Industrial Melanism The peppered moth, Biston betularia, is a speckled moth that
ID: 173734 • Letter: I
Question
Industrial Melanism
The peppered moth, Biston betularia, is a speckled moth that rests on tree trunks during the day, where it avoids predation by blending with the bark of trees (an example of cryptic coloration). At the turn of the century, moth collectors in Great Britain collected primarily light forms of this moth (light with dark speckles) and only occasionally recorded rare dark forms. With the advent of the Industrial Revolution and increased pollution, light-colored lichens on the trees died, resulting in strong positive selection for dark moths resting on the now dark bark. The dark moth increased in frequency. However, in unpolluted regions, the light moth continued to occur in high frequencies. (This is an example of the relative nature of selective advantage, depending on the environment.)
Color is controlled by a single gene with two allelic forms, dark and light. Pigment production is completely dominant, and the lack of pigment is recessive. We use the letters A and a for these alleles.
5. If p, the frequency for A, is 0.1 what is the value of q, i.e. the frequency for a? 2
6. Given the p and q values above, according to the Hardy-Weinberg principle, what are the genotypic and phenotypic frequencies?
7. In a population of 1200 individuals what would be the number of individuals of each genotype and phenotype?
8. If this population met the assumptions for Hardy Weinberg Equilibrium predict the genotypic frequencies for the next generation.
9. Now, assume that pollution has become a significant factor and that in this new population 50% of the light moths but only 10% of the dark moths are eaten (natural selection). How many moths would be left for each phenotype? What would be the p and q values of the population after predation?
What I need help with is 8 and 9
Explanation / Answer
1. Given, Frequency for A(p)= 0.1,
According to Hardy-weinberg equilibrium, p+q=1
So, 0.1+q=1. q= 0.9
Hence, the frequency of a(q)= 0.1
2. According to Hardy-weinberg equilibrium, p2+2pq+q2=1
Genotype frequency for allele A (p2)= (0.1)2= 0.01
Genotypic frequecny for allele a(q2)= (0.9)2= 0.81
2pq= 2X0.1X0.9= 0.18
A phenotype= 0.01+0.18= 0.19
a phenotype= 0.81+ 0.18= 0.99