Question

An individual homozygous to genes cd is crossed with wild type $++$ and $F_1$ crossed back with the double recessive. The appearance of the offsprings is as follows.
$++\to 903$
$cd\to 897$
$+d\to 98$
$c+\to 102$
The distance between the genes $c$ and $d$ is

• A. $20$ map units
• B. $9.8$ map units
• C. $10.2$ map units
• D. $10$ map units

Answer 1

The correct option is D $10$ map units

Recombination Frequency = (No of recombinants/Total number of progeny) * 100 map units. The total number of recombinants in this case is 98+102=200 and the total number of progeny is 200+903+897=2000. So the recombination frequency is 200/2000 X 100 = 10 map units.

So the correct option is '10 map units'.