(1). You survey a population of cotton rats at the UH Coastal center for variati
ID: 205205 • Letter: #
Question
(1). You survey a population of cotton rats at the UH Coastal center for variation at a SNP locus. You find 30 individuals that are A / A, 71, are A/G and 67 are G / G. What is the frequency of the A allele in this population?
(2). For the population in the previous question, what will the frequency of the A allele be when the population is in Hardy Weinberg equilibrium?
(3). When the population in Question 1 is in Hardy Weinberg Equilibrium, what will be the frequency of heterozygous individuals?
(4). Is the population in Question 1 in Hardy-Weinberg equilibrium? For your answer, also give the value of the chi-square statistics, the degrees of freedom, the p - value.
(5). In the current population, as described in Question 1, what would be the frequency of mating between A/A and A/G individuals?
Explanation / Answer
1) There are two A alleles in A/A and one in A/G
Frequency of A allele= 60+71/2 X 168= 131/336=0.39
Frequency of B allele= (71+2*67)/168*2= 0.61
2) The Hardy Weinberg equation is given as:
p2+2pq+q2=1
p2 = frequency of the homozygous genotype AA, q2 = the frequency of the homozygous genotype G/G, and 2pq = frequency of the heterozygous genotype A/G
Frequency of AA= 30/168= p2= 0.1785= 17.85%
p= square root of 0.1785= 0.422
3) q= 1-p= 1-0.422= 0.578
F(A/G)= 2pq= 2*0.422*0.578= 0.4878= 48.78%
F(G/G)= q2= 0.578*0.578= 0.3304
4) Null Hypothesis: there is no difference between the expected an observed values.
Observed
Expected
A/A
30
25.5528
A/G
71
79.9344
G/G
67
62.5128
Total
168
Observed
Expected
(O-E)
(O-E)2
(O-E)2/E
A/A
30
25.55
4.45
19.8025
0.775049
A/G
71
79.9344
-8.9344
79.8235
0.998613
G/G
67
62.5128
4.4872
20.13496
0.322093
Total
168
168
0.0028
119.761
2.095755
Degree of freedom= No of phenotypes-1= 3-1=2
Probability of chi square at d.f 2 = 0.3508= 35.08%
A Chi Square value between 1%-5% should allow you to reject the null hypothesis. However, 35.08% is a large P Value. Hence, the null hypothesis is accepted. The frequencies are in Hardy-Weinberg equilibrium.
As per Chegg’s rules, only question needs to be answered.
Observed
Expected
A/A
30
25.5528
A/G
71
79.9344
G/G
67
62.5128
Total
168
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.