Question
Asked Oct 27, 2019
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Genes A, B, and C lie at map positions 2.5, 7.5, and 17.5, respectively. In a three-point test cross, you observed 2 double crossovers in a total of 1000 progeny. What is the interference in this region?

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Expert Answer

Step 1

The measure of genetic linkage is termed as recombination frequency, which is used in the construction of a genetic linkage map. Recombination frequency is depicted by Ɵ (theta) and calculates the frequency by which a single chromosomal crossover occurs between two genes at the time of meiosis. Its formula is:

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recombinant offspings total number of offsprings Recombination frequency

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Step 2

Map distance between gene A is 2.5 map units.

Map distance between gene B is 7.5 map units.

Map distance between gene C is 17.5 map units.

Therefore, the distance between A and B is 5 map units.

The distance between B and C is 10 map units.

The representation for above is shown below:

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В А с 17.5 7.5 2.5 В 5 map units 10 map units A C

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Step 3

It is known that 1 map unit is equal to 1% recombination frequency.

Therefore, recombination frequency between A and...

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Frequency of double crossovers = 0.05x 0.1 -0.005 Number of expected double crossovers in a population 1000 progeny = 0.005 x1000 = 5

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