Question

5. try this problem. a homozygous normal female and a male with hemophilia have children. what are the genotype and phenotype ratios of their offspring? colorblindness is also a sex - linked trait. in order for a woman to be colorblind, she must inherit the colorblindness gene on both of her x chromosomes. for males, they only need to inherit one copy in order to exhibit the trait. look at the genotype key below and do the punnett squares. female with normal color vision (homozygous dominant) female with normal color vision (heterozygous) female with colorblindness (homozygous recessive) male with color vision male with colorblindness x^c x^c x^c x^c x^c x^c x^c y x^c y

Explanation:

Step1: Determine parental genotypes

The homozygous normal female has genotype $X^HX^H$ and the male with hemophilia has genotype $X^hY$.

Step2: Set up Punnett - square

The female can produce only $X^H$ gametes and the male can produce $X^h$ and $Y$ gametes.

	$X^H$	$X^H$
$Y$	$X^HY$	$X^HY$

Step3: Calculate genotype ratios

The genotypes of the offspring are $X^HX^h$ (heterozygous normal - female) and $X^HY$ (normal male). The genotype ratio is $2X^HX^h:2X^HY = 1:1$.

Step4: Calculate phenotype ratios

The phenotypes are normal females and normal males. The phenotype ratio is $1$ normal female : $1$ normal male.

Answer:

Genotype ratio: $1X^HX^h:1X^HY$; Phenotype ratio: $1$ normal female : $1$ normal male