Problem Set 1 Answers - MCB102 - Problem Set 1 This...

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MCB102 - Problem Set 1 This ungraded problem set covers the material discussed in Lectures 1-4. See if you can work out the answers yourself, before looking at the answers that I will post at the end of the week. Also, try the problems at the end of Chapter 3 in the Lehninger textbook. If you need help, talk to your GSIs or to me. Here is a summary of the concepts and information we have covered so far that you are expected to know for the midterm: Definitions of pH, pK a How to use the Henderson-Hasselbalch equation Single-letter and three-letter codes for the amino acids Chemical properties of the amino acid sidechains we covered in class Chemical structure of amino acids and peptide bonds The four levels of protein structure Principles of protein purification, including centrifugation, isoelectric focusing, chromatography How to determine amino acid sequences using the Sanger reagent and Edman degradation and overlapping peptides produced by enzymes and cyanogen bromide How to locate S-S bonds in proteins Peptide synthesis and mass spectrometry can be used to prepare and analyze peptides and proteins (you do NOT need to know the details of these techniques) 1. What is the pH of the following aqueous solutions? pK a s you might need: Acetic acid 4.76 Sulfuric Acid ---- 1.81 Ammonium ion 9.26 a. 0.010M HCl b. 0.010M NaOH c. 0.005M H 2 SO 4 d. 0.1 M HC 2 H 3 O 2 + 0.2 M NaC 2 H 3 O 2 ANS: 1. a. Strong acid, [H + ] = [HCl] , pH = -log(1.0x10 -2 ) = 2.00 b. Strong base, [OH - ] = [NaOH], pOH = -log[OH - ] = 2.00 , pH = 14.00-pOH = 12.00 c. Considering only the 1 st dissociation which is strong (the 2 nd will be ignored), pH = -log(5x10 -3 ) = 2.3; the pH is actually a little lower due to 2 nd dissociation. d. Acetic acid is a weak acid, acetate is its conjugate base; use the Henderson- Hasselbalch equation to get [H + ]. pH = pK a + log ([acetate]/[acetic acid]), so pH = 4.76 + log(0.2/0.1) = 5.1.
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2. How would you prepare the following buffer solutions? a. 1000mL of 0.10M Acetate buffer pH = 4.76 b. 200mL of 0.10M Phosphate buffer pH = 6.2 ANS: a. Here is one way to do it: starting with 1.0M Acetic Acid (HAc) and 1.0M NaOH, you need 100mL of 1.0 M HAc. Then you need to calculate how many total moles of acetate you need:1L x 0.10mol/L = 0.1 moles of total acetate. Then using the HH equation, plug in the pH and the pKa for acetic acid: 4.76 = 4.76 +log ([A - ]/[HA]). Since log 1=0, then ([A - ]/[HA)=1 and now you know that equal concentrations of each are needed. Since you know that you are shooting for a total of 0.1 moles acetate, and the ratio tells you that you need half of each species, you know that you will need 0.05 moles of NaOH. Using a 1.0 M NaOH, you can calculate how much of this strong base you need: 50 ml in order to have 0.05 moles total NaOH to convert 0.05 moles HAc. So now you have 100mL acetic acid + 50mL NaOH for a total volume of 150 ml but we want 1000 mL, so we add 850 mL H20. b. The easiest solution is to use 0.10 M solutions of the components of the acid-base
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Problem Set 1 Answers - MCB102 - Problem Set 1 This...

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