Reactions with enzymes play an important role in chemistry. Kinetic analyses of these reactions help — Physical Chemistry — Kinetics Chemistry Question
Enzyme Kinetics
Reactions with enzymes play an important role in chemistry. Kinetic analyses of these reactions help to understand the typical behavior of enzymes. An enzymatic reaction of substrates A and B witn an enzyme E can be described by the equations (1) – (5):
(1) E + A ⇌ EA equilibrium constant KA
(2) E + B ⇌ EB equilibrium constant KB
(3) EB + A ⇌ EAB equilibrium constant K’A
(4) EA + B ⇌ EAB equilibrium constant K’B
(5) EAB → products reaction rate v = k[EAB]
When the rate constant is small the equilibria (1) – (4) are hardly shifted due to the reaction (5). This leads to expression (6) in which Vmax is the maximal velocity of the reaction, that is reached when the enzyme is saturated with the substrates (all enzyme is bound to A and B).
Vmax / v = 1 + KA / [A] + KB / [B] + KA K'B / [A][B] (6)
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Consider the enzymatic hydrolysis of maltose by the enzyme α-glucosidase from yeast:
maltose + H2O → 2 glucose (7)
The substrate maltose is usually present in concentrations ranging from 10^-4 – 10^-1 mol dm^-3. Water is the solvent, thus its concentration is practically constant at 55.6 mol dm^-3. Expression (6) can now be simplified by letting [B] approach infinity.
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An enzymatic reaction can be strongly retarded or blocked by an inhibitor I according to the equation:
E + I ⇌ EI (8)
with equilibrium constant KI. For competitive inhibition the inhibitor competes with the substrate at the binding side of tze enzyme, thus the reaction is slowed down but leaving vmax unaffected. In the Michaelis-Menten equation KA is then multiplied by a factor (1 + [I] /Ki ), which equals 1 for [I] = 0 and is large when [I] is large. For non-competitive inhibition I does not compete with A: KA is not affected, vmax is lowered. In the Michaelis-Menten equation vmax is then divided by a factor (1 + [I] /Ki ). Inorder to investigate the hydrolysis by α-glucosidase he model substrate p- nitrophenyl-α-D-glucoside (PNPG) is used instead of maltose whereby the release of nthe yellow p-nitrofenol is monitored spectrophotometrically. The following experiment is carried out: PNPG is used in the presence of maltose to measure the activity of glucosidase.
Give the equilibrium constants KA, KB, KA’, KB’ in terms of the respective concentrations.
Model Answer
KA = [E][A] / [EA]
KB = [E][B] / [EB]
K’A = [EB][A] / [EAB]
K’B = [EA][B] / [EAB]
Give the simplified expression. NB: The simplified expression is the famous Michaelis-Menten equation for an enzymatic reaction with one substrate.
Model Answer
v = Vmax / (1 + KA / [A])
Simplify the Michaelis-Menten equation further by taking [A] as very small (thus approaching zero).
Model Answer
If [A] → 0 then KA / [A] >> 1 and v = Vmax [A] / KA.
The order n of a reaction is defined by v = k c^n. Thus, for n = 1 the kinetics are first order. What is the n of the reaction [A] → 0.
Model Answer
n = 1 (first order kinetics).
Simplify the Michaelis-Menten equation by taking [A] as very high, thus [A] → ∞, which is the case when the enzyme is completely saturated with substrate.
Model Answer
If [A] → ∞ then KA / [A] << 1 and v = Vmax.
What is the order n of the reaction for [A] → ∞.
Model Answer
n = 0 (zero order kinetics).
The constant KA is a measure for the affinity of an enzyme for its substrate. Does a high affinity correspond with a high or a low value of KA? At which velocity is [A] = KA?
Model Answer
A high affinity corresponds with a small KA. v = ½ Vmax when [A] = KA.
Draw a graph of v versus [A]. (Take [A] at the x-axis.) Indicate vmax and KA in this graph.
Model Answer
[VISUAL]
Which situation applies:
- The maltose does not influence the rate of release of p-nitrophenol.
- Maltose functions as a competitive inhibitor.
- Maltose functions as a non-competitive inhibitor.
Mark the correct answer.
Model Answer
Maltose functions as a competitive inhibitor.
Draw a graph of v versus [A] (take [A] at the x-axis) for the release of p-nitrophenol in the presence of maltose for = Kmaltose. Insert the graph made in question 18.6. Mark the points vmax and ½ vmax.
Model Answer
[VISUAL]