Question

A heptapeptide was found to hawe an anino acid composition of asp leu, lys, met.phe and tyr Trypsuin has no efect on the heptapeptide One cycle of Edman degradation randers the product whose structurs is shown below Chymatrypain treatment yields a dpapede and a tetrapaplide, as well as a bee amne acid The tetrapeptide is known to contain asp, leu lys, and met. Cyanagen bromde treatment generates a dpeptide that is determined to ariginate frone the N-taminus af the heptapetide, a letrapepoda, and hee lys The peptide sequence n sngle loter amino code is QUESTION 3 la prtenhas an isoelectnc port 예 of 102.which ofthe Olo inganno acid, ti t Ika'y lo hive a lot on o M and W o D andE o C and P Hand R o W and yY Chel Save ande and subeir.Chick Save AllAners to ve al auver

Answer 1

Ans. #2.

6 Ans. #1, #3. Trypsin treatm ent: Trypsin digest the peptide at carboxyl terminal of Arginine and Lysine in the target peptide bond. Given: Trypsin does not digest the peptide. So, Lys must be at the C-ter so that it woun't be acted upon by trypsin. So, the C-ter must be Lys. That is Position 7 = Lys #2, PTH-amino acid is the PTH derivative of Edman's degradation. The N-terminal amino acid is converted into its PTH adduct by Edman's degradation method, one amino acid per cycle. R side chain of respective amino acid The side chain shown here is Phenylalanine. Thus, N-ter is Phe So, position 1 Phe #3.CNBr treatment: Cyanogen bromide hydrolyzes the peptide at C-terminus of methionine in the target peptide bond Given, Cyanogen bromide treatment heptapetide, a tetrapeptide, and free lys To yield so, Met must be at Postion 2 and Position 6 Phe Met Met Lys So far, we have #4. Chymotrypsin Treatment: Chymotrypsin digests the peptide at the COO-terminus of a Tyr, Trp and Phe in the target peptide bond Chymotrypsin digest gives I. a dipeptide , II a tetrapeptide III. A free amino acid To yield 3 fragment, the peptide must have 2 chymotrypsin digest site 1st digest site is at Phe (position 1) to give a free amino acid To yield a dipeptide and a tetrapeptide (AND fulfil the criteia of CNBr digest) the second digestion site must be at postion 3 So, position 3 Tyi Phe Met Tyr Met Lys So far, we have Final Sequence: Postion 4 and 5 can be Asp and Leu, respectively or vice-versa. No information is provided to discriminate position 4 and 5 residues. Phe Met TyrAsp Leu Met Lys Final sequence Phe Met Tyr Leu Asp Met Lys Final sequence Final sequence = FMYDLMK or FMYLDMK

#3. In a polypeptide chain, the N- and C-terminals of amino acids are involved in peptide bond formation. So, these groups don’t contribute in the pI of a protein.

The R-group is freely exposed and contribute to the pI of the protein.

Option A. Incorrect: The side chains of methionine (M) and Trp (W) are non-ionizable. Hence, these residues don’t contribute in pI of protein.

Option B. Incorrect. Asp (D) and Glu (E) have acidic side chains. Their pKc values are 3.71 and 4.15, respectively. Abundance of these residues in protein will bring the pI of protein near pH 4.

Option C. Incorrect. Cys (C) has pKc of 8.14 whereas the side chain of proline is non-ionizable. So, a protein with abundant of these residues is likely to have a pI near 8.14.

Option D. Incorrect. Arginine (pKc = 12.10) and Histidine (pKc = 6.04) are basic residues. So, a protein with abundance of these residues is most likely to have a pI of 9.0 (the average value of 12.10 and 6.04).

Option E. Ccorrect. Tyr (Y) has pKc of 10.10 whereas side chain of Trp (W) in non-ionizable. So, only residues Y would contribute in pI. Therefore, abundance of these residues in protein is most likely to contribute pI of 10.10.