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BIOL 150 Lecture Notes - Lecture 29: Dna Polymerase, Telomerase, Reverse Transcriptase

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bronzefly406
15 Mar 2018
School
University of Kansas
Department
Biology
Course
BIOL 150
Professor
buechner

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Document Summary

Biol 150 principles of molecular & cellular biology. We reviewed the problems that cells will have in trying to replicate every base on both strands at the ends (telomeres) of a linear piece of dna such as a eukaryotic chromosome: dna. Polymerase has to bind to a primer (usually rna), and the chromosome will get shorter and shorter every time the cell replicates its chromosomes. Cells have an enzyme, telomerase, to make stretches of repeated short sequences at the end of the chromosome in order to make it longer. Primers can then bind to these repeated sequences to replicate the chromosome ends. Most of the cells of our bodies do not have active telomerase, and so can only replicate 30-50 times before the chromosome is so short that the cells die. Only stem cells, which give rise to other cell types, routinely have telomerase active.

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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.

The bacterium E. coli, which is a major model system for the study of genes and DNA, has a singular circular chromosome of about 4x106 bp. In contrast, human diploid cells have __1__ chromosomes, of __2__ different types (don’t forget that the X ¹ Y!). The total amount of DNA found in a haploid cell such as a sperm or egg is called the organism’s “C-value”; the human C-value is __3__.

Also in contrast to the E. coli circular chromosome is the fact that eukaryotic chromosomes are not circular, but instead are __4__. Eukaryotic chromosomes are complexed heavily with protein to form a DNA-protein complex generically called __5__. The proteins found can be considered to fall into two general classes: __6__ and __7__. The first class is made up of 5 major proteins called __8__, __9__, __10__, __11__, and __12__. On a weight to weight basis, these proteins make up about __13__% of the total mass of a chromosome. Since these proteins must associate relatively non-specifically with DNA, they are rich in the basic amino acids __13__ and __14__, which gives the proteins a net __15__ charge.

One goal of packaging the DNA into chromatin is to compact the genome so it will fit into a nucleus. The basic unit of chromatin structure is the __16__, which has __17__ base pairs of DNA wrapped around a “core particle” composed of __18__ copies each of the core histones. Each nucleosome is separated by a stretch of “linker” DNA so that the repeating unit is approximately 200 bp. Lets do a little arithmetic: If we consider that the nucleosome is approximately __19__ nm in diameter, that there is about 200 bp in this structure, and that 200 bp of DNA would normally occupy __20__ nm if stretched out straight (recall that each base is spaced about 0.34 nm from the next), we can see that the nucleosome alone allows for a __21__-fold compaction ratio.

The nucleosome structure is further compacted by histone __22__, which is not a part of the core particle. Instead, this histone __23__ (does what?) to cause the nucleosomal 10 nm fiber to condense into a higher-order structure referred to as the __24__ nm fiber. In the nucleus, chromosomes do not just float around freely: they are tethered to a proteinaceous structure called the __25__ at specific DNA regions called “SARs”, which stands for __26__. SARs are rich in topoisomerases, to allow the winding and unwinding of the DNA that accompanies replication and transcription to occur, despite this constraint.

The 30 nm fiber discussed above is the form in which transcriptionally active DNA is believed to exist in the nucleus; however, further condensation leads to a highly condensed inactive form called __27__; in this terminology, the active form is referred to as __28__. The inactive condensed form can be divided into two types: one type is essentially invariant between cell types, and since it is “on all t time”, it is referred to as __29__. The other type comes and goes, depending on the expression of the genes present in that region of the chromosome; it is referred to as __30__.

Eukaryotic chromosomes carry the genes, of course, but also have other functional regions. One such region is the place where the spindle apparatus attaches during mitosis and meiosis; this region of the chromosome is called the __31__. Another important part of the linear eukaryotic chromosome is the ends; these are called __32__. In humans, the end of each chromosome is composed of many repeats of the short sequence __33__. One more type of functional region on each chromosome is the place where DNA replication starts; in humans, each chromosome has several of these __34__. Thus, eukaryotic chromosomes carry several different types of functional sequences, which together allow for their __35__ during S phase, __36__ during mitosis and meiosis, and the __37__ of their genes

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

The mutation changes the number of domains in the enzyme, which makes it work more efficiently

The mutation changes the amino acid sequence of the lactase protein

The mutation increases the number of copies of the lactase gene that will be found in your genome

The mutation changes whether the lactase sequence is found in an intron or exon

The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

The substrate will increase the reaction rate by binding to the allosteric site

The substrate will increase the reaction rate by competing with the inhibitor for the active site

The reaction rate will not change unless the inhibitor can be removed

The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind

The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

Transcription regulators bind to the 5' UTR region of a gene

Regulators bind via complementary base-pairing to certain DNA molecules

Covalent bonds form between the transcription regulator and the atoms of the DNA backbone

Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene

Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

DNA -> RNA -> protein

All cells arise from pre-existing cells

DNA provides the complete instructions to create a cell

The identity of a cell is determined through gene expression patterns

All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment

Enzymes to digest lactose are only made when energy is low and lactose is available

The activator can override the inhibition of the lac operon by the repressor

The repressor can control the enhancer, while the activator can control the promoter

When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

Phosphorylation and acetylation of DNA affect its ability to be compacted

Changes to the sequence of DNA change whether DNA will wrap around histone proteins

Covalent modifications of histones affect the ability of the transcription initiation complex to form

Histones provide the codon sequences needed for translation to occur

The histone code affects which amino acids will get added to a polypeptide