1. A Lineweaver-Burk Plot is shown below.3025Curve Ay = 3.1207x + 2.49782015Curve By = 1.0003x + 2.3602105-3157111/[Catechol] (mM1)With these curves, determine the following enzyme parameters. Show all pertinent solutions.a. Km of Curve A and Curve Bb. Vmax of Curve A and Curve Bc. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curverepresents the effect of an inhibitor? Why do you say so?d. What type of inhibition is exhibited by your answer in question c? Why do you say so?1/V, (units of activity 1)

Lineweaver-Burk plot is a graphical representation used in enzymology. It can be used in the absence…

Answered: A Lineweaver-Burk Plot is shown below.…

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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Various concentrations of recombinant human insulin were prepared for use standards for an HPLC method. To verify the prepared concentrations, the samples were analyzed by measuring the absorbance at 280 nm in a short path length (5 mm) cuvette. The molar absorption coefficient for human insulin is approximately 5.875 x 10³ M-¹cm-¹. a. Calculate the extinction coefficient in mL mg-¹cm-¹. b. Calculate the concentrations of the following human insulin standards if the measured absorbances and dilutions used are: Standard 1 Standard 2 Abs. (at 280 nm) of Diluted Sample 0.305 0.685 Dilution 145.0 µL sample, 25.0 μL buffer 130.0 µL sample, 40.0 μL buffer
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 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?
Using the appropriate graph and table above, explain what the R48C mutation appears to be doing to the enzyme’s function. Discuss the kinetic parameter changes and their meaning in this context, not the structure of the enzyme, which was not given to you.
8L.3.4
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?
The table contains data about how different ASFV mutants affect enzyme activity-- for example the D29A mutant, where the D at position 29 was mutated to an A. Select three mutants at random, and for each mutant explain how that mutation could affect intermolecular force interactions (an example: hydrogen bonding could be disrupted by this mutation because…
Phosphoglycerate-mutase catalyses the isomerization of 3-phospho-glycerate to 2-phospho-glycerate. The enzyme has a molecular weight of 28,556 g/mol. In assays using 1 milligram of enzyme per assay the Km for 3-phosphoglycerate was 1.2 x 10^-3M and the Vmax was 840 μmole min^-1. What would be the Vmax in units of mole s^-1?
The hexapeptide Ala-Met-Leu-Lys-Phe-Asp is digested in the same tube with both Cyanogen Bromide and Trypsin at the same time at pH=7. Draw the structure of the product(s) that would bind to a cation exchange column. (Relevant pKa values are 2.2, 3.9, 9.5 and 10.5. Assume pKa values for any newly generated a-amino and a-carboxyl groups are 9.5 and 2.2 respectively)
A purified protein fraction has a total sample volume of 450 µL. The sample has a corrected A280 of 0.244, and the blank corrected A280 was 0.077. (Both values were measured with a path length of 1.00 cm.) If 5.50 µL of the sample was used in a reaction, calculate the mass of protein in the reaction (in µg).
The fosfomycin disc (200/50) also contains glucose-6-phosphate (G-6-P). Which ONE of the following best describes why this is the case? Select one: A. G-6-P increases the solubility of fosfomycin B. G-6-P is used to promote active transport of fosfomycin into the bacterial cell C. G-6-P is used in the formulation to correct glucose deficiencies promoted by fosfomycin D. G-6-P stabilises fosfomycin from spontaneous hydrolysis in aqueous solution E. G-6-P is an inhibitor of FosA that inactivates fosfomycin
The pKa for histidine is pKa = 6.1 while that for cysteine is pKa = 8.0 2. Assume that both histidine and cysteine are catalytic groups for a particular enzyme. Assume also that the side chain of cysteine must be in the deprotonated form. Estimate the pH at which the catalytic activity of this enzyme is the maximum and sketch a pH-activity graph.

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