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?
112
81
105
145
To determine the number of individuals expected to be heterozygous in a random sample of 250 individuals from a population at Hardy-Weinberg equilibrium, we need to use the Hardy-Weinberg equation. The equation for calculating the expected frequency of heterozygotes (pq) is:
2 X p X q X N
Where:
p = frequency of allele A
q = frequency of allele a
N = sample size (number of individuals)
Given that the frequencies of alleles A and a are 0.7 and 0.3 respectively, we can substitute these values into the equation:
2 X 0.7 X 0.3 X 250 = 105 individuals
Therefore, we can expect approximately 105 individuals in the random sample of 250 individuals to be heterozygous.
Hence, the correct answer is option 3.