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Introduction to Biochemistry and Molecu lar Biology Molecular BCMB 3100E

End of Module Assignment Summer 2020

Enzyme Module Assignment: In Insulin sulin degrading enzyme Honor Statement: This assignment adheres to the standards of UGA’s Culture of Honesty policy. The answers to these questions are my own words and all external sources have been cited. (You do not need to use proper citation formatting, just copy and paste the link.)

Take-Home Point Whether a chemical reaction occurs spontaneously depends on its G. The G of a reaction can be changed by changing the concentration of substrates or products. Enzymes can’t change the G, but they can change the rate of the reaction. Not all enzymes work at the same rate and differences among enzymes help us understand the complexity of living things. Differences in the rates of enzymes can be described by determining, for example, how well they bind substrate and how fast they produce product. This field of biochemistry is called enzyme kinetics (i.e., enzyme rates). Enzymes can be inhibited. The kinetic properties of an enzyme in the presence or absence of an inhibitor can be determined to reveal how the inhibitor works.

Learning Objective Objectivess 1. Understand the function of Insulin Degrading Enzyme (IDE) in the insulin signaling pathway. 2. Explain the relationship between Keq and G. 3. Draw reaction progress curves for spontaneous and non-spontaneous reactions, catalyzed and non-catalyzed reactions. 4. Explain how enzymes catalyze reactions, including why the transition state energy in a catalyzed reaction is lower than in an uncatalyzed reaction. 5. Explain how a graph of reaction velocity versus substrate concentration (i.e., Michaelis-Menten plot) is generated experimentally. 6. Identify Vmax and KM on Michaelis-Menten and Lineweaver-Burk plots. 7. Explain the biological meaning of Vmax and KM. 8. Define kcat and kcat/KM and explain their biological significance. 9. Predict the biological outcomes of mutations in an enzyme that affect its K M, Vmax, or kcat. 10. Summarize how enzyme catalysis relates to the thermodynamics and kinetics of a reaction. 11. Compare and contrast competitive, uncompetitive, and non-competitive inhibition. 12. Compare and contrast irreversible and reversible inhibitors. 13. Distinguish between different types of enzyme inhibitors using Lineweaver-Burk plots and calculations of V max and K M. 14. Examine the effect of mutations and synthetic molecules on the kinetics of IDE. Pa Part rt 1: Insulin Degr Degrading ading Enzyme (IDE) In the first end of module assignment, you learned about the structure of the insulin protein. In this assignment you will learn about an enzyme that cleaves the beta chain of insulin: Insulin Degrading enzyme (IDE) (IDE). The image below shows a portion of the insulin signaling pathway and the role of IDE. Insulin is released when blood glucose levels are high. Insulin binds to the insulin receptor, a transmembrane protein. This activates a signaling pathway and results in short-term changes, such as moving glucose transporters to the cell surface to uptake glucose, and long-term cellular changes, such as repressing the gene expression of enzymes in gluconeogenesis. In order for the cell to remain responsive to new

Sarah Robinson, Summer 2019. Adapted from a case by Justin Hines, Lafayette College and Marcy Osgood, University of New Mexico, and a case by Paula Lemons and Erin Dolan.

signals, the signaling pathway must be terminated. For insulin signaling, the Insulin degrading enzyme (IDE) cleaves insulin into smaller peptides that are then degraded by the lysosome.

Song et al. (2017) BBA Clinical (7): 41-54.

1. Biochemists work with the standard free-energy change at pH 7, G°’. Here are the requirements for G°’: temperature is 298K, the pH is 7, and reactants are present at 1.0 M before the initiation of the reaction. Do you think reactions occurring in cells will always match the conditions for G°’? What determines the spontaneity of a reaction, G°’ or G? Briefly explain your answer. Reactions occurring in cells will not always match the temperature, pH, and molarity requirements for G°’. Because of this, the spontaneity of a reaction is determined by G, which can be measured under any conditions. The difference between the free energy of the reactants and products in their standard state determines whether or not a reaction occurs spontaneously. It may be impacted by temperature and enzyme activity.

2. Consider a reaction A + B  C + D. The equilibrium constant at standard conditions is K’eq, equal to [C][D]/[B] [A]. Thermodynamics gives us an expression to relate the standard free energy change at pH=7.0 ( G°’) and K’eq. It is: G°’ = -RT ln Keq. We can characterize G°’ under the following conditions (see Table 6.3 in the textbook): K’eq = 1, G°’ = 0 K’eq > 1, G°’ = negative K’eq < 1 (e.g., 0.1), G°’ = positive Suppose the K’eq the reaction catalyzed by IDE is 1 × 103. Is the G°’ for this reaction zero, negative, or positive? What does that mean about the direction of the reaction under standard biochemical conditions? According to Table 6.3, at K’eq of 1 x 103 G°’ is negative. This means that the reaction is spontaneous and product-favored under standard conditions.

3. Draw free energy diagrams for the following reactions, based on your answer to Question 3 (G°’). Plot reaction progress on the X-axis and free energy on the Y-axis. Label the G°’ and the activation energy G‡. If the enzyme can catalyze that reaction, use a solid line to show the uncatalyzed reaction, and a dashed line to show the catalyzed reaction. Include a key to show what your lines mean. Take a picture of your drawings and insert it below.

a. insulin  insulin peptides

b. Insulin peptides  insulin

Notice that you have drawn these graphs for the previous question based on the G°’, not the G. These two terms are related to each other mathematically:

G = G°’ + RT ln Q Q = [products]/[reactants] and is based on the actual cellular conditions, as is the temperature (T, in Kelvin).

4. Explain how IDE is affecting the free energy of the reaction. Be specific and include a discussion of the free energy of reactants, products and X‡ (the transition state). IDE increases the rate of the above reaction by lowering the free energy of activation, G‡. This allows the reaction to more easily reach the transition state X‡ and therefore lessens the amount of energy required to degrade insulin into insulin peptides. IDE does not affect the free energy of the reactants or products: it only increases the rate of reaction.

Pa Part rt II – Kinetics of IDE Enzyme catalyzed reaction: k1 k2 E + S  ES  E + P k-1 k-2 E = enzyme S= substrate ES = enzyme-substrate complex P = product Key terms Vmax: Maximum velocity when all enzymes are saturated with substrate. Dependent of enzyme concentration (double enzyme concentration, Vmax will double) KM: Stability of the enzyme-substrate (ES) complex. Compilation of rate constants: K M = (k-1 + k2)/k1. Independent of substrate concentration. (this is a helpful video about that: https://www.youtube.com/watch?v=J2YnlDUd2LI) Kcat: Turnover number which is achieved at Vmax. It is also the rate constant of the rate-limiting step of the reaction, k2. Vmax = K2 [E]T, thus kcat is independent of enzyme concentration.

5. The structure of IDE has been solved. The space filling model is shown below, with 4 subunits D1-D4. Though the mechanism is not completely understood, it involves two confirmations of the enzyme: the closed door and the swinging door, shown below. (from: McCord et al. (2013) PNAS, 110: 13827-13832)

Scientists generated mutations in IDE to examine the role of the swinging door in catalytic mechanism. They compared the normal enzyme (WT) to a mutation in which the amino acid phenylalanine was changed to alanine at position 530 (F530A). Below is an image showing the location of F530 in the ribbon model of IDE (from: McCord et al. (2013) PNAS, 110: 13827-13832). Notice that F530 is at the interface between subunits D2 and D3.

Below are the Michaelis Menten graphs of the kinetics of the WT enzyme and the F530A mutant.

a. Estimate V max for the WT enzyme and the mutant enzyme on the graph. Vmax for WT enzyme: 140 s-1 Vmax for F530A: 1500 s-1 b. Estimate KM for the WT enzyme and the mutant enzyme on the graph. KM for WT enzyme: 12M KM for F530A: 10M

6. Below is a Lineweaver Burk plot of the data for the WT enzyme.

Using the plot above, determine Vmax and KM for the WT enzyme. Vmax for WT enzyme: 270.27 s-1 KM for WT enzyme: 95.24 M Pa Part rt 3 – Li1 7.

Scientists are working on developing molecules that affect the enzymatic activity of IDE. One of the molecules that shows promise is Li1, a synthetic peptide. Its structure is shown below. (From: Leissring et al. (2010) PLos ONE)

Leissring and colleagues tested the effect of Li1 on IDE1 in a cell culture line (CHO-IR, Chinese hamster ovary cell line). Below are the results.

a) Looking the first graph (A), how does Li1 affect IDE’s activity? Li1 decreases IDE activity.

b) Looking at the second graph (B), how does the presence of Li1 affect the levels of insulin in CHO-IR cells? The presence of Li1 does not affect the levels of insulin in CHO-IR cells, as evidenced by the solid line in graph (B).

8. Below is the Lineweaver-Burk plot of the WT enzyme (EZ) with and without the presence of Li1.

What type of inhibitor is Li1? Explain your reasoning. Include a discussion of where you would expect this inhibitor to bind (to E only, ES, or both? And where on the enzyme?) and how it affects Vmax and KM. Li1 is a competitive inhibitor since the Lineweaver-Burk plot shows it does not change the V max of the enzyme but increases the Km. Li1 would bind to the enzyme only. It would bind to the enzyme’s active site, preventing the formation of the ES complex. Li1 would not impact the Vmax because it would not affect the efficiency of the enzyme, only the total number of sites available for binding. The inhibitor increases the K m because a higher substrate concentration is required to yield ½ Vmax.

9. Thinking about this module as a whole, what concept(s) did you find the most difficult? What are you still unclear about? While working through the module, I think I found the analysis of Lineweaver-Burk plots most difficult. After watching AK lectures and reviewing my notes, I feel more confident about it! I am still a little bit confused about calculating Vmax. What work will you do to clarify it? I will find other resources on the concept and work practice problems from the textbook. Any suggestions for improvement of this module? None!...