enzyme kinetics Flashcards

1
Q

What does a catalyst do?

A

Catalyst: provides an alternative pathway
for the reaction in which the rate-
determining step has a lower Gibbs
activation energy than that of the non-
catalysed reaction

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2
Q

How does rate differ based on temperature?

A

At a given temperature, the rate
constant is greater for the
catalysed reaction, so the reaction is
faster

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3
Q

How does the reaction process differ based on if its catalysed or not?

A

Non-catalysed reaction proceeds through a single transition state,
whereas the mechanism for the catalysed reaction involves the
formation of an intermediate

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4
Q

What are enzymes?

A

Enzymes: protein molecules, catalyse a specific reaction, unchanged by
the reactions they catalyse

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5
Q

What are the two steps of an enzyme catalysed reaction?

A

There are two distinct steps in an enzyme catalysed reaction
The first step is reversible
This and other factors affect the kinetics of enzyme reactions
The second step is irreversible

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6
Q

Describe properties of enzyme bonding and process

A
  • The enzyme binds a single substrate (the model assumes cofactors are already
    bound)
  • There are two distinct steps. The substrate binds reversibly but product
    formation is irreversible
  • A steady state approximation applies (quantitative analysis of the kinetics
    of enzyme-catalysed reactions)
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7
Q

What happens after the start of an enzymic reaction (rate)?

A
  • Very soon after the start of an enzymatic reaction a ‘steady state’ situation
    is reached. This means the [ES] remains constant.
    Rate of formation of ES = Rate of consumption of ES (conversion to E + P)
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8
Q

How does E & ES change in the initial reaction phase?

A

n the initial phase of the reaction: [E] ↓ and [ES] ↑
[ES] and [E] remain steady over most of the reaction time.
Only at the end of the reaction, when [S] is exhausted, does [ES] decline

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9
Q

What is the Michaelis menten equation?

A

V (rate of reaction) = max(S)/Km+ (s)

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10
Q

What are some properties to understand of the MM equation?

A

The velocity (rate) of an enzyme catalysed
reaction increases in a non-linear fashion
Vmax or (rate)max=maximum velocity the reaction
can achieve. It is equal to K2[E]0
KM=the Michaelis constant. It is (
k2 +
k-1)/k1

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11
Q

How does the enzyme differ based on (S)? + vmax rate

A

The enzyme is said to be saturated. If the [S] increases, no more enzyme–
substrate complexes can be formed, and the rate of the reaction is independent
of [S]
Vmax=maximum rate that a given enzyme can achieve, indicates the velocity of
the reaction when the enzyme active site is saturated and indicates the
maximum number of moles of S that can be processed in a unit of time (mol s-1)

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12
Q

How is rate affected by S?

A

Rate of the reaction is independent of the [S].The rate of reaction remains constant at the maximum value, Vmax, and the reaction is zero order with respect to [S]

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13
Q

How does the equation change when S is larger than KM?

A

V = Vmax = k2

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14
Q

How does the equation change when S is very small?

A

V = Vmax (S)/Km = K2 (E)0 (S)/Km

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15
Q

What is the turnover number?

A

The number of substrate molecules converted into product by an enzyme
molecule in a unit time when the enzyme is fully saturated with substrate
It is related to Vmax and is equivalent to the rate constant
k2. It is a comparative measure of enzyme capacity or enzyme efficiency

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16
Q

What is the equation for the turnover number?

A

turnover number = K2 = Vmax/ E0 (s-1)

17
Q

What is KM?

A

KM= substrate concentration ([S]) at which the
reaction velocity is half of V maximum (Vmax/2)

18
Q

What does KM indicate?

A

KM gives an indication of how tightly the enzyme
binds its substrate

19
Q

How does km differ in weak substrates vs good substrates?

A

Weak substrate, large
KM: a high [S] is needed to
achieve Vmax/2 (and even more to reach Vmax!)
Good substrate, low
KM: Only a low [S] needed to
reach Vmax

20
Q

What is the line weaver burk equation?

A

1/V = KM/V~~~S+ 1/Vmax