Blood activates Vamp gene transcription. Therefore, blood is either an activator of a transcriptional activator (just as cAMP activates CAP in E. coli), or it represses a transcriptional repressor (just as lactose represses the lac repressor).

Since the homozygous B mutant shows constitutively high transcription whether blood is present or absent, the B protein must be this activator or repressor. So now we have to determine whether it's an activator or a repressor, and we can do that by looking the mutant phenotype to see the mutant phenotype is dominant or recessive (just as we did for the GAL4/GAL80 system):

• If B is an activator of Vamp transcription, the B^- mutant must be a gain-of-function mutation that activates transcription even in the absence of blood.
• However, if B is a repressor, then the B^- mutation is expected to be a loss of function allele (the mutant repressor cannot block Vamp gene transcription, so the presence or absence of blood becomes irrelevant).

The B⁺/B^- heterozygote shows low transcription in the absence of blood--so the B^- phenotype is recessive, loss-of-function, supporting the repressor model (B is a repressor of Vamp gene transcription, and that the repressor is itself repressed by blood). Repression can act either directly on the Vamp gene, or indirectly through gene G (see below).

What can we tell about the garlic response?

• Garlic blocks Vamp gene transcription--so it either activates a repressor or blocks an activator (gene G).
• Furthermore, garlic still blocks transcription in the repressorless B^-/B^- mutant, suggesting that regulation of transcription by garlic is either independent of the B gene pathway of regulation, or that gene G acts downstream of gene B (is epistatic to B).
• The G^-/G^- mutant shows constitutively low transcription under all conditions, so the mutant protein is either a dominant, gain-of-function repressor (it permanently represses and is resistant to garlic), or a recessive, loss-of-function activator (in the absence of activator, there is no transcription, so the presence or absence of garlic becomes irrelevant).

Again, the G⁺/G^- heterozygote indicates that the mutant phenotype is recessive. Therefore, protein G must be an activator of transcription and is repressed by garlic. In wildtype flies, Vamp gene transcription occurs only when blood is present (to remove the repressor, B) and in the absence of garlic (allowing activation by the activator, G).

Alternatively, if B and G act through the same pathway, then G acts downstream of B and is repressed by B.

Model 1 loco is an activator of krypto transcription and is normally inhibited by kryptonite; the domimant mutant allele of loco is insensitive to kryptonite, so it maintains transcriptional activation of krypto even in the presence of kryptonite.

Model 2 loco represses a repressor of krypto transcription; this repression of the repressor is normally blocked by kryptonite, but the mutant loco protein fails to respond to kryptonite, so it always blocks the repressor of krypto transcription.

The phenotype of a (recessive) maternal effect mutation is that females homozygous for the mutation have offspring that fail to develop normally regardless of their genotypes. If m is the mutant allele, mm (female) x any genotype (male) should give abnormal progeny that fail to develop correctly. In contrast, the mm genotype in males does not affect the progeny: mm (male) x M_ females will give normal, viable progeny.

[Since there is no directly observable phenotype of mm females (other than their failure to produce normal offspring), one will have to use other markers to follow the mutagenized chromosomes. For instance, one can mutagenize a stock that is heterozygous for one (or more) known recessive markers on each chromosome, mate these with non-mutagenized strains not carrying the recessive marker alleles, and cross the F1 progeny with each other. The F2 progeny of interest will be those displaying the recessive marker traits--since the only source of the recessive allele is the lone homolog (for each phenotype) that was in the mutagenized animals, we will know that we have two copies of a chromosome that had undergone mutagenesis--and therefore potentially homozygous for a new mutation. In real life, one would also use balancer chromosomes to prevent crossovers in the mutagenized chromosomes.]

A gene that is a "pure" maternal effect gene is needed only early in development, so homozygous mutant females can grow to adulthood and can produce abnormal embryos for study. If the gene is needed later in development, however, homozygous mutant females never make it to adulthood, so they cannot have progeny.

To get around this problem, we can induce mitotic recombination (using the X-ray source) in developing females that are heterozygous (m/+) for the maternal effect gene. In some of the females, mitotic recombination between the centromere and the gene will produce a clone of cells that are homozygous for the mutation in the tissues that produce the eggs (assuming that development of this tissue is not one that requires the gene to be functioning), thereby allowing the maternal effect phenotype to be manifested.

But how do we identify the females that have undergone such a recombination event? That's where the dominant female sterile mutation comes in. Assuming that the phenotype of this mutation (designated here as Fs) is that females don't produce eggs, we can start out with Fs +/+ m females with the Fs marker placed immediately centromere-proximal to the maternal effect gene, and very tightly linked to it. Because the female sterile mutation is dominant, these heterozygous females would not produce eggs. However, if there was a mitotic recombination event (between the centromere and Fs) producing Fs +/Fs + and + m/+ m cells in the appropriate tissues, the only females capable of producing eggs would also be the ones that have also become homozygous for the maternal effect mutation.

An alternative strategy is to look for conditional alleles (either arising spontaneously, or, more probably, after mutagenesis with some agenet such as EMS). In this strategy, you wouldn't need the dominant marker allele -- you could screen for female sterility at the restrictive temperature. The mutagenized animals would be raised at the permissive condition (allowing growth), then shifted to the restrictive condition to look for sterility. One control would be to leave the animals at the permissive condition throughout; that treatment should not result in female sterility.