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) 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)?

) 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)?

Anatomy & Physiology
Anatomy & Physiology
1st Edition
ISBN:9781938168130
Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Chapter2: The Chemical Level Of Organization
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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)?
← 11:57 ANSWER: H₂N HIỆNhĩ CH₂ CH₂ CH₂ Net charge +2 pka = 2.18 H.CNH: CH₂ +1 CH₂ pka = 8.95 = HỌC-NH: CH₂ pl = (10.93 +8.95)/2 = 9.94 [Answer] CH₂ pka 10.93 5G 36 √x X H₂N- HNH, CH₂ A) The stages of lysine from the most protonated to the most deprotonated form have been provided in the figure given above. CH₂ B) The pKa values for each equilibrium have been marked in the above diagram as well. C) The net charge of the Lysine molecule in each of its protonated forms are given above. D) The isoelectric pH or pl is the pH at which the net charge on the molecule is equal to 0. The isoelectric pH or pl can be calculated by taking an average of the pKa of the ionizable groups that is undergoing dissociation and is flanking the species with net charge = 0. As shown in the above figure, the pl of Lysine will be CH₂ DO
Transcribed Image Text:← 11:57 ANSWER: H₂N HIỆNhĩ CH₂ CH₂ CH₂ Net charge +2 pka = 2.18 H.CNH: CH₂ +1 CH₂ pka = 8.95 = HỌC-NH: CH₂ pl = (10.93 +8.95)/2 = 9.94 [Answer] CH₂ pka 10.93 5G 36 √x X H₂N- HNH, CH₂ A) The stages of lysine from the most protonated to the most deprotonated form have been provided in the figure given above. CH₂ B) The pKa values for each equilibrium have been marked in the above diagram as well. C) The net charge of the Lysine molecule in each of its protonated forms are given above. D) The isoelectric pH or pl is the pH at which the net charge on the molecule is equal to 0. The isoelectric pH or pl can be calculated by taking an average of the pKa of the ionizable groups that is undergoing dissociation and is flanking the species with net charge = 0. As shown in the above figure, the pl of Lysine will be CH₂ DO
11:57 ← Step1 a) The proteins are composed of twenty naturally occurring amino acids. Lysine is a basic amino acid which carries a net positive charge at physiological pH. Step2 b) ANSWER: HC-NHĨ CH₂ CH₂ CH₂ H₂N ( Net charge +2 pka = 2.18 +1 CH₂ CH₂ pka = 8.95 H.ỆNH: CH₂ CH₂ CH₂ pKa 10.93 5G 36 H₂N HỌC NH, CH₂ CH₂ A) The stages of lysine from the most protonated to the most deprotonated form have been provided in the figure given above. B) The pKa values for each equilibrium have been marked in the above diagram as well. C) The net charge of the Lysine molecule in each of its protonated forms are given above. D) The isoelectric pH or pl is the pH at which the net charge on the molecule is equal to 0. The isoelectric pH or pl can be calculated by √x
Transcribed Image Text:11:57 ← Step1 a) The proteins are composed of twenty naturally occurring amino acids. Lysine is a basic amino acid which carries a net positive charge at physiological pH. Step2 b) ANSWER: HC-NHĨ CH₂ CH₂ CH₂ H₂N ( Net charge +2 pka = 2.18 +1 CH₂ CH₂ pka = 8.95 H.ỆNH: CH₂ CH₂ CH₂ pKa 10.93 5G 36 H₂N HỌC NH, CH₂ CH₂ A) The stages of lysine from the most protonated to the most deprotonated form have been provided in the figure given above. B) The pKa values for each equilibrium have been marked in the above diagram as well. C) The net charge of the Lysine molecule in each of its protonated forms are given above. D) The isoelectric pH or pl is the pH at which the net charge on the molecule is equal to 0. The isoelectric pH or pl can be calculated by √x
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