AVPL Problem Set 8 COMPLETE

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Instructions: ✓ Download this problem set, keep the questions in the document and please type your answers in a color other than black so that your answers are easily discernable. ✓ Complete this entire problem set – give robust, detailed responses – this is practice so use it as such. ✓ If you get stuck or need clarification on any of these questions… just ask! ✓ Upload this document with your answers to the appropriate TurnItIn link in our Moodle page in PDF format ✓ Please be advised that TurnItIn is a plagiarism and similarity checking software o Make sure that you close and put away all sources of information before writing your answers so that you can be sure that your answers are in your own words and to convince yourself that you understand the concept and can explain it yourself. o If your answers are too similar to either someone else’s responses (either from this course or from previous iterations of this course) or from the internet, points will be deducted as the assumption will be made that the work you’re doing is not your own. We'll work on the following problem set questions during Discussion Section this week: Questions 3, 5, and 6. This problem set is worth 15 points: • There will be a 5-point quiz in Discussion Section based on one of the questions worked on in Discussion Section • Once you submit the entire problem set at the end of the week: o 5 points for robust completion of the entire problem set (checking for honest effort but not checking for correctness) o There will be one other question chosen at random and graded for robust answer and correctness worth 5 points Due date: Sunday, April 30 th by 11:59pm

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Question 1: Enzymes Which of the following proteins that we’ve discussed thus far in class are enzymes? Use yellow highlight to choose your answers. Then for each protein, give a brief explanation of what it does and if it IS an enzyme, briefly explain what chemical reaction(s) it catalyzes: A. RNA Polymerase RNA polymerase catalyzes transcription and the formation of mRNA. It does this by reading the separating the DNA strands and reading the DNA template strand and then matching the correct nucleotides, synthesizing an RNA strand that contains the message of the DNA. B. Lac repressor The lac repressor is a repressor that blocks the transcription of the lac operon until a lactose is present to bind the repressor and deactivate it . C. Permease (lac operon) Permease is a membrane channel protein that facilitates the transportation of lactose and other energy into the cell but is not an enzyme. D. RTK RTK is a receptor on the self- renewal signal transduction pathway. It’s job is to receive the signal and activate TK1 by phosphorylation, however it not an enzyme because it is not changing a substrate into a product. E. Beta-galactosidase (lac operon: LacZ gene product) Beta-gal is a protein that cleaves lactose to make it into usable energy (glucose) and leaves a byproduct of galactose. This is an enzyme because it catalyzes the separation of lactose for the cell to have glucose so it can use as energy. F. Ribosome The ribosome is the place where translation happens. Although it is key in the synthesis of proteins and polypeptides, it is not an enzyme.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis G. Transcription Factor from self-renewal signaling pathway Transcription factor attaches to the nucleus once it is activated by phosphorylation. It goes to the nucleus in order to start transcription of self-renewal genes. Question 2: Regulation of Protein Activity Which of the following does NOT describe a mechanism that cells use to regulate protein activity? Circle it. Cells control protein activity by phosphorylation and dephosphorylation Cells control protein activity by the binding of small molecules Cells control the rates of diffusion of substrates to enzymes or small molecules to their binding partners Cells control the rates of protein degradation Cells control the rates of enzyme synthesis Now choose three protein activity regulation mechanisms from above and give examples of these mechanisms that we’ve discussed in this class that DO happen in cells and give some molecular details describing your examples. Bullet point or number your responses. 1) Cells control protein activity by phosphorylation and dephosphorylation using kinases. We learned in the self-renewal transduction pathway that TK1 needs to be phosphorylated in order to be activated and then when it is, it goes on to phosphorylate TK2 in order to activate it. 2) We learned that cells could regulate protein activity by binding small molecules when we learned about the lac operon. For example, in the lac operon, CAP is a protein that speeds up the transcription of the lac operon but it only binds to the lac operon when it is bound to cAMP. 3) We learned that cell regulate protein activity by controlling the rates of enzyme synthesis when we learned about the lac operon. The lac operon gene codes to produce proteins and enzymes that help with the processing of lactose in the cell.

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Question 3: What do Enzymes Change? For any given chemical reaction that is catalyzed by a particular enzyme: which of the following aspects of the chemical reaction do enzymes change? Answer Yes or No for each and explain. A. Reaction rate Yes, the enzyme changes the reaction rate because they speed up the process of the chemical reaction. They do this by requiring less energy to complete this reaction and by stabilizing the reaction by providing an active site with favorable conditions for the substrate. B. Types of products generated No, the enzyme does not change the product. The enzyme only speeds up the reactants transition state into the product. They are not adding or taking anything away from the reactants. C. Activation energy Yes, enzymes greatly reduce the activation energy needed for the reaction. They do this in many ways, but usually the enzyme provides the reactants a more stable site to produce their product. The enzyme also helps with its own chemistry. D. Energy of the transition state Yes, enzymes change the energy of the transition state by reducing the activation energy required for the chemical reaction

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis E. Gibbs free energy No, enzymes do not change Gibbs free energy. That is because enzymes do not change the difference between the reactants and the product of the reaction. Question 4: Endergonic vs Exergonic Reactions Can an enzyme make an endergonic reaction exergonic? Why or why not? In your response, it would be a good idea to define the terms endergonic and exergonic. No, it cannot. An endergonic reaction is just when the reactants contain lower free energy than its products and exergonic is when the reactants contain higher free energy than the product, since the enzyme does not change the free energy of either the product or the reactant than it cannot change whether the reaction will be exergonic or endergonic. The Gibbs free energy (difference in free energy between reactant and product) is the same with or without an enzyme. What the enzyme does change is the activation energy required to induce the chemical reaction, meaning the reactants can turn into product using less energy. Question 5: Enzyme Saturation What is meant by saturation of the enzyme in conceptual terms? What kind of information does saturating substrate conditions give you about that enzyme-substrate pair? The saturation of an enzyme is the point in which even if you added more substrate to the environment, the initial rate of the reaction would not increase. This is because it means that the environment is so saturated in substrate that the enzyme is constantly catalyzing the substrate at its fastest rate. Changing the saturation of the substrate can give you information Question 6: Enzyme Interactions A. The binding interaction between enzyme and substrate is stronger than the binding interaction between enzyme and product. Explain why that makes sense based on how enzymes function. Once the substrate has become the product, its chemistry is different and therefore its affinity for the enzyme also changes, this is a good thing for the cell because once the enzyme is done catalyzing the reaction, the product leaves the enzyme, allowing it to go and catalyze another reaction.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis B. If a mutation occurred that increased the affinity of an enzyme/product association, how would that affect the overall rate of the reaction catalyzed by that enzyme? This would slow down the rate of reaction because after the enzyme has catalyzed a reaction it would be unable to go and catalyze another one if the product was still attached to its active site. In this case the product would be acting as a competitive inhibitor. Question 7: Michaelis-Menten Kineti cs The above graph is showing the Michaelis-Menten kinetics for a particular enzyme (wildtype). Two mutations are made in this enzyme (Glu205Lys means amino acid 205 is changed from glutamate to lysine and Glu206Leu means aa 206 is changed from glutamate to leucine) and the M-M kinetics are shown for those mutants as well. A. What are the V max ’s and K m ’s for the wildtype enzyme and the two mutants? (Make sure to use appropriate units in your answers) Enzymes Vmax Km Wildtype 0.02nm 0.3mM Glu205Lys 0.004nm 0.65mM Glu208Leu 0.0001nm 0.3mM B. How are the mutations affecting the affinity of the substrate for this enzyme? The Glu208Leu mutation does not affect the cells affinity but does lower the v-max, this mutation replicates the effects of a non-competitive inhibitor. The Glu205Lys mutation has lowered the cells affinity to the substrate, you can tell because the Km is higher than the non-mutant meaning that the cell needed a higher concentration of substrate for it to reach ½ v-max.

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Question 8: B-gal Kinetics As you learned in Topic 5, the enzyme B-gal, encoded for within the LacZ gene in the lac operon, catalyzes the following reaction: lactose + H20 -----> glucose + galactose To determine V max and K m of B-gal with its substrate lactose, the same amount of enzyme was incubated with a series of increasing lactose concentrations. At each lactose concentration, the initial reaction velocity was measured. The following data were obtained: Lactose Concentration (mM) Rate of lactose consumption (umol/min) 1 10.0 2 16.7 4 25.0 8 33.3 16 40.0 32 44.4 A. Plot the rate of the reaction versus lactose concentration. Make sure to label your axes.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis B. Estimate K m and V max from the plot you created. Explain how you found them. I would estimate the v-max to be somewhere around 46 umol per minute. My estimate is based on the fact that the rate of lactose consumption was growing slowly as it reaches the higher concentration of the substrate. When doubled the concentration of lactose from 16 mM to 32mM the rate only slightly increased indicating that it is near its max velocity (v-max) at 44.4 umol per minute, which is why I estimate that the v-max will be around 46 umol per minute. If the v-max is 46 umol per minute, then the Km would be somewhere around 3.5mM. I found this by looking at the lactose concentration at half the rate of the assumed v-max. Question 9: Inhibitors of Enzyme Activity A. When an enzyme is treated with a competitive inhibitor, why does Km change and Vmax remain constant? This happens because the data you would find makes it seem that the substrate has lower affinity to the enzyme. This is due to the inhibitor blocking the substrate from binding with the enzyme so it would take a higher concentration of the substrate to reach half the rate of the v-max. The v-max is the same because if you added enough of the substrate then eventually you would reach the same v-max since the enzyme would be more likely to bind to the substrate than the inhibitor, it just takes more of the substrate to be around it. B. Imagine that you’re studying B -gal activity and you decide that you want to find a competitive inhibitor of B-gal function. Take your graph for B-gal and lactose from the previous question (keep the data from the situation in which there’s no inhibitor) and draw in some data showing what the graph would look like if you added your competitive inhibitor. For the answer to this question, show the resulting graph with the data that includes the data with and without inhibitor an d then explain how it’s changing the kinetics of B -gal function.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis The competitive inhibitor changes the v-max because the inhibitor prevents the substrate from binding with the enzyme at times. This would cause the enzyme to not always bind to the reactant and therefore slowing down the rate at which it is catalyzed. However, if enough substrate was introduced to the environment the rate would eventually reach the same v-max since the enzyme is still capable of doing so. C. Now for noncompetitive inhibitors answer questions A (in this case, why does Km not change but Vmax is lower) and B (what would your Bgal data look like if you added a noncompetitive inhibitor). The non-competitive inhibitor would change the v-max but not the km because the inhibitor is causing the enzyme to change shape thus changing its chemistry and ability to function as an enzyme, the km remains the same however because even at half the velocity of the lower v-max the affinity of the substrate to the enzyme remains the same.

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis D. Now for uncompetitive inhibitors answer questions A (in this case, why Km is lower and Vmax lower) and B (what would your B-gal data look like if you added an uncompetitive inhibitor). In uncompetitive inhibitors, the v-max is lowered and the km is lowered because the inhibitor is attaching to the enzyme after it has catalyzed a reactant, what it does by attaching is that it changes the enzyme in a way that doesn’t allow the product to b e released from the enzyme thus inhibiting it from catalyzing other reactants. This is why the v-max is lower since less enzymes are available to reach the highest rate of reaction, and the km is lower because the enzymes affinity to bind to other reactants is lowered since it does not release the product to be able to.

AVPL Problem Set 8 COMPLETE

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