6. The fungal cross a b+ xat b yields 51 parental ditype, 4 nonparental ditype, and...

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Answer 1

Answer #1

The genetic distance between two genes is calculated by:

[{NPD + 1/2(TT)}/ Total number of tetrads]×100

Here NPD is non parental ditype=4

TT is tetratype =45

Total tetrads =51+4+45=100

Genetic distance = [{4+ 1/2×(45)}/100]×100= {(4+22.5)/100}×100= 26.5 m.u.

Right answer is C.

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Answer 2

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