The principle of the test cross is the same whether we have 1, 2 or more genes - you are crossing an individual of a dominant phenotype but unknown genotype to an individual that is homozygous recessive for all relevant genes.  Since the tester individual makes  one gamete, the ratios of phenotypes among the progeny of this cross indicate the type and frequencies of gametes made by the individual with the unknown genotype.  Once we know the gametes, we can "reconstruct" the genotype.

Consider again  the fruit fly Drosophila melanogaster.   Black body (b) is recessive to the normal gray body (B).   Pink eye (p) is recessive to the normal red eye (P).  You are given a male with a gray body and red eyes.  Determine its genotype.

There are now four possible genotypes that are associated with the dominant phenotype Gray body, Red eyes.  These four genotypes can produce, 1, 2, 2, and 4 different gametes, respectively.  In combination with the single gamete from the tester parent the crosses will  produce 1, 2, 2 or 4 progeny phenotypes.

For example, we carry out the test cross and obtain 400 progeny.  These progeny are sorted into phenotypes and we discover that we have 200 flies with gray body and red eyes, and 200 progeny with gray bodies and pink eyes.  These progeny must must have the following genotypes:

The tester parent has one gamete (bp).   Thus, the gray, red progeny must have be heterozygous at both loci (BbPp) due to a BP allele from the unknown parent.  The gray, pink progeny must be heterozygous at the body color locus but homozygous recessive at the eye color locus (Bbpp).  This could only happen if they progeny got a Bp gamete from the unknown individual.  Thus the fly with the unknown genotype produced two gametes BP and Bp in equal frequencies and we can reconstruct its genotype as BBPp (case #2 above).

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