Question

64. In maize, coloured endosperm (C) is dominant over
colourless (c); and full endosperm (R) is dominant over
shrunken (r). When a dihybrid of F, generation was test
crossed, it produced four phenotypes in the following
percentage:
Coloured full - 48% Coloured shrunken - 5%
Colourless full – 7% Colourless shrunken - 40%
From this data, what will be the distance between two
non-allelic genes?
(a) 48 units
(b) 5 units
(c) 7 units
(d) 12 units​

Answer 1

Answer:

12 unit.

Explanation:

Since, we get from the question that the recombinant number of alleles will be of colourless full which is 7% and coloured shrunken which is given 5%. Out of the total frequency as 100.

So, the distance between the two genes will be the recombinant frequency which is 7+5/100*100 which is 12% hence the map unit will be 12 unit.