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BIOE 185 Lecture Notes - Lecture 10: Tobacco Mosaic Virus, Chloroplast Dna, Nucleosome

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peachraccoon740
18 Nov 2017
School
Montana Tech
Course
BIOE 185
Professor
scott

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Document Summary

The eukaryotic chromosome: structure, dna classes, replication, and telomeres. In this lecture, we will compare/contrast viral and bacterial genomes to those of eukaryotes. As previously described, the viral genome might consist of dna or rna. Further, nucleic acids might be found as single strands or as double strands. For example, plant viruses (like the tobacco mosaic virus) tend to have single-stranded rna genomes, while bacteriophages tend to have double-stranded dna genomes. The shape of viral genomes can be circular or linear. In contrast to viruses, all cells use double stranded dna as their genetic material. Instead their chromosome is located in an area of the cell called a nucleoid. A typical bacterial chromosome is circular and might have a cut length of 1. 1 mm, such as in: coli. In addition to their main chromosome, e. coli and many other bacteria also have plasmids. Plasmids are smaller circular dna molecules that are replicated separately from the bacterial chromosome.

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Related Questions

QUESTION 1: The rabies virus primarily affects the nervous system. The specificity that the rabies virus has for neuronal host cells is primarily dictated by __________________________.

A. the helical shape of its viral capsid
B. the type of nucleic acid used for its viral genome (single-stranded, antisense RNA)
C. the spikes that protrude from its viral envelope
D. the segmented nature of its viral genome

QUESTION 2: Which of the following genome types has been observed in viruses? To be marked correct, you'll need to select all true statements, as there may be more than one correct answer.

A. Single-stranded RNA
B. Single-stranded DNA
C. Double-stranded RNA
D. Protein-based

QUESTION 3: Choose the correct statement about viral evolution.

A. RNA and DNA viral genomes evolve at equal rates.
B. Cellular genomes mutate at a faster rate than viral genomes due to their large size and increased chance of replicative mistakes.
C. DNA viruses mutate faster than RNA viruses because thymine is more susceptible to mutation than uracil.

D. RNA viruses mutate faster than DNA viruses due to a lack of proofreading replicative enzymes.

QUESTION 4: Your elderly patient is affected by shingles. After careful observation, you note that the virus responsible for the infection has an icosahedral capsid, is enveloped, and has double-stranded linear DNA as its genetic material. Based on this information, in which of the following viral families would you group this viral pathogen?

A. Hepadnaviridae
B. Parvoviridae
C. Coronaviridae
D. Herpesviridae

QUESTION 5: Which of the following is a key difference between lytic and lysogenic bacteriophage replication cycles?

A. Lysogenic bacteriophages release via host cell lysis, whereas lytic bacteriophages bud off from the host cell membrane.
B. The lytic replication cycle ends with host cell lysis and release of newly formed bacteriophage particles, whereas lysogenic replication leads to prophage formation.
C. Lytic replication requires an uncoating step, whereas in lysogenic replication, the bacteriophage's nucleic acid is directly injected into the host cell.

D. Whereas lytic replication enables the infection of several hosts, lysogenic replication only allows for infection of a single host cell.

QUESTION 6: Which of the following is an unsuitable culture method for an animal virus?

A. Direct inoculation of virus onto blood agar
B. Live animal inoculation
C. Tissue culture method
D. Injection of virus into embryonated eggs

QUESTION 1. Compared to eukaryotic genomes, only prokaryotic genomes:

A. contain DNA organized by histones.
B. contain usually one, circular chromosome.
C. contain plasmid DNA.
D. have chromosomes contained in a nucleus.

QUESTION 2. Azathioprine is a drug that may be prescribed to transplant recipients to reduce the chance of tissue rejection reactions. In its active state, it acts as a purine analogue, inhibiting DNA biosynthesis. This means that 6-mercaptopurine will be inserted in place of a purine containing nucleotides during DNA replication. Given this drug's mode of action, you would expect it to be incorporated into DNA in place of:

A. nucleotides containing guanine or cytosine.
B. nucleotides containing adenine or thymine.
C nucleotides containing adenine or guanine.
D. nucleotides containing guanine nucleotide or uracil.

QUESTION 3. RNA is distinguished from DNA because only RNA:

A. typically exists as a single-stranded molecule.
B. has a 3' to 5' directionality.
C. is composed of nucleotides that contain the sugar deoxyribose.
D. is composed of nucleotides that contain the nitrogen bases T, C, G, and A.

QUESTION 4. How are RNA nucleotides different from DNA nucleotides? To be marked correct, you'll need to select all true statements, as there may be more than one correct answer.

A. RNA nucleotides do not contain thymine.
B. RNA nucleotides use the sugar ribose, while DNA nucleotides do not.
C. RNA nucleotides lack phosphate.
D. RNA nucleotides incorporate pyrimidine nitrogen bases, while DNA nucleotides only incorporate purine nitrogen bases.

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.