The Expected Genotypic and Phenotypic Arrays of two unlinked genes considered jointly are  the product of the separate Genotypic and Phenotypic Arrays. This is best illustrated by example.

Example: Consider two genes A and B each with two alleles - A_{1  &} A₂: B_{1  &} B₂. The genotypic array  ratios for each gene is :

A gene        1/4A₁A₁:  1/2A₁A₂:  1/4A₂A₂    (3 genotypes)

B gene        1/4B₁B₁:  1/2B₁B₂:  1/4B₂B₂   (3 genotypes)

When the A gene and B gene are considered jointly - there will be 3 x 3 or 9 genotypes.  The genotypes and frequencies can be computed by a Punnett square method or they may be calculated directly by multiplying the genotypic arrays for the two genes.

(1/4A₁A₁:  1/2A₁A₂:  1/4A₂A₂)   x (1/4B₁B₁:  1/2B₁B₂:  1/4B₂B₂) (results in table below)

*Genotypes with the same color have the same phenotype.  See representing Phenotypes.

Two autosomal genes-complete dominance.

In the F₂ generation, the each gene would have two phenotypes in a 3:1 ratio (3/4: /4).  In the absence of linkage,  the two genes would jointly show four phenotypes (2 x 2 = 4) in a 9:3:3:1 ratio (9/16:3/16:3/16:1/16). 

This ratio can be obtained by adding up the appropriate phenotypic frequencies from the table above (color coded).

A₁B₁  = 1/16 + 1/8 + 1/8 + 1/4 = 9/16

A₁B₂ = 1/16 + 1/8 = 3/16

A₂B₁ = 1/16 + 1/8 = 3/16

A₂B₂= 1/16

A faster way to obtain the joint phenotypic array for the A and B genes is to Multiply the separate phenotypic arrays.

We can write the phenotypes for each gene as a phenotypic array.

        For the A gene the phenotypic array is  (3/4A₁+ 1/4A₂).
        For the B gene the phenotypic array is  (3/4B₁+ 1/4B₂).

The joint phenotypic array for the A and B genes is obtained by multiplying the separate phenotypic arrays:

 (3/4A₁+ 1/4A₂) * (3/4B₁+ 1/4B₂) = 9/16A₁B₁ + 3/16A₁B₂ + 3/16A₂B₁ + 1/16A₂B₂

Example 2:  Two autosomal genes - one gene shows complete dominance and the other gene shows either incomplete dominance, overdominance or codominance. That is, the heterozygote has its own phenotype.

For the A gene the phenotypic array is  (3/4A₁+ 1/4A₂).

For the B gene the phenotypic array is  (1/4B₁B₁+ 1/2B₁B₂ + 1/4B₂B₂

  Gene A has two phenotypes and Gene B has three phenotypes.  Jointly they will show 2 * 3 = 6 phenotypes.  The phenotypes and their frequencies can be summed from the Table above (not shown) or calculated from the separate phenotypic arrays.

 (3/4A₁+ 1/4A₂) * (1/4B₁B₁+ 1/2B₁B₂ + 1/4B₂B₂) = 3/16A₁B₁B₁ + 3/8A₁B₁B₂ + 3/16 A₁1/4B₂B₂ +  1/16A₂B₁B₁ +1/8 A₂B₁B₂ +1/4A₂B₂B₂

Example 3: Two autosomal genes - both genes  show either incomplete dominance, overdominance or codominance. That is, the heterozygote has its own phenotype. 

Since each genotype has a corresponding phenotype there are 3 x 3 phenotypes in the same ratio as given in the table  calculated from the separate phenotypic arrays.

For the A gene the phenotypic array is  (1/4A₁A₁+ 1/2A₁A₂ + 1/4A₂A₂)

 For the B gene the phenotypic array is  (1/4B₁B₁+ 1/2B₁B₂ + 1/4B₂B₂)

Gene A has three phenotypes and Gene B has three phenotypes.  Jointly they will show 3 * 3 = 9 phenotypes.  The phenotypes and their frequencies can be summed from the Table above (each genotype as a unique phenotype) or calculated from the separate phenotypic arrays.

 (1/4A₁A₁+ 1/2A₁A₂ + 1/4A₂A₂) *  (1/4B₁B₁+ 1/2B₁B₂ + 1/4B₂B₂)  =

    1/16A₁A₁B₁B₁+1/8A₁A₁B₁B₂+1/16A₁A₁B₂B₂+1/8A₁A₂B₁B₁+ 1/4A₁A₂B₁B₂ + 1/8A₁A₂B₂B₂+1/16A₂A₂B₁B₁+ 1/8A₂A₂B₁B₂ + 1/16A₂A₂B₂B₂