Lecture Slides II F24 PDF

First Half: Lecture Slides II Weeks 3 & 4 (Lectures 5-8)  Polypeptides and proteins: structural hierarchy, sequence. Basis of reactivity and hydrolysis A protein is a long linear chain of amino acids Myoglobin...

First Half: Lecture Slides II Weeks 3 & 4 (Lectures 5-8)  Polypeptides and proteins: structural hierarchy, sequence. Basis of reactivity and hydrolysis A protein is a long linear chain of amino acids Myoglobin O2 binding protein from muscle 153 amino acids in chain The linear chain of myoglobin is precisely folded into a compact, 3-dimensional globular structure This image is an accurate representation of the structure of myoglobin Too much detail makes it hard to comprehend The linear chain of myoglobin is precisely folded into a compact, 3-dimensional globular structure A simplified view shows distinct regular C-term patterns in folded protein heme ring The ribbon traces the path of the _________________________(peptide bonds only, no side chains) Colour coding: blue at N-terminal progressing to red at C-terminal In myoglobin, the ribbon is arranged in eight spiral or helical segments N-term Amino acid side chains fill the “spaces”, which are not empty  Protein structure is organized in three (or four) levels ▪ _______ structure: the linear sequence of amino acids ▪ __________ structure: regular repetitive patterns, such as the helical sections in myoglobin, that run along short sections of peptide chain (5-20 AAs) ▪ __________ structure: the overall pattern of 3D folding of the whole polypeptide ▪ Protein function depends ▪ in myoglobin, 8 helical sections enclose a central cavity on positioning of key amino acids close to each ▪ Some proteins also have a _____________ level of other in 3-D space, even structure though they are far apart in ▪ e.g. hemoglobin is an assembly, or complex, of the linear sequence. four globin units, each one similar to myoglobin ▪ The exact pattern of ▪ the four globins act cooperatively to improve folding is critical for oxygen transport protein function. Protein structure is organized in three (or four) levels To investigate the structure of a protein, we find out which amino acids are present, and in what sequence  Peptide bonds of protein are ________________ with help of catalyst to release individual amino acids  Acid hydrolysis: 6 M HCl 110o, 24-72 h to completion  Base hydrolysis: 4 M NaOH, 110o, 16 h to completion  Some amino acids could get destroyed under acid and base hydrolysis. Also hydrolyses the amide bonds of Asn and Gln, converting them to Asp and Glu respectively and releasing the amino group as ammonia  With digestive enzymes called __________________:  enzymes are proteins which have a catalytic function  proteases catalyse hydrolysis of peptide bonds  After hydrolysis of the protein, amino acids can be analyzed by chromatography – tells you how much of each amino acid is present  Basis of chemical reactivity in hydrolysis and other biochemical reactions Chemical reactivity arises from an unbalanced distribution of valence electrons – C-C and C-H bonds share bonding electrons equally and are both non- polar and chemically unreactive Reactions involve atoms which have unshared electrons (lone pairs), or are electron deficient, or pull electrons towards them due to electronegativity Many biochemical reactions are initiated by __________________ - A nucleophile (“nucleus-lover”) is an atom with a lone pair of electrons available to share - Nucleophiles seek out other groups that are electron-deficient (“nuclei”)- electrophiles + : : O O : N : But Not C N : Nucleophiles S : : weak strong : An atom with a lone pair may use it in different ways 1. __________________acceptor - if it simply attracts an O-H or N-H R-O: - - - H–N-R 2. __________ - if it captures H+ H+ + :NH2–R  +NH3–R 3. ______________ when it shares the lone pair with another electron-deficient atom to make a new bond + : : : : HO: + C O HO C O Nucleophilic substitution and Nucleophilic addition Nucleophilic substitution or displacement: – incoming nucleophile X attacks target atom C to displace leaving group Y which takes its bonding electrons away X: + C Y X C + :Y Nucleophilic Addition: – In cases where Y is __________ bonded to C, only one bond has to be given up, so Y does not leave X: + C Y X C Y Hydrolysis of peptide bonds (nucleophilic displacement) - : : : O H : : : O O: H H Curly arrows represent the R C N R' R C N R' R C :N R' movement of electron pairs :O: H+ :O: H :O:- : H H H Transition state C of the peptide bond is electron-deficient (an electrophile), because the electronegative O bonded to C pulls electrons away from it this lets C take up the incoming electron pair of OH in a new bond forming the transition state The transition state is semi-stable, like a compressed spring Transition state relaxes back by sharing electrons with C and re-forming the double bond Amino N acts as __________________, allowing the peptide bond to break To investigate the structure of a protein, we next 1918-2013 need to determine the amino acid sequence We can identify the first amino acid in a protein (the N- terminus) by tagging it with fluorodinitrobenzene (bright yellow) Fred Sanger, who worked on insulin from 1947-1953; won the Nobel Prize in 1958 The amino acid at the N-terminus of a protein has a ______________________ At high pH, this group deprotonates to become :NH2-, a nucleophile Reacts with fluorodinitrobenzene; HF is a good _____________________ Hydrolysis releases the N-terminal amino acid, with yellow tag attached; can identify it by chromatography Big snag: hydrolysis destroys the rest of the polypeptide chain  Determining amino acid sequence of a protein Sanger’s method can only determine the N-terminal amino acid ONCE; we can’t repeat it, since hydrolysis destroys the rest of the polypeptide chain Perh Edman (Sweden 1956) improved the Sanger method His method allows N-terminal amino acid to be reacted, removed, and identified ________________________ the peptide bonds Reaction can be _____________ to identify amino acid #2 etc  Edman Degradation 1. Coupling R1 ▪ Involves 2 steps: O R1 O NH Base NH N : 1. ____________ requires base; NH : H2N C N C reaction must be complete O R2 S H O R2 S before the next cyclization step PITC 2.Cyclization Anhydrous can take place (labels the N- CF3COOH terminal AA with PITC) O H+ + R1 Back to H3N 2. ____________ requires acid; step 1 O N reaction must be complete R2 C S before the next coupling step Shortened peptide HN can take place (cleaves the first peptide bond) Identify amino- ▪ Cycle can be repeated up to 50 terminal residue Further times, to give a 50 AA sequence Modifications using automated sequencers You will NOT be asked to reproduce this mechanism  To study long protein chains, they are cut into shorter oligopeptides by selective hydrolysis (divide and conquer) N-term end Selective hydrolysis cuts the polypeptide at specific locations, to yield a limited number of oligopeptides of definite size The digestive enzyme trypsin binds and recognizes ________________ side chains in peptides carboxylate group of the Arg or Lys is positioned next to catalytic unit of trypsin, and is target for hydrolysis Proline has wrong shape and of peptide bond won’t fit in position after Arg or Lys  Trypsin and chymotrypsin convert polypeptides into smaller fragments   X Trypsin acts on Gly---------Lys-X-------Arg-Y------Lys-Pro-----Asn to yield Gly---------Lys + X------Arg + Y------Lys-Pro-------Asn Pro residue in the position after Lys or Arg has wrong shape to fit the enzyme, so no hydrolysis occurs if target Arg/Lys is followed by Pro Note that all fragments will have Arg/Lys at the _________________, EXCEPT the fragment from the C-terminus (so can easily identify it) Chymotrypsin acts on X Gly---------Phe-X-------Trp-Y------Phe-Pro-----Tyr-Z------Asn to yield Gly-------Phe + X-------Trp + Y------Phe-Pro-----Tyr + Z------Asn All fragments (except C-terminus) have Phe/Tyr/Trp at the C-end  Cyanogen bromide is a chemical reagent which cuts polypeptide chains at methionine residues Cyanogen bromide Br-CN attacks S atom of Met Peptide chain is broken on carboxylate side and Met is converted to homoserine, Hse (serine with extra CH2)   CNBr acts on Gly------Met-X-------Met-Pro------Asn To yield Gly------Hse + X-------Hse + Pro------Asn All fragments (except C-terminus) have Hse at C-end  There are many potential cut sites in myoglobin; these yield oligopeptides of defined size Proteins digested with enzymes like trypsin yield characteristic patterns of fragments of different molar mass – can be analyzed by mass spectrometry – this is a definitive means of identifying a ______________ In experiments to determine the amino acid sequence of an ____________ protein, oligopeptide fragments from The peptide sequences are selective hydrolysis are first separated reassembled into a complete by chromatography, then each can sequence by the overlap method be sequenced by Edman’s method  Overlap method ▪ Two samples of the original polypeptide are each cut separately using two ____________ methods, each targeting different sites (e.g. trypsin, chymotrypsin) Peptides obtained Peptides obtained Using trypsin Using chymotrypsin MET ASN LYS GLN GLY ASP ILE ALA LYS MET ASN LYS GLU ALA LEU PHE GLU ALA LEU PHE ARG ARG ASP ILE ALA LYS GLU LEU TYR GLU LEU TYR GLN GLY ▪ Sequences from one set of oligopeptides are lined up to overlap with oligopeptides from the other set, to deduce how they were originally joined GLU ALA LEU PHE ARG ASP ILE ALA LYS ARG ASP ILE ALA LYS GLU LEU TYR Cut these out and build the sequence using the overlap method Met Asn Lys Gln Gly Asp Ile Ala Lys Met Asn Lys Glu Ala Leu Phe Glu Ala Leu Phe Arg Arg Asp Ile Ala Lys Glu Leu Tyr Glu Leu Tyr Gln Gly Amino acid sequences of proteins are now most often derived from the DNA sequences in genome databases ▪ Nowadays, it is common for the _______________________of a protein gene to be determined first, using cloning and molecular biology methods ▪ Can then _____________ the protein sequence, using the genetic code DNA sequence  amino acid sequence ▪ Does not involve any protein sequencing experiments – cheap and easy!  Mass spectrometry can also be used to sequence and identify proteins Tandem mass spectrometry Only tiny amounts of sample needed (spot from 2D gel) 1 2 1. Sample hydrolyzed by protease into peptides 1 2. First MS-1: sorts the different peptides and select one type to go into the collision cell 2 3. Collision cell: _______________ each peptide molecule once 3 (usually the peptide bond) in a random fashion 3 4. Second MS-2: measure fragment masses 4 4  Trypsin is the protease of choice in mass spectrometry Breakage of peptide bonds in the collision cell fragments each peptide into 2 : b-type (N-terminal fragments) and y-type (C- terminal fragments) With _____________ digestion, all y-type fragments (no matter which peptide bond is broken) will have a charged R or K at the C terminus resulting in a positive charge at the peptide C-terminus. Peptides with ____________ produce the highest signal, i.e. the ones with K or R at C-terminus R5 = Arg+ or Lys+  Deducing protein sequence through mass spectrometry ▪ The second MS - measure the masses of the C-terminal fragments (y-type) generating a set of peaks. Example Peptide: G A S S I S Y P A R The peaks are arranged in order from the smallest to the largest mass (from smallest to the largest y-type fragment) Each peak corresponds to a y-type fragment formed by breakage of a peptide bond at a different point in the peptide chain  1 2 3 4 5 6 7 8 9 G–A–S–S–I–S–Y–P–A-R Peptide Bond C-terminal Mass of Δ Mass Amino acid Broken fragment fragment None GASSISYPAR 989 989-932 = 57 Gly 1 ASSISYPAR 932 932-861 = 2 SSISYPAR 861 861-774 = 3 SISYPAR 774 774-687 = 4 ISYPAR 687 687-574 = 5 SYPAR 574 574-487 = 6 YPAR 487 487-324 = 7 PAR 324 324-227 = 8 AR 227 227-156 = 9 R 156 156-0 =  Deducing protein sequence through mass spectrometry ▪ Going from the largest to the smallest (from right to left on the chromatogram), each successive peak has one less amino acid than the peak before. ▪ The difference in mass from peak to peak identifies the amino acid that is lost in each case, revealing the sequence of the original peptide 71 57 861 932 989  Mass spec problem to work on Julie the biochemist completed mass spectrometry-based sequencing of a peptide from her protein. She found the following mass values for the peaks of the fragments (generated from the C terminal end), listed here in Daltons (Da): 128.09; 227.16; 328.21; 443.23; 629.31; 700.35; 771.38. Using the list of masses of amino acid residues (below), what is the identity of the third amino acid residue in the peptide when the sequence is written in the standard way (N- to C-terminus)? Amino acid residue masses (Da) G: 57 R:156 K: 128 V: 99 S: 87 W: 186 D: 115 A: 71 T: 101 Y: 163 N: 114 F: 147 A. Trp B. Asp C. Ala D. Thr  NCBI and BLAST NCBI database (National Center for Biotechnology Information) www.ncbi.nlm.nih.gov Vast amount of biological information: – Protein sequences – DNA & RNA sequences – Genomes – Health-related mutations – PUBMED  BLAST searching BLAST (Basic Local Alignment Search Tool) Compares input sequence to databank of all known protein sequences Produces a list of the best “hits” 1. 100% identical: positive match 2. Good identity: a ___________ 3. No identity: a new protein? Perform biochemical tests to determine identity  Homework Questions https://forms.office.com/r/u60VvtJEJG  The alpha-helix – Arts ‘n’ crafts O H C Cut this peptide chain out and bring to next class O See “Extra stuff” on CourseLink – Arts ‘n’ Crafts C H C R7 O N C H C H R6 O N C H C H R5 O N C H C H R4 O N C H C H R3 O N C H C H R2 N C H R1 N H Protein structure is organized in three (or four) levels Secondary structure: regular repetitive patterns such as helix, in short sections of the polypeptide chain The polypeptide chain forms a backbone which __________ to be linked by C- C and C-N single bonds Single bonded structures are ____________ due to bond rotation  Groups connected by single bonds can ________ about bond axis  Chain flexibility arises from bond rotation, not bond ____________  Normal 109o ____________ or 120o ____________bond angles are present  Bond rotation allows the peptide chain to adopt a variety of shapes  Conformations vs. configurations  Conformations represent states of a molecule that can be interconverted by ______________, without ____________ covalent bonds e.g., different shapes of a polypeptide chain  Configurations can only be interchanged by breaking covalent bonds, not by bond rotation e.g., ______________- forms of molecules with a CH CH 3 H 3 CH3 C C C C -C=C- double bond H H CH 3 H Cis Trans two chiral forms of amino acids (D- and L-)  For macromolecules such as proteins, we are usually concerned with conformations X-ray diffraction measures regular repeating patterns on the molecular scale NaCl protein  Crystals are ordered arrays of molecules crystals crystals  Atoms and molecules have similar ______________ to the wavelength of X-rays (0.7-1.5 Å)  when X-rays reflect off a regular ___________ structure, e.g., molecular crystal or fibre, they are deflected by an angle dependent on wavelength of X- ray and spacing of pattern  If X-ray wavelength is known, dimensions of the repeating pattern of molecules can be calculated.  Some important structural parameters were identified from early studies of diffraction patterns of fibrous proteins such as α-keratin and β-keratin.  X-ray diffraction measures regular patterns in fibrous proteins Major Pattern Minor pattern -keratin 5.4 Å 1.5 Å Protein of hair, skin, wool -keratin or fibroin 7.0 Å 3.5 Å Silk moth and spider silk The ångstrom unit, 1 Å = 1  10–10 meter, is commonly used to measure atomic structures; H atom and C–H bond are about 1 Å in size  The fibre patterns were interpreted by Linus Pauling, a physical chemist, who was an expert in molecular structure and bonding  Pauling built precise scale models of peptide chains with accurate bond lengths, bond angles and atomic radii  He found that the single-bonded peptide chain seemed too flexible, and no regular patterns would be stable.. Protein secondary structure Nobel Prize in Chemistry 1954 Linus Pauling 1901-1994 Nobel Peace prize 1962  Key finding: the peptide bond has double bond character The peptide bond has two resonance forms, one with a ________________ Pauling compared lengths of C-N bonds to correlate bond length with bond order Normal C–N is 1.49 Å Peptide C–N is 1.32 Å Normal C=N is 1.27 Å A peptide bond is rigid, fixed in __________________, because it behaves more like a double bond than a single bond  Peptide bonds form rigid planes connecting tetrahedral α-C atoms In normal peptide chain, -amino and -carboxylate are locked in rigid planar peptide bonds, only the two ______________can rotate freely  Restricted bond rotation leads to only a few possible structures  Restricted bond rotation limits freedom of motion, so that only a few regular structures can form  The peptide backbone changes direction by 109o at each tetrahedral -C, defining two possible regular repeating patterns:  in a ________ shape, every -C bond down the peptide chain turns in same direction (e.g. clockwise)  in an _________ shape, the -C bonds turn in ___________ directions down the peptide chain  If there is no regular, repeating structure, get random coil  The alpha-helix ▪ Helix forms when amino acids all have same orientation, and -C bonds turn in same direction (i.e. all clockwise) ▪ Pauling’s models showed a very stable structure with _________________ per turn of helix ▪ #1 C=O lines up with H–N #5 to form hydrogen bonds that make -helix stable ▪ Distance between each turn of helix is _______ ▪ 5.4 ÷ 3.6 = _________ is distance along the helix per amino acid ▪ These distances match -keratin patterns exactly, Looking down hence the name -helix the axis  X-ray diffraction measures regular patterns in fibrous proteins Major Minor Pattern pattern -keratin 5.4 Å 1.5 Å Protein of hair, skin, wool −keratin or fibroin 7.0 Å 3.5 Å Silkmoth and spider silk  The extended β-strand and β-sheet occur when amino acids alternate in orientation strands in _____________ directions make antiparallel β-sheet H-bonds align better in antiparallel mode strands in the _________ direction make a parallel β-sheet H-bonds connect strand to strand Dimensions of β-sheet match β-keratin patterns edge-on view of β-sheet Maximum space available for: 1. bulky or 2. awkward shaped side chains Trp, Tyr, (Phe) are _____ Val, Ile, Thr have a _____ on β-C Cys has large S atom on β-C WYF VIT C X-ray diffraction measures regular patterns in fibrous proteins Major Minor Pattern pattern -keratin 5.4 Å 1.5 Å Protein of hair, skin, wool -keratin or fibroin 7.0 Å 3.5 Å Silk moth and spider silk Formation of secondary structure is based on consensus of amino acids in immediate vicinity  Ala, Arg, Gln, Glu, His, Leu, Lys, Met, (Phe) tend to form -helix _____________ behaviour of amino acids  Trp, Tyr, (Phe), Val, Ile, Thr, Cys need _________, prefer -sheet structure  __________________determines which secondary structure forms  Preference of each amino acid for  or  secondary structure can be estimated (see Euchre card deck)  Allows secondary structure of sections of a protein to be _________ using software programs  Secondary structure breakers ▪ Gly, Pro, Asn, Asp, Ser have side chains which ___________ with secondary structure H-bonds GPNDS ▪ may disrupt secondary structure – secondary structure breakers ▪ 2 breakers in a group of 4 amino acids interrupts the secondary structure ▪ forms a ________ or ____________, allows the polypeptide main chain to change direction drastically (can be 180)  You, too, can have fun with PyMOL! A FREE educational version of PyMOL is available for download to all students in this course See “Links” tab on the top menu bar in CourseLink You will need the login and password on the PyMOL site You can import 3D-structure files (pdb files) from the Protein Data Bank (see “Links” tab) and use them to explore the protein molecule Protein structure is organized in three (or four) levels  Native and denatured states of proteins  Most proteins are folded into a unique 3D tertiary structure, which is required for their function  _________ state  _______________ unfolds proteins; unfolded form may be unstructured or aggregated  often _____________  Protein’s function is typically __________ on denaturation  Ways to denature proteins include:  heat  disruptive solvents  harsh detergents (e.g. SDS)  When purifying proteins to study them, we usually try to avoid denaturation  Tertiary structure is the overall pattern of folding of the whole polypeptide chain  The simplest possible tertiary structure is continuous secondary structure  -keratin is totally   Fibroin (a -keratin) is totally ____________________  Collagen (tendon, bone and connective tissue) has a unique ____________ structure  Collagen structure requires sequence with repeating units of three amino acids -Gly-Pro-X-  Secondary structure is rigid, so these are fibrous proteins  Most proteins are globular: this requires the polypeptide to fold back on itself Pro  Folding requires “breaks” in secondary structure, Gly which is rigid  Clusters of 2-3 secondary structure breakers (Gly, Pro, Asn, Asp or Ser; GPNDS) in a run of 4 AAs  Allows for __________________________where polypeptide can change direction to allow folding  The hydrophobic effect is a major force driving protein folding  Folding the protein encloses most of the non-polar amino acids in the ____________  Amino acids are drawn with space-filling atoms, colour- coded polar or non-polar  Non-polar AAs group together to minimize contact with H2O (hydrophobic effect)  Polar AAs form outer layer, interact well with surrounding H2O (good H-bonding) or with ions in solution PyMOL  Amino acids pack together with jig-saw puzzle fit  The side chains interlock to maximize the number of close atom-to-atom contacts  Close contacts attract by weak _________________ forces  Aka London dispersive forces  0.1 to 1 kJ/mol per contact  Good fit makes hundreds of close contacts in macromolecule, and helps to hold structure together Part of myoglobin showing two  Poor fit only makes a few contacts helices, paired like a hairpin Summary: the way a protein folds is dictated by the sequence of its amino acids  Amino acids “elect” secondary structures  “Breaker” AAs allow for folding, introduce flexible sections  Distribution of non-polar amino acids in sequence determines which parts fold inwards  Polar amino acids interact well with aqueous surroundings  Pattern of large and small side chains is arranged so that secondary structure components (e.g. helices) pair up with best possible fit A “balancing act”; many different forces in play, protein reaches the stable native state Proteins fold into a limited number of tertiary structure patterns ⚫ proteins consisting of mostly -helical segments ⚫ proteins consisting of mostly -strand segments ⚫ proteins with alternating -helical and -strand segments Problem Solving Tip: When solving complex problems, start with the information you know about that topic eg. If solving a problem on protein tertiary structure, the basic information you see here would be a good starting point. Ask yourself - What are the possible tertiary structures? A sequence with mostly groups of -helix-forming AAs will fold into an -helix bundle  Small clusters of breakers set the limits of each helix cytochrome b1  Non-polar AAs every 3 or 4 places in the helix make a _________________________, e.g. –PPNPPNNP-, which fold to inside of bundle  AAs that prefer -sheet are present, but scattered  Myoglobin is a bundle of 8 -helix sections  Bundles of 6-8 helices appear more complex because the myoglobin helices splay apart β-sheet-forming amino acids in majority fold into antiparallel β-sheet  Anti-parallel sheet is more stable because H-bonds are arranged in ___________________  Side chains _______________ of the sheet; odd on one side, even on other side  Sheet can be non-polar on one side, polar on the other side A -sheet that is polar on one side and non-polar on the other wraps around to enclose the non-polar face inside  A small sheet (3-5 strands) makes an _____________  A larger sheet (6-8 strands) wraps all the way around to form an antiparallel   This example is green fluorescent protein PyMOL  Sequences which alternate structure −−− can form parallel -sheet  Helical sections connect the strands, which all run in the ___________________  Helix lies __________ or __________ the plane of the sheet  Parallel -sheets are less stable (angled H-bonds), so must be sequestered away from H2O  Usually buried in centre of protein, thus made up of mostly non-polar amino acids If all the helices lie on one side of the sheet, the sheet wraps around to form a parallel -barrel  The -sheet forms the central _________, surrounded by the connecting   Example is the enzyme triose phosphate isomerase. pymol  Helices on both sides of the sheet give the parallel - sandwich structure  Sandwich filling is the non-polar β-sheet between two layers of a-helix  The β-sheet is often twisted for better packing  Example is part of the enzyme lactate dehydrogenase  Many larger proteins fold up in different sections called domains  Each of the structures outlined above forms from a chain of 10-20 kDa  Larger proteins fold up in 10-20 kDa sections; each section is called a ____________  A protein of 50 kDa may have 3 or 4 domains  Each domain may adopt a different folding pattern  Thus, larger proteins are often __________ in nature  Example is lactate dehydrogenase with two - sandwich domains  Protein stability and function  Covalent bonding links amino acids in a chain in a specific sequence  __________________ interactions dictate folding pattern and stability  Hydrophobic effect and van der Waals effect are the most important non- covalent interactions  ______________________ locates non-polar amino acids in core of folded protein, avoiding unfavourable interaction with H2O  contributes ~50% of total energy stabilizing native state  –5 kJ/mol per CH, CH2 or CH3 moved out of contact with H2O  ______________________– dipole-dipole interactions between atoms that are close, but not covalently bonded to each other – aka London dispersion forces  contributes significantly if many atoms are in close contact  Polar interactions may also help stabilize the correct folding of a protein ____________ may form between donors and acceptors that line up in the folded protein _________: strong electrostatic interaction of negative side chains which pair up with positive side chains that are nearby in the folded protein - salt bridge Ion pairs and H-bonds make less contribution than hydrophobic or van der Waals interactions, because most polar groups face the exterior: charged AAs pair with ions in external solution H-bonding AAs bond to H2O in surroundings Disulfide bonds help to hold together the tertiary structure of some proteins __________________ form when pairs of Cys – SH groups react with O2, releasing H2O This makes a strong covalent bond to help hold the folded protein together Few proteins have disulfide bonds; mostly proteins designed to function outside cells since O2 needed Bovine Insulin The primary structure contains all the information required for the secondary and tertiary structure of a protein Christian Anfinsen Nobel Prize 1972 The enzyme ribonuclease was denatured (catalytic activity is lost) by placing it in 8 M urea solution with 2- mercaptoethanol Urea weakens ______________________, so allows protein to unfold 2-mercaptoethanol is a _______________ that converts disulfides back to the original unlinked Cys-SH groups The denatured ribonuclease regained its biological activity once urea and mercaptoethanol were removed indicating it has regained its native structure - _______________ Showed that the amino acid sequence contains all the information required to fold the chain  Homework Questions https://forms.office.com/r/qf8BvaHYFz  Homework Questions Contd., Find “Alpha Helix and Beta Sheet: Arts N’ Crafts” on courselink. It contains paper models of α‐helix and β‐sheet. The helix model needs to be cut out and glued to form a cylinder. The β‐sheet model just needs to be folded as directed. The following represents a sequence of amino acids found in an α‐helical section of protein: Leu‐Pro‐Glu‐Met‐Phe‐Thr‐Lys‐Leu‐Gln‐His‐Ala‐Leu‐Arg a) How will the distribution of polar and non‐polar amino acids appear on the surface of the helix? b) Can you see any other potentially favourable side chain interactions arising from the three‐ dimensional arrangement? c) How many parallel strands and how many antiparallel strands are seen in the β‐sheet model? d) The α‐C atoms should be tetrahedral. Where does this place the R‐groups relative to the sheet?  BINDING AND RECOGNITION OF SUBSTRATES & BASIS OF CHEMICAL AND ENZYMATIC CATALYSIS Function of most proteins involves ability to recognize and bind other molecules Enzymes binding their specific target molecules; antibodies binding and identifying foreign molecules and tagging them for destruction Involves the same ________________________________ as protein folding  Non-polar patches on surface of target bind by hydrophobic effect  Match shapes to maximize close contact – van der Waals effect  Match of charged groups or H-bond donors and acceptors Chymotrypsin binds to polypeptides and finds aromatic amino acids Phe, Tyr and Trp  Groove in chymotrypsin binds a peptide chain by H-bonds to backbone  Side chain binding pocket is large, and surrounded by non-polar amino acids of the chymotrypsin  Phe, Tyr, Trp fit best  Binding the target positions its peptide bond next to catalytic unit Related enzymes trypsin and elastase are similar to chymotrypsin except around their side chain binding pockets  Enzymes are catalysts Enzymes bind a specific ___________________ (or molecules) and catalyze a specific chemical ________________ The target of the enzyme is called its ________________ Many enzyme names end in –ase, e.g. ribonuclease The chemical reaction must be able to occur spontaneously, but enzyme speeds it up by factor of 106 and 1017, typically 1010-fold Why are uncatalyzed reactions slow? – Without catalyst, reactions depend on ______________ events – molecules must _____________ – they must be in right ________________ – reacting molecules require a threshold ____________ – If these conditions are met, reaction may occur, but depends on chance  How do enzymes speed up reactions? The Arrhenius equation Z is the ________________frequency p is the _______________ factor, probability that collision leads to reaction – related to orientation of reactants Ea is the _________________; energy must be put into a reaction at initial steps, to break or distort bonds e–Ea/RT is the fraction of molecules at temp T (in Kelvin) which possess energy Ea Low Ea or higher T make fraction bigger, so reaction is favoured Enzymes eliminate the randomness of collision Random motion of molecules leads to close encounters but few hits between reacting pairs Enzyme binds substrates in a special pocket known as the _________________, holding them close together long enough for reaction to proceed This is the ______________________– increases Z Enzymes hold substrates in the correct orientation Two molecules may meet by random collision, but reactive groups may not be properly lined up for reaction Enzyme binds substrates, holding them in the active site so reactive groups are ideally aligned This is the __________________ – increases p ▪ Proximity and orientation can speed-up a reaction by 103 - 105-fold, maximum of 1010-fold when combined ▪ Also applies to enzymes that have only one substrate ▪ Proximity between substrate and reactive groups on enzyme ▪ Ideal alignment of substrate and reactive groups of enzyme  Enzymes decrease Ea ▪ Proximity and orientation are ____________ effects that speed up enzyme reactions ▪ Enzymes can also use ____________ catalysis to speed up reaction by lowering Ea ▪ Ea can be lowered by finding a better chemical pathway for the reaction, involving reactive groups on the enzyme X-CO-NH-Y + H2O  X-COO- + +NH3-Y Reaction shown – hydrolysis of peptide - is very slow because H2O is a very poor nucleophile, and very weak acid A chemist would use acid or base, and heat the reaction to get faster hydrolysis Cells exist close to neutral pH, and at relatively low, fixed temperature Enzymes must speed up reactions at neutral pH and normal temperature  Nucleophilic and electrophilic catalysis ________________ catalysis: enzymes can speed up reactions by providing a better nucleophile – e.g. Cys-SH, His-N:, Asp or Glu -COÖ–, – more rarely Tyr or Ser -ÖH or Lys :NH2 ________________ catalysis: an electrophile is an electron-seeking group no really good electrophilic amino acids enzyme may contain a non-amino acid helper molecule called a _____________________, as part of its structure e.g. pyridoxal phosphate (cofactor or coenzyme) with its electrophilic aldehyde group - binds at enzyme catalytic site, initiates reaction by withdrawing electrons from the substrate General acid and general base catalysis General acid: catalysis by an amino acid side chain that _________________ to the reaction General base: catalysis by an amino acid side chain that ______________ from the reaction H+ exchange takes place right at the site of reaction, so pH of surroundings is not affected Gain or loss of one H+ in a small confined volume can have same effect as strong acid or base Chemical catalysis 1. Nucleophilic catalysis 2. Electrophilic catalysis 3. General acid catalysis 4. General base catalysis These mechanisms can each contribute about 100-fold increase in reaction rate, and work together to lower Ea Stabilizing the transition state Reactions must pass through a transition state to proceed, and key atoms may change shape, e.g. trigonal planar to tetrahedral, or a bond may stretch changing shape stretching ▪ The enzyme can help by binding the substrate in the ideal shape for the transition state   Stabilizing the transition state Less activation energy is needed if the enzyme active site is ___________________ to the transition state The transition state differs chemically from the substrate. Ea can be lowered partly by choosing a different transition state, and partly by making the enzyme contribute to bond distortion that leads to the transition state  Homework Questions https://forms.office.com/r/rCv6Wni4j6  Important - prepare for Next Lecture- “Catalytic Mechanism of Chymotrypsin” Review course notes for “Catalytic mechanism of Chymotrypsin” ahead of time – challenging fast-paced material! Also review: – Chemical Reactivity – Nucleophiles – Electrophiles – Nucleophilic substitution reactions – Content from today’s lecture on “How enzymes increase reaction rates” You can use the “Chymotrypsin worksheet”: CourseLink> Content> Course Material (weeks 1-6)> Week 5-6> Chymotrypsin worksheet or your lecture slides to work through the mechanism. Please print and bring the chymotrypsin worksheet to class if you are using it.