Consider a population that is in Hardy–Weinberg equilibrium at a locus with two alleles, A and a, at frequencies of p and q, respectively. Assuming the population remains in Hardy–Weinberg equilibrium, what is the expected frequency of Aa heterozygotes after 100 generations?

a. 1
b. 2pq
c. 0.5
d. pq

B) 2pq

Explanation:

Hardy-Weinberg equilibrium refers to a model which explains the effect of evolution on the gene pool.

The model is based on the assumptions that if no evolutionary force like genetic drift, natural selection and many other will act on the population and therefore the gene pool  (gene frequency and the genotypic frequency) of a populations remains in equilibrium or constant throughout the generations.

The genotypic frequency in the model is calculated by  

P² =  genotype of a homozygous dominant trait

q² = genotype of a homozygous recessive trait

2pq = genotype of heterozygous trait.

Thus, P²+2pq+q²=1

In the given question, since the population is in equilibrium that is no evolutionary force is acting, therefore, the genotype frequency remains the same that is the frequency of Aa will remain same that is 2pq even after 100 generations.

Thus, Option-B is the correct answer.

meiosis is the correct answer

answers are:

lysogenic     viral dna merges with host cell dna

lysogenic     no symptoms

lytic               viral dna takes over

lysogenic     virus does not destroy host cell

lytic               virus destroys host cell

lytic               induces symptoms of viral infection

lysogenic       cell reproduces and spreads provirus