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MCB 2210 Lecture Notes - Lecture 3: Deconvolution, Confocal Microscopy, Small Interfering Rna

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sapphiredog102
5 Sep 2016
School
UConn
Department
Molecular and Cell Biology
Course
MCB 2210
Professor
Daggett

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Document Summary

Transfection: to infect a cell with foreign dna/rna; To cause a foreign protein to be expressed in a cell. In addition to gfp-tagged versions of molecules present in a cell, you can express (either tagged or not): Molecules that aren"t normally present in a cell. Mutant molecules that are constitutively active- i. e. active all the time. Mutant molecules that are dominant negative i. e. don"t function right and block the function of the cell"s own version of the molecule. Transfection can be transient (expression from plasmid) or stable (dna integrates into genome; heritable). Transgenic lines of animals can be generated by stable transfection of germ cells. Protein expression can be general, in all tissues, or under the control of a tissue-specific promoter (brain, heart, etc. Small interfering rna (sirna); new technologies exploit rnai mechanisms to allow us to knock out specific proteins of interest in cells by targeting mrnas.

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Related Questions

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.

Question 1

1 pts

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:

DNA

mRNA

protein

tRNA

Question 2

1 pts

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

Using an RNA probe complementary to the region not removed by the truncation.

Using an RNA probe complementary to the region removed by the truncation.

Using an DNA probe complementary to the region not removed by the truncation.

Using an DNA probe complementary to the region removed by the truncation.

Question 3

1 pts

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?

The genomic DNA of a healthy individual who does not have cystic fibrosis.

The genome of a CFTR patient known to have the specific truncation you are trying to identify.

The genome of a CFTR patient with a missense mutation but full length gene.

The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify.

The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene.

Question 4

1 pts

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

As both are composed of nucleic acids, using either would result in identical results.

An RNA probe because RNA has uracil bases.

An RNA probe because it could also be used in a translation experiment.

A DNA probe because it is more stable than RNA.

A DNA probe because RNA cannot bind to DNA.

Question 5

1 pts

Which of the following will lower the Tm of a given DNA strand?

Increasing the percentage of GC base pairs.

Raising the pH of the solution from neutral to basic.

Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.

None of the above.

Question 6

1 pts

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

To collect the bacteria at the bottom of the tube.

To break open the bacteria to release the genome.

To separate the bacteria from the bacteriophages.

To be able to detect the radioactivity.

Question 7

1 pts

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

No, because only DNA can be labeled with radioactivity.

No, because the RNA genome would not enter the bacteria upon infection.

No, because while DNA and RNA nucleotides are similar, they are not identical.

Yes, because DNA and RNA nucleotides are similar.

Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

Question 8

1 pts

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

Heat killed non-virulent bacteria is added to a live virulent bacteria strain.

Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.

A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain.

Heat killed mouse cells are added to a non-virulent bacteria.

Question 9

1 pts

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

Random DNA mutations generally won't affect RNA and protein function.

It is faster to duplicate the genome when these are present.

The existence of introns can lead to multiple variations of proteins encoded by a single gene.

It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Question 10

1 pts

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (∆ade) and histidine (∆his). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

Media containing histidine

Media containing adenine

Media lacking adenine

Media lacking histidine


Class 6: Viruses (and Cells, continued)


Making connections: viruses and microbes Match the cell biology concept (on page 1) with an important application of that concept (on page 2). Then answer the follow-up questions at the bottom of page 2.


Cell Biology concepts
1. Bacteria that cannot construct new cell wall material become fragile and break down.
2. During influenza infection, the viral enzyme neuraminidase is required to release new viruses from the host cell in which they were made. Once released, the new viruses can infect many additional cells.
3. Thermus aquaticus is a bacterium found in hot springs and can live at temperatures up to ~80°C (175°F). It has adapted so that its enzymes are most functional at the elevated temperatures, compared with many organisms’ enzymes that function best ~37°C (98°F).
4. Until the isolation and investigation of retroviruses, scientists thought that RNA could be copied from DNA, but DNA could not be copied from RNA. Research on this class of viruses led to an update to this “Central Dogma” of biology.
5. Some bacteria have adapted, so that the enzymes involved in making cell wall materials have a slightly different shape and can no longer be bound by certain drugs.
6. Rhizobia are bacteria that colonize the roots of certain plants (legumes, including beans and peas). The host plant benefits from the prokaryotes’ ability to “fix” nitrogen, converting N2 from the atmosphere into a form that is usable by the plant.
7. Many species of bacteria contain the enzyme cellulase. More than 10 billion such bacteria and other symbiotic microbes may be found in a single milliliter of material from inside the digestive system of ruminant mammals.
8. Fungi, including those that cause athlete’s foot, use ergosterol to maintain healthy membranes. Animal cells do not make ergosterol, they use a similar lipid – cholesterol.
9. During retrovirus infection, a DNA copy is made of the virus’s RNA genome. A cut is made in the host cell’s DNA and the DNA copy of the viral genome is pasted into the host chromosome by a viral enzyme called integrase.
10. Prokaryotic cells contain smaller ribosomes (“70S” ribosomes) than eukaryotic cells (“80S” ribosomes). Both types of ribosomes synthesize proteins, but the macromolecules they use to do it are somewhat different.



2
Applications
_______ Oseltamivir, also called Tamiflu, binds and inhibits the viral protein neuraminidase.
_______ Clotrimazole is a drug that is used to treat certain infections. It interferes with synthesis of ergosterol, a lipid involved in phospholipid membranes.
_______ Penicillin is a drug that interferes with synthesis of peptidoglycans (these are biological molecules that are part protein and part carbohydrate) that are present in certain cell walls.
_______ HIV is often treated with several drugs that inhibit different parts of the viral replicative cycle. Raltegravir is an example of a drug that interferes with the integration of viral genes into the host cell’s chromosome.
_______ Cows, like other vertebrates, lack the ability to breakdown cellulose. Yet they are able to live on a diet of grass and other vegetation high in cellulose.
_______ Enzymes isolated from extremophiles have been very useful in laboratory research. For example, Taq polymerase facilitates PCR, a revolutionary technique for production of DNAs of particular sequences, which requires cycling reactions through high temperatures.
_______ Plants and animals require nitrogen to make biological macromolecules, however they’re unable to convert atmospheric nitrogen (N2) into a biologically useful form.
_______ Azithromycin is a drug that binds to 70S ribosomes and blocks synthesis of any new proteins.
_______ Methicillin-Resistant Staphylococcus aureus (MRSA) can cause life-threatening infections of the skin and bloodstream. One of the reasons MRSA infections are so serious is that they are resistant to treatment by many antibiotics, including methicillin and penicillin.
_______ The enzyme reverse transcriptase, isolated from retroviruses, is an important tool in molecular biology research. Scientists interested in learning what genes are turned on in particular cells use it to make DNA copies of RNAs they are studying (The DNA copies are useful because they are more stable).


When you have completed the matching activity, answer these follow-up questions:
• What types of infections can be treated with the following drugs?
o Penicillin
o Clotrimazole
o Azithromycin
o Tamiflu
o Raltegravir
• Why are these drugs (above) safe for you to take? That is, why don’t they kill your cells?
• Cancer cells are your own human cells that have become deregulated so that they divide and pile up in an uncontrolled way. Many chemotherapy drugs treat cancer by interfering with cell division and other cell processes.
o Why are there so many more side effects for individuals undergoing chemotherapy than for individuals taking antibiotics?