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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An enzyme is found that catalyzes the reaction X ⇌ Y. Researchers find that the Km for the substrate X is 4 μM, and the kcat is 20 min−1.(a) In an experiment, [X] = 6 mM, and V0 = 480 nM min−1. What was the [Et] used in the experiment?(b) In another experiment, [Et] = 0.5 μM, and the measured V0 = 5 μM min−1. What was the [X] used in the experiment?(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In an experiment with the same [Et] as in (a), but a different [X], an amount of Z is added that reduces V0 to 240 nM min−1. What is the [X] in this experiment?(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.
A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).
An enzymatic reaction was carried out in a batch-operated reactor until exhaustion of the substrate. The formation of the product was monitored over time, obtaining the results shown in Table 1. Answer: a) Knowing that the initial substrate concentration was 0.2 M and that the stoichiometry reaction is: 1S –> 2P, estimate the substrate concentration over time from the data from [P] data in Table 1; b) Determine the parameters of the Michaelis-Menten equation, Vmax and Km, by fitting direct from the hyperbola function; c) Determine the parameters Vmáx and Km, by the linearized equation;
The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?
If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?
An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species. The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme? Explain.
An enzyme with a Km of 3 X 10-4 M was tested at an initial substrate concentration of 10-5 M. In 1 minute, 5% of the substrate was found to be utilized. (a) What percentage of the substrate was utilized in 5 minutes? (b) If the initial substrate concentration had been 8 X 10-7 M, what percentage of the substrate would have been utilized in 5 minutes? (c) Vmax =? (d) How long would it take to utilize 50% of the substrate at 8 X 10-7 M? (e) How long would it take to utilize 75% of the substrate at 10-5 M?
Enzyme A catalyzes the reaction S → P and has a KM of 50 μM and a Vmax of 100 nM s–1. EnzymeB catalyzes the reaction S → Q and has a KM of 5 mM and a Vmax of 120 nM s–1. When 100 μM ofS is added to a mixture containing equal amounts of enzymes A and B, which reaction product (Por Q) will be more abundant after 1 minute of reaction?
You plot a reaction progress curve and find the equation of the line to be y = -0.0961x + 0.7142. Calculate the U, U/μL and U/mL of lysozyme activity in this fraction. The lysozyme activity is done identical to what is indicated in the lab manual.
3. (а) 0.0050 M operate at one-quarter of its maximum rate? At what substrate concentration would an enzyme with a kcat of 30.0 s-1 and a Km of (b) trations [So]: ½ Km, 2 KM, and 10 KM. Determine the fraction of Vmax that would be obtained at the following substrate concen- (c) ( 1 (HIV-1) has been the object of innumerable studies to develop effective chemotherapeutic agents. It has been shown that p6* known as the late assembly protein is an inhibitor of HIV protease. An assay was developed using an artificial polypeptide substrate containing a p-nitrophenylalanine residue at the cleavage point that undergoes a small change in absorption at 295 nm upon bond hydrolysis that could be followed spectrophotometrically. The cleavage of the peptide bond is shown schematically on the right. Results of the assay are given in the table below. The protease of the human immunodeficiency virus- Lys NH2 Ala Nle Arg Ala Val-Nle-NH-CH- C-NH-Glu CH2 Lys NH2 Ala NO2 Nle H2OH NHIV-1 protease Ala Vo…
i) Re-arrange the Michaelis Menten equation so it involves the ratio [S]. Show all steps beginning Km noting any assumptions or required conditions. Km ii) Calculate the ratio [lo for the case when the rate of product formation is 68% of Vmax and the substrate is in great excess. d[P] dt : k₂ with = [E],[S] Km+[S]' [S]o Km iii) Explain, in a few sentences, why the ratio determines the ratio V Vmax V Vmax Begin by explaining the meaning of stating simply "it's the ratio...." is not sufficient. Include in your explanation the factors that effect v and Vmax. Consider what factors make v different from or equal to Vmax. Consider what Km represents concerning processes involving ES. " iv) Calculate KM at 310K at given the following rate constant information: k₁ = 17 s-¹M-1 at 300K with A = 7300 s-¹M-1 K-1₁ 6 s¹ at 300K with A = 14500 s -1 k₂ = 31 s¹ at 300K with A = 600 s-¹
Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)

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