Nerve Impulse

Question 1

Steps to nerve impulse

Answer 1

• Stage 1: Arrival of impulse
• Stage 2: Depolarisation
• Stage 3: Repolarisation
• Stage 4: Re-establishing original resting rate

Question 2

Stage 1 - Arrival of impulse

Answer 2

• At rest - neuron is polarised - 70mV
• When a sufficient stimulant is applied OR an impulse arrives, the cell membrane increases it’s permeability to sodium ions.
• This allows an influx of Na+ ions into the cell
• This movement is too great for the outward movement of K+ ions
• This reduces the potential differences between the ECF and ICF (inside becomes more positive)

Question 3

Stage 2 - Polarisation

Answer 3

• If the change in potential difference exceeds 15 mV (or membrane potential reaches -55 mV) - the threshold - then an action potential will occur
• ‘All or none response’
• Once the threshold is reached, voltage gated sodium channels open, allowing an even greater influx of Na+ ions into the cell
• This stimulates channels in the next section of the axon to open
• The Na+ ion movement makes the inside become positively charged (+30 mV)
• This process is knows as depolarisation - the less of the charge difference between the ECF and ICF.

Question 4

Stage 3 - Repolarisation

Answer 4

• Once the membrane potential reaches +30 mV, the sodium voltage gated channels open
• Once the membrane potential reaches +30 mV, the sodium voltage gated channels open
• K+ ions move out of the cell
• This reaches a potential difference in voltage approaches and then reaches -70mV
• There is a state where the potential difference gets below -70mV and is known as hyperpolarisation
• This is due to channels being sloe to close, more K+ leaves cell, even more negative inside (voltage gated potassium walls delay in closing)

Question 5

Stage 4 - Re-establishing original resting rate

Answer 5

• Axon returns to original resting rate (-70mV)
• Movement of ions generates the impulse
• Releases small amount of energy into

1. Adjustment nerves which create a chain reaction
2. Adjacent parts of the neuron to spread the length of the axon

Question 6

Refractory period

Answer 6

No new impulse can occur in any part that is not at rest,. This allows unidirectional propagation (prevents AP’s moving backwards)