lecture 3 Flashcards
What is responsible for the falling phase of the action potential?
Ø Hodgkin and Huxley suspected that it was the delayed outward current carried by K+
Ø They injected the cell with [42K] and measured the outward current
Ø They plotted the results as outward current vs. [42K] efflux and found a high correlation, concluding that K+does carry the outward current
Structure of the Voltage-gated Na+ channel
Three subunits: α, β1,and β2
Focus on the α-subunit which is highly conserved:
-It has four transmembrane domains, each of which is made up of 6 transmembrane α-helical segments
Unlike the K+-channel, the Na+ channel, _________________ modeled by a ball-and chain mechanism, which is a part of the protein, at the mouth of the pore.
inactivates shortly after it is depolarized,
At point A, the axon is stimulated by a micro electrode (stimulus depolarized)- this causes _______________________________
a depolarization which opens voltage dependent Na channels at point A
Opening of Na+ channels cause an influx of Na+ into thee axon down______________
its concentration gradient
Accumulation of Na+ ions at A, repels _____________________
K+ from point a forward point B and C
Migration of K charges the membrane which remember is like a capaciitor at point B, _____________________
positive inside, depolarizing. this segment of. the membrane
This causes positively charged Na+ ions on the outside of the membrane to be repelled from point B- _______________
they tend to move back toward point A where the concentration is low from the earlier influx of Na+ caused by initial stimulus
when the depolarization exceeds threshold at point B, the cycle repeats as
voltage gated Na+ channels start to open in this region
Which of the following axons will show faster signal transmission?
- the larger axon with large lamda
- the larger axon with small lamda
The larger axon will since there are a fewer local circuit currents to be generated which will take less time
__________ The action potiental is initially generated at the axon hillock near thee soma and travels unidirectionally down the axon
unidirectional flow
Why is flow unidirectional
the area of the membrane just traversed by the AP is refractory. The local circuit current excites the region of membrane just ahead of, but not behind the AP
The refractory period of the axon is due to
1) Na+ iinactivatipon
2) K+ activation which hyperpolarizes the inside of the cell
_____________- the time period following thee AP during which no other AP may be generated independent of the stimulus amplitude
absolute refractory period
-________________ time period following the absolute refractory period during which an AP may bee generated by the AP is smaller than usual and the stimulus required to generate the AP is larger than usual
relative refractory period
-The AP amplitude is smaller because some sodium channels are still inactivated; the stimulus must be larger ____________________________________
because the cells are hyperpolarized due to the delayed potassium
___________________ is responsible for unidirectional flow of ApPs (also for the upper limit of the rate of firing APs in a sensory nerve fiber- we will see this later in the course
refractory period
____________________ is in fact a cell produced by Schwann cells or oligodendrocytes
myelin
The cells form a high density, high resistance __________ around the axon
lipid sheeth
There is a high density. of _____________ at the nodes
voltage dependent Na+ channels
-the. local circuit current passes through the nodes
The axon. exhibits saltatory conduction, where the APs appear to jump ___________________
from node to node
Each layer of the membrane acts as a capacitor thus layers work like capacitors in series so that the total Cm is lowered
Cm is high at the nodes, which lowers, which allows the Vm to charge to a greater value than threshold
Myelin is only found in ______________ but not all __________) axons are myelinated
vertebrates
The voltage clamp is a classic _____________________________
electrophysiological technique to measure ion currents across the cell membrane.
Under voltage clamp conditions, voltage- gated ion channels open and close as normal, but the voltage clamp apparatus compensates for the changes in the ion current to maintain__________________
a constant membrane potential.
The investigator sets a desired holding membrane potential, also referred to as holding voltage or command potential, and the voltage clamp uses negative feedback_____________
to maintain the cell at this desired holding potential.
_______________________
Movement of charges onto (and away from) capacitor plates such as the inside and outside of the membrane is referred to as a current flow “through” the capacitor. In electrophysiology it is important to be aware that such currents flow ONLY when the voltage across a capacitor is changing with respect to time (the capacitor is being “charged”).
Capacitive Current
Capacitive current (Icap) =
C * dV/dt.
The current flow onto a capacitor equals the product of the capacitance and the rate of change of the voltage. If a constant current is injected across a lipid bilayer, the steady current flow (Icap) will produce a constant rate of
change (dV/dt) of the voltage across the capacitance (C) of the membrane, as you just saw in your simulation.
This equation also shows again that whenever the voltage is constant (dV/dT=0), there is no capacitance current.
during the voltage clamp experiment
Initial Na+ curent in absent sincère there is no driving force on Na+ when Vm=Ena
Recall Ina= gna (Vm-Ena)
The membrane was clamped at 0 mv and the resulting current trace showed an early inward component and delayed outward component. During the depolarization phase of the AP, the clamp provides ___________________ which ______________________
a hyperpolarization current which maintains Vm at a constant level
By changing (Na+)o=(Na+)I and stepping 0 mv, the early inward component was eliminated and the time course of Ik alone its result. Subtracting the trace of Ik from the total current gives the course of Ina
using ohms law they were able to calculate
gna=Ina/(vm-ena)
gk=Ik/(Vm-Ek)
