The test cross for detection of linkage is somewhat different from a  regular test cross.  In the former case, the genotypes of both of the parents are known. The parent with the dominant phenotype is known to be a heterozygote (for example,  AaBb) and the tester parent is known to be completely homozygous recessive (aabb).  The heterozygous parent with the dominant phenotype, if the genes showed independent assortment, should make four gametes AB, Ab,  aB and ab in equal frequencies.  Linkage will be detected if there are more testcross progeny who got the parental gametes from the dominant individual that you would expect by chance alone.

Consider the lowly dust rhino again. In the dust rhino one horn (A) is dominant to two horns (b) and  Wrinkled Knees (C) are  dominant to smooth knees (c).  We generate a known heterozygote from an F₁ cross.

Since the gametes that make up the double heterozygote are AC and ac, these are the parental gametes. The AaCc individual has the potential to make Ac and aC gametes. These would be the recombinant gametes.

The double heterozygote is crossed to the completely recessive individual with genotype aacc

There are more parental gametes than expected (80% vs. 50%)¹.   Therefore, the genes are linked².  The distance or map units is given by the frequency of the recombinant gametes (.10 + .10 = .20).  In this case the genes for horn number and knee wrinkles are 20 map units apart.

¹. To determine whether or not there is an excess of parental gametes it is important to have a sufficient sample size and to do the appropriate statistical tests.

². To distinguish linkage from other problems it might be necessary to carry out a series of statistical tests.

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