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Exam 1 2009-key

# Exam 1 2009-key - Biology 222 Page 1 Exam 1 Fall 2009...

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Biology 222 Exam 1, Fall 2009 Name_______KEY _______ _______ Page 1 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 HELPFUL At 20 ° C Otherwise, where R = 8.314 J mol -1 K -1 , F = 96,492 C/mol, and T = ( ° C + 273) K or alternatively, terms can be rearranged as i x =g x (V m -E x ) g= 1/R V=IR means “proportional to” V x = V o * e (-x/ λ ) e 2.7 Conduction velocity: Table of log 10 values 1) 20 points __________ 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 2009 Name_______KEY _______ _______ Page 2 Leave this page blank
Biology 222 Exam 1, Fall 2009 Name_______KEY _______ _______ Page 3 1) (20 pts) Information carries over for parts (a) – (c) a) (6 pts) A space mission to Mars has just returned to Earth but the lead scientist is unable to continue her work. Portions of her notebook have been damaged and you are responsible for reconstructing the data. Notation above the table says “Properties of Martian neuron at 45 °C.” Calculate the equilibrium potentials for each ion and give the proper units. Show your work. = = =63(-2)=-126 mV =-63(1)=-63 mV =63(2)=+126 mV =63(-3)/(-2)=+94.5 mV =63(3)/2=+94.5 mV =63(-1)/2=-31.5 mV b) (4 pts) If the Martian neuron is only permeable to 2 ions and g x /g y = 1, which 2 ions would make its V m = 0 mV? Which 2 ions would make its V m = +31.5 mV? Refer to the table in part (a) and show your work. Na + and K + make V m = 0 mV Ca 2+ and Mg 2+ or SO 4 -2 and Mg 2+ make V m = +31.5 mV K + and Cl - also make V m = +31.5 mV c) (2 pts) The notebook says that g K /g Mg = 0.25. What is the V m of the Martian neuron if it is permeable only to K + and Mg 2+ ? Show your work. = = 0 mV Ion Outside Inside E x Na + 1 100 -126 mV K + 130 1.3 +126 mV Ca 2+ 4 0.004 +94.5 mV Mg 2+ 15 150 -31.5 mV Cl - 150 15 -63 mV SO 4 -2 0.2 200 +94.5 mV

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Biology 222 Exam 1, Fall 2009 Name_______KEY _______ _______ Page 4 Information carries over for parts (d) – (g) d) (3 pts) A newly discovered single-cell Earth organism, Chlorita , has the following ionic composition in its native environment at 60 °C. Calculate the equilibrium potentials for each ion and give the proper units. Show your work. Ion Outside Inside E x Na + 5 50 -66 mV Cl - 100 1 -132 mV K + 95 .95 +132 mV = = = 66(-1) = -66 mV =66(2) = +132 mV = -66(2) = -132 mV e) (2 pts) For Chlorita at rest, P Cl = 0.25P Na and P K = 0. What is its V m at 60 ° C? Show your work. = -76.2 mV f) (2 pts) If only chloride channels were open at rest in Chlorita , what would be the direction of ion and current flow? Show your work. V m – E Cl = driving force -76.2 – (-132) = +55.8 mV Cl - flow is inward Current flow is outward g) (1 pt) You apply a chloride-channel blocking toxin to Chlorita . What is the value of V m ? At rest, V m for Chlorita depends on P Cl and P Na . Using the Goldman equation, P Cl now is 0 after toxin treatment so V m = E Na
Biology 222 Exam 1, Fall 2009 Name_______KEY _______ _______ Page 5 2) (20 pts) a) (7 pts) Identify the parts of a neuron.

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Exam 1 2009-key - Biology 222 Page 1 Exam 1 Fall 2009...

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