. In Drosophila, shaven bristles (sv) are specified by a recessive gene. A popul
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Question
. In Drosophila, shaven bristles (sv) are specified by a recessive gene. A population of fruit flies was found to consist of 66 wild type flies and 22 with shaven bristles.
a. What is the frequency of the sv allele in this population?
b. What is the frequency of the sv+ allele in this population?
c. Based upon the Hardy Weinberg formula, what proportion of the population is expected to have shaven bristles?
d. Based upon the Hardy Weinberg formula, what proportion of the population is expected to be heterozygous for the gene in question? e. Based upon the Hardy Weinberg formula, how many individuals in this population would be expected to have the genotype sv+sv+?
Explanation / Answer
sv+sv+ = 66; svsv = 22 (Total = 88 flies)
Alleles: sv+ = 66, sv+ = 66; sv = 22, sv = 22. (Total Allelle = 176)
sv+ = 66*2 / 176 = 0.75 (Frequency of sv+ allele = 0.75)
sv = 22*2 / 176 = 0.25 (Frequency of sv allele = 0.25)
p + q = 1
0.75 + 0.25 = 1 (Hence proved)
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p = 0.75
p2 = 0.5625 (Frequency of sv+sv+ homozygotes= 0.5625)
q = 0.25
q2 = 0.0625 (Frequency of svsv homozygotes= 0.0625)
2pq = 2 * 0.75 * 0.25 = 0.375 (Frequency of sv+sv heterozygotes= 0.375)
P2 + 2PQ + Q2 = 1
0.5625 + 0.375 + 0.0625 = 1 (Hence proved)
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Expected sv+sv+
p2 = 0.5625
0.5625 * 88 = 49.5
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Expected svsv
q2 = 0.0625
0.0625 * 88 = 5.5
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Expected sv+sv
2pq = 2 * 0.75 * 0.25 = 0.375
0.375 * 88 = 33.
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Q.1
Frequency of the sv allele in this population = 0.25
Q.2
Frequency of the sv+ allele in this population = 0.75
Q.3
Proportion of the population expected to have shaven bristles = 6.25 %
Q.4
Proportion of the population expected to be heterozygous for the gene in question = 37.5 %
Q.5
Proportion of the population expected to have sv+sv+ = 56.25 %
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Observed no. of svsv = 22, Expected no. of svsv = 5.5
Observed no. of sv+sv+ = 66, Expected no. of sv+sv+ = 49.5
Observed no. of sv+sv = X, Expected no. of sv+sv = 33
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