A male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophilic gene h. What proportion of his sperms will be abh :-
A
$\dfrac{1}{32}$
B
$\dfrac{1}{16}$
C
$\dfrac{1}{4}$
D
$\dfrac{1}{8}$
The answer is D.
Solution
Genotype of the man: Aa Bb $X^hY$ (heterozygous for A and B, and hemizygous for h, i.e. one $X^h$ and a Y). Assuming independent assortment, each gene pair separates independently into the sperms. Probability of a sperm getting a $= \dfrac{1}{2}$ (A or a) Probability of getting b $= \dfrac{1}{2}$ (B or b) Probability of getting the X with h $= \dfrac{1}{2}$ (the other half of the sperms carry Y) Probability of an abh sperm $= \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$ The man forms $2^3 = 8$ types of sperms in equal proportions (ABh, ABY, Abh, AbY, aBh, aBY, abh and abY), and abh is one of them. So $\dfrac{1}{8}$ of his sperms will be abh.
ConceptProbability of a gamete type
Principles of Inheritance and Variation in past papers
198 questions from this chapter have appeared across 27 exam years.
