Hw1_extra _practice AKey

# Hw1_extra _practice AKey - Homework 1 Key and has 100 times...

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Homework 1 Key 1 Problem 1 . Given : The beaker below is divided by a semi-permeable membrane that is only permeable to K+ and has 100 times the amount of potassium acetate on the left side than on the right side. (A) Under these conditions, what will the voltage of the right side of the beaker be relative to the left side? To solve this problem, we will calculate the Nernst potential for potassium (We use the Nernst equation rather than the GHK equation because potassium is the only permeable ion). We used ln ([K] l /[K] R ) because we always put the concentration of the side for which we want to know the voltage in the denominator. At 17 o C for positive monovalent ions (like potassium), 2.3RT/ZF=57.5 mV, and the equation reduces to the following: E R = E k = 57.5 mV *2 = 115 mV You now perform an experiment where you want to set (clamp) the voltage of the right side at a specific value. In this particular experiment we are going to clamp the right side of the beaker at 17 mV relative to the left side of the beaker. You do this by putting two electrodes in the right side of the beaker (see figure to left). One electrode measures the voltage. This value is fed back to electronic equipment, which then calculates how much current must be injected through the second “clamping” electrode to hold the right side voltage at 17 mV relative to the left side. Before beginning the experiment, the right side is sitting at the voltage you calculated in part (A). You then turn the voltage clamp on and within a millisecond the clamp is generating current which clamps the right side's voltage at 17 mv. (B) Shortly after clamping at +17 mV (before a large change in the concentration gradient can occur), what is the magnitude and direction of the electrical force acting on potassium ions? The electrical field force drives positive potassium ions to the left side of the beaker because the right side of the beaker is clamped at (positive) +17 mV relative to the left. Under these conditions, positive potassium ions will be repelled by the positive voltage and driven away from the right (towards the left) side of the beaker by electrical forces. The size of the electrical force can be calculated with F electrical = |zV|. So F electrical = |1*17| = 17 arbitrary force units (afu). (C) Shortly after clamping at +17 mV (before a large change in the concentration gradient can occur), what is the magnitude and direction of the chemical force acting on K + ? The chemical force will tend to drive K + ions from the high concentration of K + on the left side to the lower concentration of K + on the right side of the beaker. The size of the chemical force can be calculated with |F chemical | = 57.5* |log([I out ]/[I in ])| @17 ° C or for the beaker |F chemical | = 57.5* |log([I Right ]/[I Left ])| = 57.5* |log (100)| = 115 afus. Note: in the chemical force equation it does not matter if you

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## This note was uploaded on 12/06/2009 for the course NPB NPB taught by Professor Bales during the Fall '09 term at UC Davis.

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Hw1_extra _practice AKey - Homework 1 Key and has 100 times...

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