There are five components that must be known in order to understand a test cross. These five components are either specified by the condition of the test cross, can be directly observed or can be deduced from the available information.

These five components are:

For the following discussion I will turn to the fictional animal the dust rhino.  In the dust rhino one-horn (A) is dominant to two horns (b) and White horns  (B) are dominant to black horns (b).   In this test cross,  we need to determine the genotype of a dust rhino that has one white horn (dominant phenotype).  The dust rhino will be crossed to a complete homozygous recessive dust rhino with phenotype of two black horns.  By definition the tester parent will have the genotype of aabb.

Component 1: The gametes of the homozygous recessive or tester parent:

This is the easiest component.  By definition, the genotype of the tester parent is aabb.  Thus the tester parent can make only one gamete ab with a frequency of 100%.

Component 2: The phenotypes and frequencies of the testcross progeny:

These components can be observed.  The phenotype and the frequency of each phenotype can be directly scored.  From the nature of this type of test cross,  the test cross progeny show either  the dominant phenotype or the recessive phenotype for each gene.  For example, in a test cross of a dust rhino with one white horn to a tester parent that is completely  homozygous recessive  with phenotype of two black horns, we obtain four phenotypes (see below).

Component 3: The genotypes and frequencies of the testcross progeny: Now it is quite simple to deduce the genotypes.  The tester parent can only give a single gamete ab. Therefore, all of the testcross progeny must have at least an a and a b allele in their genotype (shown in red above).  Once we have specified half of the progeny genotype, the rest can be deduced from the phenotypes by two simple rules:

  The deduced genotypes are given in the table above.

Component 4: The gametes produced by  the unknown dominant parent.

Referring to the table above, we have now deduced the genotypes of the test cross progeny.  If we "subtract" the "red gamete (ab) from the tester parent, we are now left with the gametes produced by the unknown parent.  In this case, the unknown parent made four gametes  AB, Ab,  aB and ab in a 1:1:1:1 ratio.

Component 5: The genotype of  the unknown dominant parent.

Since the unknown parent is completely dominant, we know that its "half" genotype is A_ B_.  If this parent produces any gamete with a recessive allele for any trait (as measured by test cross progeny with recessive phenotypes) then it is heterozygous at that locus.  If it never produces a recessive gamete (and we score enough progeny) then it is homozygous at that locus.  In the example above, the unknown  parent produces gametes with both a and b alleles so its genotype must be AaBb.

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