On your house plants, you had a huge population of herbivorous spider mites with a color polymorphism. Homozygous recessive individuals are black and indicated as rr. The other allele exhibits complete dominance over the r allele. Assume the population was at Hardy–Weinberg equilibrium, with equal frequencies of the two alleles. However, then a predatory mite got into your colony, creating a population bottleneck. Only 10 individual spider mites survived: 2 red heterozygote and 8 black. However, in trying to control the spider mites, you sprayed a pesticide that killed off the predatory mite, but not the spider mites. In the absence of the predatory mites, the spider mites randomly mated and there was no evolution. Thus, within a generation, the population was back at Hardy–Weinberg equilibrium. Estimate what proportion of spider mites in your current population are red?

Initially, p = q = 0.5, and the population was in Hardy Weinberg equilibrium, that is frequency of alleles remained constant generation after generation.

Therefore, initial genotype frequencies,

freq(RR) = p2 = 0.25

freq(Rr) = 2pq = 2 x 0.5 x 0.5 = 0.5

freq(rr) = q2 = 0.25

After the predatory spider mite attack, a population bottleneck is created and genotype frequencies become as follows:

freq(Rr) = 2/10 = 0.2

freq(rr) = 8/10 = 0.8

Now, frequency of an allele is given by,

[2 x no. of homozygotes for that allele + No. of heterozygotes]/[2 x Total population], as it is a diploid species with two loci for every gene and therefore two copies of gene in every individual.

So,

freq(R) = p = [2 x no. of dominant homozygotes + heterozygotes]/[2 x Total population]

= [2 x 0 + 2]/[2 x 10]

= 0.1

freq(r) = q = [2 x no. of recessive homozygotes + heterozygotes]/[2 x Total population]

= [2 x 8 + 2]/[2 x 10]

= 0.9

Thus after population bottlenecking, new allele frequencies are give by, p = 0.1 and q = 0.9.

Now, this new population maintains the Hardy Weinberg equilibrium. Thus, here too the allele frequencies shall remain same generation after generation.

Now, the current proportion of red colored spider mites in the population is given by

= Frequency of dominant homozygotes + Frequency of heterozygotes

So, in this population of spider mites maintaining Hardy Weinberg equilibrium,

Frequency of RR mites = p2 = (0.1)2 = 0.01

Frequency of Rr mites = 2pq = 2 x 0.1 x 0.9 = 0.18

Therefore the total frequency of spider mites with red body color in the new population = 0.01 + 0.18 = 0.19

eve24great2001 eve24great2001 · Answer 2 · 2020-02-29T13:25:36+01:00

Answer:

In Hardy-Weinberg equilibrium, the model is given as:

p^2 + 2pq + q^2 = 1,

Where,

p^2 is frequency of homogeneous with dominant genotype

2pq is frequency of heterogeneous

q^2 is frequency of homogeneous that are recessive.

The recessive is black(rr)

The dominant is red.

Survived spider mites after the predatory mite enters the colony is 2 red and 8 black.

After the sprayed pesticide killed off the predatory mite, but not the spider mites, and the population was back at Hardy–Weinberg equilibrium, The estimate proportion of the homogeneous with dominant gene(That is, Red) is:

p^2 = 2^2 = 4

Explanation:

The Hardy-Weinberg model states that a population will remain at genetic equilibrium as long as five conditions are met:

(1) No change in the DNA sequence.

(2) No migration

(3) A very large population size

(4) Random mating

(5) No natural selection.