hw equilibrium problem 73% 9:53 AM POPULATION GENETICS (SELECTION) HOMEWORK (Hom

ID: 150475 • Letter: H

Question

hw equilibrium problem

73% 9:53 AM POPULATION GENETICS (SELECTION) HOMEWORK (Homework #3) Intro to Evolution, Biology 3302; Due Sept 26 in class; 10 points Problem 1 In a population of tropical frogs, a certain locus determines the size of an individual With respect to this locus, homozygous dominant and heterozygous frogs are large while homozygous recessive frogs are small. Estimated numbers of the three genotypes are: AA 604, Aa 299, and aa 97. (A) Calcul)ate the starting frequencies of the 3 genotypes and the 2 alleles. (B) What is the selection coefficient against small individuals in this population? Bigger males usually have higher reproductive success, relative to smaller males, because they are able to produce calls of lower sound frequency that are more attractive to females. Also, due to their larger size, they have a higher survival rate than small individuals. (C) Calculate the allele frequencies (p1 and q) for the next generation of frogs (next generation the offspring of the individuals in the table below). Remember, we are violating only the assumption of "no selection". So when mating occurs, it is random. Number offspring/individua survival to adulthood large (AA) 604 8 large (Aa 299 8 small (aa) 20% 20% 15% 97

Explanation / Answer

(A) p is the frequency of 'A' allele in the population.

q is the frequency of 'a' allele in the population.

p2 is the frequency of 'AA' homozygous genotype.

q2 is the frequency of 'aa' homozygous genotype.

pq is the frequency of 'Aa' heterozygous genotype.

At a particular locus, sum of the allele frequencies for all the alleles is 1, so, p+q=1

In the above example, total number of genotypes is = 604+299+97 = 1000

So, frequency of p2 (AA) will be = 604/1000 = 0.6

frequency of q2 (aa) will be = 97/1000 = 0.1

frequecy of 2pq (Aa) will be = 0.3

now, frequency of p (A) is, square root of 0.6 = 0.7

frequency of q (a) is, square root of 0.1 = 0.3

Ans: Frequency of three genotypes are : AA = 0.6, Aa = 0.3, aa = 0.1. Frequency of 2 alleles are: A = 0.7, a = 0.3.

B) Please find the chart, where survival rate is given. Reproductive rate is also given, which is the average number of offspring in a particular genotype. To determine the relative fitness, divide every survivalXreproductive rate by the highest survivalXreproductive rate.

Survival rate

20%

15%

Reproductive rate

604

299

Survival X Reproductive rate

0.2X604= 120.8

0.2X299= 59.8

0.15X97=14.55

Relative fitness (W)

120.8/120.8= 1

59.8/120.8= 0.5

14.55/120.8= 0.1

Selection coefficient is, S=1=W

For small population it is = 1-0.1 = 0.9

Ans: The selection coefficient against small individuals in this population is 0.9.

C) Total number of individuals are = (8+8+6) = 22

Frequency of AA = 8/22 = 0.36

Frequency of aa = 6/22 = 0.27

So, allele frequency A = square root of 0.36 = 0.6

allele frequency a = square root of 0.27 = 0.5

Ans: p1=0.6, q1=0.5