QuestionAugust 5, 2025

A Punnett square is drawn for parents, one of whom has phenylketonuria, a recessive genetic disorder, and the other who is a carrier of the disease. What are the odds of their children having phenylketonuria? A. 0% B. 25% C. 50% D. 75% E. 100%

A Punnett square is drawn for parents, one of whom has phenylketonuria, a recessive genetic disorder, and the other who is a carrier of the disease. What are the odds of their children having phenylketonuria? A. 0% B. 25% C. 50% D. 75% E. 100%
A Punnett square is drawn for parents, one of whom has phenylketonuria, a recessive genetic disorder, and the other who is a carrier of the disease.
What are the odds of their children having phenylketonuria?
A. 0% 
B. 25% 
C. 50% 
D. 75% 
E. 100%

Solution
4.3(271 votes)

Answer

C. 50\% Explanation 1. Identify Genotypes The parent with phenylketonuria has genotype pp (recessive). The carrier parent has genotype Pp (one dominant, one recessive allele). 2. Set Up Punnett Square Create a 2x2 grid. Place pp on one side and Pp on the other. 3. Fill in Punnett Square Combine alleles: - First cell: Pp - Second cell: pp - Third cell: Pp - Fourth cell: pp 4. Calculate Probabilities Count pp occurrences: 2 out of 4 cells. Probability = \frac{2}{4} = 50\%.

Explanation

1. Identify Genotypes<br /> The parent with phenylketonuria has genotype $pp$ (recessive). The carrier parent has genotype $Pp$ (one dominant, one recessive allele).<br /><br />2. Set Up Punnett Square<br /> Create a 2x2 grid. Place $pp$ on one side and $Pp$ on the other.<br /><br />3. Fill in Punnett Square<br /> Combine alleles: <br />- First cell: $Pp$<br />- Second cell: $pp$<br />- Third cell: $Pp$<br />- Fourth cell: $pp$<br /><br />4. Calculate Probabilities<br /> Count $pp$ occurrences: 2 out of 4 cells.<br /> Probability = $\frac{2}{4} = 50\%$.
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