Animation: pH-Dependent Electrophoresis and Microstate Composition of Glycine  Go back
  • Glycine (NH3+-CH2-COO-) has 2 ionizable groups:

  • The a-amino group (NH3+-) ionizes (i.e., loses a proton) with pK (NH2) = 9.8.

  • The a-carboxyl group (-COOH) ionizes (i.e., loses a proton) with a pK (COOH) =  2.4.

  • The pH where an equilibrium solution of glycine has zero average charge is defined as the isoeletric pH, or pI; pI (glycine) = 6.1 = 1/2 [pK (NH2) + pK (COOH)].

  • Glycine molecules potentially exist in 1 of 4 different ionization states (microstates):
    (NH3+-CH2-COO-); (NH2-CH2-COOH); (NH2-CH2-COO-); or (NH3+-CH2-COOH)

  • The equilibrium percentages or ratios of microstates vary greatly with pH and are calculated below.

  • BEFORE running each animation MAKE A PREDICTION for the migration direction or the microstate composition of glycine at the corresponding equilibrium pH value.

  • Self Assessment:  Test your predictions regarding the electrophoretic behavior and microstate distributions of glycine at different pH values.

Macrostate Dynamics: Spreading is cause by random diffusion (Brownian motion) with a pH-dependent distribution of the bulk mass over time. 

Microstate Dynamics: Individual molecule movements over time are caused by the electric field exerting its effects on charges created by pH-dependent ionization reactions.

 

Click to change pH conditions.

Click to start animation.
  • Observe the electrophoretic behavior of glycine at different pH values
    selected with the buttons at the top.

  • Use the rewind, play, and fast-forward buttons to navigate within animations. 

  • Play an animation straight through by clicking the play button.

  • Step through the animation frame-by-frame by clicking the forward or reverse buttons. 

ExampleGlycine at pH = 6.1

pKd(a-NH3+) = 9.8 pKd(a-COOH) = 2.4 pI = 6.1

Microstates

Equilibrium Microstate Fractions at pH = 6.1

(+ -):  NH3+-CH2-COO- At pH = 6.1, fraction = Ya(a-NH3+) * Yd(a-COO-) = 99.96%
(+ 0): NH3+-CH2-COOH At pH = 6.1, fraction = Ya(a-NH3+) * Ya(a-COOH) = 0.02%
(0 -):  NH2-CH2-COO- At pH = 6.1, fraction = Yd(a-NH2) * Yd(a-COO-) = 0.02%
(0 0):  NH2-CH2-COOH At pH = 6.1, fraction = Yd(a-NH2) * Ya(a-COOH) = 0.000004%

Net Charge = 0

Ya = 1 / (1 + 10pH-pKd)         Yd = 1 / (1 + 10pKd-pH)
 

© Duane W. Sears and E. M. Hildebrand
Revised: August 12, 2004

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