Given what we've discussed in class, what will be most likely outcome if you conjugate an streptomycin resistant ampicillin sensitive methionine auxotroph E. coli strain (engineered to be pir+) that is F- with a streptomycin sensitive non-HFR methionine prototroph strain that is F- and RP4+ but contains pUC18?
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Bacterial Genomics
The study of the morphological, physiological, and evolutionary aspects of the bacterial genome is referred to as bacterial genomics. This subdisciplinary field aids in understanding how genes are assembled into genomes. Further, bacterial or microbial genomics has helped researchers in understanding the pathogenicity of bacteria and other microbes.
Transformation Experiment in Bacteria
In the discovery of genetic material, the experiment conducted by Frederick Griffith on Streptococcus pneumonia proved to be a stepping stone.
Plasmids and Vectors
The DNA molecule that exists in a circular shape and is smaller in size which is capable of its replication is called Plasmids. In other words, it is called extra-chromosomal plasmid DNA. Vectors are the molecule which is capable of carrying genetic material which can be transferred into another cell and further carry out replication and expression. Plasmids can act as vectors.
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- Conjugation depends on direct cell-to-cell contactbetween a donor F+ carrying either a conjugative________ (the F plasmid) or an integrated conjugativeelement (as in Hfr strains), and a recipient lacking suchan element (F−).A pure culture of an unknown bacterium was streaked onto plates of a variety of media. You notice that the colony morphologyis strikingly different on plates of minimal media with glucose compared to that seen on trypticase soy agar plates. How can you explain these differences in colony morphology? Also, describe what happens when a nonsense mutation is introduced into the gene encoding transposase within a transposon and why is it more likely that insertions or deletions will be more detrimental to a cell than point mutations?Two auxotrophic triple Escherichia coli strains (A: met- phe- ade- val+ bio+ thr+ and B: met+ phe+ ade+ val- bio- thr-) are mixed in LB liquid medium, diluted and then spread on LB solid rich medium. Six colonies are observed: Then, replicates are performed on 6 different media (minimum medium + glucose + indicated substances). The results are shown below. Determine the genotype of the 6 colonies observed. Which ones are from strain A? From strain B? Which hypotheses can explain these results and which one do you prefer? met phe val bio Abbreviations: met ade val thr phe ade bio thr Met: methionine; Phe: phenylalanine; Ade: adenine; Val: valine; Bio: biotin; Thr: threonine.
- Eight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by wild-type bacteriophages. The mutant strains were allowed to infect the bacteria in a complementation test. A "+" indicates that lysis occurred with coinfection. A "-" indicates that lysis did not occur. GKWTMAQC G- ++++- K W T M A Q C - + +++ - ++++ - + + - + - + - - + + + A cistron is defined by no complementation in the How many genes are controlling lysis in this bacteriophage? (Use a number not a word in the space) configuration.The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was 10mm or greater, the result is listed as ++. In the lab, the MGC instructors plated all 6 of the ESKAPE pathogen safe relatives on LB agar plates. Then we patched Unknown Bacteria 5 from a Master plate onto the safe relative. The results are shown here: METRIC METRIC METRIC 1 B. subtilis S. epidermidis E. coli Complete the final column (Unknown Bacteria 5) of the table by selecting -, +, or ++ using the criteria in the…The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was greater than 10mm, the result is listed as ++. Page 50 in your research guide states: "Some antibiotics are broad spectrum, meaning that they affect a wide range of bacteria. Other antibiotics have a narrow spectrum of activity. One anatomical feature that plays a significant role in the susceptibility of a microbe to a particular antibiotic is its cell wall composition (discussed in Section 8)". Research the cell wall…
- Fifteen bacterial colonies growing on a complete medium (that means that they have all of nutrients that they need supplied in the dish, and they don't actually need to synthesize these compounds to survive) are transferred to minimal medium. Twelve of the colonies grow on minimal medium. Three colonies do not grow on minimal medium. But, if these three colonies are put on a plate that has minimal medium supplemented with the amino acid serine (min + Ser), they all What does this suggest about the three bacterial colonies (pick all that apply)? grow. They lack the ability to synthesize their own serine. O They are probably wild-type. They probably have a mutation that causes them to lack a certain protein. They probably have a mutation that causes them to be unable to perform translation. O They probably have a mutation that causes them to be unable to perform transcription. O They are able to synthesize everything that they need to grow except for serine.The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was greater than 10mm, the result is listed as ++, Page 50 in your research guide states: "Some antibiotics are broad spectrum, meaning that they affect a wide range of bacteria. Other antibiotics have a narrow spectrum of activity. One anatomical feature that plays a significant role in the susceptibility of a microbe to a particular antibiotic is its cell wall composition (discussed in Section 8)". Research the cell wall…Three pairs of bacterial cells with the given genotypes undergo conjugation. Place match the genotype of each cell after conjugation to its initial genotype. F+ x F Hfr F- F' F- Answer Bank F F F+ Hfr What is the role of the F-factor in conjugation? It contains genes necessary for replication of the donor's F plasmid. can occur. It allows auxotrophic bacterial cells to survive on minimal medium so that conjugation It contains genes that force recombination between the donor and recipient chromosomes. It contains genes necessary for the formation of the pilus. O It degrades the chromosome of the recipient cell after conjugation.
- A high cell density culture of recombinant E. coli was carried out according to the following strategy:-Step 1: single batch with exponential growth until 98% conversion of the substrate, starting from V0= 4.0 L, S0=50 g/L/ X0= 1.0 g/LStep 2: batch fed with exponential flow (SF-800 g/L, μ= 0.1 h-1) until reaching X= 50.9 g/L;Step 3: batch fed with constant flow (F= 0.1 L/h) for 4 hours (induction phase with IPTG)Note: consider that the quasi-steady state is reached in both fed-batch stages.Extra data: YX/S = 0.4 gx/gs; μmax= 0.25 h-1; Ks== 1.0 g/L a) What was the cell concentration reached at the end of step 1?b) For step 3, considering that the substrate concentration in the feed was 1/4 of that used in step 2, what was the concentration of cells reached at the end of step 3?C) In terms of cell productivity, which of the three phases of cultivation was the most productive?In Escherichia coli, four different Hfr strains, derived from the same F* strains, were mated with F strains auxotrophic for a number of nutritional requirements (Arg Bio Cys Trp Gal His Lac Mal Xyl Leu Met). Matings were interrupted at various intervals and cells were plated on minimal medium supplemented with particular nutrients to test for gene transfer. The following results show the time of entry for each of the genes in four different Hfr strains. Table 1: Time-of-Entry Mapping Data* Hfr strains Genes lac" his" arg' bio 9. Cys gal" trp' 5 11.5 2.5 mal" xyl" leu met Hfr 1 Hfr 2 Hfr 3 Hfr 4 6.5 3.5 11 15 15 4 6. 17.5 5 14.5 3 20 * The numbers denote the number of minutes elapsed before a gene enters the F cells. Draw a circular map of E. coli chromosome.Linkage maps in an Hfr bacterial strain are calculated in units of minutes (the number of minutes between genes indicates the length of time that it takes for the second gene to follow the first in conjugation). In making such maps, microbial geneticists assume that the bacterial chromosome is transferred from Hfr to F − at a constant rate. Thus, two genes separated by 10 minutes near the origin end are assumed to be the same physical distance apart as two genes separated by 10 minutes near the F −attachment end. Suggest a critical experiment to test the validity of this assumption.