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.



0G

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

.