What can contamination of reagents leads to? Unreliable results Reliable results Reproducible results Good results Catalase is the enzyme found in potatoes and It decomposes hydrogen peroxide (H202) to form water and oxygen gas. H202 + catalase - ----> H2O + 02 +E. The reaction is maintained at temperature 40 degree Celsius and there was a formation of bubbles. Based on the enzymatic reaction given, What do you think is the substance responsible for the formation of bubbles? Oxygen gas Hydrogen gas no reaction, because the enzyme catalase will be deactivated no reaction, because the enzyme catalase will be denatured Other: Based on the Fehling's test video, do you think the sugars ribose (in RNA) and the deoxyribose (in DNA) will both form a positive result (red precipitate) No Yes

What can contamination of reagents leads to? Unreliable results Reliable results Reproducible results Good results Catalase is the enzyme found in potatoes and It decomposes hydrogen peroxide (H202) to form water and oxygen gas. H202 + catalase - ----> H2O + 02 +E. The reaction is maintained at temperature 40 degree Celsius and there was a formation of bubbles. Based on the enzymatic reaction given, What do you think is the substance responsible for the formation of bubbles? Oxygen gas Hydrogen gas no reaction, because the enzyme catalase will be deactivated no reaction, because the enzyme catalase will be denatured Other: Based on the Fehling's test video, do you think the sugars ribose (in RNA) and the deoxyribose (in DNA) will both form a positive result (red precipitate) No Yes

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter24: Biochemistry

Section: Chapter Questions

Problem 35IL

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Similar questions

. Describe the lock-and-key model for enzymes. Why are the shapes of the enzyme and its substrate important in this model? What does it mean to say that an enzyme is inhibited by a particular molecule? What happens if this inhibition is irreversible? Can you think of a situation in which it might be advantageous to be able to inhibit an enzyme?
Compare the lock-and-key and induced-fit models for substrate binding to an enzyme.
The equilibrium 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 substratefrom the active site?
Reaction Velocity (mm/sec) Ma no !! L [1 1 L IM 3M FO Substrate Concentration (M) 1. IN Which statement is true? KM AN ALM 4V The graph above displays the kinetic profiles of the same enzyme catalyzing two different substrates. The red curve displays the higher turnover number (kcat). The red curve displays higher affinity between substrate and enzyme. The blue curve displays competitive inhibition. The red curve displays higher cooperativity,
An enzyme catalyzes a reaction with a KmKm of 7.507.50 mM and a VmaxVmax of 3.00 mM⋅s−1.3.00 mM⋅s−1. Calculate the reaction velocity, v0,v0, for each substrate concentration. a. [S]=4.00mM*s^-1 b. [S]=7.5mM*s^-1
The following initial rate data were obtained on the rate of binding of glucose with the enzyme hexokinase (from yeast) present at a concentration of 1.34 mmol/dm3. [C6H12O6] / (mmol dm-3) 1.00 1.54 3.12 4.02 Initial speed v0 / (mol dm-3 s-1) 5.00 7.60 15.5 20.0 use the data above to calculate the rate constant. k = ________ (mol/L/s, s-1, L/mol/s, or L2/mol2/s). 3 s.f. normal format.
The following initial rate data were obtained on the rate of binding of glucose with the enzyme hexokinase (from yeast) present at a concentration of 1.34 mmol/dm3. [C6H12O6] / (mmol dm-3) 1.00 1.54 3.12 4.02 Initial speed v0 / (mol dm-3 s-1) 5.00 7.60 15.5 20.0 (a) What is the order of the reaction with respect to glucose? (b) use the data above to calculate the rate constant. k = ________ (mol/L/s, s-1, L/mol/s, or L2/mol2/s). 3 s.f. normal format
Reversible Enzyme Kinetics: Consider the following reversible reaction: k, E + S ES P +E k_2 Show that the product will be greatly favored, i.e. [P]/[S] >> 1 only if kık2 >> k.ık.2. State all the assumptions.
A protein catalyzes the following reaction and had a Michaelis-Menten constant of Km=25 mM and a turnover number of 4.0 x 107 s-1. The total enzyme concentration is 0.012 µM and the initial substrate concentration is 4.76 µM. The enzyme has a single active site. Calculate the values of Vmax and V0 for this enzyme. 2H2O2 (aq) -----> 2H2O(l) + O2(g) Vmax= Vo=
:The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics Vo (mmol/min) S (mmol/L) 0.8 3 4 217 325 433 488 1,000 :The Km for this enzyme is approximately 647 mM 4 O mM 1 O mM 1,000 Ⓒ mM 3 O mm 6 O
The following results were obtained for an enzyme-catalyzed reaction Substrate concentration (mmol 1-¹) 5.0 6.67 10.0 20.0 40.0 Initial velocity (μmol l-¹ min-¹) 147 182 233 323 400 Calculate K and Vmax
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-Ċ–NH-Glu CH2 Lys NH2 Ala NO2 Nle H2ON HIV-1 protease Ala Vo…