Biochemistry Chapter 7
Michaelis-Menten reaction w/ constants
*we measure when [P] = 0. ignore product -> substrate reaction (k-2)
allosteric enzymes
- allow for complex metabolic pathways - quaternary structure w/ multiple active sites
negative effector (inhibitor)
- binds to T form and stabilizes it - raises the threshold concentration of substrate needed for activity - shift the curve to the right (decrease cooperativity)
positive effector (activator)
- binds to the R form at a regulatory site and stabilizes the R form - lowers the threshold concentration of substrate needed for activity - shifts the curve to the left (increase cooperativity)
(allosteric enzymes) Concerted model
- enzyme exists in two different quaternary structures, T (tense) and R (relaxed) - T and R are in equilibrium, with T being more stable state - R state is enzymatically more active than the T state - all active sites must be in the same state
For Michaelis-Menten enzymes:
- most enzymes in the cell - activity is governed by mass action
heterotropic effectors
- regulatory molecules disrupt T and R equilibrium on allosteric enzymes - shift the sigmoidal curve to the left or right depending on if activators or inhibitors
homotropic effectors
- substrates disrupt T and R equilibrium on allosteric enzymes - accounts for sigmoidal nature of the kinetics curve
(Lineweaver-Burk) x-intercept
-1/Km
Km is 0.5 µM and Vmax is 5.32 µM. What is the concentration of substrate if we need an initial velocity of 2 µM/s?
0.30 µM
assumptions
1- early reaction, [P] is negligble so the reverse reaction (k-2) can be ignored 2- steady state; [ES] remains constant over time
Allosteric; Michaelis Menten enzymes
1. quaternary structure; tertiary structure 2. response to environmental signal; not regulated in the cell 3. sigmoidal curve (cooperativity); hyperbolic shape *reaction velocity vs substrate concentration* 4. threshold affect 5. K0.5; Km 6. ATCase, hemoglobin; chymotrypsin
Michaelis constant, Km
1. related to the rate constants of the individual steps in the catalytic scheme 2. Km = [S] that yields 1/2 vmax 3. low Km = high affinity 4. k-1/k1 = dissociation constant of the ES complex
What are the advantages of Lineweaver-Burk?
1. straight line so easier to determine how well points for to a straight line than to a curve 2. effects of inhibitors on the reaction can be analyzed more easily 3. highlight differences in the types of reversible inhibition
Lineweaver-Burk plot equation
1/v0 = Km/Vmax * 1/S + 1/Vmax
(Lineweaver-Burk) y-intercept
1/vmax
diffusion limit for kcat/km values is between
10^8 and 10^9 s^-1M^-1
(second order) reaction
2A -> P or A+B -> P
kcat (k2) equation
= vmax/[E]T
What is the committed step in the allosteric enzyme pathway?
A -> B
(first order) reaction
A -> P
velocity/rate of reaction
A disappears; B appears (both as function of time)
(allosteric enzymes) A, B, C, D, E, F
A: valuable and conserved unless F is needed F: needed in limited amounts and cannot be stored B, C, D, E: chemical intermediates in the synthesis of F
How to disrupt equilibrium in concerted model
Binding of substrate to one active site traps the other active sites in the R state and removes the substrate-bound enzyme from the T and R equilibrium. The binding of S to R becomes easier. This behavior is called cooperativity and explains sharp increase in v0 of the velocity vs substrate concentration curve.
key feature of MM treatment of enzyme kinetics
ES complex is a necessary intermediate in catalysis
Final F and the enzyme they inhibit
F does not need to be structurally similar to the substrate or product of the enzyme they inhibit
Allosteric constant
L0 = T/R
(second order) units
M^-1s^-1
Allosteric effectors alter the equilibrium between the:
R and T forms of a protein
at low substrate concentration...; at high substrate concentration
T form favored; R form favored
(second order) rate equation
V = k[A]^2 or V=k[A][B]
[E] =
[E]T - [ES] *[E] is concentration of free enzyme
when v0 = 1/2vmax, Km =
[S]
To study the nature of an enzyme, Vmax is not as good a measurement as the catalytic rate constant kcat because: a. The Vmax is not a true constant since it depends on the concentration of enzyme b. The Vmax cannot be measured c. The Vmax is only valid for allosteric enzymes d. none of these
a *vmax goes forever and ever
the primary function of enzymes is to _______________ the rates, or velocities, of reactions so that they are compatible w/ the needs of an organism.
accelerate
ATP is an
activator
Threshold effect
allosteric enzymes are more sensitive to changes in substrate concentration near their Km (K0.5) values than are Michaelis-Menten enzymes w/ the same Vmax
allosteric effectors
allosteric enzymes can be inhibited or stimulated by regulatory molecules (allosteric effectors)
Homotropic effects of allosteric enzymes: a. shift the kinetics curve to the left b. are due to the effects of allosteric activators c. are due to the effects of substrates d. shift the kinetics curve to the right
are due to the effects of substrates
When enzymes have optimal kcat/km ratios, they have
attained kinetic perfection
A Lineweaver-Burk plot is useful in the analysis of enzymatic reactions because a. it is not affected by the presence of inhibitors b. it is easier to see whether points deviate from a straight line than from a curve c. all of the above d. it can be used whether or not the enzyme displays Michaelis-Menten kinetics
b
The substrate-enzyme (E-S) complex a. always breaks down to form free enzyme and substrate. b. may break down to form free enzyme and substrate, or free enzyme and product. c. always breaks down to form free enzyme and product. d. always proceeds to form the products rapidly.
b
allosteric regulators
bind to the regulatory site and disrupt the R and T equilibrium when they bind
(sequential) negative cooperativity
binding of one substrate promotes neighboring subunits to be in the T state
(sequential) positive cooperativity
binding of one substrate promots neighboring subunits to be in the R state
What is the consequence of a high KM value of an enzyme? a. This enzyme has high maximal velocity. b. The enzyme is present in a large concentration in the cell. c. High concentration of the substrate should be reached for maximal enzyme activity. d. Moderate concentration of the substrate will be enough for maximal enzyme activity.
c
When an enzyme is saturated with substrates, a. it will display first-order kinetics. b. it will display second-order kinetics. c. it will display zero-order kinetics.
c
Allosteric enzymes catalyze the __________ while Michaelis-Menten enzymes facilitate __________.
committed step; remaining steps
Which statement about Vmax is TRUE? a. Vmax is independent of enzyme concentration. b. The reaction velocity is half Vmax when the entire enzyme is bound to the substrate. c. The reaction velocity is maximal when the concentration of the substrate equals KM. d. For Michaelis-Menten enzymes, the maximal velocity is approached asymptotically.
d
When the substrate concentration is low, an enzyme will display a. zero-order b. first-order c. second-order d. cease to function
first-order kinetics
R state =
high affinity for substrate
As the T to R transition occurs, the velocity
increases over a narrower range of substrate concentration for an allosteric enzyme than for a MM enzyme *allosteric- pink, MM- blue
CTP is an
inhibitor
Inhibitors vs activators
inhibitors stabilize the T state; activators stabilize the R state
At lower substrate concentrations, allosteric enzymes are
insensitive to the changes in substrate concentration that are MM enzymes. However, once the threshold substrate concentration has been reached, the allosteric enzymes become more sensitive to changes in substrate concentration than MM enzymes.
Km =
k-1 + k2/k1
dissociation constant of ES complex
k-1/k1
what is the rate-limiting step?
k2
vmax =
k2[E]T
If [S] is less than Kcat, then [E] = [E]T so v0=
kcat/km * [S][E]T
The study of the rates of chemical reactions is called ___________ and the study of the rates of enzyme-catalyzed reactions is called _____________.
kinetics; enzyme-kinetics
(Lineweaver-Burk) slope
km/vmax
T state =
low affinity for substrate
Hexokinase catalyzes the phosphorylation of glucose and fructose by ATP. Km for glucose is 0.15mmol/L, whereas that for fructose is 1.5mmol/L. Assume V max is the same for both glucose and fructose and the enzyme displays hyperbolic kinetics. For which substrate does hexokinase have the great affinity? Also, the comparison of the two sugars for hexokinase indicates which sugar is preferred as a nutrient?
lower so glucose; glucose *lower km = higher affinity
initial velocity
measure product formation as a function of time and then determining the velocity soon after the reaction has started
Cooperativity ensures that
most of the enzyme is on (R state) or off (T state)
allosteric enzymes are regulated by...
products of the pathways under their control
zero order vs first order
rate does not depend on concentration of substrate; rate depends on concentration of substrate
Why is feedback inhibition useful?
regulate production of F to meet cellular requirements so don't make more than is needed
(first order reaction) units
s^-1
sequential model
subunits undergo sequential changes in structure
km value is approximately
the substrate concentration of the enzyme in vivo
kcat/km is an approximate value of catalytic efficiency when
under conditions of low substrate concentration *higher kcat/km = better enzyme
(first order reaction) rate equation
v = k[A]
V0 =
vmax[S]/[S] + Km
Feedback inhibition
when sufficient F is present, F can bind to a regulatory site (different from active site) on e1 (the enzyme catalyzed the committed step) and inhibit the reaction
When the substrate concentration is high, an enzyme will display a. zero-order b. first-order c. second-order d. cease to function
zero-order *or an enzyme is saturated w/ substrates.