chem help please

13

Please could I have some help with question 6e of this paper?
Paper: https://bestexamhelp.com/exam/cambridge-international-a-level/chemistry-9701/2022/9701_s22_qp_41.pdf
The ms says that Asn is nearly uncharged but I'm unsure how that is true because ph 5 is lower than 5.4?
Thank you.

Reply 1

Nitrotoluene

12

Original post

by anonymous56754

Please could I have some help with question 6e of this paper?
Paper: https://bestexamhelp.com/exam/cambridge-international-a-level/chemistry-9701/2022/9701_s22_qp_41.pdf
The ms says that Asn is nearly uncharged but I'm unsure how that is true because ph 5 is lower than 5.4?
Thank you.

We are currently examining the electrophoresis of lysine and asparagine.
Lysine has a pI of 9.8, and asparagine has a pI of 5.4. The buffer pH is 5.0.
First, remember that lysine is going to be completely protonated because the pH is much lower than its pI. So it will be fully charged and head toward the negative side.
Asparagine is fairly close to its isoelectric point and will be neutral and move little to nothing.
The lys-asn dipeptide would be positive and going toward the negative side because of the lysine but reasonably neutral because of asparagine and also not as charged as the lysine alone.
That said, after viewing the electrophoresis:
Spot E moved to the negative side, so positive.
Spot F had no movement, so it appears to be mostly non-sentinel and nothing at all.
Spot G went farther to the negative side, so it too is positive like lysine.
It can be deduced that:
Spot E likely represents the lys-asn dipeptide migrating towards the negative electrode, though its movement would be less extensive than that of free lysine.
Spot F is probably asparagine because it did not migrate very much and is very neutral.
Spot G is likely lysine because it travelled farthest toward the negative side, suggesting a charge difference because of a strong positive charge.
Here is my 2 cents!

Reply 2

anonymous56754

OP

13

Original post

by Nitrotoluene

We are currently examining the electrophoresis of lysine and asparagine.
Lysine has a pI of 9.8, and asparagine has a pI of 5.4. The buffer pH is 5.0.
First, remember that lysine is going to be completely protonated because the pH is much lower than its pI. So it will be fully charged and head toward the negative side.
Asparagine is fairly close to its isoelectric point and will be neutral and move little to nothing.
The lys-asn dipeptide would be positive and going toward the negative side because of the lysine but reasonably neutral because of asparagine and also not as charged as the lysine alone.
That said, after viewing the electrophoresis:
Spot E moved to the negative side, so positive.
Spot F had no movement, so it appears to be mostly non-sentinel and nothing at all.
Spot G went farther to the negative side, so it too is positive like lysine.
It can be deduced that:
Spot E likely represents the lys-asn dipeptide migrating towards the negative electrode, though its movement would be less extensive than that of free lysine.
Spot F is probably asparagine because it did not migrate very much and is very neutral.
Spot G is likely lysine because it travelled farthest toward the negative side, suggesting a charge difference because of a strong positive charge.
Here is my 2 cents!

thank you!!

chem help please

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