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BSCI-1510 Lecture Notes - Lecture 26: Telomerase Rna Component, Okazaki Fragments, Dna Replication

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skytoad282
10 Feb 2016
School
Vanderbilt University
Department
Biological Sciences
Course
BSCI-1510
Professor
Patton Jim

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Document Summary

In all of the lectures thus far, prokaryotic replication is a circle so there isn"t an end replication problem like in linear chromosomes. From the last okazaki fragment to the end on the lagging strand there is no way of filling it in - no dna polymerase can fill in the gap. Every round of replication that a cell undergoes, it should be losing dna at the end of each of the lagging strands. Every one of the cells undergoes numerous rounds of replication. If we are losing 200 chromosomes every time we replicate, pretty soon you"re going to lose too much of dna and your chromosome will no longer be intact. The daughter chromatids (in a mitotic spread) at the end of the sequences are telomeres. Ends of chromosomes are telomeres, we have an enzyme called telomerase (rna protein enzyme that contains both rna and protein)

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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.

Please provide correct answer and explanation to the answer,Thank you!

  1. If an enterprising graduate student decided to repeat theHershey-Chase

    experiment but neglected to perform the blending portion, wherewill the

    isotopes be found?

    a. 35S will be found in the supernatant and 32P will be found inthe pellet

    b. 32P will be found in the supernatant and 35S will be found inthe pellet

    c. Both 32P and 35S will be found in the supernatant

    d. Both 32P and 35S will be found in the pellet

    e. None of the above are correct

  2. Chargaff�s rules include:

    a. The ratio of A�s + T�s to G�s + C�s is always consistent

    b. A=G and T=C

    c. Purines/Pyrimidines = 1

    d. More than one of the above are true

    e. None of the above are true

  3. Sea Urchin DNA contains 18% guanine (which is true); how muchthymine

    does it contain?

    a. 9%

    b. 18%

    c. 32%

    d. 36%

    e. 64%

  4. E. coli DNA contains a little over 25% guanine. If samples weretaken of E.

    coli DNA and Sea Urchin DNA and denatured in water baths, whichwould

    melt at a lower temperature?

    a. E. coli DNA

    b. Sea Urchin DNA

    c. They would melt at the same temperature

    d. There is not enough information to determine the answer

    e. DNA doesn�t �melt,� it isn�t ice cream for goodness sake!

  5. If testing the second generation results in tube B, which modelcan we

    conclude is correct?

    a. Dispersive

    b. Semi-conservative

    c. Conservative

    d. Semi-dispersive

    e. We can�t conclude yet, it could still be more than onepossibility

  6. Phosphorus is required to synthesize dNTPs used in DNAreplication. Imagine

    you grow E. coli in media containing non-radioactive phosphorusfor many

    generations, then switch the E. coli to media containingradioactive

    phosphorous (32P). You then sample the E. coli after one roundof replication

    for radioactivity. Where will you find radioactive phosphorus inthe DNA?

    a. Both strands will contain 32P

    b. One strand will contain 32P

    c. Neither strand will contain 32P

    d. Drats! I knew I should have done suggested problem 12.19!

  7. Which of the following statements about chromosomes are alwaystrue?

    I. Non-histone proteins make up about 5% of a eukaryotic

    chromosome by weight

    II. Histones have a negative charge

    III. Prokaryotes have a single linear chromosome

    a. Only Statement I is true

    b. Only Statement II is true

    c. Only Statement III is true

    d. More than one of the above statements are true

    e. None of the above statements are true

  8. What portion of the human genome is made up of highly repetitiveDNA?

    a. Less than 5%

    b. 6 � 20%

    c. 21 � 30%

    d. 31 � 40%

    e. Over 40%

  9. Which of the following statements are always true?

    I. Histones are very conserved between species

    II. Archaea have nuclei

    III. Prokaryotes are haploid and eukaryotes are diploid

    a. Only Statement I is true

    b. Only Statement II is true

    c. Only Statement III is true

    d. More than one of the above statements are true

    e. None of the above statements are true

  10. An individual�s phenotype is a function of:

    a. The alleles they carry

    b. The environment they live in

    c. The interaction between an individuals genes and theenvironment

    d. Two of the above are correct

    e. All of the above are correct

  11. A homologous pair of chromosomes showing a

    crossover are shown to the right for reference. Are all

    the sperm from one human male genetically identical?

    a. Yes

    b. No, variation occurs during mitosis

    c. No, variation occurs during meiosis I

    d. No, variation occurs during meiosis II

    e. No, variation occurs during both meiosis I and II

  12. Production of gametes in humans requires which kind ofdivision?

    a. Meiotic and is reductional

    b. Mitotic and is reductional

    c. Meiotic and is equational

    d. Mitotic and is equational

    e. None of the above

Expression vectors inprokaryotes do not make functional eukaryotic gene products inbacteria very well because

Answer Not sure which one ?!

· the codon sequence for prokaryotes is differentthan the codon sequence in eukaryotes

· there are no disulfide bridges formed in proteinsnormally made in prokaryotes

· prokaryotic expression vectors cannot translateeukarytic sequences

· eukaryotic genes have introns, and prokaryotesdon't

· eukaryotic genes have exons and prokaryotesdon't


I create a knockout mouseusing the agouti/black fur system \. When I cross the agoutioffspring of the originally engineered mouse, I find a ratio of 2agouti mice to 1 black furred mouse. What is the bestexplanation?

Answer not sure which one?!

· The gene knocked out was recessive.

· The gene knocked out was dominant.

· The gene knocked out was a lethalgene.

· The knockput was integrated into a random spot,and did not knockput the original gene.

A restriction enzyme cuts DNAand leaves the following end

xxCTGCA
xxG

Which of the following could be the sequence of the correspondingend of the other end of the cut DNA?

Answer

· xxG
xxCTGCA

· xxC
xxGTGCA

· xxCCGAT
xxG

· xxGGCTA
xxC


An SNP always occurs dueto

Answer

· a one base-pair change.

· a three base-pair change.

· a single restriction site change.

· a single amino acid change.

· None of these.


I PCR out a mutated gene in apatient with Lisenbee chorea (the inability to dance in acoordinated fashion) and I compare it to another patient with thatsame disease phenotype. One subject had a mutation on chromosome 4,and the other subject couldn't rock it because of a mutation onchromosome 5. This is an exampe of

Answer

· pleitropy

· locus heterogeneity

· allelic heterogeneity

· dominant negative mutation

· penetrance

In his experiments, Mendelnoted that when two traits are involved in a genetic cross, theyare inherited independently of each other. Though Mendel didn'tknow about chromosomes, this still holds true (mostly)because

Answer

· genes on the same chromosome are often mixedthrough meiotic crossover during the formation ofgametes.

· variant alleles of the same gene do not interferewith the phenotypic expression of the gentoype in different geneticloci.

· most higher organisms are diploid, making only twoalleles available for every genetic locus, and there is a 50/50chance of geting either allele from a single parent.

· expression of phenotypes can vary from thegenotypes because of encironmantal influences.

· chromosomes often recombine

Anticipation is caused by amutation that increases in expressivity over subsequentgenerations.

Answer

· True

· False

Question 44

I have a genotype that shouldproduce a specific phenotype, but some of the individuals with thegenotype do not demonstrate any evidence of the phenotype. I wouldconsider this an example of

Answer

· allelic heterogeneity.

· locus heterogeneity.

· penetrance.

· expressitivity.

· sheer luck.

Question

Which of the following doesnot occur during the PCR reactions?

Answer

· Changes in the reaction temperature by athermocycler

· Denaturation at high temperatures

· Synthesis of oligonucleotide primers

· Extension of primers by DNA polymerase

· Annealing of oligonucleotide primers

Question

DNA markers, or variantnon-coding regions of DNA, are DNA polymorphisms that are usefulfor genetic mapping.

Answer

· True

· False

A mosaic is an organismwith

Answer

· multiple genotypes within one organism

· multiple alleles within one genotype

· more than one color of fur

· transgenes added to the zygote beforedevelopement

· a wt phenotype but a mutated genotype

Question

Genotype causesphenotype.

Answer

· No, gentoype just influences phenotype.

· Yes, genotype is the DNA sequence that createsphenotype.

Question

A true genetic chimera can becreated by

Answer

· mutating a gene early on in the development of anorganism resulting in different alleles being present in theadult

· multiple fertilized eggs or zygotes fusing to formone embryo

· adding a transgene to the genome of an organismduring fetal development only

· adding cells of a different species to an adultorganism

· adding a transgene to the genome of an animal atany stage in development

Question

The ABO blood group can bestbe explained by the concept of _______.

Answer

· dominant traits

· recessive traits

· allelic heterogeneity

· locus heterogeneity

· vampirism

If a loss of functionmutation creates a dominant phenotype, it may be becauseof

Answer

· haploinsufficiency

· penetrance

· expressivity

· allelic heterogeneity

· locus heterogeneity

Please select the best matchfor each.

Answer

· PCR: Technique used to generatecopies of a DNA fragment

· Southern blot : Screeningprocedure used to assay DNA fragments

· RFLPs : DNA fragments generatedby restriction enzymes that vary in length betweenindividuals

· Northern blot : Screeningprocedure used to assay mRNA fragments

· Electrophoresis Movement ofmacromolecules in an electrical field

Answer

A.

Screening procedure used toassay DNA fragments

B.

Movement ofmacromolecules in an electrical field

C.

Screening procedure used toassay mRNA fragments

D.

DNA fragments generated byrestriction enzymes that vary in length betweenindividuals

E.

Technique used to generatecopies of a DNA fragment

A gene mutates, and theprotein produced has a novel way of interacting with the cell, andcreates a new phenotype because of this new functionality. This iscalled

Answer

· pleitropy

· locus heterogeneity

· allelic heterogeneity

· dominant negative mutation

· gain of function dominant mutation

Question 62

Mutations in the somaticchromosomes may be inherited by the next generation.

Answer

· True

· False


A degenerate PCR primer withmany variant sequences must be used to make multiple copies of agene

Answer

· if only the protein sequence of the gene productis known to construct the primers

· if the DNA you are using is cDNA to constructthe primers

· if the DNA you are using is genomic DNA to becopied

· if the DNA you are trying to copy iscDNA

· if the vector is prokaryotic and the transformedcell is eukarytotic

Question

Please select the best matchfor each term.

Answer

· Restriction enzyme


· PCR C

· Plasmid

· YAC A

· Dideoxynucleotide

Answer

A.

Endonuclease isolated frombacteria

B.

Eukaryotic cloningvector

C.

Method for amplification ofDNA sequences

D.

Small circular fragment ofDNA

E.

Molecule used in DNAsequencing

Question

A couple goes on MauryPovitch, and the results are in: you are not the father. But noother man impregnated the female (granted, unlikely for a MauryPovitch guest, but work with me here) and he must be the father.DNA analysis claims otherwise, though the child definitely wasmom's (poor thing). What may be going on here?

Answer

· the child is parthenogenic because the motheractually impregnated herself like a shark, and the child's DNA isall mom's

· the child had a mutation that changed the genethat is used to trace paternity

· the child is a mosaic because he is actually a setof twins fused early during fetal development, and therefore camefrom 2 eggs and 2 sperm

· the dad may have germ-line mosaicism, meaning thatthe genotype of his sperm is different from his somaticgenotype

· mitochondrial DNA only comes from mom, so there isno way to tell whobthe baby's father is