BIBC 102, Metabolic Biochemistry Problem Set 1 Fall 2002

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1 Enzymology problems 1)The sweet taste of fresh corn is due to the high level of sugar in the kernal. Storebought corn that has been sitting around for a few days is not as sweet because about 50% of the sugar is converted into starch within one day of picking. To preserve the sweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes ("blanched") then cooled in ice cold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is the biochemical basis for this procedure? Like all metabolic reactions, the conversion of the delicious sugar into the much-less-sweet starch is mediated by enzymes. The brief boiling right after harvest denatures these enzymes and thus removes the catalysts that cause the loss of sugar, thus preserving the wholesome sweetness of freshly-picked corn. 2) The enzyme urease enhances the rate of urea hydrolysis at a given ph and temperature by a factor of Suppose a given amount of urease can hydrolyze a sample of urea in 5 minutes. How long would the same process take without the enzyme? the conversion without the enzyme would take 5 x minutes. To make this number more understandable, let s convert it. How many minutes in a year? days/yr x 24 hrs/day x 60 min/hr = 5.26 x 10 5 minutes. So it would take (5x10 14 )/ (5.26 x 10 5 ), that is 9.5 x 10 8 years for the uncatalyzed reaction. eeek 3) In class, we saw that the effect of changing the activation energy on a rate constant is predicted by the equation k= (kt/h)e -( G /RT) Note the left "k" is the rate constant, and the right "k" (in boldface) is the Boltzmann constant. What change in the activation energy is required for a fold increase in the rate constant at 37 degrees C (remember Lord Kelvin!)? To the left is the expression relating rate constant (in this case with a little r subscript to distinguish it from the Boltzmann constant, that, despite its fame and importance, is blue. Our problem is best addressed in two steps. First, what is the effect of a change in G on rate constant. And then, what particular change will give a increase? Let s represent the change as a big, red C. Because we know it is going to increase rate, it will be a lessening of activation energy. So the new rate constant from any change C in G is shown on the next page (because of the room required). The most important point is that the effect of the change C, because of the nature of exponents, is an independent factor that alters the old rate constant by multiplication! So it is not dependent on the original rate constant. page 1

2 So now our question is, what C will make the term e c/rt equal to It is really just a matter of algebra. (I hated when my physics professor would say that ). To solve for C in this case, C/RT = ln10 14, C = RTln10 14 I get C = 83 kj for a fold rate enhancement. 4) In general, what change in activation energy (at 37 degrees C) will result in a 10 fold change in a rate constant by the equation in 3? Why is this general? We know from question 3 above that the effect of changing the activation energy is described by a multiplicative term, e c/rt. What is particularly neat about this term is that it is symmetrical for changes in either direction. Meaning that the effect of increasing or decreasing the activation energy has the same multiplicative effect on the rate, whether it is slowed or hastend. For instance, let s compare the effect of lowering the activation energy by C, and the effect of raising the activation energy the same amount, C page 2

3 So the effect on the rate of reaction for each case is, as in the problem before, The amount of energy, in kj, needed to casue a 10 fold change in rate is that which causes e c/rt =10. That is, when C = RTln10. So at 310 K, C = (8.315)(310)2.303 = 5.93 kj/mole 5) The now-late Francis Crick made the observation (from an interview in "The Eighth Day of Creation"-a spectacular book about the researchers and research of the molecular biology revolution- by Horace Judson) the following about allosteric enzymes and allosteric regulation: ",,,I would say particularly Jack Monod's work on allostery is a very powerful theoretical concept [It] meant that you could connect any metabolic pathway with any other metabolic pathway, you see, because there was no necessary relation between what was going on at the catalytic site and the control molecule that was coming in". What does he mean by this? What Dr. Crick was getting at is that allosteric regulators, be they inhibitors or activators, do not have to look like the substrates or products of the reaction. This in not the case for competitive inhibitors, that usually closely resemble the substrate, and in that way fit into the active site and block enzyme activity. Because allosteric regulation is mediated by binding sites distant from the active site, and on separate proteins that are in quaternary association with the catalytic subunits, there are no limitations on the type of molecules that might regulate a particular enzyme. Thus, any metabolic pathway can potentially comminicate page 3

4 regulate any other metabolic pathway if the correct allosteric controls and binding sites are in operation. Very deep! 6) At what substrate concentration will an enzyme with a kcat of 30/sec and a Km of M show one quarter of its maximum rate? Determine the fraction of Vmax that would occur at the following substrate concentrations: [S]= (1/2)Km, 2Km, and 10Km. The first one is shown to the right, worked out in detail (with the cool digital pad). The other ones are done the same way. Just plug S into the algebraic equation. No need to plug in actual numbers. If S = 0.5 Km, the S/(S+Km) will equal 1/3, and so the rate will be 1/3 of Vmax. If S is equal to 2 Km, the ratio is 2/3 and the rate is 2/3 of Vmax If S is equal to 10Km, the ratio is 10/11, and the rate is 10/11Vmax. I idea is that you have to make bigger and bigger changes in S to make a difference as the rate gets nearer Vmax 7) The following experimental data were collected during a study of catalytic activity of an intestinal peptidase with the substrate glycylglycine, in which the dipeptide is cleaved into two molecules of glycine by hydrolysis. [S] (mm) rate ( mol/min) Use a Lineweaver-Burk plot to determine the Km and Vmax for this enzyme. According to your text, the correct values for the constants for this particular peptidase are Km = 2.2 mm Vmax = 0.51 µm/min page 4

5 If you are ambitious, try using a curve-fitting program with the rectangular hyperbola directly. 8) Consider the two Lineweaver-Burk plots below. Each plot shows data from three different variants of the same enzyme. That is, A, B, C are all catalyzing the same reaction, but they have different kinetic profiles, as indicated from the differences in the line positions. a) Which enzymes display Michaelis-Menton kinetics? Why do you say this? All of them do, because a straight line in this format is obeying the simple relationship between S and V. They have different Vmax or Km, depending on the specific enzyme, but all show a linear double-reciprocal plot, and so all have a constant Km and saturation behavior predicted by the M&M (as opposed to the Eminem) equation Remembering that the X intercept value is (1/K m ) and the Y intercept is 1/V max, the answers to b and c can be found by visual inspection. Convince yourselves! b) Which set all have the same Km? PLOT 1's three different enzymes all have the same X intercept, and so have the same value of Km. c) Which set have three different Vmax? Two different Vmax? PLOT 2's three enzymes all have three different X intercepts, and two distinct Y interecepts, thus, the answer is PLOT 2 d) At any concentration of S, which letter always has the highest activity in either plot? At all concentrations of S in either of the two plots, A always has the lowest values of 1/V, meaning that it has the highest values of V. Reciprocal plots are dicey that way page 5

6 9) Irreversible inhibition of an enzyme. Many enzymes are inhibited irreversibly by heavy-metal ions, such as mercury, copper, or silver cations, which can react with essential sulfhydryl groups to form mercaptides: Enz-SH + Ag + Enz-S-Ag The affinity of Ag + for sulfhydryl groups is so great that Ag + can be used to titrate SH groups quantitatively. To 10mL of a solution of 1mg/ml of a pure enzyme, an investigator added just enough AgNO3 to completely inactivate the enzyme. A total of µmol of AgNO 3 was required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in this way give only the minimal molecular weight? The Ag+ is a way to measure the number of moles of the protein associated with 10 mg. But we only get a minimum molecular weight, because with the information provided, we don t know how many Ag+ it takes to neutralize a molecule of enzyme. If it is one per enzyme, then we get the smallest molecular weight. And that is why the question asks for the minimum. For example, if it actually took two Ag+ to neutralize each enzyme molecule, then the number of moles of enz present would be half of the Ag+ used, and the estimated molecular weight would double. 10) Suppose that the substrate vs. rate plot for the enzyme in the question above showed a sigmoidal shape when rate was plotted against substrate concentration. Now what is the minimal molecular weight of the enzyme as determined from the equation. That sigmoidal plot indicates that this is an allosteric enzyme, and so it must have quaternary structure. Thus, even if every Ag+ neutralizes an individual enzyme molecule, the requirement for quaternary structure means that the in-solution molecules are AT LEAST dimers (the simplest quaternary structure) so the minimum molecular weight would be double the minimum for a protein with only tertiary structure. 11) Suppose that a 6 step, linear metabolic pathway is regulated in the following way: The third enzyme (ENZ3) in the pathway, that is, the one that catalyzes the conversion of the third molecule of the pathway into the fourth one (hint: write it out!!). ENZ3 page 6

7 is allosterically regulated. It is regulated both by the first substrate of the pathway, and by the final product. The regulation of ENZ3 makes sense: when there is too much initial substrate, the activity is increased, and more of it is used in the pathway. Conversely, when there is sufficient product, the activity of the enzyme is lowered. a) What type of modulator is the initial substrate? The final product? In all likelihood, A is an ALLOSTERIC ACTIVATOR, and G is an ALLOSTERIC INHIBITOR b) The initial substrate and the final product have (not surprisingly) very different structures. Speculate about the structural features of ENZ3 that allow it to be allosterically regulated, and the structural features that allow two totally different structures to regulate activity. Since each structure is different, and each does a different thing to the enzyme (A activates, G slows down), they most likely have completely distinct binding sites for each, and these are probably on separate proteins from the polypeptide that forms the active site. Like the picture of aspartate transcarbamoylase in the book and lecture, many allosteric enzymes have very complex structures, with multple independent regulatory proteins. c) Draw a representation of a substrate-rate plot, showing the effect of each of these two modulators on the activity of enzyme 3. BIOENERGETICS 12)Calculate the standard free energy change for each of the Keq. You can see the reactions in the book. USE 301 K (37 C) Keq 6.8 G o = -RTln6.8 = x 298 x ln 6.8 J/mole Keq G o = -RTln = x 298 x ln J/mole page 7

8 Dr. Problem Set 1 Fall 2001 Keq 254 G o = -RTln 254 = = x 298 x ln 254 J/mole 13) In general, what change in the magnitude of G' 0 will cause a 100 fold change in equilibrium constant? The mathematics are identical to those in 3 and 4. We will use the temperature of the standard conditions, 298 K. dwhat we end up with is that when DG changes by an amount C, the change in Keq is represented by e c/rt, or e -c/rt, depending on whether or not the G is made more negative, or more positive, respectively. What C value will make the first term equal to 100? C = RT ln 100 = x 298x 4.6 (4.6 is the ln 100, which is 2 x ln 10) or 11.5 kj/mole. If the G o is decreased (made more negative) by 11.5 kj/mole, the Keq increases by 100 fold. If the G o is increased (made more positive) by 11.5 kj/mole, then the Keq drops by 100 fold. 14) Draw a reaction coordinate for the simple reaction of S becoming P. Show what happens when the following changes to the diagram occur: a) The free energy of P becomes lower. b) A catalyst is adde is raisedd to the reaction mix c) the free energy of S Now, which of these changes affect the G of the reaction? Both A Both A and C will alter the DG of the reaction, which is the difference in free energy between the reactants and the products. Which affects the rate of the S to P reaction? Both B and C affect the S to P reaction rate. However, only B affects the forward and reverse rates indentially (see 3 above) and so the equilibrium is not perturbed. page 8

9 Which affects the rate of the P to S reaction? Both A and B affect the P to S reaction. But the effect in B is exactly counterbalanced by and identical to the effect on the S to P reaction, so no change in the equilibrium occurs. 15) Consider the conversion of fructose-6-phosphate to glucose-6-phosphate, that occurs in glycolysis. Fr-6P Glu-6P Keq = 1.97 a) What is the G 'o for the reaction G 'o = -RT ln Keq = x 298 ln 1.97 = x 298 x = kj/mole b) If the concentration of fructose is 1.5M and that of glucose-6p is adjusted to 0.5M, what is the G G = x 298 x ln (.333) = x 298 x (-1.1) = = -4.4 kj/mole c) Why are G 'o and G different? G 'o is a physical constant with information about the intrinsic properties of the reactants and products for the chemical reaction under study. Conversely the G is determined by the conditions that actually exist for a given reaction in specified conditions. In the case in B, the low concentration of the product (lower than standard conditions) and the high concentration of the reactant (higher than standard conditions) increased the available free energy for the reaction as shown, due to the actual condtions. 16) From data in table 14-6 calculate the G'o for the coupled reactions shown, that transfer phosphate first from phosphocreatine to ADP, and then from the resulting ATP to fructose REACTION of INTEREST is page 9

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