Calculate the fraction of Lysine that is in the following ionic form at pH 9.25.

Question

Calculate the fraction of Lysine that is in the following ionic form at pH 9.25. Use the pKs in the table. Use the given pH not the pI. Show your work.

Standard Amino Acids: Properties & Conventions Table TABLE 3-1 Properties and Conventions Associated with the Common Amino Acids Found in Proteins pH, values Amino acid proteins Nonpolar, aliphatic 2.34 5.97 0.4 Ala A 6.01 7.8 Pro P 115 1.99 10.96 6.48 9162 5.97 6.6 131 5.98 9.1 131 9.68 Met M 149 9,21 5.74 1.9 Aromatic R groups 9.13 Phe F 165 1.83 5.48 2.8 3.9 181 9.11 10.07 1.3 0.9 1.4 Polar, uncharged R groups Ser S 105 2.21 9,15 The T 119 2,11 9,62 5.87 121 1.06 10.28 8,18 2.5 1.9 Asus N 132 8.BO 5.41 146 2.17 9,13 4.2 Positively charged R groups 2.18 8.95 10.53 9.74 3.9 Lys K 146 His H 155 1.82 9,17 6.00 3.2 2.17 Arg R 9.04 12.48 10.76 4.5 5.1 Negatively charged R groups Asp D 133 1.88 3.65 2.77 Glu E 147 2.19 9.67 4.25 nger 2006 Table 3.1

Explanation / Answer

The pH 9.25 is within 1 unit of pKa2 = 8.95; therefore, we will work with pKa2.

Let HA and A- denote the protonated and de-protonated forms of Lysine at pH 9.25.

HA <=====> H+ + A-

Use the Henderson-Hasslebach equation:

pH = pKa + log [A-]/[HA]

Plug in values:

9.25 = 8.95 + log [A-]/[HA]

===> 0.30 = log [A-]/[HA]

===> [A-]/[HA] = antilog (0.3) =1.995 2.0 ….(1)

Realize that at pH 9.25, both the carboxyl group and the alpha amino group are de-protonated. The carboxyl group carries a charge of -1 while the deprotonated amino group has charge 0. The basic side chain has charge +1; the overall charge is 0. Therefore, HA denotes the neutral Lysine while A- is the ionic Lysine.

The ratio of A- and HA is 2:1; therefore,

[A-]/[HA] = 2/1

Invert,

[HA]/[A-] = ½

Add 1 to both sides,

[HA]/[A-] + 1 = ½ + 1

===> ([HA] + [A-])/[A-] = 3/2

===> [A-]/([HA] + [A-]) = 2/3

The fraction of Lysine in ionic form is 2/3 (ans).

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