Question

1. Consider a population where the following numbers of individuals have the indicated genotype: 32 AA, 56 Aa, 12 aa. This population shows…

a. positive assortative mating, b. negative assortative mating, c. cannot determine, d. random mating

2. Consider a population where the following numbers of individuals have the indicated genotype: 36 AA, 48 Aa, 16 aa. This population shows…

a. positive assortative mating, b. negative assortative mating, c. cannot determine, d. random mating

Answer 1

1) d) Random mating

Explanation- If population follow positive or negative assortative mating then it must not follow hardy weinbarg principle i.e p^2+pq+q^2 = (p+q) where p+q = 1.

For population, frequency of genotype (number of individuals with a genotype/Total population) are as follow:

P^2(AA)=0.32

q^2 (aa)= 0.12

pq(Aa)= 0.56

Which give a total 0.32+0.56+0.12= 1, hence it follow hardy weinberg principle. So, mating was random.

2)d) Random mating

Explanation- Similarity for this population too, frequency of genotypes will be as follow:

p^2(AA)= 0.36

q^2(aa)= 0.48

pq(Aa)= 0.16

Which give a total of 0.36+0.48+0.16= 1, Again here population follow hardy weinbarg principle. So, mating was random.

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