BME 201 Homework 03

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Pennsylvania State University *

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Mechanical Engineering

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Feb 20, 2024

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BME 201 Homework 3 Due on Canvas at 11:59 p.m. on Wednesday 31Jan2024 Please integrate homework into one pdf for submitting online. 1. (25 Points) Cells use the molecule ATP as a source of energy for a number of processes. When hydrolyzed, it produces ADP and inorganic phosphate: ATP ⇌ ADP + P i a) At thermodynamic equilibrium, if [ADP] = 1 M and [P i ] = 1 M, then [ATP] = 4μM. What is the equilibrium constant K eq for this reaction? K eq = [ADP][P i ]/[ATP] K eq = [1 M][1 M]/[4 x 10 -6 M] K eq = 2.5 x 10 5 M b) Use this K eq to calculate the Gibb’s free energy in a bacteria cell with [ATP] = 10 mM, [ADP] = 0.6 mM, and [P i ] = 20 mM. Which way will this reaction proceed? ∆G rxn = -RT ln(K eq ) ∆G rxn = -8.314 x 298 x ln(2.5 x 10 5 ) ∆G rxn = -30.79 kJ/mol Reaction quotient: Q = [ADP][P i ]/[ATP] = (0.6 x 10 -3 M)(20 x 10 -3 M)/(10 x 10 -3 M) = 1.2 x 10 -3 Q < K eq , so the reaction will proceed in the direction of the products, or the right side. ∆G rxn = ∆G° + RT ln(Q) ∆G° = -30.79 - (8.314 x 298 x ln(1.2 x 10 -3 )) ∆G° = -46.9 kJ/mol, or -11.2 kcal/mol

c) Motile bacteria often use flagella to move around. How much ATP (in molecules) would one bacterial cell (mass of 1 pg) need to climb from sea level to the summit of Mt. Everest (8.8 km)? Watch your conversions! Remember your dimensional analysis. Mass of cell = 1 pg = (1 x 10 -12 ) x 1kg/1000g = 10 -15 kg Height = 8.8 km x 1000m/1km = 8800 m Energy needed = potential energy = mgh = 10 -15 kg x 9.81 m/s 2 x 8800 m = 8.63 x 10 -11 J ∆G° for ATP hydrolysis = -46.9 kJ/mol -46.9 kJ/mol ---- 1 mol ATP (8.63 x 10 -11 J x 1 mol)/(46.9 x 10 3 J) = 1.84 x 10 -15 mol ATP # of ATP molecules required: 1.84 x 10 -15 mol x 6.022 x 10 23 mol -1 = 1.8 x 10 9 2. (20 Points) Casein molecules in milk have some glutamic acid residues that change their charge based on the pH of their surroundings. This is one factor why the proteins misfold and aggregate when you add vinegar to warm milk. Glutamate is the conjugate base for glutamic acid, and forms when the carboxyl group of glutamic acid loses a proton: −COOH ⇌ −COO − + H + The pK a of this reaction is 4.0. In solution glutamic acid is in rapid equilibrium, so we can think of the species as having a relative charge between 0 and -1. Use MATLAB or Excel to plot the relative charge of glutamic acid as a function of pH. (Ignore the N- and C-termini of the amino acid.) Note that the Henderson-Hasselbach equation is a ratio of [base]/[acid], and we want relative charge here ([base]/[total]). The molecule will be in either acid or base form, so for a population of molecules: [base] +[acid]=[total].

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3. (10 Points) You’ve designed an enzyme that reliably catalyzes the production of methane for use as a renewable energy source. After characterizing the enzyme, you’ve found that it has a k on = 10 μM -1 s -1 , a k off = 1 s -1 , and a k cat = 0.1 s -1 . If the concentration of substrate in your sample is 1 μM, what concentration of enzyme will you need to achieve a digestion rate of 1 μM/s?

4. (20 Points) In this exercise, we will be using Euler integration to model enzyme kinetics. The MATLAB code BME201_HW3_Q4_2024.m simulates the following system: E + S ⇌ ES → E + P by modeling the concentrations for species E, S, ES, and P. Look over the code and make sure you understand what each line is doing. Use the MATLAB file to simulate an enzymatic reaction for a range of substrate concentrations. Use the rate constants given in Problem 3 and choose E tot = 0.1 μM. Use the plots and the data cursor tool to approximate the rate dP/dt for substrate concentrations [S] = 0, 0.3, 1, and 3 μM. Hint: remember the graphical interpretation of the derivative dP/dt. Choose a value of dt and read out dP from the graph. Report the values of [S] and dP/dt in a table like the one below, comparing them to the dP/dt values reported from the Michaelis-Menten equation. [S] ∆P ∆t dP/dt ≈ ∆P/∆t Michaelis- Menten: dP/dt 0 0 3 0 0 0.3 0.0184 3 6.13 x 10 -3 7.5 x 10 -3 1 0.0259 3 8.63 x 10 -3 9.09 x 10 -3 3 0.0286 3 9.50 x 10 -3 9.67 x 10 -3

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5. (20 Points) This exercise is similar to problem 4, but now we’ll be introducing an inhibitor that binds to the enzyme. Anticancer drugs are often inhibitors that are designed to interfere with the function of an aberrant enzyme that causes the cell to divide when it shouldn’t. An inhibitor either prevents the enzyme from binding with the substrate (competitive inhibitor) or doesn’t affect substrate binding but interferes with the catalysis reaction (noncompetitive inhibitor). The two systems are modeled below: Take a second to understand what the role of the inhibitor is in each case. How do the above diagrams differ from the classic Michaelis-Menten model of enzyme kinetics? a) The provided MATLAB code BME201_HW3_Q5_CompInhibitor_2024.m models a competitive inhibitor. The larger for-loop starting at line 40 loops through a similar Euler integrator as in the last exercise, using a different starting substrate concentration for each iteration, and saves the reaction velocity dP/dt for each iteration. The Euler integrator within the forloop should look similar to the one in problem 4, just with some terms added to model a competitive inhibitor. Use this code to plot the Michaelis-Menten curve (i.e., dP/dt vs S, the output of the code) when [I] = 0 μM. Use the cursor tool to make note of the Vmax and KM. Repeat for [I] = 0.01, 0.1, and 1 μM. Use the plot to report the values of [I], V max , and K M in a table. How do V max and K M change as [I] changes?

Competitive: [I] µM V max µM/s K m 0 0.000979 10.2 0.01 0.000978 10.3 0.1 0.000976 11.8 1 0.000946 25.6 As [I] increases, V max decreases and the K m value increases. b) Now look at the MATLAB code BME201_HW3_Q5_NoncompInhibitor_2024.m, which models a noncompetitive inhibitor. It’s set up very similar to the previous code. Again, report the values of [I], v max , and K M in a table for the same range of [I]. How do v max and K M change as [I] changes?

Non-competitive: [I] µM V max µM/s K m 0 0.000979 10.2 0.01 0.000935 10.3 0.1 0.000642 11.1 1 0.000106 16.2 As [I] increases, V max decreases significantly in comparison to the competitive inhibitor, and K m increases more slowly than in the previous case. c) Notice that the parameters across the two MATLAB simulations are the same. Let’s compare competitive and noncompetitive inhibitors by plotting their simulations on the same figure. To do this, comment out the “close all” command in line 5 of both files and uncomment the “hold on” command in line 65 of the competitive inhibitor code and line 69 of the noncompetitive inhibitor code. This will allow you to plot multiple plots on top of each other the same figure. If you ever need to start over fresh, just type “close all” in the command window. Begin by running the competitive inhibitor code with [I] = 0 μM. Make note of the line color of the plot – this is the baseline Michaelis-Menten curve when no inhibitor is present. Now run the competitive inhibitor code for [I] = 1 μM. Again, make note of the

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color of the line added to the figure. Lastly, run the noncompetitive inhibitor code for [I] = 1 μM and note the color of the added line. Be sure to submit this plot with your homework. Based on these plots (and your answers for 5a and b), which seems more effective given the same concentration: a competitive inhibitor or a noncompetitive inhibitor? Why? The non-competitive inhibitor appears more efficient because it has a lower maximum velocity/ V max value, and an efficient inhibitor should result in lower V max values.