The Complementation Test: Are two mutations on the same gene?
The complementation test is a test to check if two mutants resulted from mutations in the same gene or in different genes. In the test, two mutants, which have the same phenotype but different genotypes, are crossed. If the two mutants produce normal offspring, the genes complemented; thus, the trait is controlled by two genes. Let's see some examples.
Case 1: Trait controlled by two genes.
If the trait is controlled by two genes, the mutants have been selected such that each has its mutation on a different gene. For example, let's consider a pea plant, where the mutant flowers are white and the wild-type are purple. In this case, if the trait is controlled by two genes, both mutants would appear white, but each would have a mutation in a different gene to cause this effect. Crossing these two mutants would provide one dysfunctional gene from each parent; however, the other parent's chromosome would make up for the lack of the other (we assume one functional copy of each gene is enough to develop the full wild-type phenotype). This "making up" for the lack of the other gene is called complementation.
Case 2: Trait controlled by one gene.
If the trait is controlled only by one gene, then the differing genotypes of the two mutants will be of alleles on this same gene, not of alleles in different genes as described above. If the two white mutants have mutations of different locations on the same gene, both parents' genes are defective. There is no way for complementation to occur; the functional fragments of genes cannot be transferred by themselves, because how would the body know which gene is defective or functional? Because both parents' genes are defective, there is no way for the offspring to get a good gene (unless crossover occurs in the middle of a gene but that's unlikely to happen in a perfect location). Thus complementation will not occur.
In general:
mutation caused by two genes --> genes will complement --> crossing mutants with same phenotype and different genotype yields normal offspring
mutation caused by one gene --> only one gene, so no complementing; offspring receive one copy of the gene or another --> crossing mutants with same phenotype and different genotype yields mutant offspring
(If this didn't make sense, watch the 1st video linked at the end of the article.)
Now, let's apply this to problem-solving for USABO. For complementation test problems, you'll often see this kind of table:
Each box is an individual complementation test, where two mutants were crossed. A "+" indicates that the mutants did complement (so the alleles are on 2 different genes) and a "-" indicates that the mutants did not complement (so the alleles are on the same gene). Notice that across the "N/A" line of self-crosses, both halves of the table are reflections of one another. This isn't a coincidence: for example, D crossed with C is the same as C crossed with D, so they'll obviously have the same result of complementation. Point being, you can ignore half the table because it's redundant information.
To determine how many genes are involved in this table of alleles, we need to create complementation groups, which are groups of alleles that we know are on the same gene. Try to make complementation groups for the table above. Answer is explained below.
Move from left to right on the top half of the table. A and B are on different genes. C is with both A and B. D is separate from A but on the same gene as B. Groups so far are:
AC
BCD
Next, E shares a gene with A and C, and doesn't share with B and D. Groups:
ACE
BCD
Next, F does not share with anything but C. Groups:
ACE
BCD
CF
Thus, 3 groups, so 3 genes.
Here's the full MCQ. (Source: 2020 SF Q50)
The answer is C. Also: the presence of strain C in all 3 complementation groups likely indicates that C carries a dominant negative mutation, as it doesn't allow complementation in any pair so it doesn't really belong anywhere. This is consistent with info provided in the exposition of the question (in a previous part): "However, crossing strain C with haploid wild-type yeast yields diploid yeast that fail to grow without leucine."
That's how you solve these problems! Also notice we didn't actually need to read the question lol. It's a time-saving strategy if you know how to implement it properly.
BONUS: think about how this method could be used to determine if two genes are on the same chromosome! Think about the drawbacks of using this method for that (hint: meiosis I).
Hope that was helpful! Here's some videos that helped me understand this:
https://youtu.be/uyBCMu9hdek - my fav video about this topic on the internet.
https://youtu.be/_aVUuMi3i_I - more about chromosome complementation tests, instead of genes'.
