Consider a population of sheep to be in Hardy-Weinberg equilibrium. The allele for black wool(w) has an allele frequency of 0.81 while the allele for white wool(W) has an allele frequency of 0.19. Then the percentage of heterozygous individuals in the population is

The correct option is 31%.

To find the percentage of heterozygous individuals in the population, we use the Hardy-Weinberg principle. The principle states that the frequencies of alleles and genotypes in a population remain constant from generation to generation in the absence of other evolutionary influences.

Let the frequency of the dominant allele (W) for white wool be represented by $p$ and the frequency of the recessive allele (w) for black wool be represented by $q$.

From the given data:
The allele frequency for white wool, $p=0.19$
The allele frequency for black wool, $q=0.81$

According to the Hardy-Weinberg equation for genotype frequencies:
$p^2+2pq+q^2=1$

Where:
- $p^2$ represents the frequency of homozygous dominant individuals (WW)
- $2pq$ represents the frequency of heterozygous individuals (Ww)
- $q^2$ represents the frequency of homozygous recessive individuals (ww)

We need to find the frequency of heterozygous individuals ($2pq$). Substituting the given values into the formula:
$2pq=2\times 0.19\times 0.81$

Calculating the value:
$2pq=2\times 0.1539$
$2pq=0.3078$

To convert this frequency into a percentage, we multiply by 100:
Percentage of heterozygous individuals = $0.3078\times 100=30.78\%$

Rounding to the nearest whole number gives 31%, which matches the correct option.