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BI236 Lecture 9: BI236 Week 3 - DNA Transcription II - Winter 2019

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greycaribou842
23 Jan 2019
School
WLU
Department
Biology
Course
BI236
Professor
Renuka Karunagoda

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Document Summary

General transcription factor: required to bind to its promoter and initiate rna synthesis, eukaryotes have many different transcription factors, their names include tf" (tfiia, tfiib etc. ) Elongation, termination, and rna cleavage in eukaryotic rna synthesis. Special proteins facilitate the disassembly of nucleosomes in front of the moving polymerase and their immediate assembly after the enzyme passes. If there is a dna damage, rna polymerase becomes stalled temporarily during. Termination is governed by a set of signals that differ for each type of rna polymerase. Terminated by a protein factor that recognizes an 18-nt termination signal in the growing rna chain. Termination signals include a short run of u"s (as in bacterial termination signals), and no proteins are required for their recognition. Mrnas are cleaved at a specific site, a cleavage site, before transcription is actually terminated. The cleavage site is10-35 nts downstream from a special aauaaa sequence in the growing rna chain.

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Related Questions

9. The nucleotide on the DNA template that marks the starting point for RNA synthesis is called

a. 0 (zero point)

b. +1

c. -1

d. Start codon

e. TATA

10. Which bacterial RNA polymerase synthesizes tRNA?

a. RNA polymerase I

b. RNA polymerase II

c. RNA polymerase III

d. Core enzyme

e. Promoter

11. What general transcription factor has enzymatic activity?

a. Operator

b. TFIIH

c. eIF2

d. RNAP II

e. CTD

12. What enzyme activity/activities is/are known to be associated with the transcription factor in the previous question?

a. protein kinase and a DNA helicase

b. protein phosphatase

c. ribozyme

d. synthesis of RNA

e. a and c

The TATA-binding protein (TBP) binds to the TATA box sequence in eukaryotic promoters. What is its function in transcriptional initiation?

It blocks access of RNA polymerase to the promoter, until removed by general transcription factors.

It is the subunit of prokaryotic RNA polymerase that is required to recognize promoters.

It modifies histones so nucleosomes can be removed from DNA for transcription.

It bends and partly unwinds DNA at a promoter.

The genetic code is said to be “degenerate” because

there are more codons than amino acids.
there are more amino acids than codons.
different organisms use different codons to encode the same amino acid.
some codons specify more than one amino acid.

Three general mechanisms appear to be responsible for the conversion of proto-oncogenes to oncogenes

overexpression, point mutations, deletions
inversions, translocations, methylation
familial, sporatic, phosphorylation
None of the above account for the conversion of proto-oncogenes into oncogenes

Transcriptional control of genes that acts by regulating the continuation of transcription is called

induction
attenuation
antitermination
negative inducible control

The genetic code is fairly consistent among all organisms. The term used to describe this consistency is

redundant
resilient
universal
the central dogma

The F, G, and H loci are linked in the order written. There are 30 cM between F and G and 30 cM between G and H. If a plant Ff Gg Hh is testcrossed, what proportion of the progeny will be ff gg hh, assuming no interference?

0.7
0.3
0.245
0.15

DNA synthesis is always from 5’ to 3’ because

replication must be continuous
the strands are antiparallel
primers lack 3’ to 5’ exonuclease activity
none of the above

The F, G, and H loci are linked in the order written. There are 30 cM between F and G and 30 cM between G and H. If a plant Ff Gg Hh is testcrossed, what proportion of the progeny will be ff gg hh, assuming no interference?

0.7
0.3
0.245
0.15

In the ZZ-ZW sex-determination system, if an AaBb female was crossed to an individual of genotype Aa Bb, what is the probability of a female offspring with the two dominant traits given by alleles A and B? Assume A and B are dominant alleles.

1/8
1/16
9/16
9/32

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.