The expected frequency of individuals that are heterozygous there is 48% of the heterozygous population.
According to Hardy-Weinberg equation, the sum of allele frequencies for all the alleles at the locus must be 1, so p + q = 1. Also, the Hardy-Weinberg equation is expressed as: p² + 2pq +q² = 1;
where p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population.
In the equation, p² represents the frequency of the homozygous genotype AA, q² represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa.
Here, q = 0.6. Hence, p = `1- q = 1 - 0.6 = 0.4.
Now, population of heterozygous individual will be 2pq as mentioned, that is 2 * 0.4 * 0.6 = 0.48.
It means, there is 48% of the heterozygous population.
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