from last time…
Pressure = force/area
Two major pressures in plants.
1. The positive pressure (turgor) inside living cells and that’s required
for cell and tissue growth.
2. The negative pressure (tension) that exists in the cells of the
xylem of transpiring plants.
In general we’ll use units of pressure to express the
energy status of water, the “water potential” .
How is pressure like energy/volume?
force/area
=
energy/volume?
N
m2
multiply by unity as m/m
N x m
m2 m
N .m
m3
=
force x distance = energy
volume
volume
So, we can use units of pressure to express the
energy status of water. We’ll see that water tends
to move from areas of higher to lower energy/vol,
or pressure.
a bit more review….
The gas constant, R
Remember PV = nRT?
R, the “gas constant” makes the relationship among
P, V, n, and T work.
R shows up in lots of energy equations.
Values and units for R
8.314 J mol-1 K-1
8.314 m3 Pa mol-1 K-1
We’ll use R a lot!
energy must equal
volume x pressure
How to put some numbers to all the energy
expended in doing the work of life.
•Chemical reactions - synthesizing compounds,
degrading others
•Solute transport - maintaining concentration
differences across membranes
•Maintaining electrical potentials and moving
ions across charged membranes.
How can we understand when these processes
require energy and how much?
Bioenergetics and Free Energy
Free Energy, G, is the energy available to do work.
DG is the change or difference (D) in G during a process or
reaction.
DG equations help us understand whether a reaction:
1) yields energy and can happen spontaneously (DG < 0),
2) requires energy input to occur (DG > 0),
3) or is at equilibrium (DG = 0).
We will use DG equations for understanding bioenergetics
of chemical reactions and the transport of charged and
uncharged solutes.
the free energy equations give us
values of energy per mole, J mol-1
1. DG of a chemical reaction
General equations:

DG = DG0 + 2.3 RT log (K)
 or

DG = DG0 + RT ln (K)
DG0 is the standard free energy change, defined for standardized
conditions. It allows comparisons of DG of different reactions.
K is the equilibrium constant
K = ([product]*[product])
([reactant]*[reactant])
So, DG = DG0 + 2.3 RT log (K) can be written as:
DG = DG0 + 2.3 RT log ([product]*[product])
([reactant]*[reactant])
Example using a very important reaction
DG for ATP hydrolysis: ATP  ADP + Pi
DG0 = - 33kJ mol-1
In typical cellular conditions
DG = –50 kJ mol-1 to –65 kJ mol-1,
The reaction releases energy and can happen
spontaneously.
Compare with ATP synthesis:
ADP + PI  ATP
DG > 0, requires energy, is not spontaneous.
2. Solute transport
Transport is from C1 to C2
[C1] -------------> [C2]
The standard equation: DG = 2.3 RT log [C2]/[C1]
There are 3 possibilities:
1. [C1] < [C2], so log [C2]/[C1] > 0, so DG > 0
2. [C1] > [C2], so log [C2]/[C1] < 0, so DG < 0
3. [C1] = [C2] so log [C2]/[C1] = 0, so DG = 0
Which happens spontaneously?
2. Solute transport
DG = 2.3 RT log [C2]/[C1]
transport is from C1 to C2
[C1] > [C2] means that log [C2]/[C1] < 0
so DG < 0
This can happen spontaneously, without energy input
C1
C2
------------->
2. Solute transport
DG = 2.3 RT log [C2]/[C1]
transport is from C1 to C2
[C2] = [C1] means log[C2]/[C1] = 0
So, DG = 0, equilibrium
C1
C2
<---------->
Numerical example
[C2] = 100mM, [C1] = 10mM
37 0C
How much energy to transport a mole of C?
C1
C2
------------->
DG = 2.3 RT log[C2]/[C1]
Dimensional analysis - do the units make sense?
DG
R
Units: J mol-1 = J mol-1 K-1
J mol-1 = J mol-1
T
K
log[C2]/[C1]
mol l-1/mol l-1
Now fill in the numbers
DG = 2.3 RT log[C2]/[C1]
DG = (2.3)(8.314 J mol-1 K-1)(310 0K) log(100/10)
= 2.3 x 8.314 x 310 x 1 J mol-1
= 5928 J mol-1 or 5.928 kJ mol-1
Energy is required to move a solute “up” a
concentration gradient
C1
C2
------------->
3. DG for ion transport: ions are charged solutes
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Fig. 6.4
Biological membranes are electrically polarized, like a battery.
K+
Which ion requires the most
energy to move across
the membrane, assuming the
same concentration gradient
for all four?
NO3Ca+2
SO4-2
3. DG for ion transport
DG = zF DEm
z is charge on the ion:
K+ = +1, NO3- = -1, Ca+2 = +2, SO4-2 = -2
other molecules have not net charge
F is Faraday’s constant = 9.65 x 104 J vol-1 mol-1
Em is membrane potential, volts
Example: uptake of NO3- against
-0.15 volt potential
DG = zF DEm
DG = (-1)x9.65x104 J Volt-1 mol-1 x (-0.15Volt)
= 1.45 x 104 J mol-1
= 14.5 kJ mol-1
NO3-
4. Movement along electrical and concentration gradients
DG = zF DEm + 2.3 RT log(C2/C1)
note rearrangment as DEm = -2.3 RT/zF log(C2/C1)
R = 8.314 J mol-1 K-1
z is charge of solute
F is Faraday’s constant = 9.65 x 104 J vol-1 mol-1
Em is membrane potential, volts
Enzyme kinetics
What are enzymes?
What kinds of molecules are they made of?
What do they do to reaction rates?
How do they work?
What’s the world’s most abundant enzyme?
What conditions affect the rate of enzymecatalyzed reactions?
Vmax
reaction
rate, V
(moles of
product
per second)
Substrate concentration, S
(moles/liter)
Vmax
V
Vmax x S
V= K +S
m
1/2 Vmax
Michaelis-Menten equation
Km
S (substrate concentration)
Enzyme specificity is not perfect,
other molecules can compete for the active site.
“competitive inhibition”
Enzyme structure can be modified by other molecules,
reducing enzyme activity.
“non competitive inhibition”
No inhibitor
With competitive
inhibitor
Substrate concentration
Competitive inhibitor increases Km and does not affect Vmax,
but a higher [S] is required to reach Vmax.
Mechanisms
1. Competitive inhibitor binds at same active site as substrate,
making less enzyme available to catalyze E+S reaction.
2. Competitive inhibitor binds at another site on enzyme, causing
a conformational change in active site that reduces affinity for
the primary substrate. “allosteric inhibitor”.
Conditions affecting enzyme activity.
1. pH
Enzymes have an optimum pH at which activity is maximum,
with sharp declines in activity at lower and higher pH.
pH affects enzyme activity by altering ionization state of active site or by
affecting the 3-D conformation of the active site.
2. Temperature
Enzyme activity has an optimum temperature, with sharp declines in
activity at lower and higher pH.
Reaction rates increase with temperature because enzymes and reactants
are moving faster and have higher probability of encountering one another.
Enzyme activity decreases at temperatures high enough to cause
“denaturation”, the unfolding of protein structure and loss of proper
conformation for catalysis.