Determine the net electrical charge of the following peptide atpH=10.7 Glu-Gly-V

Question

Determine the net electrical charge of the following peptide atpH=10.7 Glu-Gly-Val Amino acid pka information Glutamic acid 2.2, 4.2, 9.7 Glycine 2.3, 9.6 Valine 2.3, 9.6 The answer is -1.91 I understand that the only charges that are left are 4.2, 9.7and 2.3 but I don't understand how to find the net charge fromhere. Glu-Gly-Val Amino acid pka information Glutamic acid 2.2, 4.2, 9.7 Glycine 2.3, 9.6 Valine 2.3, 9.6 The answer is -1.91 I understand that the only charges that are left are 4.2, 9.7and 2.3 but I don't understand how to find the net charge fromhere.

Explanation / Answer

First think about which groups in the tri-peptide areionisable:     1. Terminal NH2 (on glutamicacid)     2. GLU side chain is(CH2)2-COOH, so this is ionisable (toCOO-)     3. GLY side chain is H, so not ionisable     4. VAL side chain isCH(CH3)2; this is not ionisable either     5. Terminal COOH (on valine) Now think about the partial charges on each of the ionisable groups(1, 2 and 5): 1. Terminal NH2 on glutamic acid The pKa of NH2 on glutamic acid is, from the question,2.2. At a pH of 10.7 we have from the Henderson-Hasselbalchequation:     10.7 = 2.2 + log([A-]/[HA])      8.5 = log([A-]/[HA]) 108.5 = [A-]/[HA] We want to know what [A-] and [HA] are, so for the moment we'llrewrite the above equation as:     108.5 / 1 = [A-] / [HA] Then we can equate the numerators and denominators, giving: [A-] =108.5 and [HA] = 1. Why use these values? Well, so longas the ratio of [A-]/[HA] comes out as108.5, it won't matter what values for [A-] and [HA] wechose. Obviously, having one of them equal to 1 simplifies things abit. Now that we have [A-] and [HA] we can calculate thefraction dissociated. This is equal to theconcentration of the dissociated group divided by the totalconcentration of dissociate-able group.     f = [A-] / ([A-] + [HA])       = 108.5 /(108.5 + 1)       = 1 What does this actually mean, though? Note that for NH2,the 'negative', or deprotonated, species (i.e.,A-) is NH2 and the protonated (HA) species isNH3+ = positively charged. We have calculatedthe fraction of [A-] to be 1, meaning that in this case the amountof HA present is negligible and all of it isNH2 = uncharged. This means the terminal NH2does not contribute to the tri-peptide's net charge because it isall deprotonated. (You may have realised this at the beginning, in which case youdon't need to go through with this calculation - just state thatsince pH >> pKa of the NH2 species, ~100% of itwill be deprotonated = NH2 = uncharged.) 2. GLU side chain is (CH2)2-COOH, so thisis ionisable (to COO-) We can do a similar process with the glutamic acid side chain,which has a pKa of 4.2. Using the H-H equation:     10.7 = 4.2 + log([A-] / [HA])       6.5 = log([A-] / [HA])    106.5 = [A-] / [HA] Then, as before, let: [A-] = 106.5 and [HA] = 1. Nowcalculate the fraction dissociated:     f = [A-] / ([A-] + [HA])       = 106.5 /(106.5 + 1)       = 1 Once again, since pH >> pKa, all of the group isdeprotonated. However unlike part (1), the deprotonated species isCOO-, and therefore carries a negativecharge. This group therefore contributes -1 to the net charge. 5. Terminal COOH (on valine) Valine's COOH has a pKa of 9.6. From the H-H equation:     10.7 = 9.6 + log([A-] / [HA])       1.1 = log([A-] / [HA])    101.1 = [A-] / [HA] So let [A-] = 101.1 and [HA] = 1. Then the fractiondissociated, or fraction of deprotonated COOH is:     f = 101.1 / (101.1 +1)       = 0.926 So this group contributes -0.926 to the net charge. Summing the effects of (2) and (3) we have a net charge of: -0.926+ -1 = -1.9. (Not sure where the -1.91 comes from - probablydue to significant figures somewhere along the line...)

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