7.06_2004_PS4key - 7.06 Spring 2004 PS 4 KEY 1 of 10...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.06 Spring 2004 PS 4 KEY 1 of 10 Problem Set #4 Question 1. You are studying the DNA damage response pathway in a new species of yeast that your lab studies. In Particular, you are interested in learning the molecular mechanism behind cell cycle arrest due to DNA damage. You know that the DNA damage response causes an arrest in cell cycle progression and involves the cyclin dependent kinase CDK76. In vitro studies have shown that CDK76 is active when phosphorylated at 2 Tyr’s and is inactive when phosphorylated at a Thr and a Tyr. You suspect that DNA damage causes inactivation of CDK76 which leads to cell cycle arrest. Luckily you have CDK specific antibodies that recognize both the unphosphorylated form and the Thr and Tyr phosphorylated form. You decide to do a western blot with cells that have not been treated with a DNA damaging agent and cells that have been treated with a DNA damaging agent. a) Describe a step by step procedure for running a western blot for these two cell treatments. ANSWER: 1. Take your two samples and lyse the cells. 2. Run the cell lysates on separate SDS-PAGE gels. 3. Transfer to membrane and probe blot with Ab specific for both unphosphorylated CDK and Thr and Tyr phosphorylated CDK76. 4. Add secondary Ab to each blot that allows visualization. 5. Visualize blot. Below is the result of your western blot: 7.06 Spring 2004 Ladder CDK76 No DNA Damage PS 4 KEY 2 of 10 CDK76 W/ DNA Damage * Phosphorylated Thr and Tyr CDK76 § § * § Un-Phosphorylated CDK76 Now you have evidence that CDK76 is phosphorylated in response to DNA damage. Furthermore you suspect that an inhibitor is binding CDK76 so you want to find binding partners of both active and repressed phospho-CDK76. b) How would you go about finding intracellular binding partners of CDK76? Design a step by step experiment listing any controls you would include in your experiment. ANSWER: You would use expression cloning to find binding partners of Phosphorylated CDK76. To do this you would clone a cDNA library (from WT yeast) into _-phage. Then you would infect a bacterial plate with your phage. Next you would radio-label phospho-CDK76 to use as your probe (you can do this by using _-32P labeled ATP). Once plaques start to form on bacterial plate, you transfer all the proteins on the plate to a membrane and probe the membrane with your labeled phospho-CDK76. Any spots that light up are binding partners of phospho-CDK76. You can then go back to the plaque, extract the phage and sequence the cDNA that was inserted into its genome. A good control to use would be a radio-labeled unphosphorylated CDK76 (you can label with 125I or 35S) probe. This probe would control for non-specific binding to CDK76. 7.06 Spring 2004 PS 4 KEY 3 of 10 Your screen picked up four potential binding partners of phospho-CDK76. Luckily these candidate binding partners have been previously described proteins, so you can purchase antibodies against each protein. You want more evidence that these proteins indeed bind CDK76 so you decide to IP (immuno precipitate) phospho-CDK76 using an antibody that specifically binds to all species of phosphorylated CDK76 (i.e. it will recognize both 2 Tyr and Thr and Tyr phosphorylated CDK76). So you IP cell lysates from yeast cells treated with a DNA damaging agent and from cells not treated with a DNA damaging agent. After you IP CDK76 you run a western with antibodies specific for the 4 candidate binding partners your expression cloning screen picked up. The results of this Co-IP are shown below: Ladder Protein W Protein X Protein Y Protein Z Ladder Protein W No DNA Damage Protein X Protein Y Protein Z DNA Damage c) Based on the data you have collected up to this point, what do these results suggest? ANSWER: This Co-IP suggests that proteins W, X and Z all bind to active CDK76 under non-DNA damage conditions, and that under DNA damage conditions protein Y binds to Thr and Tyr phosphorylated CDK76. 7.06 Spring 2004 PS 4 KEY 4 of 10 Luck is once again on your side because it just so happens that a mutant of protein Y exists. This protein Y mutant fails to arrest cell cycle progression in response to DNA damage. You decide to repeat your phospho-CDK76 Co-IP with wild-type cells and protein Y mutant cells treated with a DNA damaging agent and probed with an antibody against protein Y. The results of the Co-IP show the loss of the protein Y band in the mutant. d) With this data what do your results suggest? ANSWER: With this added piece of data, it seems that protein Y acts as an inhibitor of CDK76 under DNA damage conditions. Now you want to know if cell cycle arrest is in fact due to loss of CDK76 activity. e) How would you test this? Design a detailed experiment. ANSWER: You would perform a kinase assay for CDK using histone H1 as a substrate. IP active CDK and incubate with substrate and _-32P-ATP. Then separate on an SDS-PAGE gel and visualize the autoradiogram. Question 2. While performing a mutagenesis screen designed to recover yeast mutants defective in the cell division cycle, you recover a cell cycle mutant that is temperature sensitive for growth at 37°C (it has a wild-type phenotype at 25°C). a. Describe two methods by which you can try to identify the cell cycle stage where this mutant may be arrested. 7.06 Spring 2004 PS 4 KEY 5 of 10 1. FACS analysis of the DNA content would determine whether the arrest was pre – or post- S phase. 2. Immunoflouorescence of spindles can place cells in G1/S, metaphase, or anaphase. b. After arresting the mutant with alpha factor and releasing the cells from the arrest at 25°C and 37°C, you collect samples at different time points for some FACS analysis. Time after release from alpha factor 25°C 0 min 37°C 1C 2C 1C 2C 40 min 1C 1C 2C 2C 90 min 1C 2C 1C 2C 120 min 1C 2C 1C 2C 7.06 Spring 2004 PS 4 KEY 6 of 10 What conclusions, if any, can you make about where in the cell cycle the mutant has arrested? ANSWER: The only conclusion that can be made is that cells arrest with a 2N DNA content, which means that the arrest is somewhere between the end of S phase and telophase/cytokinesis. c. How could you more specifically identify the cell cycle stage at which the mutant is arrested? Why is this technique more informative than a FACS profile? ANSWER: Visualizing the spindle by immunofluorescence can determine whether cells are in G1/S, metaphase, or anaphase. This technique has higher “resolution” than FACS analysis which can only tell you if cells have passed S phase. d. You place an order for anti-tubulin antibodies from Acme Inc. so you can examine the spindle morphology of your mutants arrested at 37°C. Unfortunately, Pinky, the rat used to produce anti-tubulin antibodies has recently succumbed, creating a year-long back-log for anti-tubulin antibodies (no other companies produce this specific antibody). Can you think of another way by which one can examine whether cells are in G2, metaphase, or anaphase? Are there “molecular markers” (e.g. proteins that are produced, or degraded or morphological changes in the cell etc.) for each of these cell cycle stages that you can follow? ANSWER: There are several ways to answer this question. We know that a key morphological transition during the cell cycle is the separation of the cells’ DNA content. As a result, cells in anaphase can be identified by the presence of two separate DAPI stained masses. In addition, we know that Securin is degraded at the metaphase to anaphase transition, so we can perform some immunofluorescence (IF) on tagged Securin to determine whether cells are in metaphase or anaphase. Anaphase cell should have separate DAPI-stained masses (i.e. DNA) and no Securin staining. Metaphase cells, on the other hand, should have a single DAPI-stained mass and should show a signal for Securin IF. G2 cells would have a single DAPI mass, but these cells should not have Securin staining. Additionally, we could examine the phosphorylation status of CDK. CDK phosphorylation on Thr14 and Tyr15 by Wee1 inhibits its activity, which is required to enter into mitosis. Therefore, G2 arrested cells should have Thr14/Tyr15 phosphorylated CDK which we can detect by phospho-specific antibodies and western analysis. e. You would like to identify the gene producing the arrest phenotype at 37°C. Describe in detail how you would do this. 7.06 Spring 2004 PS 4 KEY 7 of 10 ANSWER: Transform a cDNA library from a WT strain into your mutant. Look for rescue of the arrest phenotype at 37C, sequence the plasmid to identify the gene that rescues. This should be a WT copy of the gene that is defective in your mutant. f. After isolating and sequencing the gene that is responsible for inducing a cell cycle arrest at 37°C, you attempt to determine if it is similar to other cdc genes. To your surprise, this gene has not been identified in other screens for genes involved in the cell division cycle and so you name it cdc706. You believe the product of the CDC706 gene binds to and inhibits an inhibitor of the cell cycle, CdkI (Cdk1 functions by inhibiting Cdk2). Propose a reason to explain why cdc706 might arrest at 37°C? Describe in detail how you could test your model and include diagrams of experimental results that are consistent with your model. Be sure to include appropriate controls! ANSWER: cdc706 may arrest at 37C because the protein Cdc706 may no longer bind to and inhibit the inhibitor of CDK, CdkI. This would promote constitutive inhibition of CDK, leading to an arrest. One prediction of the model is that one should be able to immunoprecipitate (IP) Cdc706 along with CDKI in the WT (wild-type) strain, but not in the cdc706 strain. Expt: Immunoprecipitate CDKI from WT (CDC706) and cdc706 cells. Probe, with an antibody against Cdc706, for the Cdc706 protein. If the model is correct, an IP against CDKI should pull down Cdc706 in the WT strain but no the mutant cells. Controls: A CDKI-deleted strain (or non-tagged) to test for stickiness of Cdc706. IP with pre-immune serum. Also test that you pulled down similar levels of CDKI. Pre-immune Serum Cdc706 delete WT (CDC706) cdc706 IP: CDKI IB(immunoblot): Cdc706 Cdc706 Pre-immune Serum CDKI Cdc706 delete WT (CDC706) cdc706 IP: CDKI IB: CDKI 7.06 Spring 2004 PS 4 KEY 8 of 10 Question 3. You are studying the interaction between the Anaphase Promoting Complex (APC) and one of it’s specificity factors, Cdc20. In wild-type yeast, Cdc20 associates with the APC and promotes the polyubiquitination of Securin. Once polyubiquitinylated, Securin is degraded, liberating Separase to initiate anaphase. a) You are interested in identifying the domain of Cdc20 that provides the specificity necessary to target the APC specifically to Securin. You make 3 mutant strains, each with different domains of Cdc20 deleted and perform the following two experiments in each mutant: i) Immunoprecipitate Securin and immunoblot using an antibody that recognizes ubiquitin. ii) Immunoprecipitate the component of the APC that interacts with Cdc20 and immunoblot using a Cdc20 antibody. Your results are as follows: Wild Type Mut 1 Mut 2 Mut 3 Wild Type Mut 1 Mut 2 Mut 3 7.06 Spring 2004 PS 4 KEY 9 of 10 Which Cdc20 mutant would you use for further study, and why? ANSWER: You would continue your studies using mutant 1, because it is defective in Securin polyubiquitination, but still interacts with the APC. Thus it is likely that the domain you deleted somehow targets the APC to Securin. Mutant 2 does not interact with the APC, so even though it is defective in Securin polyubiquitination it will not be useful in your studies. Mutant 3 has a wild type phenotype, as far as we can tell. b) Excited by these results, you quickly make a construct that expresses the wild type peptide which was deleted from the Cdc20 mutant that you selected above (let’s call this peptide A). When you overexpress peptide A in wild type yeast they arrest in metaphase. You perform a co-immunoprecipitation and obtain the following: IP: IB: Overexpression: Pep. A Pep. A Pep. B Securin Securin Securin Securin wild type Pep.A (no peptide) Pep. B Pep. C Pep. C 7.06 Spring 2004 PS 4 KEY 10 of 10 (IB=immunoblot. The Wild type and Peptide A lanes are IP’s using antibodies to Peptide A. The Peptide B and Peptide C lanes are IP’s using antibodies for the regions of Cdc20 deleted in the other two mutants from Part 3A) Based on these results, how do you think peptide A helps target the Cdc20/APC to Securin? ANSWER: Peptide A seems to be a Securin binding domain of Cdc20. It coimmunoprecipitates Securin, and when it is overexpressed it leads to metaphase arrest, presumably by preventing the endogenous Cdc20/APC from polyubiquitinating Securin. c) How would you identify the exact region of peptide A responsible for its function? ANSWER: You could delete progressively smaller regions of Cdc20 and IP with a Cdc20 antibody and immunoblot with a Securin antibody, until you identified the minimal domain necessary to co-IP Securin. ...
View Full Document

This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.

Ask a homework question - tutors are online