- 1 Student Registration/Questionnaire
- 2 M13 Genome Engineering
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20.109 Spring 2007
Course 20, minor in French, strong interest in neuroscience
Year of Graduation
emilie7 at mit dot edu
Have you taken
7.05/5.07 (Biochemistry) currently taking
7.06 (Cell Biology) no
7.02 (General Biology Lab) no
5.310 (General Chemistry Lab) no
Do you have any experience culturing cells (mammalian, yeast or microbial)?
Do you have any experience in molecular biology (electrophoresis, PCR, etc)?
Please briefly describe any previous laboratory experience
UROP in Fee Lab in McGovern Institute, BCS. February 06- present. I do brain surgery on songbirds (lesions, tracers, and electrophysiology), perform profusions, histology, immunochemistry, and microscopy.
Anything else you would like us to know?
Optional: As mentioned in lecture on 02.06.07, we would like you to consider how, as a biological engineer you might test the superstition "Spit on a bat before using it for the 1st time to make it lucky."
I would perform a double blind experiment in which both athletes testing bats and evaluators scoring performance of the bats do not know which bats have been spit upon. One could then statistically analyze the data from bats that had and had not been spit upon to see if bats with spit on them were any luckier. If, as I suspect, they are not, then I have proven that I do not need to go into molecular analysis of the spit to try to find the elements which contribute to the luck.
M13 Genome Engineering
Gene Functions and Re-engineering Ideas
|PI||Phage Secretion: C-terminals of P1 and P11 interact with N-terminal of P4 to form a complex through which mature phage are secreted.|
|PII||Replication: Initiates genome replication by nicking the double stranded phage DNA||Without P2, could the bacteria still make the phage gene products without replicating the phage DNA? If we remove P2 (and P5), could we use this virus to insert other genes into the bacteria? NOTE: Altering the second half of P2 will affect P10.|
|PIII||Rounded Tip of Filament: First protein to interact with the E. coli host pilus during infection and last point of contact as phages bud from bacteria. without P3, phage cannot fully escape host.||We might change P3 such that the phage was only capable of infecting a different host other than E. coli, if the need arose.|
|PIV||Phage Secretion: N-terminal of P4 interacts with C-terminals of P1 and P11 to form a complex (stable, barrel-shaped structure) through which mature phage are secreted.||Unless phages bottle-neck at the secretion complexes such that secretion becomes the limiting factor in phage production, it does not seem that modifying the proteins that form this complex will help expedite or even really alter our process|
|PV||Phage Packaging: sequesters + stranded phage DNA to be packaged into new phage particles||If we have a stronger promoter in front of this gene, phage packaging might occur faster, giving us more efficient production times. However, if too many + strands are taken away, this will hurt the DNA Replication process. Furthermore, no matter how many + strands are sequestered, we will still need enough coat protein to package these strands of DNA, so a strong promoter in front of this gene alone might prove useless.|
|PX||regulates number of double stranded genomes in the host. Without p10 no +strands can accumulate. Identical to C-portion of p2!||If we alter p10, we will affect p2. If we don't want phage DNA replication to occur, we can alter this gene.|
|PXI||Phage Secretion: C-terminals of P1 and P11 interact with N-terminal of P4 to form a complex through which mature phage are secreted.||
M13 Design Constraints
Would you expect the phage to tolerate p8 modifications that: 1. make the protein neutral rather than negatively charged at the C-terminus? What is the role of the C-terminus? It seems that the C-terminus may need to be charged in order to hold together the p8 proteins that make the phage coat. The charge may also be necessary for intermolecular interaction required to secrete the phage. It seems that a neutral protein will be less likely to have intermolecular interactions than a negatively charged protein, so I think that the phage would struggle to tolerate this modification. (Ideas from http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&list_uids=12217696&dopt=Abstract ) 2. encode all the Leucines with the CTA codon instead of the CTG codon? The codon sequence is not important as long as the appropriate amino acid is encoded, so the phage could tolerate this modification. 3. double the size of the protein? To answer this question, it would be helpful to know how the p8 knows how to regulate its production. Since the amount of p8 produced decreases when the genome size is decreased, this leads me to suspect that somehow the amount of p8 is being regulated such that the phage coat will always fit snugly over the phage’s DNA. Therefore, if the size of p8 is doubled, I suspect that only half the amount of protein will be used and the phage will tolerate this change just fine.
Would you expect the phage to tolerate these same modifications to p3? 1. C-terminal domain is required for release from the host cell following phage assembly, and contributes to structural stability of phage particle (See abstract http://chem.ps.uci.edu/~gweiss/WeissG-03JMB.pdf ). If we change the charge, it will alter intermolecular interactions, and it may destroy the functionality of p3, thus preventing the release of the phage and/or destabilizing the phage. 2. encode all the Leucines with the CTA codon instead of the CTG codon? The codon sequence is not important as long as the appropriate amino acid is encoded, so the phage could tolerate this modification. 3. if the C-terminal is doubled in size, I think that it will still function normally, though it’s possible that it will not be able to pass through the p1-p4-p11 complex through which mature phage are secreted.
Would you expect the phage to tolerate transcriptional terminators that are 1. 2X stronger 2. 100X stronger 3. 2X weaker 4. 100X weaker Assuming that we want the transcription to be stopped at a transcriptional terminator, it seems that it wouldn’t be a problem if the transcriptional terminator was stronger, unless an excessively strong transcriptional terminator caused the end part of the protein to be cut off. If a weak transcriptional terminator causes a protein to have a long tail that could potentially cause problems, should the tail undergo undesired intermolecular interactions of simply take up too much space in the phage. It may also be possible that a weak transcriptional terminator would not successfully stop translation of the protein, which would be a major problem. http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=556655
The Family Tree: M13's closest evolutionary relatives