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A conjugation experiment is carried out bewteen two bacteria.The original recipient is antibiotic resistant, but can not grow inthe absence of Thr, Phe and Leu. After 5 minutes of conjugation,the cells can grow without Phe or Leu. After 10 minutes ofconjugation, the cells can grow on plates without Thr, Leu or Phe.What is the correct order of these genes?

A. Thr, Phe, Leu

B. Phe, Leu, Thr

C. Leu, Phe, Thr

D. can not tell from this info

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Sixta Kovacek
Sixta KovacekLv2
28 Sep 2019

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Related questions

1. Which of the following statements about antibiotic resistancein bacteria is NOT true?

Select one:

a. Antibiotic resistance often originates from the microbes thatproduce antibiotics for their own survival.

b. The antibiotic resistance gene can be transmitted to bacteriavia transformation or transduction.

c. Antibiotic resistance cannot be conferred by conjugation asconjugation only affects the fertility of bacteria.

d. Environments where antibiotics are frequently used such ashospitals are under the higher risk of developing antibioticresistance.

e. The plasmid containing the antibiotic resistance gene canpass the genes to genetically unrelated bacteria.

2. You perform interrupted-mating experiments on three Hfrstrains (A, B, and C). Genes are transferred (from last to first)in the following order from each strain: strain A,thi-his-gal-lac-pro; strain B, azi-leu-thr-thi-his; strain C,lac-gal-his-thi-thr. How are the F factors in these strainsoriented?

Select one:

a. A and B are oriented in the same direction.

b. B and C are oriented in the same direction.

c. All of them are oriented in the same direction.

d. A and C are oriented in the same direction.

e. It cannot be determined from the information given.

3.Which of the following statements about retroviruses isfalse?

Select one:

a. All retroviruses contain gag genes whose product forms theviral protein coat.

b. Not all RNA viruses are retroviruses.

c. All retroviruses contain oncogenes, which can induce theformation of tumors.

d. All retroviruses require pol genes, which are critical forretrotranscription.

e. All retroviral genomes have gag, pol, and env genes.

4. Which of the following facts would NOT be considered as anadvantage for using bacteria and viruses for genetic studies?

Select one:

a. Low cost to maintain and little storage space required

b. Rapid reproduction and high progeny number

c. Genomes being small and readily subjected to geneticmanipulation

d. Complete absence of recombination, which maintains theintegrity of genome

e. Haploid genome for expressing mutations

Hfr Mapping Problem

An Hfr E. coli contains the following genetic traits: kans, lac+ , his+ , trp+ , ampr . It is mated with a recipient E. coli that contains the opposite genetic traits for those traits listed abov: kanr. lac- , his- , trp- , amps . Assume all other catabolic and biosynthetic genes are present in both strains. Note that kanS means the absence of the kanamycin resistance gene in the Hfr strain, which will allow growth only of the recipient cells in the presence of kanamycin. This is necessary to select against the Hfr donor strain as you are monitoring which genes transferred to the recipient.

At periodic intervals, conjugation is disrupted and cells are plated onto several types of defined media. Nothing grows at the beginning of the experiment on media containing lactose, kanamycin, and ampicillin as this is the most selective medium, but all cells grow on complex media containing no ampicillin (the least selective medium). Mating is interrupted at the times listed below and the following are some of your observations of growth characteristics on a few of the selective media plates containing the indicated nutrients/antibiotics:

15 min.

Growth on defined media containing: kanamycin, glucose, histidine (does not contain tryptophan).

No growth on defined media containing: kanamycin, lactose (sole carbon source), tryptophan, histidine.

30 min.

Growth on defined media containing: kanamycin, lactose (sole carbon source), histidine (does not contain tryptophan).

No growth on defined media containing: kanamycin, glucose, tryptophan (does not contain histidine).

No growth on defined media containing: kanamycin, ampicillin, lactose (sole carbon source), histidine, tryptophan.

45 min.

Growth on defined media containing: kanamycin, lactose (sole carbon source) (does not contain histidine or tryptophan).

No growth on defined media containing: kanamycin, ampicillin, lactose (sole carbon source), histidine (does not contain tryptophan).

60 min.

Growth on defined media containing: kanamycin, lactose (sole carbon source) (does not contain histidine or tryptophan).

No growth on defined media containing : kanamycin, ampicillin, lactose (sole carbon source) (does not contain histidine or tryptophan).

HELP ASAP; Please help me understand my lab results, so be detailed to clearly understand

Then , Please include answers to the question below to include in your response

1. what role does ampicilin (amp) and arabinose (ara) play in the experiment.

2) Because GFP was isolated from jelly fisht, what modification to the GFP DNA had to be made to result in expression in E. coli?

BACKGROUND INFORMATION

Transformation of GFP and the pGLO Plasmid GFP is a commonly used reporter protein in research labs, as the fluorescence creates a marker protein that can be used in many types of cell biology and biochemical studies. In basic research, GFP is often fused to a specific target protein of interest, creating a chimeric reporter protein. GFP has been used as a reporter protein to study blood vessel and tumor progression in mice, brain activity in mice, and malaria eradication in mosquitoes for instance (see website mentioned in notes). In the first week you will use a procedure to genetically transform bacteria with the gene that codes for Green Fluorescent Protein. Following the transformation procedure, the bacteria express their newly acquired jellyfish gene and produce the fluorescent protein, which causes them to glow a brilliant green color under ultraviolet light. During the second week you will purify the protein away from all of the other proteins produced by the bacteria using chromatography, and during the third week you will evaluate the success of the purification procedure and estimate the size (molecular weight) of GFP by using sodium dodecylsulfate polyacrylamide gel electrophoresis (SDS-PAGE). Fundamental to the process of genetic transformation of bacteria are plasmids. In addition to one large chromosome, bacteria naturally contain one or more small circular pieces of DNA called plasmids. Plasmid DNA usually contains genes for one or more traits that may be beneficial to bacterial survival. In nature, bacteria can transfer plasmids back and forth allowing them to share these beneficial genes. This natural mechanism allows bacteria to adapt to new environments. The occurrence of bacterial resistance to antibiotics is due in large part to the transmission of plasmids which contain antibiotic resistance genes. The pGLO plasmid contains the gene for GFP, and also contains the gene for the enzyme β-lactamase (bla), which provides resistance to the antibiotic ampicillin. The β-lactamase protein is produced and secreted by bacteria that contain the plasmid. β-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express β-lactamase can survive on plates that contain ampicillin. Only a very small percentage of the cells take up the plasmid DNA and are transformed. Untransformed cells cannot grow on the ampicillin selection plates. pGLO also incorporates a special gene regulation system, which can be used to control expression of the fluorescent protein in transformed cells. The gene for GFP can be switched on in transformed cells by adding the sugar arabinose to the cells’ nutrient medium. Selection for cells that have been transformed with pGLO DNA is accomplished by growth on antibiotic plates. Transformed cells will appear white (wild-type phenotype) on plates not containing arabinose, and fluorescent green when arabinose is included in the nutrient agar medium.

The transformation procedure involves three main steps. These steps are intended to introduce the plasmid DNA into the E. coli cells and provide an environment for the cells to express their newly acquired genes. 1) Cells are first treated with a solution of CaCl2, which is thought to neutralize the repulsive negative charges of the phosphate backbone of DNA and the phospholipids of the cell membrane, allowing the DNA to adhere to the cells. This is referred to making the cells ‘competent’ (i.e. the cells are capable, or competent, to take-in exogenous DNA). 2) Cells are then subjected to a heat shock, which is thought to increase the permeability of the cell membrane, allowing the cells to take-in the plasmid DNA. 3) Cells are allowed to grow in a ‘recovery’ phase in the absence of ampicillin. During this time the cells begin to produce the β-lactamase protein, which will allow them to survive when they are subsequently placed on agar plates containing ampicillin.

For the first week section of lab

To eac

h of two microfuge tubes, add 250ul of transformation buffer. Keep these tubes on ice.

Using a sterile loop, pick one colony of bacteria from the LB plate. Transfer this colony to one of the microfuge tubes.

Gently vortex the tube until the colony is dispersed. Repeat for the second tube. Add 0.08ug of DNA to one of these tubes (+DNA tube).

Incubate both tubes on ice for 10 minutes. Incubate both tubes at 42°C for 50 seconds.

Incubate both tubes on ice for 2 minutes. Add 250ul of LB broth to each tube and incubate at room temperature for 10 minutes. Flick each tube gently. Add 100ul of t

his transformation mix to the appropriate plates.

Results For plates:

+DNA: Has the E. Coli pGlo plasmid and - DNA plates do no have EColi pGLo plasmid .

Amp means ampiclin and Ara means arabinose on agar plates

+ DNA Plate LB/amp/ara--> white 9 colonies, Glow: yes

+DNA LB/amp--> 7 colonies, GLow: no

--DNA LB/AMP- 1 colony so no colony growth and no glovw

-DNA: LB--> many colonies, Glow: no

Please explain significance of result or thand the reason for the results in scientific terms based on the background information provided and your own knowledge to be detailed. Would these results be expected as wll.

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