You sample 60000 Tsetse flies in a forest savanna habitat in central Africa. In

Question

You sample 60000 Tsetse flies in a forest savanna habitat in central Africa. In this sample you determine that for a locus associated with Malathion resistance the flies have the genotype frequencies listed below.

M1M1: 15000       M1M2: 24000       M2M2:21000

After spraying the habitat with Malathion you obtain a sample of 6000 flies with the genotype frequencies listed below.
M1M1: 1500       M1M2: 1740       M2M2:2760


a. What are the allele frequencies in the Tsetse fly population BEFORE the Malathion spraying?
M1:?

M2:?

b. AFTER the Malathion spraying which allele has increased in frequency?

c. What are the relative fitnesses (W) of each genotype after the spraying?
M1M1:?

M1M2:?

M2M2:?

Explanation / Answer

Allele frequency of M1 BEFORE the Malathion spraying = (15000 X 2 + 24000) / (60000 X 2) = .45

Allele frequency of M2 BEFORE the Malathion spraying = (24000 + 21000 X 2) / (60000 X 2) = .55

Allele frequency of M1 AFTER the Malathion spraying = (1500 X 2 + 1740 X 1) / (6000 X 2)   = .395

Allele frequency of M2 AFTER the Malathion spraying = (2760 X 2 + 1740 X 1) / (6000 X 2)   = .605

M2 has increased frequqency after Malathion spraying

Absolute fitness = No. of individuals after selection / No. of individuals before selection

Relative fitness is fitness of a particular genotype relative to fitness of competing genotypes after a single generation.

Absolute fitness M1M1 = (1,500/6,000) / (15,000/60,000) = 1

Absolute fitness M1M2 = (1740/6,000) / (24,000/60,000) = .725

Absolute fitness M2M2 = (2760/6,000) / (21,000/60,000) = 1.31

We will find relative fitness of each genotype with respect to M1M2

Relative fitness M1M1 = 1/1.31    = 0.76

Relative fitness M1M2 = .725/1.31 = 0.55

Relative fitness M2M2 = 1.31/1.31 = 1.0

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