Biochemistry & Molecular Biology Amino Acid question

[Turtle]

6. Select the most CORRECT option to fill the gaps in the following statement:
An enzyme has an isoeletric point (pI) of 2. We expect it to be rich in the amino acid
residues _____ and _____ (using the one letter code) and to have a _________ net
charge at pH 7.
A. K and R, positive.
B. D and E, negative.
C. K and R, negative.
D. K and D, positive.
E. D and E, positive.

I can't figure out why C isn't the answer, and why B is the answer. I know K and R are both basic, negatively charged residues, and I know D and E are acidic, positively charged residues. I know that if an enzyme has a pI of 2, it is likely to be more basic, therefore it should be rich in K and R, and be negative.

Can someone show me where I am wrong please. Because I am obviously wrong somewhere?

[Turtle]

Also, can someone tell me why a protein with a pI of 12 wouldn't have more acidic residues than basic residues?

I thought the idea is that if a protein has a pI which is high, it will have more acidic residues, and if the protein has a low pI, then it will have more basic residues?

[simpak]

Hey Turtle I also have the Biochem MST tomorrow, yay for us, perhaps I'll see you there.
So, it's actually the other way around: if a peptide is more acidic it'll have a lower pI and if a peptide is more basic it'll have a higher pI.  This is because acidic residues are negatively charged when they're ionised and basic residues are positively charged when they're ionised.  So we have a peptide with a pI of 2, it is neutral at pH of 2 because at pH of 2 there are a lot of hydrogen molecules in solution and those hydrogen molecules are able to cling on to the negative acidic residues of the protein and neutralise the protein.  If you add more hydrogen (lower pH) then you're going to have a more positive residue and if you take some away you'll have a more negative protein because those hydrogens move off the COO- molecules of aspartate and glutamate as you take the H atoms out of solution.  So as you have correctly deduced, at a higher pH, where less H+ is in solution the protein is more negative.  I think it can be helpful to think of it the long way rather than just remembering the rule because it certainly helps you decide on the residues.

So we have C and B as you have picked out.  B says D and E (aspartate and glutamate) would be in the protein which is true; both are acidic residues and therefore are found in low pI proteins.  When we move H+ out of the solution and these are in high concentration in our protein more COO- groups appear instead of COOH groups and the protein becomes negatively charged.  It fits the description.  C says K and R which are basic residues that are positively charged and are more likely to be found in proteins with a high pI.

Similarly with the pI of 12, you would have more basic residues.  The protein is neutral at pH 12 (when there are few hydrogen molecules in solution, the solution is less acidic) and therefore when more H molecules have been taken off the NH3+ of arginine, lysine and the NH of histidine.  If you put a protein with a lot of acidic residues in an environment of pH 12 there wouldn't be any hydrogen molecules to neutralise them and you would end up with a negative, rather than a neutral protein at pH 12.

Hope this helped, if you're still confused I'll try to help more?  Just had it the wrong way round I think!

[Turtle]

Thanks Mavis for the help, I never said thank you, I feel ashamed, so thanks

I just have one more question if anyone on the forum could help me with it.
In the notes for Biochem is says:

Cation exchange: the column has a matrix that is negatively charged (anions) and at neutral pH will bind positively charged (high pI) proteins.

Anion exchange: the column has a matrix that is positively charged (cations) and at neutral pH will bind negatively charged (low pI) proteins.

I thought positively charged residues have a low PI, judging from what Mavis told me below. So why does the notes say that positively charged residues have a high PI? Thanks for the help!

[Turtle]

If someone could help me with this question too, then I would appreciate it

In my notes it says that as a protein becomes longer, and as the C terminus and N terminus become more separated, the PH range becomes narrower.

It says this is because there is less favourable stabilisation between the C terminus and N terminus. I though that this would mean the PH range becomes greater?

I thought that as the N terminus and C terminus become closer, the PH range will be less, because the N term and C term can interact favourably with each other?

Can someone explain why this happens in simple terms. Thanks

[simpak]

Hey, I shall reply again! I hope my answers help haha. Okay so for the first q don't over think this too much, this is just simple application of the rule. High pi protein has a pi greater than 7 so at a ph less than this it would have a positive charge. And to think about that theoretically, high pi proteins are neutral when there are not many protons and as you add more protons you'll begin protonating positive residues for more positive charge. And for the opposite, low pi proteins are neutral at a pi less than 7 so at a neutral ph of 7 they will be more negative, we have taken more protons out of the environment and this has pulled them off all the acidic residues to give an overall negative charge. Just think about how high pi means when the proton isn't charged and compare that to neutral ph: is neutral ph lower or higher and therefore have you increased or decreased the protons? Those have to have come from somewhere so you must be adding them on to the protein or taking them away from the protein and then what does that give you?

I feel like maybe you're only getting confused because you didn't see it as an application of the 'when the ph is greater than the pi, the protein is negatively charged rule'. A high pi protein is negatively charged at a ph higher than it's own pi but at a ph lower than it's pi it must be positive!

Question 2 has stumped me a little because I haven't revised it yet but let me read over it tomorrow and ill get back to you! Apologies for pi looking like I am reiterating calculus instead, I'm on my phone haha.

[simpak]

Okay so now I have looked over my notes, I can answer the second one (or at least attempt to).

Remember in Chemistry 2, when we learnt about how you make a strong or weak base or acid?  A strong acid will be one that can distribute its negative charge and be more stabilised in its ionic form (eg, has more resonance structures to distribute the negative charge over or more s character).  Strong acids also have a low pKa.  In this case you're not stabilising the charge of the acid by giving it more resonance structures or making atoms more electronegative but by using the nearby basic region on the amino terminal to stabilise the charge.  This in turn makes the end a stronger acid and lowers its pKa.
You can use the same logic for the basic end, except that when it is close to the carboxy terminal it would have a higher pKa and therefore be a stronger base.

Once you start pulling the ends of the protein away from one another that stabilisation is removed and both ends become less comfortable carrying charge.  They are therefore less comfortable being completely ionised in water and have a pKa closer to 7 in both cases (ie the pKa of the acid increases as it becomes a weaker acid and the pKa of the base decreases as it becomes a weaker base).  So because the difference between the pKa values is decreasing their pH range is decreasing.

The favourable interaction between the N and C terminus therefore actually makes these ends stronger bases/acids respectively because they are more happy being ionised when they can use the nearby charge of the opposing group to stabilise the charge they obtain through ionisation.

I hope that made sense! D: if it doesn't let me know and I'll try to rephrase or something...or perhaps someone else can give some input!  I feel like I'm crap at explaining chemistry clearly.

[Turtle]

Thanks Mavis, you are a genius!! 

[simpak]

Haha, not at all, but I'm glad I could help

[Starlight]

Hey guys, how do you find the turnover number for an enzyme (RE: Page 22 of exam paper 2010)

[simpak]

This stuff really confuses me 3:
I thought you would need to know the enzyme concentration to answer?

[Starlight]

This stuff really confuses me 3:
I thought you would need to know the enzyme concentration to answer?

Yep! that's what I was thinking too!
The turnover number is the thing i'm confused most about

[simpak]

It's so confusing!  Wait:
"the velocity of the enzyme-catalysed reaction in expressed in units of μmoles of product produced per second per μmole of enzyme"
Okay so I think I figured it out, but it took a shower and a revelation in the form of rereading the question haha:

Kcat = Vmax/[E]
Kcat = 2micromol sec-1/1micromol
Kcat = 2 sec-1

But that's not an option, but 120 min-1 is.
So 2 x 60 seconds = 120 so the answer is option 6.

Guess we need to make sure we read the whole question, I kept focussing on the graph without looking at what it said about enzyme concentration, doh!

[Starlight]

It's so confusing!  Wait:
"the velocity of the enzyme-catalysed reaction in expressed in units of μmoles of product produced per second per μmole of enzyme"
Okay so I think I figured it out, but it took a shower and a revelation in the form of rereading the question haha:

Kcat = Vmax/[E]
Kcat = 2micromol sec-1/1micromol
Kcat = 2 sec-1

But that's not an option, but 120 min-1 is.
So 2 x 60 seconds = 120 so the answer is option 6.

Guess we need to make sure we read the whole question, I kept focussing on the graph without looking at what it said about enzyme concentration, doh!

Thanks for clearing it up!

[Turtle]

Hey Mavis, or anyone else here, is this an okay explanation for why the PH range becomes less between NH3+ and COO- as the peptide gets longer?

As the peptide becomes longer, there will be more residues to stabilise the charges on NH3+ and COO-, which means that their Pk values will be closer together, and the PH range where they are ionized will be narrower. In contrast, a shorter peptide will have less residues to stabilise the charges on NH3+ and COO-, so their Pk values will be further apart, and the PH range at which they can be ionized will be greater.