Question

Please help it’s really important

Answer 1

Q1. The Recombinants in the cross are Yyll and yyLl.

Q2. The distance between the two genes is 20 map unit.

Step-by-step explanation:

Q1. The parents present in this cross do have a genotype of YyLl and yyll. So the gametes produced from the parents are:

A. From the 1st parent - YL, Yl, yL and yl.

B. From 2nd parent - yl.

So the four type of probable types of offsprings are YyLl, Yyll, yyLl and yyll.

Among them, the 1st and the last probable offsprings do have same genotype as that of the parents. So the Recombinant offsprings are Yyll and yyLl.

Q2. The distance between the two genes present on a chromosome is the percentage of Recombinant offsprings among the total offsprings produced.

So distance in map units =
`(Recombinant\ offsprings)/(total\ offsprings) * 100.`

Number of Recombinant offsprings = 100 + 100 = 200.

Number of total offsprings = 400 + 100 + 100 + 400 = 1000.

So the distance in map units =
`(200)/(1000) * 100 = 20`

So the distance between the two genes is 20 map units.