Exam 1 2010-key - Biology 222 Page 1 Exam 1, Fall 2010...

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Biology 222 Exam 1, Fall 2010 Name_______KEY ______________ Page 1 First Last This exam is double-sided. Put your first and last name on every sheet! Every sheet missing your name will lose 1 point. Answer all questions on pages 3-12. E QUATIONS YOU MAY FIND USEFUL At 20 ° C Otherwise, where R = 8.314 J mol -1 K -1 , F = 96,492 C/mol, and T = ( ° C + 273) K V m = 2.303 RT F log 10 P Na [ Na ] o + P K [ K ] o P Na [ Na ] i + P K [ K ] i V m = g x E x + g y E y g x + g y or alternatively, terms can be rearranged as V m = g x g y E x + E y g x g y + 1 i x =g x (V m -E x ) g= 1/R V=IR means “proportional to” V x = V o e ( x / λ ) e -2 =0.14, e -1 =0.37, e 1 =2.7, e 2 =7.4 Conduction velocity: Table of log 10 values 1) 20 points (N/18)*20 __ 2) 20 points __________ 3) 20 points __________ 4) 20 points __________ 5) 20 points __________ Total __________ log 10 [ out ]/[ in ] -3 0.001 -2 0.01 -1 0.1 0 1 1 10 2 100 3 1000
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Biology 222 Exam 1, Fall 2010 Name_______KEY ______________ Page 3 First Last 1) (20 pts) Q UESTION 1 IS NOW ONLY WORTH 18 PTS DUE TO PART D ) BEING NULLIFIED . Information carries over for parts (a) – (c) SHOW YOUR WORK! You are studying a neuron at 20 °C with the following intracellular and extracellular ion concentrations. The resting membrane potential is -65 mV. Ion [Outside] mM [Inside] mM Na + 200 20 K + 13 130 Ca 2+ 4 0.004 Mg 2+ 2 2 Cl - 1.5 150 a) (5 pts) Calculate the equilibrium potential for each ion. E Na = 58 + 1 log 10 [200] [20] = + 58 mV E Mg = 58 + 2 log 10 [2] [2] = 0 mV E K = 58 + 1 log 10 [13] [130] = 58 mV E Cl = 58 1 log 10 [1.5] [150] = + 116 mV E Ca = 58 + 2 log 10 [4] [0.004] = + 87 mV b) (5 pts) Determine the direction of ion flow and current flow for each ion. (V m -E Na ) = -65 – (+58) = -123 inward ion flow, inward current flow (V m -E K ) = -65 – (-58) = -7 inward ion flow, inward current flow (V m -E Ca ) = -65 – (+87) = -152 inward ion flow, inward current flow (V m -E Mg ) = -65 – (0) = -65 inward ion flow, inward current flow (V m -E Cl ) = -65 – (+116) = -183 outward ion flow, inward current flow c) (2 pts) You are told that g K /g Ca = 0.5. Calculate the V m of the neuron in part a) if it is permeable only to K + and Ca 2+ V m = g x g y E x + E y g x g y + 1 = (0.5)( 58) + ( + 87) (0.5) + 1 = 58 1.5 = + 38.7 mV
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Biology 222 Exam 1, Fall 2010 Page 4 d) ( 2 pts) What is the most hyperpolarized membrane potential possi ble for the neuron in part a) and at that potential, what ion(s) would have P > 0? T HIS QUESTION HAS BEEN NULLIFIED DUE TO THE V M VALUE GIVEN IN PART A ). Based on the data table, the most hyperpolarized membrane potential possible is -58 mV. At that potential, only K
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Exam 1 2010-key - Biology 222 Page 1 Exam 1, Fall 2010...

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