1. Complete the [Substrate] row 2. Determine the sucrose concentration in each tube using the total volumes combined betore addition of CuSQ. Reaction tube # 1 2 3 4 5 6 7 Acetate Buffer (mL) 0.2 0.2 0.2 0.2 0.2 0.2 0.2 Sucrose stock (0.5 M) (mL) 0.01 0.02 0.03 0.05 0.075 0.15 Distilled water (mL) 0.25 0.24 0.23 0.22 0.2 0.175 0.1 Invertase (mL) 0.05** 0.05 0.05 0.05 0.05 0.05 0.05 (Substrate] (mmol/L) |CusO4 solution (mL) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 o Heat all tubes to 100°C for 10 minutes – Cool to room temperature on ice. Arsenomolybdate sol. (mL) 0.5 0.5 0.5 0.5 0.5 0.5 0.5
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Using C1*V1 = C2*V2, please complete the calculations for the substrate row & show me how to do it as I am unsure.
*The final sample is contained in using 2mL centrifuge tubes.
answer all or else I will upvote
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- Now prepare a 500 ml media having ¼ strength MS with 5mM BAP, 5.5 mg/l Kn and coconut water 10%. Stocks are MS Macro 10X, micro 20X, FeEDTA 10X, Organic 100X, BAP and Kn 5gm/100ml. If you need anything else then you add accordingly. BAP has a MW of 225.3. Sucrose and Myo-inositol is not mentioned in the question. These you need to add. Consistency of the medium is also mentioned.3. Compute for the amount of each component of KCN broth if you were to prepare 280 ml. Express your answers using two decimal places. Follow this format in typing your answers: answerabbreviated unit (e.g. 3.00 g, 0.75 ml, 10.25 mg): Components Amount per liter Polypeptone 3.0 g N22HPO4 5.64 g (Available as NazHPO4 · 7H2O*) Monopotassium phosphate 0.225 g Sodium chloride 0.5% KCN 0.075 g (Available as 10,000 ppm solution) *MW (g/mol): Na-23; P-31; H-1; 0-16 3.3. Monopotassium phosphate /rTable 1 - Comparison of the effect of catechol concentration on the amount of product formed. Absorbance Potato extract Absorbance 0 mins after 30mins (2nd reading) (mL) 1st reading 1 Tube # la blank 2a 3a 4a 1 1 1 dH₂O Catechol (mL) (mL) 7 5 3 1 0 2 4 6 0.00 0.060 0.033 0-05-2 Q4) Give 2 reasons for adding dH₂O to these tubes in Table 1? Time for reading: 3:21 -0.11 Absorbance: Time for reading: 3.36 Q5) Tube la serves as a control, but why is this control needed? Absorbance: 0.197 Time for reading: 3.37 Based on the data from Table 1 answer these questions: Q1) What is the name of the enzyme found in potato extract? Answer: catechol Q2) What is the substrate? Answer: THO Q3) Name of product of this enzyme catalyzed reaction? Answer: Absorbance: 0.152 Time for reading: 3:39 Absorbance: . 166 ness Catechol Benzoquinone Subtract 1st from 2nd reading -0.01 0-137 0.11.19 0.119 Q6) Notice that your 1st absorbance reading in tubes 2a-4a are quite similar but it then becomes very different…
- A dialysis bag containing a 3 M solution of the sugarfructose is placed in the following solutions. In each case,give the direction in which water flows.a. 1 M sodium lactateb. 3 M sodium lactatec. 4.5 M sodium lactate0.0025 0.002 0.0015 0.001 0.0005 10 20 30 40 50 60 70 80 -0.0005 Temperature (C) 1.What trend did you observe for the rate of peroxidase compared to the temperature of the enzyme? 2.What was the optimum temperature tested for the enzyme peroxidase? 3.Did any of the temperatures tested cause the enzyme to denature? Which one? Rate (abs/sec @ 500nm)The hydrolysis of a substrate, S, by an enzyme has been studied in the lab. The following initial rates, vo, were recorded at different concentrations. [S] (M) Vo (M/min) 2.10-10-4 1.20.10-6 4.20-10-4 3.10.10-6 9.30-10-4 6.30-10-6 1.42-10-3 9.10-10-6 A. Determine the rate constants for degradation of the substrate B. What is the rate of reaction at [S] = 1.1.104 M? C. Explain, why enzymes can make reactions go faster? Does enzymes also catalyse the reverse reaction from product to substrate?
- 5.15. The following data were obtained when glucose (C6H₁2O6) was added to a batch culture of microorganisms. Determine the reaction order for the disappearance of glucose. TIME. min 0 10 20 30 40 50 CONCENTRATION, g/m³ Glucose Cells 100 67 50 40 33 29 1500 1516 1525 1530 1534 15353. Compute for the amount of each component of KCN broth if you were to prepare 280 ml. Express your answers using two decimal places. Follow this format in typing your answers: answerabbreviated unit (e.g. 3.00 g, 0.75 ml, 10.25 mg): Components Amount per liter Polypeptone 3.0 g NazHPO4 5.64 g (Available as NazHPO4 · 7H2O*) Monopotassium phosphate 0.225 g Sodium chloride 0.5% KCN 0.075 g (Available as 10,000 ppm solution) *MW (g/mol): Na-23; P-31; H-1; O-16 3.4. Sodium chloride (3. Compute for the amount of each component of KCN broth if you were to prepare 280 ml. Express your answers using two decimal places. Follow this format in typing your answers: answerabbreviated unit (e.g. 3.00 g, 0.75 ml, 10.25 mg): Components Amount per liter Polypeptone 3.0 g NazHPO4 5.64 g (Available as NazHPO4 · 7H2O*) Monopotassium phosphate 0.225 g Sodium chloride 0.5% KCN 0.075 g (Available as 10,000 ppm solution) *MW (g/mol): Na-23; P-31; H-1; 0-16 3.2. Na2HPO4 • 7H2O!'
- Absorbance of urea for a diabetic patient is 0.335 while 0.214 is the absorbance of standard kit for the same test, calculate the concentration of this test ? Concentration of standard kit = 50 mg/dl. Repl...Composition of Cell Lysis Buffer Required Concentration of Reagent Molecular Weight Volume or Weight of Reagents Needed 40 mM Tris HCl 157.59 20 mM Acetic acid 60.05 1 mM EDTA 292.24 10 mM MgCl2 from 25mM stock 95.21 5% SDS 288.37 ddH2O 18.02Describe the effect of the following changes would have on the rate of a reaction that involves the substrate UREA and the liver enzyme UREASE 1. Increasing the UREA concentration 2. Increasing the UREASE concentration 3. Increasing the temperature from its optimum value to a value 10degree higher than this value 4. Lowering the pH from the optimum value of 5.0 to a value of 3.0