User:Klare Lazor/Notebook/Chem-496-001/2012/02/07

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  • making buffer for tagging tom


  • Use of a distant reporter group as evidence for a conformational change in a sensory receptor
  • pH increases and deprotonates
  • cysteine pka=8
  • 3mg of 5-iodoacetamido fluorescien mw 515.3g/mol (4 Fold Excess)
  • how much cysteine? mw= 121.16g/mol
    • 35 cysteine in BSA


  • 35(121.16g/mol)= 4240.6 g/mol of cysteine
  • (4240.6 g/mol of cysteine)/(67,000g/mol of BSA)= 0.06329 (mass ratio)
  • (0.06329)*(0.00495 grams amount of BSA in stock)= 0.000313g/50mL of solution
  • (0.000313g/50mL)/(121.16 g/mol)= 2.5E-6mol/mL
  • (2.5E-6 mol/mL)/ (50mL)= 5.172E-7 mols of cysteine per mL solution
    • Whatever mLs of BSA we use in sample(2.767mL) X 5.172E-7mols of cysteine per mL solution=0.000001 mols of cysteine
    • To determine amount of 5-iodoacetamido fluorecien
      • .000001 mols of cytiene x (4mol of dye/1 mol cysteine) x 515.3g/mol= 0.0029 grams of 5-iodoacetamido fluoreciene

Calculations for buffer Solution

  • pka= 7.21 for phosphate
  • ph=pka+log[salt]/[acid]
  • 7.21=7.21 +log[salt]/[acid]
  • [salt]/[acid]= 1
  • so any concentration as long as they have a ratio of 1, we use 0.05mols/Liter
  • may have to increase the amount of Na2HPO4 because it is hydrated
    • 141.98g/mol Na2HPO4- * 0.050 mol/L = 7.099 g/L x .25 L =1.7747g (added extra 2 grams)
    • 137.99g/mol x 0.05 mol/L = 6.89 x .25 L = 1.725 g
    • total solution 250mL


  • mixed in 406 microliters of a pH buffer of 7.02, 2.9 mg of 5-iodoacetamido and 2.776 mL BSA from Stock
    • Ideal ph is between 7 and 8
  • after 2hours let reaction sit
  • used a dialysis bag to separate the free gold and dye from solution

Whats going on

  • Ph increases and deprotonates the cysteine, the deprotonated nitrogen(nucleophile) is then able to attack they carbonyl group of the 5-iodoactamido and attach.



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