The recombination frequency between the genes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Recombination frequency is directly proportional to the distance between genes on a chromosome. Therefore,$1\% 	ext{ recombination frequency} = 1

Vedclass · Accepted Answer

$a, c, b, d$

Vedclass · Answer

(C) Recombination frequency is directly proportional to the distance between genes on a chromosome. Therefore,$1\% 	ext{ recombination frequency} = 1 	ext{ map unit (cM)}$.Given distances:$a-c = 5 	ext{ cM}$$b-c = 15 	ext{ cM}$$b-d = 9 	ext{ cM}$$a-b = 20 	ext{ cM}$$c-d = 24 	ext{ cM}$$a-d = 29 	ext{ cM}$Step $1$: Place $a$ and $c$ at a distance of $5 	ext{ cM}$.Step $2$: Since $a-b = 20 	ext{ cM}$ and $c-b = 15 	ext{ cM}$,$b$ must be placed such that $c$ is between $a$ and $b$ $(a-c-b = 5 + 15 = 20 	ext{ cM})$.Step $3$: Check $d$. We have $b-d = 9 	ext{ cM}$ and $c-d = 24 	ext{ cM}$. Since $c-b = 15 	ext{ cM}$ and $b-d = 9 	ext{ cM}$,$d$ must be placed on the other side of $b$ such that $c-b-d = 15 + 9 = 24 	ext{ cM}$.Step $4$: Verify with $a-d = 29 	ext{ cM}$. The sequence is $a-c-b-d$. The distance $a-d = a-c + c-b + b-d = 5 + 15 + 9 = 29 	ext{ cM}$.This matches the given data. Thus,the sequence is $a, c, b, d$.

The recombination frequency between the genes $a$ and $c$ is $5\%$,$b$ and $c$ is $15\%$,$b$ and $d$ is $9\%$,$a$ and $b$ is $20\%$,$c$ and $d$ is $24\%$,and $a$ and $d$ is $29\%$. What will be the sequence of these genes on a linear chromosome?

Similar Questions

State the conclusions drawn from Morgan's experiments on linkage and recombination.

The number of linkage groups in $Pisum \text{ } sativum$ is

Two dominant non-allelic genes are $50$ map units apart. The linkage present is:

Two heterozygous parents are crossed. If the two loci are linked,what would be the distribution of phenotypic features in the $F_{2}$ generation for a dihybrid cross?

Vedclass Test Series

Exam Paper Generator

Online Exam Module