Question

13. The following genotype frequencies are observed in a population of 500 individuals. Number of individuals Genotype AA Аа 45 ad 210 245 Total = 500 Answer the following questions about this population (6 points total)

Answer 1

Answer: The condition given in the question, that is no selection, no migration, no mutation or genetic drifts are essentially the conditions for a population which is in Hardy-Weinberg equilibrium.

If the population remains in HW equilibrium, then in next generation it will exhibit same number of individuals with the dominant phenotype (in other words the frequencies of the genotypes will remain the same)

Dominant phenotype is shown by the individuals having’ AA’ (homozygous) and Aa (heterozygous) genotype, hence the number of individuals with dominant phenotype is 45+210 = 255

Out of 500, same number of individuals will be showing the dominant phenotype in next generation i.e. 45+210 =255.

(Additional information: According to the HW equilibrium   p2 +2pq+q2 =1

Where p2 is the frequency of homozygous dominant

q2 is the frequency of homozygous recessive

And 2pq is the frequency of heterozygote

Here in this example p2 =45/500 = 0.09 therefore p = 0.3, q2 = 245/500 = 0.49 therefore q = 0.7

And 2pq = 210/500 = 0.42   (It can be calculated by substituting the values of p

And q, then it is equal to (2×0.3×0.7 = 0.42)