Question

a colorblind male marries a carrier female. what is the chance that their child will be colorblind? a 0% b 25% c 100% d 50%

Explanation:

Step1: Determine the genotypes

Colorblindness is X - linked recessive. Let \(X^c\) represent the colorblind allele and \(X^C\) represent the normal allele. The male is colorblind, so his genotype is \(X^cY\). The female is a carrier, so her genotype is \(X^CX^c\).

Step2: Set up the Punnett square

The possible gametes from the male are \(X^c\) and \(Y\). The possible gametes from the female are \(X^C\) and \(X^c\). The Punnett square is:

	\(X^C\)	\(X^c\)
\(Y\)	\(X^CY\) (normal male)	\(X^cY\) (colorblind male)

Step3: Calculate the probability

There are 4 possible genotypes. The colorblind genotypes (\(X^cX^c\) and \(X^cY\)) occur 2 out of 4 times. The probability is \(\frac{2}{4}= 50\%\)

Answer:

d. 50%