Study Guide for BIOC 462A at UA
Study Guide for BIOC 462A at UA
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Date Created: 02/06/15
PROTEINS SELF STUDY QUESTIONS 1 This is the titration curve for which amino acid 1 o 2 A Ramachandran plot shows A The amino acid residues that have the greatest degrees of rotational freedom B The sterically allowed rotational angles between the side chain groups in a peptide and the peptide backbone C The sterically limited rotational angles domains where phi and psi are allowed in the protein backbone D The angles that are allowed about the bonds connecting the amide nitrogen in a peptide bond 3 Consider a peptide with the sequence GluValHisSerArg What will be the net charge at pH 4 At pH 9 4 These data relate the concentration of a ligand L and the extent of saturation of its binding site 9 for two different proteins A and B Plot 9 vs L and determine whether the proteins exhibit hyperbolic or sigmodial binding behaVior L mM 0 for protein A 0 for protein B 010 22 03 035 72 10 080 150 30 180 290 90 300 400 250 450 500 500 575 560 760 800 640 900 1300 740 970 10 The following question deals with the properties of amino acid sidechains buried in the hydrophobic interior of a protein A Would the pKa of a buried lysine be higher or lower than the pKa of a surface Lys B Would the strength of a buried hydrogen bond be stronger or weaker than a hydrogen bond on the surface C Would the strength of the electrostatic interaction between a buried ion pair be stronger or weaker than an ion pair on the surface The isoelectric point pI of an amino acid is the pH at which the net electrical charge is zero The two structures shown for Ala lil C lil Tl each have a net charge of zero A Why is the predominant form of Ala at its 3 c I c I isoelectric point the zwitterionic and not lla lla the uncharged form B Calculate the mmquot quot I I ratio of the concentration of the zwitterionic formuncharged form at the pI Explain the following observation At pH 70 polylysine which is a polypeptide in which all the residues are lysine adopts a random coil conformation ie it has no secondary structure but at pH 12 it is present as a 0L helix The binding of a ligand L to a protein P is often a simple equilibrium P L ltgt PL which is characterized by the dissociation constant Kd L P PL where L is the concentration of unbound or free ligand P is the concentration of free protein and FL is the concentration of the proteinligand complex If PT P PL total protein concentration and 9 PLPT then 9 LKd L However a plot of 9 vs L is a hyperbola see 4B and it is difficult to evaluate Kd from such a plot Derive a linear transformation of 9 LKd L plot the data from 4A and determine Kd A useful method for visualizing the properties of a protein helix is to display the sequence on a helical wheel diagram This diagram is based on a fivetum helix 18 residues Adjacent residues in the sequence are 1000 apart on the wheel and a residue occurs every 20 The sequence is written onto the wheel in the order shown Display the following sequence on a helical wheel and explain what information this representation provides AlaPheAspLysMetIleGluAsnLeuGlnArgLeu 3 TrpSerGluPheLeu Gln Myoglobin and the subunits of hemoglobin are about the same size and the structure of a subunit of hemoglobin is very similar to the structure of myoglobin Given these facts explain why hemoglobin has a higher ratio of nonpolarpolar amino acids than does myoglobin 10 17 11 This table contains the torsion angles 1 V for several residues in a protein sequence Consult a Ramachandran diagram and determine the conformation of residues 311 of residues 4351 Residue d deg 1y deg Residue q deg 1y deg Number Number 1 60 147 42 142 150 2 49 32 43 154 121 3 67 34 44 121 136 4 58 49 45 110 174 5 66 32 46 128 162 6 82 36 47 135 156 7 69 44 48 121 138 8 61 44 49 131 157 9 72 29 50 115 130 10 66 65 51 126 146 11 67 23 52 67 10 12 The binding of oxygen to hemoglobin involves an allosteric transition from a weak binding form TState to strongbinding form RState One can evaluate the affinity of the R and T States for 02 using a Hill plot in which the log0 l0 is plotted vs log p02 where 0 Fractional saturation and p02 partial pressure of 02 Use these data to construct a Hill plot and determine the p50 the p02 at which 0 05 for the T and R States p02 9 p02 9 p02 9 01 000315 288 024 1288 0969 035 00099 47 050 2951 099 079 0031 575 076 6760 0997 175 0091 794 0909 ANSWERS 1 Lys or Tyr a It requires 3 equivalents of OH39 to titrate the amino acid b the amino acid is basic because there are 2 groups with pKas gt 70 c the R group pK 10 is too high for Cys and the titration is nished at pHlt12 which eliminates Arg C At pH 4 charge is 15 0L carboxyl group completely ionized 10 side chain carboxyl 50 ionized 05 side chain of His fully protonated 10 side chain of Arg fully protonated 10 0L amino group fully protonated 10 At pH 90 charge is 7 05 0L carboxyl group completely ionized 10 side chain carboxyl completely ionized 10 side chain of His fully titrated 0 side chain of Arg fully protonated 10 0L amino group 50 protonated 05 See plot at the right Protein A exhibits hyperbolic binding behavior protein B exhibits sigmodial binding behavior 00 25 50 75 100 125 150 I39M A The pKa would be lower because deprotonation would remove a positive charge from the hydrophobic environment a favorable process B Stronger because the dielectric constant in the interior is lower than in water The hydrogen bond has a partial electrostatic character and a lower dielectric constant leads to stronger electrostatic interactions C Stronger because the dielectric constant in the interior is lower than in water and a lower dielectric constant leads to stronger electrostatic interactions A The carboxyl group pK1 23 a stronger acid than the amino group pKz 97 therefore ionization of the carboxyl group occurs before ionization of the amino group B If we represent the fully protonated form as the zwitterion as i and the uncharged form as 0 then HendersonHasselbalch equation for the ionization of the COOH group to form the zwitterion is pH pK1 logi and for the ionization of the amino group to form the uncharged species it is pH pKz logO At the pI these two equations must be equal pKl logi pKz logO log cancels from both sides which leaves logi 0 pKzpKl and 10 10ltPKZ39PK1gt 22 x 107 7 10 ll 12 At pH 70 the amino sidechains are charged which causes electrostatic repulsion between the side chains whenever the polylysine tries to adopt a secondary structure At pH 12 the sidechain amino groups are deprotonated which removes the electrostatic repulsion between sidechains thereby permitting the polylysine to adopt a 0L helical structure A linear transformation of this equation is 19 KdL l and aplot of US vs lL should be a ml1 straight line with slope Kd For this example Kd Kquot um 45mM 1 II 25 III 75 Illll 1H 1I llIII an M LEU liEI39 The helical wheel representation clearly shows that this is h an amphipathic helix with one nonpolar face amino acids 3975 E in CAPS and one polar face amino acids in lower case 9 LEU This property of the helix is not obvious from a simple gun PHE inspection of the sequence all 39I39RP set LE asp 9 LE The contacts between the subunits of hemoglobin consist of hydrophobic interactions nonpolar amino acids The analogous regions in myoglobin are in contact with water and are therefore occupied by polar amino acids Residues 311 clearly form a 0L helix whereas residues 4351 form a 5 sheet The Hill plot is shown to the right By drawing an asymptote with a slope of 10 to the points at high values of 9 and then extrapolating back to 9 05 one obtains p50 for the RState Using the same procedure with the points at low 9 gives p50 for the T State TDIIIIIIIIJIJ39 1mm
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