Final Exam B, Problem 6 | MIT 3.091SC Introduction to Solid State Chemistry, Fall 2010

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Hi I'm Jocelyn and we're going to go over fall 2009, the
final exam, problem number 6.
It says, the dipeptide shown below is derived from alanine
and serine.
And I'll write up that structure here.

And we know it's a dipeptide because we
see that it has two--
what look like--
amine groups and two kind of carboxyl groups.
So that's an indicator that we're dealing with a dipeptide
not just one amino acid.

Something to notice right off the bat is that we're given
the zwitterion form.
We have the amine group at the end here is protonated and the
carboxylic group on the other end is deprotonated.
So moving to part 1.
It asked us to indicate the position of the peptide bond.
Hopefully this is a pretty straightforward task.
We realize that the dipeptide bond is
what makes the dipeptide.
It's the bond between the amine group of one amino acid
and the carboxyl group of another.

Right there.
So circling that carbon nitrogen bond would get you
full credit on that part.
Moving to part 2.
We are asked to draw the skeletal structure of each
constituent amino acid as it would be present in an aqueous
solution of an extreme basicity, i.e. pH
greater than 12.
So before even starting to write we want to think about
what extreme basicity means.
Well it means that both our amine group and our carboxylic
acid group are going to be deprotonated.
So let's start with alanine first. We
have the central carbon.

The amine group.

And the carboxylic acid group.

Now I just drew that totally protonated.
However we want to draw it totally deprotonated because
it's in basic solution.
So I'm just going to erase the extra hydrogens.
And we have a deprotonated amine group and a deprotonated
carboxyl group.
And that is what it would look like in basic solution.
Now serine is a little different because you have a
side group with functionality.
It has not just carbon and hydrogen.
First drawing up the backbone here, we have a central
carbon, which has a methyl alcohol group off of it.
And attach the carboxylic acid.

And amine group on the left here.
So this one I drew as the zwitterion, which would be in
moderately or kind of neutral.
It depends upon the amino acid.
But again we want to deprotonate our amine group
here because pH greater than 12 will not have a protonated
amine group.
Now the question is, what do we do with
the side chain here?
It has an H attached to an oxygen.
Does that come off in super basic solution or not?
A lot of students got tripped up on this.
An alcohol actually has a pKa of about 16.
And so it won't be deprotonated
in an aqueous solution.
Therefore we want to keep this hydrogen on.
It will not be--
this will not be a charged oxygen in the basic solution
that we were asked about.
So now we are going to move on to the second
part of this problem.
The second part asks us about the tertiary structure of the
length of protein that is shown.
I am not going to draw the whole length of protein but
we're going to--
at the four specified locations--
we're going to indicate what we would call the tertiary
structure; or what functionality is causing the
tertiary structure.
So looking at that protein chain.

Professor Sadoway has indicated four different
distinct parts of the chain.
And the question is asking us, at each of the four numbered
positions name one chemical change to the environment of
the protein that would destabilize the associated
feature in the tertiary structure.
Explain the relevant chemistry so it's asking us to say what
can we do to disrupt that tertiary structure?
And the first thing we need to do is figure out what the
tertiary structure is.
So if we look at number 1, we see that it has two sulphur
atoms connected together.
So we call that a disulfide bond.

The second part, we see that we have a phenyl group, so it
has a very electron-deficient hydrogen interacting with a
deprotonated carboxyl oxygen anion.
And so we have a hydrogen bond there and that's what's
causing those amino acids to be pulled together.
So let's just say hydrogen bond.

At the third number we see that we had a deprotonate
carboxyl group and a protonated amine group
interacting.
And that's not quite a hydrogen bond but it's a
negatively charged part of the molecule interacting with the
positively charged part of a different molecule.
And so we call that electrostatic interaction.
It's similar to a hydrogen bond in that it's a
intermolecular interaction that's causing these molecules
to move closer, preference to be closer together.

And now we come to the fourth part.
Which is hydrophobic interactions.
So we see that there are a lot of just
carbon hydrogen chains.
No oxygen is around, no amine groups around.
Purely carbon and hydrogen.
And because this is an aqueous solution, those hydrophobic
parts of the molecules will preferentially go together and
kind of make a little--
almost like an-- oil droplet in water.
They're trying to decrease the amount that they experience
the polarity of the water.
And so the interactions are hydrophobic.

So now that we know what the functionality at each of these
points in the DNA structure are, we can go back to what
the question is actually asking us, which is, how can
we actually disrupt these features.
So the first one is the disulfide bond.
And in class Professor Sadoway talked about how to disrupt
the disulfide bond.
And one of the ways is to put extreme
heat on to the protein.
And that breaks the disulfide bond and you can straighten
out that portion of the DNA.
This is how you straighten your hair if it's curly or
other types of things with the disulfide bond.
So heating the disulfide bond to extreme heat will cause it
to break and thus the protein will lose that, that folding
or whatever the disulfide bond imparted to that portion of
the protein.

Now going on to number 2.
We have hydrogen bonding.
And we see that it's between a alcohol group on a ring there
and a carboxyl group.
So one way to disrupt that hydrogen bond is to make the
carboxyl group no longer deprotonated.
So if we put back on the hydrogen to that oxygen it
will become less partially negative.
It will have a lower
interaction with that hydrogen.
And so we can say, to protonate that carboxyl we
want to make the solution more acidic.
So we can add acid.

So the disulfide we're going to heat, the hydrogen bond
we're going to add acid.

Looking at number 3 we see that we have a
electrostatic reaction.
And so we want to not have that electrostatic reaction.
And one way to do that is to make one of those entities not
have a charge.
So one way to do that would be to reprotonate the carboxylic
acid, just as we did in the previous step.
Or we could deprotonate the amine group.
And that would cause it to lose its charge and thus the
electrostatic interaction would be weaker or gone.
So to deprotonate the amine group we should know that we
want to add base.

And the final part is, the portion with hydrophilic
interactions.
And here we want to figure out a way to make those
hydrophobic portions of the amino acids not hold as
tightly together.
And one thing we learned about in class is
something like soap.
Something that has a hydrophobic side and a
hydrophilic side.
And so if we put soap in with this protein it can kind of
stabilize the hydrophobic part and make it stretch out a
little bit more, not ball up so tightly.
So soap or detergent would be a good answer for this one.

And so for this question, as long as you indicated the
relevant chemistry, which is specified what tertiary
structure you were looking at and how your suggested
procedure would disrupt that tertiary structure, that's
what Professor Sadoway was looking for and that would
mean that you answered the question completely.