In a linear chromosome map distance between 4 loci is as follows a-b is 10%, a-d is 3%, b-c is 4% and a-c is 6%. The crossover frequency between c and d is

The correct option is 3%.

To determine the crossover frequency (map distance) between loci c and d, we can map the relative positions of the genes a, b, c, and d on a linear chromosome using the given map distances (where 1% crossover frequency corresponds to 1 map unit or centimorgan, cM):
- Distance between a and b (a-b) = 10%
- Distance between a and d (a-d) = 3%
- Distance between b and c (b-c) = 4%
- Distance between a and c (a-c) = 6%

Let us first determine the order of the genes a, b, and c:
The distance between a and b is the largest at 10%. We can place a at position 0.
If a is at 0, then b is at position 10.
The distance from a to c is 6%, and from b to c is 4%. Since 6% + 4% = 10% (the total distance between a and b), locus c must lie directly between loci a and b.
Therefore, the position of c is 6 (since the distance a-c is 6 and the distance c-b is 10 - 6 = 4).
The order of these three genes is: a — (6%) — c — (4%) — b.

Next, we determine the position of locus d relative to the others:
We are given that the distance between a and d (a-d) is 3%. Since the chromosome is linear, d can be in one of two directions from a (position 0):
1. If d is to the left of a, its position is -3.
In this case, the distance between c (at position 6) and d (at position -3) would be:
$6-(-3)=9$ (which represents 9%).
2. If d is to the right of a (towards c and b), its position is 3.
In this case, the distance between c (at position 6) and d (at position 3) would be:
$6-3=3$ (which represents 3%).

Comparing these possibilities with the given options, the crossover frequency between c and d is 3%.