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2. Basic Enzyme Kinetics: analysing enzymes as catalysts
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Terms in this set (52)
What are 'steady state' conditions?
involve conditions in which substrate concentration [S] is much greater than [E], so that 'v' can be measured over a reasonable period.
A consistent model has been produced, which gives us the Michaelis-Menten equation, which links:
i) kinetic data from experiments,
ii) a picture of E-S interaction,
iii) a description of this in terms of chemical, reactions and equilibria,
iv) a mathemical model.
In a V against V/[S] Eadie-Hofstee plot, what is the gradient equivalent to?
-Km
In a V against V/[S] Eadie-Hofstee plot, what goes along the y-axis?
V (rate)
In a V against V/[S] Eadie-Hofstee plot, what goes along the x-axis?
V/[S]
In a V against V/[S] Eadie-Hofstee plot, what is the x-intercept equivalent to?
Vmax / Km
In a V against V/[S] Eadie-Hofstee plot, what is the y-intercept equivalent to?
Vmax
In a V/[S] against V Eadie-Hofstee plot, what is the y-intercept equivalent to?
Vmax / Km
In a V/[S] against V Eadie-Hofstee plot, what is the x-intercept equivalent to?
Vmax
In a V/[S] against V Eadie-Hofstee plot, what is the gradient equivalent to?
-(1/Km)
Give the lineary transformation of the Michaelis-Menten equation.
V = (-km × V/[S]) + Vmax
Y = m X + C
Derive the Lineweaver-Burk equation from the mMichaelis-Menten equation, V₀ = Vmax[S] / Km + [S]
invert → ¹/V₀ = Km + [S] / Vmax[S]
factor → ¹/V₀ = (Km / Vmax[S] )+ ([S] / Vmax[S])
simplify → ¹/V₀ = (Km/Vmax[S]) + (¹/Vmax)
How do you find vmax/km from a lineweaver-burk plot?
1/gradient
What are the units of vmax / km?
mM min⁻¹
What is the basic reaction equation assumed when using the Michaelis-Menten equation?
E + S ↔ ES → E + P
What are the 4 assumptions when using the Michaelis-Menten equation?
> [S] >> [E]
> there must be no Product formed
> reversion of E + p to ES must not occur
> there must only be one substrate
How are the rates of enzyme catalysed reactions measured?
> the appearance of product (or disappearance of substrate)
> in a 'continuous' assay (e.g. spectrophotometry), not samples taken at intervals
> in conditions where [S]>>[E], allowing the enzyme to turnover many times
Why, under steady state conditions, is the concentration of product measured?
> the disappearance of substrate (concentration reduction) would be too difficult to detect, as it's concentration must be much greater than the enzyme's.
The rate declines as substrate is depleted, so product appearance is...
measured early while V is constant and the system is in 'steady state' conditions.
> this is a parameter for 'steady state' kinetics!
Once the 'steady state' is established, the reaction proceeds at...
constant rate 'v'.
> steady state 'v' is proportional to [E].
> doing the reaction with different [E] gives different a different rate 'v'.
How does the rate of a 'steady state' reaction change?
> keeping [E] constant, plot v against [S].
- lower [S], v is proportional to [S].
- higher [S], v becomes constant (it plateaus) as the enzymes becoming saturated, producing the Vmax.
At high [S], all enzymes are converted to the...
'enzyme-substrate complex' - further increases in [S] don't increase rate.
> in these conditions, the reaction is limited by how quickly the Michaelis complex can produce and liberate the product and regenerate free enzyme.
> this is Vmax.
What is the general equation for enzyme catalysed reactions?
E + S ↔ ES → E + P
> the second part is only a forward reaction (unidirectional) because the reverse reaction of the product is negligible in steady state conditions where [P] is very low.
What is the Michaelis-Menten equation?
developed in 1913 by a German and Canadian scientist.
> v = Vmax.[S]/(Km + [S])
"E + S ↔ ES" is defined by a dissociation constant 'K' where...
K = [E] [S] / [ES]
> it's also defined by rate constants, where k∨a is the forward rate reaction and k∨d is the reverse.
"E + S ↔ ES"
The rate of appearance of ES equals what?
ES = ka.[E].[S]
> K = kd / ka
"E + S ↔ ES"
The rate of disappearance of ES equals what?
ES = kd.[ES]
> K = kd / ka
"ES → E + P"
early in the reaction, [P] is very low and the reverse reaction is negligible. What are assumptions are made with this? (part 1)
> in most cases the conversion of ES to E + P is the slowest step ∴ rate-limiting
> ∴ rate constant for this reaction ∧ is kcat
"ES → E + P"
early in the reaction, [P] is very low and the reverse reaction is negligible. What are assumptions are made with this? (part 2)
> rate of appearance of P = ka2[ES]
for the general equation d[P]/dt = kcat[ES]
EQUATION 1
"ES → E + P"
What is 'v' equal to?
v = kcat.[ES]
EQUATION 1.
EQUATION 2
At vmax, what is [E total] equal to?
[ES] concentration (enzymes are saturated)
and v = kcat[ES]
∴ Vmax = kcat.[E total]
EQUATION 2
EQUATION 3
What does the dissociation constant, K, equal?
K = [E].[S]/[ES]
rearrange → [ES] = [E].[S]/K
EQUATION 3
EQUATION 4
[ES] = [E].[S]/K
Substitute the values get the 4th equation.
[ES] = ([E total] - [ES].[S total]/K
solving for [ES] →
[ES] = ([E total].[S total] / (K + .[S total])
EQUATION 4
EQUATION 4
How should you express [S] and [E]?
> [S] is in excess, relatively little of the S will be in ES complexes, so [S] at Vmax= [S] at start of the experiment.
> E is present as free E and in ES complexes,
EQUATION 4
EQUATION 5
What is the 5th equation?
EQUATION 5
v = kcat.[ES]
rearrange →
v = kcat ([E total].[S total] / (K + .[S total])
What is the lLineweaver-Burk equation
¹/V₀ = (Km/Vmax[S]) + (¹/Vmax)
If...
v = kcat ([E total].[S total] / (K + .[S total])
and...
Vmax = kcat.[E total]
then...
v = Vmax.[S]/(Km + [S])
> this is the Michaelis-Menten equation
The Michaelis-Menten equation:
v = Vmax.[S]/(Km + [S])
is in what mathematical form?
a rectangular hyperbola
> y = a.x/(b + x)
> the constants that derive the hyperbola (a and b) can be derived as Vmax and Km where X is [S]
An important numerical relationship emerges from the rectangular hyperbola, of the Michaelis-Menten equation, in the special case when...
V is exactly half of Vmax.
> Vmax/2 = Vmax [S]/(Km + [S])
Vmax/2 = Vmax [S]/(Km + [S])
Solve this equation for Km.
dividing by Vmax: ¹/₂ = [S]/(Km + [S])
solving for Km: Km + [S] = 2[S] OR Km = [S]
When V = Vmax/2...Km is given by the value of what?
[S]
What are Km and Vmax?
Km = Michaelis constant (a measure of affinity)
Vmax = maximum velocity
What is 'enzyme efficiency' presented as?
Kcat / Km
units: s⁻¹M⁻¹
Vmax will vary with [E], so...
1. Vmax is often expressed as?
per unit of protein
e.g. mol/min/mg
Vmax will vary with [E], so...
2. If one enzyme accepts two different substrates, under identical conditions and the same [E]...
Vmax is the same, but Km will vary.
> hence, the enzyme will have different affinities for different substrates.
What does a low Km value indicate?
a high affinity for that substrate, meaning it will work at a low concentration of the substrate.
> Km shows the range of [S] that the enzyme works
To determine Vmax and Km, 'v' at a series of intervals...
of [S] must be determined. This is often called a 'dose-response experiment'.
Describe the 'Lineweaver-Burk plot'?
> gives a straight line with intercepts
> on the vertical axis of 1/Vmax
> on the horizontal axis of -1/Km
> a slop of Km/Vmax
What is the 'Lineweaver-Burk plot'?
the most popular method for transforming experimental data to give values for the terms derived fromthe Michaelis-Menten equation is the Lineweaver-Burk, or 'double-reciprocal' plot of (1/V) against (1/S).
What is the y-intercept of a lineweaver-burk plot equivalent to?
1/vmax
If 1/vmax = x, what is vmax?
1/x = vmax
What is the gradient of a lineweaver-burk plot equivalent to?
km/vmax
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