Question

In experimental individual homozygous for $ab$ genes were crossed with wild type $(++)$. The $F_1$ hybrid thus produced was test crossed and progenies produced in following ration $++548,ab340,+a44,+b68$

Calculate the distance between $b$ and $a$ genes.

• A. $46m.u$
• B. $22.4m.u$
• C. $44m.u$
• D. $11.2m.u$

Answer 1

Answer: (D)

$11.2m.u$

The recombination frequency for this crossbreed is

= $\frac{numberofrecombinantprogeny}{totalnumberofprogeny}$ $\times 100$

= $\frac{44+68}{548+340+44+68}$ $\times 100$

= $11.2$ %

1% recombinant frequency is equivalent to 1 centimorgan or 1 map unit or m.u.

Therefore the distance between a and b genes is 11. 2 m.u.

Answer D is correct.