2 amino acids and proteins (1).pdf

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Unformatted text preview:                                                                                                                                                                                                                                                                        "!            "!!!!             "!                       ) (     %       # $&  #  $'"   %                 '  ) '   !                                &$      "   ( %   #                                      (     "'  "  " " - .+ "( " ,  &! "  ) *      $ +   "  "+   %   +   %             $  (//)% #   #                              !  !" ".."+",-*!                                                           In the context of polypeptide chains, amino acids are called “residues”.              Peptide bonds can link amino acids to produce long chains An “average” polypeptide is between 100 to 1500 amino acids long. (~11-165 kD) Titin: 34,500 amino acids! Polypeptide sequences are written (and numbered) from the N- to Cterminus.          % $   !                 "   #      $                                                                       "   "    $                    $     #  %    $                                       -     # $  # $  +-"          ( *!'&&     & &)!'&& '                                                               !"    !&%    !$#                                                                                                                                                                                                                       Polar, uncharged amino acids of note: cysteine forms covalent bonds in proteins                                                                                                                                                                                                       "    # "    #              !              $     % Acid/Base chemistry review Ka HA H+ + A- [H + ][A − ] Ka = [HA] Ka is the acid dissociation equilibrium constant. ⎡[H + ][A − ] ⎤ pKa = − log Ka = − log ⎢ ⎥ [HA] ⎣ ⎦      Ka Lysine side chain (free) Ka = 2.9 x 10-11 pKa = 10.54 What is the predominant form when the pH < pKa? What about when pH = pKa? What about when pH ≈ pKa?          .%,$ .-''  .'-$ .($+ .%$)( .%&(, .*$(           !%'             .,'+       .%$(* What you need to know about amino acid pKas • Be able to draw out the reaction the pKa describes for each functional group in the amino acid. • Know that pKas of the carboxyl group and amine groups that are not part of the sidechain are about 2.0 and 9.0, respectively. • Know the predominant charge state of each functional group on each amino acid at pH 7.0. • If you need to know the exact pKa values for a calculation, we will give them to you, but we may not tell you which pKa corresponds to which functional group. Example: We may ask you to do a calculation with Histidine, and tell you that the pKa values for His are 1.8, 9.3 and 6.0.                                   N ! O H   O H O H H N H &"! %$ ! O N H N "#                                                                          ( ' &"!$ 1 0.9 0.8 0.7   0.6   0.5 % # 0.4 0.3 0.2 0.1 0 0 5 10    &                        H+ + A- HA [H + ][A − ] Ka = [HA] ⎛ [HA] ⎞ [H + ] = Ka ⎜⎜ − ⎟⎟   ⎝ [A ] ⎠         − ⎛ ⎞ [A ⎟⎟ - log[H + ] = − log Ka + log⎜⎜ [HA] ⎝ ⎠ ⎛ [A − ] ⎞ ⎟⎟ pH = pKa + log⎜⎜ ⎝ [HA] ⎠                               ! Determining the fraction of a titratable group in a particular state What is the fraction of lysine side chain that is protonated at a given pH? General expression for dissociation of a weak acid: HA Ka H+  A - The fraction of the protonated form is (by definition): f HA = [HA ] [HA]+[A - ] Use the equation for Ka to get [HA]: [H + ][A - ] [HA] = Ka Combining these two equations, f HA = [HA ] [H + ][A - ] [HA] = Ka [HA]+[A - ] we get: f HA [H + ][A − ] K a = + [H ][A − ] K a + [A − ] Multiplying through by Ka, this reduces to: f HA [H + ] = K a + [H + ] A similar expression can be derived for the fraction of conjugate base f A− = Ka K a + [H + ] These two equations allow for the calculation of the fraction of acid or conjugate base at any pH for an acid with a known pKa. note that only the numerator is different f HA [H + ] = K a + [H + ] f A− = Ka K a + [H + ] How much of the unprotonated lysine sidechain exists at pH 9.54? (note that this is 1.0 pH units below the pKa) Using the Henderson-Hasselbalch equation: [A − ] pH = pKa + log [HA] [Lys-NH2] [Lys-NH3+] [A − ] 9.54 = 10.54 + log [HA] = 0.1 at pH 9.54 (1.0 pH unit less than pKa) or expressed as a fraction/percent: 0.1/1.1 = 9.1% Note that we consider a titratable group to be predominantly in only one form once the pH outside one unit of its pKa. Note that in other cases we will ask you to calculate precisely the ratio, concentration, or fraction of a species in solution. Significant concentrations of the acid and conjugate base are present when pH is near the pKa What is the pH when the fraction of unprotonated lysine is 20%? f A − = 0.2 = Ka K a + [H + ] 10 − pKa 0.2 = − pKa 10 + 10-pH pKa = −log Ka 10-pKa = Ka    pH = 9.9 Amino acids have multiple titratable functional groups For example, glycine: Gly      Glycine is a an important neurotransmitter that binds to a receptor found in the central nervous system called GlyR.       What are the possible charge states, and what are their concentrations at different pH values? At what pH values does the molecule have a net positive or negative charge?                                                   K a1 >> K a 3         1 Ka4 << 1      [b] =~ 105 [c ]                               Ka2                              &!% #  ' f NH3 +   $'  [H + ] = + [H ] + K a2 f A− =  ' "   K a1 K a1 + [H + ] f NH3+:A- = f NH3+ f A −    f NH3:A- K a1 [H + ] = + × [H ] + K a2 K a1 + [H + ] f NH3:A- = 0.66              ("      !        .*#&  .'#+ ).,#( -  )  K his [H + ] [H + ] f= + × × [H ] + K NH3 K his + [H + ] [H + ] + K COOH     / /  / % /     & /  '                    ($.! )11+'              /&       %! 0&*                                                                         !      "                       [H + ] [H + ] f+ = ⋅ K a1 + [H + ] K a2 + [H + ] f− =  K a1 K a2 ⋅ K a1 + [H + ] K a2 + [H + ]          K a1 K a2 [H + ] [H + ] ⋅ = ⋅ K a1 + [H + ] K a2 + [H + ] K a1 + [H + ] K a2 + [H + ]        pI = pKa1 + pKa2 2 The pI is often (but not always!) the average of the two pKas around the zero charge species                 ! $ #! &    & %$ ""                            !         " !   " "  " "  " " "      "  "                     '(" $             !!  #                            $ $     $ $  $  $  $ $   $  $   $      $ !   $ 2f + 2 + f +1 = f −1 + 2f − 2 $ #"  "                                                                              . )  )     )   "   #$  # $  () -, ! )' ,) !! )* ,& #$ )* +- ' ( (                        !     "  % $      #                                                                                           """""#'$"'    !                                                      $#'                                                      !                                                                                                                                                                                                                                                                   "      &#'            %!          %!          (!                                                                                                                                                                        !            "     #                                                                  !                                             !                                                                                           ...
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  • Winter '13
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  • Chemistry

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