Neuron or nerve cell
Receive stimuli and transmit
action potentials
Dendrites are input
devices
(electrically
passive)
Axons are output
devices
(actively
propagate signals)
Neuron or nerve cell
Receive stimuli and transmit
action potentials
Dendrites are input
devices
(electrically
passive)
Axons are output
devices
(actively
propagate signals)
-140
-120
-100
-80
-60
-40
-20
0
0
50
100
150
E
m
[mV]
Extracellular K
+
[mM]
Memebrane potential changes
normal
resting
potential
hyper-
polarisation
repolarisation
depolarisation
repolarisation
Equilibrium potentials
If a membrane is permeable to a single ionic
species the measured membrane potential can be
calculated from the Nernst equation.
Cell Membrane
extracell intracell
K
+
= 4mM
K
+
= 140mM
Na
+
= 144mM Na
+
= 10mM
Membrane is permeable to K
+
:
mV
V
K
95
140
4
log
5
.
61
10
mV
V
Na
71
10
144
log
5
.
61
10
Membrane is permeable to Na
+
:
mV
Cl
Cl
F
RT
V
o
i
Cl
4
.
45
]
[
]
[
ln
mV
Ca
Ca
F
RT
V
i
o
Ca
0
.
40
]
[
]
[
ln
2
2
The Goldman-Hodgkin-Katz equation.
o
Cl
Cl
i
K
K
i
Na
Na
i
Cl
Cl
o
K
K
o
Na
Na
m
c
P
c
P
c
P
c
P
c
P
c
P
F
RT
ln
Resting potential strongly
depends upon the external K
+
concentration
Resting potential is
independent of external Na
+
concentration
Graded Potentials
The cell response to stimulation (Ligands binding
to receptors, changes in membrane potential,
mechanical
stimulation,
temperature
changes,
spontaneous change in permeability).
Characteristic of the dendrites
and cell body
Decremental conduction (the
further they must travel the
weaker they become)
Graded (magnitude varies from
small to large depending on
stimulus strength or frequency).
Can summate or add onto each other
Threshold all-or-none
response
Refractory period
Moves at a constant
speed
The peak potentiaol is
independent on distance
Regenerative
propagation
An action potential
a transient depolarization from the
resting membrane potential.
Pulses are quantized all the same
pulse length 1 ms
pulse strength V = 100 mV
length and strength are
determined by kinetics of ion
channels
Voltage-gated Na
+
channels
requires two gates
Inactivation gate
Activation gate
The inactivation gate breaks the
positive feedback.
Inactivation
gate will not re-
open until the
membrane
potential
returns to the
original resting
level.
The Na
+
channels inactivate during the action potential.
The membrane potential is pulled back
towards E
K
At peak of AP, there is a very large driving
force operating on potassium and a much
smaller driving force operating on sodium
Voltage-gated K
+
channels
They open slowly in response to
depolarization.
Both
channel
s
togethe
r
Action potential propagation in myelinated axons
In nerve fibres with
diameter ≥1 µm a
myelin sheath
(Schwann cell)
surrounds the axon,
broken at intervals of
≈ 1mm.
Saltatory
Conduction (Lat.
saltere = to jump).
Synaptic transmission
Synaptic cleft
– the space
separating membranes of the
two cells making the connection
(26 – 40 nm).
Action potentials jump from one neuron to
another one through
synapses
.
According to its function, a synapse may be
EXCITATORY INHIBITORY FACILITATING
According to its way of function, a synapse may be
electric synapse (excitatory only)
chemical synapse (excitatory, inhibitory or facilitating)
There are about 10
15
synapses in a human
brain.
Fast communication betwen cells
THE ELECTRIC SYNAPSE
POSTSYNAPTIC MEMBRANE
PRESYNAPTIC MEMBRANE
PROTEIN MOLECULE
(CONNEXIN) BUILT UP OF SIX UNITS
FORMS A CHANNEL
ELECTIC SYNAPSES CONDUCTS
IN BOTH DIRECTIONS
A fish has many electric synapses, these fast working units
make possible the lightning – fast whisk of their tail. A good lot
of electric synapses may be found in some parts of the human
brain, too, (hippocampus, hypothalamus, spinal marrow, etc.),
as well as in the retina.
The Ca-concentration and the pH level controls the
function of an electric synapse
POSTSYNAPTIC
ACTION
POTENTIAL
The synaptic
time lag
is
extremly short
PRESYNAPTIC
ACTION
POTENTIAL
1 ms
The signal
transmission
through a
chemical
synapse
synaptic lag (~0,5 - 2
ms)
unidirectional
transmission
Bioenergetics
Bioenergetics
How living systems use energy to maintain
internal order and power biological
processes
Convert fuels to usable energy
Store that energy in an available form
Use that energy to build molecules and materials
Use that energy to power cellular machines
E
E
nergy transduction in
nergy transduction in
biological
biological
syste
syste
m
m
s
s
(ii)
This energy is used to synthesize
ATP.
(i)
The transduction of light and redox energy to
"free energy" stored in a trans-membrane ionic
electrochemical potential (
).
An unfavourable reaction can be driven if
it is coupled to a favourable one
Consider 2 reactions:
D
C
B
A
G°´ = +20 kJ mol
-1
E
C
G°´ = - 30 kJ mol
-1
K
C
D
A
B
eq
20,000RT
e
0.0004
For reaction (1), at
equilibrium (37°C)
This reaction, being very unfavourable, produces
very little C or D from A and B.
We could drive the reaction
by adding lots of A and B since:
G G
o
'RTln
C
D
A
B
To get the reaction to happen spontaneously, we
would need RTln([C][D]/[A][B]) < – 20 kJ mol
-1
In general, a cell does not have the capacity to
manipulate reactant concentrations in this way.
At equilibrium, the products are
favoured.
For reaction
E
C
at equilibrium (37°C),
K
E
C
eq
30,000RT
e
113500
This reaction, being very favourable, would
convert almost all of C to E at equilibrium.
The combined reaction gives:
D
E
D
C
B
A
G°’ = +20 - 30 = -10 kJ mol
-1
K
E
D
A
B
eq
10,000RT
e
47
In this case
What is going on here?
The concentration of C is always low since C converts readily to E.
D needs C to make A + B (the reverse reaction)
The likelihood of D finding C is reduced
The reaction is therefore “pulled” in the forward direction
Example of coupling
The phosphorylation of glucose catalyzed by
hexokinase.
It can be considered as the sum of two reactions
pH sensitive
This coupled process is spontaneous!!
The energy released is what defines the equilibrium
for the process.
Useful „Wasted” Energy
With a G° of -16.7 kJ mole
-1
, the equilibrium
greatly favours products K
eq
= 652.
For [ATP] = [ADP] at equilibrium there will be 652
times as much G6P as Glucose
What if
all the energy of ATP
hydrolysis
is captured in the G6P
product?
For [ATP] = [ADP] K
eq
= 1
So at equilibrium [Glc] =
[G6P]
We can never have a higher concentration of G6P
than Glc.
For steps where we want a one-way
reaction (large concentrations of
product) we MUST have a large negative
G.
Summary
Coupled processes allow an energetically
favourable reaction to power a reaction that
requires energy.
If the overall G° for the total coupled process is
negative then the process will be spontaneous.
Large negative G° values
result in high equilibrium
constants
for
product
formation. This is a „driving
force” for the equilibrium.
Coupled processes allow
for chemical energy in one
molecule to be transferred
to another.
The Need for a Common Energy
Carrier
A “high-energy” bond
is unstable and readily
hydrolyzed with a negative G
o
at pH 7.
The cell maintains ATP, ADP and P
i
levels such that:
ADP
P
i
ATP
0.002
This ensure that the free energy per mole from the
hydrolysis of ATP at 37°C is close to – 30 kJ mol
-1
.
ATP at work
Consider the unfavourable reaction:
G°´ = +17 kJ mol
-1
B
A
At equilibrium
K
B
A
eq
17,000RT
e
0.0014
(T = 37°C)
Now assume that in the cell the conversion of A to
B is somehow
coupled
to ATP hydrolysis.
Strictly we should also include the ratio of
concentrations [H
+
] /[H
2
O] in the argument
of the logarithmic term but in cells these
concentrations are always constant and so
this term is incorporated into the
G°’ term.
The reaction becomes:
H
P
ADP
B
O
H
ATP
A
i
2
G G
o
'RTln
B
ADP
P
i
A
ATP
Overall, for the reaction:
(J mol
-1
)
We can break this equation down to:
G G
AB
o'
RTln
B
A
G
ATP
o'
RTln
ADP
P
i
ATP
In a cell, ATP, ADP and P
i
concentrations are
controlled
. We will allow the A <=> B reaction to
reach equilibrium, G = 0.
0G
AB
o'
RTln
B
A
eq
G
ATP
o'
RTln
ADP
P
i
ATP
Solving for ([B]/[A])
eq
we get:
RT
RT
G
G
P
ADP
ATP
A
B
o
ATP
o
AB
i
eq
000
,
30
000
,
17
exp
500
exp
'
'
B
A
eq
135,000
So by coupling the conversion of A
to B to ATP hydrolysis in the cell,
we have shifted the equilibrium
ratio of B to A from 0.0014 to
135,000, an improvement of about
10
8
!
At death, ATP synthesis stops and so does
everything else...
Lee Harvey
Oswald at
equilibrium
There are two factors at work:
(1)
G°´ for ATP hydrolysis is large and negative
ATP
ADP
P
i
(2)
the cell maintains a high ratio of
The origins of the high-energy nature of ATP
Why does ATP have a large, negative
G°´?
(1)
The
electrostatic
repulsion caused by the
negative charges on the
terminal
phosphate groups.
(2)
The
resonance
stabilization.
The greater electron
delocalization is associated
with resonance
stabilization and a positive
contribution to the
S
o
of
hydrolysis.
The solvation energies
(25
o
C) of
H
2
PO
4
–
– 76 kcal/mole
HPO
4
2–
– 299 kcal/mole
PO
4
3–
– 637 kcal/mole.
(3)
The large positive S
o
and negative G
o
of
hydrolysis of phosphate anhydride compounds arises
from
differential solvation by H
2
O of products
and reactants
.