• You are studying a disease in which muscle becomes weak due to difficulty in generating action potentials.

• Shown on the left are the responses of a control muscle fiber and a diseased muscle fiber to various levels of depolarizing ( positive ) current injection.

• On the right are shown the responses of the same two fibers to a 10 nA ( 10-9A ) injection of hyperpolarizing ( negative ) current.

Data

https://docs.google.com/spreadsheets/d/1BXYugUqxdMbJOzF7KYZInowUu9bOUavi2shYJFBz6p4/edit?usp=sharing

1. Determine the resting membrane potential of each fiber. Also graph the current versus voltage plot for each fiber

   • Control RMP = $-70mV$$-70\ mV$

   • Disease RMP = $-90mV$$-90\ mV$

   

2. Calculate the input resistance of each fiber.

   • Let Control $V_{max}=-130mV$$V_{max} = -130\ mV$

   $$V=I\ast R$$
   $$\mathrm{\Delta }{V}_{\text{control}}=(-130mV)-(-70mV)=-60mV$$
   $$R_{\text{control}}=\frac{-60\ast {10}^{-3}V}{-10\ast {10}^{-9}A}=\frac{-6.0\ast {10}^{-2}}{-1.0\ast {10}^{-8}}=6.0\ast {10}^{((-2)-(-8))}=6.0\ast {10}^6\mathrm{\Omega }$$

   • Let Disease $V_{max}=-120mV$$V_{max} = -120\ mV$

   $$\mathrm{\Delta }{V}_{\text{disease}}=(-120mV)-(-90mV)=-30mV$$
   $$R_{\text{disease}}=\frac{-30\ast {10}^{-3}V}{-10\ast {10}^{-9}A}=\frac{-3.0\ast {10}^{-2}}{-1.0\ast {10}^{-8}}=3.0\ast {10}^{((-2)-(-8))}=3.0\ast {10}^6\mathrm{\Omega }$$

   ---

   • Then from the graphs , the slope is the same thing as conductance

   • conductance = opposite of resistance

   $$R_{\text{control}}=\frac{1}{1.67\ast {10}^{-7}}=5.988023952095808383233532934\ast {10}^6\mathrm{\Omega }$$
   $$R_{\text{disease}}=\frac{1}{3.33\ast {10}^{-7}}=3.003003003003003003003003003\ast {10}^6\mathrm{\Omega }$$

3. Measure the time constant of each fiber.

   $$-60mV\ast (1-\frac{1}{e})\approx -37.927mV$$
   $$(-70)+(-37.927)=-107.927mV$$
   $${\tau }_{\text{control}}\approx 48\text{milli seconds}$$

   ---

   $$-30mV\ast (1-\frac{1}{e})\approx -18.9636mV$$
   $$(-90)+(-18.9636)=-108.9636mV$$
   $${\tau }_{\text{control}}\approx 25\text{milli seconds}$$

4. Given the difference in resting potential and the response to hyperpolarizing ( negative ) current injection , predict what functional change might account for the reduced excitability of the disease cell.

   • the disease state has an increased amount of potassium conductance at the resting state

   • it has a shorter time constant , so it charges and discharges the membrane faster

   • they probably have upregulated potassium channels somehow