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Chapter 7: Problem 11

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.

Short Answer

Expert verified
Answer: The approximate map distance between the light and straw loci in Drosophila is 48 centimorgans (cM).

Step by step solution

01

Identify the parental phenotypes

The F1 females are phenotypically wild and have mothers with light eyes (It) and fathers with straw (stw) bristles, so their genotype can be represented as It+ st+ / It stw. The homozygous light-straw males have a genotype of It stw / It stw. The phenotypes of the offspring resulting from this cross will help us calculate the recombination frequency.
02

Identify recombinant offspring

The recombinant offspring are those that have a phenotype different from the parental phenotypes (it's the result of a crossover event). In this case, light and straw are the recombinant phenotypes with frequencies of 990 and 970/2000 respectively.
03

Calculate the recombination frequency

The recombination frequency can be calculated as the ratio of the sum of the numbers of recombinant offspring to the total number of offspring. In this case, the recombination frequency is: Recombination frequency = (number of light offspring + number of straw offspring) / total number of offspring Recombination frequency = (990 + (970/2000)) / (22 + 18 + 990 + (970/2000))
04

Compute the map distance

The map distance in centimorgans (cM) is found by multiplying the recombination frequency by 100. So, Map distance = recombination frequency * 100 First, let's calculate the recombination frequency: Recombination frequency = (990 + (970/2000)) / (22 + 18 + 990 + (970/2000)) Recombination frequency ≈ 0.48 Now we can compute the map distance: Map distance = 0.48 * 100 Map distance ≈ 48 cM The map distance between the light and straw loci in Drosophila is approximately 48 centimorgans (cM).

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Most popular questions from this chapter

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 \(y ?\)

Why is a 50 percent recovery of single-crossover products the upper limit, even when crossing over always occurs between two linked genes?

In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type \(\mathrm{F}_{1}\) female progeny were mated to fully homozygous (mutant) males, and the following progeny ( 1000 total) were observed: $$\begin{array}{lc} \text { Phenotypes } & \text { Number Observed } \\ \hline \text { spineless } & 321 \\ \text { wild } & 38 \\ \text { claret, spineless } & 130 \\ \text { claret } & 18 \\ \text { claret, hairless } & 309 \\ \text { hairless, claret, spineless } & 32 \\ \text { hairless } & 140 \\ \text { hairless, spineless } & 12 \end{array}$$ (a) Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild \(F_{1}\) heterozygote? (c) What are the map distances between the three genes? A correct formula with the values "plugged in" for each distance will be sufficient. (d) What is the coefficient of coincidence? A correct formula with the values "plugged in" will be sufficient.

Why does more crossing over occur between two distantly linked genes than between two genes that are very close together on the same chromosome?

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. $$\begin{array}{lcc} & & \text { Progeny } \\ \text { Phenotype } & \text { Plant A } & \text { Plant B } \\ \text { red, long } & 46 & 4 \\ \text { yellow, oval } & 44 & 6 \\ \text { red, oval } & 5 & 43 \\ \text { yellow, long } & 5 & 47 \\ \text { Total } & 100 & 100 \end{array}$$
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