Three Point Mapping
Embedded Files
The webpage is under continuous improvement. We welcome your feedback and suggestions.
Now that we’ve covered the basics of a two-point cross, let’s explore a more advanced technique called a three-point cross. This method allows us to map the positions of three genes on a chromosome simultaneously." Three-Point Cross: It's a genetic analysis method involving the simultaneous study of the recombination frequencies between three genes. This helps in mapping their relative positions on a chromosome in relation to one another. In a three-point cross, instead of focusing on just two genes, we consider three genes at a time. By analyzing how frequently crossovers occur between these three genes during meiosis, we can deduce their order and distance from one another on the chromosome."
Rules for a Three-Point Cross
Rules for a Three-Point Cross
The rules for conducting a three-point cross involve a series of steps and principles that guide the process of determining gene order and map distances between linked genes. Here are the key rules and steps involved:
Use Heterozygous and Homozygous Parents:
Cross a heterozygous organism for all three genes with a homozygous recessive organism. This setup ensures that all alleles can be tracked through offspring phenotypes.
Identify Phenotypes:
Offspring phenotypes will be combinations of parental and recombinant types. Record the number of offspring for each phenotype.
Classify Offspring:
Parental Types: The two most frequent phenotypes.
Single Crossover Types: Intermediate frequency phenotypes, indicating recombination between two genes.
Double Crossover Types: The two least frequent phenotypes, indicating recombination events that affect all three genes.
4. Determine Gene Order:
Compare parental and double crossover phenotypes to identify the middle gene. The gene that appears to change position in double crossovers compared to the parental types is the middle gene.
5. Calculate Recombination Frequencies:
Use the number of recombinants to calculate the recombination frequency between each pair of genes.
6. Compute Map Distances:
Convert recombination frequencies into map distances (centiMorgans, cM) using the formula:
Map Distance (cM)= (Number of Recombinant/ Total Number of Offspring) x 100
Correct for Double Crossovers:
7. Double crossovers can lead to underestimation of distances. Correct by adding double crossover counts to each single crossover calculation.
Detailed Explanation of the Rules
Detailed Explanation of the Rules
Cross Heterozygous with Homozygous Recessive
Example: AaBbCc×aabbccAaBbCc \times aabbccAaBbCc×aabbcc
Heterozygous Parent (AaBbCc): Contains dominant and recessive alleles.
Homozygous Recessive Parent (aabbcc): Provides a clear baseline for identifying recombinants.
2. Identify and Record Offspring Phenotypes
Phenotypes: Observe and count each unique combination of alleles in offspring.
Example Offspring: See Table below
Classify Offspring into Types
Parental Types: The two most frequent phenotypes.
Example: AaBbCc (400) and aabbcc (380)
Single Crossover Types: Phenotypes with one crossover between two genes.
Example: AaBbcc, aabbCc, AabbCc, aaBbcc
Double Crossover Types: Least frequent phenotypes with two crossover events.
Example: Aabbcc, aaBbCc
4. Determine Gene Order
Identify Middle Gene:
Compare the parental types with double crossover phenotypes.
The gene that differs in double crossovers is the middle gene.
Explanation: The middle gene can be determined by comparing the parental and double crossover genotype combinations. The gene that differs in the double crossovers compared to the parental types is the middle gene.
Example:
Parental: AaBbCc (400), aabbcc (380)
Double Crossover: Aabbcc (10), aaBbCc (10)
Middle Gene is B:
Reasoning: In double crossovers, B changes position, indicating it's in the middle.
Gene Order: A - B - C
5. Calculate Recombination Frequencies
Formula: Map Distance (cM)= (Number of Recombinant/ Total Number of Offspring) x 100
Example Calculation:
A to B:
Recombinants: AaBbcc (60) + aabbCc (55) + Aabbcc (10) + aaBbCc (10)
Total: 60+55+10+10=135
RF: 135/1000 x 100= 13.5%
B to C:
Recombinants: AabbCc (40) + aaBbcc (45) + Aabbcc (10) + aaBbCc (10)
Total: 40+45+10+10=105
RF: 105/ 1000 x 100= 10.5%
Correct for Double Crossovers
Double Crossover Impact:
Double crossovers can lead to underestimation of distances.
Correct by adding double crossover counts back into the calculation.
Correction Example:
A to C Calculation:
Total Recombinants: 220
Double Crossovers: Aabbcc (10), aaBbCc (10)
Correct A-C Distance= 220 + 20 / 1000 x 100 = 24. cM
Click Here: Previous Page
Click Here: Recombination Frequency
Page updated
Google Sites
Report abuse