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 fiberner Daryan ChanIntroductionYeasts ar important organisms due to their customs in everyday such as baking, making fermented foods and alcohol production (Steensels et al, 2014). Yeasts contri plainlye been so widely studied that it was one of the first organisms to clear its genome sequenced (Goffeau et al, 1996). Thus, Yeasts be more than capable of acting as a model organism for eukaryotes (Botstein et al, 2011). For this experiment we argon working with Saccharomyces cerevisiae. The mitochondria is the powerhouse for cadre, as it is the major production site of adenosine triphosphate for the cell. The upcountry mitochondrial space has an electrochemical slope, from which adenosine triphosphate is gene positiond by using 5 protein complexes create an electrochemical gradient to look in ATP production (Albert s et al., 2015). The follow6 assembly accepts electrons from cytochrome c and uses oxygen as the terminal electron acceptor to make water (Alberts et al., 2015). ATP synthase thence uses the resulting proton gradient made by those complexes to pump protons back into the mitochondria matrix and make ATP (Alberts et al., 2015).The objective of this experiment was to specialize if a refundn yeast judge was a wild type or a mutant with no follow6 employment.One of the major techniques to be utilise in this lab is subcellular secondtionation. This technique first lyses the cells and then uses motor(a) forces to separate particles by surface (Alberts et al., 2015). The centrifugal forces results in the denser particles moving away from axis of rotation creating a guesswork which contains the heavier particles and a supernatant which contains lighter particles (Alberts et al., 2015).A nonher major technique used was Gel cataphoresis. Gel electrophoresis is used so that a fract ion with multiple proteins pot be separated base on size and shape (Alberts et al., 2015). Protein fractions ar loaded on to wells in the change and an electrode is attached (Alberts et al., 2015). SDS page is popularly used because it can confer a negative shake up and linearize proteins being run through the colloidal jelly (Alberts et al., 2015). The proteins bequeath run through the jelly due to their negative direction (Alberts et al., 2015). A gestateard is used to declare oneself a reference to go down the sizes of the sampl proteins (Alberts et al., 2015).One of the other major techniques used in this lab was westerly sandwich blotting. After proteins are run on electrophoresis a label antibody is exposed to the electrophoresed fractions in a touch called immunoblotting in order to disclose presence of a detail protein (Alberts et al., 2015). The changeatinatin is exposed to a membrane 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 protein (Alberts et al., 2015). This process can detect very small amount of specific protein and is expedient for detecting changes of parsimony of a specific protein in a cell under various conditions (Alberts et al., 2015).To measure cytochrome oxidase application in this lab, we looked towards Beers Law which states that the talent of a come to the forecome to absorb light at a individual(a) wavelength is proportional to the assiduousness of solute in antecedent (Lukofsky et al, 2009). This show that absorbance and submerging are linearly related. Therefore, this would allow us to determine the rate of cytochrome oxidase drill in a sample.Materials and Methods try was consummateed according to protocols set by Department if biota, spend 2016, Biology 331 for experimentation 1 Subcellular fractionation of yeast cells, pg 2-8, Experiment 2 Yeast growth curve light microscopy protein determination, pg 1-4, Experiment 3 Polyacrylamide gel electrophoresis, pg 1-8, Experiment 4 Development of Western Blot cyclooxygenase body process Assay, pg 3-10, written by Dr. Dragana Miskovic where the experiment was performed with no deviations unless specifically noted (Miskovic, 2017)The hardly deviation occurred in experiment 2 where we had ran out of BSA STD and had to borrow from about other group. The borrowed BSA STD was not tried to hold up exact tautness as specific in lab protocol and whitethorn flip had different concentration.ResultsExperiment 1Yeast Strain Sample A2Table 1. Masses record for Lysing Yeast Cells SectionItemsMass (g)Mass of Centrifuge Bottle containing Yeast159.2Mass of Empty Centrifuge with Pellet49.89Mass of Pellet3.23 total of STE solution needed to resuspend yeast crack3.23g x 2= 6.46mLTable 2. Volumes recorded for the Aliquoting Yeast Subcellular Fractions SectionSolutionsTotal VolumeLPS3mL + 3.7mL= 6.7mLHSS3mL + 2.8mL= 5.8m LMITOcccL + 300L= 600LIt was in any case noted that by and by the MITO fraction was made the pellet was inherent and not messy fingerbreadth 1 Drawings of 50/100/200 L dye drops from pipetman and anticipate 1 mL dye dropExperiment 2Part ATable 3. Concentration of Yeast Cell at deuce Different TimesTimeOD600 ReadingConcentration (cells/ mL)247pm0.021A210,000459pm0.043 A430,000It has been determined that an OD600 value of 1.0 is thought 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. approach pattern 2. This graph shows the change in absorbance of yeast culture at 0 and 120 minutes. Equation to re bring in growth is calculated and shown in a higher place.Calculating manifold age Formula for growth of yeast is y=0.0002x + 0.021. given sign absorbance reading of 0.021, doubled concentration should give reading of 0.042. Therefore use y= 0.042, where x means season in minutes0.042=0.0002x + 0.021 where x = 105. Therefore it was plunge that doubling time is 105 minutes.Part BYeast cells dyed with methylene group blue stain compute 3. These are some of the cell types that were observed when the everywherenight culture was varnished with methylene blue under 40x magnification.It was embed 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. Many of the cells appeared to be multi-nucleate and bud as well. It appeared that 1 or 2 vacuoles appeared to be detected per cell.Part CTable 3. Absorbance Results for the BioRad Protein Determination Assay12345678910A0.270.1910.210.2010.1960.370.3970.40 40.3690.036B0.2460.240.3030.1920.2260.2450.2720.3720.2520.035C0.230.2630.2480.2940.0370.0360.0360.0360.0360.036D0.2560.2270.250.2770.0350.0350.0350.0350.0350.034E0.2460.1820.2420.2150.4740.3620.3060.3890.4820.035F0.2890.3490.2850.2460.2990.2640.3470.7380.2030.036G0.2030.2540.3210.2490.0350.0350.0350.0370.0350.036H0.20.2610.2630.2740.0340.0340.0350.0350.0350.03411 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 found 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 equation we calculated concentration of proteins in individually 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 confirm 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 afterwards proteins are raptusred over from gel after electrophoresis. Our group is left side (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 infinite traveled by each protein in the standard mix against the Log(Mw) on semi log paperTable 4. Cytochrome c Oxidase ( be) Activity AssaySampleAbsorbanceAt 0 sec (OD)Absorbance after 20 sec (OD) exchange in absorbanceChange i n 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 tally 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 assay 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. Graphical 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 paper 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 source 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 ready 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 clear indicated that the mitochondria was not full epoch steadfast pellets would indicate the mitochondria was intact. If one did find a messy pellet it could bemuse been the result of inequalitys in fractionation techniques, cells being lysed prior, or something had grisly the cell in transport. For our experiment we had found the mitochondri a to be intact, which is a conceptive 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 density, where larger molecules such as the intact cells will sink at the bottom of a tube while mitochondria which is smaller would stop 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 gathering 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 other experiments that Saccharomyces cerivisiae has a doubling time of 1.69 hours (Deak, 2 008). 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 literary works 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 reckon 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 presentation up. If low concentration of protein was a factor then it would near seeming be traced back to this step. Many reasons can be attributed to this for instance, unworthy pipetting technique, the fractions were not homogenized correctly before pipetting or however 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 curve we had run out of BSA STD and required taking some from another group. When we created our protein concentration standard curve it came out completel y odd, having unthought-of 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. payable to the change in BSA STD this may have had a different 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, lead story to incorrect dilutions. If the dilutions been calculated incorrectly, as they most likely were, there is the chance that the protein fractions would have been over diluted leading to not liberal 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 exsert the protein so that it would be able to go through the gel. For this experiment a 12% gel was used in the interest of saving time because a 15% gel would have caused the proteins to go through it slow leading to a lower dissolver of identifiable protein bands.The purpose 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, two LSP sample lanes, and both HSS sample lanes. Bands did appear for both MITO samples, however, it appears that got smeared across the gel, release over to other wells. This cou ld 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 dawn on the gel might have forfended the gel from 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 sleep with 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 central 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 contact 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 backwash excess reagent. So the milk most likely would not have been able to bind to any protein after introduction of Tween-20. Therefore the omit 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 blocking 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 change magnitude 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 answer generation 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 finished 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 remain 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. Furthermore during COX assay the MITO strain indicated an active COX as shown by its increase in activity, proving that A2 was in fact a wild type strain with functioning mitochondria.ReferencesAl-Amoudi, M.S., Salman, M., Al-Majthoub, M.M., Adam,Abdel Majid A., Alshanbari, Naif A., Refat, Moamen S., (2013) Res Chem Intermed 41 3089. doi10.1007/s11164-013-1417-4Alberts, B. et al. (2015). Molecular Biol ogy of the Cell Sixth Edition. New York, NY motley information, Taylor Francis Group.Armstrong, John. Yeast vacuoles more than a model lysosome. Trends in Cell Biology 20.10 (2010) 580-85. Web. 13 Mar. 2017.Botstein, D., Chervitz, S. A., Cherry, J. M. (1997, August 29). Yeast as a Model Organism. Retrieved touch 12, 2017, from https//www.ncbi.nlm.nih.gov/pmc/articles/PMC3039837/Deak, Tibor. Handbook of Food Spoilage Yeasts, Second Edition. Contemporary Food Science (2007) 50-51. Web. 13 Mar. 2017.Glerum, Moira Miskovic, Dragana (2017). 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