You are asked to prepare HEPES buffered saline with the following constituents: НЕРES 10mM KCI 5mM NaCI 135mM CaCl2 1mM M9SO4 0.6mM Glucose 6mM You have previously made up 100 mM stock solutions of HEPES, KCI, CaCl2, and M9SO4 and a 1M stock of NaCl. The MW of glucose is 180.16 g/mol What volume of HEPES would you require to make up 400ml of buffer?
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- Convert the experimental glucose concentration provided in below from mM (millimolar) to g/100 mL. Glucose Concentration is in mM 1.927 mM. ● Take the 1/100 dilution factor into consideration ● M will open the doors for you to get to g. ● Once at M assume 1L of solution. (Hint: this will help you when converting from M to moles) ● You will also need to calculate the molar mass of glucose which has the formula C6H12O6The isoelectric point of eIF4a is 5.02. Students are given a sample of cell lysate containing eIF4A at pH 7.4. They are also provided with two buffers of pH 7.4 to use for ion exchange chromatography. One buffer has a very high salt concentration (1 M), the other has a low salt concentration (0.1 M). Mariela decides to use an anion exchange column, while Ashok chooses a cation exchange column. Who made the better decision? Provide a detailed explanation of why one student will end up with a purer sample of eIF4A. Which buffer will you use while washing away impurities (high or low salt), and which would you use to remove eIF4A from the column? Explain your choice.Chitinase is a protein that breaks down chitin, a primary component of the cell wall in fungi, scales in fish and exoskeletons of arthropods. The activity of chitinase extracted from a plant was shown to be optimum at pH 5. You were tasked to prepare 300 mL of 150 mM buffer solution for further analysis of the extracted chitinase. REAGENTS Ka 2.5M Acetic acid Solid NaOAc•3H2O [136.08g/mol] 1.76 x 10-5 2.5M NH3 Solid NH4Cl [53.49g/mol] 5.6 x 10-10 2.5M Lactic acid Solid sodium lactate [112.06g/mol] 4.0 x 10-5 5 M HCl 5M NaOH Pls show sol'ns 1. Given the following reagents, give the moles of each component (acid & base).2. What are the mass/volume of the components needed to prepare the buffer? 3. What will the pH of the buffer be if 1mL of 5 M NaOH was added?
- 80mL of a 0.3M solution of hexapeptide Leu-His-Cys-Glu-Asn-Arg is adjusted to pH=pI. The solution is then titrated with 0.2M HCL to a final pH of 2.1. Sketch the titration curve, labelling the pH and volume axes. Indicate the volume of HCL needed to reach relevant pKa value and equivalence point(s)z Relevant pKa values are: 2.1, 4.3, 6.0, 8.3, 9.8, and 12.5.The concentration ofacetylcholine (a neurotransmitter) in a sample can be determined from the pH changes thataccompany its hydrolysis. When the sample is incubated with the enzymeacetylcholinesterase, acetylcholine is converted to choline and acetic acid, whichdissociates to yield acetate and a hydrogen ion: In a typical analysis, 15 mL of an aqueous solution containing an unknown amount ofacetylcholine had a pH of 7.65. When incubated with acetylcholinesterase, the pH of thesolution decreased to 6.87. Assuming there was no buffer in the assay mixture, determinethe number of moles of acetylcholine in the 15 mL sample.You have been provided with stock solutions of: stock A: 0.06 M sodium pyrophosphate buffer pH 8.5 stock B: 3 M ethanol stock C: 0.015 M NAD+ stock D: milli Q water Determine the volume you will need of each solution to prepare a buffer of with a final volume of 60 mL containing 10 mM sodium pyrophosphate pH 8.5, 100 mM ethanol, 1 mM NAD+. i.e. volume of stock A = _________mL volume of stock B = _________mL volume of stock C = _________mL volume of stock D = _________mL Show your calculations to arrive at your answers.
- Gentiobiose has the molecular formula C12H22O11 and has been isolated from gentian root and by hydrolysis of amygdalin. Gentiobiose exists in two different forms, one melting at 86°C and the other at 190°C. The lower melting form is dextrorotatory ([α] 16°), the higher melting one is levorotatory ([α] -6°). The rotation of an aqueous solution of either form, however, gradually changes until a final value of [α] 9.6° is observed. Hydrolysis of gentiobiose is efficiently catalyzed by emulsin and produces two moles of D-glucose per mole of gentiobiose. Gentiobiose forms an octamethylether, which on hydrolysis in dilute acid yields 2,3,4,6-tetra-O-methyl-D-glucose and 2,3,4-tri-O-methyl-D-glucose. What is the structure of gentiobiose? EXPLAIN IN DETAIL.How much caffeine can be extracted using 50.0ml of DCM if a tea bag contains 15 grams of tea leaves with 0.9% of caffeine? And has undergone solid-liquid extraction with 100ml of distilled water followed by liquid-liquid extraction using DCM. The solubility of caffeine requires 15 ml of water and 2.0ml of DCM for 40.0mg.Determine the values of KM and Vmax for the decar-boxylation of a β-keto acid given the following data. You have to plot the graph by using excel. Please include slope Substrate Concentration (mol L1) Velocity (mM min1)2.500 0.5881.000 0.5000.714 0.4170.526 0.3700.250 0.256
- The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)Convert the experimental glucose concentration from mM (millimolar) to g/100 mL. 1/100 Dilution factor, Assume 1L of Solution. Show ALL Steps 2.340 mMAcetocholinesterase is an enzyme possessing a single active site that metabolizesacetylcholine with a turn over number of 1.4 x 10^4s-1. How many grams of acetylcholine(molecular formula C7NO2H16+) will 2.16 x 10^-6 g acetocholinesterase metabolize in 60minutes? (The enzyme’s molecular mass is 4.2 x 10^4 g/mol).