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BI226 Study Guide - Vicia Faba, Photographic Emulsion, Semiconservative Replication

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salmondeer718
11 Apr 2013
School
WLU
Department
Biology
Course
BI226
Professor
Robert Boulianne

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Document Summary

Grew bacterial cells (e. coli) for many generations in medium supplemented with. 15nh4cl as the only nitrogen source: remember the nitrogenous bases have nitrogen atoms, so the dna will now be labelled with this heavy version of nitrogen. 15n contains one more neutron than naturally occurring 14n. Molecules containing 15n are more dense than molecules containing 14n. Importantly, dna containing 15n can be distinguished from dna containing 14n. Samples forced by centrifugation into a density gradient of a heavy metal salt solution. Usually cscl (cesium chloride: molecules of dna will migrate into the density gradient. The dna will stop moving when it reaches a region of cscl that matches its own density. 15n-dna will move further down the tube of cscl compared to 14n-dna. In 1957 taylor, woods and hughes provided evidence of semiconservative replication in broad bean plants (vicia faba). Can locate location of radioactive isotope in a cell.

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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.

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.

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.

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.

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.

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3’ carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP

It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide

It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP

It cannot be incorporated into a DNA strand.

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

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.

She starts by attempting to add the centromere DNA into a plasmid containing the origin of replication. Unfortunately, when adding the centromere sequence, the origin of replication is removed, thus leaving a plasmid with only a centromere and selection marker. Following plasmid transformation, what growth result will she see on her plates?

No colonies

Little colonies

Big colonies

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.

She fixes the mistakes from the previous experiment and now has a complete plasmid (selection marker, origin of replication, centromere). She then inserts telomere sequences into the plasmid. How will this impact her transformation?

It will not impact her transformation

Her transformation will no longer work because plasmids don’t require telomeres

She will now see much larger colonies

She will now see fewer but larger colonies

She will now see smaller colonies

In the Meselson/Stahl experiment, E. coli were first grown in media containing heavy nitrogen, 15N, and then transferred to light nitrogen, 14N, at the beginning of the experiment.

Imagine that their data showed that replication occurs in a conservative manner instead of semi-conservative. What fraction of the DNA helices will consist of mixed DNA after 4 rounds of replication in this case?

None

More than 75%

25-50%

51-75%

Less than 25%

Question 4

Using the terms provided, complete the statements below. Some terms may apply more than once, while others may not apply at all. (15 points)

Guanine

histone

Double helix

Adenine

conservative

cytosine

Okazaki fragments

Thymine

introns

Leading strand

RNA polymerase

5-prime to 3-prime

Uracil

chromatin

DNA ligase

Semi-conservative

promoter

Lagging strand

DNA polymerase

nucleosomes

Electron transport chain

enhancer

Watson and Crick determined that DNA exists in the form of a _______________, where two antiparallel chains of nucleotides wind around each other. The nitrogenous bases project to the interior where they hydrogen bond in specific pairs, ___________ with __________ and _____________ with ____________.Meselson and Stahl demonstrated that DNA is replicated by the _______________ model in which the parent molecule unwinds and each strand serves as a template for synthesis of a new strand. Synthesis of DNA is carried out by _________________, which builds the new strands in the ___________ direction. While the ______________ grows continuously; the _______________ is built in short sections called _________________, which will eventually be joined together by __________________. Eukaryotic __________________ is composed of DNA and _____________ proteins that bind together forming ________________________, the basic units of DNA packaging.

Question 5

In the table below, predict (yes or no) whether or not the E. coli lac operon will be transcriptionally active in the presence or absence of glucose or lactose as indicated and respond to questions "a" and "b."

Lactose

Glucose

Lac expression?

No

Yes

Yes

Yes

Yes

No

Explain each of your answers in terms of the molecular mechanisms that are known to underlie the regulation of the lac operon. Which mechanism is considered to be negative control and which is considered to be positive control? Explain.

Question 6

Use the genetic code table in your textbook to aid you in answering the questions below. (15 points)

a.) Use the genetic code table to deduce the amino acid sequence of a protein encoded by the mRNA shown here:

5’AUGAUUGGAGGUUUGAUCGGGCAAUAGGGGUUUCAGUAAAUG3’

b.) Explain how the above sequence in "a" illustrates that redundancy of the genetic code.

c.) Explain what would happen if a mutational event caused the underlined "G" to be changed to a "U"? What is the name for this kind of mutation?