The frequencies of alleles 'A' and 'a' in a population at Hardy-Weinberg equilibrium are 0.7 and 0.3, respectively. In a random sample of 250 individuals taken from the population, how many are expected to be heterozygous?
Answers
Answer:
105
Explanation:
Allele frequencies =
A = 0.7
a = 0.3
Heterozygous =
2 x A x a = 2 x 0.7 x 0.3 = 0.42 = 42%
42% of 250 = 0.42 x 250= 105
Therefore 105 individuals are expected to be Heterozygous
0.42
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.According to this equation:-
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.According to this equation:-p = the frequency of the dominant allele (represented here by A)
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.According to this equation:-p = the frequency of the dominant allele (represented here by A)q = the frequency of the recessive allele (represented here by a)
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.According to this equation:-p = the frequency of the dominant allele (represented here by A)q = the frequency of the recessive allele (represented here by a)For a population in genetic equilibrium:
0.42To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.According to this equation:-p = the frequency of the dominant allele (represented here by A)q = the frequency of the recessive allele (represented here by a)For a population in genetic equilibrium:p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)(p + q) 2= 1 so p2+ 2pq + q 2= 1
2pq + q 2= 1The three terms of this binomial expansion indicate the frequencies of the three genotypes:
2pq + q 2= 1The three terms of this binomial expansion indicate the frequencies of the three genotypes:p 2= frequency of AA (homozygous dominant)
2pq + q 2= 1The three terms of this binomial expansion indicate the frequencies of the three genotypes:p 2= frequency of AA (homozygous dominant)2pq = frequency of Aa (heterozygous)q
2 = frequency of aa (homozygous recessive).
= frequency of aa (homozygous recessive).p+q = 1
= frequency of aa (homozygous recessive).p+q = 10.3 + q = 1
= frequency of aa (homozygous recessive).p+q = 10.3 + q = 1q = 1 - 0.3 = 0.7
= frequency of aa (homozygous recessive).p+q = 10.3 + q = 1q = 1 - 0.3 = 0.7Aa = 2pq = 2* 0.3 * 0.7 = 0.42.
= frequency of aa (homozygous recessive).p+q = 10.3 + q = 1q = 1 - 0.3 = 0.7Aa = 2pq = 2* 0.3 * 0.7 = 0.42.Thus, the correct answer is 0.42