In cod, the genes for short fins (sf), rough scales (rs) and spotted body (sb) a

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 & rough scales OR wild type

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

Explanation / Answer

Answer:

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 = 0.024 = 2.4%

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

Inorder to get hort fins & spotted body & rough scales OR wild type, single crossover has to be occurred between gene rough scales and spotted body.

Single crossover frequency between rough scales and spotted body = SCO frequency – observed double crossover frequency

= 26.5 – 2.4 = 24.1%

The frequency of rough scales and spotted body population= 24.1/2 =12.05%

12.05% out of 900 people = 0.1205 * 900 = 108

The frequency of wild-type population = 24.1/2 =12.05%

12.05% out of 900 people = 0.1205 * 900 = 108

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