BZYCT-137 Solved Assignment

1. i) How is co-dominance different from incomplete dominance?
   ii) Explain the phenomenon of masking the expression of a gene by another in epistasis.
2. i) In Drosophila, the recessive, sex-linked genes, abnormal eyes facet $(fa)$$(fa)$(fa)(f a)$(fa)$ and singed bristles ( $sn$$sn$sns n$sn$ ) show 18 percent recombination.
   a) If a singed male is crossed to $\mathrm{a}\frac{{\mathrm{fa}}^{+}}{{\mathrm{fa}}^{+}}$$\mathrm{a}\frac{{\mathrm{fa}}^{+}}{{\mathrm{fa}}^{+}}$a(fa^(+))/(fa^(+))\mathrm{a} \frac{\mathrm{fa}^{+}}{\mathrm{fa}^{+}}$\mathrm{a}\frac{{\mathrm{fa}}^{+}}{{\mathrm{fa}}^{+}}$female, what phenotypes are expected in the F1?
   b) If the $F_1$$F_1$F_(1)F_1$F_1$ males and females are inbred, what phenotypic proportions would be expected to occur in $F_2$$F_2$F_(2)F_2$F_2$ males and females?
   ii) Two recessive genes, $ds$$ds$dsd s$ds$ and $mp$$mp$mpm p$mp$ are present in corn. These are linked and are 20 map units apart. From the cross:

3. a) In the following statements, choose the alternate correct word given in parenthesis.
   i) The DNA regions in chloroplast could be observed under (light microscope/electron microscope).
   ii) The regions containing $cp$$cp$cpc p$cp$ DNA are called (nucleoids/celluloids).
   iii) Each nucleoid contains (a single/a few) copies of DNA.
   iv) The $cp$$cp$cpc p$cp$ DNA and $mt$$mt$mtm t$mt$ DNA are usually (circular/linear) in nature.
   v) Both chloroplast and mitochondria contain (a few/many) copies of DNA.
   vi) The $mt\mathrm{DNA}$$mt\mathrm{DNA}$mtDNAm t \mathrm{DNA}$mt\mathrm{DNA}$ of yeast is (bigger/smaller) than $mt\mathrm{DNA}$$mt\mathrm{DNA}$mtDNAm t \mathrm{DNA}$mt\mathrm{DNA}$ of humans.
   vii) The occurrence of introns is discovered in (yeast/human).
   b) Which among the following statements are correct?
   i) Mitochondria contain 80 S ribosomes.
   ii) The mRNA encoded by nuclear genes for the smaller subunit of RuBisCo is translated in the chloroplast.
   iii) The inheritance of chloroplast genome is independent of nuclear genome.
4. i) Define mutation. Explain the following types of mutations briefly:
   a) Induced mutations
   b) Suppressor mutations
   ii) What are transposable genetic elements? How they can cause mutations?
5. i) Industrial melanism is an excellent model to demonstrate the natural selection in action. Analyze the above statement critically.
   ii) What do you understand by sexual selection? Illustrate your answer with a suitable example.
6. Define the following terms:
   i) Heterozygous
   ii) Chromosome mapping
   iii) Genetic drift
   iv) Frame shift mutation
   v) Dosage compensation
7. What is speciation? Explain the mode of speciation.
8. i) Write a note on the applications of polyploidy.
   ii) Explain the natural causes of extinction of a species.
9. Explain the Trisomy 13 – Patau syndrome in detail.
10. Explain in detail the technique used for determining the age of rocks.

Answer:

1. i) How is co-dominance different from incomplete dominance?

Answer:

1. Co-dominance

• Definition: In co-dominance, both alleles contribute equally and visibly to the organism’s phenotype. Both traits are distinct and are fully expressed, rather than blending.
• Example: One classic example of co-dominance is the inheritance of blood type in humans. The AB blood type results from the co-dominance of the A allele and the B allele. Both the A and B antigens are present on the surface of red blood cells, and neither is dominant over the other. Hence, a person with the genotype AB has both A and B antigens on their red blood cells.
• Key Feature: Both alleles are expressed equally in the phenotype without mixing, creating a distinct expression for each allele.

2. Incomplete Dominance

• Definition: In incomplete dominance, neither allele is completely dominant over the other, and the result is a mix or blend of both parental traits in the phenotype of the heterozygote.
• Example: A well-known example of incomplete dominance is the flower color in snapdragons. If a red-flowered plant (RR) is crossed with a white-flowered plant (WW), the resulting offspring (RW) will have pink flowers. In this case, the red allele does not completely dominate the white allele; instead, the offspring exhibit a blended color.
• Key Feature: The heterozygote phenotype is a mixture or intermediate form of the two homozygous phenotypes, reflecting a partial expression of both alleles.

3. Key Differences Between Co-dominance and Incomplete Dominance

4. Conclusion

1. ii) Explain the phenomenon of masking the expression of a gene by another in epistasis.

Answer:

Introduction to Epistasis

Types of Epistasis

1. Recessive Epistasis
2. Dominant Epistasis
3. Duplicate Gene Action
4. Suppressor Epistasis

1. Recessive Epistasis

• Explanation: When an individual has two copies of a recessive allele (homozygous recessive) at one locus, the effect of another gene at a different locus is masked, regardless of the alleles present at that second locus. This results in a uniform phenotype that reflects the recessive epistatic gene.
• Example: One classic example of recessive epistasis is the inheritance of coat color in mice.
  • Suppose the coat color is determined by two genes:
    • The B gene (for black or brown fur) where B (black) is dominant over b (brown).
    • The E gene (for expression of fur color) where E (allows color expression) is dominant over e (prevents color expression).
  If an individual is homozygous recessive for the e allele (ee), the mouse will have no fur color (it will be albino), even if it carries B or b alleles. In this case, the ee genotype masks the effect of the B or b alleles, and the coat color is not expressed.
• Genotypic Ratio: In a typical dihybrid cross involving recessive epistasis, the phenotypic ratio often follows a 9:3:4 pattern, where the 4 represents the epistatic phenotype (the masked expression).

2. Dominant Epistasis

• Explanation: When a dominant allele is present at one locus, it can mask the effect of another gene at a different locus. In this case, the phenotype is determined by the epistatic gene, and the expression of the hypostatic gene is overridden by the dominant allele.
• Example: A good example of dominant epistasis can be seen in the inheritance of fruit color in squash:
  • The W allele at one locus (for white fruit color) is dominant over the other allele w (for colored fruit).
  • The C gene determines the color of the fruit (where C leads to a green fruit and c leads to a yellow fruit).
  If a squash plant has at least one dominant W allele, the fruit will be white regardless of the alleles at the C/c locus. This means the W allele is epistatic to the C allele, and the presence of W prevents the color from being expressed.
• Genotypic Ratio: In the case of dominant epistasis, the typical phenotypic ratio in a dihybrid cross is 12:3:1, where 12 represents the epistatic phenotype (in this case, white fruit color).

3. Duplicate Gene Action

• Explanation: This type of epistasis involves two genes that can independently produce the same phenotype. The dominant allele from either gene can mask the expression of the other gene, resulting in a reduced number of phenotypic outcomes.
• Example: Consider the inheritance of flower color in certain plants, where two genes, A and B, control color expression. If the dominant allele from either gene is present (A or B), the plant will have a certain flower color, even if the other gene has recessive alleles.
• Genotypic Ratio: The typical phenotypic ratio in duplicate gene action is 15:1, with most individuals showing the same phenotype due to the dominance of either of the two genes.

4. Suppressor Epistasis

• Explanation: In this type of epistasis, a mutation at one locus may cause a particular phenotype, but the presence of a second gene (called a suppressor gene) can suppress or reverse the effect of that mutation, restoring the normal phenotype.
• Example: In Drosophila melanogaster (fruit flies), a mutation in the eye color gene (which typically leads to white eyes) can be suppressed by a mutation at a second locus, which causes the eyes to appear red, masking the white eye color phenotype.

5. Conclusion

2. i) In Drosophila, the recessive, sex-linked genes, abnormal eyes facet $(fa)$$(fa)$(fa)(f a)$(fa)$ and singed bristles $(sn)$$(sn)$(sn)(s n)$(sn)$ show 18 percent recombination.

Answer:

Problem Breakdown and Answer Explanation:

Given Information:

• Sex-linked genes: Abnormal eye facets $(fa)$$(fa)$(fa)(fa)$(fa)$ and singed bristles $(sn)$$(sn)$(sn)(sn)$(sn)$.
• Recombination frequency between $fa$$fa$fafa$fa$ and $sn$$sn$snsn$sn$: 18%.
• Parental Cross: A singed male $(s^n)$$(s^n)$(s^(n))(s^n)$(s^n)$ is crossed with a female with normal eyes and normal bristles $(f{a}^{+}f{a}^{+})$$(f{a}^{+}f{a}^{+})$(fa^(+)fa^(+))(fa^+ fa^+)$(f{a}^{+}f{a}^{+})$.
• The F1 progeny will be analyzed for phenotypic ratios.

a) Phenotypes Expected in F1 (First Filial Generation)

• Male Singed Parent: Since singed bristles $(s^n)$$(s^n)$(s^(n))(s^n)$(s^n)$ is a recessive, sex-linked trait, the male parent will have the genotype $s^{n}Y$$s^{n}Y$s^(n)Ys^n Y$s^{n}Y$ (one X chromosome with the singed allele and a Y chromosome).
• Female Normal Eye/Normal Bristle Parent: The female parent has normal eyes (indicated by $f{a}^{+}f{a}^{+}$$f{a}^{+}f{a}^{+}$fa^(+)fa^(+)fa^+ fa^+$f{a}^{+}f{a}^{+}$) and normal bristles (indicated by $s{n}^{+}s{n}^{+}$$s{n}^{+}s{n}^{+}$sn^(+)sn^(+)sn^+ sn^+$s{n}^{+}s{n}^{+}$), so her genotype is $f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$fa^(+)fa^(+)sn^(+)sn^(+)fa^+ fa^+ sn^+ sn^+$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$.

Gametes from the Parents:

• Male (singed): The male produces only Y and $s^n$$s^n$s^(n)s^n$s^n$ (singed bristles allele) on the X chromosome.
  • Gametes: $s^{n}Y$$s^{n}Y$s^(n)Ys^n Y$s^{n}Y$
• Female (normal): The female can only contribute $f{a}^{+}f{a}^{+}$$f{a}^{+}f{a}^{+}$fa^(+)fa^(+)fa^+ fa^+$f{a}^{+}f{a}^{+}$ and $s{n}^{+}s{n}^{+}$$s{n}^{+}s{n}^{+}$sn^(+)sn^(+)sn^+ sn^+$s{n}^{+}s{n}^{+}$ on the X chromosome.
  • Gametes: $f{a}^{+}s{n}^{+}X$$f{a}^{+}s{n}^{+}X$fa^(+)sn^(+)Xfa^+ sn^+ X$f{a}^{+}s{n}^{+}X$

F1 Progeny Genotypes:

• Female F1: $f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$fa^(+)fa^(+)sn^(+)sn^(+)fa^+ fa^+ sn^+ sn^+$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$ (from the mother) and $s^{n}X$$s^{n}X$s^(n)Xs^n X$s^{n}X$ (from the father).
  • F1 female genotype: $f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$fa^(+)fa^(+)sn^(+)sn^(+)//s^(n)Xfa^+ fa^+ sn^+ sn^+ / s^n X$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$
  • F1 female phenotype: Normal eyes and normal bristles (since the $f{a}^{+}$$f{a}^{+}$fa^(+)fa^+$f{a}^{+}$ and $s{n}^{+}$$s{n}^{+}$sn^(+)sn^+$s{n}^{+}$ alleles are dominant).
• Male F1: The male will inherit the Y chromosome from the father and the X chromosome carrying $f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$fa^(+)fa^(+)sn^(+)sn^(+)fa^+ fa^+ sn^+ sn^+$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}$ from the mother.
  • F1 male genotype: $f{a}^{+}s{n}^{+}Y$$f{a}^{+}s{n}^{+}Y$fa^(+)sn^(+)Yfa^+ sn^+ Y$f{a}^{+}s{n}^{+}Y$
  • F1 male phenotype: Normal eyes and normal bristles (as he carries the dominant alleles for both traits).

F1 Phenotypes:

• All the F1 offspring will have the normal eyes and normal bristles phenotype. Both males and females will be heterozygous for the singed trait $s^n$$s^n$s^(n)s^n$s^n$ (meaning they carry the allele but do not express it due to the presence of the dominant normal allele).
• Expected F1 phenotypes: All offspring (both male and female) will have normal eyes and normal bristles.

b) Inbreeding of F1 and Expected Phenotypic Proportions in F2

Genotypic Cross for F2:

• F1 Female: $f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$fa^(+)fa^(+)sn^(+)sn^(+)//s^(n)Xfa^+ fa^+ sn^+ sn^+ / s^n X$f{a}^{+}f{a}^{+}s{n}^{+}s{n}^{+}/s^{n}X$
  • Gametes: $f{a}^{+}s{n}^{+}X$$f{a}^{+}s{n}^{+}X$fa^(+)sn^(+)Xfa^+ sn^+ X$f{a}^{+}s{n}^{+}X$ and $s^{n}X$$s^{n}X$s^(n)Xs^n X$s^{n}X$
• F1 Male: $f{a}^{+}s{n}^{+}Y$$f{a}^{+}s{n}^{+}Y$fa^(+)sn^(+)Yfa^+ sn^+ Y$f{a}^{+}s{n}^{+}Y$
  • Gametes: $f{a}^{+}s{n}^{+}Y$$f{a}^{+}s{n}^{+}Y$fa^(+)sn^(+)Yfa^+ sn^+ Y$f{a}^{+}s{n}^{+}Y$

F2 Generation Genotypes and Phenotypes:

• Male F2 (XY):
  • The possible F2 male genotypes are:
    • $f{a}^{+}s{n}^{+}Y$$f{a}^{+}s{n}^{+}Y$fa^(+)sn^(+)Yfa^+ sn^+ Y$f{a}^{+}s{n}^{+}Y$ (normal eyes, normal bristles).
    • $s^{n}Y$$s^{n}Y$s^(n)Ys^n Y$s^{n}Y$ (singed bristles, abnormal eyes).
• Female F2 (XX):
  • The possible F2 female genotypes are:
    • $f{a}^{+}s{n}^{+}X/f{a}^{+}s{n}^{+}X$$f{a}^{+}s{n}^{+}X/f{a}^{+}s{n}^{+}X$fa^(+)sn^(+)X//fa^(+)sn^(+)Xfa^+ sn^+ X / fa^+ sn^+ X$f{a}^{+}s{n}^{+}X/f{a}^{+}s{n}^{+}X$ (normal eyes, normal bristles).
    • $f{a}^{+}s{n}^{+}X/s^{n}X$$f{a}^{+}s{n}^{+}X/s^{n}X$fa^(+)sn^(+)X//s^(n)Xfa^+ sn^+ X / s^n X$f{a}^{+}s{n}^{+}X/s^{n}X$ (normal eyes, singed bristles).

Parental and Recombinant Types:

• The parental types (no recombination) will be:
  • Normal eyes, normal bristles (fa^+ sn^+) with singed bristles, abnormal eyes (s^n).
• The recombinant types will involve crossovers between the two loci:
  • A recombination frequency of 18% means there will be 18% recombinants in total.
  • 82% will be parental genotypes.

Phenotypic Proportions in F2:

• Normal eyes and normal bristles (wild-type) will be the dominant phenotype in both males and females. These individuals will come from both parental and recombinant types.
• Singed bristles and abnormal eyes (due to the singed bristles gene) will be the second phenotype, resulting from the presence of the $s^n$$s^n$s^(n)s^n$s^n$ allele.

F2 Male Phenotypes:

• Normal eyes and normal bristles (fa^+ sn^+ Y): 82% of F2 males will show this phenotype (parental type).
• Singed bristles and abnormal eyes (s^n Y): 18% of F2 males will show this phenotype (recombinant type).

F2 Female Phenotypes:

• Normal eyes and normal bristles (fa^+ sn^+ X / fa^+ sn^+ X): 82% of F2 females will show this phenotype (parental type).
• Normal eyes and singed bristles (fa^+ sn^+ X / s^n X): 18% of F2 females will show this phenotype (recombinant type).

Summary of Expected F2 Phenotypic Proportions:

• Male F2 (82% normal, 18% singed):
  • 82%: Normal eyes, normal bristles.
  • 18%: Singed bristles, abnormal eyes.
• Female F2 (82% normal, 18% singed):
  • 82%: Normal eyes, normal bristles.
  • 18%: Normal eyes, singed bristles.

Conclusion:

• F1 Phenotypes: All offspring will have normal eyes and normal bristles.
• F2 Phenotypic Proportions:
  • Males: 82% normal eyes, normal bristles; 18% singed bristles, abnormal eyes.
  • Females: 82% normal eyes, normal bristles; 18% normal eyes, singed bristles.

2. ii) Two recessive genes, $ds$$ds$dsd s$ds$ and $mp$$mp$mpm p$mp$ are present in corn. These are linked and are 20 map units apart. From the cross:

Answer:

Problem Breakdown:

• Parental Genotype 1: $dsmp$$dsmp$dsmpds mp$dsmp$ × $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ mp^+$d{s}^{+}m{p}^{+}$
• Parental Genotype 2: $d{s}^{+}mp$$d{s}^{+}mp$ds^(+)mpds^+ mp$d{s}^{+}mp$ × $dsm{p}^{+}$$dsm{p}^{+}$dsmp^(+)ds mp^+$dsm{p}^{+}$

Recombination Frequency Interpretation:

• 20% of the gametes produced will be recombinant (have undergone a crossover between the two loci).
• The remaining 80% of the gametes will be parental (no crossover between the loci).

Parental Cross:

• $dsmp$$dsmp$dsmpds \, mp$dsmp$ is a double recessive individual (homozygous for both the $ds$$ds$dsds$ds$ and $mp$$mp$mpmp$mp$ alleles).
• $d{s}^{+}+$$d{s}^{+}+$ds^(+)+ds^+ \, +$d{s}^{+}+$ is a wild-type individual for both loci (having the dominant alleles for both traits).

Gametes Produced by the Parents:

• The $dsmp$$dsmp$dsmpds \, mp$dsmp$ parent can produce only one type of gamete: $dsmp$$dsmp$dsmpds \, mp$dsmp$.
• The $d{s}^{+}+$$d{s}^{+}+$ds^(+)+ds^+ \, +$d{s}^{+}+$ parent can produce two types of gametes: $d{s}^{+}mp$$d{s}^{+}mp$ds^(+)mpds^+ \, mp$d{s}^{+}mp$ and $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$, each with equal probability (since both loci are heterozygous).

Types of F1 Progeny:

• Parental Genotypes: The parental genotypes will be the ones that result from no recombination:
  • $dsmp$$dsmp$dsmpds \, mp$dsmp$ × $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$ = $dsmp$$dsmp$dsmpds \, mp$dsmp$ (recessive) and $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$ (wild-type).
• Recombinant Genotypes: These result from recombination between the linked genes:
  • Recombinant gametes involve a crossover between the $ds$$ds$dsds$ds$ and $mp$$mp$mpmp$mp$ genes. After recombination, the progeny could inherit:
    • $dsm{p}^{+}$$dsm{p}^{+}$dsmp^(+)ds \, mp^+$dsm{p}^{+}$ (recombined from parental $dsmp$$dsmp$dsmpds \, mp$dsmp$ and $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$).
    • $d{s}^{+}mp$$d{s}^{+}mp$ds^(+)mpds^+ \, mp$d{s}^{+}mp$ (recombined from parental $dsmp$$dsmp$dsmpds \, mp$dsmp$ and $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$).

Expected Progeny Distribution:

• Recombination Frequency: 20% recombinant offspring and 80% parental offspring.
  Therefore, the progeny are expected to be in the following proportions:
  • 80% parental:
    • $dsmp$$dsmp$dsmpds \, mp$dsmp$ (homozygous recessive for both traits).
    • $d{s}^{+}m{p}^{+}$$d{s}^{+}m{p}^{+}$ds^(+)mp^(+)ds^+ \, mp^+$d{s}^{+}m{p}^{+}$ (homozygous wild-type for both traits).
  • 20% recombinant:
    • $dsm{p}^{+}$$dsm{p}^{+}$dsmp^(+)ds \, mp^+$dsm{p}^{+}$ (recombined, showing normal alleles for $mp$$mp$mpmp$mp$, but not for $ds$$ds$dsds$ds$).
    • $d{s}^{+}mp$$d{s}^{+}mp$ds^(+)mpds^+ \, mp$d{s}^{+}mp$ (recombined, showing normal alleles for $ds$$ds$dsds$ds$, but not for $mp$$mp$mpmp$mp$).

The Question: What percentage of the progeny would be expected to be both $ds$$ds$dsds$ds$ and $mp$$mp$mpmp$mp$ in the phenotype?

• The genotype that represents the combination of $ds$$ds$dsds$ds$ and $mp$$mp$mpmp$mp$ would be the $dsmp$$dsmp$dsmpds \, mp$dsmp$ genotype.
• The $dsmp$$dsmp$dsmpds \, mp$dsmp$ genotype is a parental type and will be present in 80% of the progeny.

Conclusion:

• Expected percentage of progeny with both $ds$$ds$dsds$ds$ and $mp$$mp$mpmp$mp$ in the phenotype: 80%.

3. a) In the following statements, choose the alternate correct word given in parenthesis.

Answer:

• Correct answer: electron microscope
  Chloroplast DNA, being part of the internal structure of chloroplasts, requires the higher magnification of an electron microscope for observation. The light microscope cannot resolve these small structures in detail.

• Correct answer: nucleoids
  The DNA in chloroplasts (and mitochondria) is organized into structures called nucleoids. "Celluloids" is not a term related to DNA organization in chloroplasts.

• Correct answer: a few
  Each nucleoid typically contains a few copies of the chloroplast or mitochondrial DNA. Chloroplasts and mitochondria usually have multiple copies of their DNA, not just a single one.

• Correct answer: circular
  Both chloroplast DNA (cpDNA) and mitochondrial DNA (mtDNA) are generally circular in shape, similar to bacterial DNA. Linear DNA is typically associated with nuclear DNA.

• Correct answer: a few
  Both chloroplasts and mitochondria usually contain a few copies of their DNA. However, it is common to refer to "a few" because the exact number can vary by cell type and organism.

• Correct answer: smaller
  Yeast mitochondrial DNA is smaller compared to human mitochondrial DNA. Yeast mitochondria typically contain less genetic material than human mitochondria.

• Correct answer: human
  The occurrence of introns (non-coding regions in genes) was first identified in humans. Yeast and many other simpler organisms have fewer or no introns in their genes.

3. b) Which among the following statements are correct?

Answer:

• Incorrect.
  Mitochondria actually contain 70S ribosomes, not 80S ribosomes. The 70S ribosomes in mitochondria are similar to those found in prokaryotes. 80S ribosomes are found in the cytoplasm of eukaryotic cells.

• Correct.
  The smaller subunit of RuBisCo is encoded by the nuclear genome, and the mRNA is transcribed in the nucleus. However, the translation of this mRNA occurs in the cytoplasm, and then the subunit is transported into the chloroplast to combine with the larger subunit (which is encoded by the chloroplast genome) to form the active enzyme RuBisCo.

• Incorrect.
  The inheritance of the chloroplast genome is not independent of the nuclear genome. While chloroplasts have their own DNA, they still rely heavily on the nuclear genome for many of the proteins and other components needed to maintain their function. Moreover, chloroplasts are typically inherited maternally in most plants, meaning their inheritance follows a pattern controlled by the nuclear genome.

4. i) Define mutation. Explain the following types of mutations briefly:

Answer:

Types of Mutations:

a) Induced mutations

b) Suppressor mutations

• Intragenic suppressors: A second mutation occurs within the same gene as the original mutation and restores the gene’s function, even though the initial mutation is still present.
• Intergenic suppressors: A second mutation occurs in a different gene, which compensates for the effects of the original mutation, often by altering the function of a different protein involved in the same pathway.

4. ii) What are transposable genetic elements? How can they cause mutations?

Answer:

• Class I (Retrotransposons): These elements move through an RNA intermediate. They are first transcribed into RNA, and then the RNA is reverse-transcribed back into DNA by an enzyme called reverse transcriptase. The newly synthesized DNA is inserted into a new location in the genome. Retrotransposons are often longer than DNA transposons and can be further categorized into long terminal repeat (LTR) retrotransposons and non-LTR retrotransposons.
• Class II (DNA Transposons): These elements move directly from one location to another in the genome. They are excised from one position and inserted into a new one via a "cut-and-paste" mechanism. DNA transposons often encode a transposase enzyme, which is responsible for cutting the transposable element out of its original location and inserting it elsewhere in the genome.

• DNA Transposons (Class II): The transposase enzyme recognizes specific sequences at the ends of the transposable element, excises the element from its current location, and integrates it into a new site within the genome. This process is known as "cut-and-paste" transposition.
• Retrotransposons (Class I): The process is more complex. First, the retrotransposon is transcribed into RNA. This RNA is then reverse-transcribed into complementary DNA (cDNA) by the enzyme reverse transcriptase. The resulting cDNA is then integrated into the genome at a new location using an enzyme called integrase.

• Insertional Mutagenesis: When a transposable element inserts itself into the middle of a gene or a regulatory region, it can disrupt normal gene function. This insertion can create a mutation by either disrupting the coding sequence of the gene or interfering with its regulation. If the insertion occurs in a critical gene or regulatory region, it can lead to a loss of function or a gain of function mutation.
• Gene Disruption: In some cases, the insertion of a transposable element may directly disrupt the expression of a gene by introducing stop codons or by disrupting the promoter region that controls gene transcription. This can lead to a nonfunctional or dysfunctional protein, causing disease or developmental issues.
• Chromosomal Rearrangements: Transposable elements can also cause chromosomal rearrangements, such as inversions, deletions, or duplications, when they insert near other transposable elements. These rearrangements can alter gene sequences or regulatory elements, leading to mutations that affect gene function.
• Regulatory Effects: Some transposable elements contain their own promoters or enhancers, which, when inserted near a gene, can cause changes in the gene’s expression levels. This can result in overexpression or underexpression of certain genes, potentially leading to diseases like cancer if critical genes are affected.

5. i) Industrial melanism is an excellent model to demonstrate natural selection in action. Analyze the above statement critically.

Answer:

• Pre-Industrial Environment: Before industrialization, the tree trunks and surrounding vegetation were light-colored due to the presence of lichen. In this environment, light-colored moths were better camouflaged from predators like birds, as they blended in with the lichen-covered surfaces.
• Post-Industrial Environment: With the onset of industrialization, factories emitted large amounts of soot, which darkened the trees and killed off the lichens. In this changed environment, the light-colored moths were more visible to predators, while the darker-colored moths, which had a mutation that made them blacker in appearance, became better camouflaged. As a result, dark-colored moths had a survival advantage and reproduced more, leading to an increase in their population over time.

• Historical Data: During the 19th century, studies showed a marked shift in the frequency of melanic (dark-colored) moths. Prior to industrialization, around 95% of the moths were light-colored, and only 5% were dark. After the Industrial Revolution, the frequency of dark-colored moths increased dramatically, with some regions reporting more than 90% melanic moths by the 1950s.
• Experimental Studies: In the 1950s, researchers like H.B.D. Kettlewell conducted experiments to test the hypothesis of natural selection. Kettlewell released both light and dark moths in polluted and unpolluted environments. His results showed that dark moths were more likely to survive in polluted areas (due to better camouflage against dark tree bark), while light moths were more likely to be preyed upon.
• Genetic Evidence: Research has also identified the specific genetic mutation responsible for melanism in the peppered moth. The mutation affects the coloration of the moth by altering the production of pigments. The increase in dark-colored moths in polluted areas correlates with the genetic presence of this mutation, further supporting the idea of natural selection.

• Simplistic View of Evolution: Critics argue that the story of industrial melanism is often portrayed as an overly simplistic model of natural selection. It implies that evolution occurs in a straightforward, linear manner, driven purely by environmental pressures. However, evolution is a more complex process that involves a variety of factors, including genetic drift, gene flow, and mutations, which can all influence the outcome of natural selection.
• Role of Other Factors: The selection pressure from predation is not the only factor that could influence the population dynamics of the moths. For example, the availability of food resources, mating behavior, and other ecological interactions could also play a role in determining which moths survive and reproduce.
• Reversal of Trend: Another limitation of the industrial melanism model is that it doesn’t fully explain the reversal of the trend in the latter half of the 20th century. As air quality improved due to pollution control efforts, the lichen began to recolonize trees, and light-colored moths began to increase again in frequency. This phenomenon suggests that selection pressures are more dynamic than previously thought and that environmental factors can cause rapid shifts in population genetics.

5. ii) What do you understand by sexual selection? Illustrate your answer with a suitable example.

Answer:

• Mate Choice (Intersexual Selection): This occurs when individuals of one sex (usually females) select mates based on particular traits in the opposite sex. These traits often signal genetic quality, health, or the ability to provide resources. Examples of such traits include elaborate plumage, colorful displays, or courtship behaviors. Females typically choose mates that they perceive will increase the chances of their offspring’s survival or improve their own reproductive success.
• Competition for Mates (Intrasexual Selection): This mechanism involves competition among individuals of the same sex, typically males, for access to mates. Males may engage in physical contests, display their strength, or assert dominance to outcompete other males. The winner, or the one that demonstrates superior traits, gets access to mates. In some species, males may develop specialized structures, such as antlers in deer or large body size in elephants, to help them secure mates.

• Peacock Tail Feathers (Mate Choice): Male peacocks are famous for their large, colorful tail feathers, which they fan out during courtship displays to attract females. The peacock’s tail is considered an extreme example of sexual selection because it seems to be a disadvantageous trait in terms of survival. The heavy tail impedes the male’s ability to fly and increases his vulnerability to predators. However, females tend to prefer males with larger, more vibrant tails. This preference suggests that the tail may signal the male’s overall health, genetic quality, and ability to survive despite the handicap, making him a more attractive mate. As a result, over generations, peacocks with larger, more colorful tails have higher reproductive success, and the trait has become more pronounced.
• Elephant Seals (Intrasexual Selection): Male elephant seals engage in fierce battles for control over territories where females gather. These battles often involve physical confrontations, with the largest and strongest males winning. The dominant males get access to the most females, allowing them to sire a greater number of offspring. Over time, this selection pressure leads to the evolution of larger body size and enhanced physical strength in males, as these traits increase their chances of winning contests and securing mates.

6. Define the following terms:

Answer:

7. What is speciation? Explain the mode of speciation.

Answer:

• Ecological Isolation: Populations may exploit different ecological niches within the same environment. For example, one group might feed on a particular type of food, while another group feeds on a different food source. Over time, differences in feeding habits, mating preferences, or behavior can reduce gene flow between the groups, leading to speciation.
• Polyploidy: Polyploidy refers to the condition where an organism has more than two complete sets of chromosomes. In plants, polyploidy can result in the instant formation of a new species because the polyploid individual is unable to interbreed with the original diploid population. This form of sympatric speciation is common in plants and can occur due to errors in cell division.

• Prezygotic Isolation: This occurs before fertilization and includes mechanisms like geographic isolation, temporal isolation (mating at different times), behavioral isolation (different mating rituals), and mechanical isolation (differences in reproductive structures).
• Postzygotic Isolation: This occurs after fertilization, and includes mechanisms such as hybrid inviability (offspring do not survive), hybrid sterility (offspring are sterile, e.g., mule), and hybrid breakdown (offspring are viable but have reduced fitness in subsequent generations).

8. i) Write a note on the applications of polyploidy.

Answer:

• Creation of Larger Crops: One of the most important applications of polyploidy in agriculture is the creation of larger, more productive crops. For instance, polyploidy has been used to produce larger fruit, grains, and vegetables. Examples include polyploid varieties of wheat, corn, and strawberries, where polyploidy results in larger cells and, consequently, larger fruit or seeds.
• Increased Vigor and Disease Resistance: Polyploid plants often exhibit hybrid vigor (also known as heterosis), where the polyploid plants show superior qualities compared to their diploid counterparts. These enhanced traits include greater resistance to environmental stress, disease, and pests. Polyploid plants may also be more robust and adaptable to changing climates or less fertile soils. For example, tetraploid varieties of wheat tend to be hardier and more resistant to drought than their diploid counterparts.
• Improved Seedless Crops: In some cases, polyploidy has been used to create seedless varieties of fruits. Polyploidy can cause sterility in plants, which results in seedless varieties that are more desirable for human consumption. A well-known example is the seedless watermelon, which is often a triploid (3 sets of chromosomes) hybrid that cannot produce viable seeds.

• Enhanced Flower Size and Color: In ornamental plants, polyploidy can result in more vibrant colors and larger flowers. Gardeners and horticulturists often use induced polyploidy to create aesthetically pleasing varieties of flowers such as tulips, roses, and petunias. Polyploidy can lead to more robust plants that have higher aesthetic value, making them popular for commercial landscaping and floral industry use.
• Improved Growth Rate and Stress Tolerance: Many ornamental plants, when subjected to polyploidy, show increased growth rates and greater tolerance to environmental stressors, such as heat, drought, or poor soil. These traits make polyploid plants especially attractive for landscaping in urban or arid regions.

• Wheat and Other Cereal Crops: Polyploidy has been used extensively in cereal crops, especially wheat. The origin of modern bread wheat (Triticum aestivum) is based on hybridization and polyploidy. This species is a hexaploid (having six sets of chromosomes) formed from the hybridization of different species of wild grasses. This combination of genetic material led to higher yields, better disease resistance, and improved adaptability to different environments. Similarly, other cereals like corn and rice also benefit from polyploidy in breeding programs aimed at producing high-yielding crops.
• Hybrid Polyploids: Many commercial crops, such as certain varieties of tobacco, cotton, and potatoes, have been artificially polyploidized through hybridization, improving their performance. Polyploid hybrids can combine beneficial characteristics from two species or varieties, such as increased size, improved quality, and better stress resistance.

• Increased Production of Secondary Metabolites: In some plants, polyploidy can enhance the production of secondary metabolites, which have pharmaceutical and industrial uses. For example, certain polyploid plants produce higher levels of alkaloids, flavonoids, and other compounds that are valuable for the production of medicines, dyes, and other industrial products.
• Transgenic Plants: Polyploidy is also used in the development of genetically modified organisms (GMOs). By inducing polyploidy, scientists can create stable transgenic plants that carry specific genetic modifications, improving traits such as pest resistance, herbicide tolerance, or nutritional value. Polyploidy can provide the genetic stability required for the long-term development of such genetically engineered crops.

• Instantaneous Speciation: Polyploidy is one of the few mechanisms that can lead to instantaneous speciation, particularly in plants. When an organism undergoes a polyploidy event, it may no longer be able to interbreed with its diploid relatives due to differences in chromosome numbers. This reproductive isolation can result in the formation of a new species. Many plant species, especially those in the Genus Brassica (cabbage, kale, etc.) and Taraxacum (dandelions), have originated through polyploidy.
• Genetic Diversity: Polyploidy contributes to genetic diversity within plant populations, enabling plants to adapt to different environmental pressures. This can be crucial for understanding how species evolve and adapt over time in response to changes in their environment.

8. ii) Explain the natural causes of extinction of a species.

Answer:

• Volcanic Eruptions: Massive volcanic eruptions can release ash and gases into the atmosphere, causing widespread climate changes, acid rain, and habitat destruction.
• Asteroid Impacts: The impact of a large asteroid, such as the one that caused the extinction of the dinosaurs 66 million years ago, can generate tsunamis, wildfires, and "nuclear winter"-like conditions by blocking sunlight.
• Earthquakes and Tsunamis: These natural disasters can lead to habitat destruction and disrupt ecosystems, particularly in coastal regions.

• Ice Ages: Periods of extreme cooling can cause glaciation, which alters habitats, reduces food availability, and forces species to migrate or adapt. Many species that fail to adjust to such drastic changes go extinct.
• Global Warming Periods: Conversely, periods of warming can lead to the melting of ice caps, rising sea levels, and changes in ocean circulation. These changes can disrupt the ecosystems on which species depend.
• Volcanic Activity: Continuous volcanic eruptions can release greenhouse gases like carbon dioxide, contributing to long-term warming trends.

• Plate Tectonics: The movement of tectonic plates can reshape continents, create mountain ranges, and separate populations of species. Isolation can limit genetic diversity and adaptability, leading to extinction.
• Sea-Level Changes: Rising or falling sea levels can expose or inundate habitats, forcing species to adapt or perish.
• Continental Drift: As continents shift, climate patterns and ecosystems change, often leading to the extinction of species that cannot migrate or adapt.

• Interspecies Competition: When resources like food, water, or shelter become scarce, species may compete for survival. Stronger or more adaptable species often outcompete weaker ones, leading to their extinction.
• Introduction of Predators: Sometimes, natural migration or environmental changes introduce new predators into an ecosystem. Native species that are not equipped to defend against these predators may face extinction.
• Ecosystem Imbalances: A significant population increase or decrease in one species can disrupt the balance of an ecosystem, causing cascading effects on other species.

• Epidemics: Widespread diseases can decimate species, particularly if they lack immunity or genetic diversity. For instance, fungal infections have driven some amphibians to near extinction.
• Parasitism: Parasites can weaken hosts over time, making them vulnerable to predators or environmental changes.

• Lack of Genetic Diversity: Populations with low genetic diversity are less able to adapt to environmental changes or resist diseases, increasing the risk of extinction.
• Evolutionary Dead Ends: Some species evolve traits that make them less suited to survive in changing environments. For example, over-specialization, such as reliance on a single food source, can be detrimental when that food becomes scarce.
• Natural Inbreeding: In small populations, inbreeding can lead to genetic defects and reduced fertility, contributing to extinction.

9. Explain the Trisomy 13 – Patau syndrome in detail.

Answer:

• Types of Trisomy 13: There are three types of trisomy 13:
  1. Full Trisomy 13: This is the most common form of Patau syndrome, where every cell in the body has three copies of chromosome 13 instead of the usual two. It occurs due to a failure in cell division during meiosis.
  2. Mosaic Trisomy 13: In this form, some cells in the body have the typical two copies of chromosome 13, while others have three copies. This form can lead to a milder phenotype.
  3. Translocation Trisomy 13: In this rare form, a part of chromosome 13 breaks off and attaches to another chromosome, often chromosome 14. The individual may not have the extra chromosome but instead a rearrangement of genetic material.

• Maternal Age: One of the most significant risk factors for trisomy 13 is the age of the mother. Women over the age of 35 are more likely to have children with chromosomal abnormalities, including Patau syndrome, due to the increased risk of errors during egg cell division.
• Paternal Age: While maternal age is the most well-established risk factor, some studies suggest that older paternal age may also contribute to chromosomal abnormalities, although its role in trisomy 13 is less clear.
• Previous Children with Trisomy: If a parent has previously had a child with trisomy 13 or another chromosomal disorder, the risk of having another child with the condition may increase.
• Family History: In rare cases, there may be a family history of chromosomal rearrangements or balanced translocations involving chromosome 13, which can increase the risk of having a child with trisomy 13.

• Craniofacial Abnormalities: Children with Patau syndrome often have distinctive facial features, including cleft lip and palate, microphthalmia (small eyes), and an abnormal head shape (microcephaly).
• Central Nervous System Abnormalities: Many individuals with trisomy 13 have severe intellectual disability and developmental delays. Brain abnormalities such as holoprosencephaly (failure of the brain to divide into two hemispheres) and corpus callosum agenesis (absence of the structure that connects the two hemispheres of the brain) are common.
• Heart Defects: Heart malformations are present in the majority of cases of Patau syndrome. These can include ventricular septal defects (holes in the heart’s chambers), patent ductus arteriosus (a blood vessel that remains open), and atrial septal defects.
• Polydactyly: Some children with Patau syndrome are born with extra fingers or toes (polydactyly), which is a characteristic feature of the syndrome.
• Renal Abnormalities: Kidneys may be malformed or underdeveloped, and urinary tract problems, such as hydronephrosis (swelling of the kidneys), are common.
• Skeletal Abnormalities: Many children with Patau syndrome have a variety of skeletal abnormalities, including rocker-bottom feet, which are characterized by a rounded, downward curve of the foot’s sole.
• Other Features: Additional features may include genital abnormalities, clubbed feet, and various skin and limb deformities. Children may also suffer from feeding difficulties, poor growth, and immune system dysfunction.

• Prenatal Screening: Non-invasive prenatal testing (NIPT), amniocentesis, or chorionic villus sampling (CVS) can be used during pregnancy to detect chromosomal abnormalities. These tests analyze the DNA of the fetus to determine whether trisomy 13 is present. Amniocentesis and CVS carry a small risk of miscarriage but provide a definitive diagnosis.
• Ultrasound: Prenatal ultrasound imaging can sometimes reveal physical markers associated with Patau syndrome, such as heart defects, abnormal facial features, or other malformations. While ultrasound cannot definitively diagnose trisomy 13, it can prompt further genetic testing.
• Postnatal Diagnosis: After birth, a blood test called karyotyping can be performed to examine the baby’s chromosomes. This test will confirm if there are three copies of chromosome 13, confirming the diagnosis of Patau syndrome.

• Surgical Interventions: Some congenital defects, such as cleft lip and palate or heart defects, may be treated with surgery. However, the decision to proceed with surgery depends on the individual’s overall health and prognosis.
• Palliative Care: In many cases, the goal of treatment is to provide comfort and support for the child. Palliative care involves managing pain and addressing any discomfort caused by the physical symptoms of the condition.
• Supportive Therapies: Children with trisomy 13 may benefit from therapies such as physical therapy, speech therapy, and occupational therapy to help with developmental delays and motor skills. However, due to the severity of the condition, these interventions may not significantly alter the course of the disorder.

10. Explain in detail the technique used for determining the age of rocks.

Answer:

• Law of Superposition: This fundamental principle states that in a sequence of undisturbed sedimentary rocks, the oldest layers are at the bottom, and the youngest layers are at the top. By examining the sequence of rock layers (strata), geologists can determine the relative ages of the rocks.
• Principle of Original Horizontality: This principle asserts that layers of sediment are originally deposited horizontally. If rocks are found tilted or folded, it indicates that they were disturbed after their initial deposition, allowing geologists to infer the relative timing of the deformation.
• Principle of Lateral Continuity: According to this principle, layers of sedimentary rock extend laterally in all directions until they thin out or encounter a physical barrier. Geologists can use this information to correlate rock layers from different locations.
• Principle of Cross-Cutting Relationships: This principle states that if a rock unit or geological feature (such as a fault or an igneous intrusion) cuts through other rock layers, it must be younger than the rocks it disrupts. This helps establish the relative timing of geological events.
• Fossil Correlation: Fossils found in different rock layers can be used to correlate layers of rock from different locations. Fossils of certain species are known to have existed only within a specific time range. Thus, by identifying the fossils in rock layers, geologists can estimate the relative age of those rocks.

• Radiometric Dating: Radiometric dating is the most common technique for determining the absolute age of rocks. It is based on the principle that certain isotopes of elements in rocks decay at a constant, known rate over time. By measuring the ratio of parent isotopes to daughter isotopes in a rock sample, geologists can calculate the age of the rock. Some of the most common radiometric dating methods include:
  • Uranium-Lead Dating: This method is commonly used for dating igneous rocks. It relies on the decay of uranium isotopes (U-238 and U-235) to form stable lead isotopes (Pb-206 and Pb-207). The ratio of uranium to lead in the rock sample is used to calculate its age. Uranium-lead dating is particularly useful for dating rocks that are millions to billions of years old, making it essential for understanding the geological history of the Earth.
  • Potassium-Argon Dating: Potassium-argon (K-Ar) dating is another important method for dating volcanic rocks. Potassium-40 (K-40) decays into argon-40 (Ar-40) over time. Since argon is a gas, it escapes from molten rock, and only when the rock solidifies does the argon become trapped. By measuring the amount of argon in a rock sample, geologists can determine the age of volcanic rocks.
  • Carbon-14 Dating: Carbon-14 dating is used to determine the age of organic materials (such as wood, bone, or plant material) that are less than about 50,000 years old. This method is based on the decay of the radioactive isotope carbon-14 (C-14), which is absorbed by living organisms. When the organism dies, it stops absorbing carbon-14, and the isotope begins to decay. By measuring the remaining amount of C-14 in the sample, scientists can calculate the time since the organism’s death.
• Fission Track Dating: This method involves counting the tracks created by the spontaneous fission of uranium-238 within minerals such as apatite and zircon. The number of tracks accumulated in a mineral sample over time can be used to determine the age of the rock.
• Thermoluminescence Dating: Thermoluminescence dating is used to determine the last time minerals like quartz or feldspar were exposed to sunlight or heat. When these minerals are exposed to radiation, electrons are trapped in the mineral’s crystal structure. Heating the mineral causes the trapped electrons to release energy in the form of light, which can be measured to determine the age of the sample.

• Half-Life: The half-life of an isotope is the time it takes for half of a sample of that isotope to decay into its daughter isotope. This decay occurs at a predictable and constant rate, making it possible to calculate the age of a rock or mineral by measuring the ratio of parent to daughter isotopes. Different isotopes have different half-lives, ranging from thousands to billions of years. For example, carbon-14 has a half-life of about 5,730 years, while uranium-238 has a half-life of 4.47 billion years.
• Accuracy and Precision: The accuracy of radiometric dating depends on several factors, including the quality of the sample, the presence of contamination, and the initial conditions of the sample. Geologists typically use multiple isotopic methods on the same sample to ensure the reliability of the results.

• Stratigraphy: Stratigraphy involves the study of rock layers (strata) and their arrangement. By analyzing the sequence of rock layers, geologists can infer the relative timing of the deposition of rocks and identify major geological events such as volcanic eruptions, glaciations, or shifts in climate. Stratigraphy often uses fossil content, geochemical markers, and sedimentary structures to refine age estimates.
• Fossil Dating: Fossils found in sedimentary rock layers can also help establish the age of rocks. Fossils of certain species are known to have existed only during specific time periods, so by identifying these fossils in rock layers, geologists can approximate the age of those layers. This method is especially useful in regions where radiometric dating is difficult or impossible, such as in sedimentary rock sequences without volcanic layers.

• Contamination: Contamination of rock samples, such as the introduction of foreign carbon or parent isotopes, can skew the results of dating techniques. This is particularly an issue in carbon-14 dating, where contamination by modern carbon can lead to incorrect age estimates.
• Inability to Date Certain Rocks: Some rock types, such as sedimentary rocks, cannot be directly dated using radiometric methods. In these cases, geologists must rely on relative dating or other indirect methods.
• Errors in Initial Conditions: Radiometric dating relies on assumptions about the initial conditions of the rock sample, such as the absence of daughter isotopes at the time of formation. If these assumptions are incorrect, it can lead to inaccurate age estimates.