Chapter 6
Protein Function : Enzymes
Part 1
Enzymes
Enzyme Learning Goals: to Know
– Physiological significance of enzymes
– Catalytic power of enzymes
– Chemical mechanisms of catalysis
– Mechanism of chymotrypsin
– Description of enzyme kinetics and inhibition
Enzymes
Mostly Proteins (a few RNA’s are capable of catalysis)
Active Site: Substrate Binding + Reaction  Products
Some require Cofactors (metals) or Coenzymes (organic
cpds)
Some enzymes have other binding sites…involved in
regulation, we will see later, Part 2
EOC Problem 1 involves the sweetness of corn
affected by corn enzymes and Problem 2 calculates the
average molar concentration of enzymes in a bacterial cell:
you can take it further to find the number of molecules of
each enzyme present in a cell.
Enzyme Pioneers
First Cell Free Prep
First to Crystallize Urease
Weak bonding at active
site results in catalysis
Why biocatalysis over inorganic catalysts?
•
•
•
•
Greater reaction specificity: avoids side products
Milder reaction conditions: conducive to conditions in cells
Higher reaction rates: in a biologically useful timeframe
Capacity for regulation: control of biological pathways
COO
-
COO
NH2
O
OH
COO
OH
COO
Chorismate
mutase
COO
OOC
O
NH2
-
-
O
COO
COO
• Metabolites have many
potential pathways of
decomposition
-
• Enzymes make the
desired one most
favorable
OH
EOC Problem 4: Examines the thermal protection of hexokinase that a substrate brings to
the table: maintaining conformation under harsh conditions. Later in Part 2 of this chapter
we will see X ray data backing this up.
Enzymatic Substrate Selectivity:
Phenylalanine hydroxylase
OH
H
-
H
+
NH3
OOC
-
H
-
+
NH3
No binding
+
NH3
OOC
OOC
OH
HO
H
H
H
NH
CH3
OH
Binding but no reaction
Class Is the First Part of E.C. Number
EC 2.7.1.1 = ATP:glucose phosphotransferase or Hexokinase
2 = Transferase
7 = Phosphotransferase
1 = Transferred to a hydroxyl
1 = Glucose is the acceptor
Enzyme
Search By
Class
Bacterial Luciferase
Rxn
FMNH2 + O2 + RCHO  FMN + RCOOH + H2O + light
Continuing with the EC Numbers-1
Continuing with EC Numbers-2
NiceZyme
Enzyme with an Active Site
Active Site
Chymotrypsin
Thermodynamics of a Reaction
S + E  ES  E + P
Enzyme Catalyzed Reaction
E+S
↔
ES
↔
EP
↔
E+P
EOC Problem 3: A rate enhancement problem using Urease, the
enzyme that converts: Urea  CO2 + 2 NH3. The calculation
demonstrates how long it would take if urease were not present !
Dihydrofolate Reductase
Substrate Binds in a Fold or Pocket
NADP+
+
Folic Acid

ΔGB = binding energy
Enzyme Reactions Bind Substrate then Change
Shape to Transition State
Triose Phosphate Isomerase
Terribly Slow rate with Glyceraldehyde…phosphate
important in stabilizing binding.
Rate Enhancement Due to Proximity (Entropy
Reduction)
Acid/Base
Catalysis
Catalytic Mechanisms
– acid-base catalysis: give and take protons
– covalent catalysis: change reaction paths
– metal ion catalysis: use redox cofactors, pKa shifters
– electrostatic catalysis: preferential interactions with
Transition State.
Acid Base Catalysis – Involve Proteins R groups
Formation of a Covalent Intermediate
Michaelis Menten Curve
Michaelis Menten Equation:
L. Michaelis and Miss Maud L. Menten. 1913. "Die Kinetik der
Invertinwerkung" Biochemische Zeitschrift Vol. 49.
vo =
Vmax [S]
Km + [S]
Invertase Reaction: sucrose + H2O  glucose + fructose
Michaelis Menten Experiment
Measure Rate (v) at several concentrations of Substrate (S)
Here is one tube with one beginning concentration of S
E
SP
Calculate Δ[S]/min or Δ[P]/min.
This enzyme,
triosephosphate
isomerase is a one
substrate, one product
enzyme.
Michaelis Menten Experiment: Real Data
At each [S], the concentration of
enzyme is exactly the SAME.
Calculate Δ[S]/min for each [S]
EOC Problem 6 is about using
340nm light to measure
dehydrogenase reactions…the
classic lactate dehydrogenase.
Do this at more concentrations of S
to get a larger data set used for 
Initial Velocites are the Dashed Line
A
Michaelis Menten Plot
Michaelis Menten Equation is Non-Linear
vo =
Vmax [S]
KM + [S]
Straightened Out by reciprocals…Lineweaver Burke
Equation:
1/vo = (KM/Vmax)(1/[S]) + 1/Vmax
the Equation of a Straight Line
y = mx + b
Thus, y = 1/vo , x = 1/[S] and m (the slope) = KM/Vmax
Lets Plot this Out…next slide
Lineweaver-Burke Plot
Double Reciprocal
Data Points
are in this
range
Origin is Zero
There Are Other Equations to Convert
the Michaelis Menten Equation to a
Straight Line
Eadie Hofstie
v = -Km(v/S) + Vmax
Hanes Wolf:
S/v = (1/Vmax)(S) + Km/Vmax
all are y = mx + b
KM = is an Intrinsic Property of an enzyme
What does this mean? Intrinsic vs Extrinsic?
Vmax is an Extrinsic Property of Enzymes
At a high [S], varying only
the enzyme conc :
To get an Intrinsic Catalytic Constant from Vmax
kcat comes from Vmax and [Enz]
Vmax is [molar]/sec
[Enz] in molar
kcat = Vmax/ [Enz]
kcat/Km
Calculation of Km and Vmax
The enzyme, Practicase
Studentose  Productate
Studentose, mM
velocity, μmoles/ml/sec
1
12
2
20
4
29
8
35
12
40
Assay volume = 1 ml/tube
What’s in the tube: buffer + enzyme, then add substrate at time
Zero.
EOC Problems 11(dead easy to do by inspection) and 13 to do
by Lineweaver Burke plot
Calculation of Km and Vmax
Studentose, mM
1/[S]
Velocity,
1/v
μmoles/ml/sec
1
1
12
0.083
2
0.5
20
0.050
4
0.25
29
0.034
8
0.125
35
0.029
12
0.083
40
0.025
Now Plot this on Lineweaver Burk Plot….remember Zero is near the middle
of the graph!
Lineweaver Burke Plot of Practicase
1/
1
Practicase kcat = an Intrinsic Property
In the enzyme assay (one ml), each tube had 10 μg of
practicase. The molecular weight of practicase is 20,000 D.
Thus, each tube had
10 μg / 20,000 μg/μmole = 0.0005 μmole practicase
kcat = Vmax/ [Enz] = (50 μmole/sec)/ 0.0005 μmole = 1 x 105 s-1
Thus one enzyme reaction takes 1/ 1x 105 s-1 = 10-5 sec
or 10 μ sec.
What is Wrong with this L-B graph?
What is Wrong with this L-B graph?
Things to Know and Do Before Class
1. Principles of catalysis.
2. Enzyme naming concepts.
3. Intrinsic and Extrinsic values of Enzyme kinetics.
4. Be able to do a Michaelis Menten graph.
5. Be able to do a Lineweaver Burke graph.
6. To do EOC problems 1-6, 11, 13.