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Cell Phys > Lec 3: Enzymes (Pt. 1) > Flashcards

Lec 3: Enzymes (Pt. 1) Flashcards

1
Q

Many (most) cellular (and many extracellular) reactions in living systems are catalyzed by

A

enzymes

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2
Q

You can envision the cell as simply a ‘container’ in which millions of enzyme catalyzed reactions allow for…

A

…maintenance of function and reproduction (i.e., life)

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3
Q

enzymes are…

because they…

A

…catalysts

…speed up chemical reactions without being consumed

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4
Q

generally, enzymes are…

…although, …

A

…proteins

…ribozymes are enzymes with RNA-based functional subunits

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5
Q

Enzymes function to

A

lower the “energy of activation” (Ea) for a specific chemical reaction

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6
Q

Exergonic means: (4)

A
  1. ) The reaction is energetically (thermodynamically) favorable
  2. ) *G (Change in Gibbs free energy) is negative
  3. ) G products – G reactants < 0
  4. ) Reactants are at a higher “energy state” than the products
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7
Q

However, an exergonic reaction will not readily proceed without either: (2)

Thus, the reactants are said to be…

A
  1. ) A catalyst (in this case, any ATPase)
  2. ) Heating the reaction

….metastable

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8
Q

“metastable” reactants =

thus, they…

A

= they are “thermodynamically” unstable, but they do not achieve the activation energy needed to “react”.
…appear to be stable.

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9
Q

In order to proceed, some of the reactants must…

A

…achieve a higher energy state to reach the transition state level (Greactants + EA).

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10
Q

EA =

A

= Energy of activation

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11
Q

For an endergonic reaction

A

(*G is positive)

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12
Q

Thermal Activation =
this will allow…
is this feasible inside a cell?

A

= Adding heat to the reaction
…some reactants to reach that heightened energy state
? = no

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13
Q

In contrast, catalysts…

A

…reduce the EA enough so that many of the reactants have sufficient energy to proceed to product formation without the need to add energy (heat for example) to the system.

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14
Q

The overall rate of the reaction will depend on

A

the proportion of reactants with energy at or above the EA

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15
Q

Important properties of catalysts:

A
  1. ) Increase reaction rates by lowering the EA
  2. ) Facilitate reactions by forming transient complexes with substrates (reactants)
    - Therefore, they are not consumed during the reaction process**
  3. ) Only change the rate at which a reaction will come to equilibrium
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16
Q

Important aspects of enzymes: (5)

A
  1. ) active site
  2. ) enzymes are highly specific
  3. ) sensitivity to temperature
  4. ) sensitive to pH
  5. ) sensitive to other factors
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17
Q

active site =

A

= Domain (or domains) directly involved with formation of the transient Enzyme-Substrate (E-S) complex

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18
Q

active sites are not necessarily…

A

…one contiguous stretch of amino acids in the enzyme

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19
Q

active sites could involve…

A

…several amino acids from different regions (due to protein folding)

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20
Q

active sites are usually

A

rather small

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21
Q

Active Site shape

A

3D is shape or structure

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22
Q

What are involved in the transient formation of the E-S complex?

A

Multiple weak (non-covalent) attractions or bonds

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23
Q

Active sites generally involve

A

clefts or crevices in the overall protein

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24
Q

Active sites are highly…

which means…

A

…Specific

…only specific substrates can fit and properly align in the active site

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25
Q

Active Sites typically involve:

A

the amino acids

  • Cysteine, histidine, serine, aspartate, glutamate, arginine, and lysine
  • All are either charged or polar
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26
Q

Active sites may involve…

A

cofactors to assist in substrate binding and catalysis

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27
Q

cofactors =

A

= Non-protein components required for catalytic function

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28
Q

2 main types of Cofactors:

& 2 more from one category

A
  1. ) Inorganic = metal ions (eg., Fe2+, Mg2+, Cu+, Zn2+, etc.)
  2. ) Organic = non protein
    • Prosthetic groups: tightly bound to the enzyme (ex: Heme, biotin, flavin)
    • Coenzymes: loosely or transiently bound
      (ex: NADH, NAD, ATP) (sometimes may be considered as substrates)
29
Q

LDH active site =

A

= pyruvate binding pocket

30
Q

What allows for positioning of pyruvate substrate?

A

Two arginines, 1 histidine, and 1 aspartate

31
Q 
(Hb ex.)
Heme =
- Heme is tightly attached to... 
- via...
- The heme is held in...
- This histidine is referred to as...
A

= prosthetic group

  • …the Hb protein
  • …specific amino acid interactions in a cleft
  • …the cleft both by hydrophobic interactions and by a covalent bond between the iron and a nitrogen atom of a nearby histidine side chain
  • …the proximal histidine.
32
Q

(Enzymes with cofactors)
Holoenzyme =
Apoenzyme =

A

= complete functional enzyme including all subunits and all cofactors
= just the protein portion, without associated cofactors

33
Q

Enzymes are highly…

due to…

A

…specific

…specificity at the active site

34
Q

High specificity allows for…

& ex:

A

…high degree of regulation of cellular processes

ex: Succinate dehdrogenase (SDH)

35
Q

(SDH)

If we run the reaction in the opposite direction…

A

…the substrate pocket is so specific that the stereoisomer of fumarate, known as maleate, cannot serve as a substrate

36
Q

(SDH)

Similarly, a molecule (malonate) with…

A

…a structure similar to succinate, acts as an inhibitor of enzymatic activity

37
Q

Enzymes also have an optimum: (2)

A

temperature

pH

38
Q

Other factors enzymes are sensitive to: (3)

& examples

A
  1. ) Activators
    • Ex: AMP activates some enzymes in the glycolysis
  2. ) Inhibitors
    • Ex: ATP inhibits some enzymes in glycolysis
  3. ) Ionic strength of the medium
    • Similar to pH, ionic strength/composition could influence amino acid charges, substrate binding, and catalysis
39
Q

2 Proposed Models for Substrate Binding:

A
  1. ) Lock and Key mechanism

2. ) Induced Fit mechanism

40
Q

Lock and Key mechanism =

  • This older model has…
  • Can’t explain…
A

= substrate fits the binding pocket (active site) like a key into a lock

  • …fallen out of favor
  • …the catalytic mechanism
41
Q

Induced Fit mechanism steps: (5)

- Relatively newer model with…

A

1.) Substrate enters the binding pocket
2.) Induces a conformational change in the enzyme
3.) Leads to proper positioning of the active site
- Promotes substrate activation and the catalytic event by providing the appropriate micro-environmental conditions
amino acid positioning relative to the substrate
4.) Substrate becomes activated (reaches transition state)
5.) Catalytic event occurs

  • …series of steps
42
Q

Think of substrate activation as if…

Involving: (3)

Thus…

A

…the enzyme (due to its conformational change) is putting some ‘stress or strain’ onto the substrate
Involving:

1.) Bond distortion
Break bond or make it susceptible to catalytic attack
2.) Proton (H+) & or Hydride (H-) transfer
Enzyme may accept (or donate) protons (or hydrides) from (or to) the substrate
3.) e- transfer
Enzyme may accept (or donate) electrons from (or to) the substrate

…the Ea is effectively lowered. The substrate achieves the transition state without the need to add outside energy.

43
Q

(LDH active site (pyruvate binding pocket))
Positioning of…
allows…
to form…
What kind of transfers are occurring? –>

A

…the substrate (pyruvate) and coenzyme (NADH) in the active site
…the transfer of hydrogens from NADH and His 195 to pyruvate
…lactate
–> Hydride (H-) & Proton (H+) transfer.

44
Q

Full Catalytic Cycle =

45
Q

Enzyme Kinetics =

A

= Quantitative study of enzyme reaction rates (how fast is the reaction proceeding?)

46
Q

Enzyme Kinetics focuses on…

A

…initial reaction rates

47
Q

Many enzymes display

A

Michaelis-Menten kinetics

48
Q

Michaelis-Menten kinetics simple reaction

A

S → P

Enzyme (E)

49
Q

M-M Kinetics Reaction:

A

k1 k3
E + S ↔ ES ↔ E + P
k2 k4

E = enzyme
S = substrate
ES = enzyme substrate complex
P = product
k1 = rate constant for forward step 1
k2 = rate constant for reverse step 1
k3 = rate constant for forward step 2
k4 = rate constant for reverse step 2
50
Q

M-M Kinetics Assumptions: (3)

A
  1. ) The concentration of E is constant
  2. ) None of the product reverts back to substrate (true at initial stage, when no (or little) product is present), therefore we can disregard k4
  3. ) formation of [ES] is at equilibrium, therefore we can disregard k1 & k2
51
Q

(M-M Kinetics)

velocity (Rate) of forward reaction equation:

A

v = k3[ES]

52
Q

(M-M Kinetics)
Can we really make assumption #3?
Why?
At some point when…

A

= YES
= “The formation of ES is at equilibrium”
…we reach equilibrium, the rate of reconversion back to substrate will equal the rate of the product becoming what it was destined to

53
Q

(M-M Kinetics)

Using mathematical substitutions for [ES] and k3, we ultimately arrive at…

A

the Michaelis-Menten Equation:

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

v = initial velocity (moles/min) ***
Vmax = maximal velocity (moles/min)
[S] = concentration of substrate (M)
Km = Michaelis constant = (k2 + k3)/k1
54
Q

With the Michaelis-Menten Equation, we can…

A

…predict the initial forward rate of an enzymatic reaction, by knowing the initial substrate concentration.

55
Q

draw M-M Kinetics diagram

A

plzzzzzz don’t you want an A?!?!?!?1

56
Q

M-M Kinetics Diagram Important Features: (3)

A
  1. ) Initial velocities are greater with greater [S]
    • Particularly at low [S]
  2. ) At some [S], eventually see a plateau in initial velocity
    • Enzyme becomes saturated with S
  3. ) Km = [S] at ½ Vmax
57
Q

Importance of Km and Vmax: (3)

A
  1. ) At low [S] (Km is much greater than [S]), the [S] term in the denominator becomes irrelevant.
  2. ) At very high [S] (Km is much lower than [S]), the Km term in the denominator becomes irrelevant.
  3. ) When [S] = Km
58
Q

(1.) At low [S] (Km is much greater than [S]), the [S] term in the denominator becomes irrelevant)
- v =
- Since Vmax and Km are…
v is proportional to [S] (depends only on [S])
- order region?

A
  • v = Vmax x [S]/Km
  • …essentially constants at a given [E], v is proportional to [S] (depends only on [S])
  • 1st order region
59
Q

(2. ) At very high [S] (Km is much lower than [S]), the Km term in the denominator becomes irrelevant)
- v =
- v is = to…
- order region?
- At high substrate concentrations…
- Only way to increase v at high [S] is to…

A
  • v = Vmax[S]/[S] = Vmax
  • …Vmax (no change in rate with increasing [S])
  • … 0th order region (v is independent of increases in [S])
  • …v is ‘maxed out’
  • …increase [E]
60
Q

(Vmax and [E])
At very high [S], v =
Vmax (and v) is…
This relationship allows for…

A

= Vmax
…linearly related to [E]
…the measurement of enzyme concentrations in tissues & cells

61
Q

(3.) When [S] = Km)

v =

Therefore…

A

v = Vmax[S]/Km+[S]
= Vmax[S]/[S]+[S]
= Vmax[S]/2[S]
= Vmax[S]/[S]

…Km = [S] @ ½ Vmax

62
Q

M-M Importance to Biologists: (3)

A
  1. ) Km, relative to the [S], tells us where we are along the M-M plot
    • Gives us a rough idea of initial reaction velocity at the [S]
  2. ) Can estimate reaction velocity in vivo (if we know the [S])
  3. ) Can use Vmax to estimate
    • Turnover Number = Kcat = Vmax/[E]
      • Rate at which a single enzyme molecule can catalyze a single reaction
63
Q

Turnover Number =

Turnover Number is…

A

= Kcat = Vmax/[E]

…the rate at which a single enzyme molecule can catalyze a single reaction

64
Q

(M-M Plot)
Because the M-M plot is…
Therefore several…

A

…not linear, it can be cumbersome to use
…different linearized versions of the M-M relationship:
- Lineweaver-Burke Plot
- Eadie-Hofstee Plot
- Others (less popular)

65
Q

Lineweaver-Burke Plot is essentially the

A

Reciprocal of M-M equation

66
Q

Lineweaver-Burke Plot equations:

& now…

A

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

…and now has the basic y=mx+b format for a straight line

  • y = 1/v
  • m = slope = Km/Vmax
  • x = 1/[S]
  • b = y-intercept = 1/Vmax
67
Q

Draw Lineweaver-Burke Plot!!

A

plzzzzz you KNOW you need this A!!!

btw:
Left = higher [S]
Right = lower [S]

68
Q

Draw Eadie-Hofstee Plot!!

A

plzzz I know this one is random af but just do it hoe

Cell Phys (16 decks)

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