Exam 2 - Fal08 - BCH 3023 Fall 2008 Exam 2, Form C Name:...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
BCH 3023 – Fall 2008 Exam 2, Form C 1 Name: ANSWER KEY In class, we discussed one method to linearize the Michaelis-Menton equation. There are other methods to do this, one being an Eadie-Hofstee plot. Given the Eadie-Hofstee plot below, answer questions 1 and 2. This plot might also help with question 3. 0 20 40 60 80 100 0 10 20 30 40 50 Eadie-Hofstee Plot of Enzyme Kinetic Data v/[S], sec -1 Rearrangement of Michaelis-Menten Equation : v = V MAX [S]/(K M + [S]) v (K M + [S]) = V MAX [S] (v /[S])(K M + [S]) = V MAX (v /[S])(K M ) + v = V MAX v = -(v /[S])(K M ) + V MAX Therefore: Slope = - K M = 2.0 mM “Y” - intercept = V MAX = 100 mM/sec “X”- intercept = V MAX /K M = 50 sec -1 1. What is the value for the K M ? a. 20 mM b. 20 μ M c. 2.0 mM d. 2.0 μ M e. 0.2 mM
Background image of page 1

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

View Full DocumentRight Arrow Icon
BCH 3023 – Fall 2008 Exam 2, Form C 2 2. What is the V MAX value? a. 1.0 sec/mM b. 1.0 mM/sec c. 2.0 mM/sec d. 100 mM/sec e. 1000 mM/sec 3. What is the pattern for a pure noncompetitive inhibitor on an Eadie-Hofstee plot? a. A series of parallel lines. b. A series of lines that intersect on the v/[S]-axis (the “X”-axis). c. A series of lines that intersect on the v-axis (the “Y”-axis). d. A series of lines with different slopes that do not intersect on either axis. e. A series of lines that yield different values of K M,app Pure non-competitve inhibition has no effect on the K M and V MAX decreases. Thus, the slope does not change while the intercepts on both axes decrease as [I] increases. 4. In the binding of O 2 to myoglobin, the relationship between the concentration of O 2 (or the partial pressure of O 2 ) and the fraction of binding sites occupied, θ , can best be described as: a. sigmoidal b. linear with a negative slope c. linear with a positive slope d. hyperbolic e. concave down 5. A metabolic pathway proceeds according to the scheme R S T U V W. A regulatory enzyme, X, catalyzes the first reaction in the pathway. Which of the following is most likely correct for this pathway? a. Either metabolite U or V is likely to be a positive modulator, increasing the activity of X. b. The first product, S, is probably the primary negative modulator of X, leading to feedback inhibition. c. The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition. d. The last product, W, is likely to be a positive modulator, increasing the activity of X. e. The last reaction will be catalyzed by a second regulatory enzyme.
Background image of page 2
BCH 3023 – Fall 2008 Exam 2, Form C 3 6. You discover a mutant form of hemoglobin, Hb mut and characterize how Hb mut binds O 2 . Your studies yield a Hill number, n H , of 1.8 for the binding of O 2 to Hb mut . Which of the following is true about Hb mut , based on this result? a.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/14/2010 for the course BHC 3023 taught by Professor Harmen during the Spring '10 term at University of South Florida - Tampa.

Page1 / 9

Exam 2 - Fal08 - BCH 3023 Fall 2008 Exam 2, Form C Name:...

This preview shows document pages 1 - 4. Sign up to view the full document.

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