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Lec. 6 Intro Bio 2011 Protein structure cont. Prosthetics groups Membrane proteins Protein purification methods Ultracentrifugation Native gel electrophoresis SDS gel electrophoresis Gel filtration Chemical kinetics Enzyme kinetics Michaelis-Menton equation turnover number, Km
Last updated: 9/22/11 12:44 PM
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Riboflavin~ vitamin B2 Heme
Tetrahydrofolic acid~ vitamin B9
Most prosthetic groups are bound tightly via ONLY weak bonds.
Hemoglobin
α2β2 tetramer
heme
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Metal ions can be prosthetic groups
Zinc (Zn++)
A “zinc finger” protein domain
with two histidines and two cysteines shown
(Voet and Voet, Biochemistry)
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Membrane proteins
Hydrophobic side chains on the protein exterior for the portion in contact with the interior of the phospholipid bilayer.
Anions are negatively charged.
Cations are positively charged
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Small molecules bind with great specificity to pockets on protein surfaces
Too far
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A ligand nestled in a tight-binding pocket
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Estrogen receptor binding estrogen, a steroid hormone
estrogen estrogen
detail
Estrogen receptor is specific, does not bind testosterone
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Protein binding can be very specific
Testosterone Estrogen
Estrogen (green) binding to amino acid side chains in the estrogen receptor
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Protein separation methods
Ultracentrifugation
Mixture of proteins
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10Ultracentrifuge
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centrifugal force = mω2r
m = massω = angular velocity r = distance from the center of rotation
Causing sedimentation:
Opposing sedimentation = friction = foV.
Constant velocity is soon reached; then, no net forceSo now: centrifugal force = frictional force (balanced each other out)
And so: mω2r = foV V = mω2r/fo,
Or: V = [ω2r] x [m / fo]
fo = frictional coefficient (shape)V = velocity
V proportional to mass (MW). So higher MW moves faster
V inversely proportional to fo;V inversely proportional to “non-sphericity”
So most spherical shape moves fastest
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Intermediate:Large, +++high positive charge
Loser:Large, +low positive charge
Winner: Small, +++High positive charge
Intermediate:Small, +Low positive charge
Molecules shown after several hours of electrophoresis
Glass plates
Sample loaded here
+ + +
+++ +++ ++++
++
+++++++++
poly-acrylamidefibers
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Upper resevoir
Cut out for contactof buffer with gel
Upper resevoir
Cut out for contactof buffer with gel
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Reservoir for bufferElectrode connection
(~ 150 V)
Clamped glass sandwich
Cut out of glass platefor contact of buffer with gel
Platinum wire electrode
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Tracking dyes
Happy post-doc
Power supply
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SDS PAGE = SDS polyacrylamide gel electrophoresis
• sodium dodecyl sulfate, SDS (or SLS): CH3-(CH2)11- SO4--
• CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-SO4--
SDS
Polypeptides denatured, act as random coilsAll have same charge per unit lengthAll subject to the same electromotive force in the electric fieldSeparation based on the sieving effect of the polyacrylamide gelSeparation is by molecular weight onlySDS does not break covalent bonds (i.e., disulfides) (but treat with mercaptoethanol for that, boil for a bit for good measure)
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Reduction of protein disulfide bonds :
Protein-CH2-S-S-CH2-Protein + 2 HO-CH2CH2-SH) (mercaptoethanol)[ a reducing agent ]
Protein-CH2-SH + HO-CH2CH2-S-S-CH2CH2-OH
+ HS-CH2-Protein
Protein gets reduced; mercaptoethanol gets oxidized
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Molecular weight markers
(proteins of known molecular weight)
P.A.G.E.
e.g., “p53”
12 18
48 80 110 130 160
140 Kd
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Sephadex bead
Molecular sieve chromatography(= gel filtration, Sephadex chromatography)
Mild conditions (pH, temperature), proteins in their native state
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Sephadex bead
Molecular sieve chromatography
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Sephadex bead
Molecular sieve chromatography
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Sephadex bead
Molecular sieve chromatography
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Sephadex bead
Molecular sieve chromatography
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PlainFancy4oC (cold room)
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Non-spherical molecules get to the bottom faster
Larger molecules get to the bottom faster, and ….Non-spherical molecules get to the bottom faster
~infrequent orientation
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Most chargedand smallest
Largest and most spherical
Lowest MW
Largest and least spherical
Similar to handout 4-3, but Winners &
native PAGE added
Winners:
Winners:
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Enzymes = protein catalysts
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g l u c o s e
monomers
MacromoleculesPolysaccharides LipidsNucleic AcidsProteins
biosy
nthet
ic p
athw
ay
intermediates
F l o w o f g l u c o s e i n E . c o l i
E ac h a rro w = a sp e c ific c h em ica l re ac tio nEach arrow = an ENZYME
Each arrow = an ENZYME
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H2 + I2 2 HI
H2 + I2 2 HI + energy
“Spontaneous” reaction:
Energy releasedGoes to the rightH-I is more stable than H-H or I-I herei.e., the H-I bond is stronger, takes more energy to break itThat’s why it “goes” to the right, i.e., it will end up with more products than reactantsi.e., less tendency to go to the left, since the products are more stable
Chemical reaction between 2 reactants
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Ch
ang
e in
En
erg
y (F
ree
En
erg
y)
H2 + I22 HI
{
-3 kcal/mole
2H + 2I
say, 100 kcal/mole
say, 103kcal/mole
Atom pulled completely apart(a “thought” experiment)
Reaction goes spontaneously to the right
If energy change is negative: spontaneously to the right = exergonic: energy-releasingIf energy change is positive: spontaneously to the left = endergonic: energy-requiring
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H2 + I2 2 HI
2 HIH2 + I2
2 HIH2 + I2
Different ways of writing chemical reactions
H2 + I2 2 HI
2 HIH2 + I2
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Ch
ang
e in
En
erg
y (F
ree
En
erg
y)
H2 + I22 HI
2H + 2I
{
-3 kcal/mole
say, 100 kcal/mole
say, 103kcal/mole
But: it is not necessary to break molecule down to its atoms in order to rearrange them
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H H
+I I
H H
IIII
H H
Transition state (TS) +
H
I
H
I
(2 HI)
H H+
I I
(H2 + I2)
Products
Reactions proceed through a transition state
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Ch
ang
e in
En
erg
y
H2 + I2
2 HI
2H + 2I
{
-3 kcal/mole
~100 kcal/mole
H-H| |I-I(TS)
Activationenergy
Say,~20 kcal/mole
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Ch
ang
e in
En
erg
y (n
ew s
cale
)
H2 + I2
2 HI
{
3 kcal/mole
Activation energy
HHII(TS)
Allows it to happen
determines speed = VELOCITY = rate of a reaction
Energy neededto bring molecules together to forma TS complex
Net energy change:Which way it will end up. the DIRECTIONof the reaction, independent of the rate
2 separate concepts
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Concerns about the cell’s chemical reactions
• Direction– We need it to go in the direction we want
• Speed– We need it to go fast enough to have the
cell double in one generation time
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3 glucose’s 18-carbon fatty acid
Free energy change: ~ 300 kcal per mole of glucose used is REQUIRED
Biosynthesis of a fatty acid
So: 3 glucose 18-carbon fatty acid
So getting a reaction to go in the direction you want is a major problem(to be discussed next time)
Example
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Concerns about the cell’s chemical reactions
• Direction– We need it to go in the direction we want
• Speed– We need it to go fast enough to have the
cell double in one generation time
– Catalysts deal with this second problem, which we will now consider
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Got this far
40
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The catalyzed reaction
The velocity problem is solved by catalysts
The catalyst takes part in the reaction, but it itself emerges unchanged
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Ch
ang
e in
En
erg
y
H2 + I2
2 HI
Activation energywithoutcatalyst
HHII(TS)
TS complexwith catalyst
Activation energyWITH thecatalyst
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Reactants in an enzyme-catalyzed reaction = “substrates”
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Reactants (substrates)
Not a substrate
Active site or
substrate binding site(not exactly synonymous,
could be just part of the active site)
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Substrate Binding
Unlike inorganic catalysts, enzymes are specific
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Small molecules bind with great specificity to pockets on ENZYME surfaces
Too far
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Unlike inorganic catalysts, enzymes are specific
succinic dehydrogenase
HOOC-HC=CH-COOH <-------------------------------> HOOC-CH2-CH2-COOH +2H
fumaric acid succinic acid
NOT a substrate for the enzyme: 1-hydroxy-butenoate: HO-CH=CH-COOH (simple OH instead of one of the carboxyls)
Maleic acid
Platinum will work with all of these, indiscriminantly
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Enzymes work as catalysts for two reasons:
1. They bind the substrates putting them in close proximity.
2. They participate in the reaction, weakening the covalent bonds of a substrate by its interaction with their amino acid residue side
groups (e.g., by stretching bonds).
+
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49Dihydrofolate reductase, the movie: FH2 + NADPH2 FH4 + NADP
or: DHF + NADPH + H+ THF + NADP+
Enzyme-substrate interaction is oftendynamic.
The enzyme protein changes its 3-D structureupon binding thesubstrate.
http://chem-faculty.ucsd.edu/kraut/dhfr.mpg
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Chemical kinetics
Substrate Product (reactants in enzyme catalyzed reactions are called substrates)S PVelocity = V = ΔP/ Δ tSo V also = -ΔS/ Δt (disappearance)From the laws of mass action:ΔP/ Δt = - ΔS/ Δt = k1[S] – k2[P]
For the INITIAL reaction, [P] is small and can be neglected:Initial ΔP/ Δt = - ΔS/ Δt = k1[S]
So the INITIAL velocity Vo = k1[S]
back reaction
O signifies INITIAL velocity
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P vs. tSlope = Vo
Vo = ΔP/ Δ t
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P
t
[S1]
[S2]
[S3]
[S4]
Effect of different initial substrate concentrations on P vs. t
0.0
0.2
0.4
0.6
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P
Vo = the slope in each case
t
[S1]
[S2]
[S3]
[S4]Effect of different initial substrate concentrations
0.0
0.2
0.4
0.6
Considering Vo as a function of [S](which will be our usual useful consideration):
Slope = k1Vo = k1[S]
Dependence of Vo on substrate concentraion
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We can ignore the rate of the non-catalyzed reaction (exaggerated here to make it visible)
Now, with an enzyme:
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Vo proportional to [S]
Vo independent of [S]
Enzyme kinetics (as opposed to simple chemical kinetics)
Can we understand this curve?
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Michaelis and Menten mechanism for the action of enzymes (1913)
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Michaelis-Menten mechanism
• Assumption 1. E + S <--> ES: this is how enzymes work, via a complex
• Assumption 2. Reaction 4 is negligible, when considering INITIAL velocities (Vo, not V).
• Assumption 3. The ES complex is in a STEADY-STATE, with its concentration unchanged with time during this period of initial rates.
(Steady state is not an equilibrium condition, it means that a compound
is being added at the same rate as it is being lost, so that its concentration remains constant.)
X
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S ystem is a t equ ilib riumC onstant leve lN o net flow
S ystem is a t “steady state”C onstant leve lP lenty o f flow
Steady state is not the same as equilibrium
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E + S
ES
E + P
System is at equilibriumConstant levelNo net flow
System is at “steady state”Constant levelPlenty of flow
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Michaelis-Menten Equation(s)
[(k2+k3)/k1] +[S]
k3[Eo][S]Vo =
If we let Km = (k2+k3)/k1, just gathering 3 constants into one, then:
k3 [Eo] [S]Vo =
Km + [S]
See handout 5-1 at your leisure for the derivation (algebra, not complicated, neat)
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k3 [Eo] [S]Vo =
Km + [S]
Rate is proportional to the amount of enzyme
Otherwise, the rate is dependent only on S
At low S (compared to Km),rate is proportional to S:
Vo ~ k3Eo[S]/Km
At high S (compared to Km),Rate is constant
Vo = k3Eo
All the k‘s are constants for a particular enzyme
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At high S, Vo here = k3Eo, = Vmax
So the Michaelis-Menten equation can be written:
Vmax [S]Vo =
Km + [S]
k3 [Eo] [S]Vo =
Km + [S]Simplest form
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Understanding Vmax:( the maximum intital velocity achievable with a given amount of enzyme )
Now, Vmax = k3Eo
So: k3 = Vmax/Eo
= the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eo
k3 = the TURNOVER NUMBER
• the maximum number of moles of substrate converted to product per mole of enzyme per second;
• the maximum number of molecules of substrate converted to product per molecule of enzyme per second
• Turnover number (k3) then is: a measure of the enzyme's catalytic power.
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Some turnover numbers (per second)
• Succinic dehydrogenase: 19 per second (below average)• Most enzymes: 100 -1000• The winner:
Carbonic anhydrase (CO2 +H20 H2CO3)
600,000
That’s 600,000 molecules of substrate, per molecule of enzyme, per second.
Picture it!
You can’t.
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Km ?
Consider the Vo that is 50% of Vmax
So Km is numerically equal to the concentration of substrate required to drive the reaction at ½ the maximal velocityTry it: Set Vo = ½ Vmax in the M.M. equation and solve for S.
Vmax/2 is achieved at a [S] that turns out to be numerically equal to Km
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The equilibrium constant for this dissociation reaction is:
Consider the reverse of this reaction(the DISsociation of the ES complex):
ESk2
k1
E + S
Kd = [E][S] / [ES] = k2/k1
(It’s the forward rate constant divided by the backward rate constant. See the Web lecture if you want to see this relationship derived)
Another view of Km:
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ESk2
k1
E + S Kd = k2/k1
Km = (k2+k3)/k1 (by definition)
IF k3 << k2, then: Km ~ k2/k1
But k2/k1 = Kd (from last graphic)so Km ~ Kd for the dissociation reaction (i.e., the equilibrium constant)
(and 1/Km = ~ the association constant)
So: the lower the Km, the more poorly it dissociates.That is, the more TIGHTLY it is held by the enzyme
And the greater the Km, the more readily the substrate dissociates,so the enzyme is binding it poorly
{Consider in reverse
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Km ranges
• 10-6M is good• 10-4M is mediocre• 10-3M is fairly poor
So Km and k3 quantitatively characterize how an enzyme does the job as a catalyst
k3, how good an enzyme is in facitiating the chemical change (given that the substrate is bound)
Km, how well the enzyme can bind the substrate in the first place
.
.
.
And any enzyme can be characterized by its turnover number and its Km.