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I have some question about Colony PCR:

1. pGLO causes the transformed colonies to glow. Why do we use ampicillin selection plates when we can identify transformed bacteria optically.

A. ampicillin increases GFP expression

B. ampicillin reduces competition from untransformed bacteria

C. ampicillin is not necessary

D. ampicillin is necessary because we are not using aseptic/sterile technique

2. When picking colonies for 'Colony PCR',

A. take a large streak from the densest part of the bacterial lawn

B. choose the center of a small colony

C. dig the agarose from underneath a small colony

D. combine 2-3 small colonies

3. In colony PCR a good positive control would be ....

A. the plasmid from the transformation

B. a section of the bacterial lawn

C. a liquid culture of the same colony

D. a master mix with no PCR primer

4. If colony PCR shows a band of the correct size, when is sequencing required?

A. to make sure the correct gene is being expressed

B. to determine orientation of the gene

C. to confirm point mutations

D. both B and C

5. How do you grow the bacteria from the PCR positive colony after colony PCR?

A. You can't, PCR kills the bacteria. Just pick another colony

B. Streak the PCR products on an AMP agarose plate, the bacteria will grow just fine.

C. Don't add the entire colony to the PCR master mix

D. Re-transform bacteria assuming the same transformation efficiency

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Beverley Smith
Beverley SmithLv2
29 Sep 2019

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Related questions

HELP ASAP;

Please help me understand my lab results, so be detailed to clearly understand . INclude answers to these questions what role does ampicilin (amp) and arabinose (ara) play in the experiment. 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 (Fig. 5, not a naturally occurring 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.

for the first week section of lab

To each 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 this transformation mix to the appropriate plates.

Results:

+DNA: LB/amp--> 1 colony, Glow: uncertain

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

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

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

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.

Which of the following did Erwin Chargaff observe?

A)

Base composition changes as the organism ages.

B)

Base composition of DNA does not vary from one species toanother.

C)

In all species the number of adenines equals the number ofguanines.

D)

In all species A + T = G + C.

E)

DNAs isolated from different tissues of the same species havethe same base composition.

The melting temperature (Tm) of a DNA duplex is:

A)

the temperature at which double-stranded nucleic acids degradeinto free nucleotides.

B)

the temperature at which double-stranded DNA is toosingle-stranded to ever renature back to double strands.

C)

the temperature at which double-stranded DNA is completelymelted.

D)

the temperature at which double-stranded DNA has become 50%single-stranded.

E)

the temperature at which double-stranded DNA begins to melt.

Given your knowledge and the descriptions of each of thesemethods, which of the following does NOT rely on the ability ofnucleic acids to hybridize with each other?

A)

Southern blotting for the detection of specific DNA fragmentsthat have been separated by gel electrophoresis.

B)

Colony blot hybridization to identify a specific sequence from acollection of sequences that have been inserted into bacteria.

C)

Northern blotting for the detection of specific RNA fragmentsthat have been separated by gel electrophoresis.

D)

A DNA oligonucleotide microarray incubated with probes made fromcDNA.

E)

Western blotting for the detection of specific proteins thathave been separated by gel electrophoresis.

Which of the following is a possible reason why DNA uses thymineinstead of uracil?

A)

Cytidine is deaminated to uracil so it would be difficult forany repair mechanism to differentiate between uracil that belongsin the sequence and uracil that resulted from deamination.

B)

Adenine is deaminated to uracil so it would be difficult for anyrepair mechanism to differentiate between uracil that belongs inthe sequence and uracil that resulted from deamination.

C)

When uracil is bound to a deoxyribose it will become morereactive, forming covalent bonds with other bases.

D)

None of the above.

Which of the following is an enzyme used in cloning to breakcovalent bonds?

A)

DNA ligase

B)

Restriction endonuclease

C)

Reverse transcriptase

D)

DNA polymerase I

E)

Alkaline phosphatase

Which of the following could you use to remove the 5' phosphatesafter cleavage of a plasmid with a single type of restrictionenzyme to ensure that the plasmid would not simply fuse backtogether upon addition of ligase?

A)

Reverse transcriptase

B)

Alkaline phosphatase

C)

DNA ligase

D)

DNA polymerase I

E)

Restriction endonuclease

Which of the following would you use to create a cDNA librarythat you would not use for a genomic library?

A)

Alkaline phosphatase

B)

Reverse transcriptase

C)

Restriction endonuclease

D)

DNA ligase

E)

DNA polymerase I

A plasmid vector typically has which of the followingfeatures?

A)

An origin of replication

B)

A selectable marker to ensure that the only bacteria growingcontain the intact or recombinant vector.

C)

Unique restriction sites for insertion of DNA.

D)

Small size to facilitate entry into the cell and biochemicalmanipulation of the DNA.

E)

All of the above are features of plasmid vectors.

Which of the following is FALSE of genomic or cDNAlibraries?

A)

Genes represented in cDNA libraries include control sequencesthat direct transcription.

B)

cDNA libraries contain cDNAs made from mRNA.

C)

Genomic libraries contain fragments of genomic DNA that aregenerated by partial digests with restriction enzymes.

D)

Genomic libraries usually contain large fragments and are likelyto be constructed in YAC or BAC vectors.

E)

cDNA libraries contain sequences from expressed genes only.

Which of the following is NOT a typical component of apolymerase chain reaction (PCR)?

A)

dNTPs

B)

DNA containing the sequences to be amplified

C)

DNA ligase to connect the fragments together

D)

Primers complementary to each end of the sequence to beamplified

E)

Heat stable Taq polymerase

How can a fusion protein be formed?

A)

Site-directed mutagenesis where a fragment is replaced with asynthetic sequence containing the altered DNA sequence.

B)

Two proteins can recombine in the bacterial cell to form a newhybrid protein.

C)

Portions of two different genes are ligated together to create anew hybrid gene. When expressed the gene product is a fusion of thetwo gene products.

D)

Using oligonucleotide-directed mutagenesis to introduce specificmutations that cause the expressed protein to fuse to otherproteins.

E)

A portion of the gene can be removed to form a shorter gene andtherefore a shorter protein.

Why is green fluorescent protein (GFP) so useful in visualizingfusion proteins in eukaryotes?

A)

The GFP protein recognizes and binds to fusion proteins allowingthe location of the fusion protein to be determined.

B)

The reaction is only dependent on the presence of GFP and oneother cofactor.

C)

It is easily cloned and expressed in bacterial cells.

D)

The presence of just a few molecules of GFP fusion proteins canbe sufficient for observing the proteins microscopically allowingthe study of its location and movements in the cell.

E)

It is derived from fire flies, which are easy to cultivate inthe lab.

You wish to determine the cellular location of a protein butunfortunately it is not particularly antigenic (can t develop astrong antibody response). How could you solve this problem?

A)

Create a fusion protein with an epitope tag that can bevisualized by incubation with fluorescently tagged antibody to theepitope.

B)

Create a fusion with green fluorescent protein (GFP).

C)

Express the protein as a fusion with biotin. The addition of GFPwill cause fluorescence.

D)

A and B might work.

E)

A, B, and C might work.

F)

None of the above will work.

A positive result for the yeast two-hybrid analysis means thefollowing:

A)

The proteins that you are studying can be made into fusionproteins, but do NOT interact with each other within the nucleus ofyeast.

B)

A fusion protein containing the Gal4p activation domain, hasbound to RNA polymerase resulting increase transcription ofnumerous genes, including the reporter gene, and formation ofcolonies.

C)

Two fusion proteins, one containing the Gal4p binding sitedomain and one containing the Gal4p activation domain, haveinteracted through their Gal4p domains resulting in transcriptionof the reporter gene.

D)

Two fusion proteins interact with each other and because of theGal4p binding site domain and the Gal4p activation domain containedin each fusion protein, the reporter genes are activated and thecells die resulting in no colonies.

E)

The colonies growing are expressing the two fusion proteins, butthe two proteins have not interacted with each other.

Which of the following is the approximate size of the humangenome?

A)

3 Gm

B)

300 Kb

C)

3 Mb

D)

3 Gb

E)

3 Mm

Which of the following is correct about the structure orlocation of genes?

Question 22 options:

A)

Prokaryotic genes typically possess their own regulatorysequences that only control expression of a single gene.

B)

Eukaryotic genes often encode introns that are spliced outbefore translation.

C)

Eukaryotic genes are mostly located at the telomeres and thecentromeres of chromosomes.

D)

Eukaryotic genes are typically arranged in operons.

E)

Prokaryotic genes often encode exons that are spliced out beforetranslation

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