Question

151. In a populatin of H.W. equilibrium the recessive
individuals are 36%, find out the percentage of
heterozygous individuals in that population?
1) 60%
2) 40%
3) 48%
4)
64%​

Answer 1

Answer:

60% is the right hope it will help you

Answer 2

Answer:

2) 40%

Explanation:

Hardy-Weinberg precept offers the geneticists a device to decide while evolution is occurring. It makes use of the binomial expression p2 + 2pq + q2 to calculate the genotypic and allele frequencies of a population.

where, p2 = % homozygous dominant people

            p  =  frequency of dominant allele

            q2 =  % homozygous recessive people

            q  = frequency of recessive allele

            2pq = % heterozygous people

In the given question, darkish brown coat is dominant trait. So, % homozygous dominant people i.e. p2 = 64 % or 0.64

therefore, frequency of dominant allele i.e. p = 0.64^1/2 = 0.8

Since p+ q =1, therefore, frequency of recessive allele q = 1 - p = 1 - 0.8 = 0.2

So, q2 =(o.2)2=0.4=40%

So, the final answer is 2) 40%

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