Question

A dihybrid plant with genotype PpNn
produces four types of gametes in following
number PN = 200, pn = 200, Pn = 800, pN
800 then what is the distance between linked
genes ?
(1) 30 CM
(2) 10 CM
(3) 25 CM
(4) 20 CM​

Answer 1

Explanation:

25 Cm is the Answer means correct option is Third .

Answer 2

Given:-

A dihybrid plant with genotype PpNn produces four types of gametes as PN=200, pn=200,  Pn=800,  pN=800.

To Find:

We have to find the distance between linked genes.

Solution:-

• The plant with PpNn genotype will produce two gametes in the Parental combination of Pn and pN and it will also produce two gametes in the recombinant fashion of pn and PN after the genes cross over.

• Given that the occurrence of recombinant gametes is:-

        pn = $200$ times

        and PN = $200$ times

       The total number of recombinant gametes, therefore, sum upto  

       $200+200=400$

• And the total number of gametes = $200+200+800+800=2000$

• We can calculate the distance between linked genes is by the formula:

       $Distance\:(d)=\frac{no.\: of\: recombinant\: gametes}{no.\:of\:total\:gametes} \times 100$

                     ⇒ $d=\frac{400}{2000} \times100=20\,cm$

Therefore the distance between the linked genes is $20\,cm$.