midterm 2 Flashcards

1
Q

how to calculate Vmax

A

1/y int of lineweaver burke

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2
Q

how to calculate km

A

-1/x int of lineweaver burke

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3
Q

how to calculate Kcat

A

Vmax/[E]

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4
Q

how is Vmax affected by change in [E]

A

affected proportionally; e.g. 1/2[E] ==> 1/2Vmax

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5
Q

how is Km affected by change in [E]

A

it’s not

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6
Q

how is Kcat affected by change in [E]

A

it’s not

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7
Q

how to calculate efficiency of an enzyme

A

Kcat/Km

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8
Q

where does a competitive inhibitor bind

A

the active site

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9
Q

what changes occur during competitive inhibition (km, Vmax, etc)

A

Vmax does not change; Km increases

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10
Q

where does an uncompetitive inhibitor bind and what does it affect

A

binds to ES complex without inhibiting substrate binding and inhibits catalytic function only

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11
Q

what changes occur during uncompetitive inhibition (km, Vmax, etc)

A

decrease in Vmax, decrease in Km, no change in Km/Vmax (slopes are equal)

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12
Q

relate Vmax (+I) to Vmax (-I) for uncompetitive inhibition

A

Vmax (+I) = Vmax (-I)/alpha’

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13
Q

relate Km (+I) to Km (-I) for uncompetitive inhibition

A

Km (+I) = Km (-I)/alpha’

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14
Q

where does a mixed inhibitor bind and what does it do

A

binds enzyme (at regulatory site) with or without substrate and inhibits both substrate binding and catalysis

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15
Q

what changes occur during mixed inhibition (km, Vmax, etc)

A

decrease in Vmax, increase or decrease in Km

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16
Q

where does a noncompetitive inhibitor bind and what does it do

A

binds enzyme equally with or without substrate and both inhibits substrate and catalysis

17
Q

what changes occur during mixed inhibition (km, Vmax, etc)

A

Vmax decreases, Km stays the same –> same percent inhibition overall; MM curve shifts down and left

18
Q

Relate Vmax (+I) to Vmax (-I) for a mixed inhibitor

A

Vmax(+I) = Vmax (-I)/alpha’

19
Q

relate Km (+I) to Km (-I) for a mixed inhibitor

A

Km (+I) = (alpha/alpha’)Km (-I)

20
Q

Relate Vmax (+I) to Vmax (-I) for a noncompetitive inhibitor

A

Vmax (+I) = Vmax (-I)/alpha’

21
Q

relate Km (+I) to Km (-I) for a noncompetitive inhibitor

A

Km (+I) = (alpha/alpha’)Km (-I) …. alpha = alpha’

22
Q

relate Km (+I) to Km (-I) for a competitive inhibitor

A

Km (+I) = alphaKm (-I)

23
Q

calculate alpha

A

alpha = 1 + [I]/Ki

24
Q

calculate alpha’

A

alpha’ = 1 + [I]/Ki’

25
what does alpha relate to
free enzyme
26
what does alpha' relate to
E-S complex
27
what does Ki and Ki' tell you
affinity of enzyme to free enzyme and E-S complex, respectively
28
2 methods to control velocity of enzyme catalyzed rxns
control of enzyme activity and control of amount of enzyme
29
properties of an efficient enzyme
fast, catalyzing many reactions per second, and be able to accomplish this at a relatively low substrate concentration
30
what is the michaelis menten equation (linearized and equal to Vo)
``` 1/Vo = Km/(Vmax[S]) + 1/Vmax Vo = (Vmax[S]) / (Km + [S]) ```
31
what are the 4 assumptions of michaelis menten
1. [S] >>> [E} 2. catalysis is rate limiting: k2 <<< k1 3. Vo = k2[ES] 4. [ES] at steady state: rate of formation = rate of breakdown
32
what equation/chemical reaction is michaelis menten setup
E + S ES --> E + P
33
energy charge regulation in anabolic pathways
ATP utilizing ==> activated by ATP, inhibited by ADP/AMP
34
energy charge regulation in catabolic pathways
ATP regenerating ==> activated by AMP/ADP, inhibited by ATP
35
What does a decrease in Km tell you about inhibitor affinity (E vs E-S)
higher binding affinity to E-S complex
36
What does Ki > Ki' tell you
the inhibitor has a stronger affinity for the E-S complex (dissociation constant for E-S is lower than for free enzyme)
37
what does a Km increase tell you about inhibitor affinity (Ki)
increase in Km indicates a higher affinity for free enzyme
38
how can an enzyme with a low kcat be efficient?
have a low Km, so it can bind substrate at relatively low concentrations