# Answer to Question #65253 in Genetics for sai

Question #65253

In a random mating population oo 28,800 individuals percentyage of dominant homozygous individuals is 49.find out the percentage of heterozygous individual

Expert's answer

Solution: The frequencies of different types of genotypes in ideal population are described by the Hardy–Weinberg law: p^2 +2pq+q^2 =1 where p 2 is a frequency of dominant homozygotes, q^2 is a frequency of recessive homozygotes and 2pq is a frequency of heterozygotes. It’s based on simple equation: p + q = 1 where p and q are frequencies of dominant and recessive alleles in population. In order to find percentage of heterozygotes at first find frequencies of alleles in population:

P^2 = 49% = 0.49, so p = 0.7;

q = 1 - p = 1 - 0.7 = 0.3;

Then find frequency of heterozygote as 2pq = 2*0.3*0.7 = 0.42 = 42%

Answer: The percentage of heterozygous individuals in population is 42%.

P^2 = 49% = 0.49, so p = 0.7;

q = 1 - p = 1 - 0.7 = 0.3;

Then find frequency of heterozygote as 2pq = 2*0.3*0.7 = 0.42 = 42%

Answer: The percentage of heterozygous individuals in population is 42%.

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