In a genetic mapping study, the recombination frequency between gene A and gene B is observed to be 12%. Assuming no interference, estimate the map distance between these two loci. Further, explain why this indicates linkage between genes A and B.
Answer
12 cM; they are linked
The correct answer is 12 cM; they are linked — Therefore, genes A and B are linked with a map distance of 12 cM.
Step-by-step solution
- Recall that map distance in centiMorgans equals recombination frequency expressed as a percentage when no interference is present.
- Since the recombination frequency is 12%, the map distance is 12 cM.
- Explain linkage: Genes that are close together on the same chromosome have recombination frequencies less than 50%; 12% is much less than 50%, showing the genes are linked.
- Linkage is detected because the observed recombination is significantly lower than expected for independently assorting genes.
- Therefore, genes A and B are linked with a map distance of 12 cM.
Common mistake: Confusing recombination frequency with proportion of all progeny
Concept tested: Linkage and map distance
Published · How this solution was prepared and checked
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