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Chapter 7: Problem 18

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

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

Step by step solution

01

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.
02

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: $$Recombination\ frequency = 0.22$$
03

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.

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Most popular questions from this chapter

Explain the meaning of the term interference.

In a series of two-point map crosses involving five genes located on chromosome II in Drosophila, the following recombinant \((\sin -\) gle- crossover) frequencies were observed:$$\begin{array}{lr} p r-a d p & 29 \\ p r-v g & 13 \\ p r-c & 21 \\ p r-b & 6 \\ a d p-b & 35 \\ a d p-c & 8 \\ a d p-v g & 16 \\ v g-b & 19 \\ v g-c & 8 \\ c-b & 27 \end{array}$$ (a) If the adp gene is present near the end of chromosome II (locus 83 ), construct a map of these genes. (b) In another set of experiments, a sixth gene ( \(d\) ) was tested against \(b\) and \(p r,\) and the results were \(d-b=17 \%\) and \(d-p r=23 \% .\) Predict the results of two-point maps between \(d\) and \(c, d\) and \(v g,\) and \(d\) and \(a d p\)

In a plant, fruit color is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the results shown in the following table. Determine the location of the genes relative to one another and the genotypes of the two parental plants. $$\begin{array}{lcc} & & \text { Progeny } \\ \text { Phenotype } & \text { Plant A } & \text { Plant B } \\ \text { red, long } & 46 & 4 \\ \text { yellow, oval } & 44 & 6 \\ \text { red, oval } & 5 & 43 \\ \text { yellow, long } & 5 & 47 \\ \text { Total } & 100 & 100 \end{array}$$

Phenotypically wild \(\mathrm{F}_{1}\) female Drosophila, whose mothers had light eyes (It) and fathers had straw (stw) bristles, produced the following offspring when crossed with homozygous light-straw males:$$\begin{array}{lc} \text { Phenotype } & \text { Number } \\ \hline \text { light-straw } & 22 \\ \text { wild } & 18 \\ \text { light } & 990 \\ \text { straw } & \frac{970}{2000} \end{array}$$ Compute the map distance between the light and straw loci.

Why are double-crossover events expected less frequently than single-crossover events?
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