Enzyme Kinetics Flashcards

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1
Q

What is the velocity of a reaction?

A

It is the rate of the reaction. Change in concentration of product or reactant per unit time. It is usually measured by the rate of formation of the product(s)

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2
Q

What is a second-order reaction?

A

A chemical reaction where the rate of reaction is proportional to the square of the concentration of one reactant

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3
Q

What is a first-order reaction?

A

A chemical reaction where the rate of reaction is proportional to the concentration of one reactant

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4
Q

Explain the shape of the [Concentration] vs Time graph

A

General trend: Rate of reaction increases at a decreasing rate and eventually comes to a plateau. This is because over time, as the [substrate] decreases (converted to product), the frequency of effective collision between the substrate and the active site of the enzyme decreases.

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5
Q

How will an increase in [substrate] affect the velocity of an enzyme-catalysed reaction?

A

The increase in [substrate] will increase the frequency of effective collision between the substrate and the active site of the enzyme. This will result in an increase in the velocity of the reaction. Hence, the initial gradient of the [concentration] vs Time graph will be steeper.

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6
Q

What is Vmax?

A

The theoretical maximum velocity of a reaction attainable when Enzyme (total) = Enzyme-substrate complex -> i.e. all the enzymes are saturated by the substrate.

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7
Q

Why is Vmax theoretical?

A

To attain Vmax, [E-total] = [ES complex]. However, in order to achieve that, we need to supply a very high concentration of substrate to shift the equilibrium to the right. Furthermore, it is unlikely that all the enzymes will be saturated (i.e. bounded to a substrate)

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8
Q

What are the disadvantages of the Michaelis-Menten plot?

A

[Substrate] may be too concentrated that it forms a precipitate during the reaction. Hence, it would be difficult to estimate Vmax -> Hence difficult to determine the value of Km (since the value of Km is the [S] when the velocity of the reaction is half-maximum)

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9
Q

What is Km?

A

value of Km is the [S] when the velocity of the reaction is half-maximum

It is also a measure of the affinity of the enzyme for its substrate. It is inversely related to the affinity to the substrate.

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10
Q

What is Kd (Dissociation constant)

A

Ratio of the [free enzyme and substrate] to the [enzyme-substrate complex]

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11
Q

What is Kcat?

A

The number of substrate molecules converted to product per unit time when the enzyme is saturated with the substrate. It measures the maximal catalytic activity of an enzyme. Kcat = Vmax/[E-total]

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12
Q

What is the catalytic efficiency of an enzyme?

A

It is the ratio of Kcat to Km. The higher the catalytic efficiency, the higher the v0.

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13
Q

What are enzyme inhibitors?

A

Enzyme inhibitors are molecules that bind to enzymes and decrease their activity.

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14
Q

What are the types of inhibitors?

A

Irreversible and reversible inhibition (Competitive, Uncompetitive and non-competitive inhibition)

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15
Q

What is irreversible inhibition/ suicide inhibition?

A

Occurs when the inhibitor forms covalent bonds with functional groups at the active site of the enzyme. This effectively decreases the amount of active enzymes, hence decreases Vmax. No change in Km (affinity of the enzyme) because Km = [E][S]/[ES], and [E] and [ES] are reduced by the same extent. This is because when the inhibitor binds to the enzyme, the number of free enzymes decreases, which means that the number of ES complex that can be formed will be decreased by the same extent.

The inhibitor neither dissociates the enzyme-inhibitor complex nor restore enzyme activity upon dialysis of he solution.

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16
Q

What is competitive inhibition?

A

A competitive inhibitor competes with the substrate for the active site of the free enzyme. Hence, it reduces the number of free enzymes (enzymes with available active sites). This causes the equilibrium of E + S ES to shift to the left. This decreases the [ES], and hence Km increases.
However, Vmax remains unchanged. This is because [E] is very low compared to [S]. Hence, at high [S], the chances of an enzyme binding to a substrate is much higher than that of binding to an inhibitor. Thus, all enzymes will be bounded to a substrate to form ES complex. Therefore, Vmax will remain unchanged.

If both lines have the same intercept, the inhibitor is a competitive inhibitor.

17
Q

What is uncompetitive inhibition?

A

An uncompetitive inhibitor only binds to ES complexes. This alters the conformation of the enzyme and hence it affects the catalytic function of the enzyme. It does not affect substrate binding to the free enzyme.

Since the formation of the product is hindered, Vmax will decrease (because catalysis of the ES complex is hampered). Km will decrease (increase in affinity due to an increase in binding efficiency; [E] and [S] decreased to form ES complex) proportionally to Vmax

If the lines are parallel then the inhibitor is an uncompetitive inhibitor

18
Q

What is noncompetitive inhibition (mixed inhibition)?

A

A noncompetitive inhibitor is one that can bind to both free enzyme and ES complex. This disrupts catalysis and decreases product formation and therefore Vmax decreases.

The noncompetitive inhibitor can either cause an increase, decrease or no change in the value of Km. If the inhibitor binds more strongly to the free enzyme, it shifts the equilibrium from ES to E + S to form EI. Thus, [ES] decreases and Km increases (affinity decreases)

If the inhibitor binds more strongly to the ES complex, the equilibrium shifts from E + S to ES to form ESI. Hence, [E] decreases and Km decrease (affinity increases)

However, if the inhibitor binds to the E and ES with equal affinity, there will be no shift in the equilibrium and no there will be no change in Km.

If both lines do not have any relationship, then the inhibitor is a mixed noncompetitive inhibitor.

19
Q

What is a pure noncompetitive inhibitor?

A

A pure noncompetitive inhibitor is one that binds to free enzymes and ES complex equally -> Vmax decrease because product formation is hindered but there is no change in Km

If both lines converge to the same point, then the inhibitor is a pure noncompetitive inhibitor