Question

Answer the question based on the following data:
A man who has phenylketonuria (metabolic genetic disorder) marries a woman who is heterozygous at that locus. What is the probability that their first child will have phenylketonuria?
1
12
14
18

Answer 1

Phenylketonuria (PKU) is an autosomal recessive disorder, which means that an individual must inherit two copies of the mutated gene (one from each parent) to express the disorder.

In this scenario:
- The man with phenylketonuria has the genotype **aa** (homozygous recessive).
- The woman who is heterozygous at that locus has the genotype **Aa** (one normal allele and one mutant allele).

To determine the probability that their first child will have phenylketonuria, we can set up a Punnett square.

The potential alleles from each parent are:
- The man can only contribute **a** (from aa).
- The woman can contribute either **A** (normal) or **a** (mutant) since she is heterozygous.

The possible combinations for their child are:
- From the man **a** and the woman **A**: **Aa** (not affected, carrier)
- From the man **a** and the woman **a**: **aa** (affected)

Setting up the Punnett square:

| | A | a |
|---|---|---|
| a | Aa | aa |
| a | Aa | aa |

From this, we find:
- 2 out of 4 (50%) outcome will be **Aa** (not affected).
- 2 out of 4 (50%) outcome will be **aa** (will have phenylketonuria).

Thus, the probability that their first child will have phenylketonuria is **1 out of 4**, or 25%.

Since none of the options provided (1, 12, 14, 18) match this direct probability, if we interpret the answer as simplicity, it could correspond to **1 in 4**, but since we need to fit it to the options given, then we recognize that it should correspond to **1**, which is commonly interpreted as representing “the probability is a whole number of chance”, confirming probability rather than probability statement complexity.

Given the choices available, the answer aligns best with **1**.