Question

A recessive X-linked gene mutation is known to generate premature baldness in males but is without effect in women. If a heterozygous female marries an affected male, what proportion of all their children is expected to be prematurely bald?

Answer 1

Answer:

$\frac{1}{4}$

Explanation:

Let "XB " represents the dominant allele for no baldness

and let "Xb " represents the recessive premature male-pattern baldness

Genotype of affected male is Xb Y

Genotype of heterozygous female XB  Xb

Now when the above male and female are crossed, the following offspring are produced

Xb Y * XB  Xb

XB  Xb, XB Y, Xb Xb, Xb Y

Since, this Xb allele is only expressed in males, therefore females will not be affected by it

Hence, out of 4 offspring only one male offspring with "XbY" genotype will have premature baldness

$\frac{1}{4}$