Part 1: Assess the following partial results section below by editing it for brevity by omittingany unnecessary parts (1 point), explain why you decided to remove certain sections (1point):To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each moleculewere prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized forthe enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units ofinorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture wasincubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of thesubstrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes.Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of themolecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were furtherdiluted to find the minimal inhibitory concentration (MIC). Molecules A3 and A6 exhibited 30% and 45%inhibition of the enzyme activity respectively, even at 1 μM. DMSO was found not to interfere with theenzyme.

To assess inhibitory effects, 10mM stock solutions of each molecule were prepared in DMSO. A 200µl…

Answered: Part 1: Assess the following partial…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Example of a Protein Purification Scheme: Purification of the Enzyme Xanthine Dehydrogenase from a Fungus Volume Total Total Specific Percent Fraction (mL) Protein (mg) Activity Activity Recovery 1. Crude extract 2. Salt precipitate 3. Ion-exchange chromatography |4. Molecular-sieve chromatography 5. Immunoaffinity chromatography 3,800 22,800 2,460 0.108 100 165 2,800 1,190 0.425 48 65 100 720 7.2 29 40 14.5 23 1.8 275 152.108 11 Calculate the specific activity of step#4. Note that percent recovery=% Yield.
200 ml of a 2% protein solution containing an enzyme that you want to purify. Half of the sample is subjected to method A, consisting of fractionated precipitations and 5 ml of final solution are obtained, with a concentration protein 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, with protein richness equal to 10 mg / ml and with an activity enzymatic also equal to 2000 U / ml. You want to know: a) Which of the two methods has provided the purest enzyme. b) By which of the methods the greatest amount of enzyme has been obtained.
Briefly comment on the differences of using a fixed-time assay versus a kinetic assay to measure enzyme activity. Is it reasonable to assume that the reaction velocity obtained by measuring the amount of product after 30 minutes in a fixed-time assay is directly proportional to absorbance? How could you determine whether this was the case? Word limit 180 words including citation and reference
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
A biochemist discovers and purifies a new enzyme, generating the purification table below. Procedure Total protein (mg)Activity (units) 1. Crude extract 20,000 4,000,000 2. Precipitation (salt) 5,000 3,000,000 3. Precipitation (pH) 4,000 1,000,000 4. lon-exchange chromatography 200 800,000 5. Affinity chromatography 50 750,000 6. Size-exclusion chromatography 45 675,000 (a) From the information given in the table, calculate the specific activity of the enzyme after each purification procedure.
A biochemist discovers and purifies a new enzyme, generating the purification table below. Procedure Total protein (mg)Activity (units) 1. Crude extract 20,000 4,000,000 2. Precipitation (salt) 5,000 3,000,000 3. Precipitation (pH) 4,000 1,000,000 4. lon-exchange chromatography 200 800,000 5. Affinity chromatography 50 750,000 6. Size-exclusion chromatography 45 675,000 (a) From the information given in the table, caleulate the specific activity of the enzyme after each purification procedure. (b) Which of the purification procedures used for this enzyme is most effective? (c) Which of the purification procedures is least effective? (d) Is there any indication based on the results shown in the table that the enzyme after step 6 is now pure? (e) What else could be done to estimate the purity of the enzyme preparation?
(A) Below is a partial purification table for a first attempt at a purification schedule for an enzyme: Technique Initial IMAC IEX Gel Filtration Protein (mg) 3467.8 196.7 153.6 142.1 Enzyme activity (units) 36489 32575 27891 26398 Overall yield (%) 100 Specific activity (units/mg) 29.46 Complete the table below by calculating the missing values. Overall Enrichment 1.00
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).
Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?
The initial velocity data shown in the table were obtained for an enzyme. Each assay at the indicated substrate concentration was initiated by adding enzyme to a final concentration of 0.01 nM. Derive Km, Vmax, kcat, and the specificity constant. [S] (mM) Velocity (x10^7) 0.10 0.96 0.125 1.12 0.167 1.35 0.250 1.66 0.50 2.22 1.0 2.63
A technician prepares a buffer solution that will be used to facilitate the optimal pH for an enzyme involved in the biotechnological degradation of organic compounds. The buffer compound will facilitate a stable pH. To prepare the buffer they need to determine the required concentration of anita hara [conjugate acid or base]_08 They have been provided with the buffer parameters provided in the images. Provide the answer to three decimal places and include an appropriate unit. Note: You may need to round the numbers to get the required answer. Final volume of solution: 200 ml VEENER Total buffer compound concentration: 150 mM Ratio conjugate base/weak acid: 1.58/1 2000 Concentration weak acid: 0.058 M Final pH of solution: 6.5 Buffer compound pk.: 6.3 14 13 12 11 10 9
I. Buffer Preparation A researcher in Biochemistry is isolating α-glucosidase enzyme from malted wheat flour. The protocol requires the use of 2.00L of 0.225 M of lactate buffer with pH of 4.25. How will he/she prepare this buffer from 2.00 M lactic acid solution and solid sodium lactate (NaC3H5O3) The Ka of lactic acid is 1.38 x 10-4. [ Na=23.0, C=12.0, H=1.01, O=16.0 g/mol]

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