Question

Calculate the expected A280 of a 1.50 mg/ml solution of bovine serum albumin, based on the amino acid composition (Table 1-1, page 28 from your manual). All the cysteines in bovine serum albumin are involved in disulfide bonds. T V W Y 10 9 0 6 12 11 0 7 6 6 3 20 33 3 Table 1-1. Amino Acid Compositions of Some Selected Proteins A C D E F G H I K L M N P Q R S Bovine ribonuclease 124 amino acids M-13690 12 8 5 5 3 3 4 3 10 2 4 10 4 7 4 15 Mouse liver fatty acid binding protein 127 amino acids M=14246 2 1 5 13 5 12 1 10 17 9 5 6 3 4 3 5 Chicken lysozyme 129 amino acids M-14313 12 8 8 2 3 12 1 6 6 8 2 13 2 3 11 0 Rat glyceraldehyde-3-phosphate dehydrogenase 332 amino acids M-35705 33 5 21 13 13 31 6 21 26 18 9 21 13 6 10 19 Pig heart lactate dehydrogenase 333 amino acids M-36487 12 5 16 18 5 24 7 23 24 36 9 20 11 158 Rabbit muscle aldolase 363 amino acids M-39211 42 8 14 24 7 30 11 20 26 34 3 14 1916 15 Rat liver alcohol dehydrogenase 375 amino acids M=39514 34 16 1915 18 359 25 31 29 5 6 23 9 9 27 Rat fumarase 507 amino acids M-54464 615 25 31 18 41 13 25 3243 21 29 26 18 20 28 Bovine catalase 506 amino acids M-57586 354 39 25 31 34 21 18 27 3 10 30 38 21 31 25 Rat liver pyruvate kinase 543 amino acids M-58694 596 27 37 17 45 11 39 22 46 13 12 24 22 35 Bovine glutamate dehydrogenase 568 amino acids M-61512 506 30 37 23 52 15 38 32 39 14 22 27 12 35 36 Bovine scrum albumin 582 amino acids M=66153 46 35 41 59 27 16 17 14 59 61 4 13 28 20 23 28 Bovine gelatin (Q (III)collagen) 1049 amino acids M-93650 90 2 28 48 9 365 8 15 38 16 9 23 246 24 49 46 22 22 36 34 36 2 3 31 39 22 34 6 20 27 29 34 52 35 36 5 21 2 19 14 15 0 4

calculate the expected absorbace at 280 nm of a 1.5 mg/ml solution of bovine serum albumin based on the amino acid composition. all cysteins are involved in disulfide bonds.

Answer 1

From Beer Lambert law, Absorbance = $\varepsilon$ cl where $\varepsilon$ is the molar extinction coefficient in M^{​​​​​​-1}​​​​​​​cm ^-1, c is the concentration of the protein in M and l is the path length (usually taken as 1 cm).

Concentration of the given BSA = 1.5 mg/ml = 1.5 gram/litre = (1.5/66153) mole/litre = (2.27 x 10^-5) M

[Molecular weight of BSA given = 66153 Da. Mass/Molecular weight = Moles]

Absorbance of proteins at 280 nm takes place due to absorbance of side chains of tryptophan, tyrosine and cysteine. Each of these side chains have their own extinction coefficients, all of which need to be added to find out the total extinction coefficient of a given protein.

Extinction coefficient of tryptophan = 5500 M^-1cm^​​-1

Extinction coefficient of tyrosine = 1490 M^-1cm^{​​​​-1}

Extinction coefficient of cysteine = 125 M^-1​​​​cm^​​-1

^{​​​​​​}In BSA, number of tryptophan(W) residues = 2, number of tyrosine(Y) residues = 19, number of cysteine (C) = 35

So total molar extinction coefficient of BSA ($\varepsilon$) = 2 x 5500 + 19 x 1490 + 35 x 125 = 43685 M^-1cm^​​-1

Thus absorbance of BSA = $\varepsilon$ cl = 43685 x (2.27 x 10^-5) x 1 = 0.991

1.5 mg/ml BSA will have an absorbance of 0.991 at 280 nm.