Question

In a population of 1000 individuals, the frequency of recessive allele is 0.4. find out the number of heterozygotes in that population.

Answer 1

Answer:

As the population is in Hardy Weinberg equilibrium, the frequencies of alleles will remain the same for the next generation also.  

According to the question, the frequency of recessive allele is (q =  0.2) Frequency of dominant allele= p= 1-q= 1-0.2=0.8.  

Hardy Weinberg law states that the sum of all genotype frequencies is represented as the binomial expansion of the square of the sum of p and q.  

This sum is equal to one.  (p + q)2 = p2 + 2pq + q2 = 1.  

Hence, the frequency of homozygous dominant genotype in total population

"(p2) =  0.64 or 64%". And the frequency of heterozygous dominant genotype in total population  

"(2pq) =  2 X 0.8 X 0.2 = 0.32 or 32%".

And the frequency of dominant individuals in next population =64+32= 96%

Explanation: