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BIOL10003 Lecture Notes - Lecture 28: Small Interfering Rna, Gene Expression, Noncoding Dna

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School
University of Melbourne
Department
Biology
Course
BIOL10003
Professor
All

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Document Summary

Lecture 28 - gene expression: dna, rna and polypeptide s. There are different sizes of chromosomes and densities of genes on them. 2% of genes encode proteins; the rest are unique non-coding dna, repetitive sequences and pseudogenes (2%) No cells express all their genes at any one time, they only require certain genes at different times. Genetic information - varying levels of compaction: less compact = more accessible = more expression. = differential gene expression; e. g. different genes turned on to make stem cells specific. Hundreds of copies (dna only has one copy per cell) Single stranded, unless part is in the secondary structure (hairpin) mrna messenger rna. Carries the dna code, complementary to the dna template. Template for protein synthesis trna transfer rna. Transfers amino acids to the ribosome floats around in the cytoplasm. Trna (cid:272)a(cid:374) o(cid:374)ly (cid:271)i(cid:374)d o(cid:374)e spe(cid:272)ifi(cid:272) a(cid:373)i(cid:374)o a(cid:272)id at the 3" e(cid:374)d (cid:894)al(cid:449)ays cca(cid:895) Trna matches an anticodon to the mrna template.

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Related Questions

5. Alpha amanitin is a drug that binds RNA polymerase andprevents it from functioning in the cell (this is the compound thatis found in poison mushrooms). The reason people die when theyconsume alpha amanitin is:

a) Alpha amanitin prevents gene expression from occurring in thecell

b) DNA replication is directly inhibited by alpha amantin

c) Alpha amanitin binds tightly to the messenger RNAs that aremade in the cell, preventing translation

d) Alpha amanitin binds to nuclear pores preventing thetransport of messenger RNA into the cytoplasm

6. In order for transcriptional initiation to begin in E. colithe following elements are needed:

a) an origin and RNA polymerase

b) initiator protein and a promoter

c) the sigma subunit of RNA polymerase and an RNA template

d) RNA polymerase and a promoter

e) helicase and primase

Questions 7 and 8. You are studying two species of closelyrelated snakes, one species lives in sandy desert area and has paletan skin and the species lives in a volcanic rock area and has darkbrown skin. You a sequence the DNA from each type of snake anddiscover a gene encoding a protein that transports pigment into theskin cells. The pigment transporter protein is responsible for skincolor. When you analyze the DNA coding sequence of the pigmenttransporter from each snake you find these two DNA sequences (hashmarks have been added to indicate the reading frame):

pale tan snake: GCG/TTG/GAC/TGC/ACC (DNA template strand)CGC/AAC/CTG/ACG/TGG (complementary strand) dark brown snake:ACG/TTG/GAC/TGC/ACC (DNA template strand) TGC/AAC/CTG/ACG/TGG(complementary strand)

7. From these sequences you conclude:

a) that the amino acid sequence of the pigment transporters forthe two proteins are almost identical

b) that the amino acid sequence of the pigment transporters forthe two proteins are very dissimilar

c) that the amino acid sequence of the pigment transporters forthe two proteins are identical

8. Since the skin color is different in each snake, you concludefrom the sequence data above that:

a) the genetic code in the skin cells in each of these snakes isdifferent

b) the DNA sequence encoding the entire gene for pigmenttransporters in these two snakes is different

c) that one of the codons in the gene for the pigmenttransporter in these two snakes is different

9. You are analyzing eukaryotic cells that have been exposed tothe radioactive uracil. In most cells the radioactive uracil isinitially localized in the nucleus and then moves from the nucleusto the cytoplasm. However, you observe that in some cells theradioactive molecules stay in the nucleus and never arrive in thecytoplasm. This is most likely a defect in:

a) the transcription of the DNA

b) the processing of the mRNA

c) the translation of the RNA

d) the labeling of the RNA

e) the replication of the DNA

10. Which of the following is true of both eukaryotic andbacterial transcription? (select all that are true)
a) Transcription factors must bind the promoter before RNApolymerase initiates transcription

b) Messenger RNAs must be spliced before translation begins inorder to produce the appropriate gene product

c) During transcription only one DNA strand is used as atemplate to make the messenger RNA

d) Termination of transcription occurs when the RNA polymerasetranscribes a hairpin structure in the RNA followed by a string ofU

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.