For the Auricorus Blue-tailed Fly, that lives on the planet Trumpor, there are several recessive genes...

For the Auricorus Blue-tailed Fly, that lives on the planet Trumpor, there are several recessive genes that are all on the same autosome. In the following problems, a true breeding female fly that shows three distinct disease (gene) phenotypes is crossed to a wild-type male fly. An F1 female fly is then test-crossed to a true-breeding male fly showing all three disease phenotypes. Given the data below, what are the recombination fractions among all three loci?

Gene A Phenotype: Muppeteritus eye color

Gene B Phenotype: Farfeggnuggen wing shape

Gene C Phenotype: Farfeggnuggen wing angle

Table of Phenotype Counts
ABC 395
AB+ 51
A+C 6
A++ 46
+BC 59
+B+ 5
++C 45
+++ 405

A. A-B: 0.176 ; B-C: 0.30
B. A-B: 0.211 ; B-C: 0.145
C. A-B: 0.115 ; B-C: 0.106
D. A-B: 0.154 ; B-C: 0.111
E. A-B: 0.054 ; B-C: 0.288

Solutions

Expert Solution

Ans:

C. A-B: 0.115 B-C: 0.106

Explanation: The genotypes which are most frequent is called parental genotypes. So, ABC and +++ are parental genotypes. The genotypes which are less frequent is known as double crossover genotypes. So, A+C and +B+ are the double crossover genotypes. The calculated recombination fraction between gene A and B is 0.115 and gene B and C is 0.106.

All the calculation is given below-

All other options are incorrect because the calculated recombination fraction between A and B is 0.115 and B and C is 0.106.

gladiator answered 1 year ago

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