Midtm2007ANSW - METABOLIC BIOCHEMISTRY BIBC102 Immo E....

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
METABOLIC BIOCHEMISTRY BIBC102 Winter 2007 Immo E. Scheffler MIDTERM EXAM All answers are to be written into the Blue Book. Leave the first inside page blank for scoring. There are 10 questions. Make sure that each answer is clearly identified with the question number at the top or left side of the page. Useful Information: Avogadro's number: 6.02 x 10 23 molecules / mole 1 Faraday = 96,494 Coulomb / mole = 96,494 Joules / Volt / mole Gas constant (R) = 8.31 Joules K -1 mol -1 = 1.987 cal K -1 mol -1 = 0.082 liter atm K -1 mol -1 1 calorie = 4.184 Joules ********************************************************************************** QUESTION 1 a) (5 points) Give the complete structural formula for NAD + ; show also the reduced form with only the “business end” of the molecule. b) (5 points) Illustrate the use of this cofactor with the final reaction in glycolysis under anaerobic conditions in mammals (structural formulae required for the two substrates)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
c) (4 points) If [1- 14 C] pyruvate is given to a yeast extract (as used for fermentation), would you expect to be able to obtain radioactive ethanol? Explain your answer with the help of structural formulae In yeast the pyruvate is converted to carbon dioxide and ethanol. The CO 2 is derived from the carboxyl group of pyruvate (the #1 carbon); thus, no radioactive ethanol would be produced d) (6 points) You have learned about the participation of a Schiff base in the aldolase reaction. Illustrate the formation of a Schiff base in enzyme catalyzed reactions; use the specific example encountered in glycolysis (partial credit for a more generic explanation)
Background image of page 2
QUESTION 2 A kinetic scheme for an enzyme catalyzed reaction has been formulated by Michaelis and Menten. It makes an assumption about the steady state concentration of an enzyme-substrate intermediate. An expression for the rate (v) of the reaction can be derived as follows: k 2 [E total ] [S] V o = -------------------------- (k 2 +k -1 )/k 1 + [S] a) (4 points) show the reaction sequence and assign rate constants to the elementary steps k 1 k 2 E + S ES E + P k -1 b ) (1 point) what is the expression for K m K m = (k -1 + k 2 )/k 1 c) (2 points) what is the expression for v max v max = k 2 x [E] total where [E] total is the total enzyme concentration d ) (5 points) formulate the expression for the Lineweaver-Burk plot and show a typical plot for an enzyme that exhibits Michaelis Menten kinetics (label all axes and intercepts) 1/v o = K m /v max x 1/[S] + 1/v max
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
e) (6 points) show with the help of Lineweaver-Burk plots how one can distinguish competitive from uncompetitive inhibition In competitive inhibition (left) the intercept at 1/[S] = 0 is the same, i.e. v max does not change at infinite substrate concentrations; in un competitive inhibition (right) the inhibitor slows the reaction even at infinite substrate concentrations (the intercept changes) but the slope does not QUESTION 3 (4 points) A somewhat inexperienced researcher wants to make radioactive glucose-6-phosphate from
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/30/2010 for the course BIOLOGY BIBC 102 taught by Professor Scheffler during the Spring '10 term at CSU Northridge.

Page1 / 12

Midtm2007ANSW - METABOLIC BIOCHEMISTRY BIBC102 Immo E....

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online