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BioMedChem M3 > 29 - Basic Enzymology > Flashcards

29 - Basic Enzymology Flashcards

1
Q

Michaelis Menten Model

“E”+”S” ⇌ “E”∙”S” → “E”+”P”

1st step is REVERSIBLE

2nd step is irreversible
(unidirectional)

A

“E”+”S” ⇌ “E”∙”S” → “E”+”P”

Enzyme contacts substrate by diffusion to form the
E-S Intermediate

Which then passes over the activation barrier to yield:
Product = P
OR
E-S can FALL APART & the enzyme / substrate diffuse away

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2
Q

What is k+1

A
  • *BI-molecular** rate constant
  • (involves BOTH E & S, SEPERATELY)*

for FORMING E-S

L/mole-sec = molarity-1sec-1 = M-1-sec-1

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3
Q

What is k-1

A
  • *uni-molecular** rate constant
  • (starts with just ONE MOLECULE = E-S complex combined)*

for DISSOCIATION of E-S

sec-1

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4
Q

What is k+2

A
  • *uni-molecular** rate constant
  • (starts with just ONE MOLECULE = E-S complex combined)*

for BREAKDOWN of E-S into E + P

sec-1

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5
Q

What is this?

A

RATE LAW for the M-M Model

Km = Michaelis Constant

[E0] = original enzyme concentration

NO simple interpretation of MOLECULARITY or ORDER

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6
Q

What is Km

A

Michaelis Constant

it can be RELATED to a dissociation constant (Ks)
but is NOT quite the same

just the SAME UNITS

of micromolar

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7
Q

What is k+2 or kcat

A

TURNOVER Number

the MAXIMUM # of
substrate molecules converted / turned over into product

per unit time, per active site

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8
Q

What is the Maximal Rate of Reaction?

Rxn rate is maximixed when ALL of the active sites
on ALL of the enzyme molecules are Filled w/ substrate

A

Rate equation simplifies to:
Vmax = kcat [E0]

this occurs ONLY if the enzyme is SATURATED with substrate

  • generally requires that* [S] >>> Km
  • If [S] is not high enough to saturate the enzyme*, you can’t use this simple expression, but must instead use the original rate law
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9
Q

When can this rate equation NOT be used?

Vmax = kcat [E0]

A

if the [S] is NOT high enough to saturate the enzyme
we can not use that expression,

we MUST use the Original Rate law

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10
Q

Traditional Unit = U​

vs

katal = kat

A
  • *U** of enzyme activity is the
  • *amount of ENZYME** that can convert ONE micromole of substrate into products in ONE MINUTE
  • *kat** = SI unit of activity, the amount of ACTIVITY that converts
  • *ONE** mole of substrate into products in one SECOND
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11
Q

Equivilance of U to kat

A

1 microkat = 60 (traditional) Units of activity

Since:
kat=1mole vs U=1micromole

1 second vs 1 minute

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12
Q

What does this indicate? Km vs kcat

A

Km = micromolar
Takes A LOT of CO2 to saturate the enzyme

kcat = per second
there is a VERY FAST TURNOVER

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13
Q

What does this indicate? Km vs kcat

A

Km = micromolar
does NOT take a lot of substrate to saturate the enzyme

kcat = per second
there is a VERY FAST TURNOVER ​ –> produce pyruvate quickly

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14
Q

What does this indicate? Km vs kcat

elastase chews up on AA’s

A

Km = micromolar
Takes A LOT of substrate to saturate the enzyme

kcat = per second
there is a SLOW turnover

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15
Q

What does this indicate? Km vs kcat

fumarase is in the TCA cycle

A

Km = micromolar
does NOT take a lot of substrate to saturate the enzyme

kcat = per second
there is a VERY FAST TURNOVER ​

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16
Q

When can Km be APPROXIMATED by Ks for the E-S complex?

Km = Michaelis Constant

Ks = dissociation constant

A

Km =~ Ks
ONLY IF
k+2 / k+1 <<< Ks

rate of catalysis is MUCH SLOWER*** than ***rate of dissociation

_k<sub>+2</sub>_ = _rate of **breakdown** of E-S --\> E + P_ 
k<sub>+1</sub> = rate of ***dissocation*** of E + S \<-- E-S
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17
Q

What is Ks

A 
_k<sub>-1</sub> = rate of ***dissociation*** of **E + S** \<-- E-S_
k<sub>+1</sub> = rate of **formation** of E + S --\> **E-S**
  • *RATE of Dissociation**
  • only approximates Km if the rate catalysis is MUCH SLOWER than the rate at which E-S dissociates back to E + S*
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18
Q

How to determine the INITIAL RATE of an enzyme reaction?

and [Eo]

A

Set up the rxn with measured # of enzyme & substrate
then follow the progress by measuring appearance of PRODUCT

then PLOT = [Product] vs [time]

SLOPE = Rate of Rxn
(use the slope in the LINEAR REGION of the plot)

then TRACE BACK TO T=1 for [E0]

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19
Q

Initial Rate Determination

GRAPH

A

use the slope in the LINEAR REGION of the plot

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20
Q

How to find Half-Maximal Velocity = 1/2 Vmax

A

When [S] = Km , then v = 1/2 Vmax
the numerical value for [S0] = Km

First find the maximum (plateau) for the initial rate = v
then drop back in concentration of substrate = [S] to where v is
half the maximal value

then find the corresponding initial concentration of S

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21
Q

What is the Lineweaver-Burk Plot?

A

Double-Reciprocal Plot
1/v vs 1/[S]
(min/um) vs (M-1)

Km / Vmax = Slope
should be a STRAIGHT LINE
ONLY IF the enzyme obeys the Michaelis-Menten model

x-intercept @ -1 / Km
y-intercept @ 1 / Vmax

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22
Q

What type of plot is this?

A

Double Reciprocal

LINEWEAVER-BURK

only a STRAIGHT LINE slope if enzyme obeys M-M model

23
Q

Where is the X-Axis Intercept in the
Lineweaver-Burk (Double Reciprocal) Plot?

A

-1 / Km
(micromolar-1)

slope = Km / Vmax

24
Q

Where is the Y-Axis Intercept in the
​ Lineweaver-Burk (Double Reciprocal) Plot?

A

1 / Vmax

slope = Km / Vmax

(min / micromole)

25
Q

What plot is this?

A

Lineweaver-Burk

26
Q

What plot is this?

A
  • *Eadie - Hofstee Plot**
  • not very useful*
27
Q

What plot is this?

A

Hanes - Wolf Plot

Technically superior > lineweaver burk

because there is least distortion to experimental ERRORS

28
Q

When do we see a Non-Linear / CURVED Plot?

and WHY?

A

When the enzyme is NOT following that model

2 common possibilities of curvature:

may be MORE Kinetic STEPS in the mechanism that the M-M model allows

ENZYME may have MULTIPLE ACTIVE SITES
which may be interacting with one another
= cooperativity (pos / neg)

29
Q

What do we do with CURVED PLOTS?

A

compare to “linear” type plots to diagnose deviations
from the M-M model

USE Curve-Fitting SOFTWARE** + **full non-linear rate law
to get accurate estimates of the kinetic parameters

kcat and Km

k+2= kcat = turnover number

30
Q

Basic ways to measure the # of enzyme (concentration)

Assays

A

IMMUNOASSAY
needs specific AB’s, does NOT measure _inactive_ enzymes

ELECTROPHORESIS
stain -> scan slow also does NOT measure active enzyme

STANDARDIZED RXN
FAST, but may call for _coupling_ of 2+ sucessive rxns in order to measure the product spectrophotometrically

31
Q

Immunoassay

Positives & Negatives
(basic ways to measure enzyme concentration)

A

Need SPECIFIC ANTIBODIES to the enzyme

does NOT measure INACTIVE enzyme

only measures the ACTIVE enzyme

think immunoAssay = measures ACTIVE

32
Q

Electrophoresis

Positives & Negatives
​(basic ways to measure enzyme concentration)

A

Electrophoresis -> stain -> scan

SLOW

NOT for measuring ACTIVE enzyme

only measures the inactive enzyme

33
Q

Stantardized Reaction

Positives & Negatives
​(basic ways to measure enzyme concentration)

A

Measures the ACTIVITY of enzyme

FAST

but may call for COUPLING of two or more successive rxns in order to measure the product

SPECTROPHOTOMETRICALLY

34
Q

How do we monitor an enzyme that
does NOT use a substrate or form a product?

A

ENZYME COUPLED ASSAY

UV ABSORBANCE** or **FLUORESENCE

done DIRECTLY by coupling it to a second INDICATOR REACTION that itself, does NOT give a useful signal

need to add EXCESS SUBSTRATE & Secondary ENZYME
to ensure that the
rate limiting factor is the #enzyme in the first rxn

35
Q

Coupled Assay Time Course

Lag Phase -> up to Incubation time

time BEFORE products are being made

A

To ensure that the:
Overall Rate-limiting Factor = Amount of Enzyme 1
(1st enzyme that is or diagnostic quanitity)

we need:

EXCESS SUBSTRATES for BOTH reactions to reach SS

EXCESS ENZYME #2 the one for the indicator reaction

36
Q

Aspartate Amino Transferase (AST)

as an Example of Coupled Enzyme Assay

A

AST catalyzes this rxn:
Aspartate + a-Ketoglutarate ↔ Glutamate + Oxaloacetate

The indicator rxn #2 uses malate dehydrogenase for:

Oxaloacetate + NADH + H+ ↔ Malate + NAD+

We then follow the PRODUCTION OF NAD+ using
FLUORESCENCE** or **UV ABSORBANCE

we need excess:
reactants & 2ndary enzyme (malate dehydrogenase)

37
Q

What type of Inhibition is this?

A

Reversible COMPETITIVE inhibition

38
Q

Competitive Inhibition

A
  • *inhibitor** blocks S from binding in the active site
  • various ways to block / different points of contacts*

Formation of E-S complex is Reduced, new complex, E-I​ is formed

unlike Km,
Ki(inhibitory constant) isreally a EQ dissociation constant (Ks)

  • Competitive Inhibitor** has *_NO EFFECT_ on the catalytic step
  • *kcat = k+2** is THE SAME & Vmax is also THE SAME
39
Q

In Reversible Competitive Inhibtion

Can the addition of MORE substrate OVERCOME the inhibition?

(restore the OG rate of rxn)

A

YES

apparent Michaelis constant is numerically LARGER than Km

so it INCREASES as the concentration of INHIBITOR RISES

takes MORE substrate to HALF-SATURATE the enzyme when a competitive inhibitor is present

40
Q

Competitive Inhibition’s effects on:

kcat & Vmax & Km

(k+2) turnover rate / maximum velocity

A

kcat & Vmax STAY THE SAME

Km decreases

MORE substrate can OVERCOME inhibition

41
Q

What type of Inhibition Graph?

A

Reversible COMPETITIVE inhibition

Increasing [I] = more inhibitor leads to:

Slope = Steaper Slope

Y-axis (vertical) = Same Y-Intercept
Vmax does NOT change

X-axis = Different X-intercepts
Km decreases

42
Q

What type of Inhibition Graph?

A

UN-COMPETITIVE inhibition

Increasing [I] = more inhibitor concentration leads to:

Slope = No Change (Parallel Plot)

Y-axis (vertical) = Increase Y-Intercept
Vmax is progressively reduced

X-axis = Increase X-intercept
Km is progressively reduced

43
Q

What type of Inhibition Graph?

A

NON-COMPETITIVE inhibition

Increasing [I] = more inhibitor concentration leads to:

Slope = Increasing Slope

Y-axis (vertical) = Increase Y-Intercept
Vmax is reduced

X-axis = SAME X-Intercept
Km does NOT change

44
Q

UN-Competitive Inhibition’s effects on:

kcat & Vmax & Km

(k+2) turnover rate / maximum velocity

A

kcat idk

REDUCE BOTH Vmax & Km

NOT POSSIBLE TO OVERCOME INHIBITION

45
Q

NONCompetitive Inhibition’s effects on:

kcat & Vmax & Km

(k+2) turnover rate / maximum velocity

A

kcat ?

Vmax INCREASES

Km stays the SAME

NOT able to OVERCOME INHIBITION with more substrate

46
Q

What type of Inhibition is this?

A

NONCompetitive / MIXED Inhibition

NONCompetitive = special case of mixed
inhibitor has SAME AFFINITY for either E or E-S complex
Ki = Ki‘

MIXED
Ki & Ki’ are allowed to be different

47
Q

What type of Inhibition is this?

A

UN-Competitive Inhibition

parallel graph, vmax &km get smaller

48
Q

UN-Competitive Inhibition

A

relatively UNCOMMON, Parallel Plots

  • *Inhibitor –> E-S complex**
  • does NOT bind to the FREE ENZYME*

NO E-I complex is formed
E-S-I complex is catalytically inert

still reversible

NOT POSSIBLE to
OVERCOME INHIBITION w/ MORE SUBSTRATE

49
Q

Mixed / NON-Competitive Inhibition

A

converge to the same X-intercept (-1/km), same Km

Inhibitor –> complex with BOTH: E & E-S

E-I & E-S-I complexes are Both INACTIVE

ADDING MORE SUBSTRATE WILL NOT OVERCOME INHIBITION
since inhibition can STILL OCCUR when the substrate binds

50
Q

alcohol dehydrogenase inhibition

by 3-butylthiolane 1-oxide

A

Example of

UN-Competitive Inhibition

PARALLEL PLOTS

51
Q

NAMPT inhibtion by NAD / NADH

A

NON-Competitive Inhibition

all converge to same X-Intercept

(-1/Km)

52
Q

What is the difference between

Mixed** & **Noncompetitive Inhibition?

A

NON-Competitive Inhibition
a SPECIAL CASE of Mixed Inhibition where
inhibitor has the SAME affinity for BOTH E & E-S complex

numerically, Ki = Ki‘

  • *Mixed Inhibition**
  • *Ki & Ki‘** are allowed to be DIFFERENT
53
Q
A

BioMedChem M3 (8 decks)

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