What is the charge on lysine at pH 7.0?

           

The pKa values are as follows:

      a) carboxyl  at C terminus – 2.0

      b) amino at N terminus – 9.0

      c) Lys side chain – 10.0

     

 

pH = pKa  +  log (A-/HA)                    Charge = “# charge”/(“# charge”  +  “# neutral”)

                                                (the sign of this charge depends on the ionizable group)

 

 

for a)  7 = 2  +  log (COO-/COOH)

          5 = log (COO-/COOH)

      105 =  (COO-/COOH)

 (COO-/COOH) = 100,000/1        Charge = 100,000/(100,000 + 1)  =  1.0   ŕ  -1.0

 

for b)  7 = 9  +  log (NH2/NH3+)

          -2 = log (NH2/NH3+)

      10-2 = (NH2/NH3+)

 (NH2/NH3+) = 1/102 = 1/100        Charge = 100/(100 + 1) = 1.0              ŕ  +1.0

 

for c)  7 = 10  +  log (NH2/NH3+)

          -3 = log (NH2/NH3+)

      10-3 = (NH2/NH3+)

 (NH2/NH3+) = 1/103 = 1/1,000

            Charge = 1,000/(1,000 + 1) = 1.0                    ŕ  +1.0

 

                                                                                    ------------

                                                TOTAL CHARGE  =    + 1.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

What is the charge on lysine at pH 10.0?

           

The pKa values are as follows:

      a) carboxyl  at C terminus – 2.0

      b) amino at N terminus – 9.0

      c) Lys side chain – 10.0

     

 

pH = pKa  +  log (A-/HA)                    Charge = “# charge”/(“# charge”  +  “# neutral”)

                                                (the sign of this charge depends on the ionizable group)

 

 

for a)  10 = 2  +  log (COO-/COOH)

          8 = log (COO-/COOH)

      108 =  (COO-/COOH)

 (COO-/COOH) = 100,000,000/1

                              Charge = 100,000,000/(100,000,000 + 1)  =  1.0   ŕ  -1.0

 

for b) 10 = 9  +  log (NH2/NH3+)

          1 = log (NH2/NH3+)

      101 = (NH2/NH3+)

 (NH2/NH3+)  = 10/1          Charge = 1/(10 + 1) = 1/11 = 0.1            ŕ  +0.1

 

for c)  10 = 10  +  log (NH2/NH3+)

          0 = log (NH2/NH3+)

      100 = (NH2/NH3+)

 (NH2/NH3+) = 1  = 1/1      Charge = 1/(1+ 1) = 0.5                       ŕ  +0.5

 

                                                                                    ------------

                                                TOTAL CHARGE  =    -0.4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Consider the following peptide:

            Glu-Ala-Arg-Ser-Tyr

 

What is the net charge (to 0.1 unit) on the peptide at pH 3.0?

           

Assume the pKa values are as follows:

      a) carboxyl  at C terminus – 3.0

      b) Glu side chain – 4.0

      c) amino at N terminus – 8.0

      d) Tyr side chain – 10.0

      e) Arg side chain – 12.0

 

 

for a)  3 = 3  +  log (COO-/COOH)

          0 = log (COO-/COOH)

      100 =  (COO-/COOH)

 (COO-/COOH) = 1 = 1/1             Charge = 1/(1+1)  =  0.5                      ŕ  -0.5

 

for b)  3 = 4  +  log (COO-/COOH)

          -1 = log (COO-/COOH)

      10-1 =  (COO-/COOH)

 (COO-/COOH) = 0.1 = 1/10        Charge = 1/(1+10)  =  0.1                    ŕ  -0.1

 

for c)  3 = 8  +  log (NH2/NH3+)

          -5 = log (NH2/NH3+)

      10-5 = (NH2/NH3+)

 (NH2/NH3+) = 1/105 = 1/100,000    Charge = 100,000/(100,000 + 1) = 1.0  ŕ +1.0

 

for d)  3 = 10  +  log (-O-/-OH)

          -7 = log (-O-/-OH)

      10-7 = (-O-/-OH)

(-O-/-OH) = 1/107 = 1/10,000,000    Charge = 1/(1 + 10,000,000  = 0.0  ŕ  0.0

 

For e) 3 = 12  +  log (=NH/=NH2+)

          -9 = log (=NH/=NH2+)

      10-9 = (=NH/=NH2+)

(=NH/=NH2+) = 1/109 = 1/1,000,000,000   

Charge = 1,000,000,000/(1,000,000,000 + 1) = 1.0  ŕ +1.0

                                                                                    ------------

                                                TOTAL CHARGE  =    + 1.4

 

At pH 7, charges for a) through e) are:

-1.0, -1,0, +0.9, 0, +1.0  to give a total charge of -0.1 (close to the pI)

 

At pH 10, charges for 1) through e) are:

-1.0, -1.0, 0, -0.5, +1.0 to give a total charge of -1.5