Electricity within the body Prof. Dr. Moustafa. M. Mohamed Vice Dean

1. Electricity within the body Prof. Dr. Moustafa. M. Mohamed Vice Dean Faculty of Allied Medical Science Pharos University Alexandria Dr. Yasser khedr Department of Medical Biophysics Pharos University

2. Study Of Electricity Is Important For Two Reasons • All living organisms are controlled by electrical signals derived from sensors which respond to changes in the environment • Electrical and electronic circuits are widely used in biological measurements and nowadays highly advanced electronic systems are used in medical fields for diagnostic and therapeutic purpose.

3. In a biological system, the electric conduction is quite complicated because of the presence of different types of electric current carriers.

4. The Biological Function Of The Living Cell • Keep the composition of the cell inside. • Allows the transport of certain ions to inside and outside the cell. • Receives electrical information from other surrounding cells and passes this informationto the nucleus of the cell. • Responsible for cell to cell communications.

5. In biological cell membranes and nerves, a resting potential is caused by differences in the concentrationof ions inside and outside the biological membrane and by differences in the permeability of the cell membrane to different ions. The Cells Membranes

6. The cells membranes have a common characteristic relating to the presence of a potential difference between the interior and the exterior of the membrane. Ion concentrations

7. The Cell Membrane is Semi-Permeable

8. Nernst Equation • The equilibrium potential difference for an ion can be found from the Nernst equation: • Where • K is Boltzmann's constant (1.38x 10-23 J/K), • T is the absolute temperature of the medium [(273+t) Kelvin)] and • "e" is the charge of the electron (1.6x10-19 Coulomb).

9. So the Nernst Potential is:

10. For a nerve cell, the intracellular has a K+ ions concentration of 0.141 mol/l • whereas the extracellular fluid has K+ions concentration of only 0.005 mol/l,

11. The structure of the nerve cells (neuron)

12. Schwann Cells • These cells form a multilayered myelin sheath. • Reducing the membrane capacitance. • Increasing its electrical resistance. • This sheath allows a nerve pulse to travel further without amplification. • Reducing the metabolic energy requiredby the nerve cell.

13. Node of Ranvier • At the nodes, the amplifications of the nervepulses occur. • Thus a myelinated axon act as a cable, with the periodic amplification used to prevent the signal from becoming too weak. • By the contrast, signals in unmyelinatedaxon becomeweak in a very short distance andrequired virtually continuous amplification.

14. Cell models • K+, Ca+, Cl- ions caused the potential inside the axon to be 70 ~ 90 mV lower than outside • A stimulus causes the potential inside the axon to increase to ~ 110 mV thus is ~ 40 mV higher than outside • The raised potential rapidly decreases to its original value and the change in potential propagates alone the axon • The typical pulse propagation speed depends on axon types but is ~ 50 m/s • The duration of the pulse at any point on the axon is ~ 2x10-3 s

15. Resistance in a Metal Wire • resistance of a metal wire is directly proportional to its length, and inversely proportional to its cross-sectional area, A: • R – resistance (Ω) • ρ - resistivity(Ω m) • L – length (m) • A – cross sectional area (m2)

16. Nerve electric properties • What is R, R’, C?

17. Axon resistance and capacitance • Resistance: • Capacitance: • Leakage resistance:

18. Nerve electric properties

19. Space parameter • Space parameter, : •  = 0.05 cm (unmyelinated axon) •  = 0.7 cm (myelinated axon)

20. An axon 1 m long has nodes of Ranvier every 10-3 m. How many times is a nerve pulse amplified when it is transmitted along this axon • Solution: • No of amplification = 1 m / 10-3 m = 1000 times

21. Calculate the axoplasm resistance of 1-cm long seqment of un myelinated axon of radius 2 mm, axoplasm resistivity = 2 ohm m. • Solution: • Axoplasm Resistance R= r l/A • = 2 X 10-2/ (3.14 X (2 X 10-6)2)= 1.95 X 109 ohm

22. (Hint: axoplasm resistivity = 2 ohm m, capacitance per unit area = 5 X 10-5 F/m2 for myelinated axon and = 10-2 F/m2 for unmuelinated axon, resistance per unit area of membrane Rm = 40 ohm m2 for myelinated axon and = 0.2 ohm m2 for unmuelinated axon, Boltzmann constant = 1.38 x 10-23 J/K)

23. A myelinated segment of axon has a radius of 2 mm and a length of 1 cm. find its (a) membrane capacitance; (b) membrane leakage resistance • Solution • Membrane capacitance = 2 p r l Cm = 2 X 3.14 X1 X 10-2 X 2 X10-6 X5 x 10-5 = 6.28 x 10-12 F • Membrane leakage resistance = Rm / (2 p r l)=40/(2 X 3.14 X1X 10-2 X 2 X10-6) = 31.85 x 106 Ohm

24. An unmyelinated segment of axon has a radius of 2 mm and a length of 1 cm, capacitance per unit area 10-2 F/m2 resistance per unit area of membrane 0.2 ohm m2. find its (a) membrane capacitance; (b) membrane leakage resistance • Solution • Membrane capacitance = 2 p r l Cm = 2 X 3.14 X2 X 10-6 X 1 X10-2 X 10-2 = 1.24 x 10-9F • Rm / (2 p r l)= 0.2/(2 X 3.14 X2 X 10-2 X 2 X10-6) = 1.95 x 109 Ohm

25. Calculate the axoplasm resistance of 1-cm long seqment of myelinated axon of radius 2 mm,axoplasm resistivity = 2 ohm m • R= r l/A = 2 X 10-2/ (3.14 X (2 X 10-6)2)= 1.95 X 109 ohm

26. Find the space parameter for (a) an unmyelinated, (b) myelinated axon of radius 0.2 mm • Space parameter l • = .1 mm for unmyelinated • = 1.414 x 10-3 m for myelinated