Answer:
1000 MM individuals; 2000 MN individuals; 1000 NN individuals
Step-by-step explanation:
The Hardy-Weinberg principle states that, under certain conditions, the frequency of alleles and genotypes in a sexually reproducing population will remain constant over generations. The Hardy-Weinberg assumptions include random mating, sexual reproduction, and the absence of evolutionary forces such as mutation, natural selection, and genetic drift (as in the example above). Under these conditions, the frequency of alleles and genotypes in a population will not change and tend to the equilibrium. In this case, under Hardy-Weinberg equilibrium, the frequency of the M allele or 'p' must be equal to 0.5, and the frequency of the N allele or 'q' must be equal to 0.5 (i.e., the sum of all allele frequencies in the population must be equal to 1). Moreover, the frequencies of the genotypes will be p², 2pq, and q² >> p² (MM genotype) = (0.5)² = 0.25; q² (NN genotype) = (0.5)² = 0.25; and 2 x p x q (MN genotype) = 2 x 0.5 x 0.5 = 0.50. In consequence, under Hardy-Weinberg equilibrium, in a population of 4000 diploid individuals (8000 alleles), we have
- 4000 M alleles (M = 0.5) and 4000 N alleles (N = 0.5);
- 1000 MM individuals [p² >> (0.5)² = 0.25]; 2000 MN individuals (2pq >> 2 x 0.5 x 0.5 = 0.5) and 1000 NN individuals [q² >> (0.5)² = 0.25].