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Unformatted text preview: Name: ____________________________ Net ID: __________________ Section: _______ EXAM I BIOCHEMISTRY 462B - SPRING 2008 Write your name and UA Net ID on ALL pages! This exam has 10 questions on 6 pages (including this face page) and a total of 100 points. Read all questions carefully Show your work on mathematical problems, including units Answer briefly in the space provided Explain all answers (Filling all available space is NOT necessary for full credit!). The degree of difficulty tends to increase as question number goes up. Useful Constants: Faraday Constant (F or ) = 96,480 J / V mol Gas Constant (R) = 8.315 J / mol K Useful Equations: " G = " # G + RT ln [ Product ] [ Substrate ] G = - n E Page 1 of 6 Name: ________________________ Net ID: ______________________ Section: ________ Page 2 of 6 Question 1. (15 points) Hydrolysis of ATP to ADP and P i has a standard state G of 30.5 kJ/mol. The concentrations of ATP, ADP, and P i differ within different cell types. Therefore, the release of free energy by ATP hydrolysis will vary with cell type. (A) Using data in table below, calculate the G for ATP hydrolysis in liver, muscle, and brain. (B) The calculations should show that the G for ATP hydrolysisis the least negative in liver. This will make the G of the glucokinase and phosphofructokinase reactions in glycolysis less negative. Please provide a very short explanation for why this may be advantageous in liver but not in muscle or brain. ATP (mM) ADP (mM) P i (mM) Liver 3.5 1.8 5.0 Muscle 8.0 0.9 8.0 Brain 2.6 0.7 2.7 Answer. (A) (9 points) " G = " # G + RT ln [ ADP ][ P i ] [ ATP ] Using M (the correct way): ln [ ADP ][ P i ] [ ATP ] of liver = 5.963 " G = 45.9 kJ/mol of muscle = 7.013 " G = 48.6 kJ/mol of brain = 7.227 " G = 49.1 kJ/mol Using mM (not correct in the strict sense, but full credit will be given): ln [ ADP ][ P i ] [ ATP ] of liver = 0.9445 " G = 28.1 kJ/mol of muscle = 0.1054 " G = 30.8 kJ/mol of brain = 0.3189 " G = 31.3 kJ/mol (B) (6 points) Unlike brain and muscle, liver carries out gluconeogenesis to supply glucose to the rest of the body. Higher G values in the glucokinase and phosphofructokinase reactions mean it is easier to achieve a negative G during gluconeogenesis. Question 2. (10 points) Pantothenic acid (vitamin B5) is a component of coenzyme A. Individuals suffering from deficiency of pantothenic acid are easily fatigued. (A) Can you provide an explanation by pointing out which metabolic pathway is affected? (B) What metabolite would build up in blood as a direct result of insufficient CoA?...
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This test prep was uploaded on 04/02/2008 for the course BIOC 462b taught by Professor Grimes,tsao,borque during the Spring '08 term at University of Arizona- Tucson.
- Spring '08