Dave Gray's Session 2 Email Q&A

The following is the text of my email to Lisa Scheifele following our second "Build a Gene" session:

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Hi, Lisa. Dave Gray again.

As we discussed, I have added my class notes to the http://www.openwetware.org/wiki/BUGSS:Build-a-Gene wiki with a link from the class home page in a way that accommodates other students adding their notes. Going back through my notes has raised some more questions. If you like, you can respond or we can just catch up during a break in the next class. Here they are:

1. I noticed that hydrogen bonds join the two strands of DNA in the double helix. That made we wonder if that is the reason why they denature at approximately the boiling point of water. The bonds that cause water to stick together are also hydrogen bonds. So perhaps the boiling point corresponds to the amount of energy needed to overcome those bonds?

The energy input (temperature increase) is used to disrupt the hydrogen bonds between the bases, thereby disrupting the double helix and causing the bases to form new hydrogen bonds with water rather than the other DNA strand, thereby making the single stranded state thermodynamically favorable at high temps.

The amount of energy needed to convert the DNA from double stranded to single stranded form is not constant for all DNA, but is dependent on the length of the DNA (longer DNA=more bonds=more stable=more heat needed to separate the strands; however, even for a long DNA molecule, once a portion begins to denature it destabilizes the entire molecule, which usually denatures rapidly after that ) and the percentage of bases that are G and C (G-C bonds are stronger than AT bonds). So long DNA strands with high GC content could have denaturation temps >100C. Practically speaking, no naturally occurring DNA has that much G-C, so we conveniently use 94-98C as the denaturation temperature because that is high enough to denature the vast majority of naturally occurring DNA sequences. At temps over 99C, the other components of the reaction are affected (the nucleotide bases and the Taq protein).

2. One of the animations shows how a plasmid can naturally form in bacteria. Would it sometimes occur that in doing so, it would split portions of a vital sequence of DNA such that the bacteria would no longer be able to produce an essential protein and die?

When plasmids integrate into the bacterial chromosome, usually, they integrate at a site that has the same genes as those on the plasmid and usually the insertion is precise so that you get extra copies of the genes (plasmids want their host cells to live!). They can, however, be designed so that their integration is specifically targeted within a gene, thereby disrupting it.

3. In our diagram of the complete vector, we included a marker segment. How will this be used? I have seen visualizations that show how splicing DNA into the middle of a marker segment can be used to show what bacteria include the splice by "breaking" the featured controlled by the marker. In our diagram, however, the marker is off to the side. Perhaps it's just a marker to show which bacteria have successfully had the vector implanted in them? But then the emGFP florescence could be used for that.

We’ll definitely cover this in session 4.

4. In some of our counts we used nucleotides and in others, base pairs. I assume we just use one or the other depending on whether we are talking about a single strand of DNA vs. a double helix?

Yes, nucleotides for single-stranded, and base pairs for double stranded, though I probably slip and use them too interchangeably.

5. Bacterial DNA occurs in a loop. However, I assume it is a double-stranded loop, right?

Yes

6. I had a note regarding herculase vs taq polymerase stating that it has an error rate of 1 in 8 vs. 1 in 12 for Taq. However, I can't imagine this is talking about the error rate for adding each nucleotide. Can you clear up what this refers to?

The error rate per nucleotide is 1x10-6, while that of Taq is 8x10E-6. What this means, is that once we complete synthesis of a gene ~750 bp long, 1 in 12 molecules will have the perfect sequence when using Taq while 1 in 12 molecules will have the perfect sequence when using Herculase (remember that error also comes from synthesis of the oligonucleotides, not just from the enzyme)

7. I included in my notes the following, which I wonder if you can verify:

We have to think of the DNA we are producing at three levels, each with its own "starting point". The first level occurs during amplification. There the primers specify the starting point for replication of the DNA. By including a primer in the "forward" direction for one side of the DNA strand and a primer in the "reverse" direction for its complement, we are able to precisely replicate only the portion of the DNA we are interested in.

The second starting point is the promoter. It starts the transcription to RNA from DNA, which stops when it encounters the terminator. The promoter can be strong, weak or operate only in the presence of certain chemicals.

The third starting point begins with the Ribosomal Binding Site and is the point where translation of the RNA to a protein begins assembling amino acids.

We have to think of the <strikeout>DNA</strikeout> gene we are producing at three levels, each with its own "starting point". The first level occurs during amplification. There the primers specify the starting point for replication of the DNA. By including a primer in the "forward" direction for one side of the DNA strand and a primer in the "reverse" direction for its complement, we are able to precisely replicate only the portion of the DNA we are interested in. This amplified DNA (the emGFP gene) can then be combined with a plasmid vector, transferred to bacteria and stably maintained in cells.

To have the gene be expressed in cells, we need two more processes to happen (transcription and translation) each of which has its own starting point. The second starting point is the promoter. It starts the transcription to RNA from DNA, which stops when it encounters the terminator. The promoter can be strong, weak or operate only in the presence of certain chemicals.

The third starting point begins with the Ribosomal Binding Site and is the point where translation of the RNA to a protein begins assembling amino acids.

8. In my notes from the last session, I included a point that the 20 oligos include a 3' hydroxal group on one end, allowing the enzymes to fill in the gaps between these segments. Is there anything "tricky" about inserting the last neucleotide into the gap given that there is no "open space" after it? Or is it all the more willing to jump into place given the attractive force forces of the backbone molecules on either side?

I have never heard of any structural issues surrounding inserting the last nucleotide.

9. As I understand it, in the last session, we only annealed the oligos into emGFP coding sequences. As I understand it, we didn't amplify these first because we would need primers for both ends of each of our 60 nucleotide segments. If the primers are 20 nucleotides long, we would need 40 neucleotides of primer to amplify each 60 nucleotide segment. Is that correct?

Yes, it would be highly impractical to amplify each oligonucleotide, and in fact the molar amount of DNA that we receive from the chemical synthesis company is already quite high.

10. I see froum the session 2 writeup that we must have done a round of PCA and a round of PCR with the oligos in the last session. I remember doing PCA but not PCR. Still, that's not unusual for me these days. But if that's the case, did we add primers in the PCA phase as the instructions indicate? If so, what role would they serve if we weren't amplifying? (As I understand it, each of the oligos also act as a "primer" to fill in the blank space in between.)

Yeah, we did both! We called it template-dependent PCR and we put the reactions in right at the end. Again, I’m probably being too lax and using “primers” and “oligos” too interchangeably. In the PCA, we combined all 20 oligos, which annealed to one another and each served as a primer for the Herculase enzyme to fill in the spaces between the oligos. This is a very inefficient reaction and leads to very little of the full-length product accumulating. However, it does lead to the assembly of a small amount of the emGFP gene, which can then be used as a template in a traditional PCR. In that PCR, we added only 2 primers which anneal to the very ends of the emGFP sequence, thereby amplifying everything between them (the full-length emGFP gene). Because you need 2 primers for the exponential amplification, any molecules that began to assemble but did not reach full length will not be amplified and their relative proportion in the population will decrease as the full-length emGFP gene is amplified and accumulates.

Thanks! - Dave Gray

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To Dave Gray's Build-A-Gene Experience Notes