peptide sequencing problem\nyou are a top - notch biochemist who has isolated a polypeptide from deadly…

Explanation:

Step 1: Determine N - terminal residue

Treatment with dabsyl chloride yields dabsyl - Ser, so the N - terminal amino acid is Ser. So the sequence starts with Ser.

Step 2: Analyze CNBr treatment

CNBr cleaves at the C - terminal of Met. Since CNBr treatment has no effect, Met is not at the C - terminal of any peptide fragment that can be cleaved by CNBr, which means Met is not followed by a residue that is cleaved by CNBr (i.e., Met is at the C - terminal of the polypeptide or not in a position where CNBr can cleave). But from trypsin treatment, we get free Met, so Met is a free fragment, meaning Met is at the C - terminal of a peptide that is cleaved by trypsin, and since it's free, Met is at the end of a trypsin - cleaved peptide, and since CNBr has no effect, Met is the last amino acid? Wait, no. Trypsin cleaves at the C - terminal of Lys and Arg (except when followed by Pro). The amino acid composition has Lys₂, Gly, Met, Phe, His, Ser, Glu.

Step 3: Analyze chymotrypsin treatment

Chymotrypsin cleaves at the C - terminal of aromatic amino acids (Phe, Trp, Tyr). Treatment with chymotrypsin yielded a dipeptide and a hexapeptide. So there is one aromatic amino acid (Phe) in the sequence, and the dipeptide or hexapeptide has Phe at the cleavage site.

Step 4: Analyze trypsin treatment

Trypsin yielded a tripeptide, a tetrapeptide, and free Met. Trypsin cleaves at Lys (and Arg) C - terminal (unless followed by Pro). So the free Met implies that Met is after a Lys (since trypsin cleaves at Lys C - terminal, so Lys - Met, and then Met is free? Wait, no. If trypsin cleaves at Lys, then a peptide with Lys at the C - terminal will be cleaved. So if we have free Met, that means Met is at the C - terminal of a peptide that is cleaved from a larger peptide, so the sequence before Met is Lys, so Lys - Met, and then Met is free.

Step 5: Combine N - terminal and other info

We know N - terminal is Ser. Let's list the amino acids: Ser, Lys, Lys, Gly, Met, Phe, His, Glu (wait, amino acid composition is Lys₂, Gly, Met, Phe, His, Ser, Glu. Wait, that's 7 amino acids? Wait, Lys₂, Gly (1), Met (1), Phe (1), His (1), Ser (1), Glu (1). So total 2 + 1+1 + 1+1 + 1+1=8? Wait, maybe I miscounted. Lys₂ (2), Gly (1), Met (1), Phe (1), His (1), Ser (1), Glu (1). 2 + 1+1+1+1+1+1 = 8 amino acids.

Wait, dabsyl - Ser: N - terminal is Ser (1). Trypsin: free Met, so Met is at the end of a trypsin - cleaved peptide, so Lys - Met (since trypsin cleaves at Lys, so the peptide with Lys at C - terminal is cleaved to give Met free? Wait, no. If a peptide has sequence X - Lys - Met, trypsin will cleave at Lys, giving X - Lys and Met. So Met is free, so the sequence before Met is Lys.

Chymotrypsin: cleaves at Phe. So we have a dipeptide and hexapeptide. So Phe is at the C - terminal of the dipeptide or hexapeptide. Let's assume the dipeptide is X - Phe, and hexapeptide is the rest.

Dabsyl - Ser: N - terminal is Ser (position 1).

Let's start building:

Position 1: Ser

Now, S. protease: pentapeptide and tripeptide with no charge at pH 5. Let's think about charges. Glu is acidic (negative at pH 5), Lys is basic (positive at pH 5), His is basic (positive at pH 5), Gly is neutral, Met is neutral, Phe is neutral, Ser is neutral.

A peptide with no charge at pH 5: sum of charges is zero. So if we have a tripeptide and pentapeptide with no charge. Let's consider the amino acids.

Let's go back to trypsin: tripeptide, tetrapeptide, free Met. So the total number of amino acids: tripeptide (3) + tetrapeptide (4)+1 (Met)=8, which matches the count (Lys₂, Gly, Met, Phe, His, Ser, Glu: 2 + 1+1+1+1+1+1 = 8? Wait, no, Lys₂ is 2, so total 2+1 + 1+1+1+1+1=8.

Trypsin cleaves at Lys, so the peptides from trypsin: let's say the tripeptide has one Lys, tetrapeptide has one Lys, and Met is free (from Lys - Met).

So sequence so far: Ser -... - Lys - Met (since Met is free, so the last part is Lys - Met, Met is position 8? Wait, 8 amino acids: positions 1 - 8.

Position 1: Ser

Position 8: Met (since Lys - Met, so position 7: Lys, position 8: Met)

Now, chymotrypsin: dipeptide and hexapeptide. Chymotrypsin cleaves at Phe, so Phe is at position, say, 2, and dipeptide is Ser - Phe? No, because dabsyl - Ser is N - terminal, so Ser is position 1. If chymotrypsin cleaves at Phe, the dipeptide could be X - Phe, and hexapeptide is the rest. Let's assume the dipeptide is His - Phe (since His is basic, Phe is aromatic). Wait, no. Let's think about the amino acids: Ser, Lys, Lys, Gly, Met, Phe, His, Glu.

Wait, let's list all amino acids: Ser (S), Lys (K), Lys (K), Gly (G), Met (M), Phe (F), His (H), Glu (E).

Now, dabsyl - Ser: N - terminal is S (1).

Trypsin: cleaves at K, so we have three fragments: tripeptide, tetrapeptide, and M. So the two K's are at positions where trypsin cleaves. So one K is at position 3, and another at position 7 (since position 7: K, position 8: M). So the fragments from trypsin:

• Tripeptide: S - X - K (positions 1 - 3: S - H - K? Wait, no. Let's think about S. protease: pentapeptide and tripeptide with no charge at pH 5.

A peptide with no charge at pH 5: the sum of charges of amino acids. Glu (E) is - 1, Lys (K) is + 1, His (H) is + 0.5 (at pH 5), Ser (S) 0, Gly (G) 0, Met (M) 0, Phe (F) 0.

So for a peptide to have no charge, the number of K (+1) and E (-1) should be equal, and H (+0.5) should be even? Wait, maybe the tripeptide is S - H - K (charge: S(0)+H(+0.5)+K(+1)=+1.5, no. Wait, maybe the pentapeptide and tripeptide: let's say pentapeptide is S - H - F - G - K and tripeptide is K - E - M? No, M is at position 8.

Wait, let's start over:

1. N - terminal: Ser (from dabsyl - Ser) → position 1: Ser (S)

2. CNBr: no effect → Met is not at C - terminal of a fragment that CNBr can cleave, so Met is at the end of the polypeptide (since CNBr cleaves at Met - X where X is not Pro, but if Met is at the end, there is no X, so no cleavage. And trypsin gives free Met, so Met is at the end, and before Met is Lys (since trypsin cleaves at Lys) → position 7: Lys (K), position 8: Met (M)

3. Trypsin: cleaves at Lys, so we have two Lys. So the other Lys is at position 3 (since position 7 is Lys, position 3 is Lys). So trypsin fragments:

   • Fragment 1: positions 1 - 3: S - X - K (tripeptide)

   • Fragment 2: positions 4 - 6: X - X - X (tetrapeptide? Wait, no, 8 - 3 - 1 (Met)=4, so positions 4 - 7: X - X - X - K, but position 7 is K, so positions 4 - 6: X - X - X, position 7: K

   • Fragment 3: Met (position 8)

4. Chymotrypsin: cleaves at Phe (aromatic), so Phe is at position 2 (since position 1 is S, position 2: F). So chymotrypsin fragment 1: S - F (dipeptide), fragment 2: positions 3 - 8: X - X - X - X - X - X (hexapeptide)

5. Now, position 2: F (Phe)

6. Position 3: K (Lys, from trypsin fragment 1: S - F - K (tripeptide: positions 1 - 3: S - F - K)

7. Now, the remaining amino acids: G (Gly), H (His), E (Glu), and the other Lys is already at position 7? No, wait, Lys₂: position 3 and position 7 are Lys.

8. Now, positions 4 - 6: let's see, remaining amino acids: G, H, E.

   • S. protease: pentapeptide and tripeptide with no charge. Let's see the pentapeptide: positions 1 - 5: S - F - K - G - H (charge: S(0)+F(0)+K(+1)+G(0)+H(+0.5)=+1.5, no. Wait, maybe positions 1 - 5: S - H - F - G - K? No.

Wait, maybe the correct sequence is Ser - His - Phe - Gly - Lys - Glu - Lys - Met. Wait, let's check charges for S. protease:

Pentapeptide: Ser - His - Phe - Gly - Lys (charge: S(0)+H(+0.5)+F(0)+G(0)+K(+1)=+1.5, no.

Tripeptide: Glu - Lys - Met (charge: E(-1)+K(+1)+M(0)=0. Ah! That works. So tripeptide: E - K - M (positions 6 - 8: E - K - M? No, position 8 is M, so positions 6 - 8: E - K - M? But position 7 is K, position 8 is M, so position 6: E.

Then pentapeptide: positions 1 - 5: S - H - F - G - K (charge: S(0)+H(+0.5)+F(0)+G(0)+K(+1)=+1.5, no. Wait, tripeptide with no charge: E ( - 1)+K (+1)+M (0)=0. So tripeptide is E - K - M (positions 6 - 8: E - K - M? But position 8 is M, so position 6: E, position 7: K, position 8: M.

Then pentapeptide: positions 1 - 5: S - H - F - G - K (charge: S(0)+H(+0.5)+F(0)+G(0)+K(+1)=+1.5, no. Wait, maybe His is at position 4.

Wait, let's try again:

• N - terminal: Ser (1)

• Met at 8, Lys at 7 (Lys - Met)

• Trypsin cleaves at Lys (3 and 7), so fragments: 1 - 3 (S - X - K), 4 - 6 (X - X - X), 7 - 8 (K - M) but 7 - 8 is K - M, and Met is free, so 7 - 8 is K - M, and 4 - 6 is a tetrapeptide? No, 8 - 3 - 2 (K - M)=3, so 4 - 6 is 3 amino acids. Wait, I think I messed up the number of amino acids. Let's recount: Lys₂ (2), Gly (1), Met (1), Phe (1), His (1), Ser (1), Glu (1). 2 + 1+1+1+1+1+1 = 8. So 8 amino acids.

Trypsin: tripeptide (3), tetrapeptide (4), and free Met (1). 3 + 4+1 = 8. Correct.

Chymotrypsin: dipeptide (2) and hexapeptide (6). 2 + 6 = 8. Correct.

S. protease: pentapeptide (5) and tripeptide (3). 5 + 3 = 8. Correct.

Now, let's assign:

1. Ser (1) - N - terminal (dabsyl - Ser)

2. His (2) - let's see chymotrypsin: dipeptide could be His - Phe (positions 2 - 3: His - Phe), but chymotrypsin cleaves at Phe, so dipeptide is His - Phe (positions 2 - 3), hexapeptide is positions 1,4 - 8. No, position 1 is Ser, so hexapeptide is Ser, positions 4 - 8.

Wait, maybe the correct sequence is Ser - His - Phe - Gly - Lys - Glu - Lys - Met.

Let's check each treatment:

• Amino acid composition: Lys₂, Gly, Met, Phe, His, Ser, Glu. Correct (Ser, His, Phe, Gly, Lys, Glu, Lys, Met: Lys₂, Gly, Met, Phe, His, Ser, Glu).

• Chymotrypsin: cleaves at Phe (position 3). So dipeptide: Ser - His - Phe? No, chymotrypsin cleaves at Phe, so the bond after Phe is cleaved. So if Phe is at position 3, the dipeptide is positions 1 - 3: Ser - His - Phe (tripeptide? No, chymotrypsin gives dipeptide and hexapeptide. So Phe is at position 2, dipeptide: Ser - Phe (positions 1 - 2), hexapeptide: positions 3 - 8: His, Gly, Lys, Glu, Lys, Met.

• CNBr: no effect. Met is at position 8, so no Met - X where X is cleavable by CNBr (CNBr cleaves at Met - X where X is not Pro). Since Met is at position 8, there is no X after Met, so no cleavage. Correct.

• Trypsin: cleaves at Lys (positions 5 and 7). So fragments:

  • Tripeptide: positions 1 - 3: Ser - Phe - His (no, Lys is at positions 5 and 7). Wait, I think I made a mistake in Lys positions.

Let's start over with correct trypsin cleavage (trypsin cleaves at Lys and Arg, C - terminal, unless followed by Pro):

Amino acids: S, K, K, G, M, F, H, E (8 amino acids)

N - terminal: S (1)

Met is free, so M (8), and before M is K (7): K - M (7 - 8)

Trypsin cleaves at K, so we have two K: K (3) and K (7)

So trypsin fragments:

• Fragment 1: 1 - 3: S - X - K (tripeptide)

• Fragment 2: 4 - 6: X - X - X (tetrapeptide? No, 8 - 3 - 1 (M)=4, so 4 - 7: X - X - X - K (7 is K), so 4 - 6: X - X - X, 7: K

• Fragment 3: M (8)

Chymotrypsin cleaves at F (aromatic), so F (2)

So fragment 1 (trypsin): 1 - 3: S - F - K (S(1), F(2), K(3))

Fragment 2 (trypsin): 4 - 6: G - H - E (G(4), H(5), E(6)), 7: K(7), 8: M(8)

Now, chymotrypsin: cleaves at F(2), so dipeptide: S - F(1 - 2), hexapeptide: K(3), G(4), H(5), E(6), K(7), M(8) (positions 3 - 8: 6 amino acids, hexapeptide). Correct.

S. protease: pentapeptide and tripeptide with no charge.

Pentapeptide: 1 - 5: S - F - K - G - H (charge: S(0)+F(0)+K(+1)+G(0)+H(+0.5)=+1.5, no.