Question

In a population of 534 blue sheep, there are two alleles, P and Q, at an autosomal locus. Allele P has a frequency of 30%. The number of sheep with the QQ genotype is 262. Assuming that this population is at Hardy-Weinberg equilibrium, the expected percentage of sheep with the PQ genotype in the population is:

Hint:

In Hardy-Weinberg equilibrium, remember the basic formulas:
\( p + q = 1 \), where \(p\) is the frequency of allele P and \(q\) is the frequency of allele Q,
\(p^2 + 2pq + q^2 = 1\), representing the frequencies of the genotypes. Use these equations to calculate expected genotype frequencies in a population. Make sure to adjust the frequencies based on the given information (e.g., homozygous QQ) to calculate the other genotypes.

Answer 1

Step 1: Understand the Hardy-Weinberg equation.
The Hardy-Weinberg equilibrium states that:
\[ p^2 + 2pq + q^2 = 1 \] where:
\(p\) is the frequency of allele P,
\(q\) is the frequency of allele Q,
\(p^2\) represents the frequency of the homozygous PP genotype,
\(2pq\) represents the frequency of the heterozygous PQ genotype,
\(q^2\) represents the frequency of the homozygous QQ genotype.

Step 2: Calculate allele frequencies.
Given that allele P has a frequency of 30%, we have:
\[ p = 0.30 \quad \text{and} \quad q = 1 - p = 0.70 \]

Step 3: Use Hardy-Weinberg equilibrium to find the PQ genotype.
The frequency of the PQ genotype is given by \(2pq\):
\[ 2pq = 2 \times 0.30 \times 0.70 = 0.42 \]

Step 4: Calculate the expected number of sheep with the PQ genotype.
The total population is 534 sheep. Thus, the expected number of sheep with the PQ genotype is:
\[ 534 \times 0.42 = 224.28 \approx 224 \text{ sheep with PQ genotype} \]

Step 5: Calculate the percentage.
The expected percentage of sheep with the PQ genotype is:
\[ \frac{224}{534} \times 100 = 42.00\% \]

Final Answer:
The expected percentage of sheep with the PQ genotype is \( \boxed{42.00\%} \).