Below is a table summarizing allozyme variation in two populations of big hairy
ID: 140459 • Letter: B
Question
Below is a table summarizing allozyme variation in two populations of big hairy monsters. A total of 120 individuals were sampled in each of the two populations, the genotypes and number of individuals of each genotype are indicated in the table. Calculate the average level of heterozygosity observed in each population and contrast these populations in terms of high and low levels of genetic variation. Allozyme Loci Adh SS:96 SF:22 FF:2 Gpdh mo N-120 Xdh SS: 120SS: 3 SF: 0 FF: 0 Aldh Lap-5 Pgm SS: 20 SF: 57 FF: 43 Pop. 1 nomorphic SS: 0 SF: 0 FF: 120 SF: 25 FF: 92 monomorphic SS: 3 N-120 Pop. 2 SS: 75SS: 45 SF: 40SF: 55 FF: 5 SF: 30 FF: 87 SS: 8 SF: 45 FF: 67 SS: 120 SF: 0 FF: 0 FF: 20Explanation / Answer
So, from the above question, we should make a chart first. Resulting frequency is calculated first in each case, which is the no of individuals in each genotype, divided by the total no of individuals from all three genotypes of each gene.
In population 1, Adh gene,
Genotype
No of individuals
Resulting frequency
SS
96
0.8
SF
22
0.183
FF
2
0.016
In population 1, Pgm gene,
Genotype
No of individuals
Resulting frequency
SS
20
0.166
SF
57
0.475
FF
43
0.3583
In population 1, Aldh gene,
Genotype
No of individuals
Resulting frequency
SS
0
0
SF
0
0
FF
120
1
In population 1, Xdh gene,
Genotype
No of individuals
Resulting frequency
SS
120
1
SF
0
0
FF
0
0
In population 1, Lap-5 gene,
Genotype
No of individuals
Resulting frequency
SS
3
0.025
SF
25
0.2083
FF
92
0.7666
In population 2, Adh gene,
Genotype
No of individuals
Resulting frequency
SS
75
0.625
SF
40
0.333
FF
5
0.416
In population 1, Pgm gene,
Genotype
No of individuals
Resulting frequency
SS
45
0.375
SF
55
0.4583
FF
20
0.1666
In population 1, Aldh gene,
Genotype
No of individuals
Resulting frequency
SS
3
0.025
SF
30
0.25
FF
87
0.725
In population 1, Xdh gene,
Genotype
No of individuals
Resulting frequency
SS
8
0.0666
SF
45
0.375
FF
67
0.55833
In population 1, Lap-5 gene,
Genotype
No of individuals
Resulting frequency
SS
120
1
SF
0
0
FF
0
0
Now, we can calculate the average level of heterozygosity for a specific gene in each population. So, we need to know about the SF.
In population 1, for Adh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.8, which is 0.89. Since q = 1 - p, then q must equal 0.11. And 2pq= 2X0.89X0.11. so, SF=0.19
In population 1, for Pgm,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.166, which is 0.407 . Since q = 1 - p, then q must equal 0.593. And 2pq= 2X0.407X0.593. so, SF= 0.482
In population 1, for Aldh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0, which is 0. Since q = 1 - p, then q must equal 1. And 2pq= 2X0X1. so, SF=0
In population 1, for Xdh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 1, which is 1. Since q = 1 - p, then q must equal 0. And 2pq= 2X1X0. so, SF=0
In population 1, for Lap-5,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.025, which is 0.1581. Since q = 1 - p, then q must equal 0.8419. And 2pq= 2X0.1581X0.8419. so, SF=0.266
In population 2, for Adh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.625, which is 0.7905. Since q = 1 - p, then q must equal 0.2095. And 2pq= 2X0.7905X0.2095. so, SF=0.3312
In population 2, for Pgm,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.375, which is 0.6123. Since q = 1 - p, then q must equal 0.3877. And 2pq= 2X0.6123X0.3877. so, SF=0.474
In population 2, for Aldh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.025, which is 0.1581. Since q = 1 - p, then q must equal 0.8419. And 2pq= 2X0.1581X0.8419. so, SF=0.266
In population 2, for Xdh,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 0.0666, which is 0.2580. Since q = 1 - p, then q must equal 0.742. And 2pq= 2X0.2580X0.742. so, SF=0.382
In population 2, for Lap-5,
Since SS=p2, SF = 2pq, and FF = q2, then p (the frequency of the S allele) must be the square root of 1, which is 1. Since q = 1 - p, then q must equal 0. And 2pq= 2X1X0. so, SF=0
Now, high heterozygosity means high variation, which can be compared from the above result for each gene, in 2 populations.
So, for Adh, population 2 has high genetic variation.
For Pgm, population 1 has high genetic variation.
For Aldh, population 2 has high genetic variation.
For Xdh, population 2 has high genetic variation.
For Lap-5, population 1 has high genetic variation.
Genotype
No of individuals
Resulting frequency
SS
96
0.8
SF
22
0.183
FF
2
0.016
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