Lecture 2 - Enzyme Kinetics II Flashcards
simple kinetics I:
one simple reaction (A → B) which is referred to be “first order” - where A disappears as B is produced
simple kinetics II:
•Rate constant (k) of reaction does not change
•But with time, the velocity of the reaction reduces as the reactant A is used up and B becomes a substrate for the reverse reaction
the reaction reaches equilibrium when rates for forward and reverse reactions are EQUAL and the overall rate for the reaction is zero
simple kinetics II:
two reactants - we assume that the rate of forward reaction is linearly proportional to the concentrations of A and B, and the reverse reaction is linearly proportional to the concentration of C (A + B <—> C)
this is a second order reaction as its based on the concentration of two reactants
equilibrium:
equilibrium is reached when the net rate of reaction is zero. equilibrium constant tells us the extent of the reaction, NOT its speed.
basic problem of enzyme kinetics:
the reaction rates saturate
enzyme catalysis:
- overall this is an irreversible reaction (could be reversible)
- enzyme binds to substrate to form intermediates
- intermediates enzyme substrate then decomposes irreversibly to yield the product and the enzyme is not changed
assumption I:
if we assume [substrate] is greater than the [enzyme], then the enzyme will mostly be present as the intermediate enzyme substrate and [ES] will remain constant
rates of enzyme reactions:
•The initial rate increases with enzyme concentration [E]
•The initial rate increase with substrate concentration [S] but levels off (saturates) to a maximum value (Vmax)
assumption II:
steady state assumption - the rate of ES formation = rate of ES dissociation
(steady state is reached when the concentration of intermediate (ES) is constant)
the rate ‘v’ is given by the rate at which ES breaks down:
v = k2 [ES]
initial rate depends on:
enzyme and substrate concentrations
KM:
is the michael is constant (units of concentration)
maximum rate, V-max is:
V-max = k2 [Eo]
V-max:
maximum catalytic rate at full saturation
how can values for KM & V-max be determined?
values of V-max and KM can be determined by least-squares fitting of initial rate data to this equation
Michaelis-Menten equation:
V-max [S] / KM + [S]
Lineweaver-Burke Plot:
•More accurate estimation of Vmax than the saturation plot
•Vmax from the y intercept (1/Y)
•km from the negative x axis (-1/X) or from slope
disadvantage of Lineweaver-Burke Plots:
Disadvantage - the plot is dominated by points at low
[S] i.e. high 1/[S]
At low [S], rate measurements may be less accurate
Ki:
dissociation constant for inhibitor
how does KM & V-max change when in the presence of competitive inhibitors?
V-max remains the same but KM increases
how can inhibition be overcame with competitive-inhibitors?
through increasing [S]
example of non-competitive inhibitor:
Pyruvate Kinase catalyses final step of Glycolysis
kinetics of non-competitive inhibitor:
Increasing [S] cannot overcome inhibition
Less E available, V-max is lower, KM remains the same for available E
how does V-max & KM change with non-competitive plots:
V-max is decreased and KM is unchanged
other plots:
Eadie Hofstee Plot v against v/[S] - linear plot
GRADIENT is KM, Intercept on y-axis is Vmax
Hanes or Dixon Plot: •Plot [S]/v against [S]. Linear plot
•Gradient = 1/Vmax •Intercept on y-axis = KM/Vmax
example competitive inhibitors:
substrate “mimic”
A competitive inhibitor of dihydrofolate reductase - Used to treat cancer
Substrate for dihydrofolate reductase role in purine & pyrimidine biosynthesis.
The Active Site of an Enzyme:
1.The active site is the region that binds the substrates (&
cofactors if any)
2.It contains the residues that directly participate in the making & breaking of bonds (these residues are called catalytic groups)
3.The interaction of the enzyme and substrate at the active site promotes the formation of the transition state
- The active site is the region that most directly lowers G‡ of the reaction - resulting in rate enhancement of the reaction
common features of active sites:
1.The active site is a 3-dimensional cleft formed by groups that come from different parts of the amino acid sequence
- The active site takes up a relatively small part of the total volume of an enzyme. Why are enzymes so big? Scaffolding, regulatory sites, interaction sites for other proteins, & channels
3.Active sites are clefts or crevices - exclude H2O
4.Substrates are bound to enzymes by multiple weak attractions (electrostatic interactions, hydrogen bonds, Van der Waals forces, & hydrophobic interactions can be Involved)
5.The specificity of binding depends on the precisely defined arrangement of atoms at the active site
specificity:
- the specificity of an enzyme is due to the precise interaction of substrate with the enzyme
- this is a result of the intricate three-dimensional structure of the enzyme protein
enzyme inhibition by DIPF:
DIPF (nerve gas) reacts with Ser in acetylcholinesterase, affecting nerve function
