Complete question:
For a series of experiments, a linkage group composed of genes W, X, Y and Z was found to show the following gene combinations. (All recombinations are expressed per 100 fertilized eggs). Construct a gene map. Determine the sequence of genes on the chromosome.
- w-x = 5
- w-y = 7
- w-z = 8
- x-y = 2
- x-z = 3
- y-z = 1
Answer:
The sequence of genes on the chromosome is:
----W-------------------------X-----------Y------------Z---
Explanation:
First, we need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant one.
The recombination frequencies between two genes determine their distance in the chromosome, measured in map units. So, if we know the recombination frequencies, we can calculate distances between the four genes in the problem and we can figure the genes order out. This is:
Recombination frequencies:
1% of recombination frequency = 1 map unit (MU)
- w-x = 5 MU
- w-y = 7 MU
- w-z = 8 MU
- x-y = 2 MU
- x-z = 3 MU
- y-z = 1 MU
Now that we know the distances, we just need to analyze them to find out the correct order of the genes. First, we can look for the biggest distance, which tells us which genes are located in the extremes. w-z distance is the biggest one, so these two genes are in the extremes of the chromosome segment. ---W----------------------------------------------Z---
∫---------------------8 mu-------------------∫
The rest of the genes are located in the middle between these two.
The second biggest distance is between w-y (7 mu). Y is also 1mu distant from Y. 7 mu + 1 mu = 8 mu. So, Y is located closer to Z.
---W-------------------------------------Y------------Z---
∫-----------------------7 mu---------∫∫---1 mu--∫
∫---------------------8 mu-------------------------∫
w-x = 5 mu, and x-y = 2mu, so x is located between w and y. The sum of these distances equals the distance w-y ( 5 mu + 2 mu = 7 mu). So,
---W-------------------------X----------Y------------Z---
∫-----------5 mu -------∫∫--2mu--∫
∫-----------------------7 mu---------∫∫---1 mu--∫
∫---------------------8 mu-------------------------∫
We know that the distance between x-y equals 2, and the distance between y-z equals 1. Also, the distance between x-z equals. This leads us to assume that Y is located between X and Z.
----W-------------------------X-----------Y------------Z---
∫-----------5 mu -------∫∫--2mu--∫∫----1mu---∫
∫------ 3 mu-----------∫
∫-----------------------7 mu---------∫∫---1 mu---∫
∫---------------------8 mu--------------------------∫
