In cod, the genes for short fins (sf), rough scales (rs) and spotted body (sb) a
ID: 149557 • Letter: I
Question
In cod, the genes for short fins (sf), rough scales (rs) and spotted body (sb) are located on a single chromosome separated by the following map distances:
sf ---(10.5)---rs-------(26.5)------sb
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 & spotted body OR rough scales
Enter your answer as a whole number, round to the nearest integer.
Explanation / Answer
Answer:
The frequency of short fins and spotted body population=37
The frequency of rough scales population = 37
Explanation:
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
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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