7. Mixtures of amino acids are analyzed by first separating...

7. Mixtures of amino acids are analyzed by first separating the mixture into its components through ion-exchange chromatography. Amino acids placed on a cation-exchange resin containing sulfonate groups flow down the column at different rates because of two factors that influence their movement: 1) ionic attraction between the sulfonate residues on the column and positively charged functional groups on the amino acids, and 2) hydrophobic interactions between amino acid side chains and the strongly hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determine which will be eluted first from an ion-exchange column using a pH 7.0 buffer. a) Asp and Lys b) Arg and Met c) Glu and Val d) Gly and Leu e) Ser and Ala

Transcript

00:01 Hello students, let us solve this question.

00:03 In this question it has been given that we have a mixture of amino acid and we have to analyze them.

00:10 To analyze them we have to separate them using ion exchange chromatography.

00:21 Using ion exchange chromatography.

00:26 And in ion exchange chromatography specifically cateon exchange chromatography in which you have the sulfonate group sulfonate group right and it has been given to us that you have the buffer ph is 7 okay and we have to tell which of from the mixture of amino acid which which will be eluded first.

01:05 So actually the one with the lower net charge or a negative net charge will be eluted first.

01:12 Okay.

01:13 So now the first or a bit that is given to us is aspergene and lysine.

01:22 We have to tell between aspergene and lysine which will be eluted first.

01:25 Look at the structure of aspergene and lysine first.

01:29 So if you look at the structure of aspergene and lysine first, so this is the structure of aspegene here.

01:37 You have an o here, you have an o here, you have an o here, you have an o here, you have an o here and you have an oh here.

01:47 And when it comes to the structure of lysine, what do you have? you have this o here, oh here, this is your carboxyl group, nh2, this is your main group, nh2, right? now to see the net chart we have to know the pca values of these groups of these functional groups.

02:14 So here for this the pca value is 3 .65 for this the pca value is 5 .65 for this the pca value is what 9 .60 and for this the pka value is pqa value is 1 .80.

02:35 In lysine here the pk a value is what 6 for this the pk a value is 9 .17 for o h the pk value is 2 .18 so there can if the pk a value okay if the pk a value is greater than the ph of the solution of the buffer then protonation will happen protonation meaning hydrogen this will be added.

03:07 For example, the ph in this case is given to us a 7, right? and if you look at the pk value of the nh2 here, so what will happen? the pk value is 9 .60, which is greater than the ph, that is 7.

03:24 So it will get protonated, meaning one hydrogen will be added and this will become positively charged, right? second case scenario, what can happen? pca is, value of pkk is less than the ph.

03:38 Then what is going to happen it will lose the hydrogen h plus so in case of this when you look at the carboxyl group here it is less than the ph so what is going to happen it is going to lose its hydrogen and this will become oh minus similarly here this will also become o minus and it has lost its hydrogen so when you are calculating the net charge when we are calculating the net charge.

04:14 In this case you have one negative, one positive, one negative.

04:19 So net charge is what minus 1? in case of lysine if you see what is going to happen this will become plus 3 here and this will become oh minus as hydrogen will be removed.

04:33 So what will be the net charge here? the net charge is going to be plus 1 minus 1 that will give us 0.

04:44 So you have 0 net charge.

04:47 So from this information, we can tell that it is asper gene that will be eluded first.

04:57 Why asper gene? because it has a negative charge.

05:01 So the amino acid which has a negative charge or which has lesser charge than the other one will be eluted first.

05:08 So in part b we have to tell between arginine and methionine, which is going to be eluded first...

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