3.5 Solving Enzyme Kinetics Problems

1) Two strains of Bacterium sweetans, A and B, use sucrose (table sugar) as a sole carbon source. The first step in the process of sucrose utilization is the passage of sucrose through a sucrose transporter protein in the membrane. The characteristics of the two transport proteins are as follows (assuming [E]tot is the same in both):

Strain               A                  B
KM               1000 mM              10 mM
Vmax             1000 mmol/min        100 mmol/min

a) Assuming that the rate of sucrose uptake is the rate limiting step in growth, which strain will grow faster if the concentration of glucose is: 10 mM? 100 mM? 1000 mM?
b) One strain was isolated from the soil and the other from the floor at Toscanini's Ice Cream, which was likely to be which? Why?

Solution:
Using the Michaelis-Menten equation, you can calculate the initial velocity of sucrose uptake for each strain under the conditions listed. Note that this is the initial rate, as sucrose was depleted from the medium, the rate would decrease. To simplify matters, assume that the number of cells is small so that the sucrose concentration does not change appreciably during growth, so that the transport protein is always running at Vo.
For these calculations, substrate = sucrose.

            Vo, enzyme of A:      Vo, enzyme of B:     Faster
[sucrose]   Km=1000mM             Km= 10mM             Growing
    mM      Vmax=1000mmol/min     Vmax= 100mmol/min    strain:
-------------------------------------------------------------------------------

   10        9.9 mmol/min            50 mmol/min          B

   100       91 mmol/min             91 mmol/min        neither

   1000      500 mmol/min            99 mmol/min          A

b) It is likely that the [sucrose] levels in the soil are very low, so that strain B would be at a selective advantage there. It is likely that a soil bacterium would have transport machinery adapted for low [sucrose].
It is likely that the [sucrose] on the floor at Toscanini's has a much higher [sucrose] than the soil. Under these conditions, strain A would be at an advantage.

2) Given an enzyme with a Km = 10 mM and Vmax = 100 mmol/min.

a) If [S]=100mM, which will increase the velocity more: a 10-fold decrease in Km, or a 10-fold increase in Vmax?
b) If [S]=10mM, which will increase the velocity more: a 10-fold decrease in Km, or a 10-fold increase in Vmax?

Solution:
You can answer these questions qualitatively or quantitatively. Having a qualitative understanding of what KM and Vmax mean is more important than memorizing the Michaelis-Menten equation.
Qualitatively:

a) With [S]>>KM, the enzyme is running close to Vmax . Because of this, decreasing KM would hardly increase V at all, but increasing Vmax 10-fold would increase V 10-fold.
b) With [S] = Km, the enzyme is running at 50% of Vmax. Decreasing Km 10-fold will raise V to near Vmax, doubling V at most. Increasing Vmax 10-fold would also increase V 10-fold.

Quantitatively:

part   Km      Vmax            [S]            Vo         result
-----------------------------------------------------------------------
a      10 mM   100 mmol/min    100 mM     91 mmol/min

a      1 mM    100 mmol/min    100 mM     99 mmol/min

a      10 mM   1000 mmol/min   100 mM     910 mmol/min   greater effect

b      10 mM   100 mmol/min    10 mM      50 mmol/min

b      1 mM    100 mmol/min    10 mM      91 mmol/min

b      10 mM   1000 mmol/min   10 mM      500 mmol/min   greater effect

3)The enzyme Carbonic Anhydrase (CA) catalyzes the following reaction:
CO2 + H20 <-----------> H2CO3
Given two identical sealed tubes with CO2 gas above buffer containing H2O and CO2 - one with added carbonic anhydrase, one without. Allow the contents to come to equilibrium, then:

a) At equilibrium, which tube will have more CO2 in the gas phase, A or B?
b) Add CO2 to the gas phase of both tubes. Which tube will absorb it faster into the liquid, A or B? Why?
c) Remove CO2 from the gas phase of both tubes. Which will replenish the CO2 faster, A or B? Why?
d) Take tube A, allow it to come to equilibrium, and inject a negligible volume of highly concentrated CA into the liquid layer. Will CO2 gas be produced, absorbed, or will there be no change?
e) Take tube B, allow it to come to equilibrium, and inject a negligible volume of highly concentrated ATP. Will CO2 gas be produced, absorbed, or will there be no change?

Solution:

a) Since the tubes are at equilibrium, and since enzymes do not change the equilibrium concentrations of reactants and products (another way of saying that they do not change Delta G), both tubes will have identical [CO2] in the gas phase.
b) The CA in tube B will accelerate the absorption of CO2 by accelerating the conversion of CO2 into H2CO3.
c) The CA in tube B will accelerate the production of CO2 by accelerating the conversion of H2CO3 into CO2.
d) This is like part A, the CA cannot affect the equilibrium concentrations of reactants or products, only alter the rate. So addition of CA to a tube at equilibrium will have no effect.
{However, addition of CA to a tube that was not at equilibrium could cause rapid absorption or production of CO2. Normally, enzyme assays are carried out in reactions where the substrates have not had enough time to reach equilibrium, so that the effect of adding enzyme will be detectable.}
e) Nothing will happen. Since ATP is not a substrate of CA, the concentrations of reactants and products have not changed, so the equilibrium will not shift.
{If ATP were a substrate of CA, then adding it to the reaction would change the amount of CO2 in the gas phase.}

4) The enzyme Lysozyme is present in human nasal mucus and catalyzes the breakdown of a component of the cell wall of certain bacteria, causing them to burst. Lysozyme cleaves a polymer of N-acetyl-glucosamine (NAG)n into smaller fragments:

                                                       lysozyme
NAG-NAG-NAG-NAG-NAG-NAG-NAG-NAG-NAG-NAG-NAG-NAG + H2O --------->

	NAG-NAG-NAG-NAG-NAG-NAG	 + 	NAG-NAG-NAG-NAG-NAG-NAG

a) Given that the reaction proceeds as written, what can you conclude about the relative stability of (NAG)n?

b) Given the following information, what can you conclude about the binding of lysozyme to its substrate?


Substrate           Relative rate of hydrolysis
(NAG)2                                 0

(NAG)3                                 1

(NAG)4                                 8

(NAG)5                              4,000

(NAG)6                             30,000

(NAG)8                             30,000

c) If you treated a solution of (NAG)100 with lysozyme, what would the majority of the product be?

Solution:

a) Since lysozyme cannot effect the Delta G of the reaction, (NAG)n must be unstable (that is, the reaction where it breaks into monomers has a Delta G<0).
{But why then do bacteria use an unstable molecule as a cell wall? Because, even though the (NAG)n will eventually come to equilibrium and decompose into monomers, the uncatalyzed rate of hydrolysis is very low, so reaching equilibrium might take decades. Lysozyme increases this rate enormously, and acts as a bacteriocide.}
b) Even though the reaction catalyzed by lysozyme appears to be:
NAG-NAG + H2O --------> NAG + NAG
the substrate binding site must have a pocket designed to hold 5 or 6 NAG units before the enzyme can get into the active conformation. This is an example of substrate specificity.
c) After a short time, it would mostly be (NAG)4, since any larger fragments will be rapidly cleaved. After a long time, it will be (NAG)2 and (NAG)3.

shanec@mit.edu