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BIOL-UA 21 Lecture Notes - Lecture 9: Organism, Hepatocyte, Reference Genome

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purplelocust18
14 Oct 2016
School
New York University
Department
Biology
Course
BIOL-UA 21
Professor
Mark Siegal

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Document Summary

Lecture 9: recombinant dna iii: molecular analysis using cloned sequences: next-generation sequencing without gels: A library of dna fragments you want to sequence is obtained and ligate linkers are attached to each end of each fragment. In each ow cell, there are channels where liquid can be own through, and inside is a lawn of primers sticking up. Each primer will be complementary to one or the other side of the fragments, as they are complementary to one of the linkers. The trick is to not add too much dna, so that a single strand of dna bound to a primer is a sparse event. The strand can be acted on by dna polymerase and replicated. The produced strand is anchored to the bottom of the ow cell by a primer. If you denature and wash away the old strand, It will base pair with another primer with a free 3" end, so can get extended by dna polymerase.

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Related Questions

Please answer all

Why does a template-dependent DNA polymerase require a primer to initiate DNA synthesis?

DNA polymerase require 5' phosphate to add a new nucleotide

DNA polymerase require 3' hydroxyl group to add a new nucleotide

DNA polymerase require energy by hydrolizing the primers

Which of the following enzymes is template independent?

Reverse transcriptase

Klenow fragment

DNA polymerase

Terminal deoxynucleotidyl transferase

RNA polymerase

The natural function of a DNA ligase enzyme is:

Joining together DNA sequences with compatible cohesive ends from two different organisms to yield a recombinant DNA molecule

Sealing single stranded nicks in double stranded DNA molecules

Joining together two blunt ended fragments of DNA within a cell

Joining together two single stranded DNA molecules

The natural function of 3' -> 5' exonuclease activity of a DNA polymerase is to:

Remove the 5' end of the DNA strand that is being copied

Remove damaged nucleotide from the template strand

Remove nucleotides from the end of DNA to generate blunt ends

Remove incorrect nucleotides from the newly synthesized DNA

Which technique is used to resolve the different sizes of DNA fragments?

Gene cloning

PCR

DNA sequencing

Gel electrophoresis

Which of the following vectors does not contain bacterial origin of replication (oriC)?

plasmid

cosmid

bacterial artificial chromosome

lambda vector

Which process of bacteriophage is not used for gene cloning?

Concatemer formation

Lytic cycle

Lysogenic cycle

viral packaging

What vector would be best suited for creating a contig of bovine (cattle) chromosome 10?

lambda vector

Cosmid vector

P1 vector

Plasmid

E. coli takes up plasmid DNA by which of the following methods?

Transduction

Transformation

Conjugation

Transfection

The following times and temperatures are an example of the steps for PCR. You can use the Figure to help you answer the following questions.

Why is the first step is carried out at 94°C?

To degrade the template

To denature the template

To activate the polymerase

To facilitate the primer binding

What happens in the reaction when the temperature shifts to 55°C during cycling?

the DNA polymerase will carry out DNA synthesis by extending the annealed primers

the DNA polymerase will finish DNA synthesis

the primers anneal to the single-stranded regions of the DNA

the primers anneal to the double-stranded regions of the DNA

During cycling, what occurs when the temperature is at 72°C?

the primers anneal to the double-stranded regions of the DNA

the DNA polymerase will carry out DNA synthesis by extending the annealed primers

the primers anneal to the single-stranded regions of the DNA

the DNA polymerase will finish DNA synthesis

You designed a set of primers to amplify 1 kb DNA with polymerase chain reaction. By which cycle of polymerase chain reaction, we would expect to see the first double stranded DNA with expected size?

The end of cycle 1

The end of cycle 2

The end of cycle3

The end of cycle 4

Multiple answers to each question might be possible!

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.

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.

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.

Explain the 5’ to 3’ directionality of a DNA stand.

It is due to the fact that the free 5’ carbon is on one end and the free 3’ carbon is on the other

It is due to the fact that new nucleotide are added to the 5’ carbon of the previous nucleotide

It is due to the fact that there are 3 phosphate groups attached to the 5’ carbon

It is due to the fact the complementary strand is 3’ to 5’

More than one of the above explain the 5’ to 3’ directionality

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.

Why does DNA polymerase utilize an RNA primer?

DNA polymerase is unable to initiate strand synthesis but RNA polymerase can

DNA polymerase can only add a dNTP to an rNTP

DNA synthesis proceeds in the 3’ to 5’ when initiating strand synthesis

Chromosomal DNA contains interspersed RNA fragments

The RNA primer increases stability of the newly synthesized strand

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.