Use the following information for the next three questions: 1) Jean Manning, Cha
ID: 58135 • Letter: U
Question
Use the following information for the next three questions:
1) Jean Manning, Charles Kerfoot, and Edward Berger studied the frequencies at the phosphoglucose isomerase (GPI) locus in the cladoceran Bosmina longirostris. At one location, they collected 650 animals from Union Bay in Seattle, Washington, and determined their GPI genotypes by using electrophoresis. They counted the following # for each genotype: S1S1 = 121; S1S2 = 205; S2S2 = 324. What is the frequency of S1? Give your answer to 2 decimal places. For example, 0.12345 = .12
2) Using the information in the previous problem (to 2 decimal places), what is your calculated Chi-squared value to determine if this population is in Hardy-Weinberg equilibrium or not? Give your answer to 2 decimal places.
3) Is the population in the previous problem in Hardy-Weinberg equilibrium?
4) A farmer has a group of pigs that have an average weight of 150 lbs. He selects his heaviest pigs to breed, and their offspring in the next generation weigh 180 lbs. If the narrow-sense heritability is 0.3, what did the parents weigh that he selected to breed?
Explanation / Answer
1. Total number of population is = 650. The number of individuals with S1S1 genotype (p2) = 121; (121/650 = 0.186)
The number of individuals with S1S2 genotype (2pq) = 205; (205/650 = 0.315)
The number of individuals with S2S2 genotype (q2) = 324; (324/650 = 0.498)
The frequency of S1 allele = p= p2+ ½ (2pq) = 0.186 + 0.315/2 = 0.35
The frequency of S2 allele= q = q2 + ½ (2pq)= 0.498 + 0.315/2 = 0.65
2. Hardy-Weinberg’s Chi-square frequency:
C2 = (Observed freq- Expected freq)2 / Expected freq
Expected MM (p2) = 0.35* 0.35 = 0.122
Expected MN (2pq) = 2* 0.35* 0.65 = 0.45
Expected NN (q2) = 0.65* 0.65 = 0.422
EXPECTED NUMBER OF INDIVIDUALS WITH EACH GENOTYPE:
MM = 0.122* 650 = 79.3
MN = 0.45* 650 = 293
NN = 0.422* 650 = 274
CHI - SQUARE (X2):
X2 = (O - E)2 / E
X2 = (121-79.3)2 /79.3 + (293-205)2 /293 + (274-324)2 /274
= 21.9 + 26.4 + 9.12 = 57. 42
3. The population in the previous problem is not in Hardy-Weinberg equilibrium.