calculating the “Activity ([PNPP]) for each Supernatent.
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- Utilising the provided class data generate the following graphs: I) Michaelis Menten; II) Lineweaver-Burk; and III) Hanes-Woolf. Ensure that you clearly label each graph,and add the relevant trendlines with equations. Table 1: Class data demonstrating the Absorbance at 700nm obtained for the alkaline phosphatase enzyme reaction Table 1 tube Abs700mm 1 0.000 2 0.060 2 0.090 4 0.140 5 0.190 6 0.250 7 0.290 The equipment we used are • 20mM Tris Buffer pH 8.5 • 33mM MgCl2 • Alkaline Phosphatase (2mg/ml) in 20mM Tris Buffer pH 8.5 • 4mM Glucose-1-phosphate • Acid Molybdate pH 5.0 • Reducing Agent • Distilled Water • Glass Test tubes • Tube Rack • Cuvette • Pipettes and Tips • Water bath set to 37oC The method we used is Method/Protocol: 1. Read the protocol in its entirety before starting. Take note of any additional information that appears in subsequent steps that may influence how previous steps are performed. 2. Using glass tubes, generate the reactions mixtures…Lab: Isolation of beta-amylase from sweet potato Objective: To isolate the enzyme β -amylase using sweet potato as a source Materials Required: Sweet potato.Knife/peeler.Mortar and Pestle.A Blender.Blue capped tubes.20mM sodium Phosphate buffer at pH 7.Vortexer. Procedure: 1.Take a clean sweet potato and peel the skin off.2.Weigh the peeled sweet potato and note the weight.3.The sweet potato is cut into small pieces and transferred into a mortar and pestle.4.The pieces are crushed and then transferred into a blender.5.Add 40 ml of cold 20mM sodium phosphate buffer saline. Blend it until it forms a paste.6.Gently transfer the potato slurry into a blue capped tube.7.Allow the enzyme to extract over a 1 hour period at room temperature, with frequent vigorous stirring on a vortex mixer.8.Then the extract is filtered using a GF A glass fibre filter and the filtrate is collected in a new blue capped tube.9.Centrifuge the filtrate at 12000rpm for 20 minutes at 4 degree Celsius.10. After…Please ASAP. Thank you. Explain these results in short answer form. Why was the slope zero for blank solution? What was the optimal concentration for peroxidase activity? Do these values make sense?
- Enzyme Kinetics question Enzyme used is 10uL of a 10 ng/uL solution to a reaction mix in a final volume of 2.0 mL. (Enzyme used is 20kDa monomeric enzyme if matters) Based off lineweaver burk where 300 uM of inhibitor is used noninhibited formula is y= 4x + 0.1 (x axis is 1/S 1/mM) (y axis is 1/Vo sec/mM) inhibited formula is y = 4x + 1 I found Km as 1 mM for inhibited Vmax as 1mM/sec for inhibited How would I find Kcat? How would I find Ki?Dilution factor problem in buffer and unbuffered solution in enzyme. In factors affecting enzyme activity (pH, temeprature). In testing pH factors, 2% of 4 mL of unbuffered starch solution was added to 4 mL of buffered starch solution, plus the 2mL enzyme solution. However, in testing temperature, 1% of 8 mL buffered starch solution was added to 2mL of enzyme solution. Why use different 1% and 2% starch concentration? My teacher said that in the pH part, upon adding the volumes, it will be diluted to 1%. i don't understand how he got it? If I use M1V1=M2V2, the answer is clearly not 1%. I am very confused. Thank you.Calculation at the pre-equivalence point region of a complexation titration: What is the pSr value of a 50.00 mL solution of 0.01000 M Sr2+ (formation constant, Kf = 5.348 x 108) after the addition of 11.7 of 0.02000 M EDTA which was buffered to pH 11.00? Round your final answer to two places to the right of the decimal point.
- Calculation at the initial point of a complexation titration: What is the pSr value of a 50.00 mL solution of 0.01000 M Sr2+ (formation constant, Kf = 5.348 x 108) after the addition of 0.00 mL of 0.02000 M EDTA which was buffered to pH 11.00? Round your final answer to two decimal places to the right of the decimal point.Initial rate data for an enzyme that obeys Michaelis–Menten kinetics areshown in the following table. When the enzyme concentration is 3 nmolml-1, a Lineweaver–Burk plot of this data gives a line with a y-intercept of0.00426 (μmol-1 ml s). (a) Calculate kcat for the reaction.(b) Calculate KM for the enzyme.(c) When the reactions in part (b) are repeated in the presence of 12 μM ofan uncompetitive inhibitor, the y-intercept of the Lineweaver–Burk plotis 0.352 (μmol-1 ml s). Calculate K′I for this inhibitor.From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
- Result nad Discussion Lead Acetate Reaction: Samples: lysine, cysteine, methionine Reagents: 10% Sodium Hydroxide (NaOH) and Lead Acetate Pb(CH3COO)2 -To 1 ml of the amino acid solution taken in a test tube, add few drops of sodium hydroxide (40%) and boil the contents for 5-10 mins over a bunsen burner. Cool the contents and add few drops of 10% Lead acetate solution and observe.Mechanisms of catalysis: 2.5 Mechanism of chymotrypsin summary. These mechanisms involve several of the above-mentioned catalyses. In these summaries, do not just draw a diagram of the proposed mechanisms. It is more important to understand which reaction steps involve what kind of catalysis and how these help to reduce the activation energy needed for the reaction (e.g. a step in the reaction mechanism could be electrostatic catalysis to stabilise the transitions state)Glycolysis: Summary: Where does it occur?