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Fia Fia
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Posts: 1
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2 years ago
It’s a linkage question but I couldn’t find a topic with linkage:(

The test cross ab/ab x Ab/aB is performed. The following number of progeny of each genotype are obtained: 87 AaBb, 409 Aabb, 390 aaBb, 114 aabb.

What is the approximate distance (in map units) between the two genes in question?

A. 25 centiMorgans (CM)
B. 20cM
C. 15 cM
D. 10 cM
E. 5 cM
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Educator
2 years ago
Add all those values:

\[87+409+390+114=1000\]

Add AaBb and aabb:

\[87+114\]

Recombinant Frequency: \[\frac{201}{1000} = 0.201 \times 100% = 20.1%\]

Therefore, the distance between the two genes are 20.1 mu
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