Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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- 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 mMarrow_forwardA protein has a dissociation equilibrium constant equal to 4.0um. What is the concentration of free ligand when the fraction of bound sites is 0.7arrow_forwardAn enzyme uses a lysine and a histidine as general acid-base catalytic residues. In one direction of the reaction, the lysine and histidine residues are typically protonated (~NH3*, ~Im), but in the other direction of the reaction, the two residues are "reverse protonated" (~NH2, ~Hlm*) (see below scheme). A pH profile of the reaction at various pH values yielded estimates of 7.57 and 9.30 for the macroscopic ionization constants pk,' and pK2, respectively. A separate set of experiments with site-speciffic variants of the enzyme yielded an estimate of 8.78 for the microscopic ionization constant pKp. Given this information, what is the value of the microscopic ionization constant pKA? Report your answer to the nearest hundredth. Lys-NH3* His-HIm" KB H KA H* Lys-NH2 Lys-NH3* His-HIm" His-Im Kc H* Kp Lys-NH2 His-Imarrow_forward
- 4darrow_forwardwhat is the hydrozide ion concentration of a solution made by dissolving 4.5 g of Sr(OH)2 in 500 MI of waterarrow_forwardCalculate the mole ratio of sodium acetate and acetic acid needed to make a buffer of p H 4.68 ( see Table 15.1 for Ka values or Appendices G and H for Ka and Ksp values respectively.arrow_forward
- so for pH 1, is the net charge 3 as you said up above or 1 down in the solution?arrow_forwardpH of solution 14.00 12.00 10.00 8.00 6.00 First 4.00 equivalence point 2.00 0 First midpoint Second equivalence point Third midpoint pH = pKa=12.32 Third equivalence point HPO4(aq) + OH(aq) PO4(aq) + H2O(1) Second midpoint pH = pKa = 7.21 H2PO4 (aq) + OH(aq) HPO42 (aq) + H2O(1) Using the Henderson- Hasselback equation, show how to create 2L of a 0.1 M KPhos pH 7.5 buffer using K2HPO4 and KH2PO4. The chart to the left should help you understand what pKa to start with. Show your work. pH = pKa = 2.16 H3PO4(aq) + OH(aq) H2PO4(aq) + H2O(l) 25.0 50.0 75.0 100.0 Volume of NaOH added (mL)arrow_forwardExample 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.arrow_forward
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