Question

A woman heterozygous for haemophilia marries a haemophilic man. What will be the ratios of carrier daughters, haemophilic daughters, normal sons and haemophilic sons in F${}_1$ generation?

• A. 1: 2: 2:1
• B. 2: 1: 1: 2
• C. 1: 1: 1: 1
• D. 1: 2: 1: 2

Answer 1

The correct option is C 1: 1: 1: 1

As haemophilia is a sex-linked recessive disorder, it gets expressed when genes are altered on both the X chromosomes, allosomes (XX) in females. If out of the two X chromosomes, any one is altered and the other X chromosome is normal, then the females will not have haemophilia, but they could be the carriers and pass the disease to their offsprings.
While in males, as they have only one X chromosome in their allosomes (XY), alteration in that one X chromosome can cause the disease to get expressed.

Thus, the genotype of allosomes of a haemophilic man will be $X^{h}Y$.
The genotype of allosomes of normal man will be XY.
The genotype of allosomes of a normal woman will be XX.
The genotype of allosomes of a carrier woman will be $X^{h}X$.
The genotype of allosomes of a haemophilic woman will be $X^h{X}^h$.

According to the question, a woman heterozygous for haemophilia will be a carrier with genotype $X^{h}X$. She marries a haemophilic man whose genotype will be $X^{h}Y$.

Let’s find out their $F_1$ progenies genotypes by Punnett square method.

$X^{h}X$ X $X^{h}Y$

$\begin{array}{ccc}\text{Gametes}& X^h& \text{Y}\\ X^h& X^h{X}^h\text{(affected female)}& X^{h}Y\text{(affected male)}\\ \text{X}& X^{h}X\text{(carrier female)}& \text{XY (normal male)}\end{array}$

$\begin{array}{ccc}& \text{Genotype}& \\ \text{Carrier daughter}& X^{h}X& 1\\ \text{Haemophilic daughter}& X^h{X}^h& 1\\ \text{Normal son}& XY& 1\\ \text{Haemophilic son}& X^{h}Y& 1\end{array}$

Hence, the ratio is 1:1:1:1.