Question

color blindness is a recessive x - linked trait. what percentages of daughters and sons (separately) are expected to be colorblind if a carrier (heterozygote) mother and a normal father were to have children? use gene symbols and punnett square to demonstrate.

Explanation:

Step1: Define gene symbols

Let $X^B$ represent the normal - allele on the X - chromosome and $X^b$ represent the color - blindness allele on the X - chromosome. The mother is a carrier with the genotype $X^B X^b$ and the father is normal with the genotype $X^B Y$.

Step2: Set up the Punnett square

	$X^B$	$X^b$
$Y$	$X^B Y$	$X^b Y$

Step3: Analyze daughters' genotypes

The possible genotypes for daughters are $X^B X^B$ and $X^B X^b$. Since color - blindness is recessive, none of the daughters ($0\%$) will be color - blind.

Step4: Analyze sons' genotypes

The possible genotypes for sons are $X^B Y$ and $X^b Y$. So, $50\%$ of the sons will be color - blind.

Answer:

Daughters: 0%; Sons: 50%