[sameeralord]

1.Determine the net charge of the following peptide at PH 7.0

                     PKa values
                    C00H            NH3+      Side chain

Valine             2.29            9.72

Aspartate        1.99            10.0         3.96


2. The PKa value of the sulphydryl (-SH) group of cysteine is 8.33.  Calculate the fraction of anion to free sulfhydryl group at PH 7.O.

I'm not a homework posting type person. But today I'm really running out of time and really need help. Hope you understand and help. Thank you so much 

[renge ishyo]

1. Draw the structure of the peptide out. Start with the pH=0 situation and protonate all the carboxylic OH groups and all the NH2 groups including those that appear on side groups (caution, do NOT protonate the NH groups that are part of the peptide bonds in the center of the molecule). Now as you go to a pH above the pKa value, erase the proton from the group that that pKa value corresponds to. For instance, for aspartate you would erase the hydrogen on the nuetral carboxylic OH group of its side chain for any pH's greater than 3.96 and replace it with an O-. Do this one by one for each group and then simply count the number of positives and subtract the number of negatives to get the net charge on your picture.

2. This at least appears to be a straight application of the Henderson Hasselbach equation:

pH = pKa + log ([S^-]/[SH])

The fraction of anion to sulfhydryl group is equal to the ratio of [S^-]/[SH] found in the equation above.