Question

example #7: colorblindness is a sex - linked recessive trait located on the x chromosome. normal vision = $x^{b}$ and colorblindness = $x^{b}$. the y chromosome does not have the gene. cross a woman who is a carrier for colorblindness with a man who is normal.
genotypic ratio:
phenotypic ratio:

Explanation:

Step1: Determine Genotypes of Parents

The woman is a carrier for colorblindness, so her genotype is $X^{B}X^{b}$. The man is normal, so his genotype is $X^{B}Y$.

Step2: Set Up Punnett Square

The woman can produce gametes $X^{B}$ and $X^{b}$. The man can produce gametes $X^{B}$ and $Y$.

	$X^{B}$	$Y$
$X^{b}$	$X^{B}X^{b}$	$X^{b}Y$

Step3: Determine Genotypic Ratio

Count the genotypes: $X^{B}X^{B}$ (1), $X^{B}X^{b}$ (1), $X^{B}Y$ (1), $X^{b}Y$ (1). So genotypic ratio is $X^{B}X^{B}:X^{B}X^{b}:X^{B}Y:X^{b}Y = 1:1:1:1$.

Step4: Determine Phenotypic Ratio

• $X^{B}X^{B}$: Normal female
• $X^{B}X^{b}$: Normal (carrier) female
• $X^{B}Y$: Normal male
• $X^{b}Y$: Colorblind male

So phenotypic ratio: Normal female : Normal (carrier) female : Normal male : Colorblind male = $2:0:1:1$ (or grouped as Normal : Colorblind = $3:1$ for overall, but more precise is by sex and trait: Normal females (including carrier) : Normal males : Colorblind males = $2:1:1$; or breaking down females: normal non - carrier : normal carrier = $1:1$, males: normal : colorblind = $1:1$)

Answer:

Genotypic ratio: $X^{B}X^{B}:X^{B}X^{b}:X^{B}Y:X^{b}Y = 1:1:1:1$
Phenotypic ratio: Normal (female, non - carrier) : Normal (female, carrier) : Normal (male) : Colorblind (male) = $1:1:1:1$ (or Normal (females + normal males) : Colorblind (male) = $3:1$; more accurately, by sex - specific traits: Normal females (2, since both $X^{B}X^{B}$ and $X^{B}X^{b}$ are normal - visioned females) : Normal males (1) : Colorblind males (1), so $2:1:1$ for female normal : male normal : male colorblind)