Mutant and Wild-type Yeast Strains via Mitochondria Proteins

Mutant and Wild-type barm Strains via Mitochondria ProteinsDifferentiating between mutant and wild-type barm strains via mitochondria proteins By Jason HoangLab Partner Daryan ChanIntroductionyeasts are important organisms due to their practice sessions in everyday such as baking, making fermented foods and alcohol doing (Steensels et al, 2014). Yeasts withdraw been so widely studied that it was one of the first organisms to exhaust its genome sequenced (Goffeau et al, 1996). Thus, Yeasts are more than cap adapted of acting as a model organism for eukaryotes (Botstein et al, 2011). For this experiment we are working with Saccharomyces cerevisiae. The mitochondria is the powerhouse for carrell, as it is the major production set of ATP for the cell. The inner mitochondrial space has an electrochemical gradient, from which ATP is contractd by using 5 protein complexes create an electrochemical gradient to assist in ATP production (Alberts et al., 2015). The coxswain6 assembly accepts electrons from cytochrome c and uses oxygen as the terminal electron acceptor to make water (Alberts et al., 2015). ATP synthase then uses the resulting proton gradient made by those complexes to pump protons bum into the mitochondria matrix and make ATP (Alberts et al., 2015).The objective of this experiment was to determine if a given yeast model was a wild type or a mutant with no COX6 bodily function.One of the major techniques to be use in this lab is subcellular fractionation. This technique first lyses the cells and then uses centrifugal forces to separate particles by coat (Alberts et al., 2015). The centrifugal forces results in the denser particles moving away from axis of rotation creating a pellet which contains the heavier particles and a supernatant which contains lighter particles (Alberts et al., 2015).Another major technique employ was Gel dielectrolysis. Gel electrophoresis is utilise so that a fraction with multiple proteins can be separated establ ish on size and hurl (Alberts et al., 2015). Protein fractions are loaded on to wells in the mousse and an electrode is attached (Alberts et al., 2015). SDS page is popularly used because it can confer a negative tingle and linearize proteins being run through the gel (Alberts et al., 2015). The proteins exit run through the gel due to their negative charge (Alberts et al., 2015). A banner is used to provide a reference to determine the sizes of the sampl proteins (Alberts et al., 2015).One of the other major techniques used in this lab was western blotting. After proteins are run on electrophoresis a labelled antibody is undecided to the electrophoresed fractions in a process called immunoblotting in order to detect presence of a specific protein (Alberts et al., 2015). The gel is exposed to a tissue layer where a current is run to dive the proteins onto the membrane (Alberts et al., 2015). The membrane is then drenched in labelled antibodies to detect for a specific protei n (Alberts et al., 2015). This process can detect very small amount of specific protein and is useful for catching changes of constriction of a specific protein in a cell at a lower place various conditions (Alberts et al., 2015).To measure cytochrome oxidase practise in this lab, we looked towards Beers Law which states that the ability of a solution to absorb light at a single wavelength is proportional to the closeness of solute in solution (Lukofsky et al, 2009). This show that absorbance and concentration are linearly related. Therefore, this would allow us to determine the rate of cytochrome oxidase activity in a sample.Materials and Methods sample was performed according to protocols set by Department if Biology, Winter 2016, Biology 331 for Experiment 1 Subcellular fractionation of yeast cells, pg 2-8, Experiment 2 Yeast growth convolute light microscopy protein determination, pg 1-4, Experiment 3 Polyacrylamide gel electrophoresis, pg 1-8, Experiment 4 Development of Western Blot COX Activity Assay, pg 3-10, written by Dr. Dragana Miskovic where the experiment was performed with no deviations unless specifically noted (Miskovic, 2017)The only deviation occurred in experiment 2 where we had ran out of BSA STD and had to borrow from another group. The borrowed BSA STD was not tested to take a crap exact concentration as specific in lab protocol and may catch had different concentration.ResultsExperiment 1Yeast Strain Sample A2Table 1. spate recorded for Lysing Yeast carrels SectionItemsMass (g)Mass of Centrifuge Bottle containing Yeast159.2Mass of Empty Centrifuge with Pellet49.89Mass of Pellet3.23 sum of STE solution needful to resusp subvert yeast pellet3.23g x 2= 6.46mLTable 2. Volumes recorded for the Aliquoting Yeast Subcellular Fractions SectionSolutionsTotal VolumeLPS3mL + 3.7mL= 6.7mLHSS3mL + 2.8mL= 5.8mLMITO300L + 300L= 600LIt was also noted that after the MITO fraction was made the pellet was full and not messyFigure 1 Drawings of 50/100/200 L dye drops from pipetteman and expected 1 mL dye dropExperiment 2Part ATable 3. absorption of Yeast Cell at Two Different TimesTimeOD600 ReadingConcentration (cells/ mL)247pm0.021A210,000459pm0.043 A430,000It has been determined that an OD600 value of 1.0 is suasion to contain roughly 1 x 107 cells/ mL.An OD600 value of 0.021 A will contain a concentration of 210 000 cells/ mL.An OD600 value of 0.043 A will contain a concentration of 430 000 cells/ mL.Figure 2. This graph shows the change in absorbance of yeast culture at 0 and 120 minutes. Equation to represent growth is calculated and shown above. designing Doubling time Formula for growth of yeast is y=0.0002x + 0.021. given initial absorbance reading of 0.021, doubled concentration should give reading of 0.042. Therefore use y= 0.042, where x means time in minutes0.042=0.0002x + 0.021 where x = 105. Therefore it was prime that doubling time is 105 minutes.Part BYeast cells dyed with methylene blue stainFigure 3. These are some of the cell types that were observed when the overnight culture was stained with methylene blue under 40x gush.It was found that roughly a third had a stained positive for a nucleus. None of the cells appear to be multi-nucleate. No vacuoles were observe eitherYeast cells dyed with neutral redFigure 4. These are some of the cell types that were observed when the overnight culture was stained with neutral red under 40x magnificationIt was found that over 90% of the cells stained positive for a nucleus. M any(prenominal) of the cells appeared to be multi-nucleate and budding as well. It appeared that 1 or 2 vacuoles appeared to be sight per cell.Part CTable 3. Absorbance Results for the BioRad Protein Determination Assay12345678910A0.270.1910.210.2010.1960.370.3970.4040.3690.036B0.2460.240.3030.1920.2260.2450.2720.3720.2520.035C0.230.2630.2480.2940.0370.0360.0360.0360.0360.036D0.2560.2270.250.2770.0350.0350.0350.0350.0350.034E0.2460.1820.2420.2150.4740.3620.3060.3890 .4820.035F0.2890.3490.2850.2460.2990.2640.3470.7380.2030.036G0.2030.2540.3210.2490.0350.0350.0350.0370.0350.036H0.20.2610.2630.2740.0340.0340.0350.0350.0350.03411 12A 0.036 0.037B 0.035 0.035C 0.037 0.038D 0.035 0.044E 0.036 0.035F 0.035 0.035G 0.035 0.035H 0.035 0.034Figure 4. This graph shows absorbance readings of standard solution using BSA at different concentrations.Calculating concentrations of LSP, HSS, and Mito. Equation for concentration of solution based on absorbance reading was determined based on above graph. Equation yielded was y=0.1264x + 0.2159Average absorbance readings LSP was 0.245, HSS was 0.249, and MITO 0.289. Using found readings as y for above compare we calculated concentration of proteins in apiece sample.LSP 0.245 = 0.1264x + 0.2159Therefore x = 0.230 mg/mLHSS 0.249 = 0.1264x + 0.2159Therefore x = 0.262 mg/mLMITO 0.289 = 0.1264x + 0.2159Therefore x = 0.551 mg/mLDilution factor needed to get fraction to 2g/mLLSP (0.230 mg/mL)(0.1mL)= (0.0230 mL)(1/x mL) (10 dilution factor) = 2mg/mLTherefore x= 0.115 mLHSS (0.262 mg/mL)(0.1mL)= (0.0262g)(1/x mL) (10 dilution factor) = 2mg/ mL Therefore x= 0.131 mLMITO (0.551 mg/mL)(0.1mL)= (0.0551g)(1/x mL) (10 dilution factor) = 2mg/ mLTherefore x = 0.2755 mLExperiment 3Figure 5. PVDF Membrane after proteins are transferred over from gel after electrophoresis. Our group is remaining view (D.C, J.H)Experiment 4Figure 6. Membrane after detecting solution had been added over 10 minutes ago. Bands on right hand side are the standardFigure 7. This graph shows the distance travelled by each protein in the standard mix against the Log(Mw) on semi log typographyTable 4. Cytochrome c Oxidase (COX) Activity AssaySampleAbsorbanceAt 0 sec (OD)Absorbance after 20 sec (OD)Change in absorbanceChange in Concentration(mol/mL)COX activity (mol/ L/min)Blank0.5250.525000LSP (1)1.2591.2530.0060.21430.6429LSP (2)1.2721.2640.0080.28570.8571HSS (1)0.4930.4910.0020.07140.2143HSS (2)0.4960.4910.0050.17860.5257MITO (1)0. 5530.557-0.004-0.1429-0.4286MITO (2)0.5370.5350.0020.07140.2143Table 4. This shows theSample calculation for COX activityChange in absorbance = Absorbance at 0 sec Absorbance after 20 sec1.259-1.253 = 0.006Change in concentrationA = x b x c0.006 = 28mM-1-cm-1 x 1 cm x cTherefore c = 0.0002143 mM = 0.2143 molAssuming volume of stress is 1.0 mL, change in concentration is 0.2143 mol/mLCOX activityCOX activity = change in concentration / time0.2143 mol/mL / (1/3 min) = 0.6429 mol/mL/minFigure 8. in writing(p) representation of COX activity in LSP fractionsFigure 9. Graphical representation of COX activity in HSS fractions Figure 10. Graphical representation of COX activity in MITO fractions DiscussionGalactose was used over glucose as a carbon source for our yeast cells. This is because we wanted to determine if the mitochondria was functional in our yeast cells. Different yeast strains will use different metabolic pathways in presences of each. When glucose is used as a carbon sou rce the yeast cells will generate ATP via fermentation, whereas when Galactose is used the cell will perform oxidation. This is important to observe as different yeast strains will have varying levels of cytochrome c usage based on that.To visually determine if cytochrome c will be utilized by the cell we can look at the fractionation experiment earlier. When separating for the MITO fraction if one had found a messy pellet it would have indicated that the mitochondria was not intact while solid pellets would indicate the mitochondria was intact. If one did take place a messy pellet it could have been the result of differences in fractionation techniques, cells being lysed prior, or something had disturbed the cell in transport. For our experiment we had found the mitochondria to be intact, which is a strong indicator that the mitochondria for our sample was present.To reach cytochrome c oxidase (COX) we used diferential centrifugation which seperates objects based on size and densi ty, where larger molecules such as the intact cells will settle at the bottom of a tube while mitochondria which is smaller would remain in supernatant. This is also why we had separate centrifugations, to get samples with intact cells and samples with intact mitochondria. Density gradient centrifugation is also a widley used technique that seperates based on density. In that case we would see multiple bands form in tubes with densest molecules throng at the bottom and bands above it with less dense molecules.Experimentally we found yeast doubling time to be 105 minutes (1.75 hours) when inoculated in YPD (1% yeast extract, 1% peptone, 2% glucose). It has been determined in some(prenominal) other experiments that Saccharomyces cerivisiae has a doubling time of 1.69 hours (Deak, 2008). The difference could be attributed to many factors such as environment (amount of light, heat, and etc) and growth substrates used. But the difference is not very large and would still be considered to conform to literature results.During the methylene blue staining of yeast cells it was noted that roughly a third of yeast cells contained a nucleus but it did not seem to be multi nucleate. While the neutral red stains showed that many cells appeared to be budding with one or 2 vacuoles present per yeast cell. These findings fall in line with what is normally expected from yeast cells as they do have vacuoles in their cells (Armstrong, 2010). Furthermore results also fall in line with yeasts having nucleuses but not being multi nucleated (Roberts and Ganesan 1959)One thing that may have affected a major portion of the experiment was determining the concentration of each respective LSP, HSS, and MITO fraction and diluting it to 2mg/mL. it is important to note that during pipetting steps to get each sample that the suspensions be homogenous beforehand otherwise you may be taking up different components of the fraction and missing others depending on how deep the pipet was inserted . During the remainder of the experiment it was found that after gel transfer to PVDF membrane and during western blotting that very few to no proteins were showing up. If low concentration of protein was a factor then it would most likely be traced back to this step. Many reasons can be attributed to this for instance, poor pipetting technique, the fractions were not homogenized properly before pipetting or even the dilution factor could have been incorrect. As noted during the material and methods during the preparation of experimental samples which would to create our protein concentration standard wave we had run out of BSA STD and demand taking some from another group. When we created our protein concentration standard curve it came out completely odd, having unexpected drops in absorbance readings. The expected result was a linear curve where a higher BSA STD concentration would have led to a higher absorbance readings. Due to the change in BSA STD this may have had a differ ent concentration due to being taken from a different location in its container it could have had a different concentration. Thus causing inconsistencies for our standard curve. As the standard curve was deemed incorrect afterwards any protein concentration calculations based on it would have been flawed, leading to incorrect dilutions. If the dilutions been calculated incorrectly, as they most likely were, on that point is the chance that the protein fractions would have been over diluted leading to not enough protein to be present for visible bands for the gel electrophoresis and western blotting.For the gel electrophoresis SDS was included in solubilisation buffer to give proteins inserted into the wells a negative charge so that when a charge was applied they would run to the other end of the gel and to help unfold the protein so that it would be able to go through the gel. For this experiment a 12% gel was used in the interest group of saving time because a 15% gel would have caused the proteins to go through it slower leading to a lower occlusion of identifiable protein bands.The heading of transferring proteins from gel to PVDF membrane was to be able to visualize the movement of proteins on the gel after electrophoresis. To accomplish this we applied Ponceau stain to the membrane to increase the resolution of the bands and to ensure equal amounts of proteins are loaded onto the gel (Al-Amoudi et al., 2013). It was found that after proteins had been transferred to PVDF membrane that we had very few bands show up for the solubilizing buffer lane, both LSP sample lanes, and both HSS sample lanes. Bands did appear for both MITO samples, however, it appears that got smeared across the gel, bleeding over to other wells. This could have been the result of diluting samples in the wells for reasons noted above, the SDS gel would have been poorly constructed and contributed to the smearing, and poor electrode contact on the gel might have blocked the gel fro m having proper electrical charge. Issues could have also arisen during transfer of proteins from gel to membrane. Air bubbles could have been present during transfer which would have prevented any protein from being transferred as proteins cannot move through air. Additionally, poor folding between membrane and the gel could have attributed to smearing of MITO samples.The purpose of the western blotting was to be able specifically detect for the presence of biotinylated COX proteins. In order for a cell to express a biotinylated protein it needs to be able to take up foreign DNA, be able to properly fold COX-biotin fusion protein, the cell needs to be able to recognize the BSS signal fused to C terminus, and be able to translate COX and biotin together (moving stop codon so that it doesnt not stop middle(prenominal) across the other.)It was found that after western blotting our membrane with protein fractions that no bands had appeared even after 10 minutes of membrane being in co ntact with detecting solution. This led to Figure 7. The chart showcasing the relative distances that proteins have travelled is blank as a result. This would imply that when the blocking solution was added that it managed to block the entire membrane (and any present proteins included) from interacting with the probe. However Tween-20 was used to wash excess reagent. So the milk most likely would not have been able to bind to any protein after introduction of Tween-20. Therefore the lack of data could be attributed to low concentrations of protein on membrane for reasons as noted above. Referring to Figure 5 the only proteins that were found on the membrane after were MITO which shows that there would have been no LSP or HSS for probe to bind to, whereas for present MITO sample the concentration may not have been high enough and as a result some of it could have been washed out by the methanol step causing concentration of MITO to be so low that it could have been blocked by the bl ocking solution. There is also a possibility that our yeast samples were not able to biotinylate the COX protein at all which could explain why there were no bands occurringLooking at COX activity graphs for LSP, HSS, and MITO they seem to follow what is expected except for HSS. COX was used as an identifying marker for identifying subcellular fractions containing COX because it is an integral membrane protein for the inner membrane space. If COX activity is present then that would indicate that the mitochondria is intact and functioning. These samples should have seen increase COX activity as cytochrome c was introduced into the fractions which provide electrons to the COX protein allowing it to pump proteins and reduce oxygen to water. Both MITO and LSP experienced increased COX activity as shown by figure 8 and 10 respectively. This falls in line with what was expected with the MITO fractions experiencing higher levels of COX activity then the rest as the cytochrome c had less of a distance to travel to reach inner mitochondrial membrane space than LSP. LSP should have a signal because it would contain intact yeast cells which have mitochondria Alberts et al., 2015). Therefore LSPs rate of COX activity should be lower because the cytochrome c would have harder time reaching mitochondria. This is shown by Figure 10 having steeper reaction times than Figure 8 and 9. This reaction utilized Deoxycholate (DOC) to speed up the reaction which is why it was only done in 20 second intervals as DOC solubilizes with cytochrome c so that it can enter the mitochondria to interact with COX. If reactions were tested too long after DOC was added then the reaction would have undone before being able to measure absorbance. The one that did stand out was the HSS fraction which appeared to experience negative COX activity or none at all. This was expected as it should have all the remaining parts of the cell that werent the mitochondria, lysosomes and peroxisomes Alberts et al., 2015). This would indicate that these samples did not have an intact mitochondria with a COX protein to interact with cytochrome c. this could be explained by theIn conclusion it was found that our yeast strain A2 is the wild type strain. This is because during initial centrifugation the resulting pellet was solid indicating intact mitochondria. 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