The cis test:

    In this day rarely does one see the cis test performed. For good reason. It's a lot of work. It requires the double mutant on one chromosome and wild type on the other.   How do you get the double mutant? Recombination (rare). How do you recognize it? It would have the parental phenotype but unlike either parent - would not recombine with both parents.  This requires a massive amount of work.   If both of the parental mutants can revert to wildtype, the double mutant should also be distinguishable by a massively reduced (likely zero) reversion rate.

This whole discussion  gets at the question of:

The meaning of dominance and recessiveness.

There are two very important tenets here.

     A. The vast majority of mutants are (1) loss (nearly) of function and (2) recessive.

     B. Most gene functions show a gene dosage effect.  That is to say that the heterozygote between a wildtype and a mutant has ½ the gene product of the homozygous wildtype.

The conclusion then from points A and B is that, with recessive mutants, ½ wildtype gene function is enough for the wildtype phenotype.  In most cases, it is actually much less than 50%.  From studies of leaky mutants, etc, one can get an idea of how much actual gene product one needs to get a wildtype phenotype.  In our own studies, we know that about 15% wildtype enzyme levels of a particular starch synthetic enzyme conditions a phenotype close to wildtype. People working with anthocyanin synthesis in maize tell me that in some cases, 4 to 5% wildtype enzyme levels is enough to get the wildtype phenotype.  The amount of gene product needed to reach the threshold of going from "mutant" to "wildtype" varies from gene to gene.  It can not be predictable on theoretical grounds.  It must always be determined empirically.

(It is an interesting biological question as to why there is so much gene product present compared to how much is needed.)

To restate the point above in a more biochemical way, suppose we have the pathway:

   A => B => C  => D   => E   => F    => FINAL PHENOTYPE

and suppose that we are working with a recessive mutant in the structural gene that converts C to D ( We will call it Case) and that the enzyme shows a gene dosage effect.  Hence, in this system, we can lower the amount of enzyme ½ and still get enough D product to end up with wildtype phenotype.

        This is true under the set of conditions above.

        What if we reduce the amount of substrate (C) and then we look to see if a 50% reduction of Case gives the wildtype phenotype.  Here we reduce the amount of substrate (C) by reducing the amount of an upstream enzyme 50%.  The latter could occur by having a recessive mutant in an upstream enzyme heterozygous with its wildtype allelic counterpart.

        Will we still get the wildtype phenotype?  Usually, we do.  But this gets at the question of the threshold amount of gene product needed to go from mutant to wildtype.  Were this threshold high, it is conceivable that the phenotype could be mutant.  Likely some of the cases of non-allelic non-complementation  are explained by this logic. 

        Here, then, the negative complementation result would be misleading.  We would have heterozygosity at  two non-allelic genes (one for Case and one for the upstream enzyme) and the phenotype would be mutant.

        Note however, that this would occur regardless of whether  the mutants were cis or trans to each other.  Hence, the Benzerian definition would hold.  Here, since the cis arrangement would be mutant, we could not interpret the trans result and hence we would not say that because of the trans result, the mutants would be in the same cistron.

        Interestingly, if the two hypothetical mutants above were unlinked,  a cross of the mutant double heterozygote by one of the parents (i.e. +/m₁ +/m₂ X m₁/m₁ +/+) would give a 3 mutant to one wildtype ratio.  This is detailed below:

              Genotype at m2                    Genotype at m1                                                      Phenotype

	+ +	+ m1	(wild type)
	++	m1m1	mutant
	+ m2	+ m1	mutant
	+ m2	m1m1	mutant

        Another point about the cis - trans test is that it is not necessarily mutant and wild type. The point is whether the cis arrangement is different from the trans arrangement. In fact, this can be applied to dominant mutants and plant pathologists have historically done this with clustered disease resistance genes.  One isolates the cross over and asks if it has any properties that are unique. If so the mutants are said to fall within the same functional unit.

Cases of non-allelic non-complementation. From Elliot Meyerowitz

Winnier GE, Kume T, Deng KY, et al.
       Roles for the winged helix transcription factors MF1 and MFH1 in cardiovascular development revealed by nonallelic
       noncomplementation of null alleles
       DEV BIOL 213: (2) 418-431 SEP 15 1999

Harris MJ, Juriloff DM
       Nonallelic noncomplementation models in mice: The first arch and lidgap-Gates mutations
       GENOME 41: (6) 789-796 DEC 1998

     FIRMENICH AA, ELIASARNANZ M, BERG P
       A NOVEL ALLELE OF SACCHAROMYCES-CEREVISIAE RFA1 THAT IS DEFICIENT IN RECOMBINATION AND REPAIR
       AND SUPPRESSIBLE BY RAD52
       MOL CELL BIOL 15: (3) 1620-1631 MAR 1995

       RANCOURT DE, TSUZUKI T, CAPECCHI MR
       GENETIC INTERACTION BETWEEN HOXB-5 AND HOXB-6 IS REVEALED BY NONALLELIC NONCOMPLEMENTATION
       GENE DEV 9: (1) 108-122 JAN 1 1995

     BELANGER KD, KENNA MA, WEI S, et al.
       GENETIC AND PHYSICAL INTERACTIONS BETWEEN SRP1P AND NUCLEAR-PORE COMPLEX PROTEINS NUP1P AND
       NUP2P
       J CELL BIOL 126: (3) 619-630 AUG 1994

     BERROTERAN RW, WARE DE, HAMPSEY M
       THE SUA8 SUPPRESSORS OF SACCHAROMYCES-CEREVISIAE ENCODE REPLACEMENTS OF CONSERVED RESIDUES
       WITHIN THE LARGEST SUBUNIT OF RNA POLYMERASE-II AND AFFECT TRANSCRIPTION START SITE SELECTION
       SIMILARLY TO SUA7 (TFIIB) MUTATIONS
       MOL CELL BIOL 14: (1) 226-237 JAN 1994

        So, in summary the cistron as a unit of function represents a relationship between two mutants. Benzer equated this to a region coding for a protein, and to a great extent, this makes sense. In the trans arrangement, each of the two forms of the coding region contains a mutation and hence is non-functional. In the cis arrangement, one copy contains two blemishes while the other copy is wildtype.

        What is missing? The idea that genes are controlled in terms of expression, that these control units must be genetic (and hence judicable as units of function, recombination and mutation) and that a control region likely would be adjacent to the structural gene.

        So, what does the cistron correspond to in molecular terms? A structural gene plus any cis-dominant regulatory regions. This concept was clarified in the work on the lac operon in E. coli.

The lac operon

        While we think of E. coli as a pretty simple organism - a fair assessment compared to higher organisms - it does exhibit much of the complexity concerning gene expression as other organisms. [In fact there is a famous quote from J. Monod "What is true for E. coli is true for the elephant"]

        Early workers had found that E. coli could somehow "sense" its food supply and change its catabolic machinery accordingly. For example, if E. coli grew on the sugar lactose, a number of enzymes and proteins needed to break down lactose increased dramatically. If the bacterium were fed glucose, these enzymes and proteins were not found at appreciable levels in the E. coli cell. In the studies described in detail below, three particular enzymic activities involved in lactose catabolism were monitored closely. These were B-galactosidase, permease and transacetylase

        So, in normal E. coli cells, B-galactosidase, permease and transacetylase are at high levels when the cells are fed lactose and their levels are near zero when the cells are fed other sugars. To study the system, investigators isolated mutants.

        Mutants were fundamentally of two types: (1) One class did not produce one or more of the enzymes when wildtype did (in the presence of lactose) while the other (2) class produced the enzymes when wildtype did not (in the absence of lactose.)

        From the genetic analysis of these mutants, the model given below was proposed. Virtually all the aspects of the model were subsequently confirmed by biochemical and molecular analysis. Looking back, the model seems simple and obvious. In many ways, that is the beauty of it. In fact it was not all that simple or easy to figure out.

        Here, we will cheat. The model is presented first and then the evidence for it will be presented. Obviously, the correct chronological order is just the opposite.

The Model:
 

        Following transcription, the poly-genetic transcript is translated into three proteins: The z gene encodes B-galactosidase; the y gene encodes permease and the a gene encodes transacetylase. The P gene (for promoter) is the initiation site for transcription. It also represents the site where other proteins (not discussed in this course, but emphasized in others) also bind.

        Early in the analysis, it was clear that some single mutants affected a single enzyme while others affected all three. With a notable exception, these types of mutants fall into two classes, as described below:

Phenotypes of structural gene mutations (in contrast to regulatory mutants):

1) usually affect only one enzyme (not all three).  Exceptions to this rule are polar mutants, see below.

2) affect the structure of the resulting protein. Some mutants will be null (total knockouts) and therefore uninformative. However, among those mutants that do produce some protein, a number of parameters can be examined. These include (a) electrophoretic mobility (alteration in size and/or charge), (b) K m for one of the substrates (experimentally defined as substrate leading to ½ maximum velocity), and (c) ability to cross react with antibody made against the wild type enzyme (ie cross reaction with antibody but little to no enzymic activity)

1) affect all three enzymes.

2) can be divided into two classes: those that produce enzyme when wild type does not (o and i mutants) and those that do not produce enzyme when wild type does (p mutants). Note that the former class only exists if the operon is under negative control (ie proteins that block transcription [or translation or at any stage in gene expression for that matter]). I am not aware of any genes in plants that are the counterparts of the lac o and i genes.

3) can be divided into two classes on the ability (or inability) to produce a product. This divides i from o and p. Whereas mutants in i and in O produce enzyme in the absence of inducer, p mutants don't make enzymes in the presence of inducers.