In yeast, mating happens between distinct mating types, A and ? . Alpha ( ? ) ty
ID: 37969 • Letter: I
Question
In yeast, mating happens between distinct mating types, A and ?. Alpha (?) types release a protein hormone ?-factor, that binds to a receptor in the membranes of A yeast cells. At the end of the signal transduction cascade, A-cells show 2 distinct changes: 1, a change of shape from round, to pear shaped (this shape is called a shmoo and is the result of the formation of a germ tube. Concomitant with this is a halt to normal cell division, that is the cell cycle is stopped; Figure 1); 2, a specific gene, Fus1 is activated (this gene is important for nuclear fusion when compatible mating types come together).
You have a collection of various yeast mutants that have been transformed with a reporter gene fused to the promoter of Fus1. This reporter gene encodes beta-galactosidase (LacZ), which can be easily assayed. The assay includes extracting protein from yeast cells and incubating it with o-nitrophenylgalactopyranoside. This substrate is colorless. However, when cleaved by LacZ, galactose is released leaving free nitrophenyl, which is yellow and can be quantitated spectrophotometrically. Therefore an increase in absorbance represents an increase in Fus1 activity and is thus reflective of activation of the mating pathway.
In the experiment, each mutant strain is grown without a factor (no hormone and thus no mating) and with a factor (+ a factor which should stimulate mating). Data for wildtype is presented to show the normal response, the control. Things to look for: Notice that in wildtype, shmooing and Fus1 activity only increase when the mating factor is present indicating that the response to the mating factor is normal. When looking at mutants, look to see if shmooing or Fus1 activity increases with addition of mating factor. If it does not, then there is a break in the signaling pathway as the response is not being produced. Be careful however, as in some, the response may happen even in the absence of the mating factor, which means the pathway is being activated without the signal being received. Last note: look carefully at the data and think about real differences. For example, an increase of 5-10% is probably not significant and does not represent an increase in response. Likewise, a value above zero does not necessarily indicate a response (see WT Fus1 activity
In yeast, mating happens between distinct mating types, A and ?. Alpha (?) types release a protein hormone ?-factor, that binds to a receptor in the membranes of A yeast cells. At the end of the signal transduction cascade, A-cells show 2 distinct changes: 1, a change of shape from round, to pear shaped (this shape is called a shmoo and is the result of the formation of a germ tube. Concomitant with this is a halt to normal cell division, that is the cell cycle is stopped; Figure 1); 2, a specific gene, Fus1 is activated (this gene is important for nuclear fusion when compatible mating types come together). You have a collection of various yeast mutants that have been transformed with a reporter gene fused to the promoter of Fus1. This reporter gene encodes beta-galactosidase (LacZ), which can be easily assayed. The assay includes extracting protein from yeast cells and incubating it with o-nitrophenylgalactopyranoside. This substrate is colorless. However, when cleaved by LacZ, galactose is released leaving free nitrophenyl, which is yellow and can be quantitated spectrophotometrically. Therefore an increase in absorbance represents an increase in Fus1 activity and is thus reflective of activation of the mating pathway. In the experiment, each mutant strain is grown without a factor (no hormone and thus no mating) and with a factor (+ a factor which should stimulate mating). Data for wildtype is presented to show the normal response, the control. Things to look for: Notice that in wildtype, shmooing and Fus1 activity only increase when the mating factor is present indicating that the response to the mating factor is normal. When looking at mutants, look to see if shmooing or Fus1 activity increases with addition of mating factor. If it does not, then there is a break in the signaling pathway as the response is not being produced. Be careful however, as in some, the response may happen even in the absence of the mating factor, which means the pathway is being activated without the signal being received. Last note: look carefully at the data and think about real differences. For example, an increase of 5-10% is probably not significant and does not represent an increase in response. Likewise, a value above zero does not necessarily indicate a response (see WT Fus1 activity ?? 0.002 is too small to constitute a real response and is just background). 1.If you look at the data, you will notice that there are two GPCR??s being tested, of the two, which one serves as the receptor for the ?-factor pheromone? What is the evidence to support your conclusion? (2pts) 2. Are the two outcomes of this signaling pathway independent of each other, that is, is the signaling pathway linear, triggering both events simultaneously, or does it seem to branch so that the same signal triggers each response separately? What evidence is there to support your conclusion? (3pts) You cross the Dig1 mutant with the Kss1 mutant (the result is Dig1/Kss1 double mutant) and analyze the mating response. In this case, the new yeast strain shows a response about the same as the Dig1 single mutant. This shows that Kss1 operates upstream of Dig1, Kss1 è Dig1. Based on this data and the information provided in table 1 ?? provide a mechanism that explains these results. (3pts) Dig1 is known to bind to Ste12, based on the data from table 2, do you think that Dig1 is an inhibitor or an activator of Ste12? Explain (2pts) If you were to develop a mutant in which Dig1 was non-functional (a ?Dig1 mutant), what do you predict the phenotype would be? (2pts) Ste5 binds to the beta and gamma subunits of the active G-protein. Based on what you know, is this early or late in the signaling pathway? (1pt) If you cross ?Ste5 with Ste11, generating an ?Ste5/Ste11 double mutant, the mutant shows the same phenotype as the ?Ste5 single mutant. What does this indicate about the signaling sequence? Is it Ste5 è Ste11 or Ste11 è Ste5? (1pts) If you generate a gain-of-function mutant of Ste5 and cross it with Ste11 you get results that are the same as the Kss1 gain-of-function mutant. Given that Ste11 is a gain-of-function mutant, why do we not see these results in the Ste11 single mutant (hint: look at what Ste5 encodes and think about what it might do)? (3pts) If you look at what we learned in class about GPCR pathways, what are the ??typical?? players downstream of the GPCR? What proteins are involved here? What does that indicate to you about signaling? (3pts)Explanation / Answer
1) the two GPCR involved are ste2 and ste 11 however, ste11 is a receptor for the pheromone. because although neglible amount of fus 1 activity is seen for ste11 in absence of pheromone. but in presence of Pheromone the activity of Fus 1 and Shmooing is highly increased.
2) the two pathway are branched i.e even when there is no ste 2 (which is a transmembrane protein) still the action of shmooin is seen in ste 11 mating system.
Although Kss 1 mutant will be always phosphorylated and present in nuclues, but Dig 1 mutant can not be phosphorylated. when unphosphorylated it is always in active state thus, bound to transcription activators. thus, will always signal actively.
conclusion: Kss 1 (phosphorylated in nucleus) along with the unphosphorylated Dig 1 will thus, provide same results that Dig 1 mutant alone. as Kss1 activity doesnt affect Dig 1 when it is mutated and stays in unphosphorylated active form.
Dig 1 is an activator of Ste 12 because ste 12 then can interact with the DNA substrate in nucleus.
Dig 1 mutant can not be phosphorylated, thus unphosphorylated Dig1 will always be active to bind to Ste 12 which is a DNA binding protien thus in turn constitutively activate mating genes.
Ste 5 bnding to beta and gamma subunit of Gprotein will occur early in signalling because the whole cascade of events begins with the initial dissociation of trimeric G protein into alpha & beta-gamma subunti which then bind to ste 5.
Ste 5 and Ste 11 double mutant giving same results as ste5 single mutant indicates that Ste 5 comes before Ste 11 in the signalling pathway.
we do not see the same results in ste11 single gain of function mutants because we know from the data that Ste 5 is the central hub . thus, the further activity of Ste 11 depends on Ste5 activity. so easy to understand that single Ste 5 gain-of- function mutant is functional but, ste 11 gain-of- function mutant is non -functional.
key players in the whole signalling is , ste 5 as t binds to beta-gamma subunit of G protien and then down regulates the signalling. also,ste 12 is imp as it binds to DNA activating its transcription.