🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeClass Notes1,200,000CA670,000

BIOLOGY 1A03 Lecture Notes - Lecture 19: Phosphodiester Bond, Primase, Dna Ligase

16 views2 pages
School
McMaster University
Department
Biology
Course
BIOLOGY 1A03
Professor
Lovaye Kajiura

For unlimited access to Class Notes, a Class+ subscription is required.

doinhumolalala and 39849 others unlocked
BIOLOGY 1A03 Full Course Notes
9
BIOLOGY 1A03 Full Course Notes
Verified Note
9 documents

Document Summary

On the leading strand, only one primer is required and the dna polymerase is able to continuously add nucleotides to the new daughter strand as the replication fork progresses. In contrast, the other antiparallel parent strand has discontinuous (or fragmented) replication occur. The produced daughter strand is referred to as the lagging strand. Since dna polymerase can only add nucleotides to the 3" end of a polymerizing dna molecule, replication from this strand requires that dna polymerase replicate the dna in a direction that is away from the replication fork. The end result of replication and production of the lagging strand is that the lagging strand will contain segments or fragments of dna. These are often referred to as okazaki fragments, named after reiji. Okazaki, the scientist who discovered them in e. coli. While only one primer is required for the synthesis of the leading daughter strand, each okazaki fragment on the lagging strand is formed by separate primers.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related Documents

BIOLOGY 1A03 Lecture Notes - Lecture 2: Helicase, Okazaki Fragments, Dna Replication
erincat447
BIOLOGY 1A03 Lecture Notes - Lecture 23: Matthew Meselson, Franklin Stahl, Dna Replication
limepug436
BIOLOGY 1A03 Lecture Notes - Lecture 15: Agarose Gel Electrophoresis, Sanger Sequencing, Frederick Sanger

Related Questions

1. During DNA replication, proteins unwind the DNA duplex, and separate the DNA strands. Those proteins include:

a.) DNA polymerase and ligase

b.) Single-stranded DNA binding and protein primase

c.) Helicase and DNA polymerase

d.) Helicase and single-stranded DNA binding protein

2. What determines the nucleotide sequence of the newly synthesized strand during DNA replication?

a.) The particular DNA polymerase catalyzing the reaction

b.) The relative amounts of the four nucleotide triphosphates (ATP, GTP, CTP, and TTP) in the cell

c.) The nucleotide sequence of the template strand

d.) The primase used in the reaction

e.) Both A and D

3. The leading and the lagging strands differ in that

a.) The leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.

b.) The leading strand is synthesized by adding nucleotides to the 3’ hydroxyl (OH) end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5’ phosphate (PO4) end.

c.) The leading strand is synthesized continuously, whereas the lagging strand is synthesized in short fragments that are ultimately stitched together.

d.) Both A and B

e.) Both A and C

1.When a protein binds to DNA at a site defined by a particular nucleotide sequence, with which parts of the DNA does the protein primarily interact?

A. the phosphate and deoxyribose groups in the DNA backbone

B. the phosphate and deoxyribose groups in the DNA backbone, and hydrophobic interactions with the nucleotide bases

C. nucleotide bases within the major grooves of the DNA

D. nucleotide bases within the major and minor grooves of the DNA

E. none of the above

2. Which of the following proteins binds to DNA with sequence specificity?

A. POI Ill holoenzyme

B. DnaA

C. DnaB

D. Dam methylase

E. Both B and D

3. Which of the following is NOT a functional parallel among all known cellular systems of DNA replication?

A. Replication is initiated at specific sites

B. Replication is continuous on the leading strand and discontinuous on the lagging strand

C. DNA replication is semiconservative

D. Replication forks move away from the origin in one direction

E. Nucleases, polymerases and ligases replace RNA primers with DNA and seal the remaining nick

4. A DNA polymerase cannot synthesize a DNA strand de novo (from the beginning) from a mixture of dNTPs. Which of the following explains the role of the primer in the reaction promoted by this enzyme?

A. The primer is an initiation protein with attached nucleotides that binds to the template and to which new nucleotides are added to the 3 ' end.

B. The primer is an initiation protein with attached nucleotides that binds to the template and to which new nucleotides are added to the 5' end.

C. The primer is a preexisting strand of RNA or DNA to which new nucleotides are added to the 3' end.

D. The primer is a preexisting strand of RNA or DNA to which new nucleotides are added to the 5' end.

E. The primer is a preexisting strand of DNA synthesized by DNA pol I, prior to replicative DNA synthesis by DNA pol Ill.

5. Universal features of DNA polymerases include:

A. 5'-3' exonuclease function.

B. Formation of an 'closed' conformation of the polymerase when the template-nucleotide base pair is correct.

C. Use of a single magnesium ion to orient reacting molecules and assist in dissipation of the negative charge that develops during nucleotidyl transfer.

D. Alignment of conserved residues in the active site for attack of the DNA primer by the 3 'OH on the incoming nucleotide.

E. Continuous unbinding and rebinding of the polymerase to the DNA template as each nucleotide is added.

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.