Two mutations of T4 are c that causes clear plaques and m that produces minute p

Question

Two mutations of T4 are c that causes clear plaques and m that produces minute plaques. In an experiment c+m- is crossed with c-m+ and the parental and recombinant plaques were counted.

genotype

Number of

plaques

c+m-

c-m+

445

475

c+m+

c-m-

        36

        44

                                                Total = 1000

a) What is the recombination frequency between c and m?

b) What is the map distance between c and m?

genotype

Number of

plaques

c+m-

c-m+

445

475

c+m+

c-m-

        36

        44

Explanation / Answer

answer

Recombination frequency is given by

Number of recombinant/total number of progeny * 100

So, recombinant frequency in the given problem will be

80/1000*100 = 8%

Map units = % recombination

One map unit is the distance between genes for which one product of meiosis in 100

the distance between gene is proportional to the frequency of recombination.

Map units usually, but not always proportional to physical distance. 1 mu = 1% recombination = 1cM (centimorgan)

SO, unit in the given question will be 8 map unit or 8cM

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