22.416 SensoryMotor Physiology, Lecture 3

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22.416 Sensory-Motor Physiology, Lecture 3

• VOLTAGE-CLAMPING
• ANALYSIS OF ACTION POTENTIALS
• IN
• SQUID GIANT AXONS
• Recall that, with voltage-clamp,
• stepping VM from VMR to any VC shows how IM
  varies with time at new VM (no further change in
  VM is allowed)
• stepping to different Vcs shows how IM depends
  on VM

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V clamp can reveal channels ionic selectivity

• a) through ion substitution expts - use
  artificial saline with "normal" ions ?, ? or
  replaced- then if ?ion X ? ? I, channel must
  pass ion X at least
• AND
• b) through measurement of channel's reversal
  potential- vary VM over a wide range- find
  level where IM reverses, inward to out, or vice
  versa
• ? reflects combination of ions passed by
  channel. E.g. - "pure" K channel has rev.
  potl EK (Why?)
• - purely Na-selective channel has rev. potl
  ENa - mixed cation channels have rev. potl
  between ENa EK
• We can predict rev potl by the Goldman equation
  if we know- the membrane's permeability to each
  ion, PX, and- each ion's concentration inside
  and outside, Xi Xo .......

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Goldman equation (works for univalent ions only)

• (PKKi PNaNai PClClo)Vr -(2.3RT /F )
  x log10 ------------- (PKKo PNaNao
  PClCli)
• note similarity to Nernst equation (same R, T, F
  )- recall 2.3RT/F 61 mV at 310oK (37oC),
  58 mV at 20oC
• Goldman reduces to Nernst eq. if all Pions 0
  except one i.e., if only one ion can diffuse,
  and it reaches equilibirium
• But Goldman applies where several ions are
  undergoing net diffusion, with none in
  equilibrium, but with all their currents adding
  up to zero to maintain a unique steady-state VM
  for any set of Pion values
• Can work backwards, e.g., for Na K cation
  channel (PCl 0)- measure channels rev potl
  (say at 37oC)- then use Goldman eq. to solve for
  its ratio of PNa/PK ( r) Vr -61 mV x log10
  (Ki rNai) / (Ko rNao) ? can solve for r

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Analysis of the action potential in squid axons

• A. Variation in ap amplitude with Nao
• Hodgkin Katz reasoned that, if ap depends on
  Na entry, then
• ?Naoutside should ? ? ap height,
  (since Nernst eq says ?Nao ? ? ENa ? ? ap
  peak)
• ?Naoutside should ? ? ap height
• they recorded ap with ic wire electrode with
  various Naos
• ap height varied as predicted
  (Fig. 6.1)

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B. Effect of varying Nainside Kinside on
ap VMR (when cytoplasm is replaced)

• Baker et al. (62) -squeezed cytoplasm out of
  a giant axon -perfused axon with salines of
  varied Na K
• - they found ? Nainside ? ? ap height,
• ? Nainside ? ? ap height
• while ? Ki ? more -ve VMR, and ? Ki ?
  less -ve VMR, , also as Nernst eq predicts
• (? ? Ko ? opposite effects)
• - confirmed that membrane, not cytoplasm,
  generates VM

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C. Ion currents under voltage clamp

• Hodgkin, Huxley and Katz '52 V-clamped giant
  axons via 2 ic wire electrodes for I V
• Top traces step change in VM, from holding
  potential to new VC
• Lower traces current that flows as VM is held
  at new level
• downward (negative) deflection ? ve inward
  charge flow upward deflection ? ve
  outward current
• Found step depoln from VMR caused 1. very
  brief ve outward capacitive current as VM is
  reset,
• 2. early transient inflow of ve charge,
  followed by
• 3. late sustained net outflow of ve charge
• Note early ve inflow (2) despite depol. step
  defies Ohm's Law.

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Fig. 6.3 response of squid axon to VM step

• Total membrane current, IM

Inferred components of IM (see evidence
below)
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1. Variation of early inward current with VC

• As ve VC step increases
• - peak of early, inward current first
  increases, then decreases to
  0, then reverses ? outward
• - reverses at VC ENa, suggesting inward
  current ? Na entry I-V plot ?

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If Na carries the early, inward current, ....

• then replacing Na in bath with, e.g., choline
  should block inward current
• - Hodgkin et al. confirmed this prediction
• -later, TTX was found to block early IM by
  blocking Na channel (so only late current
  flows)
• (while TEA blocks K channel, so only early
  current flows)

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I-V relation for early current

• plot peak value of early current against VC
• find slope dI/dV for VC below 0 mV is negative
  - so dI/dV (slope conductance) is negative in
  this VC range - but how can conductance be
  negative? it implies that ve charge (depoln)
  attracts ve charge inward, which is absurd
• apparent -ve g arises because depoln opens Na
  channels Na ions are pushed inward by
  concentration gradient despite increased outward
  electrical repulsion

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Significance of negative conductance

• this negative conductance is what drives the
  ve feedback that generates the upstroke of
  the ap depoln ? more positive charge entry
  (Na) ? ? depoln
• More realistic or honest physically is the chord
  conductance, defined as gchord I / (VM -
  Vrev)
• - its (VM - Vrev) takes into account the extra
  inward force on Na due to its concentration
  gradient, so gchord is always positive
• - its increase with ? VM reflects the opening of
  Na channels
• Early inward Na surge soon stops (as Na channels
  inactivate), giving way to the ...

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2. "late" current during ve VC step

• ve outward current rises to late plateau until
  VC step ends
• - due to K outflow through V-gated K channels
  that open slowly and stay open upon
  depolarization
• later, TEA 4-AP were found to block these K
  channels so TEA during ve VC step ? early Na
  inflow only
• Both the Na entry and the K outflow inferred
  under V clamp were later confirmed by means of
  radioactive tracers
• - labelled Na entered the cell faster than it
  left during aps,
• - labelled K left cell faster than it entered

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Mathematical description of V clamp
currents(needed for precise calculation of aps
time course)

• Both Na and K currents vary with time during
  any VC step i.e., INa f(VC , t), IK
  f(VC , t)
• But I(t) g x V, and V is constant during a
  V-clamped step (since V driving voltage
  VM - Vrev VC - Eion), so gNa and gK must
  also vary with time during a step
• To obtain mathematical expressions for gNa(t)
  gK(t),
• H H separated IM(t) into INa IK during each
  VC step
• first eliminated INa by reducing Nao, leaving
  only IKthen subtracted IK from total IM (in
  normal Nao) to get INa
• Then, gNa(t) INa(t) / (VC - ENa), and
  gK(t) IK(t) / (VC - EK)