On Studybay you can order your academic assignment from one of our 45000 professional writers. Hire your writer directly, without overpaying for agencies and affiliates!
Check price for your assignment
SIGNAL TRANSDUCTION IN YEAST
For this experiment, we have chosen to address a scientific topic that is currently the focus of much research: how, at the molecular level, do cells respond to an external signal? Response to an external signal is important at all levels of biology, from bacterial chemotaxis to the development of multicellular organisms to understanding the complex events that leads to oncogenesis.
All cells have ways of responding to their environment, and the responses of cells have the same basic general components:
a) a signal = many different ones are possible like: nutrients, noxious substances in the media (bacteria) or factors released by other cells (cytokines)
b) a receptor for the signal = proteins in the cell membrane that bind the nutrients/ factors
c) transducers = proteins within the cell that change as a result of the binding of the signal to the receptor and pass the signal along to other proteins
d) a response = or a change in the behavior/metabolism of the cell (examples : changes in transcription of certain genes or the activity of enzymes that give the appropriate response to the signal)
In the case of bacteria, binding of nutrients to a receptor leads to a change in the beating of the flagella that move the bacteria. This allows the bacteria to move forward, towards the nutrients.
The response of cells to cues in the environment is called signal transduction. It is a very important area of current research. Physiological responses to hormones or drugs involve signal transduction pathways, as do the changes during development of a multi-cellular organism from a single cell. Defects in signal transduction pathways are also often found when a cell becomes a tumor cell. Understanding signal transduction pathways is critical to our understanding of biological processes from development to cancer. How do scientists find out how a signal transduction pathway works? How do they identify the proteins that act as the protein receptor and the transducers? These pathways are usually complicated and can have a number of proteins or chemical signal involved. One way scientists have dissected the components of signal transduction pathways is to use genetics. They generate mutations in organisms or cells by treating them with DNA-damaging substances, then look for cells that can no longer respond to the particular environmental cue. Characterizing the specific proteins that are affected by these mutations gives insights into the components of the entire pathway.
One pathway that has been extensively studied as a model for understanding signal transduction occurs in Saccharomyces cerevisiae (baker’s yeast). This single-celled organism can exist in 2 different haploid “cell types” called (a) and α cells. In the absence of the other cell type, each type will divide mitotically, proceeding thru one full cell cycle in 1 hour. Such vegetatively dividing cells are easily distinguished microscopically. Actively dividing cells appear as large spheres with smaller or equal-sized spheres budding from their periphery. If the 2 cell types, (a) and α cells , are in close proximity, they can mate by fusing to form a diploid cell. Mating between 2 cells requires the following things:
1) arrest of the cell cycle in the G1 phase before DNA is synthesized
2) changing cell shape (called shmooing) to bring the cells closer together
3) increasing transcription and translation of genes involved in cell fusion
How does an (a) cell know there is an α cell in the vicinity with which it can mate? The α cells release a 13 amino acid peptide “factor” called α factor. This peptide binds to a receptor, α factor receptor, on the cell surface of the a cells. Binding of the α factor to the receptor results in all 3 of the effects listed above. At the same time, a cells also produce a factor, which binds to an a factor receptor on α cells.
This exercise will examine the effect of α factor on (a) cells.
The α factor belongs to the G-protein-linked family of cell surface receptors. When α factor binds to the receptor on the surface of an (a) cell (indicating the proximity of a potential mate), a G-protein complex within the cell is activated. The activated G-protein complex subsequently turns on a protein kinase enzyme, which has the ability to phosphorylate other cell proteins. Among the targets of the newly activated protein kinase is a protein that affects cell cycle arrest, and a transcription factor that is specific for genes involved in mating. Both the cell cycle arrest factor, and the transcription factor specific for mating genes lie dormant in the cell until they are activated by the addition of phosphates (by kinases). This is an event that occurs in response to the binding of α factor at the cell surface. What we know about the signal transduction pathway for yeast mating is in the accompanying diagram.
NOTE: the signal transduction pathway branches at the end, with 1 transducer protein (the arrest factor) stopping the cell cycle, and a different transducer protein (the transcription factor) affecting transcription.
All levels of this complex signal transduction pathway involve proteins: the α factor and its receptor, the G protein, the protein kinase, the cell cycle arrest protein, the transcription factor, and several intermediates. Since the proteins are coded for on genes and these are subject to mutations (DNA damage), any mutations may render the protein products of the genes nonfunctional.
• You and your partners will be given 3 strains of yeast. One of these strains is a normal (wild type) a cell. Each of the other 2 strains contains a mutation that inactivates one of the genes involved in the signal transduction pathway for mating.
• OBJECTIVE: Figure out which gene(s) could contain the particular mutation based on what
you know about this signal transduction pathway. (STP)
Yeast strains--- a ura3- wild type
a ura3- “STP mutant X”
a ura3- “STP mutant Y”
All 3 strains contain a mutation in the URA3 gene. This gene codes for an enzyme required for the synthesis of uracil. Strains with mutations in URA3 (ura3- strains) cannot grow in media lacking uracil.
Summary of This Experiment:
A) Transformation of yeast with plasmid:
PURPOSE: IN ORDER to amplify the alpha factor response in transcribing fusion genes (FUS1)
It is a simple task to put a plasmid into a yeast cell, and select for those yeast cells that contain the plasmid using standard plating techniques. For instance, if a cell that can not grow in the absence of uracil (ura3- strain) is transformed with a plasmid containing the URA3 gene, then ONLY cells that have taken up the plasmid can grow on plates that do not contain uracil.
B) Assay for B-galactosidase enzyme:
Using a simple enzyme assay can determine quantatively how much B-galactosidase enzyme is being made by yeast cells. Because the promoter for mating genes has been linked to the gene for B-gal, an assay for B-gal activity is a good indicator of transcription of mating genes.
Assay for cell cycle progression: cell growth
The cell cycle is a combination of the stages thru which a cell passes during division. The cycles include (M) for mitosis (includes cytokinesis), and interphase. Interphase is subdivided into: a growth period prior to DNA replication (G1), a synthesis stage (S), and a 2nd growth stage that preceded cell division (G2). S. cerevisiae is a “budding yeast” that divides by growing a bud on the cell (see figure 2). These cells can be seen via the microscope. When cells are in G1, they do not contain a bud (unbudded). When they enter S phase, they begin budding. As they progress thru the cell cycle, the bud grows until, at the end of the G2, the cell looks like a “dumbell”. During M phase, the bud separates from the cell to giving two new cells. The percentage of unbudded cells is a measure of the number of cells not progressing through the cell cycle, but arrested in G1.
As you perform your experiment you should consider:
-how could you compare transcription and cell cycle arrest for yeast cells in the presence and absence of a factor?
-how would mutations in the receptor affect transcription and cell cycle arrest?
-how would mutations in the G protein affect transcription and cell cycle arrest?
-how would mutations in the protein kinase complex affect transcription and cell cycle arrest? .
-how would mutations in the transcription factor affect transcription and cell cycle arrest?
-how would mutations in the arrest factor affect transcription and cell cycle arrest?
WEEK 1- TRANSFORMATION OF YEAST WITH PLASMID DNA
Recombinant DNA molecule: PBH315 plasmid (see figure)
Plasmids are small circular molecules of DNA that are capable of being replicated in a cell. The pBH315 plasmid was made by combining the promoter region of a gene involved in cell fusion during mating (the FUS1 gene), and the gene for the enzyme B-galactosidase. (Promoters are DNA sequences of DNA that direct the beginning of transcription of a gene by serving as a binding site for RNA polymerase ). Instead of directing the transcription of the FUS1 gene, the promoter in pBH315 directs the production of B-galactosidase. The promoter is only active when α factor is detected by the cell, and in its active state, B-galactosidase is transcribed, and translated. Therefore, the amount of B-gal produced is a measurement of the amount of transcription directed by the promoter. B-galactosidase activity can be measured in a simple enzyme assay.
This plasmid also contains a functional copy of the URA3 gene, and any yeast cell that contains this plasmid, even those lacking a functional enzyme of uracil synthesis (ura3-), can grow in the absence of uracil in the media. This enables one to identify ura3- yeast cells that have taken up the plasmid.
1. get one plate of each strain of yeast cells (WT, X, Y).
2. Label three microfuge tubes with the designations for the three strains (W'T, X, Y). Scrape 3-4 colonies of yeast off each plate with a sterile wooden stick. Transfer these cells to the appropriately labeled microfuge tube by rolling the stick on the inside wall of the tube. Please try to keep this procedure as sterile as possible.
3. Add 100 ul of One-Step- Buffer to each microfuge tube and resuspend the cells by mixing with the pipette tip.
One Step Buffer contains lithium acetate and polyethylene glycol. The lithium salts change the cell envelope to make the cells permeable to DNA. The polyethylene glycol precipitates the plasmid and single stranded DNA you add in step 4 onto the yeast cell surface.
4. For each transformation,. add 5 ul of pBH315 plasmid and 5 ul of denatured (carrier DNA) salmon sperm DNA to the yeast cells.
The single stranded DNA (salmon sperm DNA) is referred to as carrier" DNA”; it is necessary for high frequency transformation of the cells with the plasmid, although the exact mechanism is unknown.
5. OPTIONAL: "no DNA" control for one lab section should also be done for each str
ain. This will contain only the salmon sperm DNA and not the plasmid DNA. What is the purpose of the "no DNA" control?
6. Incubate the transformation mixture in a 42°C water bath for 30 minutes, mixing occasionally to resuspend the yeast cells in the lithium acetate/PEG buffer.
It is thought that heating the cells creates a thermal gradient that enhances the uptake of the DNA.
7. Recover the cells by spinning them for 5-10 seconds in the microcentrifuge. Discard the supernatant ; add 0.1 ml TE (100ul) (TRIS-EDTA) buffer to each tube and Resuspend the cells thoroughly before plating The ENTIRE cell volume!!!!
Lithium salts are detrimental to yeast cell growth. Resuspending the cells in TE dilutes the remaining lithium acetate.
8. Put each of the transformation mixtures on a separate SD-ura plate and spread the cells evenly around the plate using a bent glass rod. Again keep this as sterile as possible. Flame the glass rod and allow it to cool before spreading the cells. Label the plate with your names, the yeast strain (WT , X, or Y) and +/- plasmid DNA. .
SD-ura plates contain a synthetic growth medium lacking the nucleoside uracil. Untransformed cells will not grow on these plates because they contain a mutation in the ura3 gene. However, transformed cells will grow, because the pBH315 plasmid contains a normal version of the URA3 gene which, once the plasmid enters the nucleus, will be transcribed and subsequently translated into the URA3 gene product.
9. Incubate the plates at 30·C to allow colonies to grow for approximately 2 to 3 days. STOP
GROWTH OF CELLS FOR ASSAY (PREFORMED BY LAB ASSOCIATE)
1. The day before assaying the yeast cultures, they must be transferred to liquid media. Using separate sterile wooden sticks remove a single colony from each culture plate and add to separate test tubes each containing 1 ml of SD-ura liquid media. Allow the yeast to grow overnight by placing the tubes on a roller in a 3O˚C incubator.
A single colony of yeast will generally grow "to saturation” overnight. This means that the cells will divide until they generate a culture that has as many cells as the media can support. At saturation, there are about 2 x10^8 cells/mI.
2. Approximately 5 hours before assaying the cultures, restart growth by adding 9 mls of SD-ura to each of the overnight cultures. Replace on the roller in the 3O˚C incubator.
3. Two hours before you wish to assay the cells, add ά factor as follows: remove 5 ml of each culture to a fresh culture tube; label with the culture name. Label one of each of the 5 ml cultures "+ factor" and the other "- factor". To the "+ factor" tubes, add 20 ul of the 100 uM α factor stock provided. You will now have 6 tubes.
4. Put all six cultures on the roller in the 3QOC incubator until time for the assays.
The cells require a period of time to detect and respond to the ά factor.
PROTOCOL: ASSAYING CELL DIVISION (BUDDING) :
S. cerevisiae divides by budding. We can tell something about where yeast cells are in the cell cycle by observing the size of the bud. Bud formation begins in the S phase, and the bud enlarges as the cell cycle progresses. After mitosis, the mother and daughter cells separate. In a rapidly growing yeast culture, cells will be found in all stages of the cell cycle. About 60% of cells will contain a bud, and buds of all sizes will be seen in the culture. If the cells have been depleted of nutrients or are under conditions that arrest their cell cycle in the Go or Gl phase, most cells will be unbudded, round cells. If cells are stopped right before mitosis, most cells will look like “dumbbells”, with two equal buds. Scientist who work on yeast used these observations to identify yeast strains containing mutations in genes that control the different stages of the cell cycle. These observations will also allow you to determine whether α factor is capable of arresting the cell cycle in Gl for a particular strain of yeast.
DO THIS BEFORE YOU BEGIN ANY ASSAY:!!!Label 2 sets of microfuge tubes with WT (+ and -) , X (+ and --) and Y (+and -) ; remove 0.5ml from each liquid culture and give this to the stduents doing the cell budding count;
1. After you have removed 0.5 ml of each culture into one set of labeled tube ; next from this 0.5mls remove 200 ul of each culture into the next set of labeled tubes. (Keep 300uL for any future use (in case of need to repeat). ; to this 200ul liquid cell culture then add 200 uL of 0.5 M (EDTA decreases cell clumping.)
2. For each culture, begin by vigorously vortexing it to break up clusters of cells. Place a drop on a hemacytometer slide. Observe the yeast cells using the 4OX objective. Find a square containing 16 smaller squares. Designate an the cells in these 16 squares as budded or unbudded. (This may not be as easy as it sounds; sometimes mother cells and daughter cells will not be separated). Have your partner keep tract of how many cells you say are budded and how many are unbudded. Have your partner determine the number of budded and unbudded cell for a 16 –square grid on another part of the slide. Record a combined total for the number of budded cells observed for the culture and the number of unbudded cells observed for the culture. Each student should count a minimum of 50 cells for each culture.
3. Repeat this procedure for each of the six yeast cultures.
4. Record the number of budded cells and the number of unbudded cells for all the cultures on the class data sheet
PROTOCOL: MEASURING B-GALACTOSIDASE :
(1) Estimate the cell density (TURBIDITY) of the culture by measuring the absorbance at 600 nm. If the absorbance is over 2 OD, use a portion of the culture to make a 10-fold dilution of the cells and repeat the absorbance measurement. This absorbance measurement will be used in the calculation of B-ga1actosidase. If you made the above dilution, you must multiply the absorbance measurement by 2 before using it in the equation.
Cell density can be estimated by measuring the amount of light scattering that occurs for a particular culture of microorganisms. The amount of scattering that occurs depends on the size and shape of the microorganisms as well as on their number. The relationship between light scattering and absorbance is only linear to a value of about 1.
2. return the liquid culture from the cuvettes to the original culture tube; mix again; then remove 1.5 ml to a separate microcentrifuge tube. Pellet the yeast cells by centrifuging them in the microcentrifuge for 30 seconds. Remove and discard the media.
3. Resuspend the cell pellet in 150 uL of Z buffer; then add 50 uL of chloroform and 20 uL of 0.1% SDS.
Vortex the tubes vigorously for 30 seconds.
Z buffer is a solution of buffered salts favorable for B-galactosidase enzyme activity which also contains B- mercaptoethanol (a sulfhydryl reducing agent). The chloroform (organic solvent) and SDS (detergent) act to permeabilize the yeast cells so that we can add the substrate for B-galactosidase enzyme to them.
4. Add 0.7 mL of ONPG solution (1.2 mg/ ml ONPG in Z buffer) to each tube VERY QUICKLY; close caps and VERY QUICKLY mix the ingredients (all samples at once!); QUICKLY place the tubes in a 30˚ C water bath to begin the assay, and record the exact time at which you began the assay.
O-nitrophenylgalactopyranoside (ONPG) is a colorless compound which is c1eaved by
B-galactosidase to o-nitrophenol and galactose. The resulting o-nitrophenol is yellow, which means that the amount of yellow color is proportional to the B- galactosidase activity present in the cells. If there is a lot of B-gal enzyme in the cells, the reaction will turn yellow within a minute.
5. Continue the incubations at 30˚ C until a yellow color is observed. When the reaction appears yellow, QUICKLY remove the tube and add 0.5 mL of 1 M Na2CO3 to stop the enzyme reaction. Record the exact time at which you add the Na2CO3. If the reactions do not turn yellow, stop all of the rest of them after 10 minutes.
The Na2CO3 raises the pH of the reaction to 11, which inactivates B-galactosidase.
6. Remove the cell debris from the reactions by spinning the tubes in the microcentrifuge for 2 minutes. Remove the top 1 ml from each reaction tube and add it to 2 mls of .Z buffer in a cuvette . Measure the absorbance of the reaction at 42Onm using the spectrophotometer.
7. . The amount of B-galactosidase activity is generally expressed in Miller units:
Miller units = Abs(420) x 1000 x 3
Abs(600) x (volume culture) x (time of reaction in minutes)
The factor of 3 in the numerator of this equation is to correct for the dilution made when measuring the absorbance at 42Onm in step 7 The presence of Abs(600) and the volume of culture (1.5 ml for our reactions) in the denominator allows reactions done with different numbers of cells to be directly compared. Remember to multiply the Abs (600) by 10 if you used a dilution to make a reading. The number of Miller units for gene fusions in yeast can vary from 0.1 – 1000 units or more.
RESULTS: DATA TABLE
B-galactosidase activity (Miller Units)
(-) α - factor
( +) α - factor
Mean Percent Cells budded
(-) α - factor
( + ) α - factor
Each report SHOULD contain a thorough INTRODUCTION, which includes a BRIEF review of the components involved in this experiment (yeast plasmid etc.) and a BRIEF review of the genetics of the experiment.
Each report SHOULD contain a brief MATERIAL & METHODS section which list the components of the experiment (Materials: DNA’s, cells, plates, medias etc). This section should also contain, in paragraph form, a BRIEF outline of the 2-day experiment as performed by the student: Transformation step and B-galactosidase step & Assay for cell division.
Each report SHOULD contain a BRIEF description of the RESULTS, which may includes a table.
B-galactosidase activity (Miller Units)
(-) α - factor
( +) α - factor
Mean Percent Cells budded
(-) α - factor
( + ) α - factor
Each report SHOULD contain a DISCUSSION which summarizes and explains (in terms of the cell cycle and the genetics) the results: i.e.: for WT, for X-mutant for Y mutant (budding and B-gal expression in the presence and absence of factor, how the WT reacts in the presence or absence of factor :
--- how would mutations in the transcription factor affect transcription and cell cycle arrest?
-- how would mutations in the arrest factor affect transcription and cell cycle arrest?
--- how could you compare transcription and cell cycle arrest for WT yeast cells in the presence and absence of a factor?
In addition you SHOULD EXPLAIN:
-- how would mutations in the receptor affect transcription and cell cycle arrest?
-- how would mutations in the protein kinase complex affect transcription and cell cycle arrest? .
SIGNAL TRANSDUCTION IN YEAST
For this experiment, we have chosen to address a scientific topic that is currently the focus of much research: how, at the molecular level, do cells respond to an external signal? Response to an external signal is important at all levels of biology, from bacterial chemotaxis to the development of multicellular