Question

show your work and circle your answer for the questions below.

1. colorblindness is a recessive sex - linked trait carried on the x chromosome. cross a colorblind male with a female carrier of the trait. what is the chance of them having colorblind children?

Explanation:

Step1: Determine the genotypes

Let \(X^c\) represent the color - blind allele and \(X^C\) represent the normal allele. A color - blind male has the genotype \(X^cY\). A female carrier has the genotype \(X^CX^c\).

Step2: Set up the Punnett square

Gametes of female (\(X^C,X^c\))	Gametes of male (\(X^c,Y\))
\(X^c\times X^c = X^cX^c\)	\(X^c\times Y = X^cY\)

Step3: Analyze the offspring genotypes

• \(X^CX^c\): female carrier (not color - blind)
• \(X^CY\): normal male (not color - blind)
• \(X^cX^c\): color - blind female
• \(X^cY\): color - blind male

The total number of possible genotypes of offspring is \(4\). The number of color - blind offspring (\(X^cX^c\) and \(X^cY\)) is \(2\).

Answer:

The probability of having color - blind children is \(\frac{2}{4}=\frac{1}{2}\) or \(50\%\)