Question

A protein has the amino acid sequence

DSRLSKTMYSIEAPAKLDWEQNMAL
How many peptide fragments would result from cleaving the sequence with either cyanogen bromide or trypsin? number of peptide fragments resulting from cleaving with cyanogen bromide? number of peptide fragments resulting from cleaving with trypsin?

Which of these reagents gives the smallest single fragment (in number of amino acid residues)?
(A) trypsin
(B) cyanotrypsin
(C) cyanogen bromide

Answer 1

Cyanogen bromide cleaves the amino acid sequence at methionine, resulting in two fragments, while trypsin cleaves at lysine and arginine, resulting in five fragments. Cyanogen bromide produces the smallest single fragment among the two reagents.

Step-by-step explanation:

Understanding Peptide Cleavage by Cyanogen Bromide and Trypsin

Cleaving the given amino acid sequence (DSRLSKTMYSIEAPAKLDWEQNMAL) with cyanogen bromide results in peptide fragments only where the amino acid methionine (M) is present since cyanogen bromide specifically cleaves at the carboxyl side of methionine residues. There is one methionine residue in this sequence, resulting in two fragments upon cleavage.

When cleaving the sequence with trypsin, trypsin specifically cleaves at the carboxyl side of the amino acids lysine (K) and arginine (R). The sequence has three lysine residues and one arginine residue, which would generate five fragments upon cleavage.

In terms of which reagent gives the smallest single fragment in terms of the number of amino acid residues, that would be cyanogen bromide (C) because the cleavage will only occur once, at the single methionine residue, resulting in one of the two fragments being smaller than any fragment produced by trypsin.

Answer 2

Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments

Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments

Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)

Step-by-step explanation:

Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL

Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.