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Biochemistry Final > Enzyme Kinetics > Flashcards

Enzyme Kinetics Flashcards

1
Q

What are enzymes?

A

Enzymes are biological catalysts. A catalyst is a substance that increases the rate, or velocity, of a chemical reaction without itself being changed in the overall process.

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2
Q

What is a substrate?

A

The substance that is acted on by an enzyme is called the substrate of that enzyme.

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3
Q

What is catalase?

A

Enzyme that increases the rate of H2O2 decomposition.

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4
Q

What is a first order reaction?

A
  • A reaction where n=1, meaning the reaction rate depends on the first power of the reactant concentration
  • Has units of 1/(time)
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5
Q

What is a rate constant?

A

k1 is the rate constant – provides a direct measure of how fast a reaction is. The larger value of k1, the more rapid the rate.

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6
Q

What do graphs of first-order kinetic processes look like?

A

Linear scale: exponentially decreasing, i.e. half life graph ([A] vs time)

Logarithmic scale: linearly decreasing, with a slope of –k1 (ln[A] vs time)

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7
Q

What is half life?

A

The time needed for [A] to decrease by one half. For a first order reaction, the half-life is inversely proportional to k1

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8
Q

What is k-1?

A

k-1 is the rate constant for a reverse reaction

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9
Q

What happens when k-1 = k1?

A

The observed rate becomes zero as the reaction approaches a state of equilibrium.

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10
Q

What is a second order reaction?

A
  • A reaction that occurs typically when two molecules must come together to form products

A + B -> C

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11
Q

What are the dimensions of the second-order rate constant?

A

M-1 s-1

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12
Q

What is an enzyme-substrate complex?

A

It is what is formed when a substrate binds to an enzyme

  • referred to as [ES]
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13
Q

What is the rate-limiting step?

A

The rate-limiting step is the slowest step in a multistep process. It determines the experimentally observed rate for the entire process.

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14
Q

What is a transition state?

A

The transition state is thought of as a stage through which the reacting molecule or molecules must pass, often one in which a molecule is strained or distorted or has a particular electronic structure, or in which molecules collide productively.

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15
Q

What does the activation energy represent?

A

Activation energy represents the additional free energy (above the average free energy of reactant molecules) that molecules must have to attain the transition state.

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16
Q

What is transition state theory?

A

The theory that specific, complementary binding interactions between the transition-state structure and the enzyme active site accounts for the rate enhancements achieved by enzymes.

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17
Q

What does transition state theory assume?

A

Assumes that a reactant molecule that attains the transition state rapidly decomposes to a lower energy state such as the product state, or to an intermediate state.

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18
Q

What is rate enhancement?

A
  • Ratio of the rate constants for the catalyzed and noncatalyzed reactions for a given set of conditions.
  • Indicates how much faster the reaction occurs in the presence of an enzyme.
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19
Q

Do catalysts have an effect on the position of the equilibrium?

A
  • No, because delta G is the same whether a catalyst is present or not.
  • Thus, K, which equals k1/k-1 remains unchanged by a catalyst.
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20
Q

How is enthalpy related to enzymes?

A

Enzymes lower delta H# by increasing the number of bonding interactions between the catalyst and the transition state.

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21
Q

How is entropy related to enzymes (second order)?

A

Enyzmes make entropy of activation (delta S#) less negative (increase) by, for example, binding two reacting molecules in proper mutual orientation.

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22
Q

How is entropy related to enzymes (first order?)

A

Parts within a molecule are re-oriented so that a transition state may be reached.

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23
Q

The thermodynamic cost of making the entropy of activation more positive is paid by _______

A

Favorable enthalpy interactions between the catalyst and substrate

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24
Q

What is an intermediate state?

A

An intermediate state resembles the transition state but is of lower energy

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25
Q

Why are intermediate states useful?

A

Intermediate states reduce the activation energy requirement by altering the reaction pathway.

The result is that two lower activation energy barriers replace the single higher barrier.

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26
Q

What is the initial rate of an enzyme-catalyzed reaction?

A

The rate before a significant concentration of P appears

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27
Q

How is the steady state used in measurements?

A

Normally, we measure rates only after the steady state been established and before [ES] has changed much.

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28
Q

What variables are involved in expressing the steady state?

A

Substrate [S], free enzyme [E], enzyme-substrate complex [ES], and product [P]

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29
Q

What is the steady state?

A

After a brief initial period of a simple enyme-catalyzed reaction, [ES] reaches a steady state in which ES is consumed approximately as rapidly as it is formed

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30
Q

[E] + [ES] = ?

A

[E] + [ES] = [E]t

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31
Q

What happens to [E] and [ES] as they’re consumed?

A

[ES] falls very slowly as substrate is consumed, while [E] accordingly rises

32
Q

What is k1[E][S]?

A

Formation of ES complex

33
Q

What is k-1[E][S]?

A

Dissociation of ES complex

34
Q

What is kcat[ES]?

A

Breakdown of ES to E + P

35
Q

Formula for rates of formation and breakdown in steady state?

A

k1[E][S] = k-1[E][S] + kcat[ES]

36
Q

What is the Michaelis Constant?

A

KM is numerically equal to the substrate concentration at which the reaction velocity has attained half of its maximum value

Thus, KM is a measure of the substrate concentration required for effective catalysis to occur.

37
Q

What does a high KM mean? Low KM?

A

An enzyme with a high KM requires a higher substrate concentration to achieve a given reaction velocity than an enzyme with a low KM but the same kcat.

38
Q

Michaelis Menten general rules, what are the two assumptions?

A

1) a two-step reaction in which kcat <

39
Q

What is the Ks?

A

Ks = (k-1/k1) = [E][S]/[ES]

40
Q

What is the turnover number?

A
  • The number of substrate molecules turned over per enzyme molecule per second
  • Kcat is the symbol for it
41
Q

What determines catalytic efficiency?

A

(Kcat)/(KM)

42
Q

What is maximum velocity?

A

The reaction approaches maximum velocity at high substrate concentrations, where [S] is much greater than KM, because the enzyme molecules are saturated; every enzyme molecule is bound by substrate.

43
Q

What determines the maximum possible value of a rate constant?

A

The frequency with which enzyme and substrate collide in solution.

44
Q

What is the limit of catalytic efficiency?

A
  • Limited by diffusion: every encounter leads to reaction, so nothing but the rate of molecular encounters limits the velocity.
  • Diffusion theory predicts that Kcat/KM will attain a maximum value of about 10^8 to 10^9
45
Q

What is a Lineweaver-Burk plot used for?

A
  • Also known as a double reciprocal plot
  • Provides a quick test for adherence to the MIchelis-Menten kinetics and allows easy evaluation of the critical constants.
46
Q

What are the variables for Lineweaver-Burk plot?

A

(y=1/v), (m=KM/Vmax), (x=1/[S]), (b=1/Vmax)

47
Q

How do you determine kcat from Vmax?

A

Vmax= kcat[E]t

48
Q

What is an Eadie-Hofstee plot?

A

(y=v), (x= v/[S]), (m= KM)

Vmax at (V/[S]) = 0

Km = slope of the line

49
Q

What is the most likely explanation for a mutation that affects only kcat?

A

An amino side chain involved in catalysis (the active site nucleophile) has been mutated, but not one involved in binding to the substrate in the ground state

50
Q

What is the most likely explanation for a mutation that affects only KM?

A

An amino side chain that binds to the substrate but is not involved in the transition state

51
Q

Is it possible for a mutation to affect both kcat and KM?

A

Yes, it happens frequently.

This would occur for a residue that binds substrate and then makes stronger interactions with the transition state as a result of an altered geometry in the transition state that optimizes the interaction.

52
Q

What are multisubstrate reactions?

A

Reactions involving more than one substrate.

Chymotrypsin, for example, cleaves peptide bonds by using two substrates: (1) a polypeptide and (2) water

53
Q

What is random substrate binding?

A

In random substrate binding, either substrate can be bound first, although in many cases one substrate will be favored for initial binding, and its binding may promote the binding of the other.

An example is the phosphorylation of glucose by ATP, with hexokinase as the enzyme

54
Q

What is ordered substrate binding?

A

In ordered substrate binding, one substrate must bind before a second substrate can bind significanty.

An example is the oxidation of substrates by the cofactor NAD+, which
is related to the cofactor NADP+ in the dihydrofolate reductase mechanism.

55
Q

What is the ping-pong mechanism?

A

One substrate is bound, one product is released, a second substrate comes in, and a second product is released.

Chymotrypsin: first substrate is protein, first product released is amino side of the alpha carbon. Second, second substrate is water, second product released is carboxyl complex.

56
Q

What is the pre-ready state?

A

?
Also known as the burst phase.

K3 is smaller than k2, so intermediates form quickly on each enzyme molecule, with accompanying release of 1 mole product A. After this, more A can be formed only after each intermediate breaks down and the enzyme becomes available again.

57
Q

What do single-molecule studies of enzyme activity tell us?

A

Each state of the enzyme (E, ES, EP) has multiple conformations and each conformation is associated with a different free energy.

Energy barriers along both dimensions could give rise to the variations in kcat observed in these experiments.

58
Q

What is a reversible enzyme inhibitor?

A

Reversible enzyme inhibitors involve noncovalent binding of the inhibitor to the enzyme and can always be reversed, at least in principle, by removal of the inhibitor.

59
Q

What is an irreversible enzyme inhibitor?

A

Irreversible enzyme inhibitors involve covalent binding of the inhibitor to the enzyme

Seen in toxins, poisons, and sometimes medicines

60
Q

What does a competitive inhibitor do?

A

A type of reversible enzyme inhibitor

The inhibitor competes with the substrate by binding to the same site on the enzyme.

Increases apparent KM

61
Q

How can you tell that an inhibitor is competitive by looking at a Lineweaver-Burk plot?

A

The lines of I- (without inhibitor) and I+(with inhibitor) intersect the y axis at the same place

62
Q

What is KMapp?

A

It is the apparent KM, used when measuring inhibitor interactions

63
Q

How do you determine KI?

A

(y=KMapp), (x= [I]), (m = KM/KI), (b=KM)

64
Q

What happens to Lineweaver-Burk plots and Eadie-Hofstee plots when a competitive inhibitor is involved?

A

Still linear graphs, with KM (but not Vmax) hanged by the presence of the inhibitor.

65
Q

How can you reverse a competitive inhibitor?

A

Increase the concentration of substrate

66
Q

What does an uncompetitive inhibitor do?

A

Binds to a second site on an enzyme surface (not the active site) in such a way that it modifies kcat.

Does not interfere with substrate binding, but completely or partially prevents the catalytic step.

Reduces Vmax as well as KM

67
Q

How can uncompetitive inhibition be distinguished from competitive inhibition?

A

(1) uncompetitive inhibition cannot be reversed by increasing [S]
(2) the Lineweaver-Burk plot yields parallel rather than intersecting lines

68
Q

What is mixed inhibition?

A

Occurs when a molecule or an ion can bind to both the free enzyme and the ES complex.

69
Q

What is noncompetitive inhibition?

A

A type of mixed inhibition where KI = KI’

Where KI = inhibitor binding to E

KI’ = inhibitor binding to ES

In most cases, the inhibitor has a greater affinity for the free enzyme than for the ES complex, thus alpha is typically greater than alpha’

70
Q

By what factor is KM increased in competitive inhibition?

A

KM is increased by a factor of alpha/alpha’

71
Q

By what factor is Vmax reduced in uncompetitive inhibition?

A

Vmax is reduced by a factor of 1/alpha’

KM is increased by a factor of alpha/alpha’

72
Q

At what values of [S] do mixed inhibitors effectively reduce v?

A

Mixed inhibitors effectively reduce v at low and high values of [S]

73
Q

What does a Lineweaver-Burk plot look like for mixed inhibition kinetics?

A

(y=1/v), (x=1/[S]), (m=alpha*KM/Vmax), (b= alpha’/Vmax)

X intercept= -alpha’/alpha*KM

74
Q

What are affinity labels?

A

Affinity labels are substances that can be used to label the active site of an enzyme.

75
Q

What is a suicide inhibitor?

A

Affinity labels that are unreactive until acted on by the enzyme, at which point it binds irreversibly.

Called suicide inhibitor because the enzyme “kills” itself by processing the inhibitor from a benign to a reactive form.

Biochemistry Final (4 decks)

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