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BIOCHEM > CH 5 > Flashcards

CH 5 Flashcards

1
Q

Enzymes

A

reactions are 10^3 to 10^20 times faster

  • -> speeds up the attainment of equilibrium but does not change the position of equilibrium
  • they function on specific substrates
  • they are known for specificity, sometimes even stereospecificity (only a specific stereoisomer)
  • they contain reaction specificity: purity of product ~100%, saves energy, inhibits build-up up toxic by-products

20,000 genes -> 3,000 encode for enzyme subunits
–> our genome encodes for 1,000 different enzymes

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2
Q

PCR Plateau effect

A
  • depletion of substrate (dNTPs or primer)
  • stability of reactants (dNTPs or enzyme)
  • end-production inhibition (PPi, dsDNA)
  • competition of a reactants by nonspecific products or primer-dimer
  • reannealing of specific product at concentrations above 10-8 M
  • incomplete denaturation/strand separation of product at high product concentration
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3
Q

PCR plateau effect

A

The term plateau effect is used to describe the attenuation of the normally exponential rate of product accumulation in PCR. The attenuation occurs during the late PCR cycles when the accumulation of product reaches 0.3 to 1 picomole. Depending on reaction conditions and thermal cycling, one or more of the following may influence when the plateau is reached:
Depletion of substrates (dNTPs or primers)
Stability of the reactants (dNTPs or enzyme) particularly at the denaturation temperature
End-product inhibition (pyrophosphate, duplex DNA)
Competition for reactants by nonspecific products or primer-dimer
Reannealing of specific product at concentration above 0.8 M (may decrease the extension rate or processivity of Taq DNA polymerase or change branch-migration of product strands and displacement of primers)
Incomplete denaturation/strand separation of product at high product concentration.
An important consequence of reaching plateau is that nonspecific products resulting from mispriming events, initially present at low concentration, may continue to amplify preferentially. Optimizing the number of PCR cycles is the best way to avoid amplifying background products.

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4
Q

commercial use

A

rennet (rennin) is used to separate components of milk

  • contains protease (chymosin) that cleaves casein [protein] between Phe-Met
  • hydrophobic fragments are produced
  • collected as curd
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5
Q

coupling and units

A
  • many times enzymes can couple reactions

- most are oligomeric, meaning they’ll have separate binding sites for substrate and effectors

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6
Q

Naming and category

A

aside from historical names (e.g. trypsin), enzymes are named:

  • with an -ASE suffix
  • after the substrate they work on
  • and/or the reaction they catalyze
  • -> enzymes are categorized under one of the six general class of organic chemical reactions they catalyze
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7
Q

Six general class of organic chemical reactions

A

1) oxidoreductases
2) transferases
3) hydrolases (most abundant)
4) lyases
5) isomerases
6) ligases (least abundant)

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8
Q

oxidoreductases

A
  • enzymes that catalyze oxidation-reduction reactions
  • most typically referred to as dehydrogenases
  • also includes oxidases, peroxidases, oxygenases or reductases
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9
Q

transferases

A

enzymes that catalyze group transfer reactions

  • many require co-enzymes, with substrates having partial covalent bonding to enzyme or co-enzyme
  • includes enzymes like kinases
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10
Q

hydrolases

A
  • enzymes that catalyze hydrolysis

- use water as the acceptor of the group transferred

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11
Q

lyases

A
  • enzymes that catalyze lysis of a substrate generating a double-bond in non-hydrolytic, non-oxidative, elimination reactions
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12
Q

isomerases

A
  • enzymes that catalyze structural changes within a molecule
  • only one substrate, only one product
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13
Q

ligases

A
  • enzymes that catalyze ligation (joining/gluing) of two substrates
  • require input of energy
  • normally referred to as synthetases
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14
Q

kinetics

A

enzyme kinetics deal with the amount of product of a reaction per unit of time
-> the rate (velocity of a reaction) varies directly with the concentration of reactants (substrate)
–> simplified (non-enzymatic) conversion of substrate to product can be expressed as:
Δ[P]/ Δt = v = k[S]

k is rate constant - speed and efficienct
As P increases, S decreases -> reflected in a kinetic curve of reaction

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15
Q

Velocity (v)

A

slope of the progress curve over a particular time interval
- In a simple reaction, S is depleted so P would level off

Slope = k = ΔV/Δ[S]

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16
Q

Determining Vo (initial velocity)

A

We want to find initial velocity because concentration of substrate and product changes during an experiment

17
Q

Determining the k of a reaction

A

If you take the Vo of multiple interactions, you should get a straight line

–> taking the slope of this curve will give us the K of the reaction

18
Q

Equilibrium

A
  • when the reaction curve flattens out, v is zero IF reaction is NON-reversible
  • -> if reversible, it means that the reaction has reached equilibrium where net [substrate] and [product] are in dynamic equilibrium
19
Q

Velocity with more than 1 substrate

A

for a simple reaction with more than one substrate: S1 + S2 –> P1 + P2

  • k is determined by the concentration of both substrates
  • if both are found at similar levels, then:

v = k[S1][S2]

20
Q

Lock and Key model

A
  • Emil Fischer
  • proposed the lock and key model for enzymes: only specific substrates (S) can fit into an enzyme (E), binding and forming an enzyme-substrate complex (ES)
  • they interact transiently on the way to the product (P)

E + S –> ES –> E + P

21
Q

enzyme-substrate complex (ES)

A

E + S –> ES –> E + P
this formula simplifies sthe reaction 1 enzyme-1 substrate

the reaction takes 2 steps:

1) formation of the ES complex
2) the chemical reaction and dissociation of E and P

22
Q

Rates

A

the rate of a reaction depends on both the amount of substrate and enzyme
- given equal amounts of substrate: Vo will increase if the amount of enzyme [E] is increased (linear relationship_

23
Q

Enzymatic reactions

A

in an initial reaction, there are two steps governing the reaction: k1 and k-1 for formation and dissociation of the ES complex
K1: forward, formation
k-1: reverse, dissociation

  • then k2 govern the actual chemical reaction that makes product
    why is there no reverse reaction for k-2?
    –> this is describing the beginning of a reaction, where Vo is measured, conversion of EP –> ES is negligible

E + S ES –k2–> E + P

24
Q

Progress curve

A

in this curve, we plot product (y axis) vs time (x axis)
–> there are two initial slopes because there are two different [E]

–> change in [E] = change in Vo

–> reaction doubles when double enzyme (2E) is added

initial slope = Vo = Δ[P]/ Δt

25
Q

initial velocity and rates

A

ES –> E + P at the start of reaction gives us initial velocity (Vo)

(E + P –> EP) is negligible at the start of the reaction

  • rates differ with [enzymes] as opposed to simple chemical kinetics because reactions depend on [E]
  • -> S has to bind E before P can form
26
Q

Vo vs [S]

A

can we describe enzyme-catalyzed reactions mathematically?
–> yes, one way is to examine the effects of [S] variation on Vo:

E + S ES –k2–> E + P

you see two very different effects at low [S] vs high [S]

27
Q

Vo vs [S]

A

at low [S], S is the limiting factor in the reaction, not enough S to form ES leads to a limit on P

–> What happens to the curve at higher and higher concentrations?
E becomes the limiting factor in the equation; only so many active sites on a finite number of E molecules available

28
Q

Hyperbolic Curve

A

As with myoglobin, a hyperbolic curve indicates processes involving simple dissociation.
–> E and S forms ES complex which leads to P

29
Q

Equation for the hyperbola

A 
Y = Slope
a = asymptote of the curve (line which is tangent to a curve at infinity
x = substrate
b = 1/2 a = 1/2 Vmax

Formula: y = ax/(b+x)

–> the enzyme is half-saturated when S = Km

30
Q

Michaelis-Menten equation

A

describes the relationship between initial velocity of a reaction and substrate concentration
–> used to derive various other equations regarding these relationships

FORMULA:

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

km = michaelis constant

31
Q

Steady State Derivation

A
  • Postulates a period of time during which the ES complex is formed at the same rate it decomposes
  • what does this indicate about the concentration of the ES complex? it is constant –> called the steady state
  • -> it is been observed to be a common condition for metabolic reactions in cells
  • -> if in steady state, rate of P formation depends on k2

ES —k2—> E + P

Vo = k2[ES]

What would be the rate limiting step then in this kind of enzymatic reaction? ES –> E +P

At steady state then, the Vo in this case depends on the k2 constant and [ES]

32
Q

Derivations

A
  • the rate of formation of ES from E + S depends on the concentration of free enzyme (assuming [S] is not limiting):

[E]total - [ES]
–> The [ES] is also constant until consumption of S causes [S] to approach [E]total

Rate of ES formation = Rate of ES decomposition

k1([E]total - [ES])[S] = (k-1 + k2)[ES]

–> Michaelis constant (Km) can also be defined as:

(k-1 + k2)/k1 = Km = ([E]total - [ES]) [S] / [ES]

33
Q

Measurable [ES] at steady state

A

These parameters can be measure in experiments and can be used to derive [ES] at steady state

[ES] = [E]total[S] / (Km + [S])

–> substituting [ES] into Vo equation gives us:

Vo = k2[ES] = (k2[E]total[S]) / (Km + [S])

BIOCHEM (8 decks)

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