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MCB 2610 Midterm: MCB 2610 Reading Notes_ Exam 2

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carminepug60
25 May 2018
School
UConn
Department
Molecular and Cell Biology
Course
MCB 2610
Professor
Daniel Gage

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Chapter 7 on DNA, RNA and protein--skip parts that cover eukaryotes ; Genes and
Genomes
Bacterium Deinococcus radiodurans
●Can withstand exposure to UV light, hydrogen peroxide and desiccation and can survive
huge amounts of radiation.
●Its gram positive but contains an outer membrane that is associated with gram negative
bacteria.
Radiation kills organisms by causing extensive damage to DNA and so the cell cannot repair the
damage.
7.1 The Role of DNA
How do we know DNA is the hereditary molecule?
●DNA is a polymer of nucleotides, each containing 4 nitrogenous bases.
●DNA is complementary, antiparallel, the base pairs of one strand interact with the base
pairs of the other and they have base-stacking interactions.
Griffith Experiment
Started with the bacterium Streptococcus pneumoniae, either S or R were injected into mice. If
injected with Smooth they died. But if S was heat killed and then injected into mice they didn't
die indicating that it was pathogenic or virulent. And the R strain was non pathogenic or virulent.
But if both heat killed S bacteria and R bacteria were injected the mouse still died.
A factor from the smooth cells transformed the phenotype of rough cells, producing some
smooth cells!
Defined DNA as genetic material. It causes pneumonia but also sinus and ear infections, and
meningitis.
The Avery, MacLeod, and McCarty experiment
What was this transforming agent?
They purified the transforming factor from S cells and determined that it was mostly DNA but
they thought possibly it could have contain some protein or RNA.
S. pneumoniae contains a gene that results in the production of a polysaccharide capsule which
makes the bacterial colonies appear smooth and helped the bacteria evade the host immune
response and making the strain pathogenic. So the R strain doesn’t have that gene.
The Hershey-Chase Experiment
Bacteriophage T2 binds to e.coli and injects its genetic material and then directing the E.coli to
produce new viral particles. It proved that the radioactive phosphorus entered the cells, while the
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radioactive sulfur did not. DNA and not protein entered the host bacterium and directed the
synthesis of new phage particles.
Structure of DNA
Watson, Crick and Franklin figured out the structure of DNA.
DNA consists of 2 chains and each chain is a polymer of nucleotides and each nucleotide
consists of a
●5-carbon sugar, 2-deoxyribose
●A phosphate group that links the 3’ carbon to the 1’ carbon of sugar
●One of four nitrogen containing, or nitrogenous bases attached to the 1’ carbon of the
sugar.
Nitrogenous bases are classified as purines, which are adenine, guanine double ring structure or
pyrimidines which are cytosine and thymine containing a single ring structure.
Each linear DNA has a 5’ end and 3’ end. And two chains when aligned are in antiparallel and
complementary to one another. Hydrogen bonds hold the 2 chains together with two H bonds
linking A to T and three H bonds for C to G.
In bacteria and archaea, they usually have a single copy of their chromosome and of all their
genes, they contain plasmids and small circular extrachromosomal DNA molecules.
7.2 DNA Replication
How does DNA replicate?
Semiconservative replication
Origins of replication
Replicon model. 2 main components: replicator and an initiator. The replicator would be the site
on the DNA molecule where the replication started. The initiator was a protein that could interact
with the replicator and begin the process.
In bacteria, DNA replication starts at a specific site on the chromosome, the origin of replication
or oriC.
When oriC is added to an artificially constructed closed circular DNA molecule, then that
molecule can be replicated within an E.coli cell.
DnaA - DNA binding protein produced by the dnaA gene.
DnaA protein binds to the dnaA boxes(9bp repeats) and interact with the 13 bp repeat regions
resulting in strand separation and formation of a single stranded region.
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●DnaA is an essential protein, without it DNA replication does not begin.
● IT IS THE INITIATOR
→ It begins separating or melting the DNA
→ Then DnaB (helicase) and DnaC (helicase loader) are recruited to the site to help with
unwinding
→ DnaG (primase) synthesizes short segments of RNA needed to prime DNA replication.
→ Single stranded DNA binding protein (SSB) attach to the newly formed single stranded DNA
to keep the strands from reannealing and a single stranded replication bubble forms.
→ Then the DNA polymerases then begin replicating the DNA at both ends of this bubble, the
replication forks.
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Document Summary

Chapter 7 on dna, rna and protein--skip parts that cover eukaryotes ; genes and. Can withstand exposure to uv light, hydrogen peroxide and desiccation and can survive huge amounts of radiation. Its gram positive but contains an outer membrane that is associated with gram negative bacteria. Radiation kills organisms by causing extensive damage to dna and so the cell cannot repair the damage. Dna is a polymer of nucleotides, each containing 4 nitrogenous bases. Dna is complementary, antiparallel, the base pairs of one strand interact with the base pairs of the other and they have base-stacking interactions. Started with the bacterium streptococcus pneumoniae, either s or r were injected into mice. But if s was heat killed and then injected into mice they didn"t die indicating that it was pathogenic or virulent. And the r strain was non pathogenic or virulent. But if both heat killed s bacteria and r bacteria were injected the mouse still died.

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Related Questions

magine that the year is 1928 and the scientific world is in an uproar trying to determine what component of the cell isresponsible for genetic inheritance. Is it DNA, protein, perhaps even RNA? You are busy in your lab, trying to help solve this crucial mystery that still plagues (then) modern day biology. Your work focuses on a microorganism, a bacterium named Streptococcus pneumoniae, and its effects on mice. There are two strains of S. pneumoniae (which is responsible for pneumonia), the “Rough” strain (R) and “Smooth” strain (S). The strains are very similar except the S strain has a smooth polysaccharide coat that enables the microorganism to avoid the organism’s immune system and cause illness (and death in mice). The R strain has a rough coat which is detected and dealt with by the immune system (and as a result causes no death in mice). You decide to perform a series of experiments using the R and S strain of S. pneumoniae involving injections of these strains in mice. Your results are documented in the figure below: living S strain of S. pneumoniae mouse dies of infection living R strain of S. pneumoniae mouse lives mouse lives S strain heat-killed living R strain mouse dies of infection living, pathogenic S strain recovered S strain heat-killed From these results, you figure out that something from the heat killed S strain is transforming the R strain into the virulent strain. What is this “Transforming principle”? Is it related to the cellular component that was passed from one generation to the next? As you know, this was the initial experiment performed by Dr. Frederick Griffith that eventually led to the discovery of the true cellular component that is the genetic inheritance molecule as well as the “transforming principle” (Hint: It’s DNA!).

1. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells?

a. Griffith b. Watson and Crick c. Avery, MacLeod, and McCarty d. Hershey and Chase

2. From your results, which one of the cellular components can be determined NOT to be the component responsible for genetic inheritance? Why?

a. Proteins because these macromolecules become denatured with heat. b. DNA because these macromolecules form a single helix in solution. c. RNA because it's double stranded 100% of the time d. Sugars because glycolysis happens.

3. Which one of the following experimental trials below are considered to be controls for this experiment? a. S Strain grown in 10% ethanol medium. b. Living S strain; Living R strain; Heat Killed S Strain c. Living R Strain with heat killed S strain; Living S Strain d. Heat Killed S strain; Living R strain; R strain grown in 10% ethanol medium

4. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule? a. Griffith b. Watson c. Crick d. Hershey and Chase

5. Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information. What radioactive material was used to track DNA during this experiment?

a. 3H b. 14C c. 35S d. 32P

6. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.

a. True b. False

7. The polarity of a DNA strand results from the polarity of the nucleotide subunits.

a. True b. False

8. There are five different nucleotides that become incorporates into a DNA strand.

a. True b. False

9. Hydrogen bonds between each nucleotide hold individual DNA strands together.

a. True b. False

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