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  • Two linked genes a and b show recombination. The individuals of dihybrid cross between <img alt="++/++\times" s

Two linked genes a and b show 20 \% recombination. The individuals of dihybrid cross between ++/++\times ab/ab shall show gametes

Option: 1

++80: a b: 20


Option: 2

++50: a b: 50


Option: 3

++40: a b \, \, 40:+a \, \, 10:+b: 10


Option: 4

++30: a b \, \, 30:+a \, \, 20:+b: 20


Answers (1)

Given that two linked genes, a and b, show 20% recombination, it means that 20% of the gametes produced will be recombinant (carrying a combination of alleles different from the parental arrangement), and the remaining 80% will be non-recombinant (carrying the same combination of alleles as the parental arrangement).

The correct gametic ratio for the dihybrid cross between ++/++x ab/ab is:

40: a b \, \, 40:+a \, \, 10:+ \text { b } 10

In this cross, the gametic ratio is as follows:

Non-recombinant gametes  \text { (++ and ab): } 80 \%+20 \%=100 \%

+: 40 \% ( 40 \% of the gametes will be non-recombinant for the a and b alleles)

ab: 40 \% ( 40 \% of the gametes will be non-recombinant for the a and b alleles)

Recombinant gametes (+a$ and $+b): 10 \%+10 \%=20 \%

a: 10% (10% of the gametes will be recombinant for the a allele)

b: 10% (10% of the gametes will be recombinant for the b allele)

Again, the correct gametic ratio is:

40: a b \, \, 40:+a \, \, 10:+b \, \, 10

This means that 40% of the gametes will be non-recombinant for both a and b, 40% will be non-recombinant for either the a or b allele, and 10% will be recombinant for both a and b. Hence, the correct answer is option 3.

 

Posted by

Sumit Saini

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