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BIOL 2421 - Cellular Biochem > 9. Enzyme kinetics > Flashcards

9. Enzyme kinetics Flashcards

1
Q

define exergonic reaction

A

a spontaneous reaction

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2
Q

what value of delta G does an exergonic reaction have

A

negative

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3
Q

describe the speed of exergonic reactions when catalysts aren’t used

A

very slow

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4
Q

are most chemical reactions exergonic or endergonic

A

exergonic

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5
Q

what is a catalyst

A

a substance that speeds up attainment of equilibrium of a chemical reaction

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6
Q

T or F: a catalyst changes the position of the equilibrium when it’s used

A

false! it only speeds up the process

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7
Q

what happens to a catalyst after a reaction is complete

A

it is recycled to participate in future reactions

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8
Q

T or F: a catalyst may be temporarily changed during a reaction

A

true; but it remains unchanged in the overall process

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9
Q

define catalysis

A

the increase in the rate of a chemical reaction due to the participation of a catalyst

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10
Q

what are biological catalysts

A

enzymes

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11
Q

what key properties do enzymes have that you don’t see in inorganic catalysts

A

high degree of substrate specificity, substantial reaction acceleration, and they function in aqueous solutions at physiological pH

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12
Q

why are enzymes control points for metabolism

A

they can accelerate chemical reactions by a factor of a million or more

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13
Q

what is enzyme kinetics

A

the study of reaction rates and how they respond to changes in reaction parameters

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14
Q

what letter is rate represented by

A

V (ie velocity)

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15
Q

in the reaction of S –> P, what does the rate represent

A

the quantity of S that disappears in a specific unit of time

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16
Q

what are the units of V

A

M/s

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17
Q

T or F: V (the rate) should equal the rate of appearance of P

A

true

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18
Q

for a first order reaction, what is the rate formula

A

V=k[S]
where V=rate
k=rate constant
and [S]=concentration of substrate

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19
Q

what are the units of k for a first order reaction

A

s^-1

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20
Q

describe the reaction rate (V) in regards to k

A

higher k value = faster reaction

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21
Q

for a second order reaction, what is the rate formula

A

V=k[S1][S2]

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22
Q

what are the units of k for a second order reaction

A

(M^-1)(s^-1)

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23
Q

what are pseudo-first order reactions

A

a second order that appears as a first when one substrate is present in far excess of the other

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24
Q

what is the rate formula for a pseudo-first order reaction

A

V=k[S1][S2], but S2 is negligible

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25
Q

how do we plot reaction rates?

A

[P] vs time

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26
Q

from a graph of product formed over time, what formula do we use to find V

A

delta P/delta t = V = k[S]

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27
Q

what is V0

A

initial velocity

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28
Q

how do we find V0 from a [P] vs time graph

A

calculate the slope at time=0 (tangent line)

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29
Q

describe the slope of V0 depending on different amounts of [S]

A

at low [S], V0 increases almost linearly with increases in [S]

at higher [S], V0 increases by smaller and smaller increments in response to increases in [S]

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30
Q

what is Vmax (in regards to [S] and V0)

A

the point where increases in [S] no longer increases V0

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31
Q

what is Vmax (in regards to the reaction)

A

Vmax is the fastest rate at which it can convert substrate to product. Adding more S doesn’t increase the rate

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32
Q

what are k1, k2, k-1, k-2

A

k1: E+S –> ES
k2: ES –> E+ P
k-1: ES –> E + S
k2: E+P –> ES

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33
Q

which step in the formation of a product from E and S is the slow step

A

ES –> E+P

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34
Q

what is another name for the slow step of a reaction

A

the rate limiting step

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35
Q

what is the M-M equation

A

V0=Vmax[S]/(Km+[S])

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36
Q

what are the assumptions of the M-M equation

A

there is an ES complex, each step has distinct rates, very early in the reaction there is no product (so k-2 is negligible), the second step is rate limiting, [Etotal]=[Efree]+[ES], and [ES] is constant (formation=breakdown)

37
Q

what is Km

A

the Michaelis constant

38
Q

what are the units for Km

A

M

39
Q

what formula represents Km

A

(k-1 + k2)/k1

ES breakdown/ES formation

40
Q

what does Km represent

A

it summarizes the rates of a reaction

it’s ES breakdown over ES formation

41
Q

what happens to Km when V0=1/2 Vmax

A

Km=[S]

42
Q

how do we plot using the M-M equation

A

V0 on the y axis, [S] on the x axis

the curve is hyperbolic

43
Q

T or F: we can use the M-M equation for second order reactions

A

false; only first order

44
Q

when does Km=Kd

A

when k2 is limiting and can be ignored, then Km=(k-1)/k1, which is equivalent to Kd

45
Q

reminder: what is Kd

A

dissociation constant of the ES complex

46
Q

what does a low Km represent

A

strong binding affinity between E and S

47
Q

what is another name for the L-B plot

A

double reciprocal plot

48
Q

how do we derive the L-B equation

A

by inverting the M-M equation

49
Q

what is the L-B equation

A

1/V0 = (Km/Vmax)(1/[S]) + 1/Vmax

50
Q

what are the axes for the L-B plot

A

y: 1/V0
x: 1/[S]

51
Q

what are the L-B plot intercepts

A

x=0: 1/Vmax

y=0: -1/Km

52
Q

what is the benefit of using an L-B plot vs an M-M plot

A

LB plot allows a better approximation of Vmax (because it’s linear, nit hyperbolic)

53
Q

what is the turnover number of an enzyme

A

the number of substrate molecules converted into product by enzyme in a unit of time when the enzyme is fully saturated with substrate

54
Q

what variable reveals the turnover number of an enzyme

A

Vmax

55
Q

what variable is the turnover rate equal to

A

k2 (which is the rate limiting step)

56
Q

what is the rate constant k2 often referred to as

A

kcat

57
Q

how can you adjust V0=k2[ES] to solve for kcat?

A

kcat=Vmax/[Etotal]

58
Q

what does kcat tell us

A

the amount of product we produce per second

59
Q

what does 1/kcat tell us

A

the time each individual reaction takes

60
Q

what rate constant is called the specificity constant

A

kcat/Km

61
Q

what does the specificity constant measure

A

the catalytic efficiency, because it takes into account both the rate of catalysis with a particular substrate (kcat) and the nature of the enzyme-substrate interaction (Km)

62
Q

why do you need both kcat and Km for the specificity constant

A

two enzymes catalyzing different reactions may have the same kcat (turnover number) but the rate of uncatalyzed reactions may be different and this the rate enhancements brought about by the enzymes may differ greatly

63
Q

what is the equation for kcat and Km

A

V0=kcat/Km x [Etotal][S]

64
Q

T or F: there is an upper limit to kcat/Km

A

true

65
Q

why is there an upper limit to kcat/Km

A

it’s imposed by the rate at which E and S can diffuse together in an aqueous solution and form ES

66
Q

what is the upper limit value for kcat/Km

A

10^8 to 10^9, in the units M^1s^1

67
Q

define kinetic perfection

A

when the catalytic velocity of enzymes is only restricted by the rate at which they encounter substrate in the solution

68
Q

what are the two types of reversible inhibition

A

competitive and uncompetitive

69
Q

in competitive inhibition, where does the inhibitor bind

A

to the active site

70
Q

how do we overcome competitive inhibition

A

by adding more substrate

71
Q

if we add more substrate in competitive inhibition, will the same Vmax be reached as without the inhibitor?

A

yes

72
Q

in competitive inhibition, since the [S] needs to be increased to reach Vmax, what else is increased

A

the Km at 1/2Vmax

73
Q

by which factor will Km be increased in competitive inhibition when more S is added

A

alpha

74
Q

how does the MM equation change to account for Km being increased by a factor of a (in competitive inhibition)

A

V0= Vmax[S]/aKm + [S]

75
Q

on a graph, how can we tell if inhibition is competitive

A

using the LB plot

76
Q

what are the signs of competitive inhibition on an LB plot

A

steeper slope=more inhibitor, more inhibitor=closer x intercept is to zero

77
Q

how do you overcome uncompetitive inhibition

A

you can’t. Adding more S has no effect

78
Q

T or F: adding more S to uncompetitive inhibition will overcome it

A

false; adding more S has no effect

79
Q

why can’t we overcome uncompetitive inhibition by adding more S

A

the inhibitor binds at a site distinct from the active site, forming an ESI complex

80
Q

what value is reduced in uncompetitive inhibition

A

Vmax

81
Q

by which factor is Vmax reduced by in uncompetitive inhibition

A

alpha’

82
Q

which way does the equilibrium shift in uncompetitive inhibition and why

A

shifts towards ES, because the inhibitor is removing ES from the population as ESI is formed

83
Q

in uncompetitive inhibition, what is the result in the equilibrium shifting towards ES

A

it makes the enzyme appear as if it has a higher affinity for S than it really does

84
Q

in uncompetitive inhibition, what else (other than Vmax) is reduced (and why which factor)

A

[S] required to reach Vmax is reduced by alpha’

85
Q

in uncompetitive inhibition, what effect does the reduced [S] and Vmax have

A

we have a decreased Km^app value

86
Q

how does uncompetitive inhibition appear on an LB plot

A

parallel lines

87
Q

why are the lines on an LB plot parallel when there’s uncompetitive inhibition

A

each line has an equally reduced slope, so all lines have the same slope = parallel

88
Q

what happens to the intercepts on an LB plot of uncompetitive inhibition when inhibition is increased

A

absolute value of the intercepts increases (ie they are more up and to the left)

BIOL 2421 - Cellular Biochem (21 decks)

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