Assume the length of an axon membrane of about 10 cm is excited by an action pot

Question

Assume the length of an axon membrane of about 10 cm is excited by an action potential (length excited = nerve speed pulse duration = 50 m/s 2.0 ms = 10 cm). In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and opposite charge of negative organic ions, as shown in the figure below. Model the axon as a parallel-plate capacitor and take C = ?eoA/d and Q = C?V to investigate the charge as follows. Use typical values for a cylindrical axon of cell thickness d = 1.0 10-8 m, axon radius r = 20.0 µm, and cell-wall dielectric constant ? = 3.0.

(a) How much energy does it take to restore the inner wall of the axon to -70 mV from +30 mV?
J

(b) Find the average current in the axon wall during this process. (Hint: The time required to return to the resting state is 3.0 ms.)
µA

Explanation / Answer

Pnhÿ6="DataGridMessage_ctl02_lblMessageDescription"> length of the axon membrane is l = 10 cm = 0.1 m radius of the axon is r = 18 µ m = 18*10-6m Then area of surface of axon membrane is A = l*2pr = 0.1 m *2p* 18*10-6m = 1.8*2p*10-6m2 cell thickness d = 1.1 10-8 m dielectrci constant of the cell wall is k = 2.0 Then capaciutance is C = oA/d = 2.0*(8.85*10-12C2/Nm2)*1.8*2p*10-6m2/1.1 10-8 m =1.8189*10-8F Voltage is V = 70 mV = 0.07 V Then charge is Q = CV = 1.8189*10-8F *0.07 V = 1.27*10-9C Then charge per unit area is s = Q/A = 1.27*10-9C/1.8*2p*10-6m2 = 1.123*10-4C/m2 Then charge per unit area in terms of the number of square angstroms (2) per electronic charge is s/e = (1.123*10-4C/m2 /1.6*10-19C )(10-20m2/1 2) = (1/142475.52 ) e = 1e/(377.459oA)2 Then number of K+ ions are on the outside of the axon is N=Q/e = 1.27*10-9C/1.6*10-19C = 0.79375*1010 b) Total charge on the inner membrane is Q' = CV' = 1.8189*10-8F *0.03V = 0.54*10-9C Then net positive charge q = Q' -(-Q) = 0.54*10-9C+ 1.27*10-9C = 1.81*10-9C Now number of sodium ions is N' = q/e = 1.81*10-9C/1.6*10-19C = 1.13125*1010 c) TIme t = 2 ms = 0.002 s Then current I = Q/t = 1.81*10-9C/ 0.002 s = 0.905*10-6A = 0.905µA d) energy does it take to raise the potential of the inner axon wall to +30 mV is U = (1/2)CV2 = (1/2)*(1.8189*10-8F )(0.03volts)2 = 8.185*10-12J nbsp;Û707Pnhÿ6iv> The mass of H2O produced = 0.ÑAo*Pnhÿ6l &nÀR&bPnhÿ6 &nºX&bPnhÿ6 &nbsªhbpPnhÿ6bsp;= 13.885 g p; = ‰‰1mPnhÿ6 The mass of H2O produced = 0.771mol* 18 g/mol = 13.885 g QÁ *hÿ6xk5hÿ6xk5d Part getPart(String name,int start){ String find = boundary+EOLN+"Content-Disposition: form-data; name=""+name+"""; int block_start = dataString.indexOf(find,start); if(block_start == -1) throw new IllegalArgumentException("Part named "+name+" not found");*è/ hÿ6xk5// Data block ends here: int block_end = dataString.indexOf(EOLN+DOULE_DASH+boundõbohÿ6xk5EOLN.length()); if (block_end == -1) throw new IllegalArgumentException("Part named "+name+" not found"); // Parameter block should contain an empty line between index and index1 int data_start = dataStriýnehÿ6xk5_LINE,block_start) + EMPTY_LINE.length(); if (data_start == -1 || data_stñ!&thÿ6xk5hÿ6xk5>throw new IllegalArgumentException("Pé)nmxk5˜

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.