Question

Part I. You are investigating the evolution of cuticle color in a scarab beetle inhabiting a desert. Coloration is determined by a single gene A with two alleles: A₁ codes for a dark brown color, and A₂ codes for a grey color. Since the alleles are incompletely dominant, heterozygotes are tan. Therefore the heterozygotes are almost invisible on the sandy desert floor, while homozygotes are more visible and thus more susceptible to predation.

Questions: (a) Based on the given information, what type of selection is most likely to occur to coloration in this population? (2 points), hint: think directional, balancing, and disruptive selection). (b) How do you know? Briefly explain by comparing the relative fitness of the three genotypes (use s and taccordingly) (3 points).

Part II. You want to know wheter selection is in fact acting or coloration, and wheter the beetle population is evoling at the A locus. In two consecutive years, you sampled bettles from the population to examine the change in number of individuals with the three genotypes. After genetic analyses, you find that the frequency of the A₂ allele if q= 0.20 at the initial generation.

Question: You come back to the population after a generstion, and sample 1,000 scarabs. Calculate p, p², 2pq, q² and the expected number of brown individuals , tan individuals, and grey individuals in the population, assuming that the population in Hardy -Weinberg equilibrium (5 points). Hint: Remeber A₂ = q = 0.20

Part III. The population sampled actually was made up of 650 brown, 340 tan, and 10 grey individuals. Question: (a) Find ALL the values needed to complete the table below and write them down on your answer sheet. Round all values to two decimals. (Hint: The expected values are the ones you estimated for the scarabs in Part II.) (4 points)

phenotype	brown	tan	grey	total
genotype	A1A1	A1A2	A2A2	-
expected
observed	650	340	10	1000
(O-E)2/E				x2=

Answer 1

Answer 1

Population of scarab beetles homozygous to coloration gene, which is clearly visible to the predators so the environmetal condition is against the survival of these homozygous population, and the heterozygous population is Tan is colour which is not visible. The intermediate genotype is selected by the natural selection so it is disruptive selection.

Answer 2

Frequency of A2 = q = 0.2

According to Hardy-Weinberg Law p+q = 1

so, here frequency of A1 = p = 1- q = 1- 0.2 = 0.8

Expected frequency of each genotype will be p2, 2pq and q2

p2 = 0.8x0.8 = 0.64 Expected number = 0.64x1000= 640

2pq = 2x0.8x0.2= 0.32 Expected number = 0.32x 1000= 320

q2 = 0.2x0.2 = 0.04 Expected number = 0.04 x1000= 40

phenotype	brown	tan	grey	total
genotype	A1A1	A1A2	A2A2	-
expected	640	320	40	1000
observed	650	340	10	1000
(O-E)2/E	0.15	1.17	90	x2=

(O-E) 2/ E = (650-640)² /640 =