Question

3/31
look at this cross: x^b x^b x x^b y
what proportion of the male children are colorblind?
75%
25%
100%
50%

Explanation:

Step1: Analyze the cross

The mother is \(X^{B}X^{b}\) (carrier for colorblindness, an X - linked recessive trait) and the father is \(X^{B}Y\) (normal male). For male children, they inherit the Y chromosome from the father and one X chromosome from the mother. The mother can pass on either \(X^{B}\) or \(X^{b}\) to her sons.

Step2: Determine the genotypes of male children

The possible genotypes of male children are \(X^{B}Y\) (normal) and \(X^{b}Y\) (colorblind). The probability of getting \(X^{b}\) from the mother (to have a colorblind son) is \(\frac{1}{2}\) or 50%.

Answer:

50% (The option with "50%")