Question

In a linear chromosome map, distance among 4 loci is 20% from a to b, 6% from a to d, 8% from b to c and 12% from a to c. What will be the crossover frequency between c and d?

• A. 12%
• B. 6%
• C. 9%
• D. 3%

Hint:

To construct chromosome maps, start by placing the two genes with the largest map distance at the outer ends, then place the remaining genes in between by checking additive distances.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Genetic linkage mapping determines the relative positions of genes along a chromosome.
Map distances are measured in centimorgans (cM), where \(1\text{ cM}\) corresponds to a \(1\%\) crossover (recombination) frequency.
For closely linked genes on a linear chromosome, these map distances are additive.

Step 2: Detailed Explanation:
Let us analyze the given map distances to determine the linear order of the four loci (\(a, b, c, d\)):
The distance between \(a\) and \(b\) is \(20\text{ cM}\). Let us place \(a\) at position \(0\) and \(b\) at position \(20\): \[ a(0) \rule[5pt]{40pt}{5pt} b(20) \]
The distance between \(a\) and \(c\) is \(12\text{ cM}\).
The distance between \(b\) and \(c\) is \(8\text{ cM}\).
Since \(12 + 8 = 20\), locus \(c\) must lie between \(a\) and \(b\).
Placing \(c\) at position \(12\): \[ a(0) \rule[5pt]{24pt}{5pt} c(12) \rule[5pt]{16pt}{5pt} b(20) \]
The distance between \(a\) and \(d\) is \(6\text{ cM}\). Locus \(d\) must be located at position \(6\) along this line: \[ a(0) \rule[5pt]{12pt}{5pt} d(6) \rule[5pt]{12pt}{5pt} c(12) \rule[5pt]{16pt}{5pt} b(20) \]
Now, we can find the distance between \(c\) and \(d\): \[ \text{Distance } (c - d) = \text{Position of } c - \text{Position of } d \] \[ \text{Distance } (c - d) = 12\text{ cM} - 6\text{ cM} = 6\text{ cM} \]
A map distance of \(6\text{ cM}\) corresponds to a crossover frequency of \(6\%\).

Step 3: Final Answer:
The crossover frequency between c and d is 6%.