QUESTION 3In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kt of theinhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures anscientific notation. For example, 0.000023434 is 2.34eQUESTION 4Approximately how many domains are in an 82.5 kDa protein?O a. 2O b.3O C. 4O d. 12.5QUESTION 5How long is an a-helix that contains 36 amino acid residues?6.0 nm5.4 nm78 Mostly clody a nClick Save and Submit to save and submit. Click Save All Answers to save all ansuersP Type here to search

A competitive inhibitor generally reduces the rate of catalysis by decreasing the proportion of…

Answered: QUESTION 3 In the presence of 0.000005…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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You have an initial solution in which you added quantities of “A” and “B” such that there is 4.5 M “A” and 2.5 M “B” and no complex (“AB”) at time 0. After equilibrium, you are able to isolate and quantitate the “AB” complex, and find its concentration is 1.5 M. Given that RT is 0.59 kcal/mol, what is the delta Go’ for the association reaction?
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For the following reaction, 4.91 grams of water are mixed with excess chlorine gas. The reaction yields 12.5 grams of hydrochloric acid.chlorine (g) + water (l) hydrochloric acid (aq) + chloric acid (HClO3) (aq) What is the theoretical yield of hydrochloric acid ? grams What is the percent yield of hydrochloric acid ? %
How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s
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+ Young Saguaro Cacti Under another plant NOT under another plant Total Expected Percentage 35% 65% 100% Expected Number X^2= 11.44 Degree of freedom: 1 p value: <.019 85.75 159.25 245 Observed Number 111 134 245 (245) (35) = 85.75 245-85.75= 159.25 X^2 for associated = ((111-85.75) ^2)/85.75 = 7.44 X^2 for not associated = ((134-159.25) ^2)134 = 4.00 Ex^2 = 7.44 +4.00 11.44 (OBS-Exp) 2/EXP 7.44 4.00 Σ=11.44
8L.16.2
4.2
A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction: HAPPY-SAD The researchers begin to characterize the enzyme. The researchers determined the kcat of the enzyme to be 452 s-1. In a separate experiment with [E;) at 1.5 nM and [HAPPY) at 38 µM, the researchers find that Vo is equal to 320 nM s-1. What is the measured Km of happyase* for its substrate HAPPY? Make sure to include units in the answer.
If the enzyme lactase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, what is its Vmax, Km and the slope on a linewver-burk plot?
make a Gibbs free energy graph as free energy as y-axis a for the transition state as a function of [Gdn HCl] (x-axis) with a line of best fit

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QUESTION 3 In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kt of the inhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures an scientific notation. For example, 0.000023434 is 2.34e