1A. The phage λ derivative used to deliver TnlacZ/in to cells is defective for replication (rep-) and for lysogeny (int-). Imagine that an unwanted mutation making the phage rep+ accumulated in the phage stock (The int- allele is unaffected by this mutation.) How would your results change if you attempted to produce transposon insertion mutants with these phage?

int- λ phages have non-functional integrase, and will not integrate with the host chromosome.

rep+ λ phages have functional lytic genes, and may take advantage of the host to become lytic.

Phages with the phenotype (int- rep+) will not integrate into the host chromosome, and will behave like plasmids. This phage, with its rep+ gene, will use the host’s resources to replicate itself, and eventually lyses the host cell.

Since (int- rep+) λ is co-grown with the E. coli lawn, it will lyse most of the cells before they can form colonies, which would make the X-Gal aspect of the transposition assay rather difficult to spot. How does this work? Observe this model:

Remember that the intent of this experiment is to assay for transposon insertion through antibiotic resistance! Having rep+ makes this experiment difficult, as the same λ phage vectors responsible for carrying the Tn5 transposon, will lyse most of the cells.

1B. Imagine that an unwanted mutation making the phage int+ accumulated in the phage stock. (The rep- allele remains unaffected.) How would your results change if you attempted to produce transposon insertion mutants with these phage?

int+ λ phages have non-functional integrase, and will integrate with the host chromosome.

rep- λ phages have non-functional lysogeny genes, and will not take advantage of the host to become lytic.

Phages with the phenotype (int+ rep-) will integrate into the host, carrying the full transposon gene into the host chromosome and conferring both chloramphenicol and kanamycin resistance.

Observe the transposon model:

The chloramphenicol resistance (cat) assay is asking whether the insertion of either transposons were successful.

The kanamycin resistance (neo) assay is asking whether the full Tn5 gene, or the ISlacZ/in region, was successfully transposed.

Since each of the lysogenic cells have received the full Tn5 gene piggybacking the phages, they all have chloramphenicol and kanamycin resistance, so the kanamycin assay in this case tells us very little about the specific regions that were transposed.

In the possible scenario that the ISlacZ/in transposon undergoes intramolecular transposition and successfully lands downstream of another promoter region (and the host survives), the host chromosome will still possess both the cat and neo gene, which still confers resistance of both antibiotics, and rendering the kanamycin assay irrelevant.

2. As mentioned, the Tn5 transposase gene (tnp) is regulated such that only about one in 10,000 copies of TnlacZ/in will actually transpose into a cell’s chromosome. Theoretically, one could engineer the tnp gene to be over expressed, such that the transposition frequency could be increased from 0.0001 to greater than 0.5, thus generating thousands more mutant cells. What would the experimental disadvantage to having transposition occur this frequently?

The multiplicity of infection (MOI) is 10 in our experiment. There are roughly 10 λ phages for every E. coli cell. The probability of a transposition occurring can be expressed by the following function:

The normal probability of infection per cell can be calculated as seen below:

According to this calculation, a transposition of Tn5 occurs once in every 1,000 cells. We can extrapolate that a two transpositions occurs once in every million cells, and three transpositions every billion cells.

Given the numbers of cells we are working with, the probability that a cell has beyond two transposition (which is already rare) becomes highly unlikely. The above probability of infection allows us to distinguish between the ISlacZ/in and TnlacZ/in insertions through the kanamycin assay.

Suppose the transposition frequency is now raised to 0.5:

According to this calculation, there is likely 5 transpositions occurring per cell. With multiple insertions of the transposons, there is a high probability that BOTH ISlacZ/in and TnlacZ/in are inserted into the host, conferring both chloramphenicol and kanamycin resistance to virtually all cells and invalidating our assay.

Note that with 0.5 transposition frequency, each of the transposon has a 50% chance of jumping after every chromosomal replication, which makes the results like β- galactosidase production highly unstable and difficult to track.

3. What feature of the colonies resulting from TnlacZ/in mutagenesis indicates that the transposon(s) have inserted at different sites in the E. coli chromosome?

The truncated ‘LacZ gene, which lacks a promoter and ribosome binding site (RBS), will start transcribing β- galactosidase after the transposon has inserted correctly after said sites.

Not all promoters express their genetic codes with the same frequency, some will rarely undergo transcription, while others are constitutive. Observe two possible cases of transposon insertions:

The promoter activates the LacZ gene that then transcribes β- galactosidase, which is then is responsible for cleaving X-Gal, which turns blue upon being cleaved. With this process, the blueness of the colonies assay for the strength of the promoter activity.

In the experimental results, the colonies have differing hues of blue (or sometimes just white), which indicates varying frequencies in promoter activity, which may be attributed to insertion into different sites in the E. coli chromosome.

4A. 1/6 of all tnlacZ/in or ISlacZ/in insertions within a gene are expected to result in gene fusions encoding hybrid proteins. Explain why this particular fraction is expected.

Suppose that the orientation of the Tn5 insertion is random, there is a 1/2 chance that Tn5 is inserted with ‘LacZ in the correct orientation with respect to the promoter.

Suppose that the Tn5 insertion into the reading frame is random, there is a 1/3 chance that Tn5 is inserted between the correct reading frame.

Of the 1/2 chance of getting the insertion orientation correct, there is only a 1/3 chance of getting the reading frame insertion correct as well. In order to have a functional fusion protein, both the orientation and reading frame needs to be correct. 1/2 * 1/3 = 1/6.

A few possible scenarios of Tn5 insertions can be seen below:

Note that in all cases, the antibiotic resistance genes are still functional because they are expressed from their own promoters, regardless of reading frames or orientation.

4B. Imagine that you actually observe a much smaller fraction than one-sixth blue colonies on X-Gal when you carry out the mutagenesis. What is the most likely explanation for this result?

One-sixth only accounts for the successful insertions with respect to the promoter and RBS sites. Assuming random insertion chances anywhere on the genome, there will be insertions that will just miss the promoter target altogether and land elsewhere on the chromosome.

Below are just a few of many possible insertion scenarios that can cause non-functional protein transcriptions:

Assuming that Tn5 insertions have many chances of missing the promoter sites, we should expect much less than 1/6 of the colonies turning blue.

5. If you continue to incubate the chloramphenicol-resistant colonies from the original plating, some of them that were originally white will eventually exhibit dark blue sectors. (Such colonies will usually be kanamycin resistant as well as chloramphenicol resistant.) How might you explain the occurrence of these blue sectors?

The white colonies which have dark sectors that turns blue over time, probably has cells that carry the full Tn5 transposon in their chromosomes. Once the transposon jumps to the correct promoter target, it can produce β- galactosidase, which gives that cell the ability to cleave X-Gal, utilizing it as a carbon source and giving it a growth advantage and turning it blue. Below is one of the many possible scenario of a white colony cell turning blue as a result of intramolecular transposition of the Tn5 gene:

Notice that in the case of intramolecular transposition, the tnp gene must be present for it to occur.

If the tnp gene is present, it is logical that both the chloramphenicol (cat) and kanamycin (neo) resistance gene are present in the host as well, which would confer resistance to both antibiotics.

Note that this dual antibiotic resistance is conserved regardless of whether ISlacZ/in or TnlacZ/in intra-transposes, since both the (cat) and (neo) genes are on the host chromosome before and after the intramolecular transposition.