LECTURE 2 - voltage gated ion channels Flashcards
What are voltage gated ion channels?
Na+ and K+ channels
- closed at -ve Vm (membrane potential)
- opens by depolarisation (increased Vm = stimulus)
- controlled Vm = ability to study channel properties
What is voltage clamp?
A method to record the flow of ions (current) through channels
Why does Vm need to be fixed during voltage clamp?
To enable a stable measurement of channel activity, the flow of ions leads to a change of Vm
- change in number of active V-gated channels and
- change in driving force
can both lead to change in current
What is driving force?
The difference between membrane potential and the equilibrium potential for an ion
Vm-Eion
What is the voltage clamp methodology?
- Electrode measures and controls Vm (some Na channels are opened by voltage change, Na enters)
- Im (current of membrane) injects current to keep voltage constant (at -30mv)
* an influx of 2 +ve charges requires injection of 2 -ve charges - Measuring Im allows us to see how much current was required to keep voltage constant because that = how much voltage has flown through
Equal and opposite current is injected (via Im) to maintain Vm
What does a voltage clamp recording of current show?
2 depolarisations are essentially occurring
- Small - leak current only
- Larger - sodium and potassium current summed
What does the voltage clamp experiment show?
Shows that the channels are only opening in response to a change in voltage and nothing else
HOWEVER the experiment does not identify which ion is involved but direction of flow determines if current is +ve or -ve –> gives a good clue
What does the direction of ion flow mean in terms of ion movement?
Positive (up) = +ve ions leaving cell or -ve ions entering cell
OUTWARD
Negative (down) = +ve ions entering cell or -ve ions leaving cell
INWARD
Describe the voltage clamp recording of current
There are 2 depolarisations
Small: leak current only, at resting membrane potential (-60mV) there is no NET FLOW, at -50mV there is NET FLOW
Large: total ionic current, summed current of V-gated K and Na channels
How can voltage gated K+ channels be pharmacologically isolated?
TTX = tetrodotoxin
- blocks Na+ channels leaving only K+ channels remain
- K+ channels are slower opening and do not inactivate
- graph shows sigmoidal curve with IK moving outward (up) as there is K+ efflux
How can voltage gated Na+ channels be pharmacologically isolated?
TEA = tetraethylammonium
- blocks K+ channels leaving just Na+ channels
- Na+ channels open much faster (= sharp start to graph) and then inactivate (= levels off at 0)
- graph shows sharp inward movement of ions (down), then gradually increases to 0
Describe the current vs voltage graph for V-gated K+ current
(Each data point is the max. current at that Vm)
x axis = voltage
y axis = current
- slightly curved upwards line starting at -50mV
- more depolarisation = bigger current and more channels opening –> increased driving force
- driving force = the difference between voltage and equilibrium potential for an ion
- Ek ~ -80mV, moving to the right of graph (upwards) = moving away from Ek therefore there is a bigger driving force
Describe the current vs voltage graph for V-gated Na+ current
(upside down bell curve)
x axis = voltage
y axis = current
- starts decreasing at -50mv
- inwards movement of ions (therefore down on graph)
- initially there is a bigger current but it gets smaller
- above +60mV (Vm) there is an outward current because more channels are opening but you are moving closer to ENa
- ENa = line crosses axis
Why does Vm influence the size of current (current voltage graphs)?
- Increased Vm (increased depolarisation) = more channels opened as they are V-gated
2. Increased Vm = changes in driving force (Vm-Eion) For Ik (Ek =-80mV) - depolarisation moves AWAY from Ek = bigger driving force = increased Ik
For INa (ENa = +55mV) - depolarisation moves CLOSER to ENa = smaller driving force = decreased INa
Final current amplitude depends on combination of 1 and 2
What does the size of IK and INa depend on (current)?
- Number of open channels - indicated by conductance
2. Driving force
What is conductance?
A direct measure of channel activity
Ohms law: V = R x I
Conductance: g=1/R
so I = g x V
calculate conductance:
gK = Ik/ (Vm -Ek)
How can you get conductance vs time graph from current?
Divide the current by the driving force at each time point
Explain the conductance vs time graph
- gNa is positive because both Ina and (Vm-ENa) are -ve (-ve/-ve = +ve)
- increased depolarisation = more channels open and they open faster
- K+ channel opening is delayed
- Na+ channels: activate then inactivate
Explain the conductance vs voltage graph
both gK and gNa sigmoidal curve with gK finishing slightly higher than gNa
- shows how the activity of channels varies according to Vm
- x axis = vm
- y axis = conductance (g)
- calculated from current-voltage data
- maximal g = all channels open
- at +60mV, gNa is large but Ina is 0 because Vm-ENa= 0
- conductance is calculated from current and driving force but is NOT explained by either
- slope of line reflects sensitivity to Vm, NOT SPEED OF OPENING
What are A-type K+ channels (IA) and what are they important for?
- another type of V-gated K+ channels
- the previous ones were DELAYED RECTIFIER K+ CHANNELS (IK)
- important for controlling onset and frequency of firing APs
What are the properties of A-type K+ channels?
- conduct K+ efflux - REDUCE excitability of neurones
- VOLTAGE-GATED = opened by depolarisation
- activate quickly (unlike delayed rectifier)
- inactivate (like Ina, unlike DR) - proportion of channels that are inactivated depends on Vm
- mostly inactivated at resting membrane potential (unlike INa which inactivates at more depolarised Vm)
- inactivation removed by hyper polarisation before channels are opened by depolarisation
Explain the experimental recording of A-type K+ channels starting at both -40 and -80mV and ending at -5mV
Starting at -40mV:
- result only due to IK (delayed rectifier) ALONE
Starting at -80mV:
- result due to BOTH delayed rectifier and A-type as -80mV is a hyper polarisation voltage so A-type get activated
How does IA delay initiation of AP firing?
- hyperpolarisation removes the inactivation of IA
- depolarisation opens IA channels
- K+ efflux decreases cell excitability which means there is a delay in reaching threshold
