Question

A backcross was set up between two homozygous laboratory mouse strains \(A\) and \(B,\) with the \(F_{1}\) backcrossed to \(B\). The \(F_{2}\) were typed using SNPs \(x\) and \(y,\) which varied between strains \(A\) and \(B\left(x^{A}, x^{B}, y^{A}, y^{B}\right) .\) Out of 100 mice, 38 were \(x^{A} y^{A}, 40\) were \(x^{B} y^{B}\) 11 were \(x^{A} y^{B},\) and 11 were \(x^{B} y^{A} .\) What is the genetic distance between SNPs \(x\) and \(y ?\)

Short answer

Answer: The genetic distance between SNPs x and y is 22 centiMorgans.

Step-by-step solution

Identify Double Cross-Over Offspring and Total Offspring

In the problem, we are given the number of mice with each genotype: - 38 \(x^{A} y^{A}\) mice - 40 \(x^{B} y^{B}\) mice - 11 \(x^{A} y^{B}\) mice - 11 \(x^{B} y^{A}\) mice Double cross-over offspring are the ones with \(x^{A} y^{B}\) and \(x^{B} y^{A}\) genotypes, which amounts to a total of 22 mice (11 + 11). The total number of offspring in the experiment is 100.

Calculate Recombination Frequency

We will use the recombination frequency formula to calculate the genetic distance between SNPs x and y: $$Recombination\ frequency = \frac{Number\ of\ double\ cross - over\ offspring}{Total\ number\ of\ offspring}$$ Substitute the numbers we found in step 1: $$Recombination\ frequency = \frac{22}{100}$$ Simplify: $$Recombination\ frequency = 0.22$$

Calculate Genetic Distance

Genetic distance, which is measured in centiMorgans (cM), is calculated by multiplying the recombination frequency by 100: $$Genetic\ distance\ (cM) = Recombination\ frequency \times 100$$ Substitute the recombination frequency that we calculated in step 2: $$Genetic\ distance\ (cM) = 0.22 \times 100$$ Simplify: $$Genetic\ distance\ (cM) = 22$$ The genetic distance between SNPs x and y is 22 centiMorgans.