examII answer key Biochemestry

examII answer key Biochemestry - BioSci 1000 exam II...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
BioSci 1000 exam II – Schwacha page 1 Student number ________________ Exam II – Schwacha section 50 points total 10/28/08 1. Turn off and stow all electronic gizmos 2. Put away all incriminating study material and notes 3. Print your name on each page 4. Read each question carefully 5. Answer each question in ink only and write legibly, otherwise no regrade 6. Remember to sign the class list when you turn in your exam. 5. Good luck! Calculators are NOT allowed or required. __________________________________________________________________ Information that may or may not be helpful Boltzmann Constant, k = 1.381 X 10 -23 J/K Avogadro’s number, N =6.022 X 10 23 mol -1 Faraday’s constant, F = 96.48 kJ V -1 . mol -1 Gas constant = R=8.315 j/mol . K 25 o c = 298 K ln10 = 2.3; log 10=1; log 1 = 0 Murphy’s law of thermodynamics: Things get worse under pressure Dobie's Dogma: If you are not thoroughly confused, you have not been thoroughly informed (educated…).
Background image of page 1

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

View Full DocumentRight Arrow Icon
page 2 Multiple guess – circle the best answer 1. For the reaction A B, Δ G' ° = –60 kJ/mol. The reaction is started with 10 mmol of A; no B is initially present. After 24 hours, analysis reveals the presence of 2 mmol of B, 8 mmol of A. Which is the most likely explanation? (2 pts) A) A and B have reached equilibrium concentrations. B) An enzyme has shifted the equilibrium toward A. C) B formation is kinetically slow; equilibrium has not been reached by 24 hours. D) Formation of B is thermodynamically unfavorable. E) The result described is impossible, given the fact that Δ G' ° is –60 kJ/mol. 2. The standard reduction potentials ( E' °) for the following half reactions are given. (2 pts) Fumarate + 2H + + 2e succinate E' ° = +0.031 V FAD + 2H + + 2e FADH 2 E' ° = –0.219 V If succinate, fumarate, FAD, and FADH 2 , all at l M concentrations, were mixed together in the presence of succinate dehydrogenase, which of the following would happen initially ? A) Fumarate and succinate would become oxidized; FAD and FADH 2 would become reduced. B) Fumarate would become reduced; FADH 2 would become oxidized. C) No reaction would occur because all reactants and products are already at their standard concentrations. D) Succinate would become oxidized; FAD would become reduced. E) Succinate would become oxidized; FADH 2 would be unchanged because it is a cofactor, not a substrate. 3. The conversion of 1 mol of fructose 1,6-bisphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of: (2 pts) A) 1 mol of NAD + and 2 mol of ATP. B) 1 mol of NADH and 1 mol of ATP. C) 2 mol of NAD + and 4 mol of ATP. D) 2 mol of NADH and 2 mol of ATP. E)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

examII answer key Biochemestry - BioSci 1000 exam II...

This preview shows document pages 1 - 3. Sign up to view the full document.

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