A female with short fins and a rough scales was mated to a male with a spotted b

Question

A female with short fins and a rough scales was mated to a male with a spotted body (assume that these traits are mutant phenotypes, and that the mutant alleles are recessive).

The resulting phenotypically wild-type females were mated with males that had the mutant phenotype for all three traits, producing 900 offspring.

Question: Assume that interference for these genes is 0.16. How many offspring are expected to have the following phenotypes?

short fins & rough scales OR spotted body

Enter your answer as a whole number, round to the nearest integer.

Explanation / Answer

Answer:

Number of short fins and spotted population = 37

Number of rough scales population = 37

Explanation:

Some information is missed. But I know the question. Hence, I could able to answer as follows.

Interference = 1- Coefficient of Coincidence

0.16 = 1- Coefficient of Coincidence

Coefficient of Coincidence = 1-0.16 = 0.84

Coefficient of Coincidence = Observed double crossover frequency / Expected double crossover frequency

Expected double crossover frequency = 10.5% * 26.5% = 0.028

0.84 = ?/ 0.028

Observed double crossover frequency = 0.84 * 0.028 = 0.0235 = 2.35%

sf rs Sb / Sf Rs sb ---Wild type female genotype

Inorder to get short fins, spotted body OR rough scale, single crossover has to be occurred between gene short fins and rough scales.

Single crossover frequency between short fins and rough scales = SCO frequency – observed double crossover frequency

= 10.5 – 2.35 = 8.15%

The frequency of short fins and spotted body population= 8.15 / 2 =4.075%

4.075% out of 900 people = 0.04075 * 900 = 36.675 = 37

The frequency of rough scales population = 8.15 / 2 =4.075%

4.075% out of 900 people = 0.04075 * 900 = 36.675 = 37

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