Be sure to answer all parts.Tris [tris(hydroxymethyl)aminomethane] is a common buffer for studying biological systems.(K, = 5.01 x 109 and pk, 8.3)%3D%3D(a) Calculate the pH of the Tris buffer after mixing 10.5 mL of 0.20 M HCI solution with 25.0 mL of0.10 M Tris.7.06(b) This buffer was used to study an enzyme-catalyzed reaction. As a result of the reaction, 0.00020mol of H* was produced. What is the pH of the buffer at the end of the reaction?(c) What would be the final pH if no buffer were present?

Molarity is a measure of the concentration of a solution, expressed as the number of moles of solute…

Answered: Tris [tris(hydroxymethyl)aminomethane]…

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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calculate the volume of stock solutions required to make up the buffer solutions that will be used for protein purification. The solutions you need to prepare for purification are: i. Binding Solution A: make up 50 mL 50 mM HEPES buffer (pH 7.5), 300 mM NaCl, 5mM imidazole, 5% (v/v) glycerol ii. Wash Solution B: make up 50 mL 50 mM HEPES buffer (pH 7.5), 300 mM NaCl, 75mM imidazole, 5% (v/v) glycerol iii. Elution Solution C: make up 10 mL 50 mM HEPES buffer (pH 7.5), 300 mM NaCl, 500 mM imidazole, 5% (v/v) glycerol please show your working . Thnk you
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A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = H+ (aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.220 M, [H+] = 3.00 × 10−4 M, and [A¯] = 3.00 × 10−4 M. Calculate the value of pKa for the acid HA. pKa =
Paper electrophoresis of Asn, Ala, Asp, Lys and Ser mixtures at pH = 7 shows the fastest movement towards the anode The fully saponified 10g oil sample consumed 1.87g KOH, and the iodine value of the oil was 28.3. So how many unsaturated double bonds are there in each oil molecule on average? (lodine is known to have an atomic weight of 127) Given that 100g cellulose sample is completely hydrolyzed to obtain 78g glucose, then the percentage of cellulose in the sample is _% Given that the Km of catalase is 20mmol/L and the substrate concentration is 80mmol/L, then the percentage of catalase bound to the substrate is %
A purified protein fraction has a total sample volume of 450 µL. The sample has a corrected A280 of 0.244, and the blank corrected A280 was 0.077. (Both values were measured with a path length of 1.00 cm.) If 5.50 µL of the sample was used in a reaction, calculate the mass of protein in the reaction (in µg).
N-(2-hydroxyethyl)piperazine-N'-(2-ethanesulfonic acid) buffer at pH 7 with 400 mM NaCl. A A purified protein is in a Hepes dialysis membrane tube holds a 5.0 mL sample of the protein solution. The sample tube floats in a beaker containing 0.50 L of the same Hepes buffer, but with 0 mM NaCl, for dialysis. Small molecules and ions (such as Na*, CI", and Hepes) can to diffuse across the dialysis membrane, but the protein cannot. Assume there are no sample volume changes during the dialysis. Calculate the final concentration of NaCl in the protein sample once the dialysis has come to equilibrium. Calculate the final NaCl concentration in the 5.0 mL protein sample after dialysis in 300 mL of the same Hepes buffer, with 0 mM NaCl, twice in succession. [NaCl] after a single dialysis: [NaCl] after a double dialysis: mM mM
A purified protein fraction has a total sample volume of 360 µL. The sample has a corrected A280 of 0.484, and the blank corrected A280 was 0.052. (Both values were measured with a path length of 1.00 cm.) If 5.00 µL of the sample was used in a reaction, calculate the mass of protein in the reaction (in µg).
Example The reference range for blood pH is 7.35 – 7.42. What is this range expressed as [H+] in nmol l-1? Calculation: pH = - log [H+] 7.35 = - log [H+] [H+] = antilog -7.35 = 4.47 x 10-8 mol.l-1 = 44.7nmol.l-1 similarly: pH 7.42 = 38.0 nmol.l-1 Range [H+] = 38 – 44.7 nmol.l-1 If the blood pH decreases in an acidosis from 7.42 to 7.15, what is the change in [H+] in nmol.l-1? so this is example on the worksheet, but i still understand how to answer the question.

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