Can someone clarify some DNA related questions to me

[September24]

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I have some "notation questions" related to to 5' 3'.

1. How is the phosphodiester bond made. I understand it is the backbone of the dna. But is it between the phosphate group on the 3' nucleotide and the sugar of the 5' sugar?? Or is it the phosphate group of the 5' nucleotide and the sugar of the 3' nucleotide?


2. Replication, transcription and translation. Is this correct?:

Replication:
Read: 3'-->5'
Replicated(new DNA is built): 5'-3'

Transcription:
Read: 3'-->5'
Added:5'-->3'

Translation:
mRNA is read: 3' to 5'
Amino acid bond is made: Between 5' N terminus and 3' carboxy terminus

 

[thethreeheadeddragon]

1. The phosphodiester bond is formed in a condensation reaction (loss of H2O) when the 3' hydroxyl OH group from one nucleotide's sugar group reacts with the 5' phosphate group from a second nucleotide. Basically, you'll be left with a backbone that looks like a free 5' phosphate, sugar, (phosphodiester bond, sugar)repeating, free 3' OH.

2. Your translation description is not quite right. When the mRNA attaches to the ribosome, it attaches 5' end first. The reading frame moves towards the 3' end, until it hits a start codon, at which point the tRNA's will come along and attach amino acids until it hits the stop codon near the 3' end.

 

[September24]

Oh okay. For the PD bond, I read a question which listed two nucleotides. One was labeled as a 5' nucleotide and the other was 3'. the explanation stated that the bond would be formed between the phosphate of the 3' nucleotide and the sugar of the 5' nucleotide. Does this make sense? I thought it was the 5' of phosphate and 3' of sugar.

 

[September24]

Also, for translation. I understand that the ribosome moves along the mRNA in the 5' to 3' direction, but I can't seem to find out the details of the peptidyl transferase reaction. Is the peptide bond made between the 3' carboxy terminus of the peptide at the P site and the 5' amino group of the peptide in the A site. Or is it opposite. The peptide bond is made between the 5' amino terminus of the peptide at the P site and the 3' carboxy terminus of the peptide at the A site.

 

[thethreeheadeddragon]

Yes, so you really have to be looking at the whole molecule first when you're talking 5' and 3' ends. The 5' prime nucleotide will be the first one in the sequence. At it's 3' OH end, it will react with the 5' PO4 end of the 3' molecule. Weird wording, but hopefully that makes some sense. And as for the peptide, it has a 5' amino NH2 end and a 3' carboxyl COOH end. So what happens during amino acid elongation is another condensation reaction forming an amide bond between the 3' COOH end of the 5' molecule and the 5' NH2 end of the 3' molecule.

 

[September24]

I'm still a bit lost on both but here's what I learned. I'll try to get translation first.

For translation: MY BR book says the amino nitrogen of the amino acid on the A site (which is closer to the 3' end of mRNA) attacks the carboxy terminus of the AA on the P site (closer to 5' end of mRNA). So basically, the peptide bond is made from 5' amino end of A site AA to the 3' carboxy of the AA on P site.

You said its between "between the 3' COOH end of the 5' molecule and the 5' NH2 end of the 3' molecule". I think thats what I'm getting to as well. By " 5' " molecule, you mean the molecule that is closer to the 5' end of the mRNA. Closer to the 5' end would be the P site. The peptide bond is made between the carboxy end of the P site and the amino end of the A site.

My head is spinning. But if that sounds correct, I may be getting it.

 

[thethreeheadeddragon]

Yep, that's correct!

 

[September24]

Okay. Thanks! I'm gonna sleep on the phosphodiester bond question but I was wondering if you could answer this translation question. I got this from another thread.

"

The question is basically asking, if you change everything from the middle to the end of an amino acid sequence, which end changes and which end would remain the same?

It basically asks, when amino acids are assembled into proteins, are they built from the amino to the acid direction, or the acid to the amino direction?

The answer is that amino acid sequences are always assembled N to C, much like nucleotide sequences are built 5 to 3."


Its about a random question (if you want me to post the entire thing, I can). The key is the last sentence 'amino acid sequences are always assembled N to C". This really clarifies the whole translation debacle I had. However, I thought from this post, we decided that the carboxy terminus in the P side is added to the amino terminus in the A site. That's Carboxy to amino. But the post I just read said that it was N to C.

 

[thethreeheadeddragon]

Well, yes at a smaller scale there's always going to be a 3' COOH attaching to a 5' NH2, but the point is that the very first amino acid has a free 5' NH2 (and the very last one has a free 3' COOH), and thus, the start of the amino acid sequence is N and the end is C.

 

[camaryal]

Here are some tips for when you're thinking about 5' and 3' and translation.

1) When you're thinking about translation, don't refer to it as the 3' COOH or 5' NH2. When you're synthesizing protein, there's no 3' or 5' on the amino acid. All that has to do with is at the nucleotide level, not protein level. The ribosome sits down on the mRNA and elongation occurs from 5'-->3' and peptide synthesis occurs from N-->C. The amine will attack the carbonyl to make the peptide bond on the new amino acid on the tRNA in the A site, then the ribosome will move over to the next codon, moving that tRNA to the P site so the next one can move into the A site.

2) With the phosphodiester bond, think about it like this: you always have to synthesize from 5' to 3'. What's the basic structure of a nucleotide? First you have a sugar with a hydroxyl group off C3 of the ring. C4 is bonded to C5, which is bound to the phosphate. Why do we always have to add to the 3' end? Because when new nucleotides come in during replication or transcription, their 5' end has a triphosphate that can react with the already present 3' hydroxyl to form the phosphodiester bond. What would happen if we synthesized from 3'-->5'? As we're synthesizing, we'll have a triphosphate end to which we can attach a hydroxyl. Is that possible? Sure. But what happens when you put in the wrong nucleotide? You could take off that wrong nucleotide, but they you're left with a monophosphate end and you can't synthesize from that. So you're basically screwed and synthesis wouldn't occur past that point. That's why you always synthesize from 5'-->3' and why the phosphodiester bond is formed with a newly attached 5' phosphate and ending with a 3' hydroxyl.