In a population of 1000 fruit flys the allele for normal size wings... Styles de
ID: 174191 • Letter: I
Question
In a population of 1000 fruit flys the allele for normal size wings... Styles deal with other population genetic problems. Some of these problems are easy; some are hard The Paragraph equations are given at the end. These are the equations that will be supplied on the final, so you should know which to use, as well as how to use them. 1. In a population of 1000 fruit flies, the allele for normal size wings (l) is present at a frequency of 0.7. The recessive allele (y is present at a frequency of 0.3. For every 12 flies that successfully reproduce, 8 flies reproduce and 5 u flies reproduce. A. What are the relative finesses of each genotype? B. What is the mean fitness of the population? C What are the genotypic frequencies after 1 generation of selection? D. What are the allelic frequencies after 1 generation of selection? E. What is the new mean fitness? 2. In a population of 900 zeebogs, there are 75 sane critters (homozygous for the recessive allele c) A. Assuming the population is in Hardy Weinberg equilibrium, frequencies? what are the allele B. After a cataclysmic change to their environment (the zeebog stock market crashes), sane zeebogs become more fit than the other genotypes. For every 5 sane zeebogs that reproduce, 3 heterozygotes reproduce and 1 crazy homozygote reproduces.Explanation / Answer
Answer 1.
l+ = p = 0.7; l = q = 0.3 (According to HW equilibrium)
So, frequency of "l+l+" genotype = (0.7*0.7) = 0.49 = p2
q2 = (0.3*0.3) = 0.09 = frequency of "ll" genotype
2pq = frequency of "l+l" genotype = (2*0.7*0.3) = 0.42
Total genotypes which reproduce = (12+5+8) =25
A.
Relative fitness of "l+l+" = (12/25) = 0.48 = WAA
Relative fitness of "l+l" = (8/25) = 0.32 = WAa
Relative fitness of "ll" = (5/25) = 0.2 = Waa
B.
Mean fitness = p2 WAA + 2pq WAa + q2 Waa
= (0.49*0.48) + (0.42*0.32) + (0.09*0.2)
= (0.23 + 0.13 + 0.018)
= 0.378
c.
Genotype frequencies after one generation of selection:
Frequency of "l+l+" genotype = p2 WAA = (0.49*0.48) = 0.23
Frequency of "l+l" genotype = 2pq WAa = (0.42*0.32) = 0.13
Frequency of "ll" genotype = q2 Waa = (0.09*0.2) = 0.018
d.
Allelic frequencies after one generation of selection
p = 0.23 + (1/2*0.13) = 0.295
q = 0.018 + (1/2 * 0.13) = 0.083
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