One type of natural selection is balancing selection. One scenario that can lead to balancing selection is when a heterozygote phenotype has a fitness advantage. Which of the following are reasonable predictions of balancing selection due to heterozygote advantage? Assume that after 10 generations under the balancing selection, the f(A1) = 0.5, where "A1" is one of two alleles of a given gene; "A2" is the other allele.

1. The genotype, A1A2, will become fixed in the population.
2. The f(A2) will increase, while the f(A1) will decrease.
3. One of the two alleles will become fixed in the population.
4. The f(A1A2) will increase, while the f(A1A1) and f(A2A2) will decrease.
5. The f(A1) will remain roughly the same over the following generations.

The genotype, A1A2, will become fixed in the population.

Explanation:

Balancing Selection is a process that can be achieved in many ways, most of the times, when the heterozygotes for the alleles being analyzed have a better fitness than the homozygote. More importantly, bàlancing selection explain the different ways of selective processes by which various alleles, that is, different types of a gene are constantly maintained in the gene pool of a population at rate higher than the usual from genetic drift alone.

Using calculation to understand better

The Hardy-Weinberg principle can be used in this case to calculate genotype frequencies from allele frequences.

From assumption A1 and A2 are two alleles at the same locus,

p is the frequency of allele A 0.5 =< p =< 1

q is the frequency of allele A20.5 =< q =< 1,

and p + q = 1

where the distribution of allele frequencies is the same in men and women, i.e.:

Male(p,q) female (p,q)

if they procreate : (p + q)2 = p2 + 2pq + q2 = 1

where:

p2 = frequency of the A1 A1 genotype <-- HOMOZYGOTE

2pq = frequency of the A1 A2 genotype <-- HETEROZYGOTE

q2 = frequency of the A2 A2 genotyp <-- HOMOZYGOTE

these frequencies remain constant in successive generations.

The Allele frequency in the 10th generation

A1A1A1A2A2A2

1st Generation p2 2pq q2

10th Generation F(A1) = W+ Y/2 = p2 +1/2 (2pq) = p (p+q) = p

F(A2) = Z + H/2 = q2 +1/2 (2pq) = q (p+q) = q

The result is that, there is no change in allele frequencies:

in the 1st/current generation, we have p and q

in the 10th generation, we have p and q

Note: The frequencies of genotypes F(G) be called W, Y, and Z

with 0 =< [D,H,R] = < 1

and D + H + R = 1

The frequencies of alleles F(A) be called p, and q

with 0 =< [p,q] =< 1

and p+q = 1

The Hardy–Weinberg principle, otherwise known as the Hardy–Weinberg equilibrium, model, theorem, or law, "states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences."

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explanation:

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