# User:Floriane Briere/Notebook/CHEM-496/2012/02/08

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## Objective

Today's objective is to prepare the refolding buffer and the BSA control. The refolding buffer will be used next week to higher the pH after the Gold NPs/fibers synthesis; higher the pH is a mandatory step if we want to tag our Gold NPs/fibers with the dye. In fact, the dye we are using is able to tag deprotonated lysine (whose pKa = 8.95 for the amino group); thus pH of the solution is going to be an important characteristic that we'll have to control during the entire experiment.

The BSA control is made with the same amount of acid but we are using HCl instead of HAuCl4 (the HCl solution has a similar concentration to the HAuCl4 solution we're using for Gold NPs synthesis); so the solution is identical to the normal one but Gold is absent. Moreover, we are also going to determine how much dye we'll use to tag our 10mL solutions.

## Protocol

• BSA control (Gold/BSA ratio = 166)
1. 8mL of water + 1mL of HCl (2.84mM) + 1mL of BSA (17.7µM)
2. 2 hours in the oven (80°C)
• Refolding buffer
1. 1.514g of Tris
2. 3.51mL of HCl (0.948M)
3. Fill with water up to 250mL
4. Measure the pH using pHmeter; obtain a pH = 8.73

## Data

• Tris buffer calculations

8.5 = 8.06 + log[salt]/[acid] [salt]/[acid] = 2.754 (50-x)/x = 2.754 --> x = 13.318 mM = [HCl] (0.013318 M)(0.25 L) = (0.948 M)(x L) x = 3.512 mL HCl (0.050 M Tris)(0.25 L)(121.14 g/mol) = 1.514 g Tris

• Mass of dye for reaction calculations

It's been established that one molecule of BSA is composed of 60 lysine residues (K. Hirayama, S.Akashi, M. Furuya and Ken-ichi, "Rapid Confirmation and Revision of the Primary Structure of Bovine Serum Albumin by ESIMS and frit-FAB LC/MS", Biochemical and Biophysical Research Communications 173, 2(1990) 639 - 946). Moreover, to tag our proteins we are going to use an excess of three dye molecules for every one lysine residue (Protein-protein interactions: a molecular cloning manual; google books, p195).

MW of dye = 573.51 g/mol, MW BSA = 66463 Da

1.77 x 10-7 M BSA x 0.001 L = 1.77 x 10-8 mol BSA 1.77 x 10-8 mol BSA x 6.022 x 1023 molecules/mol = 1.066 x 1016 molecules BSA 1.066 x 1016 molecules BSA x 60 lysine residues = 6.393 x 1017 total lysine residues 6.393 x 1017 total lysine residues x 3 = 1.918 x 1018 dye molecules needed 1.918 x 1018 dye molecules x 1 mol/(6.022 x 1023 molecules) = 3.186 x 10-6 mol dye 3.186 x 10-6 mol dye x 573.51 g/mol = 1.83 mg dye