2002 AIPMT Principles of Inheritance and VariationLinkage Maps Medium Question 2 of 2 in Linkage Maps

There are three genes a, b and c. The percentage of crossing over between a and b is 20%, b and c is 28% and a and c is 8%. What is the sequence of genes on chromosome

A b, a, c
B a, b, c
C a, c, b
D None

The answer is A.

Solution

Map distance is proportional to the frequency of crossing over: a–b = 20 units, b–c = 28 units, a–c = 8 units.
The largest distance (b–c = 28) is between the two genes at the ends.
The third gene, a, lies between them: b–a + a–c = 20 + 8 = 28 = b–c.
[add image: linkage map b ---20--- a --8-- c]
So the sequence of genes on the chromosome is b, a, c.

Concept Gene order from recombination frequencies

Linkage Maps in past papers

2 questions from this chapter have appeared across 2 exam years.