2. In wild petunias, the allele for red flower color (R) is dominant over the al
ID: 189045 • Letter: 2
Question
2. 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 750 plants have white flowers and 270 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 270 plants that have red flowers are homozygous and how many are heterozygous at the flower color locus? Show your work and round allele frequencies to two decimal places. 3. A zoologist is investigating a population of squirrels whose coat color is controlled by a single gene whose two alleles (B, & B2) are co-dominant. B,B1 individuals are black, B,B2 individuals are brown, and 8282 are tan. She surveys the population and finds 600 black squirrels, 256 brown squirrels, and 112 tan squirrels. Is this population in HWE? If not, what might explain this pattern? Make sure to show your work.Explanation / Answer
2.
The frequency of R (red flowers) = 270/1020 = 0.261 = 27%
The frequency of r (white flowers) = 750/1020 = 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 270, 68% are homozygous and 32% are heterozygote.
3.
Hardy - Weinberg equation is (p +q)2 = p2 + 2pq + q2 =1
According this equation, it can be identified that whether a population is in equilibrium or not.
In the given problem B1B1 - Dominant (homozygous for black coat colour of squirrels) =600
B2B2 --- Dominant (homozygous for tan coat colour of squirrels) = 112
B1B2 --- Dominant (heterozygous and produce brown coat colour of squirrels) = 256
Frequency of B1 = 600/968 = 0.61
Frequency of B2 = 112/968 = 0.11
Frequency of B1B2 = 256/968 = 0.26
Estimated frequency of B2 (q) = square root of (B2 or q)2 = (0.11)2 = 0.012
Estimated frequency of B1 (p) = 1- q or 1 - B1 = 1 - 0.012 = 0.987
Estimated frequency of B1B2 = 2pq or 2 B1B2 = 2 (0.012) (0.987) = 0.027.
Now (p +q)2 = p2 + 2pq + q2 ?=1; (B1B2)2 = (0.987)2 + 0.027 + (0.012)2 = (0.974) + 0.027+(0.0001) = 1.00
The value is one hence the alleles are in HWE (Hardy - Weinberg Equilibrium).
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