A pka that is unable to bind camp will therefore not

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Unformatted text preview: to bind cAMP. cAMP binding liberates the active version of PKA from its psuedosubstrate. A PKA that is unable to bind cAMP will therefore not be activated upon epinephrine stimulation and glucose 1-phosphate concentration will not increase. ii) The cell contains an unphosphorylatable version of Inhibitor of Phosphoprotein Phosphatase. Inhibitor of Phosphoprotein phosphatase will not inactivate phosphoprotein phosphatase in its unphosphorylated form. Thus phosphoprotein phosphatase will be active even in the presence of active PKA, and glucose 1-phosophate concentration will not increase. iii) The cell contains a mutant version of Gsa that cannot hydrolyze GTP . Gsa that cannot hydrolyze GTP will remain active once turned on. This will constitutively activate adenylyl cylase and all the downstream targets of PKA, which will lead to increased glucose 1-phosphate concentration. 4) a) Given the following values, calculate the electrochemical equilibrium potential of each ion (F=23, 062cal/(mol V), R = 1.987 cal/K mol, T = 293K). K+out = 3 mM Na+out = 117 mM Cl-out = 120 mM 7.06 Spring 2004 K+in = 90 mM PS 2 KEY Na+in = 30 mM 5 of 7 Cl-in = 4 mM E = (RT/zF) ln [ionout/ionin] EK = - 85.9 mV ENa = 34.4 mV ECl = - 85.9 mV b) What does the electrochemical equilibrium potential reflect/what is true at the electrochemical equilibrium potential? At the electrochemical equilibrium potential, the concentration gradient for the ion is balanced by the electrical gradient. Hence, there is not net flux of ions across the membrane. That is, the rate (notice rate, not concentration) of movement of the ion in both directions is equal. c) If one reduces the magnitude of the K+ concentration gradient acro...
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This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.

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