Genetics 2019
3. Answer any two questions from the following.
a) Define spontaneous mutation. What is a base analogue? A base analogue can induce mutation -explain with suitable examples.
Ans→ A spontaneous mutation is a genetic change that occurs naturally without any external influence, typically due to errors during DNA replication or repair.
A base analogue is a chemical compound that closely resembles a natural DNA base and can be incorporated into DNA, potentially causing mutations.
Base analogues can induce mutations by resembling natural nitrogenous bases and being incorporated into the DNA during replication. Once incorporated, they may pair incorrectly, causing base substitutions. Here are two examples:
• 5-bromouracil (5-BU): This is a base analogue of thymine. It can pair with adenine (like thymine) but also with guanine (like cytosine) under certain conditions. If 5-BU pairs with guanine, it can lead to a transversion mutation (thymine to guanine pair swapped for adenine to cytosine pair) during DNA replication, causing a permanent mutation.
• 2-aminopurine (2-AP): This is an analogue of adenine. It typically pairs with thymine, but it can also pair with cytosine. If 2-AP pairs with cytosine, it causes a mutation when the DNA replicates, resulting in a base substitution (adenine to guanine, for example).
b) The mechanism of dosage compensation in Drosophila is opposite than in human’- Justify the statement.
Ans→Drosophila:
In Drosophila, dosage compensation is achieved through hyperactivation/hyper transcription of the single X chromosome in males to equalize the gene expression levels with females (which have two X Chromosome).
The male Drosophila has one X chromosome and in order to compensate the difference is gene expression dosage with females, the single X chromosome involves increased transcription:
Human
In humans, dosage compensation is achieved by through the inactivation of one X chromosome in females through a process known as X-inactivation results Barr body.
This means, each somatic cell of a female, one X-chromosome becomes transcriptionally inactive, balancing the gene expression levels between males and females.
So the key difference lies in which sex undergoes dosage compensation. In Drosophila, it’s male through hyperactivation of X chromosome and in human its female through X-inactivation
Hence, the dosage compensation in Drosophila is opposite in human.
c) Describe the genetic and biochemical basis of PKU.
Ans→ Genetic Basis of PKU
- Gene Mutation: PKU is caused by mutation in the PAH gene located on chromosome 12. This gene provides instructions for making enzymes called phenylalanine hydroxylase (PAH).
- PAH Gene Function: PAH is crucial for breaking down the amino acid-phenylalanine (Phe) found in protein containing foods. In individuals with PKU, the PAH enzyme is either deficient or non-functional due to mutations.
- Inheritance: PKU is typically inherited in an autosomal recessive manner. To develop PKU, an individual must inherit a mutated PAH gene from both parents.
Biochemical Basis of PKU
- Phenylalanine Accumulation. In individuals with PKU, the lack of function, PAH enzyme leads to accumulation of phenylalanine in the blood and tissues.
- Neurological impact: Excessive phenylalanine levels are toxic to the developing brain particularly during infancy and early childhood result neurological problems.
- Tyrosine Depletion: Phenylalanine is normally converted into tyrosine by PAH enzyme. In PKU, this conversion is impaired, leading to reduced levels of tyrosine,. Tyrosine is important for the production of neurotransmitter & melanin.
Genetics 2020
a) Define spontaneous mutation. What is a base analogue? A base analogue can induce mutation- explain with suitable examples.
Ans→ A spontaneous mutation is a genetic change that occurs naturally without any external influence, typically due to errors during DNA replication or repair.
A base analogue is a chemical compound that closely resembles a natural DNA base and can be incorporated into DNA, potentially causing mutations.
Base analogues can induce mutations by resembling natural nitrogenous bases and being incorporated into the DNA during replication. Once incorporated, they may pair incorrectly, causing base substitutions. Here are two examples:
• 5-bromouracil (5-BU): This is a base analogue of thymine. It can pair with adenine (like thymine) but also with guanine (like cytosine) under certain conditions. If 5-BU pairs with guanine, it can lead to a transversion mutation (thymine to guanine pair swapped for adenine to cytosine pair) during DNA replication, causing a permanent mutation.
• 2-aminopurine (2-AP): This is an analogue of adenine. It typically pairs with thymine, but it can also pair with cytosine. If 2-AP pairs with cytosine, it causes a mutation when the DNA replicates, resulting in a base substitution (adenine to guanine, for example).
b) What is dosage compensation? Briefly explain the role of alternative splicing in sex determination in Drosophila.
Ans→ Dosage compensation is a mechanism that balances the expression of X-linked genes between males (XY) and females (XX) to ensure equal levels of gene products, despite the difference in the number of X chromosomes.
Alternative splicing plays a crucial role in sex determination in Drosophila by regulating the expression of sex-specific proteins through the differential processing of mRNA transcripts. The key gene involved is Sex-lethal (Sxl):
- Sex-lethal (Sxl):
- In females, the Sxl gene is transcribed and spliced in a way that produces a functional Sxl protein, which then directs the splicing of downstream genes.
- In males, Sxl is transcribed but spliced differently, leading to a non-functional protein.
- Transformer (tra):
- In females, Sxl protein ensures that the tra pre-mRNA is spliced to produce functional Tra protein.
- In males, absence of functional Sxl protein results in incorrect splicing of tra, leading to a non-functional Tra protein.
- Doublesex (dsx):
- Functional Tra protein in females promotes splicing of dsx pre-mRNA to produce a female-specific Dsx isoform.
- In males, the absence of Tra protein results in a male-specific Dsx isoform.
These sex-specific isoforms of Dsx regulate the expression of target genes responsible for the development of male or female phenotypes..
c) What is Xist RNA? Write down the steps of X Chromosome inactivation. Enumerate the ways in which Xist blocks transcription.
Ans→Xist RNA (X-inactive specific transcript) is a long non-coding RNA responsible for silencing one of the two X chromosomes in female mammals during X-chromosome inactivation.
X-Chromosome Inactivation
- Initiation: X-chromosome inactivation occurs early in embryonic development.
- Xist RNA Production: Xist (X-inactive specific transcript) gene on one X-chromosome is activated.
- Spreading of Xist RNA: Xist RNA spreads along the X-chromosome.
- Chromatin Modification: Chromatin on the inactive X-chromosome undergoes modifications.
- Barr body formation: The inactive X chromosome condenses into a dense structure known as Barr body.
- Maintenance of Inactivation: X-chromosome inactivation is stably maintained through cell division.
Xist blocks Transcription
- Xist RNA coats the entire X-chromosome undergoing inactivation.
- Xist RNA recruits proteins that modify chromatin structure
- Xist associated proteins induce histone modifications.
- Xist-mediated recruitment of DNA methyl transferases result DNA methylation.
- Xist RNA and associated proteins physically access the transcriptional machinery to the X-chromosome and block the transcription.
- The transcriptional repression induced by Xist is stably maintained through cell divisions.
Genetics 2021
a) Define epistasis. Differentiate between dominant and recessive epistasis.
Ans→ Epistasis is the interaction between genes, where the effect of one gene is modified or masked by one or more other genes.
| Features | Dominant Epistasis | Recessive Epistasis |
| 1. Definition | A dominant allele at one gene locus masks the expression of alleles at another locus. | A recessive allele (in homozygous form) at one gene locus masks the expression of alleles at another locus. |
| 2. Genotypic Interaction | The dominant allele of one gene (e.g., A) suppress the effect of another gene (e.g., B/b) | The homozygous recessive genotype (e.g., aa) suppresses the effect of another gene (e.g., B/b). |
| 3. Phenotypic Ratio | 12:3:1 in the F2 generation. | 9:3:4 in the F2 generation. |
| 4. Mechanism | A single copy of the dominant allele is enough to suppress the expression of the other gene. | Both copies of the recessive allele are required to suppress the expression of the other gene. |
| 5. Example | In squash color, the allele W (white) is epistatic to G (green) and g (yellow). | In Labrador retrievers, the allele ee (recessive) masks the expression of B (black) or b (brown). |
b) Describe the genetic and biochemical basis of PKU
Ans→ Genetic Basis of PKU
- Gene Mutation: PKU is caused by mutation in the PAH gene located on chromosome 12. This gene provides instructions for making enzymes called phenylalanine hydroxylase (PAH).
- PAH Gene Function: PAH is crucial for breaking down the amino acid-phenylalanine (Phe) found in protein containing foods. In individuals with PKU, the PAH enzyme is either deficient or non-functional due to mutations.
- Inheritance: PKU is typically inherited in an autosomal recessive manner. To develop PKU, an individual must inherit a mutated PAH gene from both parents.
Biochemical Basis of PKU
- Phenylalanine Accumulation. In individuals with PKU, the lack of function, PAH enzyme leads to accumulation of phenylalanine in the blood and tissues.
- Neurological impact: Excessive phenylalanine levels are toxic to the developing brain particularly during infancy and early childhood result neurological problems.
- Tyrosine Depletion: Phenylalanine is normally converted into tyrosine by PAH enzyme. In PKU, this conversion is impaired, leading to reduced levels of tyrosine,. Tyrosine is important for the production of neurotransmitter & melanin.
c) Briefly describe the generalized transduction process with suitable diagrams.
Ans→ (1) Phage Infection: A bacteriophage (virus) infects the bacterial cell surface, and injects its genetic material (DNA/RNA) into the bacterium
(2) Bacterial DNA degradation: The phage takes control of the bacterial cellular machinery, and produces new phage replicas.
(3) Packing Error: During assembly of new phage replicas a packing error may occur. The phage encapsulates the fragment of bacterial DNA.
(4) Transducing Phage formation: The resulting phage now carrying bacterial DNA, is termed a transducing plage.
(5) Infection of new bacteria: The transducing phage infects a new bacterial host cell.
(6) Integration into Host Genome: The transferred bacterial DNA may integrate into the genome of the new bacterial host.
(7) Homologous recombination: Homologous recombination may occur between the transduced bacterial DNA and recipient’s bacterium chromosome.
(8) Gene transfer and genetic Diversity: The transduced bacterial DNA can introduce new genetic traits into the recipient’s bacterium.
Genetics 2022
a) Define spontaneous mutation. What is a base analogue? ‘A base analogue can induce mutation’ -Explain.
b) Discuss the role played by Y-chromosome in human sex determination.
Ans→ Y-chromosome, particularly the SRY gene is initiating the male developmental pathway during embryogenesis. It leads the formation of male reproductive organs and the secondary sexual characteristics in humans.
(1) The SRY gene is located on the Y chromosome. During embryonic development the SRY gene is expressed in the undifferentiated gonadal tissues
(2) SRY triggers the development of testes, The presence of testes leads to the secretion of testosterone.
(3) Testosterone is a key male sex hormone that directs the development of male reproductive structures.
(4) Testes also produces Anti-Mullerian Hormone (AMH) which suppresses the development of female reproductive structures (Mullerian ducts).
(5) Testosterone promotes the development of Wolffian ducts into male reproductive structures (epididymis, vas deferens).
(6) Beyond primary sexual characteristics, the Y chromosome influences secondary sexual characteristics.
c) The following numbers were obtained from test cross progeny in Drosophila.
| Phenotypes | Number |
| 1. + m + | 218 |
| 2. w + f | 236 |
| 3. + + f | 168 |
| 4. w m + | 178 |
| 5. + m f | 95 |
| 6. w + + | 101 |
| 7. + + + | 3 |
| 8. w m f | 1 |
| 1000 |
Ans→
| Phenotypes | Number | Type |
| 1. + m + | 218 | Parental combination |
| 2. w + f | 236 | Parental combination |
| 3. + + f | 168 | SCO product between w & m |
| 4. w m + | 178 | SCO product between w & m |
| 5. + m f | 95 | SCO product between m & f |
| 6. w + + | 101 | SCO product between m & f |
| 7. + + + | 3 | DCO product |
| 8. w m f | 1 | DCO product |
| 1000 |
Recombination frequency between gene w & m
= {( 168 +178 + 3 + 1 )/1000 } × 100
= (350/1000) × 100
= 35.0 %
The distance between w & m is 35 cM.
Recombination frequency between gene m & f
= {( 95 +101 + 3 + 1 )/1000 } × 100
= (200/1000) × 100
= 20.0 %
The distance between m & f is 20 cM.
Recombination frequency between w & f
= 35 % + 20 %
= 55 %
Distance between gene w & f is 55 cM.
