If the hybrid is short, the mutants are, in all likelihood, allelic. By allelic, we mean that they are in the same gene (here defined as the unit of function ). If the hybrid is tall, the mutants are non-allelic. (There are rare exceptions to this which we will discuss latter in this section. For the time being, we will ignore these rare exceptions.)

        This cross of the two dwarf mutants is called an allelism test or a complementation test. It is a test for function.

        If the two recessive mutants lie in the same unit of function, then that function is not restored in the hybrid and the plant is still dwarf. We do not get complementation and the mutants are said to be allelic.

    If, in contrast, the mutants are in different units of function, then the hybrid will be tall. We observe complementation and the mutants are said to be non-allelic.

    Why is this so? Again, it is best to think in terms of a pathway. One mutant (lets say the Austrian one) lacks, say, the ability to convert compound B to C while the other mutant (lets say South Florida) lacks the ability to convert D to E.
 

The pathway:       A => B => C => D => E => Plant height.

        In a stock homozygous for the Austrian mutant, C can not be synthesized and hence the plant will not be tall. Compounds A and B will build up whereas compounds C, D, and E would be depleted, relative to wildtype. In a stock homozygous for the South Florida mutant, E can not be synthesized and hence the plant will not be tall. Compounds A, B, C and D will build up whereas compound E would be depleted, relative to wildtype.

        In the diploid hybrid, the ability to convert B to C is contributed by the South Florida mutant while the ability to convert D to E comes from the Austrian mutant. Hence the pathway is complete and the plant is tall.

        (A personal note: The information above is the basis for the field of Biochemical Genetics and underlies how we think about genes, alleles, enzymes and pathways. If you have trouble with this concept, see me. From years of experience, I know that students who do not grasp this have problems with the rest of the course.)

    That there can exist an almost infinite number of alleles per gene requires precise nomenclature. How we label genes and alleles can cause almost total confusion if not done properly or precisely. Compounding the problem is the fact that not all fields of genetics use the same rules. Normally names of genes reflect the phenotype of the recessive mutant.. In Mendel's work, we called the gene for plant height, dwarf and used the gene symbol, D. Dwarf refers to the mutant phenotype. Note that we do NOT use the name "tall" or the gene symbol T. Because dwarf is recessive as seen in the F₁ generation, the allele for it is given the symbol small d or d while the dominant allele contained in normal peas is dominant and is termed large D or D. Genes and their symbols are usually italicized.

    (I work with corn genes which, when mutant, give rise to a seed with inadequate starch levels and the seed appears shrunken at maturity. These are termed shrunken genes and the recessive alleles are symbolized sh. As is usually the case, there are several genes which when mutant give rise to the shrunken phenotype and these are distinguished by a number, i.e., sh1, sh2, sh3, sh4, etc. Furthermore, since many mutants can occur in one gene, one can find many mutant forms of a particular gene. To distinguish among these alleles, the number is followed by a dash and then a descriptor. For example, the first mutant that described the shrunken-2 gene is called sh2-Reference or sh2-R. Reference refers to the fact that it was the first mutant allele and hence described the gene. Other alleles are given different descriptors. The policy now is to simply give a mutant a number, usually involving the year. For example, 99001 would be the first shrunken mutant found in 1999. Note that one cannot determine genotype by phenotype. Hence, just by looking at the mutant, we can tell that it is shrunken but we can't tell which gene it belongs to. Because of this, we would initially call this mutant sh*-99001. This descriptor would remain until is was shown to be allelic to one of the genes already described or possibly allelic to a non characterized gene. For example, if this turned out to be an allele of Sh4, our new mutant would be called sh4-99001. If the new mutant is not allelic to any of the genes in the literature, it would be given the next number in the shrunken genic series, say 15, and would be termed sh15-99001. It would also be appropriate (preferable as far as I am concerned) to call it sh15-R since it would be the first mutant of the gene and hence would describe the gene.)

    Returning to the example of the dwarf pea plants from Florida and Austria, there are two outcomes from the cross between them. Either the hybrid is tall or it is dwarf.

    If the hybrid is tall, we obtained complementation, the mutants are not allelic and they reside in different complementation groups or genes. Given this result, we would write the genotypes involved in the cross as follows:

        D₁ / D₁ d₂ / d₂  X   d₁/d₁ D₂/ D₂      and the resulting F₁ would be D ₁/d₁ D₂/d₂

    If the result of the cross is mutant, we would say that the mutants are in the same function or gene, are allelic and we would write this something like this:

       d ₁-x /d₁-x      X      d₁-y/d₁-y       and the F₁ would be: d₁-x/d₁-y.

        Here, because we are working with two different mutant alleles of the same gene (D₁₎, we must give descriptors to identify the two different alleles (_-x and _-y).
 

As the number of mutants to cross increases linearly, the number of crosses increases geometrically.
 

        In the example above, we were concerned whether two mutants were allelic in the functional sense. All that was required was one cross. Suppose we find a third mutant with this phenotype and wish to determine if it is allelic to either of the first two. This would require two additional crosses (mutant 3 with mutant 1 & mutant 3 with mutant 2) or a total of three crosses to check all three mutants. The addition of a fourth mutant would require three additional crosses (plus the first three) for a total of six. With five mutants, the # of crosses becomes 10, with six it is 15, etc. I think you can see the pattern and realize the amount of work involved if the number of mutants becomes large.

        The formula for the number of non reciprocal crosses equals  n (n -1) divided by 2  where n equals the number of mutants. Reciprocal refers to using the two stocks as both male and female in the cross. For purposes here, we will assume that there is no difference although, as covered in a later section of the course, this is not always the case.

    Note that this formula is quite close to n²/2 or almost exponential. In other words, as the number of mutants increases linearly, the number of crosses increases exponentially.

    Given below are a number of possible data sets derived from crosses of 10 mutants. Study these and make sure you understand this. I guarantee you will see some form of this on the exam.

    How many non-reciprocal crosses would be required?    45

    What would be the results if all the mutants were in one gene? All 45 crosses would yield  mutant offspring.

    What if each mutant  is in a separate gene?  All 45 crosses would yield wildtype offspring.

   What if one is in one gene and other nine are in one other gene? 9 crosses yield wild types and the remaining 36 crosses      yield mutants.

    What if two in one gene and eight in the other? 16 wild type and rest mutant.

    What if three in one gene and seven in the other? 21 wild type and rest mutant.

    What if two in one gene, one in another and the rest in a third gene? 23 wild type, rest mutant.
 

Now, the gene as a unit of transmission. As Mendel taught us, alleles segregate. (Law of Segregation.)

    Remember the data from Mendel that in the F₂, we got only tall or short, round or wrinkled. We did not get intermediate phenotypes or non-parental or recombinant phenotypes.

    Let us suppose that we again start with our two recessive dwarf mutants from Austria and from South Florida and test whether the two mutants are in the same gene as defined by Mendel's Law of Segregation.

    Let us take the two possibilities.

(a) If the two mutants fall in the same gene, as defined by the Law of Segregation, what would be expected in the F₂? We expect no non-parental types and hence we expect all mutant.

(b) If the two mutants fall in different genes, as defined by the Law of Segregation, we would expect recombinant, non-parental (and hence non-mutant or tall) types.

    Do the two methods give the same answer?
        Yes!! (at least initially) This was one of the first questions that the early geneticists asked. If one took say two mutants for a seed mutant in corn, look at the hybrid between them and observed the mutant phenotype, self pollination of such seed would produce an ear in which all resulting kernels were mutant.

    Given the total ignorance of the nature of the gene at the time, the fact that both methods gave the same result provided satisfaction to early workers.

    However, when investigators started looking at large populations derived from heterozygous individuals, rare exceptions were found. An example:  A paper appeared in Genetics in 1944 (This is even before I was born!!) in which a gene in the fruit fly, Drosophila meloganster was examined. The gene, lozenge (Lz) affects aspects of the eye structure and shape.
 

Two independently-derived recessive alleles of this gene, 1_z^g and 1_z^s , were made heterozygous in females and these were crossed to males containing the first allele.    Among 8393 resulting flies, 8379 exhibited the expected mutant phenotype.
However, 14 exceptional wildtype flies (0.16% of the total population) were observed.

    How could this be? Several explanations were considered.
 

1) One or both of the two mutant alleles, (1_z^g &1_z ^s)  was reverting back to wildtype. If this is true, then the reverting allele should be able to revert to wildtype in a homozygous conditions.  As a test, comparably-sized progenies from homozygotes for each allele were examined in an identical manner.

      The outcome of these crosses:  No exceptional wildtype individuals were found. Production of the exceptions required a heterozygous condition between the two mutant alleles.

2) That heterozygosity was required suggested that rare recombinational events occurred within the unit of function. If the exceptional wildtypes were due to recombination, two testable predictions were made:
 

(a) Because recombination only occurs in female fruit flies, heterozygous males should NOT produce exceptional wildtypes. Comparably sized progeny from heterozygous males were examined and no exceptional wildtypes were found.

(b) If recombination, one should find, at high frequency, exchange of outside genetic markers, i.e.,

a                                            1_z^s                                                            b
 
 

The rare exceptional wildtypes, if caused by recombination within the lozenge functional unit, should carry the large A marker to the left and the small b marker to the right at high frequency or they should carry the small a marker to the left and the large B marker to the right at high frequency.

A note on life cycles and the time of recombination:  In higher organisms, the next generation begins at the time of fertilization.  Two haploid gametes fuse to form the diploid zygote.  The resulting single diploid cell then undergoes mitosis to produce two identical diploid cells. Genetic information is passed on to the next two cells and the resulting two cells are genetically identical (barring mutation).   This series of mitotic events continues to occur to produce virtually all of the cells of the organism.  A few diploid cells will give rise to the gametes that  fuse to form the next generation.  In the process of gamete formation, diploids cells undergo meiosis to form haploid cells.  It is during meiosis that recombination occurs.  For example in our fly experiment above, the results of fertilization between the two homozygous parents (lz^g /lz^g   and lz ^s/lz ^s )  is lz ^g/lz ^s .   All diploid cells in this heterozygous individual are of that genotype.  BUT  a few resulting haploid gametes arising from this heterozygote can be wildtype because of recombination.  Because we cannot look at gametes to score this phenotype, we must wait until the next generation to score the eye phenotype.