Below you will find a series of enzyme kinetics problems - you do not have to hand these problems in, but be sure you complete and understand these problems. You will find that exam questions in this area will be very similar to the problems in this section. If you do not understand the methods for solving these problems, please consult with the members of your team, or feel free to contact me. Good luck!!!
1. In the first order reaction where substrate A is converted to product B, the concentration of A at time = 0 is 0.50 mM. After 2 seconds, it is 0.25 mM. What will it be after 5 seconds?
2. a. If the half-time of a first-order reaction is 0.3 seconds, what is its rate constant k?
b. How long will it take for 95 percent of the reactant to disappear?
3. Transaminase catalyzes the reaction:
Pyridoxal phosphate (PP) acts as a coenzyme in this catalytic process. Calculate Km for the apoenzyme-coenzyme complex from the following data, obtained when the concentration of PP was varied while concentrations of glutamate and oxaloacetate and other conditions were held constant:
| Glutamate disappearing per minute, mg | PP added, micromolar |
|---|---|
| 0.17 | 0.30 |
| 0.27 | 0.50 |
| 0.43 | 1.0 |
| 0.65 | 2.0 |
| 0.73 | 3.0 |
| 0.78 | 4.0 |
| 0.79 | 5.0 |
| 0.81 | 10.0 |
4. Salicylate inhibits the catalytic action of glutamate hydrogenase. a. Determine the type of inhibition by graphical analysis of the following data. Assume that the salicylate concentration is held constant at 40 mM. b. Also calculate Km for the enzyme and c. Ki, the dissociation constant for the enzyme-inhibitor constant.
| Substrate Concentration, mM | Product per min (no salicylate) | Product per minute (salicylate) |
|---|---|---|
| 1.5 | 0.21 | 0.08 |
| 2.0 | 0.25 | 0.10 |
| 3.0 | 0.28 | 0.12 |
| 4.0 | 0.33 | 0.13 |
| 8.0 | 0.44 | 0.16 |
| 16.0 | 0.40 | 0.18 |
5. From the following data from an enzymatic reaction, determine a) the type of inhibition, b) Km for the enzyme and c) Ki for the enzyme-inhibitor complex.
| Substrate Concentration, mM | Product per hr, ug (no inhibitor) | Product per hr, ug (6 mM inhibitor) |
|---|---|---|
| 2.0 | 139 | 88 |
| 3.0 | 179 | 121 |
| 4.0 | 213 | 149 |
| 10.0 | 313 | 257 |
| 15.0 | 370 | 313 |
6. Glycerokinase catalyzes the reaction:
The following results were obtained upon the addition of the chromium salt of ATP to the reaction mixture:
| Substrate Chromium ATP concentration, M | Vu | Vu | Vu | Vu |
|---|---|---|---|---|
| ---- | 0.100 | 0.050 | 0.033 | 0.025 |
| 0.0 | 50.0 | 40.0 | 33.3 | 33.3 |
| 10 | 6.25 | 6.06 | 6.06 | 5.88 |
| 20 | 3.45 | 3.13 | 3.08 | 3.23 |
| 30 | 2.38 | 2.27 | 2.35 | 2.33 |
a) Classify the inhibition according to type.
b) Calculate a value for the inhibitor constant Ki.
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