Answer:
The frequency of red spider mites is 0.19, that is 19 out of every 100 spider mites have red body color.
Explanation:
According to the question:
Dominant phenotype: Red body color (RR or Rr).
Recessive phenotype: Black body color (rr).
Let allele frequency of R be p and that of r be q.
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
