In zebrafish three genes have been mapped on the third chrom

In zebrafish, three genes have been mapped on the third chromosome, in the order

fiz---6 m.u.---myob-------14 m.u.-------sufi

The wild-type allele of each gene is dominant.

A strain that is
fiz myob sufi / + + +
is testcrossed, and there are 1000 offspring fish.

PLEASE EXPLAIN ALL STEPS TO EACH QUESTION PART:
a) How many wild-type offspring would you expect to see in these 1000 offspring?

b)How many
fiz + +
offspring would you expect to see in these 1000 offspring?


c) How many
+ + sufi
offspring would you expect to see in these 1000 offspring?


d) How many
+ myob +
offspring would you expect to see in these 1000 offspring?

Solution

1. Since it is given that there are 6 mu between given su and myob. This means that there is a 6% chance of crossover between the two. Now 6% chance means that there must be 60 offsprings of s++ or +myob su. But since this is a double crossover then we have to halve the chance. This means that there will be 30 offsprings of s++ or +myob su and s++ would be half which means 15.

2. Now there are 14 mu between myob and su. This means there is 14% chance of croosover. so there\'s a 14% chance of a crossover between them. This crossover will give either su myob + or + + su. From 140 individuals deducted in a double crossover this means 140 i.e. 70 of su myob + 70 of + + su.

3. As per the give mapping number of wild type offspring will be 400.

4. Number of + myob + offspring will be 4.