Question

the proteins containing the following amino acids

61 LYS     12 SER      25 MET

22 HIS      75 GLU     56 ILE

29 ARG     45 PRO     83 LEU

77 ASP   67 GLY   26 TYR

24 TR       65 ALA      47 PHE

71 VAL

A) what its net charge at PH 1

b) What is net charge at ph 13

c) calculate the pI?

Answer 1

C) pI of the protein = Average of the pKa's of aminoacids in a given protein..

[(61*10.53)+(22*6.10)+(75*4.07)+(29*12.48)+(77*3.86)+(26*10.07)+(1*9.21)+(1*2.10)]/292 = ~ 6.89

(Note: number of aminoacids * pKa of carboxyl or amino group of  terminal or side chain of the aminoacids)

FYI: The net charge on the molecule is affected by pH of its surrounding environment and can become more positively or negatively charged due to the gain or loss, respectively, of protons (H⁺). At a pH below their pI, proteins carry a net positive charge; above their pI they carry a net negative charge.

Here is the formula to calculate the net charge of a protein at a given pH

Q_protein =Q^- +Q⁺

Q^- = (-1)/(1 + 10 ^{-(pH-pKa (carboxyl group))})

Q⁺ = (+1)/(1 + 10 ^{(pH-pKa (amino group))})

A) Net charge of the protein at pH 1 would be..

Q_protein= (+1)/(1 + 10 ^(1-9.21)) + (+61)/(1 + 10 ^(1-10.53)) + (+22)/(1 + 10 ^(1-6.10))) + (-75)/(1 + 10 ^-(1-4.07)) + (+29)/(1 + 10 ^(1-12.48))) + (-77)/(1 + 10 ^-(1-3.86)) + (+26)/(1 + 10 ^(1-10.07))

Q_protein = (+1)/(1 + 10 ^-8.21)) + (+61)/(1 + 10-9.53⁾) + (+22)/(1 + 10 ^(-5.10)) + (-75)/(1 + 10 ^(+3.07)) + (+29)/(1 + 10 ^(-11.48))) + (-77)/(1 + 10 ^(+2.86)) + (+26)/(1 + 10 ^(-9.07)) = (+1) + (+61) + (+22) + (-0.064) + (+29) + (-0.1) + (+26) = +138.83

B) Net charge of the protein at pH 13 would be..

Q_protein= (-1)/(1 + 10 ^-(13-2.10)) + (+61)/(1 + 10 ^(13-10.53)) + (+22)/(1 + 10 ^(13-6.10))) + (-75)/(1 + 10 ^-(13-4.07)) + (+29)/(1 + 10 ^(13-12.48))) + (-77)/(1 + 10 ^-(13-3.86)) + (+26)/(1 + 10 ^(13-10.07))

Q_protein = (-1)/(1 + 10 ^-10.9)) + (+61)/(1 + 10^(+2.47)) + (+22)/(1 + 10 ^(+3.6)) + (-75)/(1 + 10 ^(-8.93)) + (+29)/(1 + 10 ^(+.52)) + (-77)/(1 + 10 ^(-9.12)) + (+26)/(1 + 10 ^(+2.93)) = (-1) + (+0.206) + (+0.005) + (-75) + (+6.7) + (-77) + (+0.03) = -146.05