Preparation of Gold and Bovine Serum Albumin Stock Solutions
 Using a metal spatula, Bovine Serum Albumin (BSA) was prepared by dissolving .0249 g of white crystalline powder of BSA into .025 L of water to obtain 15 μM of BSA.
 The amount of .0249 g of BSA was acquired from the following calculation:
.025 L of water × (15×10^{6} mM of BSA ÷ 1 L of water) = 3.75×10^{7} mM of BSA
3.75×10^{7} mM of BSA × 66430 g (formula weight of BSA) = .0249 g of BSA
 Finally, the molarity of the BSA stock solution was accounted for to be 1.5 × 10^{8} m/L
 Using a metal spatula, Gold (Au) stock solution was made by dissolving .085 g of orange, granular solid of hydrogen tetrachloroaurate (HAuCl_{4}) in .025 L of water to obtain 10 mM HAuCl_{4}.
 The amount of .085 g of HAuCl_{4} was acquired from the following calculation:
.025 L of water × (10×10^{3} mM of HAuCl_{4} ÷ 1L of water)= .00025 mM of HAuCl_{4}
.00025 mM of HAuCl_{4} × 339. 79 g (formula weight of HAuCl_{4})= .085 g of HAuCl_{4}
 Not all weighed HAuCl_{4} was dissolved into solution. The remaining HAuCl_{4} material from the weighing paper amounted to .0194 g.
 The amount of .0194 g HAuCl_{4} residue was subtracted from the theoretical value .085 g of HAuCl_{4} to obtain the amount of HAuCl_{4} dissolved in solution.
 The actual HAuCl_{4} present in solution is .0656 g. The molarity of the HAuCl_{4} stock solution was .007722417 m/L or 7722 μM.
Preparation of different mole ratios of Au/BSA
 It was decided to prepare solutions of mole ratios; 60, 80, 100, 120, 128, 130, 132, 133, 134, 136, 138, 140, 160, and 170. The total volume to be prepared for each mole ratio solution is 6 mL.
 After allowing some time for the stock solutions of Au and BSA to completely dissolve, calculations were made to verify the volume needed of each constituent, Au, BSA, and water, to obtain the appropriate mole ratio.
 Using the dilution equation, M_{1}V_{1} = M_{2}V_{2}, the following calculations were made:
M_{1} = 15 μM
V_{1} = Volume needed from 15 μM BSA stock solution to complete ratio
M_{2} = 1.5 μM (concentration of BSA for mole ratio 60)
V_{2} = 6 mL (total volume of final solution)
 V_{2} for 60 = (1.5 μM × 6 mL) ÷ 15 μM = 0.6 mL of 15 μM BSA stock solution
 After making this calculation, it was decided by our group to keep the BSA concentration constant. As a result, 0.6 mL of 15 μM BSA stock solution was distributed to all other remaining mole ratios in their respective test tube containers.
 Calculation of the required Au concentration for all other mole ratios was divided amongst other group members, Michael Nagle and Puja Moody(see their calculation section). Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below:
1. First obtained the mole concentration needed for Au in correspondence to its mole ratio using the following factors:
X= mole concentration needed for Au
1.5 μM is the mole concentration of BSA used since this is held constant
X_{Au}= 1.5 μM × 133 (mole ratio) = 199.50 μM of Au
X_{Au}= 1.5 μM × 134 (mole ratio) = 201 μM of Au
X_{Au}= 1.5 μM × 136 (mole ratio) = 204 μM of Au
X_{Au}= 1.5 μM × 138 (mole ratio) = 207 μM of Au
X_{Au}= 1.5 μM × 140 (mole ratio) = 210 μM of Au
2. After obtaining the mole concentrations of Au, these mole concentrations were applied into the dilution equation used earlier to obtain the volume of Au needed. The following volumes for Au are:
V_{Au 133}= (199.5 × 6 mL) ÷ 7722 μM of Au = 0.155 mL of Au
V_{Au 134}= (201 × 6 mL) ÷ 7722 μM of Au = 0.156 mL of Au
V_{Au 136}= (204 × 6 mL) ÷ 7722 μM of Au = 0.158 mL of Au
V_{Au 138}= (207 × 6 mL) ÷ 7722 μM of Au = 0.160 mL of Au
V_{Au 140}= (210 × 6 mL) ÷ 7722 μM of Au = 0.163 mL of Au
 To obtain the exact volume of water needed for each mole ratio, the corresponding volume of Au for the given mole ratio and 0.6 mL of BSA was subtracted from 6 mL. The volume of water needed for mole ratios 133, 134, 136, 138, and 140 are listed below:
Volumes of water needed
V_{133} = 6 mL − .155 mL of Au − 0.6 mL of BSA = 5.245 mL
V_{134} = 6 mL − .156 mL of Au − 0.6 mL of BSA = 5.244 mL
V_{136} = 6 mL − .158 mL of Au − 0.6 mL of BSA = 5.242 mL
V_{138} = 6 mL − .160 mL of Au − 0.6 mL of BSA = 5.240 mL
V_{140} = 6 mL − .163 mL of Au − 0.6 mL of BSA = 5.237 mL
 Using calibrated micropipets, the appropriate amounts of Au, BSA, and water were distributed to their respective mole ratio containers. The test tubes were capped with metal tops and wrapped in foil.
 Upon placing the Au/BSA solutions into the oven, the solutions appeared as clear and colorless liquid solutions.
 The rack containing the 14 mole ratios was placed inside an oven set to 85 °C for 4 hours.
 Lab area was cleaned and the hygroscopic chloroauric acid was returned to its dessicator. BSA was returned back to the refrigerator for proper storage.
