Transcript o - Columbia University

1
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
All the polypeptides are denatured and behave as random coils
All the polypeptides have the same charge per unit length
All are subject to the same electromotive force in the electric field
Separation based on the sieving effect of the polyacrylamide gel
Separation is by molecular weight only
SDS does not break covalent bonds (i.e., disulfides) (but can treat with
mercaptoethanol for that) (and perhaps boil for a bit for good measure)
2
Disulfides between 2 cysteines can be cleaved in the laboratory by reduction, i.e.,
adding 2 Hs (with their electrons) back across the disulfide bond.
One adds a reducing agent:
mercaptoethanol (HO-CH2-CH2-SH).
In the presence of this reagent, one gets exchange among the disulfides and the
sulfhydryls:
Protein-CH2-S-S-CH2-Protein + 2 HO-CH2CH2-SH --->
Protein-CH2-SH + HS-CH2-Protein + HO-CH2CH2-S-S-CH2CH2-OH
The protein's disulfide gets reduced (and the S-S bond cleaved), while the
mercaptoethanol gets oxidized, losing electrons and protons and itself forming a
disulfide bond.
3
P.A.G.E.
e.g., “p53”
Molecular weight
markers
(proteins of known
molecular weight)
4
Molecular sieve chromatography
(= gel filtration, Sephadex chromatography)
Sephadex bead
5
Molecular sieve chromatography
Sephadex bead
6
Molecular sieve chromatography
Sephadex bead
7
Molecular sieve chromatography
Sephadex bead
8
Molecular sieve chromatography
Sephadex bead
9
Fancy
Plain
4oC (cold room)
Larger molecules get to the bottom faster, and ….
Non-spherical molecules get to the bottom faster
~infrequent
orientation
Non-spherical
molecules get to
the bottom faster
10
11
Handout 4-3: protein separations
12
Winners: Largest and
most spherical
Lowest MW
Largest and
least spherical
Similar to handout 4-3,
but Winners &
native PAGE added
Winners:
Most charged
and smallest
13
Enzymes =
protein catalysts
Each arrow = an
ENZYME
14
Flow of glucose in E. coli
Macromolecules
Polysaccharides
Lipids
Nucleic Acids
Proteins
yn
th
e
tic
pa
t
hw
ay
monomers
bi
os
intermediates
glucose
Each arrow = an ENZYME
Each arrow = a specific chemical reaction
15
Chemical reaction between 2 reactants
H 2 + I2
2 HI
H 2 + I2
2 HI + energy
“Spontaneous” reaction:
Energy released
Goes to the right
H-I is more stable than H-H or I-I here
i.e., the H-I bond is stronger, takes more energy to break it
That’s why it “goes” to the right,
i.e., it will end up with more products than reactants
i.e., less tendency to go to the left, since the products are more stable
16
say, 100
kcal/mole
say, 103
kcal/mole
H2 + I2
2 HI
{
Change in Energy (Free Energy)
2H + 2I
Atom pulled completely apart
(a “thought” experiment)
-3 kcal/mole
Reaction goes spontaneously to the right
If energy change is negative: spontaneously to the right = exergonic: energy-releasing
If energy change is positive: spontaneously to the left = endergonic: energy-requiring
17
Different ways of writing chemical reactions
H 2 + I2
2 HI
H 2 + I2
2 HI
H 2 + I2
2 HI
H 2 + I2
2 HI
H 2 + I2
2 HI
18
say, 100
kcal/mole
But: it is not necessary to break
molecule down to its atoms in order
to rearrange them
say, 103
kcal/mole
H2 + I2
2 HI
{
Change in Energy (Free Energy)
2H + 2I
-3 kcal/mole
19
Reactions proceed through a transition state
I
I
+
H H
I
I
+
H H
I
I
H H
I
I
H
H
Transition state
(TS)
(H2 + I2)
I
H
+
I
H
(2 HI)
Products
20
Change in Energy
2H + 2I
~100 kcal/mole
H-H
| |
I-I
(TS)
Say,
~20 kcal/mole
2 HI
{
H 2 + I2
-3 kcal/mole
Activation
energy
Allows it to happen
Energy needed
to bring molecules
together to form
a TS complex
determines speed =
VELOCITY =
rate of a reaction
H 2 + I2
2 HI
{
Change in Energy (new scale)
21
HHII
(TS)
Activation
energy
3 kcal/mole
Net energy change:
Which way it will end up.
the DIRECTION
of the reaction, independent of the rate
2 separate
concepts
22
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
23
Example
Biosynthesis of a fatty acid
3 glucose’s
18-carbon fatty acid
Free energy change: ~ 300 kcal per mole of glucose used is REQUIRED
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)
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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
– Catalysts deal with this second problem, which we will now
consider
25
The velocity problem is solved by catalysts
The catalyzed reaction
The catalyst takes part in the reaction,
but it itself emerges unchanged
26
Change in Energy
HHII
(TS)
Activation
energy
without
catalyst
TS
complex
with
catalyst
H 2 + I2
2 HI
Activation
energy
WITH the
catalyst
27
Reactants in an enzyme-catalyzed reaction = “substrates”
28
Reactants (substrates)
Active site
or
Not a substrate
substrate binding site
(not exactly synonymous,
could be just part of the active site)
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Unlike inorganic catalysts,
enzymes are specific
Substrate Binding
30
Small molecules bind with great specificity to pockets on ENZYME surfaces
Too far
31
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
32
+
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).
Dihydrofolate reductase, the movie: FH2 + NADPH2  FH4 + NADP
or:
DHF + NADPH + H+  THF + NADP+
33
Enzyme-substrate
interaction is often
dynamic.
The enzyme protein
changes its 3-D structure
upon binding the
substrate.
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P
Velocity = V = ΔP/ Δ t
So V also = -ΔS/ Δt (disappearance)
From the laws of mass action:
back reaction
ΔP/ Δt = - ΔS/ Δt = k1[S] – k2[P]
For the INITIAL reaction, [P] is small and can be neglected:
ΔP/ Δt = - ΔS/ Δt = k1[S]
So the INITIAL velocity Vo = k1[S]
O
signifies INITIAL velocity
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Vo = ΔP/ Δ t
P vs. t
Slope = Vo
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0.6
P
Effect of different
initial substrate
concentrations on
P vs. t
[S4]
[S3]
0.4
[S2]
0.2
[S1]
0.0
t
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Vo = the slope in each case
Effect of different
0.6 initial substrate
concentrations
Dependence of Vo on
substrate concentraion
[S4]
[S3]
P
0.4
[S2]
0.2
[S1]
0.0
t
Considering Vo as a function of [S]
(which will be our usual useful consideration):
Vo = k1[S]
Slope = k1
Now, with an enzyme:
We can ignore the rate of the noncatalyzed reaction (exaggerated here
to make it visible)
38
Enzyme kinetics (as opposed to simple chemical kinetics)
Vo independent of [S]
Vo proportional to [S]
Can we understand this curve?
39
40
Michaelis and Menten mechanism for the action of enzymes (1913)
41
Michaelis-Menten mechanism
X
• 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.)
Steady state is not the same as equilibrium
System is at equilibrium
Constant level
No net flow
System is at “steady state”
Constant level
Plenty of flow
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43
System is at equilibrium
Constant level
No net flow
E+S
ES
E+P
System is at “steady state”
Constant level
Plenty of flow
44
Michaelis-Menten Equation(s)
See handout 5-1 at your leisure for the derivation
(algebra, not complicated, neat)
Vo =
k3[Eo][S]
[(k2+k3)/k1] +[S]
If we let Km =
= (k2+k3)/k1, just gathering 3 constants into one, then:
Vo =
k3 [Eo] [S]
Km + [S]
All the k‘s are constants for a particular enzyme
45
Rate is proportional
to the amount of
enzyme
Otherwise, the rate is
dependent only on S
k3 [Eo] [S]
Vo =
Km + [S]
At low S (compared to Km),
rate is proportional to S:
At high S (compared to Km),
Rate is constant
Vo ~ k3Eo[S]/Km
Vo = k3Eo
46
At high S, Vo here = k3Eo, == Vmax
So the Michaelis-Menten equation can be written:
Vo =
k3 [Eo] [S]
Km + [S]
Vo = Vmax [S]
Km + [S]
Simplest
form
47
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.
48
Some turnover numbers (per second)
• Succinic dehydrogenase:
• Most enzymes:
• The winner:
Carbonic anhydrase (CO2 +H20
19 (below average)
100 -1000
H2CO3)
600,000
That’s 600,000 molecules of substrate, per molecule of enzyme, per
second.
Picture it!
You can’t.
49
Km ?
Consider the Vo that
is 50% of Vmax
Vmax/2 is achieved at a [S] that turns out to be numerically equal to Km
So Km is numerically equal to the concentration of substrate
required to drive the reaction at ½ the maximal velocity
Try it:
Set Vo = ½ Vmax in the M.M. equation and solve for S.
50
Another view of Km:
Consider the reverse of this reaction
(the DISsociation of the ES complex):
k2
ES
E+S
k1
The equilibrium constant for this dissociation reaction is:
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)
{
51
Consider in reverse
k2
ES
E+S
Kd = k2/k1
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
52
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
53
Got this far
Enzyme inhibition:
competitive, non-competitive, and allosteric
Competitive:
A competitive inhibitor resembles the substrate
54
55
A competitive inhibitor can be swamped out at high
substrate concentrations
Handout 5-3b
56
+
Vo
Apparent (measured) Km increases
Substrate concentration
Inhibitor looks like the substrate
And, like the substrate, binds to the substrate binding site
Biosynthetic pathway to cholesterol
57
58
Zocor
(simvastatin)
59
½ Vmax
w/o inhibitor
½ Vmax with
yet more
inhibitor
Km remains unchanged. Vmax decreases.
60
Substrate
Non-competitive inhibitor
Example: Hg ions (mercury) binding to –SH groups in the active site
61
Non-competitive inhibitor example
Substrate still binds OK
But an essential participant in the reaction is blocked
(here, by mercury binding a cysteine sulfhydryl)
Hg++
62
63
Allosteric inhibition
Inhibitor
binding
site
+
Active
Active
= allosteric inhibitor
Inactive
= substrate
Allosteric inhibitor binds to a different site than the substrate,
so it need bear no resemblance to the substrate
The apparent Km OR the apparent Vmax or both may be affected.
The effects on the Vo vs. S curve are more complex and ignored here
64
Allosteric inhibitors are used by the cell for
feedback inhibition of metabolic pathways
Feedback inhibition of enzyme activity, or “End product inhibition”
PQRSTUV
End product
End product
End product
First committed step is usually inhibited
protein
65
Thr deaminase
glucose ...... --> --> threonine -----------------> alpha-ketobutyric acid
A
Substrate
B
C
protein
isoleucine
(and no other aa)
Allosteric inhibitor
Also here: Feedback inhibitor
(is dissimilar from substrate)
Rich medium = provide glucose + all 20 amino acids and all vitamins, etc.
20 minutes !, in a rich medium
60 minutes, in a minimal medium
66
67
Direction
Flow
of glucose
in E. coli
of reactions
in metabolism
Macromolecules
Polysaccharides
Lipids
Nucleic Acids
Proteins
yn
th
e
tic
pa
t
hw
ay
monomers
bi
os
intermediates
glucose
Each arrow = a specific chemical reaction
68
Energy
}Freedifference
determines the direction
of a chemical reagion
69
For the model reaction A + B
C + D,
written in the left-to-right direction indicated:
Consider the quantity called the change in free energy
associated with a chemical reaction, or: Δ G
Such that:
• IF Δ G IS <0:
THEN A AND B WILL TEND TO PRODUCE C AND D
(i.e., tends to go to the right).
• IF Δ G IS >0:
THEN C AND D WILL TEND TO PRODUCE A AND B.
(i.e., tends to go to the left)
• IF Δ G IS = 0:
THEN THE REACTION WILL BE AT EQUILIBRIUM:
NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY.
ΔG = Δ
Go+
70
RTln([C][D]/[A][B])
• where A, B, C and D are the concentrations of the reactants and the
products AT THE MOMENT BEING CONSIDERED.
(i.e., these A, B, C, D’s here are not the equilibrium concentrations)
• R = UNIVERSAL GAS CONSTANT = 1.98 CAL / DEG K MOLE (R =~2)
• T = ABSOLUTE TEMP ( oK )
0oC = 273oK; Room temp = 25o C = 298o K
(T =~ 300)
• ln = NATURAL LOG
• Δ Go = a CONSTANT:
a quantity related to the INTRINSIC properties of A, B, C, and D
71
Also abbreviated form:
Δ G = Δ Go+ RTlnQ (Q for “quotient”)
Where Q = ([C][D]/[A][B])
Qualitative term
Quantitative term
Josiah
Willard
Gibbs
(1839 1903)
72
Δ Go
STANDARD FREE ENERGY CHANGE of a reaction.
If all the reactants and all the products are present at 1 unit
concentration,
then:
Δ G = Δ Go + RTln(Q) = Δ Go + RTln([1][1] / [1][1])
Δ G = Δ Go + RTln(Q) = Δ Go + RTln(1)
or Δ G = Δ Go +RT x 0,
or Δ G = Δ Go,
when all components are at 1
….. a special case
(when all components are at 1)
“1” usually means 1 M
So Δ G and Δ Go are quite different,
and not to be confused with each other.
Δ Go allows us to compare all reactions under the
same standard reaction conditions that we all
agree to, independent of concentrations.
So it allows a comparison of the stabilities of the
bonds in the reactants vs. the products.
It is useful.
AND,
It is easily measured.
73
74
Because,
• at equilibrium, Δ G = Δ Go + RTln(Q) = 0
and at equilibrium Q = Keq =
(a second special case).
[C]eq [D]eq
[A]eq [B]eq
• So: at equilibrium,
Δ G = Δ Go + RTln(Keq) = 0
• And so: Δ Go = - RTln(Keq)
• So just measure the Keq,
• Plug in R and T
• Get: ΔGo, the standard free energy change
E.g., let’s say for the reaction A + B
Keq happens to be:
[C]eq[D]eq
[A]eq[B]eq
75
C + D,
= 2.5 x 10-3
Then Δ Go = -RTlnKeq = -2 x 300 x ln(2.5 x 10-3)
= -600
x
-6
= +3600
3600 cal/mole (If we use R=2 we are dealing with calories)
Or: 3.6 kcal/mole
3.6 kcal/mole ABSORBED (positive number)
So energy is required for the reaction in the left-to-right direction
And indeed, very little product accumulates at equilibrium
(Keq = 0.0025)
76
Note:
If ΔGo = +3.6 for the reaction A + B < --- >C + D
Then ΔGo = -3.6 for the reaction C + D <--- > A + B
(Reverse the reaction: switch the sign)
And:
For reactions of more than simple 1 to 1 stoichiometries:
aA + bB <--> cC + dD,
ΔG = ΔGo + RT ln [C]c[D]d
[A]a[B]b
Some exceptions to the 1M standard condition:
Exception #1:
• 1) Water: 55 M (pure water) is considered the “unit”
concentration instead of 1M
The concentration of water rarely changes during the course of
an aqueous reaction, since water is at such a high
concentration.
• So when calulating Go, instead of writing in “55” when water
participates in a reaction (e.g., a hydrolysis) we write “1.”
• This is not cheating; we are in charge of what is a “standard”
condition, and we all agree to this: 55 M H20 is unit (“1”)
concentration for the purpose of defining Go.
77
Exception #2:
In the same way,
Hydrogen ion concentration, [H+]: 10-7 M is taken as unit
concentration, by biochemists.
since pH7 is maintained in most parts of the cell despite a
reaction that may produce acid or base.
This definition of the standard free energy change requires the
designation ΔGo’
However, I will not bother.
But it should be understood we are always talking about ΔGo’ in
this course.
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Summary
ΔG = Go + RTln(Q)
This combination of one qualitative and one
quantitative (driving) term tell the direction of a
chemical reaction in any particular circumstance
ΔGo = - RTln(Keq)
The ΔGo for any reaction is a constant that can be looked up
in a book.