Textbook Question
Why are double-crossover events expected less frequently than single-crossover events?
1465
views
1
rank
Ch. 5 - Chromosome Mapping in Eukaryotes
Klug12th EditionConcepts of Genetics ISBN: 9780135564776
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Introduction to Genetics17
- Ch. 2 - Mitosis and Meiosis36
- Ch. 3 - Mendelian Genetics51
- Ch. 4 - Extensions of Mendelian Genetics74
- Ch. 5 - Chromosome Mapping in Eukaryotes51
- Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages46
- Ch. 7 - Sex Determination and Sex Chromosomes33
- Ch. 8 - Chromosome Mutations: Variation in Number and Arrangement40
- Ch. 9 - Extranuclear Inheritance31
- Ch. 10 - DNA Structure and Analysis43
- Ch. 11 - DNA Replication and Recombination44
- Ch. 12 - DNA Organization in Chromosomes30
- Ch. 13 - The Genetic Code and Transcription54
- Ch. 14 - Translation and Proteins47
- Ch. 15 - Gene Mutation, DNA Repair, and Transposition41
- Ch. 16 - Regulation of Gene Expression in Bacteria29
- Ch. 17 - Transcriptional Regulation in Eukaryotes41
- Ch. 18 - Post-transcriptional Regulation in Eukaryotes33
- Ch. 19 - Epigenetics33
- Ch. 20 - Recombinant DNA Technology44
- Ch. 21 - Genomic Analysis36
- Ch. 22 - Applications of Genetic Engineering and Biotechnology36
- Ch. 23 - Developmental Genetics37
- Ch. 24 - Cancer Genetics34
- Ch. 25 - Quantitative Genetics and Multifactorial Traits54
- Ch. 26 - Population and Evolutionary Genetics45
Chapter 5, Problem 9
The genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes?
dp–ap: 42
dp–cl: 3
ap–cl: 39
Verified step by step guidance1
Identify the three genes involved: dumpy (dp), clot (cl), and apterous (ap), and note the given pairwise genetic distances: dp–ap = 42 map units, dp–cl = 3 map units, and ap–cl = 39 map units.
Recall that genetic distances represent the recombination frequency between two genes and that genes that are closer together have smaller distances, indicating tighter linkage.
Start by placing the two genes with the smallest distance on the chromosome, which are dp and cl with 3 map units apart. This suggests dp and cl are very close neighbors.
Next, consider the distances involving ap: dp–ap is 42 and ap–cl is 39. Since dp and cl are close, and ap is far from both, ap must be located on the chromosome such that the distances to dp and cl match the given values.
Arrange the genes in a linear order and verify that the sum of the smaller distances equals the largest distance, i.e., check if the distance between dp and cl plus the distance between cl and ap equals the distance between dp and ap, or if dp is between cl and ap, to determine the correct gene sequence.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
51sWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Genetic Linkage and Gene Mapping
Genetic linkage refers to genes located close together on the same chromosome that tend to be inherited together. Gene mapping uses recombination frequencies between linked genes to estimate their relative positions on a chromosome, measured in map units or centimorgans (cM). Lower recombination frequencies indicate closer proximity.
Recommended video:
Guided course
09:09
Mapping Genes
Two-Point Mapping
Two-point mapping involves crossing organisms to measure recombination frequency between two genes at a time. The percentage of recombinant offspring reflects the distance between those genes. By comparing distances from multiple two-point crosses, the relative order of genes can be inferred.
Recommended video:
Guided course
13:26QTL Mapping
Determining Gene Order from Recombination Distances
To determine gene order, compare pairwise distances: the smallest distance indicates the closest genes, and the largest distances help place the third gene relative to the others. The sum of two smaller distances should approximate the largest distance if genes are in a linear sequence.
Recommended video:
Guided course
03:37Recombination after Double Strand Breaks
Related Practice
Textbook Question
What is the proposed basis for positive interference?
1051
views
Textbook Question
What two essential criteria must be met in order to execute a successful mapping cross?
681
views
Textbook Question
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 the 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 testcross 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 arrangement of the alleles on the homologs.
1306
views
Textbook Question
In the cross shown here, involving two linked genes, ebony (e) and claret (ca), in Drosophila, where crossing over does not occur in males, offspring were produced in a 2 + : 1 ca : 1 e phenotypic ratio:
These genes are 30 units apart on chromosome III. What did crossing over in the female contribute to these phenotypes?
1054
views
Textbook Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed:
Given that the adp gene is near the end of chromosome II (locus 83), construct a map of these genes.
1407
views