Question

3) (30 pts) The small protein lysozyme is an important first line of defense against bacterial infection. It has 129 amino acids and contains both tyrosine and tryptophan. The absorbance of a 1 mg/ml sample of lysozyme was measured in a cuvette of thickness 1.00 cm and was found to be 3.19 at 240 nm and 2.58 at 280 nm. The molar absorption coefficients of tryptophan and tyrosine at 240 nm are 2.00 x 103 dm3 mol-1 cm-1 and 1.12 x 104 dm3 mol-1 cm-1, respectively, and at 280 nm they are 5.40 x 103 dm3 mol-1 cm-1 and 1.50 x 103 dm3 mol-1 cm-1, respectively. What are the molar concentrations of the two amino acids in the 1 mg/ml protein sample? If the molar mass of the protein is 14,400 g/mol, how many tryptophans and tyrosines are present in the sequence of this protein? (looking for the molar ratio, moles aa/mol protein)

Answer 1

The small protein lysozyme has 129 amino acids and contains both Hyrosine and Amuptophan. Thickness of the Cuvette 21.00 cm (8) at 240mm, absorbance of a Ing/mL sample of lysozyme, A240² 3:19 (given, but very high does not follow the - Been's law exactly, it should be Orq-0.8) Molaro absorption coefficient (e)s Eltroyptophan) = 2.00x10$dmo malomt t(typosine) 128109 dm3 montent . Absombance of the solution can be expressed as a sum of the absorbance of the amino acid torypto phaw and Approsina. ...A290 2 € (treyptophan) Litryptophana] + € (ty prosina) Aymeringe om, 3197 2x 103 (tryptophan] el + l-12% 104 tep posire] <1 chip motor Similarly at 280nm, absorbance of the solution. T A280 = 2.58 (given) ,molano absorption Coefficient . € (sryptophan) 2 5.40x 10$dm? mot emne Elterosin) a 1.50x 103 dm?nditioned

" namer Agu € (tryptophan) Estryptophan] 2+E (teprotira) Etemosineje or, 2:98 > 5-90x 10 %[-tryptophan]«1 + 1.50x100 [typoging xl am meaf 1 To solve egeation@and , egr 0 x 4 = ep@x 2 we get 17.126 3 10.8x10ºftreptophan] + 6-048x109[tyrosine] 516 > 1098x103 Ltryptophan] + 3x108 Etang teprossin] 12.066 - 57480 [tyrosine] 4) [tyrosine] > 2.099x10-4 cmI Now, putting this concentratiow down into egr © Weget; 3.19 = 210% (tryptophan] + 112x107x 2.099x1054 om, 2x10$ (-ryptophan] > 3:19 – 2.381 0.839 1 --. [-raptophong » 4.198810-4 (1)

Sample! f protein The molarr mass of the protein is o 144009/mol It is mentioned that ImL of the solution conta . Numbers of moles of prokin protein contains by the sample a 1x100g - 14400gmot! Now, we calculated the 26.944x10-8 mol/rol the molar Concentration of two amino acids in the protein Samples, [tyrosine 2 2.099410-4 (M) 22.099x18-9 mohl si lood ml. protein Sample has 2 2.099x10-4 mol typogine r. AmL actual a a " 22.0998109 x 1 mol typrosine 1000 = 2.099x10-7mo tyrosine [toryptophan] » 4.195x10–9(M) 24. 195x107 mill! i loco mL protein Sample has : 4.195x1024 mol toptophan * Lmb a " 24.195x10-4 1071 mol tryptophane 74.125x10 - Foro tropptophan Molaro ratio o móles of teyrosine aminonch for tyrosine moles of protein 21099x10-7 7 3.023. 1000 - ; 6.944x10-8 for tryptophan ? molematus up to open amino cucid - 4.195x10* | Holansnatio moles of tryptophan amino acid istop! 4. 195x10 6.944408 6.041