phgy 311 - review
PHGY 311 - REVIEW Hosted by Vladimir Grouza VP Academic - PULS
PURPOSE • No theory • Specific application • Problem solving
© Physiology Undergraduate League of Students, 2011.
MIDTERM – 2009 issued by the Department of Physiology, McGill University, Montreal, QC.
“Neuron A has a resting potential of -70 mV and recieves a +35 mV EPSP on one of its dendrites. The decay of the EPSP along the dendrite is shown in Figure A. The threshold for an AP on Neuron A is – 49 mV.” 1. 2.
What is the length constant of the dendrite? For this synapse to evoke an AP in Neuron A, the distance between the synapse and the soma must be shorter than X um. What is X?
© Physiology Undergraduate League of Students, 2011.
• Recall: 𝑉 𝑥 =
;𝑥 𝑉𝑜 𝑒 𝜆
𝑏𝑢𝑡 𝑤𝑒𝑛 𝑥 = 𝜆 𝑉 𝜆 = 𝑉𝑜 𝑒 ;1 𝑉 𝜆 𝑖. 𝑒. ≈ 0.37 𝑉𝑜 Means we can read the space constant right off the graph. © Physiology Undergraduate League of Students, 2011.
© Physiology Undergraduate League of Students, 2011.
© Physiology Undergraduate League of Students, 2011.
• Therefore, 𝝀 = 𝟔𝟎 𝝁𝒎
Easy.
© Physiology Undergraduate League of Students, 2011.
• For this synapse to evoke an AP in Neuron A, the distance between the synapse and the soma must be shorter than X um. What is X? • We know that RMP = -70 mV and that threshold is -49 mV • Therefore, need -70 - 49; ΔV = +21 mV • Simply substitute into the previous formula!
© Physiology Undergraduate League of Students, 2011.
𝑉 𝑥 =
;𝑥 𝑉𝑜 𝑒 𝜆
+21𝑚𝑉 = +35𝑚𝑉 ×
;𝑥 𝑒𝜆
21 𝑥 = −𝜆 × ln 35 𝑥 = −60 𝑢𝑚 × −0.510826 …
𝒙 = 𝟑𝟎. 𝟔𝟓 𝝁𝒎
© Physiology Undergraduate League of Students, 2011.
“Neuron A makes a glutamatergic synapse on Neuron B. Neuron B has a RMP of -70 mV and the input resistance and capacitance of Nueron B were determined from the response to a 0.1 nA hyperpolarizing current pulse shown in Figure B. The postsynaptic receptors are a homogenous population of AMPA receptors and the single channel current records for thses receptors at different membrane potentials are shown in Figure C. Stimulating Neuron A produces an EPSP on Neuron B; the EPSCs at the A-B synapse last for 2.5 ms and can be considered as a square pulse because they activate and decay very quickly. The quantal content for the A-B synapse is 20. When recording from Neuron B in voltage clamp at -40 mV in the absence of stimulation the average mEPSC at the A-B synapse is 0.5 nA.”
WHOA.
© Physiology Undergraduate League of Students, 2011.
Motivational Interlude
© Physiology Undergraduate League of Students, 2011.
1. Draw the single channel IV curve for the AMPA receptors.
© Physiology Undergraduate League of Students, 2011.
• Just mind the scale and then take out the useful information, s.t. you don’t have to make reference to this graph in the future. • Making a table helps:
© Physiology Undergraduate League of Students, 2011.
• Now plot out the IV curve in the given space.
• Note that Erev for these AMPAr is +10 mV. This makes sense since AMPAr are permeable to monovalent cations, so you can feel good about your IV curve.
2. What are the input resistance (Rin) and membrane capacitance (Cm) for Neuron B?
© Physiology Undergraduate League of Students, 2011.
Here’s what we know 𝑉 𝑡 = 𝑉𝑓 × (1 −
;𝑡 𝑒𝜏)
𝑎𝑠 𝑡 → ∞ 𝑉 𝑡 → 𝑉𝑓
𝑤𝑒𝑟𝑒 𝑉𝑓 = 𝐼 × 𝑅𝑖𝑛 • This just means that as time passes, the current through the membrane “charges the membrane capacitor,” ceases to flow, and consequently there is no more depolarization. • Therefore, when the hyperpolarizing pulse plateaus, you can get the input resistance. You can also get the time constant of the exponential growth/decay of voltage by looking at the time course. © Physiology Undergraduate League of Students, 2011.
•
Furthermore, because the membrane properties aren’t changed during the hyperpolarizing pulse (assume no ion channels have been opened/closed) the dynamics are the same for both the growth and the decay of the hyperpolarization. Therefore, you can read the time constant off either one:
𝑤𝑒𝑛 𝑡 = 𝜏, 𝑉 𝜏 = 𝑉𝑓 (1 − 𝑒 ;1 ) 𝑉 𝜏 𝑖. 𝑒. ≈ 0.63. 𝑉𝑓 • Let’s have another look at the figure. © Physiology Undergraduate League of Students, 2011.
• We see that the hyperpolarizing pulse plateaus at -5 mV. • We also see that the it reaches 0.63 x -5 mV = -3.2 mV within 20 ms. • The exact same thing is happening on the decay phase, so we know we’re not changing membrane properties by applying this pulse.
© Physiology Undergraduate League of Students, 2011.
• From here, the rest comes with a little algebra: 𝑉𝑜 = 𝐼 × 𝑅𝑖𝑛 𝑉𝑜 𝑅𝑖𝑛 = 𝐼 5 × 10;3 𝑉 7 𝛺 = 50 𝑀𝛺 𝑅𝑖𝑛 = = 5 × 10 0.1 × 10;9 𝐴 • The membrane capacitance Cm is obtained from the time constant, which we can read from the graph. 𝜏𝑟𝑒𝑠𝑡 = 𝑅𝑖𝑛 𝐶𝑚
𝜏𝑟𝑒𝑠𝑡 𝐶𝑚 = 𝑅𝑖𝑛 20 × 10;3 𝑠 ;9 𝐹 = 0.4 𝑛𝐹 𝐶𝑚 = = 0.4 × 10 5 × 107 𝛺
Once again, really straightforward. © Physiology Undergraduate League of Students, 2011.
3. What is the conductance change associated with the EPSC at synapse A-B? ( i.e. find ΔgEPSC ).
© Physiology Undergraduate League of Students, 2011.
Here’s what we know • Firstly, we know single-channel properties of AMPAr from the IV curve we drew. This gives us two main observations: – Reversal Potential. – Direction of current at physiological potentials (important later).
• Next, look at the question again: “When recording from Neuron B in voltage clamp at -40 mV in the absence of stimulation the average mEPSC at the A-B synapse is 0.5 nA”
• This means we can find Δg for the mEPSC using our favourite formula. 𝐼 = 𝑔 × (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) © Physiology Undergraduate League of Students, 2011.
• Finally, we know that mEPSCs are brought about by the spontaneous release of a single vesicle of neurotransmitter. • If quantal content is the number of such vesicles released (on average) for a synaptic event, simply multiplying the conductance change associated with the mEPSC by the quantal content will yield the conductance change associated with a full EPSC. • Look again at the question: “The quantal content for the A-B synapse is 20.”
Well, what are we waiting for?
© Physiology Undergraduate League of Students, 2011.
• Right away, let’s get the gmEPSC: • Erev is 10 mV, and for negative potentials we have negative currents. 𝐼𝑚𝐸𝑃𝑆𝐶 = 𝑔𝑚𝐸𝑃𝑆𝐶 × (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) 𝑔𝑚𝐸𝑃𝑆𝐶
𝐼𝑚𝐸𝑃𝑆𝐶 −0.5 × 10;9 𝐴 = = = 10 𝑛𝑆 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) (−40 − 10) × 10;3 𝑉
Now just multiply by 20: 𝑔𝐸𝑃𝑆𝐶 = 𝑔𝑚𝐸𝑃𝑆𝐶 × 𝑁 = 10𝑛𝑆 × 20 = 200 𝑛𝑆 © Physiology Undergraduate League of Students, 2011.
4. What is the input resistance during the rising phase of the EPSP? • Recall that when we open ion channels, we change membrane properties for the duration of the EPSP. • Every time we increase conductance, we lower total membrane resistance. • An easy way to conceptualize this is to draw the circuit diagram.
© Physiology Undergraduate League of Students, 2011.
Simplified Cable Properties of the Neuron Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Glutamate 1
Cm Rin(rest)
𝑔𝐸𝑃𝑆𝐶
Intracellular
transmembrane current
© Physiology Undergraduate League of Students, 2011.
•
Recall from E&M that resistors in a circuit are added in parallel.
•
The resistance through all of the AMPAr channels activated during the EPSP is 1 𝑔𝐸𝑃𝑆𝐶
•
=
1 200 𝑛𝑆
= 5 × 106 𝛺
Having this, we can proceed to calculate the effective membrane resistance during the EPSP: 1 𝑅𝑡𝑜𝑡
•
=
𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
+ … + 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑜𝑡𝑒𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑡𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡!
Here’s a little shortcut for adding two parallel resistors: 𝑅𝑡𝑜𝑡 = 𝑅𝑡𝑜𝑡
•
1
𝑅𝑟𝑒𝑠𝑡 ×𝑅𝐸𝑃𝑆𝐶 𝑅𝑟𝑒𝑠𝑡 :𝑅𝐸𝑃𝑆𝐶
= (𝑅
1
𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
)
;1
50MΩ × 5MΩ = = 4.5454 … × 106 ≈ 4.55MΩ 50MΩ + 5MΩ
As we shall see, this method (not the algebraic shortcut!) can be used to add any number of parallel resistors, i.e. find the summation of any number of synaptic inputs. © Physiology Undergraduate League of Students, 2011.
5. What is the peak amplitude of the EPSP? Draw an accurate figure of the EPSP time course at the A-B synapse and specify the rising and decaying time constants. Assume Neuron B is at rest before the synapse fires.
© Physiology Undergraduate League of Students, 2011.
What does it all mean?! • Definitely confusing at first. • Isn’t current voltage-dependent?! • However, the body of the question allows you to make certain assumptions: “…the EPSCs at the A-B synapse last for 2.5 ms and can be considered as a square pulse because they activate and decay very quickly.” This means we can calculate the magnitude of the current from rest and assume that it does not change in a voltage-dependent manner as the membrane is depolarized. © Physiology Undergraduate League of Students, 2011.
• We can use our favourite formula to quantify the EPSC: 𝐼𝑒𝑝𝑠𝑐 = 𝑔𝑒𝑝𝑠𝑐 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) • Note that we use gepsc instead of 1/Rtot because the current magnitude comes from the IV curve. 𝐼𝑒𝑝𝑠𝑐 = 200𝑛𝑆(−70 − (+10))𝑚𝑉 𝐼𝑒𝑝𝑠𝑐 = -1.6 x 10-8A = 16 nA • Note that the net negative current corresponds to positive charges flowing into the cell, by convention. • Now we can find our Vf using Ohm’s Law 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡
𝑉𝑓 = 16nA × 4.55MΩ 𝑉𝑓 = 72.8 mV © Physiology Undergraduate League of Students, 2011.
• Finally, all we need now is to calculate the new time constant of the EPSP: 𝜏 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚 𝜏 = 4.55MΩ × 0.4 nF 𝜏 = 1.8 ms • Now we have all of the information we need to calculate the progression of the EPSP.
© Physiology Undergraduate League of Students, 2011.
• As before, calculate how far the EPSP will go. • Simply plug in the appropriate values: 𝑉 𝑡 = 𝑉𝑓 × (1 −
;𝑡 𝑒𝜏)
𝑉 2.5 = 72.8mV × (1 −
;2.5 𝑒 1.8 )
• Therefore, ∆𝑉 = 54.6 mV • Final absolute voltage is -70 + 54.6 = -15.4 mV
• Now we can draw our EPSP and EPSC. © Physiology Undergraduate League of Students, 2011.
−𝑡 1.8𝑚𝑠 −𝑡 20𝑚𝑠
Rising phase: 𝑉 𝑡 = 72.8 × (1 − 𝑒 Decay phase: 𝑉 𝑡 = −15.4 × (𝑒
) − 70mV (from rest)
)
© Physiology Undergraduate League of Students, 2011.
(from EPSP)
What we did • Started by calculating gEPSC from IV curve and quantal content. • Re-calculated membrane resistance Rtot during the EPSP by linear summation of parallel resistors. • Calculated EPSC knowing gEPSC and Erev of AMPAr channels, using the assumption that current is a square pulse and therefore voltageindependent, as well as information from our IV curve. • Used this information to calculate the final voltage Vf if the EPSP is allowed to progress for an infinite time. • Finally, re-calculated the membrane time constant τEPSP using Rtot and our calculated value of membrane capacitance. • Plugged in all of these values into the equation for time-dependent exponential growth of membrane voltage. • Drew a nice picture, noting time constants.
N B D. © Physiology Undergraduate League of Students, 2011.
6. Neuron B also receives an inhibitory synapse (Synapse C-B) that causes a 180 nS increase in Cl- conductance in Neuron B that lasts 20 ms. If the C-B synapse fires 10 ms before the A-B synapse, what is the peak amplitude of the EPSP at the A-B synapse?
© Physiology Undergraduate League of Students, 2011.
• This is definitely everyone’s least favourite question. • However, we can quickly show that it’s really not bad. • Let’s return to our circuit model and reexamine summation of resistances.
© Physiology Undergraduate League of Students, 2011.
Circuit model revisited Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Cm
𝟏
1
𝒈𝑰𝑷𝑺𝑪
𝑔𝐸𝑃𝑆𝐶
Intracellular
Rin(rest)
transmembrane current © Physiology Undergraduate League of Students, 2011.
Synaptic summation • What we have in this case is the inhibitory shunting effect. • As membrane conductance is increased, the magnitude of the depolarization must be decreased because total resistance is smaller for the same depolarizing current. • Let’s see the math:
© Physiology Undergraduate League of Students, 2011.
Adding Resistances - revisited 1 𝑅𝑡𝑜𝑡 1 𝑅𝑡𝑜𝑡
=
1 𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
+
1 𝑅𝐼𝑃𝑆𝐶
= 20nS + 200nS + 180 nS
𝑅𝑡𝑜𝑡 = 2.5 MΩ ≪ 4.5 MΩ 𝑏𝑒𝑓𝑜𝑟𝑒 𝑡𝑒 𝐼𝑃𝑆𝐶 Now we all know what to do from here!
© Physiology Undergraduate League of Students, 2011.
• The current IEPSP is not affected by the IPSC, so that stays the same: 𝐼𝑒𝑝𝑠𝑐 = -1.6 x 10-8A = 16 nA • The final voltage Vf needs to be adjusted for the new resistance: 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡 𝑉𝑓 = 16 nA × 2.5 MΩ 𝑉𝑓 = +40 mV • The time constant τEPSP also needs to be adjusted 𝜏 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚 𝜏 = 2.5MΩ × 0.4 nF 𝜏 = 1.0 ms • As the EPSP falls within the outlined timecourse of the IPSC (10 ms before EPSP, lasting 20 ms), we need not consider any further interactions. © Physiology Undergraduate League of Students, 2011.
• Now we can just plug these values into the same formula: ;𝑡 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 𝜏 ) ;𝑡 𝑉 𝑡 = 40 mV × (1 − 𝑒 1.0 ms ) ;2.5 ms 𝑉 2.5 = 40 mV × (1 − 𝑒 1.0 ms )
• Therefore, ∆𝑉 = +36.7 mV which is less than the prior value, as expected. • Finally, -70 mV + 36.7 mV = -33.3 mV is the peak value of the EPSP with inhibition. © Physiology Undergraduate League of Students, 2011.
Congratulations! • We stuck to the same procedure, and arrived at a reasonable result. • The trick: always mind the total membrane resistance and the time constants. The rest is just algebra. • Remember: every time there is a synaptic input, there is a conductance change in the post-synaptic membrane, leading to a decrease in total resistance. © Physiology Undergraduate League of Students, 2011.
MIDTERM – 2010 issued by the Department of Physiology, McGill University, Montreal, QC.
• • • • • • • • •
Let’s speed things up: A neuron has an RMP of – 70 mV and Cm of 30 pF; governed by resting Na+, K+ and Cl- conductances. The G-V curve shows voltage-dependent Na+ conductance (not shown here) ENa is +50 mV The K+ conductance is independent of voltage EK is – 90 mV Threshold is determined only by INa and IK. The neuron receives and excitatory glutamatergic synapse and an inhibitory GABAergic synapse. Glutamatergic synapse kinetics shown in the next figure.
© Physiology Undergraduate League of Students, 2011.
PART A
• Given time constant of rising to be 6 ms. • Can read off falling time constant as ~24 ms (can approximate). • This is the time it took for the EPSP to decay from +28 mV to +10.3 mV • i.e.
𝑉 24 = 28 mV × (1 − 𝑒
−24 ms 24 ms
) 10.3 = 28x(1-e-1).
© Physiology Undergraduate League of Students, 2011.
1. Draw the neural circuit straightforward.
© Physiology Undergraduate League of Students, 2011.
• 2. What is the threshold for the neuron? • Use the GV curve for Na+ and find the voltage at which |INa| > |IK| • For this case it’s found to be -50 mV.
• 3. What is the input resistance for the neuron at rest? • Use falling time constant read from graph • 𝑅𝑖𝑛 =
𝜏𝑑𝑜𝑤𝑛 𝐶𝑚
=
24 𝑚𝑠 30 𝑝𝐹
= 800MΩ
© Physiology Undergraduate League of Students, 2011.
4. What is the conductance change GEPSP for the glutamatergic synapse? • Find Rtot during the EPSP: • 𝑅𝑡𝑜𝑡 =
𝜏𝑢𝑝 𝐶𝑚
• As before,
•
GEPSP =
=
6 𝑚𝑠 30 𝑝𝐹
1
=
𝑅𝑡𝑜𝑡
1 200𝑀Ω
= 200MΩ 1
𝑅𝑟𝑒𝑠𝑡
−
+
1 𝑅𝐸𝑃𝑆𝐶
1 800𝑀Ω
=
1 𝑅𝑟𝑒𝑠𝑡
+ GEPSP
= 3.75 × 10;9 𝑆 = 3.75 nS
5. What is the amplitude and duration of the synaptic current? Draw the synaptic current under the EPSP assuming it’s a square pulse (as before). • From the figure, we see that the current pulse is delivered for 5 ms. • In order to figure out IEPSC, we can use Ohm’s Law, but first we need Vf for the EPSP: −𝑡 𝜏
−5 6
• 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 ) → 28 = 𝑉𝑓 × (1 − 𝑒 ) • 𝑉𝑓 = +49.5 mV
•
𝑉𝑓 = 𝐼𝐸𝑃𝑆𝐶 × 𝑅𝑡𝑜𝑡 → 𝐼𝐸𝑃𝑆𝐶 =
49.5 mV 200𝑀Ω
= 247.5 pA
© Physiology Undergraduate League of Students, 2011.
that’s 45 marks out of 100. © Physiology Undergraduate League of Students, 2011.
PART B
“The postsynaptic GABAA receptors at the inhibitory synapse are highly selective for chloride ions. Figure 3 shows the IV relationship for the inhibitory IPSC. The single channel conductance of the GABAA channel is 20 pS and one quantum of GABA at this synapse opens 20 GABA channels.”
© Physiology Undergraduate League of Students, 2011.
Note that Erev for Cl- is -70 mV.
© Physiology Undergraduate League of Students, 2011.
1. When the inhibitory synapse, what is the amplitude of the IPSP? • Recall that the neuron rests at -70 mV. • This is the same as the reversal potential for Cl-. • Therefore, there is no driving force for Cl- to cross the membrane. • amplitude of IPSP is ZERO.
© Physiology Undergraduate League of Students, 2011.
2. What is the quantal content of the inhibitory synapse? • Re-examine the question: • “The single channel conductance of the GABAA channel is 20 pS and one quantum of GABA at this synapse opens 20 GABA channels.” • So right away we can get gmIPSC by multiplying: • 𝑔𝑚𝐼𝑃𝑆𝐶 = 𝑄 ∗ 𝛾𝐼𝑃𝑆𝐶 𝑔𝑚𝐼𝑃𝑆𝐶 = 20x20pS = 400pS. • We know that quantal content, in the most apparent form, is given by: 𝐺𝐼𝑃𝑆𝐶 𝑁= 𝑔𝑚𝐼𝑃𝑆𝐶 • But we have the IV curve, so let’s go find GIPSC.
© Physiology Undergraduate League of Students, 2011.
• The conductance is given by the slope of the IV curve. • Therefore, we can pick any point on the IV curve and calculate the slope, which is constant. © Physiology Undergraduate League of Students, 2011.
• Once again, we use our favourite formula: 𝐼𝐼𝑃𝑆𝐶 = 𝐺𝐼𝑃𝑆𝐶 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) +3.5 nA = 𝐺𝐼𝑃𝑆𝐶 (0 − (−70))mV +3.5 nA 𝐺𝐼𝑃𝑆𝐶 = = 50nS +70 mV • Now just divide the conductance to get quantal content: 𝐺𝐼𝑃𝑆𝐶 50 nS 𝑁= = = 125 𝑔𝑚𝐼𝑃𝑆𝐶 0.4 nS
© Physiology Undergraduate League of Students, 2011.
3. If the excitatory glutamatergic and the inhibitory GABAergic synapses fire together and have the same time course, what is the amplitude of the synaptic potential? 4. Draw the time course of the synaptic potential and explain how you arrived at your answer for the rising and falling phases. BATTLE PLAN: • Linear addition of the parallel resistances obtain new Rtot’ • Use this to calculate new time constant obtain new τup • Remember that the Glu-mediated current remains the same, use Ohm’s law to calculate new final voltage obtain new Vf • Use this to calculate V(t) for t = 5 ms, as before. • Draw the curve!
© Physiology Undergraduate League of Students, 2011.
Circuit model revisited Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Cm
𝟏
1
𝒈𝑰𝑷𝑺𝑪
𝑔𝐸𝑃𝑆𝐶
Intracellular
Rin(rest)
transmembrane current © Physiology Undergraduate League of Students, 2011.
• Calculate new resistance: • •
1 𝑅𝑡𝑜𝑡 ′ 1 𝑅𝑡𝑜𝑡 ′
= =
1 𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
1 + 800MΩ
+
1 𝑅𝐼𝑃𝑆𝐶
3.75nS + 50 nS
• 𝑅𝑡𝑜𝑡 ′ = 18.2MΩ
© Physiology Undergraduate League of Students, 2011.
• Calculate new time constant for rising phase: ′ • 𝜏𝑢𝑝 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚
• 𝜏𝑢𝑝 = 18.2MΩ × 30 pF • 𝜏𝑢𝑝 = 0.55 ms
© Physiology Undergraduate League of Students, 2011.
• Calculate new final voltage using Ohm’s Law: • 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡 ′ • 𝑉𝑓 = 247.5 pA × 18.2MΩ • 𝑉𝑓 = 4.5 mV
© Physiology Undergraduate League of Students, 2011.
• Calculate new peak membrane potential given a time course of t = 5 ms −𝑡 𝜏
• 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 ) −5 0.55
• 𝑉 5 = 4.5 mV × (1 − 𝑒 ) • ∆𝑉 = 4.5 mV < 24 mV before IPSC, as expected.
© Physiology Undergraduate League of Students, 2011.
Now we can just draw the new EPSP!
© Physiology Undergraduate League of Students, 2011.
• that’s another 35 marks. You’re now at 80/100 for the electrophysiology section. • Study well the theory, the questions can be a little tricky. • Furthermore, the ability to apply these quantitative methods is contingent on the understanding of the theory. • The models are just an heuristic device for understanding basic electrophysiology. • Take your time. Write slowly and neatly. Make sure your answer makes sense. © Physiology Undergraduate League of Students, 2011.
Acknowledgements • Thanks to PULS for helping me set up this review. • Thanks to everyone for advice in the preparation of the review. • Thanks to Profs. Cooper and Haghighi for the reference material. • Most importantly, thank you all for coming.
GOOD LUCK. © Physiology Undergraduate League of Students, 2011.
Q&A
© Physiology Undergraduate League of Students, 2011.
PURPOSE • No theory • Specific application • Problem solving
© Physiology Undergraduate League of Students, 2011.
MIDTERM – 2009 issued by the Department of Physiology, McGill University, Montreal, QC.
“Neuron A has a resting potential of -70 mV and recieves a +35 mV EPSP on one of its dendrites. The decay of the EPSP along the dendrite is shown in Figure A. The threshold for an AP on Neuron A is – 49 mV.” 1. 2.
What is the length constant of the dendrite? For this synapse to evoke an AP in Neuron A, the distance between the synapse and the soma must be shorter than X um. What is X?
© Physiology Undergraduate League of Students, 2011.
• Recall: 𝑉 𝑥 =
;𝑥 𝑉𝑜 𝑒 𝜆
𝑏𝑢𝑡 𝑤𝑒𝑛 𝑥 = 𝜆 𝑉 𝜆 = 𝑉𝑜 𝑒 ;1 𝑉 𝜆 𝑖. 𝑒. ≈ 0.37 𝑉𝑜 Means we can read the space constant right off the graph. © Physiology Undergraduate League of Students, 2011.
© Physiology Undergraduate League of Students, 2011.
© Physiology Undergraduate League of Students, 2011.
• Therefore, 𝝀 = 𝟔𝟎 𝝁𝒎
Easy.
© Physiology Undergraduate League of Students, 2011.
• For this synapse to evoke an AP in Neuron A, the distance between the synapse and the soma must be shorter than X um. What is X? • We know that RMP = -70 mV and that threshold is -49 mV • Therefore, need -70 - 49; ΔV = +21 mV • Simply substitute into the previous formula!
© Physiology Undergraduate League of Students, 2011.
𝑉 𝑥 =
;𝑥 𝑉𝑜 𝑒 𝜆
+21𝑚𝑉 = +35𝑚𝑉 ×
;𝑥 𝑒𝜆
21 𝑥 = −𝜆 × ln 35 𝑥 = −60 𝑢𝑚 × −0.510826 …
𝒙 = 𝟑𝟎. 𝟔𝟓 𝝁𝒎
© Physiology Undergraduate League of Students, 2011.
“Neuron A makes a glutamatergic synapse on Neuron B. Neuron B has a RMP of -70 mV and the input resistance and capacitance of Nueron B were determined from the response to a 0.1 nA hyperpolarizing current pulse shown in Figure B. The postsynaptic receptors are a homogenous population of AMPA receptors and the single channel current records for thses receptors at different membrane potentials are shown in Figure C. Stimulating Neuron A produces an EPSP on Neuron B; the EPSCs at the A-B synapse last for 2.5 ms and can be considered as a square pulse because they activate and decay very quickly. The quantal content for the A-B synapse is 20. When recording from Neuron B in voltage clamp at -40 mV in the absence of stimulation the average mEPSC at the A-B synapse is 0.5 nA.”
WHOA.
© Physiology Undergraduate League of Students, 2011.
Motivational Interlude
© Physiology Undergraduate League of Students, 2011.
1. Draw the single channel IV curve for the AMPA receptors.
© Physiology Undergraduate League of Students, 2011.
• Just mind the scale and then take out the useful information, s.t. you don’t have to make reference to this graph in the future. • Making a table helps:
© Physiology Undergraduate League of Students, 2011.
• Now plot out the IV curve in the given space.
• Note that Erev for these AMPAr is +10 mV. This makes sense since AMPAr are permeable to monovalent cations, so you can feel good about your IV curve.
2. What are the input resistance (Rin) and membrane capacitance (Cm) for Neuron B?
© Physiology Undergraduate League of Students, 2011.
Here’s what we know 𝑉 𝑡 = 𝑉𝑓 × (1 −
;𝑡 𝑒𝜏)
𝑎𝑠 𝑡 → ∞ 𝑉 𝑡 → 𝑉𝑓
𝑤𝑒𝑟𝑒 𝑉𝑓 = 𝐼 × 𝑅𝑖𝑛 • This just means that as time passes, the current through the membrane “charges the membrane capacitor,” ceases to flow, and consequently there is no more depolarization. • Therefore, when the hyperpolarizing pulse plateaus, you can get the input resistance. You can also get the time constant of the exponential growth/decay of voltage by looking at the time course. © Physiology Undergraduate League of Students, 2011.
•
Furthermore, because the membrane properties aren’t changed during the hyperpolarizing pulse (assume no ion channels have been opened/closed) the dynamics are the same for both the growth and the decay of the hyperpolarization. Therefore, you can read the time constant off either one:
𝑤𝑒𝑛 𝑡 = 𝜏, 𝑉 𝜏 = 𝑉𝑓 (1 − 𝑒 ;1 ) 𝑉 𝜏 𝑖. 𝑒. ≈ 0.63. 𝑉𝑓 • Let’s have another look at the figure. © Physiology Undergraduate League of Students, 2011.
• We see that the hyperpolarizing pulse plateaus at -5 mV. • We also see that the it reaches 0.63 x -5 mV = -3.2 mV within 20 ms. • The exact same thing is happening on the decay phase, so we know we’re not changing membrane properties by applying this pulse.
© Physiology Undergraduate League of Students, 2011.
• From here, the rest comes with a little algebra: 𝑉𝑜 = 𝐼 × 𝑅𝑖𝑛 𝑉𝑜 𝑅𝑖𝑛 = 𝐼 5 × 10;3 𝑉 7 𝛺 = 50 𝑀𝛺 𝑅𝑖𝑛 = = 5 × 10 0.1 × 10;9 𝐴 • The membrane capacitance Cm is obtained from the time constant, which we can read from the graph. 𝜏𝑟𝑒𝑠𝑡 = 𝑅𝑖𝑛 𝐶𝑚
𝜏𝑟𝑒𝑠𝑡 𝐶𝑚 = 𝑅𝑖𝑛 20 × 10;3 𝑠 ;9 𝐹 = 0.4 𝑛𝐹 𝐶𝑚 = = 0.4 × 10 5 × 107 𝛺
Once again, really straightforward. © Physiology Undergraduate League of Students, 2011.
3. What is the conductance change associated with the EPSC at synapse A-B? ( i.e. find ΔgEPSC ).
© Physiology Undergraduate League of Students, 2011.
Here’s what we know • Firstly, we know single-channel properties of AMPAr from the IV curve we drew. This gives us two main observations: – Reversal Potential. – Direction of current at physiological potentials (important later).
• Next, look at the question again: “When recording from Neuron B in voltage clamp at -40 mV in the absence of stimulation the average mEPSC at the A-B synapse is 0.5 nA”
• This means we can find Δg for the mEPSC using our favourite formula. 𝐼 = 𝑔 × (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) © Physiology Undergraduate League of Students, 2011.
• Finally, we know that mEPSCs are brought about by the spontaneous release of a single vesicle of neurotransmitter. • If quantal content is the number of such vesicles released (on average) for a synaptic event, simply multiplying the conductance change associated with the mEPSC by the quantal content will yield the conductance change associated with a full EPSC. • Look again at the question: “The quantal content for the A-B synapse is 20.”
Well, what are we waiting for?
© Physiology Undergraduate League of Students, 2011.
• Right away, let’s get the gmEPSC: • Erev is 10 mV, and for negative potentials we have negative currents. 𝐼𝑚𝐸𝑃𝑆𝐶 = 𝑔𝑚𝐸𝑃𝑆𝐶 × (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) 𝑔𝑚𝐸𝑃𝑆𝐶
𝐼𝑚𝐸𝑃𝑆𝐶 −0.5 × 10;9 𝐴 = = = 10 𝑛𝑆 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) (−40 − 10) × 10;3 𝑉
Now just multiply by 20: 𝑔𝐸𝑃𝑆𝐶 = 𝑔𝑚𝐸𝑃𝑆𝐶 × 𝑁 = 10𝑛𝑆 × 20 = 200 𝑛𝑆 © Physiology Undergraduate League of Students, 2011.
4. What is the input resistance during the rising phase of the EPSP? • Recall that when we open ion channels, we change membrane properties for the duration of the EPSP. • Every time we increase conductance, we lower total membrane resistance. • An easy way to conceptualize this is to draw the circuit diagram.
© Physiology Undergraduate League of Students, 2011.
Simplified Cable Properties of the Neuron Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Glutamate 1
Cm Rin(rest)
𝑔𝐸𝑃𝑆𝐶
Intracellular
transmembrane current
© Physiology Undergraduate League of Students, 2011.
•
Recall from E&M that resistors in a circuit are added in parallel.
•
The resistance through all of the AMPAr channels activated during the EPSP is 1 𝑔𝐸𝑃𝑆𝐶
•
=
1 200 𝑛𝑆
= 5 × 106 𝛺
Having this, we can proceed to calculate the effective membrane resistance during the EPSP: 1 𝑅𝑡𝑜𝑡
•
=
𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
+ … + 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑎𝑛𝑦 𝑜𝑡𝑒𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑡𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡!
Here’s a little shortcut for adding two parallel resistors: 𝑅𝑡𝑜𝑡 = 𝑅𝑡𝑜𝑡
•
1
𝑅𝑟𝑒𝑠𝑡 ×𝑅𝐸𝑃𝑆𝐶 𝑅𝑟𝑒𝑠𝑡 :𝑅𝐸𝑃𝑆𝐶
= (𝑅
1
𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
)
;1
50MΩ × 5MΩ = = 4.5454 … × 106 ≈ 4.55MΩ 50MΩ + 5MΩ
As we shall see, this method (not the algebraic shortcut!) can be used to add any number of parallel resistors, i.e. find the summation of any number of synaptic inputs. © Physiology Undergraduate League of Students, 2011.
5. What is the peak amplitude of the EPSP? Draw an accurate figure of the EPSP time course at the A-B synapse and specify the rising and decaying time constants. Assume Neuron B is at rest before the synapse fires.
© Physiology Undergraduate League of Students, 2011.
What does it all mean?! • Definitely confusing at first. • Isn’t current voltage-dependent?! • However, the body of the question allows you to make certain assumptions: “…the EPSCs at the A-B synapse last for 2.5 ms and can be considered as a square pulse because they activate and decay very quickly.” This means we can calculate the magnitude of the current from rest and assume that it does not change in a voltage-dependent manner as the membrane is depolarized. © Physiology Undergraduate League of Students, 2011.
• We can use our favourite formula to quantify the EPSC: 𝐼𝑒𝑝𝑠𝑐 = 𝑔𝑒𝑝𝑠𝑐 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) • Note that we use gepsc instead of 1/Rtot because the current magnitude comes from the IV curve. 𝐼𝑒𝑝𝑠𝑐 = 200𝑛𝑆(−70 − (+10))𝑚𝑉 𝐼𝑒𝑝𝑠𝑐 = -1.6 x 10-8A = 16 nA • Note that the net negative current corresponds to positive charges flowing into the cell, by convention. • Now we can find our Vf using Ohm’s Law 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡
𝑉𝑓 = 16nA × 4.55MΩ 𝑉𝑓 = 72.8 mV © Physiology Undergraduate League of Students, 2011.
• Finally, all we need now is to calculate the new time constant of the EPSP: 𝜏 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚 𝜏 = 4.55MΩ × 0.4 nF 𝜏 = 1.8 ms • Now we have all of the information we need to calculate the progression of the EPSP.
© Physiology Undergraduate League of Students, 2011.
• As before, calculate how far the EPSP will go. • Simply plug in the appropriate values: 𝑉 𝑡 = 𝑉𝑓 × (1 −
;𝑡 𝑒𝜏)
𝑉 2.5 = 72.8mV × (1 −
;2.5 𝑒 1.8 )
• Therefore, ∆𝑉 = 54.6 mV • Final absolute voltage is -70 + 54.6 = -15.4 mV
• Now we can draw our EPSP and EPSC. © Physiology Undergraduate League of Students, 2011.
−𝑡 1.8𝑚𝑠 −𝑡 20𝑚𝑠
Rising phase: 𝑉 𝑡 = 72.8 × (1 − 𝑒 Decay phase: 𝑉 𝑡 = −15.4 × (𝑒
) − 70mV (from rest)
)
© Physiology Undergraduate League of Students, 2011.
(from EPSP)
What we did • Started by calculating gEPSC from IV curve and quantal content. • Re-calculated membrane resistance Rtot during the EPSP by linear summation of parallel resistors. • Calculated EPSC knowing gEPSC and Erev of AMPAr channels, using the assumption that current is a square pulse and therefore voltageindependent, as well as information from our IV curve. • Used this information to calculate the final voltage Vf if the EPSP is allowed to progress for an infinite time. • Finally, re-calculated the membrane time constant τEPSP using Rtot and our calculated value of membrane capacitance. • Plugged in all of these values into the equation for time-dependent exponential growth of membrane voltage. • Drew a nice picture, noting time constants.
N B D. © Physiology Undergraduate League of Students, 2011.
6. Neuron B also receives an inhibitory synapse (Synapse C-B) that causes a 180 nS increase in Cl- conductance in Neuron B that lasts 20 ms. If the C-B synapse fires 10 ms before the A-B synapse, what is the peak amplitude of the EPSP at the A-B synapse?
© Physiology Undergraduate League of Students, 2011.
• This is definitely everyone’s least favourite question. • However, we can quickly show that it’s really not bad. • Let’s return to our circuit model and reexamine summation of resistances.
© Physiology Undergraduate League of Students, 2011.
Circuit model revisited Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Cm
𝟏
1
𝒈𝑰𝑷𝑺𝑪
𝑔𝐸𝑃𝑆𝐶
Intracellular
Rin(rest)
transmembrane current © Physiology Undergraduate League of Students, 2011.
Synaptic summation • What we have in this case is the inhibitory shunting effect. • As membrane conductance is increased, the magnitude of the depolarization must be decreased because total resistance is smaller for the same depolarizing current. • Let’s see the math:
© Physiology Undergraduate League of Students, 2011.
Adding Resistances - revisited 1 𝑅𝑡𝑜𝑡 1 𝑅𝑡𝑜𝑡
=
1 𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
+
1 𝑅𝐼𝑃𝑆𝐶
= 20nS + 200nS + 180 nS
𝑅𝑡𝑜𝑡 = 2.5 MΩ ≪ 4.5 MΩ 𝑏𝑒𝑓𝑜𝑟𝑒 𝑡𝑒 𝐼𝑃𝑆𝐶 Now we all know what to do from here!
© Physiology Undergraduate League of Students, 2011.
• The current IEPSP is not affected by the IPSC, so that stays the same: 𝐼𝑒𝑝𝑠𝑐 = -1.6 x 10-8A = 16 nA • The final voltage Vf needs to be adjusted for the new resistance: 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡 𝑉𝑓 = 16 nA × 2.5 MΩ 𝑉𝑓 = +40 mV • The time constant τEPSP also needs to be adjusted 𝜏 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚 𝜏 = 2.5MΩ × 0.4 nF 𝜏 = 1.0 ms • As the EPSP falls within the outlined timecourse of the IPSC (10 ms before EPSP, lasting 20 ms), we need not consider any further interactions. © Physiology Undergraduate League of Students, 2011.
• Now we can just plug these values into the same formula: ;𝑡 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 𝜏 ) ;𝑡 𝑉 𝑡 = 40 mV × (1 − 𝑒 1.0 ms ) ;2.5 ms 𝑉 2.5 = 40 mV × (1 − 𝑒 1.0 ms )
• Therefore, ∆𝑉 = +36.7 mV which is less than the prior value, as expected. • Finally, -70 mV + 36.7 mV = -33.3 mV is the peak value of the EPSP with inhibition. © Physiology Undergraduate League of Students, 2011.
Congratulations! • We stuck to the same procedure, and arrived at a reasonable result. • The trick: always mind the total membrane resistance and the time constants. The rest is just algebra. • Remember: every time there is a synaptic input, there is a conductance change in the post-synaptic membrane, leading to a decrease in total resistance. © Physiology Undergraduate League of Students, 2011.
MIDTERM – 2010 issued by the Department of Physiology, McGill University, Montreal, QC.
• • • • • • • • •
Let’s speed things up: A neuron has an RMP of – 70 mV and Cm of 30 pF; governed by resting Na+, K+ and Cl- conductances. The G-V curve shows voltage-dependent Na+ conductance (not shown here) ENa is +50 mV The K+ conductance is independent of voltage EK is – 90 mV Threshold is determined only by INa and IK. The neuron receives and excitatory glutamatergic synapse and an inhibitory GABAergic synapse. Glutamatergic synapse kinetics shown in the next figure.
© Physiology Undergraduate League of Students, 2011.
PART A
• Given time constant of rising to be 6 ms. • Can read off falling time constant as ~24 ms (can approximate). • This is the time it took for the EPSP to decay from +28 mV to +10.3 mV • i.e.
𝑉 24 = 28 mV × (1 − 𝑒
−24 ms 24 ms
) 10.3 = 28x(1-e-1).
© Physiology Undergraduate League of Students, 2011.
1. Draw the neural circuit straightforward.
© Physiology Undergraduate League of Students, 2011.
• 2. What is the threshold for the neuron? • Use the GV curve for Na+ and find the voltage at which |INa| > |IK| • For this case it’s found to be -50 mV.
• 3. What is the input resistance for the neuron at rest? • Use falling time constant read from graph • 𝑅𝑖𝑛 =
𝜏𝑑𝑜𝑤𝑛 𝐶𝑚
=
24 𝑚𝑠 30 𝑝𝐹
= 800MΩ
© Physiology Undergraduate League of Students, 2011.
4. What is the conductance change GEPSP for the glutamatergic synapse? • Find Rtot during the EPSP: • 𝑅𝑡𝑜𝑡 =
𝜏𝑢𝑝 𝐶𝑚
• As before,
•
GEPSP =
=
6 𝑚𝑠 30 𝑝𝐹
1
=
𝑅𝑡𝑜𝑡
1 200𝑀Ω
= 200MΩ 1
𝑅𝑟𝑒𝑠𝑡
−
+
1 𝑅𝐸𝑃𝑆𝐶
1 800𝑀Ω
=
1 𝑅𝑟𝑒𝑠𝑡
+ GEPSP
= 3.75 × 10;9 𝑆 = 3.75 nS
5. What is the amplitude and duration of the synaptic current? Draw the synaptic current under the EPSP assuming it’s a square pulse (as before). • From the figure, we see that the current pulse is delivered for 5 ms. • In order to figure out IEPSC, we can use Ohm’s Law, but first we need Vf for the EPSP: −𝑡 𝜏
−5 6
• 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 ) → 28 = 𝑉𝑓 × (1 − 𝑒 ) • 𝑉𝑓 = +49.5 mV
•
𝑉𝑓 = 𝐼𝐸𝑃𝑆𝐶 × 𝑅𝑡𝑜𝑡 → 𝐼𝐸𝑃𝑆𝐶 =
49.5 mV 200𝑀Ω
= 247.5 pA
© Physiology Undergraduate League of Students, 2011.
that’s 45 marks out of 100. © Physiology Undergraduate League of Students, 2011.
PART B
“The postsynaptic GABAA receptors at the inhibitory synapse are highly selective for chloride ions. Figure 3 shows the IV relationship for the inhibitory IPSC. The single channel conductance of the GABAA channel is 20 pS and one quantum of GABA at this synapse opens 20 GABA channels.”
© Physiology Undergraduate League of Students, 2011.
Note that Erev for Cl- is -70 mV.
© Physiology Undergraduate League of Students, 2011.
1. When the inhibitory synapse, what is the amplitude of the IPSP? • Recall that the neuron rests at -70 mV. • This is the same as the reversal potential for Cl-. • Therefore, there is no driving force for Cl- to cross the membrane. • amplitude of IPSP is ZERO.
© Physiology Undergraduate League of Students, 2011.
2. What is the quantal content of the inhibitory synapse? • Re-examine the question: • “The single channel conductance of the GABAA channel is 20 pS and one quantum of GABA at this synapse opens 20 GABA channels.” • So right away we can get gmIPSC by multiplying: • 𝑔𝑚𝐼𝑃𝑆𝐶 = 𝑄 ∗ 𝛾𝐼𝑃𝑆𝐶 𝑔𝑚𝐼𝑃𝑆𝐶 = 20x20pS = 400pS. • We know that quantal content, in the most apparent form, is given by: 𝐺𝐼𝑃𝑆𝐶 𝑁= 𝑔𝑚𝐼𝑃𝑆𝐶 • But we have the IV curve, so let’s go find GIPSC.
© Physiology Undergraduate League of Students, 2011.
• The conductance is given by the slope of the IV curve. • Therefore, we can pick any point on the IV curve and calculate the slope, which is constant. © Physiology Undergraduate League of Students, 2011.
• Once again, we use our favourite formula: 𝐼𝐼𝑃𝑆𝐶 = 𝐺𝐼𝑃𝑆𝐶 (𝑉𝑚 − 𝐸𝑟𝑒𝑣 ) +3.5 nA = 𝐺𝐼𝑃𝑆𝐶 (0 − (−70))mV +3.5 nA 𝐺𝐼𝑃𝑆𝐶 = = 50nS +70 mV • Now just divide the conductance to get quantal content: 𝐺𝐼𝑃𝑆𝐶 50 nS 𝑁= = = 125 𝑔𝑚𝐼𝑃𝑆𝐶 0.4 nS
© Physiology Undergraduate League of Students, 2011.
3. If the excitatory glutamatergic and the inhibitory GABAergic synapses fire together and have the same time course, what is the amplitude of the synaptic potential? 4. Draw the time course of the synaptic potential and explain how you arrived at your answer for the rising and falling phases. BATTLE PLAN: • Linear addition of the parallel resistances obtain new Rtot’ • Use this to calculate new time constant obtain new τup • Remember that the Glu-mediated current remains the same, use Ohm’s law to calculate new final voltage obtain new Vf • Use this to calculate V(t) for t = 5 ms, as before. • Draw the curve!
© Physiology Undergraduate League of Students, 2011.
Circuit model revisited Constructed using Falstad Circuit Simulator v1.5n
Extracellular
Cm
𝟏
1
𝒈𝑰𝑷𝑺𝑪
𝑔𝐸𝑃𝑆𝐶
Intracellular
Rin(rest)
transmembrane current © Physiology Undergraduate League of Students, 2011.
• Calculate new resistance: • •
1 𝑅𝑡𝑜𝑡 ′ 1 𝑅𝑡𝑜𝑡 ′
= =
1 𝑅𝑟𝑒𝑠𝑡
+
1 𝑅𝐸𝑃𝑆𝐶
1 + 800MΩ
+
1 𝑅𝐼𝑃𝑆𝐶
3.75nS + 50 nS
• 𝑅𝑡𝑜𝑡 ′ = 18.2MΩ
© Physiology Undergraduate League of Students, 2011.
• Calculate new time constant for rising phase: ′ • 𝜏𝑢𝑝 = 𝑅𝑡𝑜𝑡 × 𝐶𝑚
• 𝜏𝑢𝑝 = 18.2MΩ × 30 pF • 𝜏𝑢𝑝 = 0.55 ms
© Physiology Undergraduate League of Students, 2011.
• Calculate new final voltage using Ohm’s Law: • 𝑉𝑓 = |𝐼𝐸𝑃𝑆𝐶 | × 𝑅𝑡𝑜𝑡 ′ • 𝑉𝑓 = 247.5 pA × 18.2MΩ • 𝑉𝑓 = 4.5 mV
© Physiology Undergraduate League of Students, 2011.
• Calculate new peak membrane potential given a time course of t = 5 ms −𝑡 𝜏
• 𝑉 𝑡 = 𝑉𝑓 × (1 − 𝑒 ) −5 0.55
• 𝑉 5 = 4.5 mV × (1 − 𝑒 ) • ∆𝑉 = 4.5 mV < 24 mV before IPSC, as expected.
© Physiology Undergraduate League of Students, 2011.
Now we can just draw the new EPSP!
© Physiology Undergraduate League of Students, 2011.
• that’s another 35 marks. You’re now at 80/100 for the electrophysiology section. • Study well the theory, the questions can be a little tricky. • Furthermore, the ability to apply these quantitative methods is contingent on the understanding of the theory. • The models are just an heuristic device for understanding basic electrophysiology. • Take your time. Write slowly and neatly. Make sure your answer makes sense. © Physiology Undergraduate League of Students, 2011.
Acknowledgements • Thanks to PULS for helping me set up this review. • Thanks to everyone for advice in the preparation of the review. • Thanks to Profs. Cooper and Haghighi for the reference material. • Most importantly, thank you all for coming.
GOOD LUCK. © Physiology Undergraduate League of Students, 2011.
Q&A
© Physiology Undergraduate League of Students, 2011.
