sing Figure 1.3 of the Introduction as an example, a) draw all the structures of the tribasic amino acid lysine involved in the equilibrium reactions that would take place during titration against NaOH, starting with the fully protonated form below (draw the R-group in full). HAN+-CH- COOH (CH2)4 NH°+ b) indicate the numerical pa value of each equilibrium reaction, and which ionizable group is being dissociated in each step. c) indicate the net charge of the amino acid at each step and identify the zwitterion. d) Calculate the pI of this amino acid (show the calculation). e) What would be the predominant ionization states of this amino acid at physiological pH (7.4) and at this pH, what would the ratio of these two states be (show the calculation)?

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Chapter1: Biochemistry: An Evolving Science
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Using Figure 1.3 of the Introduction as an example, a) draw all the structures of the tribasic amino acid lysine involved in the equilibrium reactions that would take place during titration against NaOH, starting with the fully protonated form below (draw the R-group in full). HAN+-CH- COOH (CH2)4 NH°+ b) indicate the numerical pa value of each equilibrium reaction, and which ionizable group is being dissociated in each step. c) indicate the net charge of the amino acid at each step and identify the zwitterion. d) Calculate the pI of this amino acid (show the calculation). e) What would be the predominant ionization states of this amino acid at physiological pH (7.4) and at this pH, what would the ratio of these two states be (show the calculation)?
en average the two pra s
carboxyl 2.4 and pKa a-amino 9.8) to determine the pI (6.1). Since pH 7.0 is above the pl, there would be no
cationic form present, only zwitterionic and anionic forms. The zwitterion would be the conjugate acid
while the anion would be the conjugate base. The ratio of anion to zwitterion can then be calculated
using the pKa a-amino and the Henderson-Hasslebalch equation as shown below the figure.
H
H
I
I
NH2-CH-COO™
NH3 -CH-COOH
Net charge:
cation (+1)
2.4=pKa a-crboxyl
Calculation of ratio at pH 7.0:
H
I
NH3-CH-COO™
zwitterion (0)
9.8=pKa a-amino
pl= (2.4+9.8)/2 = 6.1
Figure 1.3 Dissociation reactions of the dibasic amino acid glycine to illustrate pKa net charge and
calculation of pl. At pH 7.0, only the zwitterionic and anionic forms would be present in solution.
pH =pKa + log A/HA
7.0 9.8+ log (anion/zwitterion)
10-2.8
anion/zwitterion
anion/zwitterion = 0.00158 or 1/633
anion (-1)
So, the ratio is 0.00158 anion: 1 zwitterion, or simply 1 anion: 633 zwitterion. From this we can see
that the majority of glycine would be in the zwitterion form at a pH of 7.0.
Amino acids with an
Transcribed Image Text:en average the two pra s carboxyl 2.4 and pKa a-amino 9.8) to determine the pI (6.1). Since pH 7.0 is above the pl, there would be no cationic form present, only zwitterionic and anionic forms. The zwitterion would be the conjugate acid while the anion would be the conjugate base. The ratio of anion to zwitterion can then be calculated using the pKa a-amino and the Henderson-Hasslebalch equation as shown below the figure. H H I I NH2-CH-COO™ NH3 -CH-COOH Net charge: cation (+1) 2.4=pKa a-crboxyl Calculation of ratio at pH 7.0: H I NH3-CH-COO™ zwitterion (0) 9.8=pKa a-amino pl= (2.4+9.8)/2 = 6.1 Figure 1.3 Dissociation reactions of the dibasic amino acid glycine to illustrate pKa net charge and calculation of pl. At pH 7.0, only the zwitterionic and anionic forms would be present in solution. pH =pKa + log A/HA 7.0 9.8+ log (anion/zwitterion) 10-2.8 anion/zwitterion anion/zwitterion = 0.00158 or 1/633 anion (-1) So, the ratio is 0.00158 anion: 1 zwitterion, or simply 1 anion: 633 zwitterion. From this we can see that the majority of glycine would be in the zwitterion form at a pH of 7.0. Amino acids with an
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