Skip to main content
Pearson+ LogoPearson+ Logo
Ch. 5 - Genetic Linkage and Mapping in Eukaryotes
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
Chapter 5, Problem 30b

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.
Table displaying phenotypes and counts from a Drosophila genetic linkage experiment, including wild-type and recessive traits.
Calculate the recombination frequencies between each of the gene pairs.

Verified step by step guidance
1
Step 1: Understand the problem and the data. This problem involves calculating recombination frequencies between three X-linked genes in Drosophila: echinus (eye mutant), crossveinless (wing-vein mutation), and scute (bristle mutation). The phenotypes and their corresponding numbers are provided. Recombination frequency is calculated as the number of recombinant offspring divided by the total number of offspring, multiplied by 100 to express it as a percentage.
Step 2: Identify parental and recombinant phenotypes. Parental phenotypes are the most frequent phenotypes in the progeny, while recombinant phenotypes are less frequent. From the data, the parental phenotypes are 'Echinus' (8576) and 'Scute, Crossveinless' (8808). The remaining phenotypes are recombinant types.
Step 3: Group the recombinant phenotypes based on single and double crossovers. Single crossovers involve recombination between two genes, while double crossovers involve recombination between all three genes. For example, 'Scute' (977) and 'Echinus, Crossveinless' (1002) represent single crossovers between specific gene pairs, while 'Scute, Echinus' (681) and 'Wild type' (1) represent double crossovers.
Step 4: Calculate the recombination frequency for each gene pair. To calculate the recombination frequency between two genes, sum the number of progeny representing recombinants for that gene pair (including single and double crossovers), divide by the total number of progeny (20,765), and multiply by 100. For example, to calculate the recombination frequency between echinus and scute, sum the progeny numbers for 'Scute' (977), 'Echinus, Crossveinless' (1002), 'Scute, Echinus' (681), and 'Wild type' (1), then divide by 20,765 and multiply by 100.
Step 5: Repeat the calculation for the other gene pairs. Perform the same process for the other two gene pairs: echinus and crossveinless, and scute and crossveinless. For each pair, sum the relevant recombinant progeny numbers, divide by the total progeny, and multiply by 100 to determine the recombination frequency. This will give you the genetic distances between the genes in map units (centimorgans).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Genetic Linkage

Genetic linkage refers to the tendency of genes located close to each other on the same chromosome to be inherited together during meiosis. This phenomenon occurs because linked genes are less likely to be separated by recombination events. Understanding genetic linkage is crucial for analyzing inheritance patterns and calculating recombination frequencies, which help determine the relative positions of genes on a chromosome.
Recommended video:
Guided course
07:33
Chi Square and Linkage

Recombination Frequency

Recombination frequency is a measure of the likelihood that two genes will be separated during meiosis due to crossing over. It is calculated by dividing the number of recombinant offspring by the total number of offspring and multiplying by 100 to express it as a percentage. This frequency provides insights into the genetic distance between genes, with lower frequencies indicating closer linkage.
Recommended video:
Guided course
03:51
Recombination after Single Strand Breaks

Phenotypic Ratios

Phenotypic ratios represent the relative frequencies of different phenotypes observed in the offspring of a genetic cross. In this experiment, the phenotypic ratios of the progeny can be used to infer the genetic relationships between the traits studied. By analyzing these ratios, one can identify the presence of recombination events and calculate the recombination frequencies between the linked genes.
Recommended video:
Guided course
10:48
Mutations and Phenotypes
Related Practice
Textbook Question

A 2006 genetic study of a large American family (Ikeda et al., 2006) identified genetic linkage between DNA markers on chromosome 11 and the gene producing the autosomal dominant neuromuscular disorder spinocerebellar ataxia type 5 (SCA5). The following lod score data are taken from the 2006 study:                  

Based on the available information, is DNA marker A linked to the gene producing SCA5? Explain your answer.

439
views
Textbook Question

A 2006 genetic study of a large American family (Ikeda et al., 2006) identified genetic linkage between DNA markers on chromosome 11 and the gene producing the autosomal dominant neuromuscular disorder spinocerebellar ataxia type 5 (SCA5). The following lod score data are taken from the 2006 study:

Based on available information, is DNA marker B linked to the gene for SCA5? Explain your answer.

592
views
Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Determine the gene order and identify the alleles on the homologous X chromosomes in the trihybrid females.

497
views
Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Compare the recombination frequencies and speculate about the source of any apparent discrepancies in the recombination data.

424
views
Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Use chi-square analysis to demonstrate that the data in this experiment are not the result of independent assortment.

499
views
Textbook Question

A genetic study of an early onset form of heart disease identifies 10 families containing members with the condition. No clear dominant or recessive pattern of inheritance is evident, but an analysis of SNP markers for five families detects a strong association with a marker on chromosome 12, and genetic linkage analysis for the marker produces a lod score of 2.2.


What do the association and lod score results suggest about this genetic marker?

530
views