Question

In an experiment, individuals homozygous for 'ab' genes were crossed with wild type (++). The F${}_1$ hybrid thus produced was test crossed and progenies were produced in the following ratio:
++/ab 896
ab/ab 880
+a/ab 108
+b/ab 116

Calculate the distance between a and b genes.

• A. 11.2 m.u.
• B. 22.4 m.u.
• C. 2.8 m.u.
• D. 4.6 m.u.

Answer 1

The correct option is A 11.2 m.u.

Given: Test Cross
Parents: ++/ab X ab/ab
(F${}_1$ hybrid parent) (Recessive parent)

Recombinant: +a/ab and +b/ab

To find: Distance between a and b genes

Calculation:
The recombination frequency for this test cross

$=\frac{\mathrm{Number}\mathrm{of}\mathrm{recombinants}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{offsprings}}$ ×100
$=\frac{108+116}{896+880+108+116}\times 100=\frac{224}{2000}\times 100=0.112\times 100=11.2\%$

1% recombinant frequency is equivalent to 1 map unit (m.u.)
Therefore, the distance between a and b genes is 11. 2 m.u.