How was it established experimentally that the frequency of recombination (crossing over) between two genes is related to the distance between them along the chromosome?
Table of contents
- 1. Introduction to Genetics51m
- 2. Mendel's Laws of Inheritance3h 37m
- 3. Extensions to Mendelian Inheritance2h 41m
- 4. Genetic Mapping and Linkage2h 28m
- 5. Genetics of Bacteria and Viruses1h 21m
- 6. Chromosomal Variation1h 48m
- 7. DNA and Chromosome Structure56m
- 8. DNA Replication1h 10m
- 9. Mitosis and Meiosis1h 34m
- 10. Transcription1h 0m
- 11. Translation58m
- 12. Gene Regulation in Prokaryotes1h 19m
- 13. Gene Regulation in Eukaryotes44m
- 14. Genetic Control of Development44m
- 15. Genomes and Genomics1h 50m
- 16. Transposable Elements47m
- 17. Mutation, Repair, and Recombination1h 6m
- 18. Molecular Genetic Tools19m
- 19. Cancer Genetics29m
- 20. Quantitative Genetics1h 26m
- 21. Population Genetics50m
- 22. Evolutionary Genetics29m
4. Genetic Mapping and Linkage
Mapping Overview
Problem 9
Textbook Question
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

1
Step 1: Understand that the problem involves determining the sequence of three linked genes (dumpy, clot, and apterous) based on genetic distances provided from two-point mapping crosses.
Step 2: Recognize that genetic distances are measured in map units (centimorgans, cM), which represent the frequency of recombination between genes. Smaller distances indicate closer proximity between genes.
Step 3: Analyze the genetic distances provided: dp–ap = 42 cM, dp–cl = 3 cM, and ap–cl = 39 cM. These distances suggest that dp and cl are very close, while ap is farther away from both dp and cl.
Step 4: Use the distances to infer the sequence of the genes. Since dp–cl is the smallest distance (3 cM), dp and cl must be adjacent. The larger distances (dp–ap = 42 cM and ap–cl = 39 cM) indicate that ap is farther away, placing it on the opposite side of cl relative to dp.
Step 5: Conclude that the sequence of the genes is dp–cl–ap, based on the genetic distances provided and their relative positions on the chromosome.

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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 a chromosome to be inherited together during meiosis. This phenomenon occurs because linked genes are less likely to be separated by recombination events. Understanding linkage is crucial for determining the relative positions of genes on a chromosome.
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Two-Point Mapping
Two-point mapping is a method used to determine the distance between two genes on a chromosome based on the frequency of recombination between them. The genetic distance is expressed in centimorgans (cM), where 1 cM corresponds to a 1% chance of recombination occurring. This technique helps in constructing genetic maps and understanding gene order.
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Recombination Frequency
Recombination frequency is the proportion of offspring that exhibit a recombinant phenotype compared to the total number of offspring. It is used to calculate genetic distances between linked genes. A lower recombination frequency indicates that genes are closer together, while a higher frequency suggests they are further apart, aiding in the determination of gene sequences.
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