Michaelmas Term - Experimental Physiology Background and Debrief Sheets

8/22/2019 Michaelmas Term - Experimental Physiology Background and Debrief Sheets

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Michaelmas Term Experimental Physiology Background and Debrief Sheets

Michaelmas 5:1

Membrane Electrophysiology

This practical is composed of two parts. In the first part, you will measure the voltages

generated by an ion concentration gradient that appear across an artificial membrane made

of ion exchange resins. This membrane is selectively permeable either to cations (positive

ions) or to anions (negative ions). Most bioelectric potentials are in the range you will be

measuring and are generated by similar ion gradients. In the second part, you will introduce

capacitors and measure the resulting time constants.

Background reading

Equilibrium potentials: from your lectures if you have had them already, or from a good text-

book. If you are unfamiliar with electrical circuits, read An Introduction to Electrical Circuits,on the HOM website.

Learning objectives

By the end of today's class you should be able to:

Understand how ion gradients can be used to generate a potential difference across amembrane, and the factors which determine the voltage achieved.

Understand how resistors and capacitors behave in an electrical circuit. Know how to measure a time constant. Understand how these electrical concepts relate to the behaviour of neurones.

Background: electrical circuits, resistors and capacitors

This class requires a basic knowledge of electrical circuits (current, voltage, resistance and

Ohms law). If you have not covered this kind of thing in school, or if you have forgotten the

details, you should look at the web-pages entited An Introduction to Electrical Circuits,

available on the Homeostasis CamTools website, under Course Resources and then

Practicals.

We shall also be looking at capacitors today, which you are unlikely to have come across

unless you studied physics to A2 level (or equivalent). You can read about capacitors on the

same web-pages, but this material will be covered by a demonstrator within the class itself.

The notes to go with this discussion are presented in the Appendix to this class (see later).


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Michaelmas 5:2

Experimental set-up

The chamber that you will be using as an artificial cell has a piece of ion exchangemembrane clamped between two compartments, each containing KCl solution. Onecompartment is referred to as the reference side, the other is the test side. Powerlab is

being used as a voltmeter today and the potential is always reported as test side relativeto reference, i.e. if you record +20 mV the test side is 20 mV more positive than thereference side. The small sub-compartments to each side have Ag/AgCl electrodes sealed inthem with agar, which will be used to measure the potential difference between thereference and test sides (they feed in to PowerLab, via a connection board which is used insubsequent experiments today). The electrodes are sensitive to chloride levels in thesolution with which they are in contact and this might affect the potential differencemeasured: the agar allows us to change the solutions in the chamber without affecting theimmediate environment of the electrodes, thus eliminating this potential problem.

Experiment 1: Comparing theoretical and measured voltages

The first experiment today will involve changing the solution on the reference side while thetest side is kept at 100 mM KCl. You will use PowerLab to measure the voltage developedacross the membrane. We shall be comparing these measured values to the theoreticalvalues calculated according to the Nernst equation. You will learn about the Nernstequation in your lecture course, and it will also be discussed in the practical class itself.

The form of the Nernst equation that you will use is the following:

([][])

where Eion is the equilibrium potential for a given ion at 20C, expressed in mV, and z is thecharge on the ion.

You will be comparing measured and theoretical results graphically, and attempting tointerpret the reasons behind any differences that you might observe.
http://labtutornav.openpopup%28%27extras/AgCl_electrodes.html')http://labtutornav.openpopup%28%27extras/AgCl_electrodes.html')


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Michaelmas 5:3

Experiment 2: time constants (theory)

Because it generates a voltage, your cell works effectively as a little battery. In this

experiment (unlike the previous one) you will have 100 mM solution in the reference side

and 10 mM solution in the test side. We expect to record a positive voltage, for reasons

which should become clear within the class itself.

This initial positive voltage, i.e. the potential across the membrane which is generated by our

battery, we shall call VT (total voltage). The movement of

the ions both within the solutions and across the membrane

is not unimpeded, and so we can speak of an internal

resistance (Ri) to ion movement (much of this being the

resistance of the artificial membrane). We can therefore

represent our membrane as a voltage source and an internal

resistance in series, as shown on the right.

In your first experiment, you will simply be connecting avoltmeter (PowerLab) across the membrane and recording

the voltage. Because there is no complete circuit, no

current flows and so there is no voltage drop across the

internal resistor. This means that we record the "total"

voltage, VT: the internal resistance, although present, is not

affecting the result and can be ignored. Note that the

voltmeter is only recording and can be assumed not to

represent part of the circuit.

Imagine that you then connect in a capacitorC, from your connection board, in the position

shown in the diagram below-right.

When you first do this, your voltmeter will no longer be

recording VT but instead will be recording the voltage

across an uncharged capacitor...which is zero! However,

you have completed a circuit and so current can now flow.

As it does so the capacitor will charge up and you will

record a rising voltage. When the capacitor is fully charged,

current will stop flowing and you will once again be

recording VT.

During this period of changing voltage, the voltage at time tafter plugging in the capacitor

will be given by:

Vt= VT(1-e-t/)

...where (Greek symbol tau, in the superscript to the right

of the equation) is the time constant, given by the product of

the capacitance Cand the internal resistance, Ri.


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Michaelmas 5:4

Experiment 2: Measuring time constants

In this experiment, you will connect a capacitor from our connection-board into our circuit, asdescribed previously, and you will measure the resulting time constant (). The timeconstant of a real nerve cell membrane will affect the velocity of action potential conduction:

see the Appendix for more on what a capacitor does, and its biological relevance.

You will use two capacitors in this experiment, labelled A (10 F) and B (22 F) on theconnection-board. Having plugged in the discharged capacitor, you will be measuringchanges in voltage using PowerLab.

As explained previously, the initial voltage, before the

capacitor is plugged in, equals the total voltage VTdeveloped across the membrane. Having connected in adischarged capacitor, the measured voltage dropsinstantly to zero. Then, as charge builds up on thecapacitor, you should see a slow return to the originalvoltage. During this recovery period the voltage at time tisgiven by Vt = VT(1-e

-t/), where the time constant is givenby the product of the capacitance C and the internalresistance, Ri.

When tequals , Vt = VT(1- e-1) = VT(1- 1/e), where e is the constant 2.718.so Vt = 0.63VT.

In other words, we can work out from a trace like the one shown above, by noting the time

at which Vhas reached 63% of its original value. Having measured in this way, we shall beusing the known values of the capacitors A and B and the equation = RiCto calculate thevalue of the internal resistance of the artificial membrane, Ri.


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Michaelmas 5:5

Appendix: An introduction to capacitance

1) Capacitance is an important property of cell membranes.To understand this phenomenon, we shall look first at thekind of capacitor found in an electrical circuit. This kind ofcapacitor is made from two sheets of electrical conductor,

separated by a thin layer of insulator.

2) A current, I, is a flow of charged particles. In an electrical

circuit, these particles are electrons (in a solution, they areions). If we send a current, charge (electrons) builds up onone plate of our capacitor, but the electrons cannot crossto the other plate because of the layer of insulator. Theleft-hand plate becomes negatively charged. (Note: theconvention in circuits is to draw current as if it were theflow ofpositive charge, but for clarity it is drawn here as ifnegative charge were flowing as is actually the case,since electrons are negatively charged).

3) As negative charge builds up on the left plate of thecapacitor, it repels electrons from the right side (of course,this actually happens at the same time as step 2, ratherthan in two stages). Although no electron actually crossesthe insulator layer, the flow of electrons onto the left plateand off of the right plate results in what you mightconsider to be a virtual current (a capacitative current)across the capacitor.

4) When the capacitor is fully charged, no more electronscan flow onto it, and so no more can be repelled.Therefore, there is no more current flowing across thecapacitor. The amount of charge that it carries is Qcoulombs. (Note that the maximum amount of charge thatit can take depends upon the voltage driving current ontoit in the first place: it can store more charge with a bigger

driving voltage).

5) If you connect a voltmeter across the charged capacitor(PowerLab is essentially a voltmeter), you can measure avoltage across it, calculated as V=Q/C. The bigger thecapacitance, C (farads), the more charge can be held for agiven voltage. You would get a bigger capacitance if yourcapacitor plates had a larger surface area, or if theplates were closer together(thinner insulating layer).

KEY POINT 1: A CAPACITOR STORES CHARGE.


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Michaelmas 5:6

6) At this point, lets go back to something more basic. If wehad a resistor (R) in a circuit, and we put some current (I)

through it, we would immediately register a voltage acrossthat resistor.

7) As soon as you switch on the current, you would instantlyrecord the voltage V, and this would stay constant as longas you maintained the current. The voltage that you would

measure is given by V = IR, Ohms Law. When youswitch off the current, the voltage falls instantly to 0.

8) Now, lets stick a capacitor into the circuit, in parallel withthe resistor. Initially, it is easier for electrons to

accumulate on the capacitor than to flow through theresistor. However, as electrons continue to pile onto thecapacitor, the developing charge makes it increasinglydifficult for more to join in. The rate of chargeaccumulation on the capacitor is therefore related to theamount of charge already there...which suggests anexponential relationship.

9) If we were to measure the voltage while the capacitor ischarging, we would find a relationship given by theequation:

)e(1VV t/maxt

where Vmax is the final (maximum) voltage when the

capacitor is fully charged (= Q/C, = IR), t is time and is

the time constant. The time constant is equal to RC

(resistance capacitance). The equation tells us that

when t = infinite, Vt = Vmax, i.e. the capacitor is fully

charged. IC represents the current that passes through

the capacitor.

10) If we look carefully at the current trace, we can see that

current only flows through the capacitor while thecapacitor is charging. When the capacitor is fully charged

(and the voltage across it has reached its peak value), no

more charge can accumulate on it and IC drops to zero.

All the current now flows through the resistor instead.

11) If we then turn off our current source, the capacitor will

discharge. Electrons will flow through the wire from the

negatively-charged plate to the positively-charged plate

via the resistor, until the charge on the capacitor is zero

once more. Of course, the current flowing through the

capacitor while discharging is flowing in the opposite

direction to when it was charging.


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Michaelmas 5:7

12) Since the current flowing during the discharge of the

capacitor is in the opposite direction to when the capacitor

was charging, it is represented as being negative, which is

downwards on the trace to the right. The voltage during

discharge decays exponentially, following the relationship

t/

maxt eVV

When t = infinite, the voltage has dropped to zero.

So a capacitor introduces a time-course to any change in

voltage, whether it is charging or discharging. This would

not be so if you only had a resistor in the circuit, in which

case the voltage would change instantly (see 7).

KEY POINT 2: A CAPACITOR TAKES TIME TO CHARGE

OR DISCHARGE, AND THEREFORE SLOWS ANYCHANGE IN VOLTAGE.

13) So what is the point to all this? Well, a nerve cell (or indeedany cell) is surrounded by a plasma membrane, made ofphospholipids. We have two electrically-conducting regions(the cytoplasm and the extracellular fluid, both electrolytesolutions), separated by a thin layer of insulator (the plasmamembrane). The membrane therefore acts as a capacitor!For this reason, we portray charge as lined up on eitherside of the membrane, just as charge builds up on eachplate of a capacitor.

14) Because of the capacitance of the membrane, any changein voltage across the membrane, as for example thedepolarisation associated with an action potential, will taketime to occur: the membrane has a time constant, . Thiswill affect the velocity at which the action potential ispropagated.

KEY POINT 3: THE CAPACITANCE OF AN EXCITABLE

CELL MEMBRANE WILL AFFECT THE VELOCITY OF

ACTION POTENTIAL CONDUCTION.

15) If you think of myelin as effectively increasing the thicknessof the cell membrane, you would expect this to decreaseits capacitance (see 5). Although this is a grossoversimplification of what is really happening, myelinationdoes indeed have the effect of reducing membranecapacitance. However, myelination also increases themembrane resistance, so overall the time constant (= RC)

might not actually change much. Myelination increasesconduction velocity mainly because the increasedmembrane resistance increases the length constant, .The length constant is another story...


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Michaelmas 5:8

A data-handling question to try at home

The following question was set in the Tripos exam some years ago, with a time allocation of

30 minutes. It is an interesting and instructive question, which is well worth your while

attempting. You may want to ask your supervisor to go over it with you.

The diagram shows a cell in Ringer at steady-state. Na+, K+ and Cl+ are the only significant

diffusible ions. The membrane potential is -35 mV, inside negative, and the temperature is20C.

1. Using the Nernst equation, calculate the equilibrium potentials for Na+ and K+ ions.

2. Calculate the internal Cl- concentration, assuming that this ion is passively

distributed.

3. Assuming electrical neutrality inside the cell, determine the concentration of the

indiffusible ion A-.

4. Sugar is added to the Ringer and the cells shrink. The volume of the cells ismeasured optically, and for an average cell the results at different times are:

(a) Plot the data and determine the shrinkage.

(b) Estimate the final sugar concentration in mM.

Data-handling exercises of this kind are to be found at the end of some of the practicals;

answers do notneed to be submitted. There are answers to these questions on the

CamTools website!

Na+ 100 mM Na+ 15 mM

K+ 2 mM K+ 90 mM

Cl- 80 mM Cl- ?

HCO3- 22 mM A- ?

Time(sec)

0.5 1.0 2.0 4.0 6.0 10.0 15.0 20.0

Volume(m3)

4060 3930 3690 3400 3190 2980 2860 2810

CellRinger


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Michaelmas 5:9

Membrane Electrophysiology - debrief sheet

PowerLab was used in this experiment as a voltmeter, measuring the potential difference

between two electrodes. The value that you record is (always) the potential of the test

electrode relative to the reference electrode. If you recorded a negative voltage at most test

concentrations, that means that the test side of the membrane is negative relative to the

reference side, i.e. the test side is at a negative potential, the reference at a positive

potential. N.B. When reporting potentials, you should always make it clear whether they are

positive or negative by putting a + sign by any positive values.

Except when the two solutions were identical, the test side contained a more concentratedconcentration of KCl than the reference side (which you varied...it was only used as an

electricalreference). This means that the concentration gradient was from the test side to

the reference side. Ions would thus tend to flow from test side to reference side across the

membrane, and this movement of charge results in the potential difference across the

membrane which you record. If the potential difference becomes so great that the electrical

gradient is exactly equal and opposite to the concentration gradient, there will be no further

net flux of the ion in question. A dynamic equlibrium has been reached, and the measured

voltage represents the equilibrium potential for the ion that is moving.

Given that ions initially move from test to reference side, and given that you should have

measured a negative voltage (i.e. test side is negative relative to reference side), it must

have been positive ions that were moving across the membrane. The membrane was

therefore selectively permeable to cations, in this case potassium.

The advantages of a semi-log plot

In this experiment, graphs were plotted with a normal y-axis but a logarithmic x-axis (a

semi-log plot). There are many different reasons why one might use a logarithmic scale. In

this case, it relates to the nature of the Nernst equation:

[][]


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Michaelmas 5:10

Given the laws of logarithms, this can be re-written as:

[][]

If we were to plot voltage (EK, in the case of our theoretical relationship) on the y-axis and

[K+]ref on the x-axis, we would expect a curve because of the log10 function in the equation

above. It is rather difficult to compare theoretical and measured curves visually.

If, however, we plot voltage on the y-axis against log10[K+]ref on the x-axis, the equation

above would give us a straight line, of gradient +58 and intercept -58log10[K+]test. It is much

easier to compare our measured data points with such a line, so as to see whether our

relationship is also a simple, logarithmic one.

Rather than calculating out the logarithms of the [K+]refvalues used, we can alternatively plot

the raw [K+]refvalues on a logarithmic axis and get exactly the same-shaped graph which

is easier to read.


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Michaelmas 5:11

Why was the measured voltage smaller than predicted?

The voltages that most people measured were consistently smaller (closer to zero) than the

equilibrium potential values predicted from the Nernst equation, deviating most at low values

of [K+]ref. A curve is usually a better fit to the measured data than a straight line.

Some people suggested that the reason for this might be that the laboratory temperature

was not exactly 20C, which is assumed in the 58 multiplier within the form of the Nernst

equation used. However, the full version of the Nernst equation uses temperature in Kelvins,

and the small difference between 293K and whatever the lab temperature was (maybe

297K) will result in a negligibly small difference and a straight line of different slope rather

than a curve.

Other people suggested that the solutions used might have become contaminated over the

course of the experiment by poor washing technique, or perhaps the 100 mM solution was

losing ions and so its concentration progressively changed. These represent testable

hypotheses: you would be expected to see for yourself if you are right by trying lots of

washing and fresh solutions! It turns out that this makes very little difference to the results.

The most likely reason for the discrepancy between measured and theoretical lines is that

the membrane is partially permeable to chloride as well as potassium. A small chloride

permeability results in a deviation from the predicted line, this deviation being increasingly

large at low test concentrations. In this case, the voltage measured represents the combined

equilibrium potential for potassium and chloride taken together, and will depend on their

relative permeabilities. The voltage generated would always be less negative than EK

because of the (smaller number of) anions moving in the same direction across the

membrane, i.e. towards the reference side. If the membrane is permeable to more than just

one ion species, the Goldman equation rather than the Nernst equation would provide a

more accurate model of what is happening (see graph below).

From data collected in the 2012 class, neglecting obvious outliers, the mean relative Cl-

permeability of our membranes was 0.04 (n=119, S.D. 0.02), where K+ permeability is 1.


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Michaelmas 5:12


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Michaelmas 5:13

The internal resistance of the membrane

Because it generates a voltage, your cell is effectively a little battery. What we recorded in

the first experiment we shall call VT (total voltage), the equilibrium potential across the

membrane which is generated by this battery.

The movement of the ions both within the solutions and across the membrane is not

unimpeded, and so we can speak of an internal resistance (Ri) to ion movement (much of

this being the resistance of the artificial membrane). In fact, no battery is a pure voltage

source (however they might be represented in GCSE physics!): even a normal, flashlight

battery has an internal resistance resulting

from the internal movement of charge from one

terminal to the other when current is flowing. Areal battery can be thought of as a voltage

source and an internal resistance in series, as

represented on the right. We shall estimate its value in the next experiment.

Capacitors and time constants

The first thing that you did in the capacitor experiment is to change your solutions such that

the reference side now contains 100 mM KCl, and the test side 10 mM. This should have

generated apositive voltage (make sure you understand why!).

Firstly, you measured that total positive voltage, VT. Then you connected in a discharged

capacitor, C. The voltage at the instant when you plug in the capacitor is zero, as there is no

potential difference across an uncharged capacitor: V = Q/C, where Q is charge, initially

zero, and C is capacitance, a constant. Due to the effects of sampling, PowerLab may not

have recorded the instant where zero was hit, but your trace should get close, as shown.Because plugging in the capacitor completes a circuit, current starts to flow once again and

the capacitor slowly becomes charged. As it charges, the voltage Vthat you measure slowly

increases back to what it was before, reaching VT when the capacitor is fully charged

Ri


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Michaelmas 5:14

(whereupon no more current can flow).

The point of this experiment was to demonstrate that the capacitor introduces a time-course

to a change in voltage it takes a while for the voltage to build back up again. The time that

this takes relates to the time constant, (= RiC). From the equation for how voltage builds

up over time (see background sheet), at time = , the voltage is 63% of the final (= initial)

voltage, VT. A larger time constant (caused, for example, by a bigger capacitance C) means

that it takes longer for the voltage to build back up.

The internal resistance values

The next stage of the experiment was to use your measured time constants, together with

the known values of capacitors A and B (10 and 22 F), to calculate the value of the internal

resistance of the membrane, Ri. This is easily done using the equation = RiC.

From 2012 class data, ignoring obvious outliers and using only paired data where two Ri

values were calculated for the same membrane, the mean value for Ri determined using

capacitor A was 6.8 k (n=107, S.D. 2.9 k) and using capacitor B it was 7.6 k (n=107,

S.D. 4.8 k). Because the data are not normally distributed (so a paired t-test would not be

appropriate), a Wilcoxons signed-ranks test was used on the paired data to see if the

difference between values ofRi determined using capacitors A and B was significant: it was

(p < 0.001). This is peculiar, because we would expect the membrane resistance to be the

same in both cases. A hypothesis which might explain this is presented below.

Why did the two Ri values differ? (this part represents extra material, not examinable!)

We found in this experiment that the membrane potential was notequal to the equilibrium potential for potassium,

EK. Therefore, by definition, potassium was not at equilibrium and it would thus experience an electrochemical

driving force which would cause it to continue moving across the membrane. Although the number of ions that

move is small, over time the concentrations on each side of the membrane, especially immediately adjacent to

the membrane, would be expected to become more equal (in a real cell the sodium-potassium ATPase pump

makes sure that this does not happen, maintaining concentrations constant, but there was no pump in our

artificial membrane). If reference and test concentrations are becoming more similar over time, the membrane

voltage VT will drift towards zero, and indeed many students did notice a slow drift in that direction in their first

experiment.

IfVT is declining slowly over time, this would mean that our capacitor would charge up to a value less than theoriginal VT. That will flatten the shape of the curve somewhat, such that the time constant that you measure is

greater than it should be. With a larger capacitance and consequently a slower return to VT, this effect would be


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Michaelmas 5:15

more pronounced. This leads us to predict that measured values and consequently calculated Ri values using

capacitor B would be larger than those found using capacitor A, as was indeed the case.

Conclusion

Why do we need to know so much physics? You know the answer to this by now. A nerve

cell membrane (or any membrane) represents a thin layer of insulator separating two

conducting solutions, and as such it has a capacitance. There are ion channels in the

membrane, representing a resistance. The membrane therefore has a time constant, and the

value of this time constant (proportional to the size of the capacitance) will affect how quicklyany voltage change will occur and thus the velocity of action potential conduction. If you

want to know what effect a giant axon has, or what the effect of a demyelinating disease like

multiple sclerosis might be, you need to understand (among other things) what effect, if any,

capacitance might have on the time constant of the membrane.

Although many people tend to separate biology and physics in their mind, perhaps

because they are often taught as distinct subjects, all sciences inter-relate and it is very

important that we de-compartmentalize them. We have to understand something of all

sciences, if we want to address real questions in the real world.

MJM, 25/10/12


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Michaelmas 5:16

The Frog Nerve-Muscle Preparation

Today you will record the electrical responses both of the gastrocnemius muscle of an edible

frog, Rana esculenta, and of the sciatic nerve that innervates it, making both extracellularrecordings of action potentials and examining neuromuscular transmission. PLEASEBRING A LAB COAT TO THIS CLASS!

Background reading

Nerve conduction, neuromuscular transmission, muscle contraction.

Learning objectives

By the end of today's class you will be able to: Perform delicate surgery necessary to remove a living nerve and muscle from an

organism. Make extracellular recordings from a live nerve, and understand the basis of its

excitation. Understand the differences between extracellular and intracellular recordings, and the

nature of a compound action potential. Calculate nerve conduction velocity. Demonstrate properties of neuromuscular transmission, and muscle contraction.

Background: nerves and muscles

The sciatic nerve contains many thousands of

individual nerve fibres, and the gastrocnemius

muscle many thousands of individual muscle fibres.

Each individual nerve fibre innervates very many

muscle fibres, the neuromuscular junctions (end-

plates) generally being located near the centre of

each muscle fibre. The whole structure - the nerve

fibre and all of its attached muscle fibres - is

called a motor unit. Each motor unit obeys the "all-

or-none law" - when the nerve fibre is stimulated, the

whole unit produces either a full sized twitch or

nothing at all.

In the time interval between the nerve being

stimulated and the muscle starting to contract, three

events occur:

1) The action potential is set up in the nerve fibre and then travels down to the synapse.

2) The nerve activates the end plate and sets up an action potential in the muscle fibre.

3) The action potential propagates over the whole muscle fibre and makes it contract.


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Michaelmas 5:17

The origin of a diphasic, extracellular recording

The action potentials which you will be recording today in the sciatic nerve are compound

action potentials (the result of activity in lots of nerve fibres). They are recorded

extracellularly, and they are diphasic in nature. The following explains the nature of a

diphasic, extracellular recording. Imagine that an action potential is sweeping from left toright along an axon. The voltage recorded is the potential difference of recording electrode B

relative to A (i.e. B is the test electrode and A is the reference electrode).

In order to get a monophasic extracellular recording, electrode B would have to be placed

on the moist cotton attached to the nerve on the right. Alternatively, the nerve could be

crushed between A and B so that the action potential could not pass. In either case, the

potential at B would remain constant throughout. What shape would the recording be then?

What shape would it be if electrodes A and B were swapped around?

In fact, the considerable length of the active zone of the nerve fibre relative to the distance

between the electrodes means that the above represents an over-simplification. A more

accurate explanation can be found within your LabTutor experiment, as a pop-up at the

bottom of the Background page.


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Michaelmas 5:18

The preliminary leg dissection (done for you already)

The following, included for your interest, explains how the frog has been dissected inpreparation for the class. You do not need to perform these steps yourself!

Step 1. The skin wasstripped from the hind part of theanimal by pulling it firmlydownwards, rather like a pair oftights. From now on it is vital tokeep all exposed tissue wet at alltimes, using the frog Ringersolution provided. (Ringersolution is similar in ioniccomposition to frog extracellularfluid).

Step 2. So as to obtain a longlength, the whole of the sciaticnerve was dissected together withits spinal roots. Taking a dorsalapproach,, the spinal roots of thenerve were exposed by movingthe urostyle (the bonyprolongation of the vertebralcolumn) and cutting through themuscle and fascia that connectthe urostyle to the side membersof the pelvic girdle, on the left andthe right.

Step 3. Using scissors, the urostylewas cut through at its base andremoved. The three roots of thesciatic nerve (two thick and onethin) were identified. A thread waspassed under all three, as close aspossible to the vertebral column,and the nerve was tied off andsevered above the tie.

The leg, with sciatic nerve attached and a cotton thread tied to this for easy handling, is thengiven to you for the next stages of the dissection.

The leg dissection performed in the class

Step 4. The fixing pin is pushedright through the top edge of theknee joint, near the thigh. A 15 cmcotton thread is tied tightlyaroundthe Achilles tendon, just above thesesamoid bone.

Step 5. The tendon is cut distal tothe sesamoid bone, and themuscle is freed up to itsattachment at the knee.

Step 6. The leg bones are cutthrough just below the knee jointso as to leave the gastrocnemiusmuscle attached only at the knee.


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Michaelmas 5:19

Step 7. The nerve must now be freed from the thigh tissueswhich surround it. The two large thigh muscles are grasped andpulled apart, so that they separate along their natural lines ofcleavage. This reveals the sciatic nerve (yellowish, floppy andslightly corrugated: do not confuse it with tendons, which are

whiter, straighter, and show fine longitudinal striations, or withthe femur!). The tissues are teased apart with forceps until thenerve lies free in the space between the muscles.

Starting from the hip end of the preparation, any tiny nervebranches are snipped away so as to free the main sciatic nervetrunk, to within about 5 mm of the knee. Dissecting any furtherwill weaken the nerve and make it more vulnerable tomechanical damage and drying out.

Step 8. Having freed the nerve almost to the knee,the stronger scissors are used to cut through thethigh just above the knee joint. The gastrocnemiusmuscle, attached to the knee, is now isolated,

together with its nerve supply.

Step 9. The nerve-muscle preparation is now readyfor mounting in the organ chamber. The pin throughthe knee-joint will be used this to anchor the musclein position within the organ chamber.

Setting up the apparatus

You will use Input 1 of the PowerLab 26T to record contraction of the muscle via themechanical transducer, and Input 2 to record electrical activity either from the nerve or fromthe muscle (you will change where you plug in the pair of recording electrodes accordingly,over the course of the experiment). The input from the recording electrodes actually goesinto PowerLab via a bridge amplifier, which increases sensitivity and allows you to zero thetrace.

The pin is placed in the white pin-holder in themiddle of the chamber, to anchor the muscle inplace. The nerve drapes loosely over the set ofeight electrodes below the pin-holder. Thecotton loop attached to the muscle is tied tothe notch on the short arm of the mechanicaltransducer.

The preparation must be irrigated from time totime with Ringer solution, to prevent dryingout. Ringer solution is similar in ionic

composition to frog extracellular fluid.However, no blobs of fluid must be left on the


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Michaelmas 5:20

electrodes, as these may short-circuit the potentials you are trying to record.

In these experiments you will initially be stimulating the nerve with the lowest pair ofelectrodes, separated from the recording electrodes by an earth. The electrodes are colour-coded: anodes (positive electrodes) are red, cathodes (negative electrodes) are black and

the earth electrode is green. The vertical order indicated in the diagrams is important, andyou should understand why. However, as long as this vertical order is right, it does notmatter which of the eight electrode sockets are used; there is no difference between rightand left positions.

Experiment 1: Recording a nerve action potential

You will initially set the stimulus amplitude to a relatively low value (100 mV), and the nerveis stimulated just once to record an action potential. You may see a stimulus artefact, anelectrical response that precedes the action potential, coinciding in time with the stimulus. It

is caused by direct, passive conduction of the stimulating current to the recording electrodes,travelling within the salty Ringer solution which bathes the nerve.

Experiment 2: Effect of increasing stimulus amplitude

In this experiment, you will be stepping up the stimulus amplitude from a low value where noaction potential is seen (around 10 mV), in small steps, generating traces each time. Abovesome threshold value an action potential should appear as a diphasic deflection of the trace(see Appendix).

Experiment 3: Recording muscle action potentials

The recording electrodes are moved as shown, so as to record electrically from the musclerather than the nerve.

You will be comparing the different latencies(i.e. the times, relative to the time ofstimulation) of the electrical and mechanicalactivities of the muscle, and interpreting thereasons for the differences.


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Michaelmas 5:21

Experiment 4: Conduction velocity

In this experiment, you will compare the muscle action potentials obtained when stimulatingthe nerve in a low position, far from the muscle, to those obtained when the stimulatingelectrodes are moved closer to the muscle. By dividing the difference in distance by the

difference in latency, you can calculate conduction velocity of the sciatic nerve. Using this,together with the results from experiment 3, you will be able to estimate the delay at theneuromuscular junction.

Properties of muscle contraction

These extra experiments can be performed, if youhave time, to make maximum use of yourpreparation.

First, move your two stimulating electrodes to thepositions where they touch the muscle itself; themuscle is thus directly stimulated rather than beingstimulated via the nerve.

Experiment 5: How does contraction depend on the size of the stimulus?

In this experiment you will stimulate the muscle with single shocks of varying voltage (fromzero) and measure the amplitude of the resulting contraction. As you increase stimulus

amplitude, you will eventually find a threshold at which you just get a response. As youincrease further, there comes a point where further increases have no extra effect: thestimulus is then said to be supra-maximal.

You will be looking at the response around the threshold value, to see whether you get alinear increase in contraction with an increase in voltage, or a more complicated relationship.

Experiment 6: Summation and refractoriness

In real life, single action potentials rarely occur inisolation: information is carried in the form of trains ofaction potentials, in which the frequency at any momentconveys the message. In muscle, different frequenciesresult in different degrees of tonic contraction. Thereason is that because the twitch lasts so long, asecond twitch generated before the first twitch has dieddown will result in summation of the two. If thefrequency is very high, the result will be a gradual build-up of contraction or tension to a more-or-less steadylevel.


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Michaelmas 5:22

You will investigate this process by looking at the effects of pairs of shocks, with differentintervals between them ranging from 150 msec down to 2 msec. You should get a picturelike the one shown.

Over much of this range, you should observe linear summation, i.e. the total response is

nearly (but not quite exactly) the sum of what you would have got with each stimulusseparately, i.e they are being added on linearly.

As you reduce the time interval between the shocks, something odd happens: there comes apoint where the additional effect of the second shock starts to get smaller. The reason is thatthe second stimulus is being applied before the end of the refractory period after the firstaction potential. The refractory period will vary among the different muscle fibres, but if thetime between stimuli is very short, all fibres will be refractory and so there will be only onetwitch. The muscle is thus incapable of responding to a shock delivered too soon after it hasbeen stimulated: it needs time to recover. You should therefore see that refractoriness sets alimit on the frequency with which a muscle can generate twitches.

Experiment 7: Repetitive stimulation

The smooth contraction produced by

repetitive stimulation of a muscle is

called tetany (This differs from the

bumpy record produced by

summation). You will demonstrate the

approach to tetany by presenting the

muscle first with 5 stimulus shocks, allof the same amplitude and 80 msec

apart, then 10 shocks with 40 msec

separation, then 20 shocks with 20

msec separation. The response

should resemble the one shown here.


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Michaelmas 5:23

Some questions and calculations to try at home

1. Why is it the cathode that stimulates a muscle or nerve? (Workout what is needed to depolar isethe membrane).

The outside of a resting nerve (or muscle) fibre is at a positive

potential relative to the inside. In order to set off an action potential,

the membrane must be depolarised, i.e. the difference in potential

between inside and outside must be reduced. This will be achieved at

the position where the negative stimulating electrode the cathode

is on the outside of the fibre.

2. When the stimulating electrodes are arranged on a nerve

supplying a muscle as on the right, the action potential set up by the

cathode can sometimes fail to propagate past the anode to the

muscle (a condition known as anode block). Why?

Because under these circumstances, the membrane under the anode will

be hyperpolarised (i.e. the converse of what was described above). This

can make it difficult for an arriving action potential to depolarise this

section of membrane.

3. Suppose that in the arrangement above the whole bath were filled

with saline, and the anode were removed from the nerve and just left

dangling in the surrounding fluid. Would an action potential still be

set up? If so, in which direction would it propagate?

In principle, action potentials would be set up at the cathode and would propagate in both

directions. The current flow pattern round two electrodes in a conducting medium like saline

resembles magnetic field lines: some leave the anode in a direction almost diametrically

opposite to the cathode, and all parts of the medium experience some electrical field. This

means that in practice, the voltage required to initiate the action potentials would be relatively

high, because of the alternative pathways for current flow between anode and cathode which

do not involve the high-resistance nerve.

4. What events contribute to the latent period between musclestimulation and muscle

contraction?

Propagation of the action potential across the surface of each muscle fibre and down the T

tubules; opening of the Ca2+

channels in the sarcoplasmic reticulum; diffusion of Ca2+

to its

sites of action; activation of the contractile mechanism; cross-bridge cycling; stretching of theseries elastic component, leading finally to the generation of measurable tension.

5. The nerve membrane potential changes by about 100 mV during an action potential. How

much charge must flow per unit area to accomplish this? (Take the membrane

capacitance to be 1 F cm-2

).


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Michaelmas 5:24

V=Q/C (voltage = charge/capacitance). Rearranging, Q=VC. We will need to use SI units of

volts and Farads m-2

. 1 F cm-2

is equal to 110-6

F cm-2

, or 0.01 F m-2

. Therefore:

Q = 0.1 V 0.01 F m-2

= 110-3

coulombs m-2

6. If the axon diameter is 10 m, what would be the increase in intracellular [Na+] if all this

charge is carried by sodium ions? (Necessary data: Faradays constant - the number of

coulombs per mole - is approximately 105).

Imagine a length of cylindrical nerve axon, length = L metres, radius = r metres:

The surface area of this tube (we can ignore the ends) is given by 2rL metres2. Its volume is

given by r2L metres3.

We know from Question (5) that 110-3

coulombs of charge flows across the membrane for

every metre2of surface area. Given Faradays constant, this equates to (110

-3)/10

5moles per

metre2. The surface area of our section of nerve is 2rL metres

2, so the amount of sodium that

enters is given by:

5

3

10

rL.2101

moles (1)

Now, this number of moles has entered a volume of r2L metres

3. This means that the increase

in concentration is given by:

Lr.10

rL.210125

3

moles metre-3

(2)

which simplifies to:

r.10

1025

3

moles metre-3

(3)

L

r


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Michaelmas 5:25

Conveniently, the length L cancels out on both top and bottom...so it would not matter what

length of axon we were to choose! Normally, we would want our concentration to be expressed

in moles per litre, and there are 1000 litres in every metre3, so our final expression for the

increase in concentration is:

r.10102 8

3

moles litre-1 (4)

In our particular case, the radius r is 5 m, or 510-6

metres. Putting this into Equation (4)

above gives our final concentration increase as:

)105.(10

1026-8

3

= 410-6

moles litre-1

, or 4 M.

7. What would the change in intracellular [Na+] be for a 1 m diameter axon?

Our axon diameter (and therefore radius) is 1/10 what is was in Question (2). Because the

radius appears in the denominator of Equation (4) above, our concentration increase will be ten

times as great... the answer is40 M.

8. Use your answer to (6) to set an upper bound on the number of action potentials that an

axon can carry when the sodium pump is not working. Assume that an increase in

intracellular [Na+] of 100 mM is the limit before the action potential fails.


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Michaelmas 5:26

The Frog Nerve-Muscle Preparation - debrief sheet

As with most of our practicals, the purpose was to learn experimental skills more than

specific facts. Many people will have taken some time to iron out the various problems that

can occur when preparing to record from live tissue, and others may have gone in different

directions with their own experiments. For those of you who did not get to the end of the

formal practical class, dont worry you certainlywont lose marks forthis, and the important

points needed for the practical paper in the exam are below.

The most common problems experienced in this practical were as follows:

1. Incorrect placement of electrodes. In order to initiate an action potential (AP), a nerve

fibre must be depolarized: the stimulating electrodes produce an electrical current that, if

large enough, can achieve this. A negative electrode placed externally will reduce the

potential difference between the inside and outside of a resting axon membrane,

depolarizing the membrane and initiating an AP which will travel in both directions from

this point. It is therefore the cathode (black electrode) where the AP is expected to

originate. The cathode should always be placed nearer to the recording electrodes than

the anode: if not, the hyperpolarization produced by the anode might prevent the AP

from passing this point on the nerve (anode block). The green earth electrode shouldbe connected between the stimulus electrodes and the recording electrodes, so as to

reduce the stimulus artefact (see below). Which way around the recordingelectrodes are

relative to each other does not matter, as explained below.

2. Expecting yourcompound, extracellular, diphasic AP to look like an intracellular AP,

as you see in the textbooks. If your AP looks upside-down, i.e. a dip followed by a peak,

you can always swap the positions of the recording electrodes to invert your trace...but

for an extracellular, diphasic AP, this does not really matter there is no convention like

there is for an intracellular recording (where the test electrode must be inside the cell).See the background material for this class for more on this.

3. Mistaking the stimulus artefact (see later) for an AP.

4. Too much fluid on the nerve allows the stimulating current to spread far from the

stimulating electrodes. This is likely to result in a large stimulus artefact, especially if a

large stimulus voltage is used, but additionally it may result in direct excitation of thenerve some way away from the electrodes. This can result in no apparent effect on the


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Michaelmas 5:27

muscle when you change the position of the stimulus electrodes on the nerve. If you

have a length of Ringer-saturated cotton also draped over the electrodes, this can also

act as an alternative route for the stimulating current, or a short-circuit.

5. Spontaneous twitching activity in the muscle. Some students investigated this, and most

found that it was associated with spontaneous APs recorded from the muscle, but not

from the nerve. This suggests that the activity (which is associated with a slowly

deteriorating preparation) might have originated in the neuromuscular junctions, perhaps

because of vesicles of acetylcholine being spontaneously released by dying nerve

endings.


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Michaelmas 5:28

The stimulus artefact

The stimulus artefact, if present,

is due to the stimulating current

travelling directly to the

recording electrodes through the

external Ringer fluid, which

conducts electricity because it is

an electrolyte solution. This

direct transmission is much

faster than an AP, so the

artefact appears very close to

the point of stimulation (the

small spike in the lower trace

represents only the beginningof

the stimulus, which actually

lasted 400 s). The artefact can

vary a lot in shape, but not in

time of arrival. The earth wire

helps to channel some of this

unwanted current to earth, so

that less reaches the recording

electrodes. Without the earth wire in place, the artefact is very large. With the earth wire

placed between the stimulating and recording electrodes, the stimulus artefact is greatly

reduced but the AP should be largely unaffected. If you didnt see a stimulus artefact in yourexperiment, you were lucky!

The compound action potential in the nerve

You were recording a compound

action potential the summed

result of APs in each of the

many nerve fibres that make up

the sciatic nerve trunk. You will

have noticed that on increasing

the stimulus voltage from a very

low value, there was a graded

increase in the size of the

compound AP. The reason forthis is that although an AP in an


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Michaelmas 5:29

individual axon is all-or-none, the whole sciatic nerve consists of many axons, each with a

slightly different threshold and at a different distance from the stimulating electrodes. As the

stimulus intensity is increased, the size of the response increases as progressively more

nerve fibres are excited.

Note that, in the example shown above, there is a recording artefact right at the beginning of

the trace, before the stimulus was applied: this can be ignored.


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Michaelmas 5:30

The muscle action potential

The muscle AP which you recorded is also a compound AP. It often looks quite strange

because of the position of the neuromuscular junctions (NMJs), most of which are near the

centre of the muscle between the two recording electrodes. Action potentials generated at

the NMJs travel in both directions and reach the electrodes at different times; different

relative positions of the NMJs result in different times of arrival at the two electrodes and the

net result is a compound trace which, depending on the muscle shape and size, may not

show the classically diphasic shape. Some people (not many) also noticed a small degree of

variability between successive recordings of the muscle AP, even at the same stimulus

intensity. The reason for this is simple: the muscle moves each time it is stimulated and may

come to rest in a slightly different position relative to the electrodes each time. The exact

response that you record is quite dependent upon the fibres that are nearest the recording

electrodes. However, as always in practicals, you should report what you see: please do not

worry about what you think you should have seen!

Nerve conduction velocity

In order to calculate nerve conduction

velocity, you shifted the position of the

stimulating electrodes by e.g. 9 mm, and

calculated the difference in time of arrival of

the muscle AP (see right). Under these

circumstances, the time difference

(represented by the horizontal distance

between equivalent points on the two traces)

must represent the extra time it took the AP

to travel the extra 9 mm along the nerve.

Why not simply divide the distance between

nerve and muscle by the time from stimulus

to response? Because that time would

include neuromuscular transmission time,

and would therefore be longer than the time

for nerve conduction only, yielding an

erroneously low conduction velocity.


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Michaelmas 5:31

Some typical values

Nerve conduction velocities typically 20-50 ms-1. Start of muscle AP 2.5 to 3.5 ms after nerve

stimulation, depending of course on where the nerve is stimulated; muscle contraction 10-25

ms after stimulation.

In order to calculate the delay at the NMJ, you must know that the AP starts at the

stimulating cathode (see earlier), and assume that the nerve fibres extend to the middle of

the muscle belly, where most neuromuscular junctions are located. Say that this distance is

about 30 mm in a given muscle. If the conduction velocity was 35 m s -1, that means that it

would take 0.86 ms for the AP to travel from the cathode to the NMJ (0.03 m divided by 35 m

s-1, the result converted into milliseconds by multiplication by 1000). If it takes 3 ms to see a

muscle AP after nerve stimulation, that would mean that delay at the NMJ would be around

2.14 ms (3 - 0.86 = 2.14). If it takes 15 ms to see the first signs of a contraction in the

muscle, following nerve stimulation, the delay associated with excitation-contraction coupling

would be around 12 ms (15 - 3 = 12). So the greatest delay overall, by far, is in excitation-

contraction coupling! You should be able to describe the physiological events which

contribute to this.

Your results should be similar to these, but remember that there will always be individual

variation.

Properties of muscle contraction

Typical values: latency around 8 ms; size around 1 mm; rise-time around 20 ms; rate of

contraction around 100 mm s-1

.

How does contraction depend on the size of the stimulus?

A wide range of different graphs can be obtained here. Results may not be very repeatable

since (1) the preparations deteriorate over time, (2) different amounts of Ringer solution can

affect the electrical contact between muscle and electrodes, and (3) the muscle moves whenit contracts, resulting in a different position with respect to the stimulating electrodes each


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Michaelmas 5:32

time. The main observation should have been that the whole muscle shows a graded

increase in twitch size as the stimulation voltage increases: more and more fibres are

recruited with increasing stimulation intensity. The relationship is not linear though nor

would you expect it to be!

Most people found threshold voltages between 200 and 800 mV. This is more than ten times

greater than the voltage change needed inside the cell to cause an AP. Clearly, the

resistance of the pathway between the electrodes and the muscle fibres (not all of which are

equidistant from the electrodes) is critical.

Response of the muscle to more than one shock

The traces on the next page represent textbook results in the experiments where the

muscle was exposed to more than one stimulus. On the left, bringing two stimuli closer

together in time resulted initially in summation, with the second peak being superimposed on

the first, and therefore of greater amplitude. When the two stimuli were brought even closer

in time, the second peak got smaller and finally disappeared. This is when the second

stimulus falls into the refractory period of the first AP, so the muscle sees only one AP. In

the right-hand trace, increasing the frequency of APs results eventually in the smooth,

sustained contraction known as tetanus.

If you got this far in the experiment, did your results look as good as these? Perhaps not. But

then, we could find textbook results in a textbook! We can often learn more interesting

things from the peculiarities of our own preparation...which is what practicals are all about.

How can smooth contraction be obtained at much lower frequencies in vivo?


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Michaelmas 5:33

In the live animal, a motor neurons axon divides near the gastrocnemius muscle to

innervate around 200 separate muscle fibres (this number is variable and depends on the

muscle). Each of these fibres has just one neuromuscular junction. The group of muscle

fibres all innervated by the same neuron is known as a motor unit. Although all fibres in themotor unit contract simultaneously when the neuron is stimulated, there are many, many

motor units in the whole muscle! These motor units are stimulated asynchronously in life,

allowing for smooth contraction, even though each individual motor unit is stimulated at a

relatively low rate by its neuron. Contraction of one motor unit follows immediately after

contraction of another has finished, so tension in the muscle as a whole is sustained .

In your preparation, however, all muscle cells are stimulated at once by the external

electrodes, leading to a jerky response as they all contract and then all relax together.

50 Hz electrical noise

The figure below shows a phenomenon that several groups noticed. There is a strange

wiggle in the upper trace, which represents an electrical recording from the muscle.

This wiggle appears to be a regular (if noisy) sinusoid, which makes us suspect that it might

not have a biological cause. A quick check of the frequency tells us that the interval between

peaks is 20 ms, which means that the frequency is 50 Hz (50 cycles per second). That

frequency should ring loud bells in the ears of experimental physiologists, at least in the UK,

since it is the frequency with which the AC voltage of our mains electricity oscillates. This

noise is probably due to either a poor earth connection or a mains power lead too close by!

You could first look around the setup for a likely mains power source and move your set-upaway from this. If that does not fix it, you could test PowerLab by connecting both of the


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Michaelmas 5:34

recording electrodes to the earth wire and seeing if the noise goes. If it does, then the

electronics are probably fine if not, you might try rebooting PowerLab.

The aims of this practical

We do not expect you to get perfect results every time in these practical classes...you can

see those in your text-books. The point is to see how science is done first-hand, to work with

live tissue, to recognize the potential experimental errors and shortcomings (which may have

been made in the experiments that generated the published results of others), to problem-

solve and to think critically. Problem-solving is great fun and you learn a lot. Dont get

disheartened by things not working first time - rise to the challenge!

Some of you found time within the practical to perform experiments of your own, which we

very much encourage. If you did so, you should include a write-up of this as part of your

report, especially if you did not finish the default experiments as a result.

If you do an extra experiment which you think others would benefit from seeing, please send

the results, with an explanation of what you did, to me, Dr. Matt Mason ([email protected]),

and I will make it into an extra debrief sheet for the CamTools website.

MJM 8/11/12


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Michaelmas 5:35

Mechanical Properties of Muscle

Using the frog gastrocnemius muscle preparation, you are going to measure three important

properties of muscle contraction. All of these can be determined from a single set ofrecordings of the effect of stimulation upon the length of a muscle which is made to lift aseries of different weights:1. The passive length-tension curve, i.e. the relationship between length and tension when

the muscle is not contracting.2. The total length-tension curve when the muscle is tetanically stimulated. This will enable

you to determine how active tension depends on length, a relationship which has to dowith the sliding filament mechanism of the contractile machinery.

3. The relation between maximum rate of shortening and load, which will tell you about thepower that the muscle can exert and how this power is related to length and tension - animportant topic in applied muscle physiology (e.g. in athletics).

Measurements of this kind form the basis of our understanding of the mechanical properties

of muscles. PLEASE BRING A LAB COAT TO THIS CLASS!

Background reading

The cross-bridge cycle. Muscle mechanics.

Learning objectives

By the end of today's class you will be able to:

Understand the difference between active and passive tension. Understand how active tension varies with resting length of skeletal muscle. Understand the relationships between force, velocity and power, in muscle contraction.

Background: Active and passive tensions within a muscle

An unstimulated muscle is elastic, like a rubber band. If you hang different weights from it, itstretches by different amounts; you can then plot a length/tension curve that describes itspassive behaviour. Although a perfect elastic element would obey Hooke's Law, whereextension is directly proportional to tension, the behaviour of a muscle will deviate somewhatfrom this and the relationship will be curved. We speak of the tension in an unstimulatedmuscle as passive tension. The extracellular collagen, the plasma membranes andespecially the intracellular protein titin are the elements which take the passive tension; thecross-bridges are not involved. If the muscle is stimulated at a high frequency (tetanized) itdevelops extra tension, due to the activation of the contractile elements. This extra tension,produced by the power-strokes within the cross-bridge cycle. requires energy in the form ofATP and is hence referred to as active tension. The active tension plus inevitable passivetension in a muscle at a given length can be added together to give the total tension in thatmuscle.


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Michaelmas 5:36

An isometric experiment to determine active tension

We want to measure the active tension in our muscle, andsee how this varies with muscle length: this will give us someinsight into the sliding filament mechanism which underlies

muscle contraction. In principle, this is easy to do, if we couldconduct an isometric experiment. In an isometric (= "samelength") experiment, the muscle would be clamped at bothends so its length could not change, and it would then bestimulated to contract. Although it could not shorten, the extratension developed due to contraction could be recorded bytransducers in the clamps.

An isotonic experiment to determine active tension

We do not have the apparatus for theisometric experiment outlined above, but we

can obtain the same information indirectlyfrom an isotonic experiment (isotonic ="same tension"). This has the advantage ofallowing us to measure other things too, suchas contraction velocity and therefore power.In our isotonic experiment, we use themuscle, clamped only at the bottom, tosupport a given mass m via a see-saw leverapparatus.

Because the mass is not dropping under the influence of gravity, the tension in the string

supporting it must equal the force downwards, i.e. mg, where g is the acceleration due togravity, 9.81 msec-2. Because the muscle is supporting this, the muscle must alsoexperience the same tension, mg(things are a little more complicated than this if lever armsdiffer in length, as in the practical, but we'll keep things simple for now). Since the muscle isnot contracting, this tension must be the passive tension, developed because the musclehas been stretched beyond its resting length. We can measure the length of the muscle, x,at this point.

If we now stimulate the muscle to contract,it will shorten to length y, lifting up themass by the same amount. Whentetanically contracted, the muscle is still

supporting the same load against gravity,so the tension within the muscle will still bemg. This is why we refer to this as an"isotonic" experiment. However, thetension at length yis total tension, madeup of an active component due to cross-bridge cycling, and a passive componentdue to the muscle still being stretchedbeyond its resting length. Because themuscle is shorter, the passive tensionwithin it is less than before; the differencehas been made up by the active

component.


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Michaelmas 5:37

If we were to repeat this experiment withdifferent loads on the muscle (stretching it todifferent starting points before stimulating),we would get lots of horizontal pairs ofreadings.

We can then join up the "passive tension"points and the "total tension" points to gettwo curves. The passive tension relationshipis curved, not linear, because the muscledoes not obey Hooke's law; below a certainlength, there is no passive tension in themuscle, so the passive tension curve followsthe x-axis.

Now comes the clever part. Imagine amuscle of length z mm. We can read offfrom our graph what the total tension of amuscle at this length would be (i.e. thetension in a muscle that, whencontracted, was z mm long). We can alsouse the graph to establish the passivetension of a z mm long muscle (i.e. anunstimulated muscle which was weighted so

that it stretches to this length), by lookingvertically downwards from our total tensionpoint. The active tension must represent thevertical distance between these two points,i.e. the difference in tension between totaland passive for a muscle of that particularlength.


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Michaelmas 5:38

By measuring the vertical distance betweenthe total and passive curves at regular

intervals, we can produce an active tensioncurve. We should find that it forms a peaksomewhere in the middle of the length range.This is exactly what we shall be attempting toproduce in the practical today.

Leg measurements

You will be provided with the leg of an edible frog, Ranaesculenta, which you will skin and keep moist using Ringersolution. You will then use dividers to determine the maximumand minimum possible lengths of the gastrocnemius muscle,which runs between knee and ankle joints at the back of the leg.The maximum muscle length is obtained when the leg ispositioned as in Figure 1 below; the minimum length is whenpositioned as in Figure 2.

These measurements give you the total range of movement, but

might not be representative of the muscle lengths during normalactivity in vivo. To simulate this, imagine that the frog is first crouched on its lily-pad (leg fullyflexed; Figure 3) and then jumping (leg fully extended; Figure 4). You will also measure thediameter of the gastrocnemius muscle when in an intermediate position.

Figure 1 Figure 2 Figure 3 Figure 4

The leg dissection is identical to what you performed in the frog nerve-muscle experimentlast week, except that today we will be stimulating the muscle directly, so we are notinterested in the sciatic nerve.


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Michaelmas 5:39

Setting up the apparatus

As last week, the gastrocnemius muscle is mounted in aorgan chamber with a pin through the knee joint, and itsfree end is attached to the short lever arm of a

mechanical transducer. A weight carrier is attached tothe long arm; over the course of the experiment, you willadd weights to this and see how this changes themuscle contraction properties. Stimuli are applieddirectly to the muscle this week, rather than via thesciatic nerve. PowerLab is set up to generate a burst of50 Hz stimulation lasting one second, starting half asecond after the beginning of the sweep, in order totetanize the muscle.

You will initially put the muscle and string under a lot oftension, using a heavy weight, to tighten knots and

make sure that there is no slippage.

PowerLab must then be calibrated using the length of the muscle measured within theorgan chamber, so that the computer knows what the resting length of the muscle is, andcan then represent lengths in absolute values.

The lengths that you will be measuring include the resting length, when the muscle is undertension from the weight it is supporting, but is not actively contracting, and the tetanisedlength, when it is fully contracted.

The dL/dt trace (middle panel, above, calculated automatically by PowerLab) is the

derivative of length (from the upper trace) with respect to time, i.e. velocity of contraction. If


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Michaelmas 5:40

the length is not changing, dL/dt will be zero, whereas if it is changing then dL/dt will be thegradient of the upper trace.

We are interested in the maximum velocity (rate of shortening) attained during the initialcontraction, because this will ultimately allow us to work out the power of the contraction.

Because the length trace drops when the muscle is contracting, the gradient dL/dt will takea negative value, but you should ignore the negative sign!

Calculating the load that the muscle sees

Because the load lever arm is twice as long asthe muscle lever arm on your mechanicaltransducer, the muscle is working at a mechanicaldisadvantage. If the load that you added is Mgrams, the muscle actually has to develop enough

tension to suspend 2M grams. For this reason, allof the loads will eventually be multiplied by two,when it comes to calculating muscle tension andpower.

Calculating the tension in the muscle

Tension is a form of force, and as such is equal to mass x acceleration. The mass inquestion is the load seen by the muscle, 2M, as above. The acceleration in question is the

acceleration due to gravity, 9.81 msec

-2

.

Calculating power

You must also understand how to calculate the power developed during shortening, at eachload.

Power (rate of working) = work / time, and work = force x distance.

Therefore power = force x velocity.

The force in question is the tension experienced by the muscle, as above. The velocity inquestion is the maximum rate of shortening, dL/dt, which you will obtain from the traces.

Some of these calculations are performed automatically for you in this practical, but you are

expected to understand how you would perform them "manually", for example if you were

given a trace recorded from a practical class.


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Michaelmas 5:41

The experiment

The order of the experiment is as follows. You will run a trace, and from it record restinglength, tetanised length and maximum velocity of contraction. Placing an additional 20 gweight on the weight carrier and having allowed the muscle time to adjust to this, you will run

another trace and take the same measurements. This procedure is repeated until the weightis so high that the muscle is unable to contract.

The measurements will be used to calculate active tension for any given length, along thelines explained earlier. You will also be calculating the relationship between force andvelocity, and how power relates to these variables.

Analyzing your data

Passive and total tension will be plotted for you by LabTutor, but unfortunately LabTutor isnot capable of working out active tension from these curves. You therefore need to use thegraph paper on the next page to draw out these curves, including the active tension curve,manually!

To do this, plot tension (on the vertical axis) against muscle length in both the resting andthe tetanised state. Tension should be in Newtons (units of force: multiply the mass seen bythe muscle, in kg, by 9.81 msec-2). If you prefer, to save time you can simply plot your mass(2M) in grams on the y-axis, since this will be proportional to tension.

Subtract the passive tension at a given length from the total tetanic tension and plot theresulting active tension as a function of length, on the same graph. See the Background

section, earlier, for information on how to do this.


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Michaelmas 5:42

Graph paper for active tension calculations

Mark on the graph the maximum and minimum muscle lengths observed before dissection.

Does the working range between these lengths coincide with the peak in the active tension

curve?


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Michaelmas 5:43

Some questions and calculations to try at home

The tension in the muscle is generated by crossbridges between the two kinds of sliding

filament, myosin and actin. Each crossbridge is believed to act cyclically, exerting a certain

tension while it contracts through a certain distance, and

then letting go.

1. The tension in the muscle is generated by

crossbridges between the two kinds of sliding filament,

myosin and actin. Each crossbridge is believed to act

cyclically, exerting a certain tension while it contracts

through a certain distance, and then letting go. Use your

data to work out the tension in each myosin filament

when the muscle is generating maximum active tension.

(Necessary d ata: the fi laments are arranged in a

tr iangular lat t ice with approxim ately 50 nm between

nearest neighb ours . In cros s-section , each fi lamen t is assoc iated with a certain area -

imagin e each o f them as th e centre of a rectangular ti le. Loo k at the answers at the end if

you s t i l l can' t see how to w ork ou t the area per f i lament. You need also to est imate the

diameter of the muscl e).

In a myosin triangular lattice of unit distance a, each filament is associated with an area ofa

23/2 (see diagram above). If a is 50 nm this area is 2.1710

-15m

2. Use r

2to work out the

cross-sectional area of your muscle. Given a cross-sectional area of 25 mm2

= 2.510-5

m2,

then the number of filaments in this cross-section is (2.510-5

)/(2.1710-15

) = 1.151010

filaments.

[Cross sectional area of one muscle filament = 2.1710-15

m2

Cross sectional area of muscle = 25 mm2

= 2.510-5

m2

Number of filaments = 1.151010

filaments.]

Lets say that maximum active tension is equivalent to 200 grams. Tension is a unit of force,

and F=ma, where F is force, m is mass (in kg) and a is acceleration due to gravity,

approximately 10 msec-2

. This means that the tension in the muscle must have been 0.210 = 2

N.

The tension per filament is simply this value divided by the number of filaments in parallel,

which comes to 2/(1.151010

) = 1.7410-10

N per filament.


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Michaelmas 5:44

Note that the length of the muscle, and so the number of sarcomeres in series, does not make

any difference to the tension developed it is only the number of sarcomeres in parallel

(and hence the number of myosin filaments in parallel) which contributes to the active

tension developed in the muscle. [A longer muscle means that the same active tension

occurs over a longer distance the absolute max value of the active tension is the same.]

2. Use this value to estimate the force generated by each crossbridge. Assume that each

crossbridge is attached half of the time during generation of maximum active tension.

(Necessary data: a my osin fi lament is 1.6 m long , of w hich the central 0.2 m is bare o f

the headgroup s that form th e crossb ridges. There are 9 headgrou ps per 42.9 nm repeat

distance on the myosin f i lament. Remember that the tota l force generated by a

sarcom ere is equal to the force generated by all the active cross -bridges at one end).

The arrangement of a myosin filament is represented diagrammatically below, using length

data from the question.

At each end, there is 0.7 m or 700 nm length occupied by head-groups. At 9 per 42.9 nm, this

gives us a total number of (7009)/42.9 = 146 head-groups at each end. Of these, half are

active: 73 active head-groups. Using the total force per filament from Question (1), the force per

cross-bridge is (1.7410-10

)/73 = 2.3810-12

N per head group.

3. Use this value to calculate the maximum work per crossbridge cycle, assuming that the

stroke length is 5 nm. Compare your answer with the energy available from the

hydrolysis of one molecule of ATP.

(Necessary data: hyd rolys is of 1 mo le of ATP to ADP releases 30.5 kJ. Avo gadro 's

numb er - the numb er of molecules per mole - is 61023).

Work is equal to force distance, so using our force per head-group from above, work per

cross-bridge per 5 nm cycle is given by (2.3810-12

)(510-9

) = 1.1910-20

J.

1.6 m

0.7 m 0.7 m0.2 m


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Michaelmas 5:45

1 mole ATP generates 30.5 kJ of energy, so 1 molecule will produce a total given by

(3.05104)/( 610

23) = 5.0810

-20

J.

The % efficiency is calculated as 100(1.1910-20

)/( 5.0810-20

) = 23%.

Questions 4, 5 and 6

These ones you should think about yourself, and discuss with your supervisor!

4. The muscle which closes a crab's claw has exceptionally long sarcomeres. How would

this affect (a) the maximum velocity of shortening, and (b) the maximum force which

can be developed by the muscle? (Hint: consider each actin-myosin crossbridge as a

unit which can generate a fixed maximum force or pull at a fixed maximum rate).

5. Why do bicycles have gears? One of your curves tells you the answer.

6. What does the frog use the gastrocnemius for? What kind of fibres do you think it is

likely to contain, and why?


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Michaelmas 5:46

Mechanical Properties of Muscle - debrief sheet

Looking at the

Documents

Michaelmas Term - Experimental Physiology Background and Debrief Sheets