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In a complementation test involving 5 monogenic recessive mutants with the same phenotype, a, b, c,...

In a complementation test involving 5 monogenic recessive mutants with the same phenotype, a, b, c, d, ande, only the cross b X c produces the F1’s that do not resemble wild-type, while the F1’s from all other crosses do resemble wild-type. What conclusion is correct?

A.

These mutants represent four genes.
B.

These mutants represent three genes.
C.

These mutants represent two genes.
D.

These mutants represent five genes.
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Answer #1

Similar phenotypes may be a result of mutations in genes.These phenotypes do not give idea about the number of genes that are mutated.Number of mutants in a gene can be found by genetic assay and is called complementation test.This test can identify if mutations are located in the same genes or diffrent genes..The complimentation test is performed by crossing homozygous recessive mutants.It is intercrossed two at a time.Then observe if there is wild type phenotype.If recessive mutations are alleles of same gene it will not give complementary result.But if the two recessive mutation are of two diffrent genes then the cross would have wild type provided by wild type alleles.So the right answer is option c these mutants represent two genes.

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