Biomolecules: lecture 9
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1 Biomolecules: lecture 9 - understanding further why amino acids are the building block for proteins - understanding the chemical properties amino acids bring to proteins - realizing that many proteins are posttranslationally modified before they are functional in cells - how amino acids bond together to a chain by peptide bonds - how the peptide bond determines the way how proteins (i.e. polypeptide chains) can fold and form a 3D structure
2
3 ph and pka ionization of an amino acid: ph = -log 10 [H + ] pka is the ph at which 50% dissociation occurs Henderson-Hasselbalch equation: ph = pka + log ([A - ] / [HA]) T.A. Brown Biochemistry Scion Publishing Limited 2017
4 Isoelectric point pi is the ph at which the net charge of the molecule is zero For a 2 pka system: pi = ( pka1 + pka2) / 2
5 Typical pka values for naturally occurring amino acids: a-amino Arg (d-guanido) 12.0 Lys (e-amino) His (imidazole) a-carboxyl Asp (b-carboxyl) Glu (g-carboxyl) Cys (thiol) Tyr (phenolic hydroxyl) Question: Why is histidine often found in the active site of enzymes which catalyze redox reactions?
6 >> the chemical environment affects the pka of an amino acid This is a very important detail for many enzymatic reactions, i.e. the surrounding amino acids in a protein structure can have a big effect.
7 Example: the side chain of lysine is positively charged in physiological conditions and mostly also in conditions of typical protein purifications NOTE: in a protein only the side chain ionizable groups are considered, bacause aminoand carboxy groups are bound in the main chain (except amino and carboxy termini)
8 Post translational modifications a) Glycosylation b) Phosphorylation c) Proteolytic processing d) Disulphide bond formation e) Amidation f) Hydroxylation g) Acetylation h) Lipid attachment i) Sulphation j) ADP-ribosylation
9 Proteolytic processing
10 Disulphide bond formation
11 Modified amino acids
12 Peptide bond connects amino acids >> polypeptide amino acid amino acid Peptide bond amino acid residue amino acid residue (aminohappotähde)
13
14 NOTE: Peptide bond between N and C atoms is shorter than the bond between N and C-alpha. What can you expect based on this?
15 Peptide bonds exhibit double bond characteristics (due to resonance hybridization) including: a) Dipolar character b) All atoms in the peptide bond lie in a plane c) Geometrical isomerization with the trans isomer being much more stable than the cis Exercise: Draw the peptide bond such that to show the double bond character it has.
16 Cα - in between two peptide bonds - side chain begins at alpha-carbon!
17 Amide plane: phi = ϕ and psi = ψ angles Amide plane rotation in proteins:
18 Fig 3.15 phi and psi angles can only have certain values in a polypeptide chain of a folded protein Reproduced from: Biochemistry by T.A. Brown, ISBN: Scion Publishing Ltd, 2017
19 Ramachandran plots for permitted phi / psi angles Alanine Glycine
20 Forces involved in protein structure a) Steric repulsion b) Covalent bonds c) Electrostatic attraction/repulsion d) Dipole-dipole interactions, including hydrogen bonds e) Van der Waals forces
21 How protein structure forms (here example of non-covalent bonds) Alberts et al. Essential Cell Biology, 3rd edition
22 What is a domain? domeeni Alberts et al. Essential Cell Biology, 3rd edition Mathews et al. Biochemistry, 4th edition Hox-hox: domains are separate parts, but still parts of the SAME popypeptide chain
23 Hox-hox: subunits are separate popypeptide chains
24 Organisation of protein structure a) Primary Structure (1 ) is the amino acid sequence of the protein s polypeptide chain(s) + all covalent bonds! b) Secondary Structure (2 ) is the local spatial arrangement of a polypeptide s backbone atoms c) Tertiary Structure (3 ) is the three dimensional structure of the entire polypeptide d) Quaternary structure (4 ) is the spatial arrangement of the subunits in oligomeric proteins
25 One of the main driving forces for protein structure formation is the formation of a hydrophobic core in which the non-polar side chains can pack tightly together to exclude water. Any unsatisfied hydrogen bond donors or acceptors would prefer to be in an environment where they can form hydrogen bonds e.g. with water. Thus to form a stable hydrophobic core all backbone hydrogen bonding must be satisfied. This requires all backbone CO and NH groups to be involved in hydrogen bonding, this leads to regular secondary structure
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