Explore BrainMass
Share

Genetic Linkage and Crossover

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

Hi. Here's the problem:

In Zebrafish, 3 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.

1. How many fiz + + offspring would you expect to see in these 1000 offspring?

2. How many + + sufi offspring would you expect to see in these 1000 offspring?

I got two other questions regarding this problem correct: that there would be 400 wild-type offspring (a parental class) and only 4 "+ myob +" offspring (a double crossover class). But I can't seem to figure these two out.

My approach: If there are 6 map units between fiz and myob, then there's a 6% chance of a crossover between them. That is, a 6% chance of getting either "fiz + +" or "+ myob sufi". 6% x 1000 = 60. And the "fiz + +" class should be approximately half of that, which is 30.

But that's the wrong answer, and I'm not sure why.

Same thing with question number 2.

There are 14 m.u.'s between myob and sufi, so there's a 14% chance of a crossover between them, a crossover that would result in either "fiz myob +" or "+ + sufi." 14% x 1000 = 140. 140/2 = 70.

But again, that's not the right answer.

Can you explain to me how to do this problem, and the correct answers?

© BrainMass Inc. brainmass.com October 24, 2018, 5:11 pm ad1c9bdddf
https://brainmass.com/biology/diagnostics-of-linkage/genetic-linkage-and-crossover-3763

Solution Preview

The reason you were getting the wrong answer is that you were not taking into account the double cross-overs. That is, the progeny that are: f++ AND +ms AND .5(f+s AND +m+) equals 60.
<br><br>You halve the double cross-overs ...

$2.19
See Also This Related BrainMass Solution

Genetic problems regarding gene mapping of linked genes

Problem 1

A female mouse with yellow-mottled fur, eye-ear reduction and regular teeth was mated to a male mouse with dark gray fur, normal eye-ear size and irregular teeth. The F1 females were all wild type, that is, dark gray fur, normal eye-ear size and regular teeth. The F1 males had yellow-mottled fur, eye-ear reduction and regular teeth. A cross of F1 females to F1 males produced the following offspring in the F2 generation.

Phenotype for Number of
Fur color Eye-Ear size Teeth Males Females
Dark Gray Normal Regular 11 142
Yellow Mottled Reduced Irregular 18 0
Dark Gray Reduced Irregular 47 0
Yellow Mottled Normal Regular 54 57
Dark Gray Reduced Regular 3 51
Yellow Mottled Normal Irregular 4 0
Yellow Mottled Reduced Regular 135 150
Dark Gray Normal Irregular 128 0

a. Draw a genetic map showing the distances between these genes.
b. Express the differences in map units.
c. What is the coefficient of coincidence?
d. What is the degree of interference?

Problem 2

A female rabbit with hemophilia, rickets, and a tail was mated to a male rabbit that did not have hemophilia or rickets but lacked a tail. The F1 females were all wild type, that is, no having hemophilia or rickets but having a tail. The F1 males had hemophilia, rickets, and tails. A cross of F1 females to F1 males produced the following offspring in the F2 generation:

Hemophilia Rickets Tails Males Females
Yes No No 5 0
No Yes Yes 3 84
Yes Yes No 30 0
No No Yes 27 160
Yes No Yes 86 91
No Yes No 81 0
Yes Yes Yes 128 165
No No No 140 0

a. What are the genotypes of the original parents?
b. Write Each genotype to show which alleles are linked together on the same chromosome.
c. What are the genotypes of the F1 males and females? Write each genotype to show which alleles are linked together on the same chromosome.
d. Draw a genetic map showing the distances between these genes. Express the distances in map units.
e. What is the degree of interference?

View Full Posting Details