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Q.176 DOB Genetics.

Q.176 DOB Genetics.

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Grade:12th pass

1 Answers

Raheema Javed
156 Points
7 years ago
An individual with cd genes crossed with wild type ++ will produce two types of gametes- +c and d+. On crossing +c and d+ we will get the following offsprings- ++, +c, +d and cd. Among these offsring +c and +d are parental offspring and ++ and cd are recombinant offsprings.
The linkage map distance between the c and d gene loci can be calculated by calculating the percentage of recombination between two genes. This is done by adding the number of recombinanat offspring and dividing it by total number of offspring and multiplying by 100.
Distance between two genes= No. of recombinant offspring/Total number of offspring*100
In the above example,
No. of cd offspring=880
total number of offspring=2000
Distance between cd gene =880/2000*100=44%=44 map units

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