QUESTION 8 1 pe tryptophan and 4 tyrosine residues and no disulfide crosslinks g

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QUESTION 8 1 pe tryptophan and 4 tyrosine residues and no disulfide crosslinks gives an absorbance (A) of 0 2 in cuwete of path length (0) tem Using beers law AsClu1 and theexinction coeficie fr Trp l0 1cm-1 and Tyr liw 1490 M-1 cm-1, calculate the concentration of the protei. 14 milli Molar 2.8 milli Molar O 140 micro Molar 14 micro Molar O 2.8 micro Molar O QUESTION 9 1 poir Proteins A- a 70 kD monomer, B-a 40 kD monomer and X a protein of unknown size and subunit composition, were run on an SDS than either A or B on SDS-Page and eluted later than A but before B on the SEC column. Which characterization of subunit size and composition of X is consistent with these results X is an 80 kD monomer O X is a homo-dimer of 60 kD subunits X is a 40 kD monomer O X is a 20 kD monomer O X is a homo-dimer of 30 kD subunits Click Save and Submit to save and submit. Click Save All Answers to save all answers Save All Answers Sa

Explanation / Answer

8). The net Absorbance (A) as per Beer's law will be given by the equation,

(A) =[C x (extinction coeff for Trp IW x no. of tryptophan residues + extinction coeff for Tyrosine IW x no. of tyrosine residues) x l].

A= 0.2

l= 1cm

Trp IW= 5500 M-1 cm-1

Tyr IW= 1490 M-1 cm-1

No. of tyrosine = 4

No. of tryptophan= 2

Therefore,

0.2 = [C x (2x5500+4x1490) x 1] = [C x (11,000+2,980) x 1]= C x 13980

So, C= 0.2/13980= 14.30 x 10-6 M = 14.3 M (as 1M= 106 M) = 14 micromolar (Ans)

9). Protein X gave a single band in SDS-PAGE, this means it is either a monomer or a homodimer (composed of two identical subunits). However, it ran faster than both Protein A, 70kD and Protein B, 40kD. Therefore, this means that the molecular weight of protein X or its individual subunits is less than Protein A, 70kD and Protein B, 40kD. This means it is either a 20kD monomer or a homodimer of two 30kD subunits (as in SDS-PAGE the two subunit will separate). Now, in size exclusion chromatography the first one to come out is Protein A, 70kD, second protein X and last Protein B, 40kD. This means the overall MW of protein X is more than that of Protein B, 40kD but less than that of Protein A, 70kD.

This means Protein X is homo-dimer of 30kD subunits. So that it’s total molecular weight is 60kD. During SDS-PAGE, the two dimer separate as run as identical bands of 30kD and thus travel fast as they are smaller than protein A and B. In size exclusion chromatography, the protein A being largest comes out first, and then our 60kD protein X followed by Protein B, 40kD.

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