DOC PREVIEW
UA BIOC 460 - Exam Guide

This preview shows page 1-2 out of 6 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

WRITE YOUR NAME AND I.D. NUMBER LEGIBLY ON EVERY PAGE – PAGES MAY BE SEPARATED FOR GRADING! CHECK TO BE SURE YOU HAVE 6 PAGES, NAME (print): INCLUDING COVER PAGE. 4-digit course ID # I swear/affirm that I have neither given nor received any assistance with this exam. Signature: Date: BIOCHEMISTRY 460 FOURTH HOUR EXAMINATION FORM A (YELLOW) ANSWER KEY May 5, 2004 A CALCULATOR MAY BE USED ON THIS EXAM. IF YOUR CALCULATOR IS PROGRAMMABLE, ALL MEMORY BANKS MUST BE CLEARED BEFORE YOU ENTER THE EXAM, AND WE MAY CHECK & CLEAR CALCULATORS DURING THE EXAM. SHOW YOUR WORK FOR ALL CALCULATIONS, AND BE SURE TO STATE UNITS OF ANY NUMERICAL ANSWERS. If the reasoning, calculations, or answer are shown anywhere other than in the space provided, make a note in the space provided and put answer on back of SAME PAGE so the grader for that page will have it. USEFUL CONSTANTS: R (gas constant) = 8.315 x 10–3 kJ•mol–1•Kelvin–1 F (Faraday constant) = 96.48 kJ•Volt–1•mol–1 If temperature = 25 °C, absolute temperature T = 298 ºKBiochemistry 460, Exam #4 Form A ANSWERS NAME May 5, 2004 4-digit course ID# page 2 1. (3 pts) What metabolic intermediate is both a product of the glycogen phosphorylase reaction and is also involved in glycogen synthesis? Glucose-1-phosphate (glucose-1-P) 2. (7 pts) A. (4 pts) The conversion of phosphoenolpyruvate to pyruvate in the cytosol is an exergonic reaction catalyzed by the enzyme pyruvate kinase (ΔGº' = -31.4 kJ/mol). Since the reverse of this reaction is highly unfavorable (ΔGº' = +31.4 kJ/mol), explain how it is possible that the conversion of pyruvate to phosphoenolpyruvate in the gluconeogenic pathway is found to be exergonic (ΔGº' = -22.6 kJ/mol)? The conversion of pyruvate to phosphoenolpyruvate utilizes two different enzymes (pyruvate carboxylase and phosphoenolypyruvate carboxykinase), each of which require hydrolysis of a high energy phosphate bond in a coupled reaction with ATP (GTP). B. (3 pts) Does this part of the gluconeogenic pathway occur in the mitochondrial matrix or the cytosol? One of the reactions takes place in the mitochondria (pyruvate carboxylase) and one occurs in the cytosol (phosphoenolypyruvate carboxykinase). 3. (5 pts) A. (3 pts) Name the gluconeogenic enzyme that catalyzes the bypass reaction in glycolysis catalyzed by phosphofructokinase (i.e., what gluconeogenic enzyme opposes phosphofructokinase). Fructose 1,6-bisphosphatase B. (2 pts) Does AMP activate (+) or inhibit (-) this gluconeogenic enzyme? AMP inhibits fructose 1,6-bisphosphatase 4. (5 pts) Von Gierke's disease is due to a deficiency in the enzyme glucose-6-phosphatase. Why do individuals with this disease release a small amount of glucose into the blood after injection with a high dose of glucagon? Glucagon stimulates glycogen breakdown and the product of debranching enzyme is free glucose which is released into the blood.Biochemistry 460, Exam #4 Form A ANSWERS NAME May 5, 2004 4-digit course ID# page 3 5. (4 pts) Glycogen phosphorylase b activity is regulated by: (circle the one best answer) a. phosphorylation of serine residues in the dimer, shifting the equilibrium to the R state. b. low energy charge leading to allosteric activation, shifting the equilibrium to the R state. c. dephosphorylation of serine residues in the dimer, shifting the equilibrium to the T state. d. high ATP levels leading to allosteric inhibition, shifting the equilibrium to the T state. e. All of the above. f. None of the above. 6. (6 pts) Fill in the blanks in the two sentences below with the most correct words: Epinephrine (hormone) activates a signaling pathway in liver cells controlled by protein kinase A (enzyme) that results in the phosphorylation of enzymes leading to the stimulation of glycogen degradation (process). Insulin (hormone) signaling activates a pathway in liver cells controlled by protein phosphatase 1 that results in dephosphorylation (modification) of enzymes leading to the stimulation of glycogen synthesis (process). 7. (5 pts) When the energy charge in liver cells is low, and oxaloacetate levels are limiting, what is the fate of excess mitochondrial acetyl CoA? Name a physiological condition that causes this. Excess acetyl CoA is converted to ketone bodies. Ketone bodies are produced as a result of starvation (or diabetes). 8. (5 pts) By what general mechanism of enzyme regulation does citrate control acetyl CoA carboxylase activity? Does citrate activate or inhibit acetyl CoA carboxylase ? Explain your answer in terms of metabolic needs of the cell. Citrate activates acetyl CoA carboxylase by the mechanism of allosteric control. This makes sense because when cytosolic citrate levels are high, it signals that the mitochondrial matrix is not utilizing excess citrate (or acetyl CoA) and therefore it is time to store energy as fat.Biochemistry 460, Exam #4 Form A ANSWERS NAME May 5, 2004 4-digit course ID# page 4 9. (5 pts) As shown in the diagram at the right, Olestra is a fat substitute composed of a sucrose molecule covalently linked to several fatty acids. A. (2 pts) Why is this fat substitute considered a zero calorie fat molecule? Because pancreatic lipase does not recognize Olestra as a substrate and it passes through the body without giving up its fatty acid components for energy storage. B. (3 pts) Why does Olestra prevent the dietary absorption of fat soluble vitamins? Fat-soluble vitamins associate with the hydrophobic fatty acids and are excreted. 10. (5 pts) The hump of a camel contains over 10 kilograms of triglycerides, much of which is metabolized during long journeys across the desert. Explain why the camel's hump permits it to survive for extended periods of time without drinking water. Oxidation of fatty acids generates large amounts of H2O. 11. (5 pts) What is the biochemical basis for improved anti-inflammatory action and reduced stomach bleeding in individuals taking the COX-2 inhibitor Celebrex as compared to aspirin or ibuprofen? Celebrex has a high affinity for the active site of COX-2, but is too large to bind to COX-1. Aspirin is a relatively small molecule that binds to and inactivates both COX-1 and COX-2. COX-1 is an enzyme that protects the stomach lining from damage due to the acidic


View Full Document

UA BIOC 460 - Exam Guide

Documents in this Course
Exam 4

Exam 4

3 pages

Enzymes

Enzymes

4 pages

Enzymes

Enzymes

14 pages

Exam 4

Exam 4

6 pages

Enzymes

Enzymes

10 pages

Enzymes

Enzymes

5 pages

Enzymes

Enzymes

19 pages

Load more
Download Exam Guide
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam Guide and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam Guide 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?