Question

1. in humans colorblindness (b) is an example of a sex - linked recessive trait. in this problem, a male with colorblindness marries a female who is not colorblind but carries the (b) allele. using a punnett square, determine the genotypic and phenotypic probabilities for their potential offspring.

genotype of offspring:

phenotype of offspring:

Explanation:

Step1: Determine Genotypes of Parents

Colorblindness is X - linked recessive. Male (XY) with colorblindness has genotype $X^{b}Y$. Female (XX) not colorblind but carrier has genotype $X^{B}X^{b}$.

Step2: Set Up Punnett Square

Male gametes: $X^{b}$ (from X) and $Y$ (from Y). Female gametes: $X^{B}$ and $X^{b}$.

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

Step3: Analyze Genotypes

• $X^{B}X^{b}$: 1/4 (25%)
• $X^{b}X^{b}$: 1/4 (25%)
• $X^{B}Y$: 1/4 (25%)
• $X^{b}Y$: 1/4 (25%)

Step4: Analyze Phenotypes

• $X^{B}X^{b}$: Female, carrier (not colorblind)
• $X^{b}X^{b}$: Female, colorblind
• $X^{B}Y$: Male, not colorblind
• $X^{b}Y$: Male, colorblind

Each phenotype has 25% probability? Wait, no:

• Female carrier: 25%
• Female colorblind: 25%
• Male non - colorblind: 25%
• Male colorblind: 25%? Wait, no, let's count:

From Punnett square, 4 cells.

• $X^{B}X^{b}$: female, carrier (1)
• $X^{b}X^{b}$: female, colorblind (1)
• $X^{B}Y$: male, non - colorblind (1)
• $X^{b}Y$: male, colorblind (1)

So genotypic probabilities: Each genotype has 25% (1/4). Phenotypic:

• Female carrier: 25%
• Female colorblind: 25%
• Male non - colorblind: 25%
• Male colorblind: 25%? Wait, no, the number of each phenotype:

Wait, for females: 2 cells ($X^{B}X^{b}$ and $X^{b}X^{b}$), so 50% of females. 50% of males ($X^{B}Y$ and $X^{b}Y$).
Among females: 50% carrier, 50% colorblind.
Among males: 50% non - colorblind, 50% colorblind.
Overall phenotypic probabilities:

• Female carrier: 25% (1/4 of total offspring)
• Female colorblind: 25% (1/4)
• Male non - colorblind: 25% (1/4)
• Male colorblind: 25% (1/4)

Genotypic:

• $X^{B}X^{b}$: 25%
• $X^{b}X^{b}$: 25%
• $X^{B}Y$: 25%
• $X^{b}Y$: 25%

Answer:

Genotypic Probabilities:

• $X^{B}X^{b}$: 25% ($\frac{1}{4}$)
• $X^{b}X^{b}$: 25% ($\frac{1}{4}$)
• $X^{B}Y$: 25% ($\frac{1}{4}$)
• $X^{b}Y$: 25% ($\frac{1}{4}$)

Phenotypic Probabilities:

• Female, carrier (not colorblind): 25% ($\frac{1}{4}$)
• Female, colorblind: 25% ($\frac{1}{4}$)
• Male, not colorblind: 25% ($\frac{1}{4}$)
• Male, colorblind: 25% ($\frac{1}{4}$)

(Or in terms of Punnett square fill - in:
Genotype cells: $X^{B}X^{b}$, $X^{b}X^{b}$, $X^{B}Y$, $X^{b}Y$ (each in one cell)
Phenotype cells: Female carrier, Female colorblind, Male non - colorblind, Male colorblind (each in one cell))