bibc 102, metabolic biochemistry randy hampton...

BIBC 102, Metabolic Biochemistry Randy HamptonProblem Set 1 Fall 2002

Enzymology problems

1)The sweet taste of fresh corn is due to the high level of sugar in the kernal. Store-bought corn that has been sitting around for a few days is not as sweet because about 50% of thesugar 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 icecold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is thebiochemical basis for this procedure?

Like all metabolic reactions, the conversion of the delicious sugar into the much-less-sweetstarch is mediated by enzymes. The brief boiling right after harvest denatures theseenzymes and thus removes the catalysts that cause the loss of sugar, thus preserving thewholesome sweetness of freshly-picked corn.

2) The enzyme urease enhances the rate of urea hydrolysis at a given pH and temperatureby a factor of 1014 Suppose a given amount of urease can hydrolyze a sample of urea in 5minutes. How long would the same process take without the enzyme?

the conversion without the enzyme would take 5 x 1014 minutes. To make this numbermore understandable, let’s convert it. How many minutes in a year?

365.25 days/yr x 24 hrs/day x 60 min/hr = 5.26 x 105 minutes.

So it would take (5x1014

)/ (5.26 x 105), 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 1014

fold increase in the rate constant at 37 degrees C (remember Lord Kelvin!) ?

To the left is the expression relating rateconstant (in this case with a little r subscript todistinguish it from the Boltzmann constant, that,despite its fame and importance, is blue. Ourproblem is best addressed in two steps. First,what is the effect of a change in ∆G‡ on rateconstant. And then, what particular change will

give a 1014 increase? Let’s represent the change as a big, red C. Because we know itis going to increase rate, it will be a lessening of activation energy. So the new rate constantfrom 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 ofexponents, is an independent factor that alters the old rate constant by multiplication! So itis not dependent on the original rate constant.

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So now our question is, what C will make the term ec/RT equal to 1014. It is really just amatter of algebra. (I hated when my physics professor would say that…). To solve for Cin this case,

C/RT = ln1014,

C = RTln1014

I get C = 83 kJ for a 1014 fold rate enhancement.

4) In general, what change in activation energy (at 37 degrees C) will result in a 10 fold change in arate constant by the equation in 3? Why is this general?

We know from question 3 above that the effect of changing the activation energy isdescribed by a multiplicative term, ec/RT . What is particularly neat about this term is that it issymmetrical for changes in either direction. Meaning that the effect of increasing ordecreasing the activation energy has the same multiplicative effect on the rate, whether it isslowed or hastend. For instance, let’s compare the effect of lowering the activation energyby C, and the effect of raising the activation energy the same amount, C

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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

ec/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"-aspectacular book about the researchers and research of the molecular biology revolution- by HoraceJudson) the following about allosteric enzymes and allosteric regulation: ",,,I would say particularlyJack Monod's work on allostery is a very powerful theoretical concept…[It] meant that you couldconnect any metabolic pathway with any other metabolic pathway, you see, because there was nonecessary relation between what was going on at the catalytic site and the control molecule that wascoming in". What does he mean by this?

What Dr. Crick was getting at is that allosteric regulators, be they inhibitors or activators, do nothave to look like the substrates or products of the reaction. This in not the case forcompetitive inhibitors, that usually closely resemble the substrate, and in that way fit into theactive site and block enzyme activity. Because allosteric regulation is mediated by bindingsites distant from the active site, and on separate proteins that are in quaternary associationwith the catalytic subunits, there are no limitations on the type of molecules that mightregulate a particular enzyme. Thus, any metabolic pathway can potentially comminicate

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regulate any other metabolic pathway if the correct allosteric controls and binding sites are inoperation. Very deep!

6) At what substrate concentration willan enzyme with a kcat of 30/sec and a Km of0.0005 M show one quarter of its maximumrate? Determine the fraction of Vmax that wouldoccur at the following substrate concentrations:[S]= (1/2)Km, 2Km, and 10Km.

The first one is shown to the right, worked out indetail (with the cool digital pad). The other onesare done the same way. Just plug S into thealgebraic equation. No need to plug in actualnumbers.

If S = 0.5 Km, the S/(S+Km) will equal 1/3, andso the rate will be 1/3 of Vmax.

If S is equal to 2 Km, the ratio is 2/3 and the rateis 2/3 of Vmax

If S is equal to 10Km, the ratio is 10/11 , and therate is 10/11Vmax.

I idea is that you have to make bigger and bigger changes in S to make a difference as the rate getsnearer Vmax

7) The following experimental data were collected during a study of catalytic activity ofan intestinal peptidase with the substrate glycylglycine, in which the dipeptide is cleaved into twomolecules of glycine by hydrolysis.

[S] (mM) rate ( mol/min)

1.5 0.212.0 0.243.0 0.284.0 0.338.0 0.4416.0 0.45

Use a Lineweaver-Burk plot to determine the Km and Vmax for this enzyme.

According to your text, the correctvalues for the constants for this particularpeptidase are

Km = 2.2 mM

Vmax = 0.51 µM/min

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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 differentvariants of the same enzyme. That is, A, B, C are all catalyzing the same reaction, but they havedifferent 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 relationshipbetween 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 saturationbehavior predicted by the M&M (as opposed to the Eminem) equation

Remembering that the X intercept value is –(1/Km) and the Y intercept is 1/Vmax , theanswers 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 samevalue 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 Yinterecepts, 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…

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9) Irreversible inhibition of an enzyme. Many enzymes are inhibited irreversibly byheavy-metal ions, such as mercury, copper, or silver cations, which can react with essentialsulfhydryl groups to form mercaptides:

Enz-SH + Ag+ à 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 justenough AgNO3 to completely inactivate the enzyme. A total of 0.342 µmol of AgNO3 wasrequired. Calculate the minimum molecular weight of the enzyme. Why does the value obtained inthis way give only the minimal molecular weight?

The Ag+ is a way to measure thenumber of moles of the proteinassociated with 10 mg. But we onlyget a minimum molecular weight,because with the informationprovided, we don’t know how manyAg+ it takes to neutralize a moleculeof enzyme. If it is one per enzyme,then we get the smallest molecularweight. And that is why the questionasks for the minimum. For example,if it actually took two Ag+ to neutralizeeach enzyme molecule, then thenumber of moles of enz presentwould be half of the Ag+ used, andthe estimated molecular weightwould double.

10) Suppose that the substrate vs. rate plot for the enzyme in the question above showed asigmoidal shape when rate was plotted against substrate concentration. Now what is the minimalmolecular weight of the enzyme as determined from the equation.

That sigmoidal plot indicates that this is an allosteric enzyme, and so it must have quaternarystructure. Thus, even if every Ag+ neutralizes an individual enzyme molecule, therequirement for quaternary structure means that the in-solution molecules are AT LEASTdimers (the simplest quaternary structure) so the minimum molecular weight would bedouble theminimum for aprotein withonly tertiarystructure.

11) Supposethat a 6 step,linearmetabolicpathway isregulated in thefollowing way: The third enzyme (ENZ3) in the pathway, that is, the one that catalyzes theconversion of the third molecule of the pathway into the fourth one (hint: write it out!!). ENZ3

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is allosterically regulated. It is regulated both by the first substrate of the pathway, and by the finalproduct. The regulation of ENZ3 makes sense: when there is too much initial substrate, the activityis increased, and more of it is used in the pathway. Conversely, when there is sufficient product, theactivity 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 ALLOSTERICINHIBITOR

b) The initial substrate and the final product have (not surprisingly) very differentstructures. Speculate about the structural features of ENZ3 that allow it to be allostericallyregulated, 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 theseare probably on separate proteins from the polypeptide that forms the active site. Like thepicture of aspartate transcarbamoylase in the book and lecture, many allosteric enzymeshave very complex structures, with multple independent regulatory proteins.

c) Draw a representation of a substrate-rate plot, showing the effect of each of these twomodulators on the activity of enzyme 3.

BIOENERGETICS

12)Calculate the standard free energy change for each of the Keq. You can see thereactions in the book. USE 301 K (37 C)

Keq 6.8

∆G’o = -RTln6.8 = -8.315 x 298 x ln 6.8 J/mole

Keq 0.0475

∆G’o = -RTln0.0475 = -8.315 x 298 x ln 0.0475 J/mole

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BIBC 102, Metabolic Biochemistry Dr. Randy HamptonProblem Set 1 Fall 2001

Keq 254

∆G’o = -RTln 254 = = -8.315 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 thestandard conditions, 298 K. dWhat we end up with is that when DG changes by an amountC, the change in Keq is represented by

ec/RT, or e-c/RT

, depending on whether or not the ∆G is made more negative, ormore positive, respectively. What C value will make the first term equal to 100?

C = RT ln 100 = 8.315 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 by100 fold.

If the ∆G’o is increased (made more positive) by 11.5 kJ/mole, then the Keq drops by100 fold.

14) Draw a reaction coordinate for the simple reaction of S becoming P. Show what happens whenthe following changes to the diagram occur:

a) The free energy of Pbecomes lower.

b) A catalyst is adde is raiseddto the reaction mix

c) the free energy of S

Now, which of these changes affect the ∆G ofthe reaction?

Both A Both A and C will alter the DG ofthe reaction, which is the difference in freeenergy between the reactants and theproducts.

Which affects the rate of the S to P reaction?

Both B and C affect the S to P reactionrate. However, only B affects the forwardand reverse rates indentially (see 3above) and so the equilibrium is notperturbed.

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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 byand 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 occursin glycolysis.

Fr-6P à Glu-6P Keq = 1.97

a) What is the ∆G'o

for the reaction

∆G'o

= -RT ln Keq = -8.32 x 298 ln 1.97 = -8.32 x 298 x 0.678 = -1.68 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 = -1.68 + 8.32 x 298 x ln (.333) = -1.68 + 8.32 x 298 x (-1.1) = -1.68 – 2.72 =-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 standardconditions) 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, thattransfer phosphate first from phosphocreatine to ADP, and then from the resulting ATP to fructose

REACTION of INTEREST is

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