Bio151 Sp23 Topic 8 Problem Set Answer Key

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Instructions: ü 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. ü Complete this entire problem set – give robust, detailed responses – this is practice so use it as such. ü If you get stuck or need clarification on any of these questions… just ask! ü Upload this document with your answers to the appropriate TurnItIn link in our Moodle page in PDF format ü 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 RNAP is an enzyme. RNAP synthesizes RNA and catalyzes the reaction that links nucleotides together – e.g. forms phosphodiester bonds. B. Lac repressor Lac repressor is not an enzyme. Lac repressor is a regulatory transcription factor that blocks RNAP from binding to promoters. C. Permease (lac operon) Not an enzyme – just serves as a protein channel that sits in the membrane. D. RTK RTK is an enzyme. It is a kinase, which takes phosphate off of ATP and sticks that phosphate on a tyrosine on its substrate. E. Beta-galactosidase (lac operon: LacZ gene product) B-gal is an enzyme. It catalyzes the breakage of the glycosidic bond between glucose and galactose in lactose. F. Ribosome Ribosome is an enzyme and more specifically, the RNA component of the ribosome has enzymatic activity. It catalyzes formation of peptide bonds between amino acids. G. Transcription Factor from self-renewal signaling pathway TF is not an enzyme. It binds DNA and helps recruit RNAP to promoters.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis Question 2: Regulation of Protein Activity Which of the following does NOT describe a mechanism that cells use to regulate protein activity? Circle it. see green highlight 5 pts – one for correctly say yes or no (circling/choosing or not circling/choosing) for each option below: 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 – for the most part cells don’t have the ability to control diffusion rates, and so they are dependent on diffusion rate-limiting kinetics. This means a lot of what happens (and how fast) depends on concentration of these molecules because of this. 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. There are many to choose from – if you are not sure of an example, feel free to ask me or a TA to check it. 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.

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis A. Reaction rate Yes, reaction rate does change (they increase it). This is because the enzyme is making it “easier” for the substrates to obtain the transition state and thus easier for the reaction to happen. This means that the reaction will happen more frequently if catalyzed by an enzyme. B. Types of products generated No, enzymes don’t change the types of products generated. The energy and nature of the reactants (substrates) doesn’t change nor does the energy or nature of the products. The enzyme just makes it more likely that the reaction will happen. C. Activation energy Yes, enzymes change the activation energy – they lower the activation energy. They do so by making it easier for the transition state to happen and thus the difference between the energy of the transition state and the energy of the reaction (which is how the activation energy is defined) is lower as a result. D. Energy of the transition state Yes, enzymes change the energy of the transition state (they lower it) – they basically somehow stabilize the transition state or make it easier for the substrates to achieve the normally unstable transition state. E. Gibbs free energy

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis No, Gibbs free energy (the change in energy between the reactants and products) does not change when an enzymes is used to catalyze a chemical reaction. The energy of the products and the energy of the reactants don’t change and therefore the difference between the two don’t change. 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, an enzyme cannot make an endergonic reaction exergonic. The free energy associated with the reactants and the products in a reaction are what determine whether a reaction is endergonic or exergonic. Since enzymes don’t change the free energy associated with reactants and products, enzymes cannot change the “gonic” properties of the reaction (ok, I totally made that word up). 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? When substrate concentrations are high enough, the enzyme no longer has to wander around to find substrate – there is so much substrate that as soon as the enzyme pushes out the reaction product, another substrate binds to the enzyme active site. Thus, you can get a sense of catalytic rate at saturating substrate concentrations because it takes away the issue if random diffusion being rate-limiting. This is also the point where even if you add more substrate, the rate of the reaction does not change. 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. You would expect the Kd of the enzyme-substrate pair to be lower than enzyme- product. This is because the enzyme and substrate need to have affinity for each to make the initial interaction. However, once the reaction occurs and products are

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis formed, the enzyme should quickly dissociate from the products so that the enzyme can bind another substrate and do its thing all over again. 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? A decrease in the Kd of the enzyme-product association means that there is an increased affinity of the enzyme for the product, and therefore it would take longer for the enzyme to dissociate from the product. That means that the overall rate of the reaction would decrease because the enzyme can’t do the next reaction with the next substrate until it releases the product and allows for another substrate to bind. Question 7: Michaelis-Menten Kinetics 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) Vmax Km Wt 0.02 change Abs/min 0.25mM 205 mutant 0.004 change Abs/min 0.8mM 206 mutant 0.001 change Abs/min 0.4mM (approx.)

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis B. How are the mutations affecting the affinity of the substrate for this enzyme? The mutation leads to a higher Km which means the mutant enzymes have lower affinities for the substrate. Although Glu206Leu just looks like a dead enzyme so it’s a little hard to tell affinity when the enzyme can’t catalyze its reaction. 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. Something like this except with some numbers – axes should be labeled.

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. Km is probably around 4mM and Vmax around 48-50umol/min (answers should have units). Question 9: Inhibitors of Enzyme Activity A. When an enzyme is treated with a competitive inhibitor, why does Km change and Vmax remain constant? Competitive inhibitor binds in the active site of the enzyme and competes with substrate for binding to the enzyme’s active site. This changes the apparent Km of the enzyme for the substrate – not because it’s changing the actual binding affinity, but because competitive inhibitor makes is look like the enzyme is having a hard time finding the substrate. But, Vmax can be reached, eventually, because you can keep adding substrate and eventually out-compete inhibitor for the enzyme’s active site. 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 and then explain how it’s changing the kinetics of B-gal function.

Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis 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). Noncompetitive inhibitors bind somewhere other than the active site and (by definition) change the ability of the enzyme to catalyze the chemical reaction but do not change the ability of the enzyme to bind substrate – therefore Km does not change but Vmax is lower. See answer to B for graph. D. Now for uncompetitive inhibitors answer questions A (in this case, why Km lower and what does that mean and Vmax lower) and B (what would your B-gal data look like if you added a noncompetitive inhibitor). Uncompetitive inhibitors will only bind once an enzyme is bound to substrate. They bind somewhere other than the active site and only bind the E-S complex. Then they inhibit catalytic activity. This actually makes Km lower and makes the affinity of the enzyme for the substrate look better (when in actuality it’s just preventing the enzyme from releasing the substrate). Vmax is lower because the inhibitor is inhibiting the ability of the enzyme to catalyze the reaction. See below for the graph: 1 is the reaction without inhibitor and 2 is the reaction with inhibitor.

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Bio151 Sp23 Topic 8 Enzyme Kinetics Problem Set Francis

Bio151 Sp23 Topic 8 Problem Set Answer Key

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