Integration in B.subtilis
We have investigated the use of double crossover events to integrate our genetic devices into B.subtilis. In addition, we want to design integration biobricks, that allow site specific integration of parts and devices into the B.subtilis chromosome. We aim to create a series of integration bricks that will direct integration into neutral endogenous genes. By inserting into neutral genes our B.subtilis will still be viable and in addition, we will have a means of phenotypic screening to identify successful integration events.
|PyrD - dihydroorotate dehydrogenase 1B||Growth Media needs to be supplemented with uracil - Uracil needs to be added to a final concentration of 40μg/ml to allow growth. This means that we would need to do a form of replica plating and compare which colonies cannot grow without this supplement ||Sequence|
|GltA - glutamate synthase||Growth Media needs to be supplemented with glutamate - Glutamate needs to be added to a final concentration of 500μg/ml to allow growth. This means that we would need to do a form of replica plating and compare which colonies cannot grow without this supplement ||Sequence|
|SacA - sucrase-6-phosphate hydrolase||Growth Media needs to contain glucose as a carbon source - Transformants can only grow on minimal media containing glucose as a carbon source not sucrose. This means we would need to be careful what type of media we culture our cells in ||Sequence|
|AmyE - alpha-amylase||Disruption of α-amylase - Transformants have to be grown on plates overnight and then stained with Grams iodine solution and incubated for a few minutes. Clear rings will form around colonies containing the amylase gene.Colonies without significant zones of clearing are amylase negative, and therefore presumed to be amy knockouts ||Sequence|
|ThrC - threonine synthase||Threonine auxotropy Threonine needs to be added to a final concentration of 50μg/ml to allow growth. This means that we would need to do a form of replica plating and compare which colonies cannot grow without this supplement  (N.B.Find More e.g. Protocols)||Sequence|
How long does the homologous sequence need to be for integration?
The integration vectors that are currently available contain two regions of homology to allow for double crossover events. These regions are typically made from the whole gene sequence, i.e. the 5' homology sequence corresponds the 5' half of the gene and the 3' homology sequence corresponds to the 3' half of the gene. This means that typically the regions of homology for a vector are between 500bp to 2000bp. Because we will be using DNA synthesis to construct our integration parts we are limited to size and use of integration parts greater than a few hundred bp is unfeasible.
An early paper claimed that the minimal length of homology required was only 70bp . However, a more recent paper has shown that 150bp was the minimal length of homology required for integration 
How do we maintain the correct orientation?
For the integrated devices to be expressed we need to make sure that they are integrated correctly in the 5' to 3' direction. Because we are using the double crossover integration orientation is easy to maintain because the two flanking integration sequences ensure correct orientation of integration.
What is the effect of integration site on gene expression?
A study using the PyrD, GltA and ThrC integration sites has shown that there is no significant difference in the levels of gene expression between these different sites of integration .
Maximum length of DNA that we can integrate?
There have been various estimations over the maximum length of DNA that can be stability incorporated into the B.subtilis chromosome. There estimations vary from 10kb to 16kb for double homologous cross over events . For us it is unlikely that our design will contain more than a few genes in tandem and so we should not exceed 2-3kb for each construct.
Transcriptional Read through
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