Preliminary text The complementation test involves placing two different mutations that usually have the same or similar phenotype in a common cytoplasm and then through some means determine whether they function to produce a wild type phenotype. This allows genes to be classified rapidly according to functional characteristics.
Question 1 Eight independently isolated mutants of E. coli all of which are unable to grow in the absence of histidine (they are his-), were examined in all possible cis and trans-heterozygotes (partial diploids). All the cis-heterozygotes were able to grow in the absence of histidine. The trans heterozygotes yielded two different responses: some of them grew in the absence of histidine and some did not. The experimental results, using + to indicate growth and 0 to indicate no growth, are given in the following table. How many genes are defined by these eight mutations? Which mutant strains carry mutations in the same gene(s)?
GROWTH OF TRANS HETEROZYGOTES (- HISTIDINE) MUTANT: 1 2 3 4 5 6 7 8 8 0 0 0 0 0 0 + 0 7 + + + + + + 0 6 0 0 0 0 0 0 5 0 0 0 0 0 4 0 0 0 0 3 0 0 0 2 0 0 1 0Question 2. In Drosophila, white, cherry, and vermillion are all sex-linked mutations affecting eye color. A white-eyed female crossed with a vermillion-eyed male produces white-eyed male offspring and red-eyed (wild type) female offspring. A white eyed female crossed with a cherry-eyed male produces white eyed sons and light cherry eyed daughters. Do these results indicate whether or not any of thgese mutations affecting eye color are functionally allelic? Is so, which mutations are functional alleles?
Question 3 The rosy (ry) gene in Drosophila encodes the enzyme xanthine dehydrogenase; the active form of xanthine dehydrogenase is a dimer containing two copies of the rosy gene product. Mutations ry2 and ry42 are both located within the region of the rosy gene that encodes the rosy polypeptide gene product. However, ry2 ry42 trans heterozygotes have wild type eye color. How can the observed complementation between ry2 and ry42 be explained given that these two mutations are located in the same gene.
Question 4. Seven mutants of Neurospora are unable to grow on minimal medium unless it is supplemented with one or more of the metabolites A through G. On the basis of the data given below, where "+" indicates growth and "0" indicates no growth, draw a pathway for the synthesis of the seven substances A through G and show where the mutants 1-7 are blocked.
Growth in the Presence of Metabolites
Answer to Question 1
Question 2 Consider the very first test in the upper left corner of the table. This is his1 and his8. The problem states that the cis heterozygote his1+ his8+/his1 his8 functions. The 0 in the table for the trans heterozygote means that his1 his8+/his1+ his8 does not function. So his1 and his8 are defective in a single non complementing function. This is also true for his8 and all the other his mutants except his7. So his8 and his7 complement one another in trans and therefore must be in separate functional units. The data in the next row confirm this. His7 complements all the mutants (except itself of course). None of the other mutants complement one another. They must therefore all be deficient in a common function. Thus there are two groups of mutants: his7 is all there is in one group. his1 through his6, and his8, comprise the other group.
Answer to Question 2 Consider the first cross. Recalling Morgan's experiments, the white eyed female must be homozygous. The vermillion eyed male must be hemizygous. Each individual produced in the cross will have the white allele on one X chromosome. Thus, all the males should be and are white eyed. However, the daughters in this cross all have red eyes, when we know that they have the white gene on one X chromosome and the vermillion gene on the other. Apparently white and vermillion have complemented one another to produce a red eyed individual. Now look at the second cross. The situation is similar for the males. For the females, we get light cherry eyes in those individuals having the white allele on one X chromosome and the cherry allele on the other. The data suggest that vermillion and white are functionally separate, even though they both affect eye color. However, cherry and white may be allelic in the more conventional sense, because an intermediate phenotype is obtained, as if white were a functionless form of the gene that produces the same product as the cherry gene. In other words, cherry and white are functionally allelic.
Answer to Question 3 This is a case of two wrongs making a right. This can be accomplished because sometimes a defect in one polypeptide of a dimeric enzyme can be compensated by a DIFFERENT defect in the other polypeptide in the same enzyme. This is called intragenic complementation.
Answer to Question 4. This takes a long time to solve, but it is worth going through it. It helps to have a system of notation, for example consider mutant 1. This mutant can be made to grow by feeding substance A. This can be written in shorthand as A (1), indicating that A is the product of the enzyme coded by the gene that is defective in mutant 1. In this notation, the biochemical pathway is being written with the final product (let us call it P) on the far left. Looking at the table, then, one can see that feeding this mutant a mixture of substances D and G, or a mixture of substances B,C, and E, also permits the mutant to grow. This means that the final product of the pathway P, is separated from intermediate A by intermediates (D or G) and by intermediates (B or C or E). This leaves open a lot of possible pathways, but it also eliminates a large number.
We can write P <===== [D|G+B|C|E]__A(1) to indicate that.
Next consider mutant 2. Notice that there are very few things that can support the growth of this mutant. In fact, it looks like only B can do so (C does not work by itself, neither D nor G work separately or together, etc). So the final product P is separated from intermediate B by none of the other known substances in the experiment. We can write:
P <====B (2) [everthing else] to indicate that.
Mutant 3 is much like mutant 2, but it is auxotrophic for C instead of B.
P <======= C(3) [everything else]
[everything else] here means only that none of the intermediates are to the left of C or B.
Mutant 4 is very interesting. Both D and E can allow it to grow, either when provided separately or in combination with other substances. This clearly suggests that D and E lie between the product P and the location of mutation 4.
P <======= D,E (4).
It is not useful to write out the entire solution; you need to go through the exercise. I found I tried several different pathways before finding one that fit the data. These hints as to method should suffice. The pathway I derived is as follows:
P <==== B(2)+C(3)G(7)+E(5)D(4)A(1)F(6), that is B, C and E are all required for P formation, G is needed by the product of gene 7 to make C, and E is made by the conversion of F to A to D to E, using enzymes 5, 4, 1, and 6 respectively. The cool thing about this analysis is that it does biochemistry without actually doing any chemistry. This is not however the only possible pathway consistent with the data. For example, G could be needed for B formation instead of C formation or in addition to C formation.