(Solved) For the oligonucleotide pApGpC, what will its net charge be at pH 7.6

spliceofslife

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13 years ago

For the oligonucleotide pApGpC, what will its net charge be at pH 7.6

For the oligonucleotide pApGpC, what will its net charge be at pH 7.6 You may round off to the nearest tenth of a formal charge.

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duddy

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13 years ago

I don't know how to get started on this? Any hints Crying Face

Could you show me what you've done, I'm sure I can pick up on it once I see an example!

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spliceofslife Author

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13 years ago

Ah yeah I don't really have any examples and it's not in the textbook but I know you'll need the Henderson-Hasselbalch equation. And the pKas which I'm also confused on which to use since they are not listed in my textbook! Thanks for responding to my questions I really appreciate it! I think the acid dissociation constant also plays a role

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duddy

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Staff Member

13 years ago

No problem, I want to learn how to do this too...

Well, the henderson-hasselbach equation says:

pH = pKa + log(A/HA)

Using this, we substitute 7.6 into pH, right?

Where A is the concentration of the conjugate base (this case, it's a deprotonated, negatively charged nucleotide); HA is the concentration of the conjugate acid (unprotonated neutral nucleotide). The nucleotides are A G and C... The first p will be negative (so -1); but would that be the overall charge?

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spliceofslife Author

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13 years ago

I guess I'll need to use the equation for each of the nucleotides. I'll be finding the answer out this afternoon. I'll make sure to post the work/answers if your curious!

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duddy

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Staff Member

13 years ago

Quote from: spliceofslife (13 years ago)

I guess I'll need to use the equation for each of the nucleotides. I'll be finding the answer out this afternoon. I'll make sure to post the work/answers if your curious!

Thanks man, I appreciate that.

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spliceofslife Author

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13 years ago

There was also a part b&c but thats what the key says. im still kinda confused though

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duddy

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13 years ago

Quote from: spliceofslife (13 years ago)

There was also a part b&c but thats what the key says. im still kinda confused though

Thanks for getting back at me, I'd just like to know where they got those fractions. I'm guessing they were given Confounded Face

Did he/she say anything?

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spliceofslife Author

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13 years ago

Ill show you how they got the +.4 for adenine at ph 4

So adenine pka is 3.8

pH=pKa+ log[A-/HA]
4.0=3.8 + log[A-/HA]
log[A-/HA]=0.2
[A-/HA]=10^0.2
[A-/HA]=1.58/1 (this is what confuses me)
HA=1/(1.58+1)=.38=~0.4

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spliceofslife Author

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13 years ago

I understand it completely now, i can walk you through a problem if you would like

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duddy

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#10

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13 years ago

Quote from: spliceofslife (13 years ago)

I understand it completely now, i can walk you through a problem if you would like

Yes please, if you wouldn't mind Undecided

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blogniva

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#11

13 years ago

I understand . Thank Face with Open Mouth

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spliceofslife Author

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#12

13 years ago

You plug everything into the formula, an for the last step(of the math i posted) you add in the formal charge that molecule would have at the pH. I'm not very good at explaining it, mostly because im not comfortable with it still. luckily it was not on my test which i got an 89% on with a class average for 45%. I'm pretty stoked. Thanks for always helping me out when i have questions.

If you have any specific questions about this problem ill try and help out. I have my next midterm tuesday so ill probably be posting lots of questions today!

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lakisch

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#13

13 years ago

Good post

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