Earthworm Action Potentials – Lab report PDF

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BIOL 4510 - BIOL 4510 Animal Physiology Lab

Lesson

Earthworm Action Potentials – Lab

Student email [email protected]

Evoked action potentials – Challenge

Select Annotate to indicate the stimulus artifact and the action potential on the recordin 

Evoked action potentials – Activity

Stimulus 6 4 2

Evoked action potentials – Analysis Fiber Type

Latent Period (ms)

Median giant ber

1.55

What is the relationship between stimulus strength and response amplitude in a single axon There is norelation between stimulus strength and response amplitude in a single axon.

An intracellularly recorded nerve action potential approximates 80 mV. Why is your recorde smaller? The recorded action potential is much smaller because this was an extracellular nerve actio Extracellular recordings record whether or not an action potential occurred thus through th information may be lost, resulting in a smaller action potential.

The typical shape of the action potential is biphasic with two peaks of opposite polarity. Can y The action potential is biphasic becuase there are 2 electrodes, one negatively charged and charged. As the stimulating electrode sends negative electrons down, when it comes in con recording electrode, the action potential will point down wards as similar charges oppose an contact with the positive recording electrode, the action potential will point upwards as opp

6

Stimulus

4

2

0 0.2

Action Potential (mV)

0.1

-0.1

750 mV

0 Time

-0.2 0.01

Record

1

2

0.02

3

4

5

6

7

8

Lateral axon recruitment – Analysis Fiber Type

Latent Period (ms)

Lateral giant ber

3.55

Why do you suppose the latency diered between the median and lateral bers? TheMedialGiantFiberis larger and therefore allows the transmission of action potentials d than theLateralGiantFibers.

The distance from the cathode (negative stimulating electrode) to the recording electrode is and D2 from the second recording.The time from stimulation to arrival of the action potent recording electrode is T1,and tothe second recording electrode is T2. The conduction velocity (CV) along the nerve can be most accurately calculated using which e CV = D1/T1 CV = D2/T2 CV = (D2 - D1)/(T2 - T1) CV = (T2 - T1)/(D2 - D1)

Conduction velocity – Activity

6

Stimulus

4 2

0.2

Distance 1=39mm, 2 V

0 Action Potential (mV)

0.1 0 -0.1

Time

-0.2 0.005

Record

1

2

0.01

0.015

0.02

0.0

It is usually assumed that an action potential begins immediately at the cathode. If this were calculating conduction velocity would provide the same answer. However, when a strong sti action potential may begin some distance away from the cathode. Under these conditions, t be more accurate. Did you observe a dierence in the conduction velocity values produced by the two method Yes, adierencewas observed between thetwo methods.

Refractory period – Activity

6

Stimulus

4 2 0 1

Action Potential (mV)

0.5 0 Time

10 ms

-0.5 1

Record

0.005

1

2

3

0.01

4

5

0.015

6

7

0.02

8

0.0

Briey describe the cellular events responsible for the refractory period (Hint: Discuss the m During the absolute refractory period, Na+ channels are open but inactivated, not allowing the relative refractory period, some Na+ channels are open allowing Na to enter the cell wi voltage, fully activating those channels.

Bidirectionality – Activity

6

Stimulus

5 4 3 2 1 0 1

Action Potential (mV)

0.5 0 -0.5

Time

1

Record

0.002

1

0.004

0.006

0.008

2

Bidirectionality – Analysis

0.01

0.012

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