An enzyme is found that catalyzes thereaction X = Y. Researchers find that the Km for the substrate X is 4 µM, and the kcat is 20min-1.(a) In an experiment, [X] = 6 mM, and Vo = 480 nM min¬1. What was the [E] used inthe experiment?(b) In another experiment, [E] = 0.5 µM, and the measured Vo = 5 µM min1. What wasthe [X] used in the experiment?(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme,with an a of 10. In an experiment with the same [E] as in (a), but a different [X], anamount of Z is added that reduces Vo to 240 nM min. What is the [X] in this experiment?

A catalyst that accelerates the chemical reaction occurring in the cells is referred to as an…

Answered: An enzyme is found that catalyzes the…

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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Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme (lactase). (enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM) 0.125 per minute 0.25 per minute 0.50 per minute 1.25 per minute 5.0 per minute
You run an experiment using 0.5 µM of an enzyme. The kinetic parameters for your enzyme are kcat = 1800 s-1 and KM = 25 µM. (a) If k-1 is 240 s -1, what is the value of k1? (Report your answer in units of M-1 s -1) (b) What is Vmax for the enzyme (Don't forget to include units in your answer)? (c) You set up an experiment with the following initial concentrations of substrate. Calculate the rate of the reaction at each substrate concentration (report your answer in units of µM sec-1 [S](μM) 1 100 1000 [S][μM) 1 100 1000 v (μMsec ¹) (d) You now add 10 nM of a competitive inhibitor with a KI = 2 nM. What is the rate of the reaction (v) for the following substrate concentrations and how does it compare to the rate without inhibitor (VO)? Vinhib (uMsec-¹) Vinhih/V
Consider the two reactions below. They are fast and have reached equilibrium by the end of the time shown. At time zero (t = 0) which is true? [Product] (M) I 2 1 Time The AG of reaction 1 is greater (more positive) than the AG of reaction 2. The AG of reaction 1 is less (more negative) than the AG of reaction 2. ○ The AG of reaction 1 is equal to the AG of reaction 2.
You set up the following two tubes: Tube 1: 1mL of alkaline phosphatase + 4mL of 400 µM p-NPP Tube 2: 1mL of alkaline phosphatase + 4mL of 200 µM p-NPP Which tube would have a faster reaction rate? Explain your reasoning (why does concentration of p-NPP matter? Relate it back to week 4 lab concepts). Week 4 lab concepts: - concentration gradient - differentially permeable membrane - solute, solvent, diffusion, osmosis, hypotonic, isotonic, hypertonic, molarity, osmolarity
Two experiments were performed with the enzymeribonuclease. In experiment 1 the effect of increasingsubstrate concentration on reaction velocity was measured.In experiment 2 the reaction mixtures were identical tothose in experiment 1 except that 0.1 mg of an unknowncompound was added to each tube. Plot the data accordingto the Lineweaver-Burk method. Determine the effect of theunknown compound on the enzyme’s activity. (Substrateconcentration is measured in millimoles per liter. Velocity ismeasured in the change in optical density per hour.)
you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?
For an enzyme with a substrate Km of 0.001 M. a) What substrate concentration for an initial rate equal to one-half maximaly velocity? and b) for an initial rate equal to 0.9 maximal velocity> * a) [S} = 0.001 M, b) [S] = 0.009 M a) [S} = Km, b) [S] = 0.09 M a) [S] = 0.01 M, b) [S] = 0.9 M a) [S] = 1 M, b) [S] = 9 M
The commercial sample of carboxyl esterase contained 15.0 mg of enxyme in 0.30 mL of suspension. In your experiment, you add 0.090 mL of this suspension to the sample tube that contains a total volume of 3.0 mL. How many mg of carboxyl esterase enzyme do you have in the sample tube?
Concentrations in biochemical systems are often very dilute. Consequently, scientific notation and logarithms are often used to express concentrations. In scientific notation, numbers are expressed as coefficient x 10" To convert a number to scientific notation, proceed as follows: 1. Move the decimal place so that there is one digit in front of the decimal. 2. Account for the moved decimal in the value of x. If the decimal moved to the right, x is negative; if it moved to the left, x is positive. A logarithm is basically an exponent. Unless otherwise indicated, a logarithm is the a of 10". The numbers after the decimal point are significant; the number before the decimal just identify the location of the decimal point for the number. Notice that it is easy to estimate a logarithm from scientific notation; it's the exponent! Logarithms are commonly used to express the concentration of H. The pH is defined as pH= log (In), where the base number is 2.303. The same general rules as logs…
in food microbiology, how do you compute for concentration (M) and absorbance (A) of peroxidase activity on hydrogen peroxide at different pH values? (wavelength used = 415nm). can you please explain the calculations thank you if given: Temperature = 75 C Molar absorptivity coefficient = 10.5 Path length (cm) = 2 I = transmitted light = 0.45
Suppose you were running out of substrate (PNPP) and decided to reduce the concentration used in your reactions (25 µM, 50 µM, 100 µM, 200 µM, 300 µM) by a factor of 100. How might this affect your kinetic analysis?
In a alkaline phosphatase kinetics experiments a studnet first made a standard curve with different amounts of PNP. In a reaction carried out with 10nM alkaline phosphatase and 100 uM PNPP substrate the student measured an absorbance value of 0.76 at 410nm after 30 seconds of reaction. Assuming the 30 second time point is in the lienar range of the reaction and the substrate has not been depleted at this time point, calculate the Vo at 100uM PNPP (in units of concentration per time).Beer's law formula given is y=0.0323x (y axis is Absorbance) (x Axis is PNP uM)

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