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BISC 202 Lecture Notes - Lecture 3: Complementary Dna, Primase, Helicase

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School
Simon Fraser University
Department
Biological Sciences
Course
BISC 202
Professor
Michael Hart

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Document Summary

Bisc 202 lecture 3 - discovering dna structure and understanding dna replication. Chromosomes carry genetic information, dna and protein. Through several experiments in the 20th century: Discovered dna holds the genetic information and not proteins. 5" end = location with a phosphate group on 5" carbon. 3" end = location with no phosphate group and can add new nucleotides to the 3" carbon. Each complementary strand is antiparallel to each other. One strand is 5" to 3" while the other is 3" to 5" (important to know for. Dna has a spiral shape and a constant width. # pyrimidine bases = # purine bases. Semi-conservative: dna replication produces 2 copies that contain one new strand and one original strand. Nucleotides arrive with 3 phosphate groups (atp), to provide energy. A bond is formed between the 3" carbon and one phosphate group on atp, the other 2 phosphate groups are released, which provides enough energy for the reaction to occur.

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Related Questions

Respond to each sentance with true or false regarding the replication of DNA:

Complementary base pairing relies on the number of hydrogen bonds that each base can make.

In a single nucleotide, the phosphate group is attached to the 5' carbon of the sugar unit.

The antiparallel arrangement of double-stranded DNA is due to the phosphate group being bonded to the 3' carbon on one strand and the 5' carbon on the complementary strand.

The phosphate attached to the 5' carbon of a given nucleotide links to the 3' -OH of the adjacent nucleotide.

The -OH group on the 3' carbon of the sugar unit is the attachment site for the nitrogenous base.

1. Unlike DNA, RNA contains

deoxyribose

a double strand

phosphates

uracil

Nucleic acid structure

2. In transcription, uracil will bond with:

thymine

cytosine

adenine

guanine

Transcription

3. The first team of scientist to document the structure of DNA was:

Franklin and Wilkins

Watson and Crick

Pauling and Chargoff

None of the above

DNA structure

4. Organelles that synthesize protein are:

lysosomes

ribosomes

endoplasmic reticulum

golgi bodies

5. Most of the time in a cell's life, it's DNA is in the form of:

Chromatin

Chromatids

Replicated chromosomes

Spindle fibers

6. Uracil is a nitrogen base that pairs with adenine. Using this information, one could conclude that uracil is a:

Purine

Pyrimidine

Neither a purine or pyrimidine

Not enough information

7. a nucleotide does not consist of a:

polymerase

5-carbon sugar

phosphate group

nitrogen base

8. The genetic code in DNA of an individual depends on the sequence (order) of:

nucleotides

purines only

phosphates

sugars

9. Before a cell divides, it must duplicate its DNA in a process known as:

translation

transformation

transcription

replication

10. The basic unit (building block) of a protein is a (an):

carbohydrate

amino acid

lipid

hydrogen bond

11. Unlike DNA, RNA contains:

deoxyribose

a double strand

phosphates

uracil

12. The structure of DNA was first documented in:

1945

1953

1956

1962

13. In DNA replication, the new molecules produced are:

of unequal length

both assembled in the same direction

built with purines as the backbone

assembled in opposite directions

DNA replication

14. The "backbone" of DNA is composed of:

ribose and phosphates

deoxyribose and phosphates

ribose and uracil

deoxyribose and thymine

DNA structure

15. RNA differs from DNA in which of the following ways?

RNA is single-stranded

RNA contains ribose

RNA does not contain thymine

All of the above

DNA structure

16. Which of the following is NOT required for the transcription process itself?

RNA polymerase

DNA

uracil-containing nucleotides

ribosomes

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.