🤓 Based on our data, we think this question is relevant for Professor Singh's class at KSU.
(i) FDNB analysis determines the amino acid present in the N terminal - C
(ii) Trypsin digetsion cleaves the C-side of basic amino acids, namely: K and R. This shows that Fragment 2 must be the peptide in the C-terminal as it is the only sequence that do not have a cleavage point for trypsin.
(iii) Cyanogen Bromide digestion cleaves the C-side of methionine and methionine can ebd up forming a homoserine lactone. The 7mer would thus, start with R up until the C-terminal.
A single 14-amino acid peptide chain is to be assigned from the following experimental information below. Your job is to provide the full triple letter code sequence (if possible), calculate its isoelectric point and show your logic behind your answer.
(i) Amino acid analysis: Cx5, Dx1, Rx3, Nx1, Mx1, Wx1, Yx1, and Sx1
(ii) Dinitrofluorobenzene (FDNB) analysis indicated the presence of C on reaction of the native peptide.
(iii) Trypsin Digestion of the intact peptide produced 4 peptide fragments that were then exposed to Edman degradation analysis, they provided the following sequence information:
Fragment 1: CYSR
Fragment 2: CC
Fragment 3: DWMR
Fragment 4: CNCR
(iv) Cyanogen bromide digestion of the original intact peptide provided 2 peptide fragments, one 6 mer + homoserine lactone and one 7mer.
(v) The resulting 7mer was HPLC isolated and subjected to Pepsin digestion followed by amino acid analysis. The Pepsin digested peptide fragments A and B, were assigned the following amino acid composition:
Fragment A: S, Y, C, R Fragment B: C and R
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Based on our data, we think this problem is relevant for Professor Singh's class at KSU.