In a linear chromosome, map distances between four loci are as follows: a-b-10, b-c-4, a-d-3, a-c-6. the expected cross over frequency between c and d is-

a) 3%

b)9%

c) either 3%or 9%

d)4% to 12%

plz provide the solution

Here a-c = 6 m.u i.e 6%

and a-d = 3 m.u i.e 3%

There frequency for c-d = a-c + a-d = 6% + 3% x 100

$\frac{6}{100}+\frac{3}{100}\times 100=9\%$

The correct answer is b.

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