FST 123 Problem Set 3 Spring, V o at [S] 0 = 10 mm (mm/min)

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1 FST 23 Problem Set 3 Spring, 2009 Name. A student needed to know the activity versus ph profile for an enzyme that he was going to use in a later application. He collected the appropriate data and determined K M and Vmax at ph 6.8. They were about 0.mM and 800mM/min, respectively. He then measured V o at [S] 0 = 0.002mM and at [S] 0 = 0mM, and obtained the data given in the table below. ph V o at [S] 0 = 0 mm (mm/min) V o at [S] 0 = mm (mm/min) 2 precipitated precipitated a. Graph the data so that you can interpret how the velocity changes with ph. The optimum ph is around 7, so divide the data in each column by the value at ph 7. Then, plot the result (V ph /V opt ) vs ph. This is a normalized plot of V vs. ph. b. Interpret the data in terms of ionizations of the various forms of the enzyme that alter binding or catalysis. At high [S], there is an acid transition, so it must correspond to a decrease in V max the catalytic step. At low [S], there is an acid transition, which we already know corresponds to a decrease in V max and an alkaline transition, which must correspond to an increase in K M, because the rate at low [S] is (V max /K M )[S] and since V max is approximately constant in the alkaline region.

2 So, you could make a diagram like this: EH 2 + S EH 2 S X EH 2 + P EH - + S EHS - EH - + P E -2 + S X ES -2 E -2 + P c. What are the pk a s involved? Look at the midpoint of the transition or plot as described in lecture: ph vs (V/(-V)) for the acid transition or ((-V/V) for the alkaline transition, where V is the fraction in part A (i.e., V ph /V opt ). Plot Log of this vs ph. You should get a slope of.000, and the line should cross the X-axis at ph = pk a. Slope Y-intercept X-intrcept Acid side Alkaline side So, the pka s are 4.79 and c. What other situations other than effects on K M and V max could give rise to similar observations? I can think of three possibilities:. There is an ionization of the substrate that causes the change 2. There is an ionization of the buffer or some other component that causes the change 3. The enzyme denatures at the acid side (probably not the basic side, because the enzyme seems to work there, as long as there s enough substrate.

3 d. Indicate how you could rule out each of the possibilities you listed in part c (or, find that it, rather than an ionization in the active site, controlled the behavior).. Look up the pka of the substrate 2. Look up the pka of the buffer or 3. Carry out a preincubation experiment in which the enzyme is incubated at the test ph for a fixed time. The enzyme is assayed at the ph optimum or other ph at which it is apparently stable. If the activity vs. ph of-incubation profile looks like the ph vs. activity profile, then instability contributes to the apparent activity profile. e. The ΔHionization was found to be 0.2 kcal/mole for the acidic ionization and 0 kcal/mole for the basic ionization observed for the enzyme. Suggest what groups on the enzyme are the most likely to be responsible for the ph behavior. The acid side sounds like a carboxylate: Asp, Glu, or the C-terminus The alkaline one sounds like a primary amine: the N-terminus or a Lys sidechain. 2. One of the major biochemical discoveries of the 990s was that the inorganic molecule NO is important in the regulation of many biological processes. NO was declared the Molecule of the Year by the American Chemical Society, and three scientists were awarded a shared Nobel Prize in 998 for elucidating its function. But, NO gas has been known for many years. It can be formed at high temperature by the reaction N 2 (g) + O 2 (g) 2NO(g) The equilibrium constant for this reaction is 4.08 x 0-4 at 2000K and 2.5 x 0-3 at 2400K. Yeah, I said high temperature. a. From this information, calculate ΔG o, ΔΗ o and ΔS o for the reaction under these conditions (use 2000K if the parameter is dependent on temperature). ΔG = -RTln K eq. ΔG = -.98 cal/mole K 2000K ln 4.08 x 0-4. ΔG = 30.9 kcal/mole The slope of ln K eq vs /T equals ΔH/R, so you can ) do the plot, or 2) pretend to do the plot by calculating the slope based on rise over run, (ln K T2 ln K T )/ ( T 2 " T ) = ΔH/R, or or 3) plug the data into the equation derived from differentiation and integration (vant Hoff equation), ln K eq2 = - "H K eq R [T -T 2 ] T T 2 which turns out to be the same as doing 2). All approaches give ΔH = 43.2kcal/mole (remember to multiply the slope by minus R) ΔS = (ΔH - ΔG )/T = (43,60 cal/mole 30,900 cal/mole)/2000k = 6.cal/mole K b. Calculate the value of the equilibrium constant at 2500K.

4 The direct relation is between DG and K eq, but, unfortunately, we don t know DG at 2500K. So, we ll have to go back to vant Hoff: ln K eq2 K eq = - "H R [T -T 2 T T 2 ], We ll let T 2 be 2500K, and we ll solve for K 2, knowing the DH we got in part a. T could be either 2000 or 2400K since we knoe Keq at both temperatures. Let s use 2000K and its respective K eq of 4.08 x 0-4. ln K eq "0 = "03 cal /mole #4.98 cal/mole K [ 2000 " ] = 2.8, K eq "0 #4 = 8.85, so K eq2 = 3.6 x At saturating levels of substrate, an enzyme-catalyzed reaction was found to be 3.6 times faster at 30 C than at 20 C. a. Assuming ln k is linear with /T in this temperature range, calculate E a for the reaction. The assumption means that denaturation is not an issue and the temperature dependence arises from the activation energy, only. k = A 0 e -Ea/RT ln k = ln A 0 - Ea RT So, the slope of ln k vs /T is E a /R, and the slope can be written ln(k2)! ln(k)! T 2 T But, you don t know absolute rate constants, so, it would be convenient to write the slope in a different form, so that you can plug in the ratio of rate constants. Actually, you don t have rate constants, just ratios of rates, but taking ratios causes the concentration terms in the rate equation to cancel out. V max is k 2 E T, so taking ratios of V max s obtained at different temperatures with the same enzyme concentration is equal to the ratio of k 2 s: ln( k2 ) T 2 k! T = ln(3.6) 303 " 293 = - Ea R E a = -R ln(3.6) 303 " 293 E a = -R.28 =.35 kcal/mole "4 ".3x0 b. Calculate Q 0 for the reaction. 20 C and 30 C are 0 apart, so we just need to take the ratio of the rates, which is 3.6 c. Calculate ΔH for the reaction. According to Eyring theory.

5 k = k T B h e("#h +T#S )/ RT or ln k T = ln k B h - "H RT + "S R The slope of ln(k/t ) vs. /T ought to be ΔH /R. The slope is rise/run: (ln k 2 - ln k # )/ " & % ( = ln k 2T # / " & % ( T 2 T $ T 2 T ' k T 2 $ T 2 T ' # 3.6" 293& # ln% (/ $ 303 ' 303 " & % ( = -.07 x 0 3 $ 293' ΔH =.07 x 0 3 R = 2.9 kcal/mole 4. Describe three distinctly different ways that a change in temperature can alter the velocity of an enzyme-catalyzed reaction. a) Increase in T causes increase in rate according to the Arrhenius equation. b) K M could change. If the enzyme is not saturated with substrate, the rate would also change. c) Instability of the enzyme at high temperature could decrease the rate with time. d) Temperature-dependent change in the pka of a group on the enzyme whose protonation state affects KM or the catalytic step