Question

what percentage of the female offspring will be a carrier for this recessive disease? 0% 50% 25% 100%

Explanation:

Step1: Identify female offspring genotypes

From the Punnett square, female offspring genotypes are $X^aX^a$, $X^aX^A$, $X^A X^a$

Step2: Count carrier female offspring

Carriers have one recessive allele ($X^a$) and one dominant allele ($X^A$): $X^aX^A$, $X^A X^a$ (2 total)

Step3: Calculate percentage of carriers

Total female offspring = 3. Percentage = $\frac{2}{3} \times 100\% \approx 66.7\%$ — correction: re-examine Punnett square (full grid: top row $X^a$, $X^A$; left column $X^a$, $Y$, $X^A$). Female offspring: $X^aX^a$, $X^aX^A$, $X^A X^a$, $X^A X^A$? No, correct female genotypes: $X^aX^a$, $X^aX^A$, $X^A X^a$ (3) → carriers are 2. Wait, no: standard carrier is heterozygous. Wait, re-express:
Wait, correct step:

Step1: List all female genotypes

From the square, female offspring are $X^aX^a$, $X^aX^A$, $X^A X^a$ (3 individuals)

Step2: Define carrier (heterozygous)

Carriers = $X^aX^A$, $X^A X^a$ (2 individuals)

Step3: Compute percentage

$\text{Percentage} = \frac{\text{Number of carriers}}{\text{Total female offspring}} \times 100\% = \frac{2}{3} \times 100\% \approx 66.7\%$ — but wait, the options are 0%, 50%, 25%, 100%. Re-examine the Punnett square: top row is $X^a$, $X^A$; left column is $X^a$, $Y$, $X^A$. Female offspring:

1. $X^a \times X^a = X^aX^a$
2. $X^A \times X^a = X^aX^A$
3. $X^a \times X^A = X^A X^a$
4. $X^A \times X^A = X^A X^A$ (missed earlier, total 4 female offspring)

Step1: Correct total female offspring

Total female offspring = 4: $X^aX^a$, $X^aX^A$, $X^A X^a$, $X^A X^A$

Step2: Count carrier females

Heterozygous carriers: $X^aX^A$, $X^A X^a$ (2)

Step3: Calculate percentage

$\text{Percentage} = \frac{2}{4} \times 100\% = 50\%$

Answer:

50%