Log out Go to App
Go to App

Learning Materials

Features

Discover

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

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 \(y ?\)

Review the Chapter Concepts list on page \(136 .\) Most of these center on 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.

Are sister chromatid exchanges effective in producing genetic variability in an individual? in the offspring of individuals?

Colored aleurone in the kernels of corn is due to the dominant allele \(R\). The recessive allele \(r,\) when homozygous, produces colorless aleurone. The plant color (not kernel color) is controlled by another gene with two alleles, \(Y\) and \(y\). The dominant \(Y\) allele results in green color, whereas the homozygous presence of the recessive \(y\) allele causes the plant to appear yellow. In a testross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained: colored, green 88 colored, yellow 12 colorless, green 8 colorless, yellow 92 Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the precise association of the two genes on the homologs (i.e., the arrangement).
See all solutions

Recommended explanations on Biology Textbooks

Reproduction

Read Explanation

Biological Processes

Read Explanation

Biological Organisms

Read Explanation

Substance Exchange

Read Explanation

Ecology

Read Explanation

Cell Communication

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free