Question

#6. if a colorblind son from the previous problem has children with a woman with normal color vision (homozygous dominant). what percentage of the sons is colorblind? make a punnett square to find out.

Explanation:

Step1: Determine the genotypes

Color - blindness is a sex - linked recessive disorder. Let \(X^c\) represent the color - blind allele and \(X^C\) represent the normal allele. Males have the genotype \(XY\). A color - blind son has the genotype \(X^cY\). A woman with normal color vision (homozygous dominant) has the genotype \(X^CX^C\).

Step2: Set up the Punnett square

For sons, the father contributes the \(Y\) chromosome (since the sex of the child is determined by the father's gamete). The mother contributes \(X^C\) gametes.
The Punnett square for the cross between \(X^cY\) (father) and \(X^CX^C\) (mother) for sons:

Father's gametes	Mother's gametes (\(X^C\))

Answer:

0%