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BIOLOGY 1A03 Chapter Notes - Chapter 4: Genetic Code, Ovarian Cancer, Semiconservative Replication

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magentazebra667
8 Mar 2016
School
McMaster University
Department
Biology
Course
BIOLOGY 1A03
Professor
Lovaye Kajiura

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BIOLOGY 1A03 Full Course Notes
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BIOLOGY 1A03 Full Course Notes
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Document Summary

Chapter 4: nucleic acids and the rna world. 5 possible nitrogenous bases (cytosine, adenine, guanine, thymine, and uracil) Phosphate group at c #5: pentose sugar, provides attachments for nitrogenous bases and phosphates. C#3 oh is involved in bonding with the next nucleotide. The type of bond formed between adjacent nucleotides is a 5"3" phosphodiester linkage. In rna and dna, one end has an unlinked 5" carbon and the other has an unlinked 3" carbon (sugar-phosphate backbone is directional: by convention, base sequences are written in the 5"--> 3" direction (direction of dna. 4. 2 dna structure and function hydrogen bondings are weak, or reproduction would not be ef cient. Dna"s secondary structure was discovered in 1953 by watson and crick (with the aid of research done by rosalind franklin) Watson and crick"s model explained some already known facts: The total number of purines and pyrimidines in dna is the same; the number of a"s and.

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Related Questions

1. Which of the following is the correct sequence of events (from start to finish) in cellular fatty acid oxidation?

A. Fatty acid activationà carnitine shuttleà beta-oxidation

B. Beta-oxidationà carnitine shuttleà Fatty acid activation

C. Fatty acid activationà carnitine shuttleà beta-oxidation

D. Carnitine shuttleà Fatty acid activationà beta-oxidation

2. What reaction is catalyzed by lipoprotein lipase, or lipase?

A. Hydrolysis of TAGs into fatty acids and glycerol

B. Hydrolysis of TAGs into amino acids and glycerol

C. Hydrolysis of cholesterol in to fatty acid and water

D. Hydrolysis of cholesterol into acetyl-CoA

3. How do fatty acids of different carbon lengths cross into the mitochondrial matrix for oxidation?

A. Small FAs of 10 carbons or fewer can freely diffuse through the membrane

B. Large FAs are cleaved into two smaller units that are actively transported across the membrane

C. A membrane-bound transferase catalyzes the conversion of long-chain FAs and FA-carnitine esters, which are transported across the membrane via a specific carrier

D. The breakdown of large FAs starts outside the matrix, allowing for the smaller particles to pass easily through the membrane

4. How can any RNA nucleotide be easily distinguished from a DNA nucleotide?

A. RNA nucleotides have a unique phosphodiester bond linkage

B. RNA nucleotides have uracil nitrogenous bases

C. RNA nucleotides have an -OH on the 2’ carbon of the ribose

D. RNA nucleotides cannot be easily distinguished from DNA nucleotides

5. Which of the following statement is true concerning the sequence pTACGTAACGTA?

A. This is an RNA sequence

B. In order to have a more rigid structure in a double-stranded molecule, the sequence given would need more adenosine

C. If the complementary strand were present, the number of purines would be equal to the number of pyrimidines

D. The amount of purines is equal to the amount to pyrimidines

6. Complementary base pairing is considered which level of structure?

A. Secondary

B. Tertiary

C. Quaternary

D. Primary

7. Which of the following is/are required for the polymerization reaction catalyzed by DNA polymerase III?

I. RNA primer

II. DNA primer

III. RNA template

IV. DNA template

V. NTPs

VI. dNTPs

VII. Magnesium

VIII. Calcium

A. II, IV, V, and VII

B. II, IV, VI, and VII

C. I, IV, V, and VIII

D. I, IV, VI, and VII

8. DNA polymerase I, from E. coli, has three different catalytic sites. Which of the following is NOT one of them?

A. RNA-dependent RNA polymerase

B. RNA-dependent DNA polymerase

C. DNA-dependent RNA polymerase

D. DNA-dependent DNA polymerase

9. Eukaryotes solve the problem of replicating the ends of their linear chromosomes by:

A. Using a protein as primer

B. Enzyme activity that uses an RNA template and reverse transcription in a series of repeat cycles

C. All of the listed solution

D. Terminal redundancy whereby each end of the chromosome is duplicated allowing recombination

Match each term with its definition. Each term or definition is used only once

- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

right-spiraling common DNA form
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

referencing the opposite orientation of hydrogen-bonded DNA strands
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

viral particle whose natural host is a prokaryote
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

universal holding that the number of purines always equals the number of pyrimidines
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

nucleic acid base sequences able to form a double-stranded structure by virtue of matching base pairs
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

5’→ 3’ or 3’→ 5’
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

nuclease able to cleave phosphodiester linkages in the DNA backbone
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

the structure formed by two DNA strands twisted around a common axis in a spiral
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

bacterial cycle of phage- infected cells resulting in cell destruction and viral release
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

main access site for proteins that recognize specific DNA sequences
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

a cytosine modification favoring Z DNA formation
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

nucleic acid monomer consisting of a sugar, a phosphate group, and a nitrogenous base
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

connects two nucleic acid sugar molecules via an ester bond
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

hydrolyzes peptide bonds linking amino acids
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

U or C in RNA
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

a pentose sugar
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

an enzyme that breaks down RNA
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

either of the two nucleotide chains of a double helix
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

a pyrimidine able to hydrogen bond with adenine
- A. B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T.

a DNA configuration favored by high ionic strength and GC repeats
A.

protease
B.

lysis
C.

double helix
D.

bacteriophage
E.

complementary
F.

Chargaff’s rule
G.

B DNA
H.

major groove
I.

nucelotide
J.

antiparallel
K.

RNAse
L.

directionality
M.

phosphodiester linkage
N.

ribose
O.

Z DNA
P.

pyrimidine
Q.

methylation
R.

DNAse
S.

uracil
T.

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.