{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

g&g_ch_13_p_1-4 - G&G Ch 13 Problem 1(pg 438 This...

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

View Full Document Right Arrow Icon
G&G Ch. 13 Problem 1 (pg. 438) This problem requires simple manipulation of the Michaelis-Menten equation in order to determine the ratio we’re being asked for: V max [S] K m + [S] = v 0 [S] K m + [S] = v 0 V max We are given that [S] = 4K m . Use this to substitute for [S]: 4K m K m + 4K m = v 0 V max 4K m 5K m = v 0 V max = 0.8 v 0 V max Problem 2 (pg. 438) Let’s use the Michaelis-Menten equation again. We’ll simply plug values in to the equation to get the v 0 . Remember to convert all your units to a common form!!! V max [S] K m + [S] = v 0 (0.1mmol/mL•s)(0.02mmol/mL) 0.002mmol/mL + 0.02mmol/mL = v 0 = 0.091mmol/mL•s = 91μmol/mL•s v 0
Image of page 1

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

View Full Document Right Arrow Icon
G&G Ch. 13 (cont’d) Problem 3 (pg. 438) Remember that K s refers only to the rate constant for dissociation of the ES complex in to E and S, while K m refers to the rate constant for dissociation of the ES complex into both E and S and E and P: k –1 k 1 = K s k –1 + k 2 k 1 = K m ES E + S k –1 k 1 ES k –1 k 1 E + P E + S k 2 Use these equations to calculate K s and K m : k –1 k 1 = K s = 1x10 3 s -1 7x10 7 M -1 s -1 = 1.43x10 -5 M 1x10 3 s -1 + 2x10 4 s -1 7x10 7 M -1 s -1 = 3x10 -4 M
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern