Question

genetic inheritance: predict occurence of color blindness in offspring
if a female was a carrier for sex - linked color blindness, what percentage of her male children would also be color blind?
multiple choice
25%
50%
0%
100%

Explanation:

Step1: Determine the female carrier's genotype

A female carrier for sex - linked color blindness has the genotype \(X^BX^b\) (where \(X^B\) is the normal allele and \(X^b\) is the color - blind allele).

Step2: Analyze the male children's inheritance

Male children receive the \(Y\) chromosome from the father and an \(X\) chromosome from the mother. The possible \(X\) chromosomes from the mother are \(X^B\) or \(X^b\). There are 2 possible \(X\) - chromosome contributions from the mother (\(X^B\) and \(X^b\)) for male children.

Step3: Calculate the probability

The probability of a male child receiving the \(X^b\) (color - blind) allele is \(\frac{1}{2}\) or \(50\%\) since there are 2 equally likely \(X\) - chromosome contributions (\(X^B\) and \(X^b\)) from the mother for male offspring.

Answer:

50%