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 coupling.

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:

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 = [(1/2 -1/2r)AB + 1/2rAb + 1/2raB + (1/2 -1/2r)ab] * [(1/2 -1/2r)AB + 1/2rAb + 1/2raB + (1/2 -1/2r)ab]

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.

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.

• If there is complete recombination (no statistical linkage)  then the joint genotypic array  looks exactly like independent assortment. 
   
• If there is no recombination (complete linkage) then the two genes are essentially inherited as if they were a single gene. 
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• If there is some recombination (incomplete linkage) then all genotypes are seen but there is significant deviation from the genotypic array seen in independent assortment.

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    (1/4-1/2r+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 + 1/4r²  = 1/2r - 1/4r²

aB    1/2r - 1/2r² + 1/4r²  = 1/2r - 1/4r²

ab     1/4-1/2r+1/4r²