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Biochem > Kinetics of enzyme reactions > Flashcards

Kinetics of enzyme reactions Flashcards

1
Q

what is kinetics

A

study of the rate at which compounds react

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2
Q

the rate of an enzymatic reaction is affected by what 5 things

A

enzyme, substrate, effectors, temperature, pH

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3
Q

What is the unit for the rate of change in concentration

A

Moles/L Sec-1

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4
Q

Write out the Rate equation and explain what each thing

A

picture

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5
Q

Describe the relationship between the reactant and time for first order reactions

A

The amount of reactant decays exponentially

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6
Q

The slope for first order reactions for graph ln vs time is what?
How can you tell when a reaction is first order?

A

-k

Rate depends on the first power of the concentration

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7
Q

What is the equilibrium constant

A

Ratio of forward and reverse rate constants

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8
Q

Most biochemical reactions are what order

A

second order

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9
Q

what is the unit for second order reactions

A

(Mol/L)-1 s-1

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10
Q

Rate equations for second order

A 
V= K2[A]^2 
V= K2[A][B]
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11
Q

What allows enzymes to find their substrates

A

Thermal motion

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12
Q

How is initial velocity determined?

A

Bymeasuringproductformationasafunctionoftime velocity of reaction),andthendetermining thevelocitysoonafterthereactionhasstarted

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13
Q

What apparatus is used to measure initial velocity

A

Stopped flow apparatus

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14
Q

what is the michaelis menten equation?

A

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

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15
Q

What did Michaelis and Menten speculate about the enzyme, substrate and rate determining step? And show this in simple form

A

Substrate forms to an enzyme to make a complex in a fast and reversible step.
This complex then breaks down in a slower step to yield the free enzyme and product so is the rate limiting step.
E+S ES-> E + P

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16
Q

Assumptions made in deriving the M&M equation about k-2

A

k-2 = 0, so reverse reaction between P and E is negligible

17
Q

why is the overall rate proportional to the concentration of Enzyme Substrate

A

The slower second reaction limits the overall rate

18
Q

When [S] is small most of enzyme is in what form and so the rate is proportional to what?

A

Is in uncombined form and the rate is proportional to [S] (pseudo first order)

19
Q

The maximum initial rate is observed when? (assumptions made in deriving MM equation)

A

when all the E is in the ES complex at high [S] i.e.

E is saturated (zero order)

20
Q

when there is excess substrate the reaction quickly achieves what state

A

Steady state in which enzyme substrate remains approximately constant over time

21
Q

How else V0 can be written

A

= v = Kcat [Etot][S]/ Km + [S]

22
Q

What is Kcat the same as

A

K2

23
Q

What is Kcat

A

turnover number

The number of substrate molecules one enzyme molecule can convert per second

24
Q

Km is what

A

Michaelis constant

Approx measure of substrates affinity for an enzyme and the substrate conc which gives half maximal velocity (Vmax)

25
Q

At high [S], velocity doesnt depend on what (Saturation Kinetics)

A

[S]

26
Q

When Vo = 0.5Vmax, what is [S]

A

Km, Km is the substrate concentration that yields .5Vmax

27
Q

The ideal rate is the M&M equation but deviations from this equation is due to what 4 things

A
  • Limitation of measurements
  • Substrate inhibition
  • Substrate prep contains inhibitors
  • Enzyme prep contains inhibitors
28
Q

What is Km in terms of velocity

A

Conc of substrate giving half-maximal velocity

29
Q

What is defined as the dissociation constant Kd of the ES complex or affinity of the enzyme for its substrate

A

When k2 is rate limiting k2

30
Q

What is Km in terms of K-1, k2 and K1

A

(K-1+K2)/K1

31
Q

Turnover number Kcat has units of sec-1, describe what is shows

A

Number of substrate molecules turned over per enzyme molecule per second

32
Q

k1 (first order) has what unit and what does a large k1 mean?

A

K1 is the rate constant unit s-1

Large K1 means a fast reaction

33
Q 
Define
Vmax
Km
V0
and which two contribute to measuring enzyme perfomance?
A

Vmax is maximumm velocity
Km is the substrate concentration at which V0 is equal to one half of Vmax
V0 is the initial velocity
Vmax and Km measure enzyme performance

34
Q

How is Catalytic/Enzyme efficiency calculated? And what value must the result be in for it to be efficient?

A

Kcat/ Km
DeterminedbythefrequencywithwhichSandEcollide
108‐109M‐1s‐1 (The diffusion-controlled limit)

Biochem (34 decks)

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