Expected Results if Linkage Occurs - Repulsion |
If two genes are linked, the gametic and phenotypic arrays can no longer be obtained by multiplying the individual arrays for each gene. However, we can still develop mathematical tools to do this.
Consider a cross between two homozygous individuals. For simplicity at the A locus there are two alleles A and a and at the B locus there are two alleles B and b. These genes are on the same chromosome and recombination occurs with a frequency of r.
For this example, assume that the P1 parent has genotype aaBB and the P2 parent has genotype aaBB. The F1 progeny have genotype AaBb (often written as aB/Ab) This is a situation that is often called repulsion.
The F1 progeny produce four gametes AB, Ab, aB and ab but they will not be in equal frequencies. The parental gametes (aB, aB) will be the most frequent and the recombinant gametes will be the least frequent (AB, ab). With a recombination rate of r, the gametic frequencies will be:
Gamete | Freq. |
AB | 1/2r |
aB | 1/2 -1/2r |
aB | 1/2 -1/2r |
ab | 1/2r |
ab |
The F2 generation is generated by the cross F1 x F1.
For autosomal traits, all four F2 generations give the same genotypic arrays for both males and females. The genotypic array in the F2 is obtained by multiplying the gametic array in females by the gametic array in males.
Genotypic array = [1/2rAB + (1/2 -1/2r)Ab + (1/2 -1/2r)aB + 1/2rab] * [1/2rAB + (1/2 -1/2r)Ab + (1/2 -1/2r)aB + 1/2rab
For two linked genes with a recombination frequency of r the genotypes in each generation for independent assortment, complete linkage, and one possible incomplete linkage are given below.
Genotype |
Freq. F2 |
Independent Assort r= .5 |
Complete Linkage r = 0 |
Incomplete linkage r = .25 (1/4) |
AABB | 1/4r2 | 1/16 | 0 | 1/64 |
AABb | 1/2r - 1/2r2 | 1/8 | 0 | 6/64 |
AAbb | 1/4-1/2r+1/4r2 | 1/16 | 1/4 | 9/64 |
AaBB | 1/2r - 1/2r2 | 1/8 | 0 | 6/64 |
AaBb | 1/2-r+r2 | 1/4 | 1/2 | 20/64 |
Aabb | 1/2r - 1/2r2 | 1/8 | 0 | 6/64 |
aaBB | 1/4-1/2r+1/4r2 | 1/16 | 1/4 | 9/64 |
aaBb | 1/2r - 1/2r2 | 1/8 | 0 | 6/64 |
aabb | 1/4r2 | 1/16 | 0 | 1/64 |
The recombination rate can vary from .5 (independent assortment) to 0 (non recombination - complete linkage).
Values of r less than .5 but greater than 0 occur for genes that are incompletely linked.
The phenotypic array for a two completely dominant genes can be obtained by adding up the appropriate genotypes (color coded). If the recombination rate (r) is known, the expected phenotypic ratios can be calculated directly.
AB 1/4r2 + (1/2-r+r2) + (1/2-r+r2) + (1/2-r+r2) = 3/4 -1/2r + 1/4 r2
Ab (1/2r - 1/2r2) + (1/4-1/2r+1/4r2) = 1/4 - 1/4r2
aB (1/2r - 1/2r2) + (1/4-1/2r+1/4r2) = 1/4 - 1/4r2
ab 1/4r2
Quick Links F2 Cross |
|
1 | Basic F2 cross - autosomal genes |
2 | Basic F2 cross- X-linked gene |
Expected Phenotypes of F2 Cross - Single Autosomal Traits | |
Expected Phenotypes of F2 Cross - X-linked gene | |
4 | F2 cross - two unlinked autosomal genes |
5 | F2 cross - one autosomal - one X-linked gene |
6 | Linkage - Coupling and Repulsion |
7 | F2 cross - two linked autosomal genes -coupling |
8 | F2 cross - two linked autosomal genes - repulsion |
9 | Measuring the recombination rate from F2 data |
Recombination from F2 data - Table of Values | |
10 | Basic F2 cross- Z-linked gene (inactive) |
12 | Expected Phenotypes of F2 Cross - Z-linked gene (inactive) |
13 | F2 cross - one autosomal - one Z-linked gene (inactive) |
Also See |
|
Detection of Linkage - Basic Info - Two Autosomal Traits - Complete Dominance | |
Detection of Linkage - More complex examples | |
Detection of Linkage - One More Example | |
Detection of Linkage - Two Autosomal Traits - Heterozygote has Unique Phenotype | |
Home page | |
Detailed List for Mendelian Genetics |