Question

The distance between the genes a, b, c and d in mapping units are a – d = 3.5; b – c = 1; a –b = 6; c – d = 1.5; and a – c = 5. Find out the sequence of arrangement of these genes. Find out the sequence of arrangement of these genes
A. acdb
B. abcd
C. acbd
D. abdc
E. adcb

Answer 1

Hint: The distance between two genes is calculated by the unit mu or mapping unit or centimorgan (cM). There are four genes and five values are given. We will do a trial and error method for calculating the sequence.
Step-by-step explanation:
There are five values given in the question. The values are as follows –
a – d = 3.5;
b – c = 1;
a –b = 6;
c – d = 1.5; and
a – c = 5
The distance between a and b is 6 and it is the maximum value given in the data. So, we can say that a and b are located farthest to each other. Thus, we can write a_______b.
The distance between a – c is 5 and b – c is 1. If we add these two values then 6 will come. c is the common gene that is present in the two values. So, a and b will be present on both sides of c. Thus, we can write a_______c___b. c is close to b as the distance given is less than the distance between a and c.
The distance between a and d is 3.5 and c and d is 1.5. If we add these two values then 5 will come. d is the common gene that is present in the two values. So, a and c will be present on both sides of d. Thus, we can write a________d_____c. c is close to d as the distance given is less than the distance between a and d.
So, finally we have to arrange the sequence. So the sequence of genes will be a_______d_____c___b.

So, the final answer is option E. adcb.

Note: The distance between the genes also help us to understand the percentage of crossing over and the recombination frequency. The lower the value of the distance between two genes, the greater the chance for the genes to recombine and cross over.