To determine: The reason behind not being the concentration of finding Km and Vmax on the constant spaced x-axis of the graph.
Introduction: Enzymes are “biocatalysts” which takes part in a process and alters the
To suggest: An improvement in the graph.
Introduction: Enzymes are “biocatalysts” which takes part in a process and alters the rate of reaction. The enzymes work on the entity known as “substrates” and after having interaction with the substrate is converted into the “product.” The Km of an enzyme can be described as the amount of substrate required for the purpose of saturation. Whereas, when the rate of reaction is at the top, the condition is known as Vmax.
Want to see the full answer?
Check out a sample textbook solutionChapter 6 Solutions
Becker's World of the Cell (9th Edition)
- The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.arrow_forwardThe KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?arrow_forwardIn an uncatalyzed reaction, KF(uncatalyzed) = 2.5 X 10-8/s and Keq is 5 X 10². For the same reaction in the presence of an enzyme, KF(catalyzed) = 2.5 X 10²/s. Remember that Keq = KF/KR = k₁/k-1. (a) What is the rate enhancement of the catalyzed reaction in the forward direction? (b) What is the value for KR(uncatalyzed)?arrow_forward
- The KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while the KM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of the following statements are likely true? Chymotrypsin has a higher apparent affinity for Substrate A. The Vmax would be higher in the presence of Substrate A. The kcat would be higher with Substrate A. The V max would be higher in the presence of Substrate B. Two of the above are true.arrow_forwardThe enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.arrow_forwardWhich of the three graphs below has the largest kcat? Explain. Which of the three graphs below has the tightest subtrate binding? Explain. Which of the three graphs has the largest specificity or efficiency constant? Explain. Not all graphs have to be used as answers. Some graphs may qualify for more than one question. Explain.arrow_forward
- The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.arrow_forwardA biochemist is trying to determine the type of proteases they have isolated from walrus blubber. The three proteases and their relative activities in the presence of the indicated non-specific irreversible inhibitors are shown in the table below: Protease a Protease B Protease y Given this data, please answer the following question: The catalytic site of Protease ß contains an important: C R H + lodoacetate Normal Activity Normal Activity No Activity U + Tetranitromethane Normal Activity No Activity Normal Activityarrow_forwardThe effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.arrow_forward
- Proline racemase catalyzes the conversion between L-proline and D-proline. The Km and kcat for this reaction are 0.15 M and 550/sec respectively. If the enzyme concentration is 1.45 X 10-5 mmole/ml what is the Vmax of this reaction?arrow_forwardFor a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.arrow_forwardDuring the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity-versus-substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. This is an allosteric enzyme and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly. This is an enzyme that displays Michaelis-Menten kinetics and you purify away an inhibitor. This is an allosteric enzyme and during purification you purify away an activator. This is an allosteric enzyme displaying a double-displacement mechanism and during purification you purify away one of the substrates: This is an enzyme that displays Michaelis-Menten kinetics, and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly.arrow_forward
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education