5. Let's revisit the question from last week's problem set with the cellulase enzymediscovered by Colby students. A small molecule was also identified that inhibits thenewly discovered cellulase. The data are shown below.[S]2 x 10-6 M4 x 10-61 x 10-52 x 10-54 x 10-51 x 1042 x 10-31 x 10-²4 x 10-²Vo (μm/sec)28407595Vmax=112128139140141C. What is the Ki of the inhibitor?Vo (μm/sec) + Inhibitor (2 µM)16A. Use the data in the table above to estimate KM and Vmax for cellulase in thepresence and absence of inhibitor.KM =KM (+ Inhibitor) =Vmax (+Inhibitor) =B. What is the mechanism of inhibition? Briefly explain.2745587285898990

Parameters such as Km and Vmax are used for comparing enzyme activities. If we know the initial rate…

Answered: 5. Let's revisit the question from last…

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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You gathered your data from Lab 4, and found the following values: Total Activity of lysozyme in HEW: 14 units Total Activity of lysozyme in carb 1: 21 units Total amount of protein in HEW: 70 mg Total amount of protein in Carb 1: 15 mg Calculate the extent of purification of lysozyme in Carb 1. 150% 7 fold 70% 15 fold
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Electrophoresis is performed at PH 6.8 on a mixture of mutated hemoglobin that differ from normal haemoglobin (Hb) only by the substitution of one amino acid- Hb X: Val replaced par Glu - Hb Y: Asp replaced by Leu - Hb Z: Glu replaced by Lys What will be the order of migration between cathode and anode of these mutated Hb compared to normal Hb? Justify your answer.
200 ml of a 2% protein solution are available, containing an enzyme to be purified. Half of the sample is subjected to method A, consisting of fractionated precipitations, and 5 ml of final solution are obtained, with a protein concentration equal to 5 mg / ml and enzymatic activity equal to 2000 U / ml. The other half is subjected to method B, consisting of ion exchange chromatography, and a final solution of 10 ml is obtained, with a protein richness equal to 10 mg / ml, and with an enzymatic activity equal to 2000 U / ml. You want to know: a) Which of the methods has provided the purest enzyme. b) By which of the methods has the greatest amount of protein been obtained.
Last week in lab you plated various E. coli mutants. This week we will be analyzing those plates. To help interpret those results, predict the growth/color pattern you'd expect for a Lacz- mutant. Table 1: Options are blue or white for NA plates; Table 2: Options are: growth or none for minimal plates). Table 1 [Select] Table 2 [Select] Xgal glucose Xgal + lactose [Select] [Select] Minimal lactose ΝΑ [Select] Xgal + IPTG [Select] maltose > [S
The bacteria that you are growing to express biochemisfunase are resistant to ampicillin, so you keep ampicillin in your media to prevent contamination from other bacterial strains. You find a tube of ampicillin in the lab freezer that says “2000x AMP” on the lid, which indicates that it is 2000 times more concentrated than what you need (i.e., a concentration of 1x ampicillin in your media). How much of the 2000x ampicillin stock would you add to your 1.0-L culture to get a final concentration of 1x ampicillin?
1) The Nde1 enzyme comes at a concentration of 40 U/ul. According to the manufacturer, the container has enough for 10,000 reactions of 2 ug of DNA, assuming that 1 U/ug of DNA is recommended. How many units in total does the packaging contain? At what concentration in U/ml does the enzyme come? If I had to use 10 U for a reaction of 10 ugs of DNA, how would I serve it? 2) A serially diluted phage solution is used to inoculate an E coli plate in the double-layer plate assay. In 24 hours you count 148 plaques. Considering that the original stock was 0.1, and that you used the sixth dilution of the series, what is the PFU for that phage in this strain of E coli? What is the dilution factor for this plate?
An Fab fragment binds to lysozyme with a dissociation constant of Ka = 10-11 M. A 1 nM (10-9 M) solution of lysozyme is treated with increasing concentrations of the Fab fragment. At what concentration of added Fab will half of the lysozyme be bound to the Fab? [Fab] 9.9 Incorrect nM
Follow up question:... The answer given in problem 31. 17 is inadequate for me, and I simply don't understand the explanation for the answer... The problem is: Ribosoms were isolated from bacteria grown in a "heavy" medium (13C and 15N) and from bacteria grown in "light" medium (12C and 14N). These 70S ribosoms were added to an in vitro system engaged in protein synthesis. An aliquot removal serveral hours later was analyzed by density-gradient centrifugation. How many bands of 70S ribosoms would you expect to see in the density gradient? Is it possible to give a more comprehensive explanation? :) Expert Solution Step 1 Introduction :- A ribosome is a sophisticated molecular machine that produces proteins from amino acids during the process of protein synthesis, also known as translation. Protein synthesis is a fundamental operation that all live cells must carry out. Step 2 Answer :- One light band, one heavy band, a hybrid of light 30S and heavy 50S, and a hybrid…
You have an aqueous solution containing: Alaine (a mono-amino, monocarboxylic acid) Fructose (a non-ionic monosaccharide) Glycogen (a non-ionic large polysaccharide) Ribose-5-phosphate (an anionic monosaccharide phosphate) t-RNA (a polyanionic nucleic acid, MW~ 30,000). Assuming you have a distinctive assay for each of these compounds, what procedures would you use to obtain gram quantities of each of these compounds free of each of the other compounds
You have an extract of BL21(DE3) cells containing a soluble, 75 kDa recombinant protein that has a histidine tail at its C-terminus. This extract is still in raw form and you need to purify this protein without losing its solubility. Check the option below that represents the best purification methodology in the order of execution. * a) (1) isoelectric precipitation; (2) microfiltration; (3) affinity chromatography; (4) anion exchange chromatography b) (1) centrifugation; (2) microfiltration; (3) isoelectric precipitation; (4) affinity chromatography c) (1) centrifugation; (2) microfiltration; (3) affinity chromatography; (4) ultrafiltration d) (1) microfiltration; (2) jumping out; (3) centrifugation; (4) affinity chromatography
To approximate the concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterial cell is a cylinder (diameter 1.0 μm, height 2.0 μm), that the cytosol (specific gravity 1.20) is 20% soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.

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