9% of Hardy-Weinberg population of 800 individual recessives. How many of this population are heterozygous?
9% of Hardy-Weinberg population of 800 individual recessives. How many of this population are heterozygous?
The correct option is A 336
According to the Hardy-Weinberg equation, the sum of the 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, q2 = 9% = 0.9. From this, q = 0.3. Hence, p = `1- q = 1 - 0.3 = 0.7. Now, population of heterozygous individual will be 2pq as mentioned, that is 2 * 0.7 * 0.3 = 0.42. It means, there is 42% of the heterozygous population. So, from 800 individuals, there will be 800 * 42 /100 = 336 individuals will have heterozygous genotype. Thus, the correct answer is option A.
According to the Hardy-Weinberg equation, the sum of the 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, q2 = 9% = 0.9. From this, q = 0.3. Hence, p = `1- q = 1 - 0.3 = 0.7. Now, population of heterozygous individual will be 2pq as mentioned, that is 2 * 0.7 * 0.3 = 0.42. It means, there is 42% of the heterozygous population. So, from 800 individuals, there will be 800 * 42 /100 = 336 individuals will have heterozygous genotype. Thus, the correct answer is option A.
