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2/24 Ex2L1: Intro to Protein Structure & Finding Total Charge

by: Allie Bartlett

2/24 Ex2L1: Intro to Protein Structure & Finding Total Charge Chem 349

Marketplace > Biochemistry > Chem 349 > 2 24 Ex2L1 Intro to Protein Structure Finding Total Charge
Allie Bartlett

GPA 3.0
General Biochem
Dr. Runquist

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This is the introduction to protein structure, introduces how to find total charge of a polypeptide and a brief explanation of pKa and pH values.
General Biochem
Dr. Runquist
Class Notes
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This 10 page Class Notes was uploaded by Allie Bartlett on Friday February 27, 2015. The Class Notes belongs to Chem 349 at a university taught by Dr. Runquist in Fall. Since its upload, it has received 59 views.

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Date Created: 02/27/15
EXAM LECTURE i List of Amino Acids that we need to know amp their charges Charged o Arginine Arg o Lysine Lys o Aspartic acid Asp Glutamic acid Glu PolarHydrophilic may participate in hydrogen bonds Glutamine Gln Asparagine Asn Histidine His Serine Ser Threonine Thr Tyrosine Tyr Cysteine Cys Methionine Met NonPolar Hydrophobic normally buried inside the protein core 0 Alanine Ala o Proline Pro Glycine Gly Amino Acids pH is the inverse logarithm of hydrogen ion conc ogH o The lower the pH is 16 the HIGHER concentration of Hydrogen ions H o The higher the pH is 814 the conc of Hydrogen is LOWER Is the isoelectric point This is the point where the overall charge of the protein or molecule is equal to ZERO 9K4 Is the dissociation constant of an acid o Is equal to log ka 0 It is related to ionization of an acid a The smaller the pKa the STRONGER the acid a It is found in the HendersonHasselbach equation When pHpKa the ratio between protonated amp deprotonated COOICOOH is equal to 1 PHPKIDam P leg HA conjugate aeld conjugate baee E the ratio of COO deprotonatedprotonated basic form acidic form COOH ex lfthe pKa 2 and the pH 1 then there is more of the acidic COOH because the pH is below pKa The biological pH range is 68 if the pH is 4 GroupAmino StructureRgro pKa pHltpKa pHgtpKa acid up Cterminus H2NCHCOOH 33 1 1 Nterminus H2NCHCOOH 90 Neutral1 1 Asp amp Glu CH2 COOH 40 1 1 His 300 60 Neutral 1 1 I HEN E H C I H Cys CH2 SH 84 Neutral 0 1 Tyr 105 1 1 Lys 105 Neutral 1 1 Arg 125 Neutral 1 1 carving When the pHlt pKa there are excess H protons therefore the molecule or amino acidprotein will be protonated which ADDS protons 1 When the pHgt pKa then there are excess OH hydroxide anions and will deprotonate the molecule which steals the protons 1 and give the more basic form of the amino acid Some examples ex1 Find the total charge of glutamate at pH 70 pKa 33 C quot pH 7gt3 pKa 90 1 pH 7lt9 HEM C H 1 I CHE pKa 40 I pH7gt4 1 a C0 Total charge of glutamate with pH70 is 1 pl 33402 pH 365 Find the isoelectric point by starting from the lowest pKa group then as soon as the amino acid surpasses 0 add those two pKa numbers I added 33 and 40 because first the amino group comes out to 1 and the carboxylic group is 1 equaling 0 then when you take in consideration the R group 1 you know that by starting from the lowest pKa till we hit 0 and one past it are the numbers we add and divide by 2 Nonpolan aliphatic side groups H CH3 I JH i1 l 376 39 r e 3 7 ie iii HEN CG HgN G032 HEN Glycine Alanine Valinu My 11quot Ala ll Vol V HI quot cogquot liH a 39 v e Haw DICE1 H3N 1 y ngN 5393 Lcu l h39llclhloninc lsolcucjac Lou L M Ill I lPolar uncharged side groups DH H D I an quot Haw coil 39 co e HEM I Tl 39sleil 1c 139 1 Immune Scrum V r r it l Eur T hm T HEN CI W H ID co l EEH do HEN quotN quot cog a H Halal l ll i gt5 l mll l AhP Hdlh llilannune Pm P Asp ID Lila J Proteins Aromatic side groups He g V v Nquot IHI HEN coil HEN quot 39 clogj Phenylalanine 39Tl39yroslnc quotlfiry 39 oplmh F hc F Tyr 1 quot Trp W Positively charged side groups NH G02 H3N col Lysine lli sllidi39nc Aw mghm Lys l5 lins ll Am N Negatively charged side groups NHI do 33 a El H3Nl HEN films mate l sparlate Glu E 1 sp I 0 Once proteins are synthesized they must foldup in the correct orientation folding actually begins as translation occurs 0 Proper folding is absolutely critical Cystic fibrosis is an example of folding gone wrong Due to improper folding ofjust one amino acid there is inability to regulate chlorine in the body Structure 1 Primary All peptides have primary folding The N gt C sequence of AA linked together by an amide bond 2 Secondary only some peptides are folded in the secondary structure Arrangement of AA close in sequence stabilized by hydrogen bonding backbone of the peptide 36 reaiduesfturn a Alpha helices Hydrogen bonding is what holds the alpha helix together Rigid peptide bonds between the amino acids allow for folding in the helix PLANAR alpha helix Rgroups are in the transorientation i Carbonyl oxygen in each amino acid Hydrogen bondsCO with the amino NH group of an Amino acid 4 residues ahead ii The Amino acids almost never found in alpha helices are Proline too rigid glycineneither L nor R bc no asymmetric carbon iii common example is myoglobin primarily composed of alpha heHces 1 13 4 IMiLIQIS IMjVIS ILILIQIS ILIVIS LI II lQl P Key Le Met Group Coloring Key Len Nnrlpular Len Met Polar Unchalged 9 Acidic Ser V i f 11 Em C Leu ser 0 y 0 l K Gin y lgtLeu 139 E39Te ik Val 3 On Val E G Gm Ser Typical arrangement helical representation looking down at an alpha helix Amphipathic helices b Beta sheets i Secondary structure ii Pleated sheets iii Antiparallel Sheets Run in opposite directions 1 N gtCC gtNC gtN iv Parallel sheets Run in the same direction N gt C N gt C N gt C v Hydrogen bonding is not as uniform as antiparallel bonding vi Hairpin turns Help change the direction of peptide tight turn made up of 4 amino acids They are highly bent and mostly contain PROLINE amp GLYCINE vii Irregular loops also help the direction of peptides Tertiary Reserved for proteins that are made of more than 1 polypeptide ie dimers monodimers trimers Quaternary is the arrangement of more than one protein molecule in a multisubunit complex The nomenclature here can get a bit confusing because we call a single polypeptide chain a protein if it can function on its own Intro to Beta sheets amp continue alpha helices Ex Given the pKa values estimate to charge of the hexapeptide at pH 75 START FULLY PROTONATED FORM AT THE LOWEST pH 90NH3alaOhis60arg125 serOcys84lys105 COOH 33 1 First we fully protonate the peptide chain which means that every spot that accepts H will be fully saturated 2 We then apply the desired pH 75 3 Determine which AA will lose gain or remain in the neutral state depending on the pHpKa value a b c NH3 75lt9 gt 1 adds proton aa O his 75gt60 gt 0 its zero because the slightly more basic solution TAKES a proton away his is already at a protonated therefore equals 0 arg 75lt125 gt remains 1 neutral because it is already protonated fully and cannot add anymore protons ser O cys 75lt84 gt remains at 0 because the Sulfur R group is fully protonated at 0 charge lys75lt105 gt cannot add anymore protons in the fully protonated form therefore the charge is 1 COOH 75gt33 gt 1 basic steals the proton i TOTAL CHARGE 1oo1oo11 We make alpha helices Beta Sheets To compact the peptide by folding amp bending on either side of the peptide bond Myoglobin is 92 composed of alpha helix The main alpha helix reflects the backbone of the peptidethe amino acid side chain sticks out outside the helix Hydrogen bonds involving the NH amp carbonyl oxygen in the backbone stabilize the alpha helix 0 Secondary structure 0 Pleated sheets 0 Antiparallel chains 0 0 Parallel sheets Run in the same direction N gt C N gt C N gt C Hydrogen bonding is no as uniform as antiparallel bonding o Hairpin turns Help change the direction of peptide tight turn made up of4 amino acids They are highly bent and mostly contain PROLINE amp GLYCINE o Irregular loops also help the direction of peptides Tertiary Structure Arrangement of secondary structure o Nonpolar inside and polar outside Arrangement of secondary structure to each other 0 Hydrophobic interactions are what drives tertiary protein structures The structure to a unique shape is reason to get away from H20 Forces that Stabilize tertiary structure 0 The R groups stabilize tertiary structure 0 hydrophobic interactions drive it amp stabilize o Covalent bonds gt Disulfide linkages and cysteine residues can be oxidized to make disulfide linkages Hydrogen bonding Amino Acids that have charged groups can help stabilize ionic interaction Quaternary Structure Arrangement of tertiary structure to each other 0 Only one polypeptide can be quaternary structure 0 Ex Triose phosphate isomerase has two subunits o Deoxygenated hemoglobin has four subunits 2 alpha amp 2 Beta igG Heavy chins are held together by disulfide linkages amp heavy chains are linked to light chains by disulfide linkages 0 Just because they are held together by disulfide bonds DOES NOT mean that they are ALL Quaternary held together by these type of linkages gt Protein structure relates to its function Protein Jobs 1 Molecular Regulators of transcriptiontranslation 2 Cell Component Cytoskeletal proteins strong 3 Transporters Transporters of fatty acidsO2 Ca NO 4 Enzymesinvolved with biological processes carbohydrate lipid nitrogen utilization Q What is the function of hemoglobin Myoglobin Hemoglobin has 4 polypeptides Myoglobin has 1 polypeptide amp 1 Oxygen binding site


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