1) a) Calculate the net charges on the following peptides at...

1) a) Calculate the net charges on the following peptides at pH=7.0? Show steps. Show individual charges on amino acids to explain how you arrived at the net charge for each of the peptide. (9 points) Peptide 1: Ala-Glu-Lys-Phe-Asp Peptide 2: Asp-Glu-Ser-Phe-Arg Peptide 3: His-Val-Gly-Asp-Phe (b) Based on the net charges of the 3 peptides, what would be the order of elution of the peptides from a cation exchange column? Explain your reasoning? (4 points)

Transcript

0:00 Hello everyone.

00:01 In this question we have been given the polypeptide chain and the ph given to us is 7 and we need to let we need to tell the net charge of each of the polypeptide chain.

00:17 So here this peptide 1 has alenium glutamate lysine, phenylalanine and spartate.

00:29 So the alenine here will have the amine group and the spartate here will have the carboxylic group.

00:39 The alenine has the functional group of ch3, the glutamate has c -o -o -h, and the lysine has the amine group, that is the upsyline amine group.

00:52 And the phenyl aline has the nh2 and the espartate also has the carboxylic group.

00:59 As its functional group.

01:03 So here we can see that at ph 7 which is neutral ph.

01:08 So here this cooh will be be protonated because its pca value is 2.

01:18 So here this will have the negative charge and the pca value of the amine group here is 9.

01:27 So it will remain in its protonated form because here the ph is less than the pk value of the amine group and the coh value of the cooh on the glutamate is 5 so that means this here will get deprotonated so here we will have the minus charge then the pk value of the spartate here is 4 and so this will also get deprotonated.

02:04 Here the pk value of this lysine is 11.

02:08 So that means this will remain in the protonated state.

02:12 And the phenylalanine will also remain in the protonated state.

02:16 So here we can see we have 1 plus 1 then minus 1 plus 1 sorry here on the phenylalanine there is no functional group.

02:28 So we will cancel this.

02:29 Here we have minus 1 on the spartate and then we have minus 1 here...