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3.5 Solving Enzyme Kinetics Problems

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

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.

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.


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:


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:




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