Please help me if your gifted in genetics! My last chance to submit tonight!! PL
ID: 176117 • Letter: P
Question
Please help me if your gifted in genetics! My last chance to submit tonight!! PLEASE AND THANK YOU SO MUCH!!!
A population of chickadees was sampled and individuals were genotyped for the Adhlocus. The following data were collected:
A1A1 12
A1A2 17
A2A2 5
A2A3 8
A1A3 10
A3A3 3
Part A
What are the allele frequencies of the A1, A2, and A3 alleles (in that order)?
What are the allele frequencies of the A1, A2, and A3 alleles (in that order)?
A) 0.709, 0.545, 0.382
B) 0.355, 0.273, 0.191
C) 0.464, 0.318, 0.217
D) 0.355, 0.318, 0.327
Part B
What are the genotype frequencies of A1A1, A1A2A1A3 (in this order)?
What are the genotype frequencies of A1A1, A1A2A1A3 (in this order)?
A) 0.464, 0.318, 0.218
B) 0.218, 0.309, 0.182
C) 0.091, 0.145, 0.309
D) 0.218, 0.318, 0.464
Part C
Assuming that this locus is neutral, if this population suddenly became isolated and inbreeding became common, what might happen to the allele and genotype frequencies after many generations?
Assuming that this locus is neutral, if this population suddenly became isolated and inbreeding became common, what might happen to the allele and genotype frequencies after many generations?
A) Allele frequencies may stay the same but homozygosity would increase and heterozygosity would decrease.
B) Allele frequencies may stay the same but heterozygosity would increase and homozygosity would decrease.
C) Allele frequencies would go up and genotype frequencies would go down.
D) Allele frequencies would go down and genotype frequencies would go up.
A) 0.709, 0.545, 0.382
B) 0.355, 0.273, 0.191
C) 0.464, 0.318, 0.217
D) 0.355, 0.318, 0.327
Explanation / Answer
PART A: Total population size= (12+17+5+8+10+3)=55
Frequency of A1= (12x2 + 17 + 10)/55x2= 0.464
Frequency of A2= (17 + 5 x 2 + 8)/55x2= 0.318
Frequency of A3= (8 + 10 + 3x2)/ (55x2)= 0.218
Correct answer is C
PART B: genotype frequencies of A1A1= 12/55= 0.218
genotype frequencies of A1A2= 17/55=0.309
genotype frequencies of A1A3=10/55=0.182
Correct answer is B
PART C: A: Inbreeding always increases homozygosity.