Download E = enzyme, S= substrate • The key does not fit the lock quite

E = enzyme, S= substrate
• The key does not fit the lock quite perfectly, but
rather exercises a certain strain on it.
1. Both the substrate and the enzyme have to change their
shape for binding.
2. When bound to the enzyme the bond is stretched such that
it is ready to break.
3. After the bond breaks, a new bond forms somewhere else
in the substrate.
4. This changes the shape of the substrate and it can easily
unbind from the enzyme.
Without the enzyme
The interaction with the
enzyme lowers the
energy
When bound, entropy
lower
Sum of contributions
The enzyme has reduced ∆Gǂ , but cannot change ∆G.
•
•
•
•
•
This is the picture for a system with a single enzyme and a
single substrate molecule.
If we have many substrate molecules, the chemical
potential of the reactants grows (kBT ln cS), pulling up the
left side of the free energy landscape.
∆Gǂ decreases, ∆G increases.
If we have many product molecules, the chemical potential
of the reactants grows (kBT ln cP), pulling up the right side of
the free energy landscape.
∆G decreases.
•
Generalize the reaction coordinate to count the substrate
molecules -> cyclic machine
Many enzymes can be regarded as simple cyclic machines; they
work by random-walking down a one-dimensional free energy
landscape. The net descent of this landscape in one forward step
is the value of ∆G for the reaction S -> P.
• Enzymes display saturation kinetics – the reaction velocity –
d[S]/dt – is not linear in [S], but saturates at large [S].
• In the ratchet picture, large [S] -> large ∆G -> perfect ratchet
-> velocity saturates
• In the enzyme picture, large [S] -> the enzyme is occupied most
of the time.
• Another way to make the ratchet perfect (irreversible) is to
lower [P] -> the kBT ln cP becomes strongly negative.
• In the enzyme picture, there is no product, P, to bind to the
enzyme to go back to the substrate, S.
Mechanochemical motors – 2D energy landscape
Macroscopic world
1:1 gear
w1 = load, w2 = machine
If rigid with N teeth:
 =  + 2n/N
If slippery, rubber teeth (N=3, w1 = w2 = 0):
Preferred motion along the valleys.
 =  = 2  bump along the valley  slipping (from one valley
to the next)
What is the effect of applying a load, w1 > 0, and a driving torque,
w2 > 0 (w1 < w2 )?
Like in 1D, if the load is
too large, the machine
could get stuck at  = 
= 2.
In 2D, there is also the
option of slipping to
the next valley.
If slipping becomes too large, the machine doesn’t work!!
Microscopic world
• A mechanochemical machine that catalyses a reaction S->P and
has a second binding site that can attach to any point along a
periodic track  a ratchet on a 2D free energy landscape
• Model for kinesin: catalyses ATP->ADP and moves along
microtubule
• the two coordinates are: the number of substrate molecules
processed and the position along the microtubule.
If the valleys on the energy landscape sufficiently deep  motion
effectively only along valleys = tight coupling  effectively 1D
(coordinate along the valley)
Real Enzymes
• In metabolism (glycolysis), a derivative of glucose,
phosphoglycerate, transfers one of his phosphate groups to ADP
 ATP
ADP
ATP
Phosphoglycerate
kinase
1,3-Bisphosphoglycerate
3-Phosphoglycerate
• If phosphoglycerate would transfer its phosphate group to water
 no ATP
• To avoid that, phosphoglycerate kinase engulfs
phosphoglycerate and ADP protecting them from water.
the
• Phosphoglycerate kinase consists of two domains, connected
with a hinge. Some of the amino acids required for catalysis are in
one domain and some in the other. When the hinge closes these
amino acids come in contact with each other and the substrates,
phosphoglycerate and ADP .
Phosphoglycerate kinase
ATP
The enzyme will stretch the substrates into their transition state.
The kinetics of enzymes (Michaelis-Menten)
• Most often we don’t know the energy landscape
• Instead, assume:
1. almost no product  abrupt drop of G before products  no
back step
2. EP -> E + P is very rapid, neglect EP as an intermediate state
3. There is only one molecule of enzyme
• The reaction is
E+S
cs k1
k1
k2
ES 
 E+P
• The fraction of time that the enzyme is unoccupied is PE ,
and occupied, PES = 1 - PE . Then
d
PE  k1cS (1  PES )  (k1  k2 ) PES
dt
In steady state:
k1cS
PES 
k1  k2  k1cS
The rate at which a single enzyme creates the products is k2PES.
Then, if cE is the concentration of enzyme and
k1  k2
KM 
k1

;
v  vmax
vmax  k2cE
cs
K M  cs
Michaelis-Menten
law
• KM = Michaelis constant; cS = KM  v = vmax/2
• linear at small cS , saturation at large cS
• The Lineweaver-Burk plot (linearized plot)
1
1

v vmax
 KM 
1 

cs 

• E.g.: pancreatic carboxypeptidase  cleaves amino acids from
the end of a polypeptide.
Pancreatic carboxypeptidase acting on a peptide with only two
amino acids. KM = 6.4 mM, vmax = 0.085 mM s-1