Question

a non colorblind male marries a carrier female. if they have a male child, what is the chance he will be colorblind? a 25% b 0% c 100% d 50%

Explanation:

Step1: Determine the genotypes

Let \(X^B\) represent the normal - vision allele and \(X^b\) represent the color - blind allele. Males have the genotype \(XY\). Since the male is non - colorblind, his genotype is \(X^BY\). Females have the genotype \(XX\). Since the female is a carrier, her genotype is \(X^BX^b\).

Step2: Analyze the inheritance for male children

For male children, they get the \(Y\) chromosome from the father. From the mother (\(X^BX^b\)), the possible alleles that can be passed to the male child (who gets \(Y\) from the father) are \(X^B\) or \(X^b\). The probability of getting \(X^b\) from the mother (since there are 2 possible \(X\) - linked alleles (\(X^B\) and \(X^b\)) that the mother can donate to the male child) is \(\frac{1}{2}\) or \(50\%\).

Answer:

D. 50%