Question

In a population of anteaters in South Africa, T₁ and T₂ are autosomal, incompletely dominant alleles that control tongue length. The alleles are polymorphic in this population, with f(T₁) = 0.95 and f(T₂) = 0.05. Anteaters that have long tongues are T₁T₁, T₁T₂ individuals have medium-length tongues, and T₂T₂ individuals have short tongues. A disease that wipes out the ants with deep nests occurs in this ecosystem, exerting strong natural selection on the long-tongued anteaters (they are ineffective at eating ants from shallow nests). As a result, 100% of the short-tongued anteaters survive this change in food supply, 40% of the medium-tongued anteaters survive, and 10% of the long-tongued anteaters survive.

A) Assuming the population begins in Hardy-Weinberg equilibrium and consists of 1,000 individuals, how many long-, medium-, and short-tongued individuals would you expect to be present in the original population (before selection)?

B) Assuming the population begins in Hardy-Weinberg equilibrium, what are the allele frequencies after one round (the initial round) of natural selection?

C) Assuming random mating takes places among the survivors of this first round of selection, what are the genotype frequencies of their offspring (the second generation)?

Answer 1

HARDY-WEINBERG EQUILIBRIUM:

p + q = 1,

p2 + 2pq + q2 = 1

Where,

p = frequency of T1 allele in the population
q = frequency of T2 allele in the population
p2 = percentage of T1T1 individuals (long-tongued)
q2 = percentage of T2T2 individuals (medium-tongued)
2pq = percentage of T1T2 individuals (short-tongued)

A) Total population = 1,000

p = 0.95, q = 0.05

p2 = 0.95 * 0.95 = 0.9025

Number of long-tongued individuals

= 0.9025*1,000 = 902.5 = 902

q2 = 0.05 * 0.05 = 0.0025

Number of short-tongued individuals

= 0.0025*1,000 = 2.5 = 3

2pq = 2 * 0.95 * 0.05 = 0.095

Number of medium-tongued individuals

= 0.095*1,000 = 95

B) After selection:

let, p' and q' be new allele frequencies after selection

Now,

100% of the short-tongued anteaters survive

Number of short-tongued individuals after selection = 3

40% of the medium-tongued anteaters survive

Number of medium-tongued individuals after selection = 38

10% of the long-tongued anteaters survive.

Number of long-tongued individuals after selection = 90.2 = 90 (= p'2)

New total population = 3 + 38 + 90 = 131

p'2 = 0.69 , q'2 = 0.02 , 2p'q' = 0.29

p' = 0.83 , q' = 0.14

C) Here, p'2 + q'2 + 2p'q' = 1

Also, p' + q' = 0.97 (approximately 1)

Therefore, the population is still in equilibrium

Hence, the allele frequency will stay the same in the next generation