Chromosome mapping is the process of determining the relative positions of genes on a chromosome. It provides a genetic framework showing gene order and approximate distances between genes based on how frequently recombination (crossing over) occurs between them. Mapping genes helps in understanding inheritance patterns, locating genes associated with traits or diseases, and forms the basis for modern genomics.
The concept of gene mapping began with the work of Thomas Hunt Morgan and his students (especially Alfred H. Sturtevant) using Drosophila melanogaster. Sturtevant realized that recombination frequencies between genes could be used as a measure of genetic distance, producing the first genetic maps.
Linear arrangement: Genes are arranged linearly along chromosomes. Closely located genes tend to be inherited together (linked).
Linkage: When two or more genes are located on the same chromosome and do not assort independently, they are said to be linked.
Recombination: During meiosis, homologous chromosomes exchange segments (crossing over), producing recombinant chromatids and new allele combinations.
Map units (centimorgans): Recombination frequency (RF) is expressed as a percentage. 1% RF = 1 map unit (1 centimorgan, cM). RF is proportional to the physical distance (for relatively short distances).
The probability that a crossover occurs between two gene loci increases with the physical distance separating them. Thus, by measuring how often recombination separates two linked alleles in gametes or progeny, we can estimate how far apart those genes lie on the chromosome.
Principle: Cross a double heterozygote (A a; B b) with a double homozygous recessive (a a; b b) and score the progeny for parental and recombinant phenotypes.
Procedure:
- Make the test cross:
Aa Bb × aa bb. - Count the number of parental (non-recombinant) and recombinant progeny.
- Calculate recombination frequency:
RF (%) = (Number of recombinant progeny / Total progeny) × 100
Suppose progeny counts are: Parental = 860, Recombinant = 140, Total = 1000.
RF = (140 / 1000) × 100 = 14% → distance = 14 cM
- Gives only distance between two genes, not their order relative to other genes.
- Cannot detect double crossovers involving more than two genes.
- For large distances (>20–30 cM), RF underestimates actual physical distance due to the occurrence of multiple crossovers.
The three-point test cross is more informative: it determines gene order, allows detection of double crossovers, and yields more accurate map distances.
General procedure:
- Cross a female (or male) triple heterozygote with a triple homozygous recessive tester:
A a; B b; C c × a a; b b; c c. - Score the progeny phenotypes and classify them into 8 classes (2^3) and count their frequencies.
- Identify parental (most frequent) classes and the rare double-crossover classes.
- Use the double crossover to determine the gene order; the gene that differs in the double crossover class compared to parental is the middle gene.
- Calculate recombination frequencies between each pair using single and double crossover classes.
Suppose a cross involving three genes A, B and C yields the following progeny counts (phenotypes abbreviated):
| Phenotype | Count |
|---|---|
| Parent 1 (ABC) | 420 |
| Parent 2 (abc) | 400 |
| Recomb (Abc) | 50 |
| Recomb (aBC) | 45 |
| Recomb (ABc) | 30 |
| Recomb (abC) | 25 |
| Double crossover (AbC) | 10 |
| Double crossover (aBc) | 20 |
Step 1 — Identify parental classes: The most frequent classes are ABC (420) and abc (400) → these are parental.
Step 2 — Identify double crossovers (least frequent): AbC (10) and aBc (20). Compare a double crossover phenotype with parental to find the middle gene. For example, parental ABC vs double AbC differ in the second position (B ↔ b) so B is the middle gene.
Step 3 — Calculate distances:
Distance A—B = (single crossovers between A and B + double crossovers) / total
Here, single crossovers between A and B are classes where A and B are recombinant but C is parental: (Recomb (ABc) + Recomb (abC) = 30 + 25 = 55). Add double crossovers (10 + 20 = 30). So total recombinants for A—B = 55 + 30 = 85.
If Total progeny = 1000 (sum of all above), RF (A—B) = (85 / 1000) × 100 = 8.5% → 8.5 cM
Similarly calculate B—C and A—C distances.
Gene order: In three-point mapping, the gene that changes in double crossover progeny relative to parental gives the middle gene.
Interference: When one crossover reduces the probability of another nearby crossover. Interference (I) is measured as:
I = 1 - (observed double crossovers / expected double crossovers)
Where expected double crossovers (if crossovers occur independently) = (RFAB × RFBC × total progeny)
- Genetic map: Based on recombination frequencies (centimorgans).
- Cytogenetic map: Based on banding patterns seen under a microscope (e.g., G-banding).
- Physical map: Based on molecular distances (base pairs) obtained by DNA sequencing and molecular markers.
Molecular markers allow high-resolution mapping. Common markers include:
- RFLP — Restriction Fragment Length Polymorphism
- SSR / Microsatellites — Simple Sequence Repeats
- SNP — Single Nucleotide Polymorphism (most abundant)
- RAPD, AFLP, etc.
Modern approaches: linkage mapping with molecular markers, association mapping (GWAS), and whole-genome sequencing provide precise localization of genes/QTLs.
- Identifying genes responsible for inherited diseases.
- Marker-assisted selection in plant and animal breeding.
- Constructing genetic maps for new species.
- Comparative genomics and evolutionary studies.
- Positional cloning (map-based cloning) to isolate genes.
- Recombination frequency does not always scale linearly with physical distance for distant loci because of multiple crossovers.
- Interference and chromosomal features (heterochromatin, recombination hotspots) affect crossover rates.
- Mapping resolution depends on population size and marker density.
- Chromosome mapping arranges genes in linear order and estimates distances using recombination frequencies.
- Two-point and three-point test crosses are classical genetic tools for mapping.
- Three-point crosses reveal gene order and detect double crossovers — increasing accuracy.
- Molecular markers and sequencing now allow high-resolution physical maps.
Exercises for practice (try solving):
- Perform a two-point mapping calculation: given parental = 720, recombinants = 280, calculate RF and distance.
- Given a three-point test cross data set (create your own counts), identify gene order and compute map distances.
- Explain how interference would affect expected double crossover numbers.
