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BIOL 221 Chapter Notes - Chapter 5: Contact Geometry, Chromosome, Bacteriophage

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UPenn
Department
Biology
Course
BIOL 221
Professor
Bonini

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5.3 Bacterial Transformation
-transformation =take up DNA fragments from external medium
-source of DNA can be other cells of same species/ cells of other species
-if DNA is of different genotype from that of recipient, genotype of recipient can be permanently
changed
-in conjugation, DNA is transferred from one living cell to another through close contact
-transformation = isolated pieces of external DNA are taken up by a cell through the cell wall
and plasma membrane
-transformation measures how closely 2 genes are linked on bacterial chromosome
-when DNA (bacterial chromosome) extracted for transformation experiments, breakage into
smaller pieces is inevitable
-if 2 donor genes close together on chromosome -> carried on same piece of transforming DNA
(double transformation)
-genes widely separated on chromosome -> carried on separate transforming segments
-if widely separated genes, frequency of double transformants = product of single transformant
frequencies
-if genes are linked, proportion of double transformations > product of single transformant
frequencies
-not all cells in population can be transformed
-bacteria can take up DNA fragments from surrounding medium
-inside cell, DNA fragments can integrate into chromosome
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Document Summary

Transformation =take up dna fragments from external medium. Source of dna can be other cells of same species/ cells of other species. If dna is of different genotype from that of recipient, genotype of recipient can be permanently changed. In conjugation, dna is transferred from one living cell to another through close contact. Transformation = isolated pieces of external dna are taken up by a cell through the cell wall and plasma membrane. Transformation measures how closely 2 genes are linked on bacterial chromosome. When dna (bacterial chromosome) extracted for transformation experiments, breakage into smaller pieces is inevitable. If 2 donor genes close together on chromosome -> carried on same piece of transforming dna (double transformation) Genes widely separated on chromosome -> carried on separate transforming segments. If widely separated genes, frequency of double transformants = product of single transformant frequencies. If genes are linked, proportion of double transformations > product of single transformant frequencies.

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Related Questions

2.Conclusion:

What have you learned from your study?

Interpret your data/ results:

Do your results support of disprove your hypothesis?

Compare and contrast your result with those of previous research.

Significance of the results: “Big Picture”

What is the global impact of your results?

What questions remain unanswered?

What kinds of experiments can come out of your study?

i have lab for biology and i need help with the Lab Report

the lab name is Bacterial Transformation Lab

1. Introduction:

Introduce the background and the nature of the problem your experiment is investigating.

Summarize the current knowledge of your subject.

Use examples of other research that have examined the same or a similar topic as your experiments.

Ending your introduction:

State your hypothesis

State your predictions

Give a brief statement about how you tested your hypothesis.

please help me with it is important for my lab class tomorrow

that is the lab i put the info please help me for the intro and for number 2 please be clear please

i would be happy if you help me it is due tommorrow i realy need your help on the intro and number 2 please be clear and show clear work hope you do it because it is due tomorrow

Bacterial Transformation Lab

Genetic engineering is one of the newest applications of basic knowledge in biology. It involves the isolation of specific genes from one organism and the insertion of those genes into a second organism of the same, or even different, species. Remember that a gene is a piece of DNA which codes for a product, usually a protein, which gives an organism a particular trait. The development of these techniques has generated considerable excitement and answered many questions, but it has also raised questions. Many scientists and others see genetic engineering as offering opportunities to make life better, for example, by exploring how genes are regulated, by curing diseases of genetic origin, by adding desirable traits to agricultural crops such as the ability to fix nitrogen in corn and pest resistance in cotton, and by genetically altering bacteria with genes enabling them to digest oil spills.

For almost 100 years, it has been possible to isolate DNA from cells, but only in the last 40 years or so has its function been understood. Your textbook discusses the work of such investigators as Fred Griffith and Oswald Avery, who demonstrated that if DNA was isolated from a pneumonia-causing strain of bacteria and added to cultures of nonvirulent bacteria, the recipient cells were transformed into virulent types that caused pneumonia. These transformation experiments were the first to show that genes could be artificially transferred in the laboratory. Since the time of those experiments (Avery's work was done in the 1940s), our knowledge of gene structure, function and control has increased astronomically. The application of this knowledge has created the field of genetic engineering.

Modern genetic engineering techniques depend on three critical factors:

A suitable host organism in which to insert the foreign gene;

A vector to carry foreign genes into the host; and

A means of isolating the host cells that have taken up the foreign gene.

The host organism

The most commonly used host organism is the gut bacterium Escherichia coli (E. coli). Its chromosome is a single large circular DNA molecule, called the genomic DNA, containing about 4 x 106 base pairs. Not every cell will take up foreign DNA when exposed to it, but E. coli grows rapidly, dividing in less than 20 minutes, so that a single cell can give rise to a billion genetically identical descendants in a matter of 10 hours. Thus, a few transformed cells can provide a large number of offspring within a day. The strain we are using is K-12.

The vector

A vector is a DNA molecule used as a vehicle to carry genes from one organism to another. This DNA molecule can be modified by gene-splicing techniques using restriction enzymes and ligases so that it carries the genes of interest (essentially a cut-and-paste operation). Sometimes viruses are used as vectors, but for E. coli, plasmids are often used. A plasmid is a small circular DNA molecule containing 1000 to 200,000 base pairs. Plasmids exist naturally in many strains of E. coli and other bacteria. Plasmids replicate as the cell grows, independent of replication of the genomic DNA. These plasmids are transmitted to each of the product cells during cell division. Plasmids carrying genetic information that is beneficial to the host cell are maintained in a given population by natural selection. Plasmids frequently carry genes that confer antibiotic resistance, an obvious benefit.

When E. coli cells are exposed to plasmids or any other foreign DNA, some of the cells, called competent cells, will take up the DNA. We can induce cells to become competent by treating them with divalent cations, such as Ca++, followed by a brief exposure to high temperature, called a heat shock treatment. Exactly how this treatment causes cells to become competent is not understood, but it is described as making the membranes "leaky" so that large DNA molecules (like plasmids), which would not normally pass through the cell membrane, can enter the cell. This treatment does not make all cells competent, but it greatly increases the number of competent cells. Once in a cell, the plasmids replicate and are passed to the product cells during cell division.

The selective medium

The last requirement for genetic engineering is a means of isolating host cells that have taken up the gene of interest. The techniques used depend on the growth requirements of the host organism. For example, each species of bacteria has particular nutrient requirements. The composition of the growth medium (agar or broth) can be modified to allow growth of only certain bacteria -- for example, by adding or removing nutrients or other chemicals from the medium mixture. This modified medium is called a selective medium, because it selects for the growth of only some species. By introducing an antibiotic such as ampicillin into the medium in which the bacteria are grown, you will select for growth of only those bacteria carrying the antibiotic-resistance gene on newly acquired plasmids. In a matter of days, billions of antibiotic-resistant cells, all with the plasmid, can be grown.

Genetic engineers use a modification of this technique to insert a foreign gene into E. coli. In simplified terms, the plasmid (already carrying an ampicillin resistance gene) is modified by inserting the gene of interest. The gene of interest has been isolated from another source; for example, the gene for human insulin has been isolated from the human genome. The modified plasmid, carrying the inserted gene, is then mixed with competent E. coli and the bacteria are placed in a medium containing ampicillin. Only those cells that have the plasmid will grow in the medium, and they are also the cells carrying the insulin gene (in their plasmids). These selected cells can now be used to manufacture insulin, which can be used for insulin therapy by diabetics.

The genes

In the following experiment, you will insert a plasmid into E. coli. The plasmid you will use carries the gene bla, which codes for a protein (beta-lactamase) that is secreted by the bacteria, inactivating the ampicillin in the agar and allowing for bacterial growth. The plasmid also carries an indicator gene. The indicator gene we are using was isolated from a bioluminescent jellyfish, Aequorea victoria. It codes for Green Fluorescent Protein (GFP).

The GFP gene has been modified to include a regulator system that can be used to control the expression of the gene. Gene expression in all organisms is carefully regulated to allow for adaptation to differing conditions and to prevent wasteful overproduction of unneeded proteins. The genes involved in the breakdown of different food sources are good examples of highly regulated genes. For example, the sugar arabinose is both an energy source and a carbon source for bacteria. The bacterial genes that make digestive enzymes to break down arabinose for food are not expressed when arabinose is not in the environment. But, when arabinose is present, these genes are turned on. When the arabinose runs out, the genes are turned off.

Arabinose initiates transcription of these genes by promoting the binding of RNA polymerase. In the genetically engineered pGLO DNA, some of the genes involved in the breakdown of arabinose have been replaced by the jellyfish gene GFP. When bacteria with this plasmid are grown in the presence of arabinose, the GFP gene is turned on and the bacteria glow green in UV light. When arabinose is absent form the growth medium, the GFP gene remains turned off and the colonies appear white. This is an example of the central molecular framework of biology in action: DNA ? RNA ? protein ? trait.

In summary, you will render E. coli cells competent to take up plasmid DNA. The plasmid carries two identifiable phenotypic markers:

The beta-lactamase or bla gene codes for resistance to the antibiotic ampicillin. We will use this to select for bacteria that have taken up the plasmid.

The pGLO gene codes for Green Fluorescent Protein (GFP), which is switched “ON” if arabinose is present.

Safety Procedure

For your own safety and the safety of others, it is essential that you follow all instructions about handling bacteria and disposal of materials. When you finish work for the day, swab your table with disinfectant. WASH YOUR HANDS thoroughly with soap before leaving the lab for any reason, and at the end of the day after you have cleaned your table. Report any spills to your instructor.

List of Materials

Prepared agar plates (2 LB, 2 LB/amp, 1 LB/amp/ara)

Transformation solution (50 mM CaCl2, pH 7.4)

LB broth

1 pack of inoculation loops

Sterile transfer pipets

1 foam microtube holder w/ tubes

E. coli starter plate

1 red biohazard bag for loop disposal (common use – Loops/pipets only - NO PAPER!!!)

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

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.