Question

A protein has the amino acid sequence:

DSRLSKTMYSIEAPAKLDWEQNMAL

How many peptide fragments would result from cleaving the sequence with....

a: cyanogen bromide

b: trypsin

c: Which of these reagents gives the smallest single fragment (in number of amino acid residues)?

Please explain how to work out this problem. Thanks!

Answer 1

General guidance

Concepts and reason

Proteins with long peptides can be cleaved into smaller fragments by using certain chemicals that cleave at specific sites on the amino acid residues. Some of these chemicals include Cyanogen Bromide $\\left( {{\\rm{CNBr}}} \\right)$ that cleaves proteins after methionine side chains.

Fundamentals

In chemical fragmentation, certain chemicals cause proteolysis Cyanogen bromide $\\left( {{\\rm{CNBr}}} \\right)$ is used to specifically cleave the methionine residues at the C-terminal.

Step-by-step

Step 1 of 6

(a)

Cyanogen Bromide will cleave the carboxyl terminal of methionine in the peptide. The given protein has amino acid sequence DSRLSKTMYSIEAPAKLDWEQNMAL.

Explanation | Common mistakes | Hint for next step

The peptide when treated with Cyanogen Bromide, it would make cuts at the carboxyl group of methionine forming a homoserine lactose.

Step 2 of 6

Chemical fragmentation with Cyanogen bromide would give the following fragments:

$\\begin{array}{l}\\\\{\\rm{DSRLSKTM}}\\mathop {\\rm{|}}\\limits^{\\rm{1}} {\\rm{YSIEAPAKLDWEQNM}}\\mathop {\\rm{|}}\\limits^{\\rm{2}} {\\rm{AL}}\\\\\\\\{\\rm{ 1 2 3}}\\\\\\end{array}$

Part a

Thus, number of peptide fragments formed after cleaving with cyanogen bromide is $3$.

Explanation

The proteolysis with Cyanogen bromide would make two cuts at the C-terminal of Methionine in the peptide and gives 3 fragments DSRLSKTM, YSIEAPAKLDWEQNM, and AL.

Step 3 of 6

(b)

Trypsin will hydrolyze the peptide bonds after lysine and arginine. The given protein has amino acid sequence DSRLSKTMYSIEAPAKLDWEQNMAL.

Explanation | Common mistakes | Hint for next step

The peptide when treated with Trypsin, it would make cuts at the carboxyl group of arginine and lysine unless followed by proline.

Step 4 of 6

Enzymatic proteolysis with Trypsin would give the following fragments:

$\\begin{array}{l}\\\\{\\rm{DSR}}\\mathop {\\rm{|}}\\limits^{\\rm{1}} {\\rm{LSK}}\\mathop {\\rm{|}}\\limits^{\\rm{2}} {\\rm{TMYSIEAPAK}}\\mathop {\\rm{|}}\\limits^3 {\\rm{LDWEQNMAL}}\\\\\\\\{\\rm{ 1 2 3 4}}\\\\\\end{array}$

Part b

Thus, the number of peptide fragments formed after cleaving with Trypsin is $4$.

Explanation

The fragmentation with Trypsin would make three cuts at the C-terminal of Arginine and Lysine in the peptide and gives 4 fragments DSR, LSK, TMYSIEAPAK, and LDWEQNMAL.

Step 5 of 6

(c)

Fragmentation with Cyanogen Bromide would yield peptide with 3 fragments. The protein DSRLSKTMYSIEAPAKLDWEQNMAL when treated with Cyanogen Bromide it makes a cut after the methionine group, forming the smallest fragment with only two amino acids.

Explanation

When Cyanogen Bromide cuts the protein after methionine, then a small fragment is formed having amino acids A and L, which is as following:

DSRLSKTM-YSIEAPAKLDWEQNM-

Step 6 of 6

When the peptide is treated with Trypsin, it would make cuts after amino acids lysine and arginine giving 4 fragments.

Part c

Thus, the smallest single fragment with only 2 amino acid residues is formed by Cyanogen Bromide.

Explanation

The protein when treated with Trypsin, it would give the following fragments:

DSR-LSK-TMYSIEAPAK-LDWEQNMAL

Here four fragments are formed and the small fragment has three amino acids. When compared to the peptide cleaved with Trypsin, the peptide hydrolyzed by Cyanogen Bromide, has smallest fragment with only two amino acids.

Answer

Part a

Thus, number of peptide fragments formed after cleaving with cyanogen bromide is $3$.

Part b

Thus, the number of peptide fragments formed after cleaving with Trypsin is $4$.

Part c

Thus, the smallest single fragment with only 2 amino acid residues is formed by Cyanogen Bromide.

Answer only

Part a

Thus, number of peptide fragments formed after cleaving with cyanogen bromide is $3$.

Part b

Thus, the number of peptide fragments formed after cleaving with Trypsin is $4$.

Part c

Thus, the smallest single fragment with only 2 amino acid residues is formed by Cyanogen Bromide.