When we look at the a cross involving two different genes we can derive the expected phenotypic distribution for the two genes if we assume independent assortment. The expected joint phenotypic distribution is the product of the phenotypic distribution at each individual locus.

Consider the following four genes involving the mythical dust rhinos (creatures that live under dormitory beds in genetics courses).

Now we need to figure out the expected phenotypic ratios in the event of either (1) a monohybrid cross and/or (2) a test cross.

* Note  for penetrance calculation)

3/4 of dust rhinos are either DD or Dd genotypes. Only 80% or 4/5 of those individuals who have the DD or Dd genotype show the shyness phenotype. That is 3/4 * 4/5 = 12/20 = 3/5.

1/4 of the dust rhinos have the dd genotype and have the boldness phenotype. In addition, 3/20 (3/4 *1/5) of the progeny have the DD or Dd phenotype but the shyness phenotype did not penetrate so they actually show the boldness phenotype. The total progeny showing the boldness phenotype is:

1/4 + 3/20 = 5/20 + 3/20 = 8/20 = 2/5.

This gives us the 3:2 ratio seen in the table above:

You can use this information to construct expected ratios for two or more loci.

Example 1: Consider the following cross:

A homozygous gray, bicorn dust rhino is crossed to a homozygous white, unicorn dust rhino. The F₁ are all gray, bicorn. The F₁ progeny are intercrossed to get an F₂. What are the expected progeny ratios in the F₂?

As previously calculated, we expect to get a ratio of gray to white o r 3:1 (3/4:1/4)  for body color and a ratio of bicorn to unicorn of 3:1 (3/4:1/4) for horn number. For the joint distribution of genotypes we simply multiply the ratios. This assumes independent assortment.

(3 gray:1 white) * (3 bicorn: 1 unicorn) for a total of:

3 * 3 = 9 gray, bicorn
3 * 1 = 3 gray, unicorn
1 * 3 = 3 white, bicorn
1 * 1 = 1 white, unicorn

Example 2: We cross a homozygous gray, shy dust rhino to a homozygous white, bold dust rhino. The F₁ are all gray, shy and gray, bold in a 4:1 ratio. (They have the same genotype, AaDd, but the penetrance of shyness is only 80%). The F₁ progeny are intercrossed to get an F₂. What are the expected progeny ratios in the F₂ (assuming independent assortment)?

We can again multiply the separate expected ratios:

(gray: white = 3/4:1/4) * (shy: bold = 3/5:2/5) for a total of:

3/4 * 3/5 = 9/20 gray, shy
3/4 * 2/5 = 6/20 gray, bold
1/4 * 3/5 = 3/20 white, shy
1/4 * 2/5 = 2/20 white, bold

Additional Problems: (answers available)

Now you try it with the following crosses (using the phenotypes from the table above):

Problem 1:    BbCc x BbCc

Problem 2:    AaDd x aadd

Problem 3:    CcDd x CcDd

Additional  Problems and Discussion

Detection of Linkage

More complex examples

Still one more example

Calculating Expected Ratios From a Cross Involving Two or More Genes.