25.
2 answers:
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Answer:
Please find the punnet square to this question as an attachment
F1 generation:
genotype = BW
Phenotype = Erminette offsprings
F2 generation:
genotype = BB (1): BW(2): WW(1)
Phenotype = 1 Black, 2 Erminette, 1White
Explanation:
This question involves a gene coding for feather color in chickens. The allele for black feathers (B) is codominant with the allele for white feathers (W) to form an erminette chicken (black and white speckles).
According to this question, a cross between a chick with black (BB) feathers and chicken with white (WW) feathers will result in an all erminette chicken (BW) in the F1 generation (see attached image)
Also, in the F2 generation got by self-crossing the Erminette genotype in the F1 generation (BW), the following genotypic and phenotypic ratios are observed:
Genotypic ratio = BB (1): BW(2): WW(1)
Phenotypic ratio = 1 Black, 2 Erminette, 1White
Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
