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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 \(\operatorname{SNPs} x\) 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

Determine parental and recombinant genotypes

First, let's identify the parental and recombinant genotypes in the given data: Parental genotypes: - 38 mice with \(x^{A}y^{A}\) - 40 mice with \(x^{B}y^{B}\) Recombinant genotypes: - 11 mice with \(x^{A}y^{B}\) - 11 mice with \(x^{B}y^{A}\) In total, there are 100 mice in the F2 generation.
02

Calculate the recombination frequency

The recombination frequency is the proportion of recombinant genotypes among all genotypes. To find this, we will divide the total number of recombinant offspring (mice with \(x^{A}y^{B}\) and \(x^{B}y^{A}\)) by the total number of offspring (100 mice). Recombination frequency = \(\frac{11 + 11}{100} = \frac{22}{100} = 0.22\)
03

Convert the recombination frequency to genetic distance in centimorgans (cM)

To convert the recombination frequency to the genetic distance in centimorgans (cM), multiply the recombination frequency by 100. Genetic distance = \(0.22 \times 100 = 22\) cM The genetic distance between SNPs \(x\) and \(y\) is 22 centimorgans.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recombination Frequency
Understanding the recombination frequency is essential for genetic mapping and discovering how genes and genetic markers are inherited. It represents the likelihood of a crossover occurring between two genetic loci during meiosis, the process that leads to the production of gametes such as sperm or eggs. When two genetic markers, such as SNPs, are more frequently inherited separately rather than together, it indicates that they are more likely to be separated during recombination due to their physical distance on a chromosome.

In our exercise, we calculate the recombination frequency by taking the sum of the offspring that have recombinant genotypes (i.e., the genotypes that are different from the parent strains) and dividing this number by the total number of offspring. This frequency tells us how often recombination occurs between two points, in this case, SNPs, on a chromosome. The formula used is:
\( \text{Recombination frequency} = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \).

It's important to make this concept relatable by considering it like a measure of how 'close' or 'far' two markers are on the DNA strand—closer markers recombine less frequently and thus have a lower recombination frequency.
Centimorgans
The unit centimorgan (cM) is named after the American geneticist Thomas Hunt Morgan, who made significant contributions to the field of genetics. It is used to measure genetic distance and one centimorgan is equivalent to a 1% chance that a marker at one genetic locus will be separated from a marker at another locus due to recombination in a single generation.

In the context of our exercise, to convert the calculated recombination frequency into centimorgans, we simply multiply the frequency by 100. Therefore, a recombination frequency of 0.22 corresponds to 22 cM, indicating that there is a 22% chance that the two SNPs will be separated during the formation of gametes. This measurement provides a more tangible representation of genetic distance.

When explaining centimorgans, it's helpful to visualize a chromosome as a roadmap, with centimorgans as the 'miles' measuring the distance between various 'locations' or genes. This can make the concept of genetic linkage and distance more approachable for students.
SNPs
Single Nucleotide Polymorphisms, commonly known as SNPs (pronounced 'snips'), are the most common type of genetic variation among individuals. A SNP represents a difference in a single nucleotide—A, T, C, or G—in the DNA sequence. For instance, a SNP may replace the nucleotide cytosine (C) with thymine (T) at a specific location in the genome. Although most SNPs do not have any observable effect on an individual’s health or physical characteristics, some can be very informative in linking certain traits or diseases with specific genetic variations.

In the given exercise, SNPs 'x' and 'y' are used as genetic markers to explore inheritance patterns in mice. The difference in the SNPs between strains A and B allow researchers to track which strain the offspring have inherited particular genetic material from. Understanding how frequently recombination occurs between these SNPs helps in mapping their position on the chromosome and can be vital for genetic research, such as determining the likelihood of inheriting a genetic disorder.

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Most popular questions from this chapter

An organism of the genotype \(A a B b C c\) was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are presented in the following table. (a) If these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring, and in what proportion, assuming simple dominance and recessiveness in each gene pair? (b) Answer part (a) again, assuming the three genes are so tightly linked on a single chromosome that no crossover gametes were recovered in the sample of offspring. (c) What can you conclude from the actual data about the location of the three genes in relation to one another?

The gene controlling the Xg blood group alleles \(\left(X g^{+} \text {and } X g^{-}\right)\) and the gene controlling a newly described form of inherited recessive muscle weakness called episodic muscle weakness \((E M W X)\) (Ryan et al., 1999 ) are closely linked on the X chromosome in humans at position \(\mathrm{Xp} 22.3\) (the tip of the short arm \() .\) A male with EMWX who is \(\mathrm{Xg}^{-}\) marries a woman who is \(\mathrm{Xg}^{+}\), and they have eight daughters and one son, all of whom are normal for muscle function, the male being \(\mathrm{Xg}^{+}\) and all the daughters being heterozygous at both the \(E M W X\) and \(X g\) loci. Following is a table that lists three of the daughters with the phenotypes of their husbands and children. (a) Create a pedigree that represents all data stated above and in the following table. (b) For each of the offspring, indicate whether or not a crossover was required to produce the phenotypes that are given.

Review the Chapter Concepts list on page \(138 .\) Most of these center around the process of crossing over between linked genes. Write a short essay that discusses how crossing over can be detected and how the resultant data provide the basis of chromosome mapping.

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

In Drosophila, a female fly is heterozygous for three mutations, Bar eyes \((B),\) miniature wings \((m),\) and ebony body \((e)\) Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows. Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.

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