Recessive phenotype(q^2) =40/250.q^2=40/250.q=2/5.q=0.4...we know that;p+q=1.p+0.4=1.p=0.6...it's asked about no. of heterozygotes.=2pq..2pq=2×0.6×0.4...=0.48...This is the frequency of heterozygotes ...it's asked about no. of heterozygotes in the population=2pq×given population...=0.48×250.=120....120 is the no. of heterozygotes in the population..

Calculation of Heterozygotes in a Hardy-Weinberg Population

In a Hardy-Weinberg population, the frequency of alleles and genotypes in a population remains constant from generation to generation. This is based on several assumptions, including that the population is large, mating is random, there is no migration or mutation, and there is no natural selection.

Given Data:
- Total population = 250 individuals
- Recessive phenotype = 40 individuals

Using the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1

where p is the frequency of the dominant allele, q is the frequency of the recessive allele, p^2 is the frequency of the homozygous dominant genotype, 2pq is the frequency of the heterozygous genotype, and q^2 is the frequency of the homozygous recessive genotype.

Step 1:
Calculate the frequency of the recessive allele (q)
q^2 = 40/250
q^2 = 0.16
q = √0.16
q = 0.4

Step 2:
Calculate the frequency of the dominant allele (p)
p = 1 - q
p = 1 - 0.4
p = 0.6

Step 3:
Calculate the frequency of the heterozygous genotype (2pq)
2pq = 2(0.6)(0.4)
2pq = 0.48

Step 4:
Determine the number of heterozygotes in the population
Number of heterozygotes = 2pq x Total population
Number of heterozygotes = 0.48 x 250
Number of heterozygotes = 120

Therefore, there are approximately 120 heterozygotes in the given population.