2. The activity of serine proteases is severely suppressed at low pH < 6. Why might this be the case? Draw the structure of the serine protease active site (just the main sidechains) at pH < 6 and at pH 8 to support your answer.
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- 3. Solve the sequence of an oligopeptide 7 residues long which gave: Asp Leu Lys Met Phe Tyr The following facts were observed: a. Trypsin treatment had no apparent effect b. The PTH derivative from Edman degradation was c. Brief chymotrypsin treatment yielded several products including but not limited to a dipeptide and a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Lys, and Met. d. Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and a free Lys. Instructions Make use of the table below to determine the sequence of the mystery protein.2. The mature form of TEM-1 ß-lactamase, an enzyme of 290 amino acid residues that hydrolyzes penicillin antibiotics. The protein has the following polar amino acid side chains: 18 Arg; 11 Lys; 6 His; 4 Tyr; 16 Asp; 20 Glu; 8 Asn; 5 Gln. Approximate pKa values 1.5 Arginine 12.5 Lysine 10.0 Histidine 6.5 OD 280nm Tyrosine 10.0 Aspartate or Glutamate 4.0 Terminal NH3+ 8.0 Terminal COOH 4.0 0.0 A 0.01 0.10 KCI B Volume (mL) The chromatogram above shows a typical result in the purification of TEM-1 ß-lactamase overex- pressed in E. coli cells. The protein is eluted from an anion exchange column at pH 6.25 with an increasing gradient of KCI. Fraction A is shown to correspond to native enzyme. Since Fraction B has the same amino acid sequence for the first 10 residues, the conclusion is that this fraction represents enzyme in which side chains of glutamine or asparagine residues have been hydrolyzed (deamidated). (a) ( ) What is the isoelectric point of the native enzyme? (b) ( ) Compared to…2. The mature form of TEM-1 ß-lactamase, an enzyme of 290 amino acid residues that hydrolyzes penicillin antibiotics. The protein has the following polar amino acid side chains: 18 Arg; 11 Lys; 6 His; 4 Tyr; 16 Asp; 20 Glu; 8 Asn; 5 Gln. Approximate pKa values 1.5 Arginine 12.5 Lysine 10.0 Histidine 6.5 OD 280nm Tyrosine 10.0 Aspartate or Glutamate 4.0 Terminal NH3+ 8.0 Terminal COOH 4.0 0.0 Α - 0.01 0.10 KCI B Volume (mL) The chromatogram above shows a typical result in the purification of TEM-1 ß-lactamase overex- pressed in E. coli cells. The protein is eluted from an anion exchange column at pH 6.25 with an increasing gradient of KCI. Fraction A is shown to correspond to native enzyme. Since Fraction B has the same amino acid sequence for the first 10 residues, the conclusion is that this fraction represents enzyme in which side chains of glutamine or asparagine residues have been hydrolyzed (deamidated). (a)( ) What is the isoelectric point of the native enzyme? (b)( ) Compared to…
- 1. Depict the structure features in the active site of cysteine protease 2. provide a step-wise reaction mechanism to show the hydrolysis of the peptide bond catalyzed by a cysteine protease2. Amino acid analysis of the a heptapeptide gave the following residues: Asp Glu Leu Lys Met Tyr Trp NH4+. The following facts were observed: Trypsin treatment had no effect. The phenylthiohydantoin released by Edman degradation was OH H C-C-CH₂ H. Brief chymotrypsin treatment yielded several products including a dipeptide and a tetrapeptide. The tetrapeptide contained Glu, Leu, Lys and Met is some order. Cyanogen bromide treatment afforded a tetrapeptide that had a net positive charge at pH 7 and tripeptide that had a zero net charge at pH 7. What is the amino acid sequence for this heptapeptide?1. In the mechanism of action of triose phosphate isomerase, Lys12 is an essential amino acid in the flexible phosphate gripper loop that undergoes a large conformational change to trap the substrate in the active site excluding water and holding the phosphate group in the correct geometry for isomerization rather than dephosphorylation. What would be the effect of the mutation of Lys12 to the unusual amino acid shown below? Explain. protein backbone
- wnich snows tne specinicity pockets. The S pocket nas a Ra glutamic acid in the bottom, the S2 pocket is small and hydrophobic, and the S,' pocket is deep and hydrophobic. Suggest a 3-amino acid sequence that this protease would R2 H cleave and indicate between which sites the peptide bond would be broken. S2 Which sequence would this protease cleave? Val-Lys-Phe Phe-Lys-Val Lys-Phe-Val Val-Phe-Lys Phe-Val-Lys O Lys-Val-Phe The peptide bond that is broken is between which sites? O S2 and S,' OS, and S,' O S2 and S1 O S2 and S1, and S and S,' IZ2. As we've seen previously, peptide bonds involving the amino acid proline can populate both "trans" and "cis" isomers. trans-Pro cis-Pro This isomerization reaction can be the rate-limiting step in protein folding. Depending on the protein sequence, the relative favorability of the trans and cis states can change. For Pro-117 in the protein staphylococcal nuclease, the rates of the forward and reverse reactions are: ktrans→cis = 3.1 × 10-2 and kcis→trans = 2.5 x 10-3 s-1. (a) What is AG°' for this reaction? (b) A prolyl isomeriase enzyme increases the rate of the forward reaction to kcat.trans->cis = 2.6 s-1. What is the new rate constant for the reverse reaction (kcat,cis→trans)? (c) This prolyl isomerase is a Michaelis-Menten enzyme with kcat = 2.6 s!, KM = 1.9 µuM. If you have 1 nM enzyme, what is Vo for [S]o = 0.7 µM? What about [S]o = 30 µM? For each condition, what limits the reaction speed?10. Chymotrypsin is a serine protease enzyme. The Km for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8*102, and the Km for the reaction of chymotrypsin with N- acetyltyrosine ethyl ester is 6.6*10“ M. catalytic triad Ser 195 His 57 Gly 193 N-H OH R-N- Ca N-H O-C- Asp 102 N-acetyl valine N-acetyl tyrosine Chymotrypsin Active Site a. What is the nucleophile here and how is it activated? b. Which substrate has an apparent higher affinity for the enzyme. c. Propose a reason for the difference in affinity based on the shape of each of the substrates (see active site figure, cleaves on the C-side of aromatic residues).
- 15. An amino acid binding protein in cell membranes has been shown to bind histidine, but only histidine molecules with a net charge of +1. Calculate [His+] in 20 mM histidine solution at pH = 6.4. The pka values for histidine are a-CO2H = 1.82, side chain = 6.00, a-NH3* = 9.17. Show all your work. NH3 HO Histidine (shown in fully protonated form)9. Shown below is a binding pocket for a protein with a ligand bound. The ligand interacts with a serine and a valine side chain. ligand a. Briefly explain, in terms of Ka, the effect of a mutation that replaces the serine residue with a valine residue. b. Briefly explain, in terms of Kd, the effect of a mutation that alters the ligand (as seen below) binding to the original, unmutated, binding site. :0: ligand 213. Amino acid analysis of a seven residue long peptide gave Asp Leu Arg Met | Phe Met | Phe | Tyr (i) The PTH amino acid derivative released by Edman degradation was Tyr-PTH (ii) Trypsin treatment had no effect (iii) Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and free Arg (iv) Brief chymotrypsin treatment yielded several products, including a dipeptide and a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Arg, and Met What is the amino acid sequence of this seven residue long peptide, and explain how you reached that conclusion?