BSCI 230 Today, 2/8/01
Lecture #4
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Todayís topic: ENZYMES
Reviews
Two factors determine the spontaneity
of a reaction:
Enthalpy: DH
Entropy: DS
The spontaneity of a reaction is expressed
as the Gibbs Free Energy or DG
Gibbs Free Energy
DG = DH
- T DS
DG and Reactions
C6H12O6
<---> 6CO2 + 6 H2O
To the right, entropy is increasing,
bonds are broken to give off energy, DG is negative
(metabolism; catabolic)
To the left, entropy is decreasing,
bonds are being formed, DG is positive (synthesis,
anabolic)
Reactions:
CH4 + 2O2 <-->
CO2 + 2 H2O
ATP <---> ADP + Phosphate
2 H2 + O2 <-->
2 H2O
Why doesnít this just happen without
a spark??
Energy of Activation with a Catalyst
Proteins as Catalysts
Chemical reaction can occur if it
releases free energy
Due to high activation energies, biological
reactions occur slowly without catalysts
Enzymes (catalysts) do not alter the
equilibrium or the change in free energy
Catalysts decrease activation energy
Proteins as Catalysts
Enzymes are proteins (rare exceptions)
and end in ìaseî
They act on substrate molecules
An enzyme is not irreversibly changed
by the reaction
Enzymes bind substrates to ëactive
siteí (see Fig 6-2)
Properties of Enzymes
-
Specificity for substrates
-
inhibition by similar molecules
-
Saturation
-
maximum rate of product formation at fixed [enzyme]
-
Predictable kinetics (reaction speeds under different conditions)
Experiment with an Enzyme
Series of test tubes with constant
pH, temperature, and [Enzyme]
Each tube has a different [S] (substrate
concentration)
We will measure the amount of product
formed over time (i.e., the rate of product formation) at each [S]
mmoles/min
What would happen at a higher initial
[S] ??
What happens at a very high [S]?
Experimental Results at One VERY HIGH
Initial [S]
Experimental Results at One [S]
Experimental Results: Rate vs. [S]INITIAL
Km = [S] at 1/2 Vmax
Km is a measure of the
affintiy of the enzyme for the substrate
[S] vs. Velocity plot is not linear
and thus not easy to plot or gather information from
Therefore plot
1/[S] vs. 1/Vi
Lineweaver-Burk Plot:
1/Vi = Km/Vmax
(1/[S]) + 1/Vmax
(note: a straight line!)
y = m (x) + b
Can you explain this result? (PH curve)
Can you explain this result? (temperature)
Inhibition
Is Possible
Similar molecules interfere or compete
with substrate for active site of enzyme
Less product is formed in competitive
inhibition
Other molecules may inactivate enzyme
molecules reducing maximum rate of product formation
Noncompetitive
inhibition
Other terms:
Allosteric (other site)
inhibition or activation by binding
and changing shape of enzyme protein
Feedback Inhibition
product builds up and decreases enzyme
activity
Metabolism
Catabolic reactions:
Breaking bonds, liberating energy,
increasing entropy
a negative DG
Anabolic reactions
Making more complex molecules, requiring
energy, decreasing entropy
a positive DG
Metabolic Reaction
C6H12O6
<---> 6CO2 + 6 H2O
To the right, energy is liberated
and carbon (C) loses electrons (and protons) and is thus oxidized
To the left, energy is required, C
gains electrons (and protons) and is reduced
Capturing the energy form metabolic
reactions: