Q1 A zoologist is investigating a population of squirrels whose coat color is co
ID: 32196 • Letter: Q
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Q1 A zoologist is investigating a population of squirrels whose coat color is controlled by a single gene whose two alleles (B1 & B2) are codominant. B1B1 individuals are black, B1B2 individuals are brown, and B2B2 are tan. She surveys the population and finds 126 black squirrels, 238 brown squirrels, and 49 tan squirrels. Is this population in HWE? If not, what might explain this pattern?
Q2 In wild petunias, the allele for red flower color (R) is dominant over the allele for white flower color (r). We sample a population and find 240 plants have white flowers and 85 have red flowers. If the population is in HWE, what are the allele frequencies of R and r in this population? How many of the 85 plants that have red flowers are homozygous and how many are heterozygous at the flower color locus?
Q3 Phenylketonuria is a severe form of mental retardation caused by a recessive allele. Assume that the condition affects about 3 in 25,000 newborn babies. In a population of 450,000 people, how many are expected to be heterozygote 'carriers' for this disease?
Explanation / Answer
1)
Hardy - Weinberg equation is (p +q)2 = p2 + 2pq + q2 =1
According this equation, it can be known that whether a population is in equilibrium or not.
In the given problem B1B1 --- Dominant ( homozygous for black coat color of squirelles) =126
B2B2 --- Dominant ( homozygous for tan coat color of squirelles) = 49
B1B2 --- Dominant ( heterozygous and produce brown coat color of squirelles) = 238
Frequency of B1 = 126/413 = 0.30
Frequency of B2 = 49/413 = 0.12
Frequency of B1B2 = 238/413 = 0.57
Estimated frequency of B2 (q) = squaere root of (B2 or q)2 = (0.12)2 = 0.014
Estimated frequency of B1 (p) = 1- q or 1 - B1 = 1 - 0.014 = 0.986
Estimated frequency of B1B2 = 2pq or 2 B1B2 = 2 (0.014) (0.986) = 0.027.
Now (p +q)2 = p2 + 2pq + q2 ?=1; (B1B2)2 = (0.0986)2 + 0.027 + (0.014)2 = (0.972) + 0.027 + (0.0001) = 0.999
The value is very near to one hence the alleles are in HWE (Hardy - Weinberg Equilibrium).
2)
The frequency of R (red flowers) = 85/325 = 0.261 = 27%
The frequency of r (white flowers) = 240/325 = 0.738 = 73%
Frequency of the genotype RR = (0.261)2 = 0.068
Frequency of genotype rr = (0.738)2 = 0.544
Frequency of genotype Rr = 2 (0.261) (0.738) = 0.385
Out of 85, 68% are homozygous and 32% are heterozygote.
3)
Frequency of the allele (q) = 3/25000 = 0.00012
Frequency of the allele (p) = 1 - q = 1 - 0.00012 = 0.99988
Frequency of heterozygotes or carriers (2pq) = 2 (0.00012) (0.99988) = 0.00024. = 0.02% would be carriers of the disease.
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