mathrm{C}+\mathrm{G} \text {, Given } \mathrm{A}=96 \text { so } \mathrm{T}=96 \text {, and } \mathrm{A}+\mathrm{T}=192\)

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

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:

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

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.

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-.

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).

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

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

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

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.

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

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.

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)

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.

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.

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:

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:

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

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.

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.