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Enzymes > Enzymes as catalysts > Flashcards

Enzymes as catalysts Flashcards

1
Q

In an enzyme kinetics experiment what are you studying?

A
  • initial velocity at varied initial concentrations of substrate
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2
Q

Initial velocity equation

A

initial change in substrate concentration/ change in time

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3
Q

When does the initial rate depend on substrate concentration?

A

At low substrate concentrations

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4
Q

At low substrate concentrations what limits how much ES can form and what is the result?

A

[S] limits

-v (initial rate) is dependent on [S]

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5
Q

At high substrate concentrations, what limits how much ES can form and what is the result?

A

[E] limits

- v is dependent on enzyme concentration, so v (initial rate) does not change as [S] changes

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6
Q

when is Km approximately equal to Kd?

A

when the rate of ES dissociation is much greater than the rate of product formation
k2»k3

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7
Q

what is the equation for the dissociation constant (Kd) for the ES complex?

A

Kd=k2/k1

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8
Q

what is the benefit of having a low Km value?

A

it would allow an enzyme to show significant levels of catalysis at low concentrations of substrate (would be important for tissues that have very small amounts of substrate)

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9
Q

what is the benefit of having a high Km value?

A

it would allow sensitive regulation of enzyme activity in response to small changes in [S] in the range of 2-5 mM substrate
this would be critical for enzyme regulation in tissues with high level of [S]

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10
Q

what is another term for k3?

A

Kcat or turnover number

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11
Q

v=

A

Vmax [S] / ( [S] + Km)

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12
Q

Vmax =

A

K3 [E]t

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13
Q

what does a higher ratio of Kcat/Km imply??

A

higher efficiency

means you are getting more conversions to product for a given amount of ES complex

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14
Q

What does km mean?

A

The [S] at which v=1/2 Vmax

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15
Q

When is km approximately equal to the dissociation constant - kd?

A

when k2»k3

-when the substrate binds and dissociates before it goes to form product

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16
Q

When is the catalytic rate most sensitive to [S]?

A

When [S] «< km

17
Q

fES will be higher for a low km or a high km?

A

low km

18
Q

fES =

A

v/vmax = [s] /( [s] +km)

19
Q

what are the units of Kcat?

A

inverse time s^(-1)

20
Q

what is the definition of Vmax and what are its units?

A

maximal velocity achievable for a SPECIFIC concentration of enzyme
units= concentration/time

21
Q

what does a higher ratio of Kcat/Km imply??

A

higher efficiency

22
Q

when would you use a lineweaver-burk plot?

A

to accurately determine Km and Vmax

23
Q

definition of an irreversible inhibitor

A

binds very tightly to an enzyme and inactivates an essential functional group on that enzyme

24
Q

definition of a reversible inhibitor

A

binds reversibly to enzyme and gives a temporarily inactivation when it is bound

25
Q

how can a competitive inhibitor be over come?

A

by increasing [S]

26
Q

what is the definition of Ki?

A

the strength at which an inhibitor binds to an enzyme

it is the disassociation constant for the enzyme-inhibitor complex

27
Q

what does a higher Ki imply?

A

weaker inhibition (it disassociated easily)

28
Q

how will a competitive inhibitor effect the y and x intercepts on a lineweaver-burk plot?

A

the y intercept will NOT change but the x intercept will change

29
Q

how will a non-competitive inhibitor effect the y and x intercepts on a lineweaver-burk plot?

A

the y intercepts will change and the x intercept will NOT

30
Q

what value will appear to change in the presence of a competitive inhibitor?

A

Km

31
Q

what value will appear to change in the presence of a non-competitive inhibitor?

A

Vmax

32
Q

What is the equation for Kd- the dissociation constant in terms of concentration of substrates and products?

A

Kd = koff/kon = [A][B]/[AB]

33
Q

At equilibrium what two values are equal?

A

association constant and dissociation constant

Enzymes (4 decks)

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