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I'm having difficulty with the following problem:

In yeasts, genes MEC1 and SGS1 favor survival in response to HU (hydroxyurea). In the figure below, Δ indicates homozygosis for the mutant allele that causes loss of function. mec1-100 yeasts are heterozygous (one MEC1 allel is wild and the other is mutant).

a) What genetic interaction do genes SGS1 and MEC1 maintain?
b) How do you explain the results at the molecular level?
c) How do you explain the difference in phenotype between the homozygous and heterozygous yeasts for the allel that causes loss of function for the MEC1 gene?
d) Why would you combine the sgs1Δ mutant with mec1-100, instead of combining it with mec1Δ?

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My attempt to solve this was as follows:

Observing the sgs1 Δ mutant in homozygosis for MEC1, we see that the phenotype that is produced is within a scale close to the WT phenotype. Further, in heterozygosis for mec1-100 without mutation we also see a phenotypical scale similar to the WT. When we see the heterozygotes for mec1-100/sgs1Δ the scale drops abruptly and the phenotype passes to be very far from the WT phenotype. This last one can be explained as an incomplete dominance in which the genotype of mec1 must present both WT allels to achieve its effect, because just one of the WT alleles cannot fulfill its function, so what we see in comparison with the sgs1Δ/MEC1 homozygote dominant is different from mec1-100/sgs1Δ. When we mutate the gen mec1Δ we see that the viability of the cells is reduced, even though the gen sgs1 is still present.

What has been observed and described previously could be explained as a recessive epistasis where regardless of what happens with gen SGS1, if MEC is mutated (double recessive) it masks the effect of Sgs1. That is clearly seen because when mec1 is present, the viability is close to the WT scale while when mec is mutated or in heterozygosis (which would work as a mutant by the incomplete dominance) it does not matter what happens with sgs1 (if it is mutated or not mutated) the phenotype of the molecular pathway is not produced (a phenotype of low viability is produced instead).

At the molecular level, it could be explained as follows:

A substrate exists that uses the enzyme produced by MEC1 to activate the viable molecular pathway. In absence of MEC1 the pathway is not activated, SGS1 uses MEC1 to carry out its function that also promotes viability, therefore if MEC1 does not exist SGS1 does not have an enzyme to interact with.

enter image description here

But turns out that the correct answer for this problem is that the genetic interaction is a redundancy between SGS1 and MEC1.

My question here is:

Can this problem, as expressed in the first paragraph, be taken as a recessive epistasis? Why can't it be considered a recessive epistasis? In cases where we have a logarithmic scale, if the correct answer is a genetic redundancy, can we take it as a viability generated by both genes anyway? Is the viability generated by each gene comparable at all?

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    $\begingroup$ This is a well-formulated question, but the prompts suggest that it is an exam question or other similar coursework. If that is true, you should add the "homework" tag. (You have clearly shown your attempt at an answer, which means that this question is appropriate for the SE Biology community.) $\endgroup$ Feb 6, 2022 at 21:19
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    $\begingroup$ @MaximilianPress thanks! Will add it now. Yes, this is from a set of prep exercises for an exam that is driving me a little bit crazy. $\endgroup$ Feb 6, 2022 at 23:47

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Based on the graph of experimental results, WT diploid yeast are able to grow in the presence of HU (hydroxyurea). Deleting both copies of the MEC1 gene results in cells that stop growing in the presence of HU; therefore the wild-type product of the MEC1 gene is involved in resistance to HU (as described at the start of the homework question). Deleting both copies of the SGS1 gene has a slight effect on growth in the presence of HU. From this we infer that the wild-type product of the SGS1 gene has some role in resistance to HU, but these mutant cells can still grow, and therefore contain some other activity (or activities) that provide resistance.

The MEC1 gene is dose-sensitive; when you remove one functional copy with the mec1-100 loss of function allele, the cells resistance (or growth) is intermediate between wild-type cells and the mec1 homozygous null (knockout, or deletion).

mec1-100 heterozygotes, that also lack both wild-type copies of the SGS1 gene have much worse viability/growth/resistance to HU (when compared to either the sgs1 homozygous loss-of-function allele alone OR the mec1 heterozygous loss-of-function allele). In other words, it appears "more" mutant, or closer in phenotype to the mec1 homozygous mutant. Therefore the wild-type function of the SGS1 gene overlaps with the wild-type function of the dose-sensitive MEC1 gene. Two genes with overlapping genetic functions are by definition redundant.

I would've described sgs1 as a recessive enhancer of a mec1 hypomorph. Historically this can be a powerful way to identify genes that may be involved in the same cellular process. When I see the term 'recessive epistasis' I think of trying to order genes in a genetic pathway. As in, which gene is 'upstream' and which gene is 'downstream' in the genetic pathway. It doesn't make sense (to me) to try to "order" two genes with redundant function. Sometimes redundant genes encode proteins with homologous sequence (e.g., different classes of tubulins), but sometimes redundant genes indicate that there are parallel pathways at work (and one compensates for the lack of the other). I suspect that you are using a different definition of 'recessive epistasis'.

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