packback discussions Flashcards

The His- E. Coli strain cannot produce histidine, making it an auxotroph for Histidine. The Lac- strand cannot consume lactose, but can still grow on other sugars and carbon sources. The strain that is auxotrophic will be the His- strain. This is because it cannot synthesize histidine. The Lac- strain will not be auxotrophic, because it's only deficiency is lactose.

Selectable mutations are mutations that are easy to "select", they typically provide some sort of advantageous mutation that gives it an edge over other bacteria, for example antibiotic resistance would be an example of a selectable mutation. Nonselectable mutations on the other hand are mutations that arent as easy to select for. For example a mutation that turns a bacteria into an auxtroph would be nonselectable, as the only way to be able to differenciate it, is by putting the cells on a replica plate without the auxtroph nutrient and a control, and seeing which cells die.

I would expect rifampin-resistant cells to be isolated at a higher frequency.

Rifampin resistance occurs through multiple possible point mutations in the rpoB gene. This alters RNA Polymerase so rifampin can no longer bind while the enzyme still functions. Because there are so many different nucleotide changes that can lead to this outcome, the chance of isolating resistant mutants is pretty high.

On the other hand, reversion of an arginine auxotroph usually requires a specific back-mutation to restore the original reading frame or codon, which is much rarer. So refampin resistance is more commonly observed due to the greater number of mutational “targets” that can generate the phenotype.

The SOS regulatory system is a response in prokaryotes when there is a mistake or damage in DNA. This system is highly regulated, a transcriptional, and global response.

This system is activated when damage or an error is detected, typically through the accumulation of single-stranded DNA. During this time, the cell cycle will arrest as DNA mechanisms join in to repair any errors. The system will then deactivate once the accumulation of single-stranded DNA lowers, as it indicates repair is successful.

Ultimately, the SOS regulatory system has a role in maintaining DNA integrity and stability in prokaryotes.

or

The SOS regulator system is a response in bacteria to DNA damage. This pathway is triggered when damaged DNA is detected and a set of genes is turned-on to repair the DNA. The system is initiated when there is an accumulation of single-stranded DNA containing lesions. Proteins, LexA and RecA, play a crucial role in regulating the SOS pathway. LexA repressor binds to the promoter region of SOS and prevents expression. The RecA assists in the removal of LexA and promotion of the SOS gene. After the damage is repaired a signal tells LexA to begin accumulating again leading to the repression of the SOS gene again.

a. If mutants are resistant to the antibiotic coumermycin, we could use selective plates. For this, we could use a plate medium that contains this specific antibiotic. In this case, bacterial cells that lack the resistance genes would be killed and not grow into colonies. This then would result in grown colonies having the resistance factor being selected for.

b. For the selection of revertants of a trp mutation that makes tryptophan, we would need to deplete the medium of tryptophan. This would allow for the revertants that can produce tryptophan to be able to form colonies; however, the bacterial cells that cannot produce tryptophan would not be able to divide into colonies.

c. For mutants with a suppressor in the araA gene that relieves the sensitivity to arabinose due to a mutation in araD, we could use selective plates. In these plates, we could use medium with arabinose. In this case, the mutants would not show sensitivity to arabinose and thus would grow on this selective medium.

Introducing Genetic Variation

The product of homologous recombination without crossing over is the repair of DNA without the exchange of genetic material. The products of homologous recombination using crossover is the result of homologous chromosomes that exchange genetic material between the two. This results in new combinations of alleles in chromosomes. However, non-crossover is more likely to result in mutations due to the mission of mismatch repair and gene conversion wich can introduce more mutations.

and

Since we know that recombination is the physical exchange of DNA between genetic elements, adding homologous to this term means that the genetic exchange between the homologous DNA sequences comes from two different sources. Furthermore, the RecA protein (which functions in genetic recombination) is vital for this type of recombination. The two homologous DNA sequences are pretty much identical to each other in regard to sequence, allowing the bases to pair and facilitate genetic exchange. Furthermore, the crossover products are when there is genetic information an exchange between the chromatids in meiosis, making new combinations of alleles and genetic variation. Non-crossover products of homologous recombination do not utilize the physical exchange of genetic information between chromosomes, which is less likely to result in mutations because there is no actual process of breaking and rejoining DNA as seen in crossover products. Non-crossover is more prevalent when there is a repair of a double-strand break in DNA.

To score recombinant colonies from the cross between Hfr (thr⁺ leu⁺ gal⁺ strˢ) and F⁻ (thr⁻ leu⁻ gal⁻ strʳ) strains of E. coli, two main methods can be used. The first method is direct selection on defined media with donor counterselection. After mating the strains for about 30–60 minutes, the mixture is disrupted to stop further conjugation and plated on minimal medium containing streptomycin, which kills the strˢ Hfr donors. The medium composition determines which recombinants are selected: for example, minimal medium lacking threonine and leucine selects for thr⁺ leu⁺ recombinants, while minimal medium with galactose as the sole carbon source (plus threonine and leucine) selects for gal⁺ recombinants. Only F⁻ cells that acquired the corresponding Hfr genes will grow. The second method is replica plating after nonselective growth. Here, the mated culture is plated first on rich medium with streptomycin so that only F⁻ derivatives survive, and then individual colonies are transferred by replica plating onto a series of selective plates (each lacking a specific nutrient or containing galactose as the sole carbon source). Growth on a given plate identifies which gene(s) the F⁻ strain acquired (e.g., thr⁺, leu⁺, or gal⁺). Comparing these patterns allows identification and scoring of different recombinant classes while ensuring donors are excluded by streptomycin.

The recipient cells won't become F+ conjugation with Hfr cells because of how the F factor is arranged.

In Hfr cells, the F factor is not a separate plasmid, rather it is integrated directly into the bacterial chromosome. During conjugation, DNA transfer begins from the integrated F factor and then continues into nearby chromosomal genes. However, because the bacterial chromosome is so large, the mating bridge usually breaks before the entire F factor sequence can be transferred. As a result of this, the recipient may gain some new genes, but rarely gets the full F factor. Because the recipient cell does not have the complete F factor, the recipient cell cannot make it's own pilus, so it stays F-.

To find the genes required for making the red pigment, you can use transposon Tn5 (kanamycin resistant) for random mutagenesis. First, introduce Tn5 into the bacterial cells using a non-replicating (suicide) plasmid or through conjugation with a donor strain carrying Tn5. After transfer, plate the cells on medium containing kanamycin to select only those that have the Tn5 insertion. Each colony will have Tn5 randomly inserted into a different gene. Next, screen the Kanᴿ colonies on pigment-producing plates and look for mutants that lose the red color—these are likely disrupted in pigment-related genes. Finally, identify which gene was hit by sequencing from the Tn5 insertion site or PCR using primers at the Tn5 ends. This will tell you which genes are required for red pigment production.

How can we find where to insert insulin gene?

To solve this, we could introduce the Tn5-insulin construct into E. coli.

First, we will have to insert the Tn5-insulin construct into the E.coli as a chromosome, and not a plasmid. This is because a plasmid can replicate within the E.coli host. Then we will allow Tn5 to insert the transposon randomly into the bacterial genome. Afterward, we can select for the kanamycin-resistant colonies by plating cells on medium that contains kanamycin. Lastly, the colonies will be screened for the insulin gene.

The forward primer is 5'-atggaccagt-3' and the reverse primer is 5'-ttacggcagt-3'.

Primers in the PCR amplification process are short pieces of DNA used by Taq DNA polymerase to synthesize and extend the DNA strand, therefore it must be designed to amplify specific genes, such as this FtsZ gene. The start codon for the FtsZ protein gene is ATG, and the end codon is TAA. Since the primers are 10 nucleotides each, the forward primer is 5'-atggaccagt-3'. The last 10 nucleotides, 5'-actgccgtaa-3', can be used to design the reverse primer. First, you design the complementary strand, which would be 3'-tgacggcatt-5', then set it in the 5'-3' direction, thus giving you the final reverse primer as 5'-ttacggcagt-3'.

Here’s a minimal 16-nt primer pair that adds EcoRI (5′) and HindIII (5′) sites to amplify E. coli ftsZ:

• ftsZ-EcoRI-F (16 nt): 5′-GAATTCATGGACCAGT-3′ (EcoRI site + first 10 nt of ftsZ; includes start codon ATG)
• ftsZ-HindIII-R (16 nt): 5′-AAGCTTTTACGGCAGT-3′ (HindIII site + reverse-complement of the last 10 nt of ftsZ)

PCR with these, then digest the product and pUC18 with EcoRI/HindIII and ligate.

host cell needed

The host cell used for white/blue screening is typically E. coli, however, other bacterias can work. The host cell being used needs to have a mutation in the LacZ gene, meaning it cannot make beta-galactosidase by itself. When the plasmid is added, it either takes up the foreign DNA to break the enzyme and turn the colonies white, or it does not take up the foreign DNA and the colonies turn blue. The host cell used just need to be any cell that can provide the correct LacZ mutant.

and

The host cell must align to the general conditions to be successful, so E. coli is great example.

Host cells for cloning should follow these factors: capable of rapid growth in an inexpensive medium, be non-pathogenic or the virulence genes deleted, able to incorporate DNA and produce competent cells, genetically stable in culture, and be equipped with appropriate enzymes to allow replication of the vector. A couple of hosts that meet these criteria and do bacteria include E. coli and Bacillus subtilis. For blue/white screening, the kind of host cells that can be used are specific strains of E. coli that have the mutated lacZ gene (lacZdeltaM15), which lacks the alpha fragment of B-galactosidase protein. Overall, the absence or deletion of the lacZ gene is critical because these cells will be transformed into the plasmid due to the disruption of the lacZ-alpha coding sequence.

blue/white screening

The plasmid contains the gene LacZ and has a gene for antibiotic resistance, as well as the foreign DNA you inseted. You give the bacteria the plasmid with the genes of interest then grow them on a plate containing antibiotics. If the bacteria grow in the presence of the antibiotic, then they took up the plasmid, but they didn't necessarily take up the foreign DNA. Once you have the colonies grown, you can look at the color of them to determine if you have recombinant DNA. Blue colonies mean the plasmid was taken up, however the insert DNA was not, so there is no recombinant DNA. White colonies mean the plasmid was taken up as well as the foreign DNA inserted. White colonies are what you are looking to get.

RT-PCR just adds the reverse transcription step to make RNA measurable and amplifiable.

Regular PCR is pretty straight forward. RT-PCR adds a twist because it is designed to work with RNA, not DNA. Since DNA polymerase can’t directly copy RNA, scientists first use an enzyme called reverse transcriptase to convert the RNA into complementary DNA. Once that is made, the PCR steps are basically the same as usual where the DNA is amplified to be analyzed.

RT-PCR is useful because it starts with RNA and can be used to study gene expression by looking at which mRNAs are in a cell. It can also be used to detect RNA viruses like SARS-CoV-2 during testing for Covid. Without the reverse transcription step, regular PCR wouldn’t work on RNA based targets.

and

RT-PCR is unique because of the amplification of RNA.

Both of these PCR methods have the similarity of being endpoint reactions to amplify DNA and RNA sequences, yet there are some distinctions between the two. While PCR starts with DNA, RT-PCR starts with RNA and uses reverse transcriptase to convert RNA into complementary DNA. This is then activated by a DNA polymerase to feed into the amplification process. Additionally, this process is commonly used as the initial step in qPCR, which is used to measure the amount of DNA and cDNA in a sample.

E.coli DH5a can technically receive chromosomal DNA from Hfr cells, but it isn’t ideal because of the genetic modifications it carries.

DH5a has mutations that improve the plasmid uptake and reduce recombination, which helps maintain cloned DNA without rearrangements. Since DH5a is still an E.col strain, it can act as the F-recipient in conjunction when paired with an Hfr donor, allowing some genes to be transferred. But, because DH5a has mutations in recombination pathways the incoming DNA usually doesn’t integrate correctly into its chromosome. The transferred fragments would most likely be degreased or remain unincorporated.

Yes, you can clone this gene and express it in E. Coli.

You can clone the E. coli gene and express the enzyme, but you have to determine the full sequence of the gene's cDNA. When the cDNA is fully sequenced, you design the forward primer based on nucleotides and a reverse primer. The reverse primer will amplify and clone the complete gene into an E. Coli expression vector. A potential issue is a codon bias. The mammalian gene sequence may contain codons that are rarely used by E. coli.

and

Yes, you can clone the gene, but first we need the full sequence

To clone the gene, we need to amplify the entire gene using PCR and also determine the ending sequence to create the reverse primer. First, you'll use the primer based on the known sequence from the 5'-3' to identify the unknown 3' end of the gene from a cDNA sequence. Next, you sequence the resulting PCR to reveal the gene's complete sequence from its start codon to its stop codon. Once we have the full sequence, we can design a new pair of primers to amplify the entire coding region for cloning into an E. coli expression.