🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeClass Notes1,200,000US530,000

BIOLCHEM 415 Lecture Notes - Lecture 31: Exonuclease, Dnab Helicase, Chromosome

37 views7 pages
purplebeaver856
21 Jul 2016
School
U of M
Department
Biological Chemistry
Course
BIOLCHEM 415
Professor
Raymond Trievel

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

Document Summary

The reaction of a dna polymerase is rather simple; it is the formation of a phosphodiaster bond. You need to separate the two stands to excess the information content in dna. We need a helicase to unwind the double helix strand. The strands are antiparallel with one running 5"-3" and the other with 3"-5". Dna synthesis will only proceed in a 5"-3" direction. Since the information content of the genome is present in the dna molecules, Watson and crick proposed that the two strands could separate and that each parent strand would serve as a template for the synthesis of the daughter strand in a semi-conservative fashion. In the course of dna replication, dna can become damaged via mistakes that occur during the replication process. Mutation in raas is a single nucleotide mutation and leads to very dramatic affects. The polymerase plays an important role in synthesizing and proofreading it to prevent errors.

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

BIOLCHEM 415 Lecture Notes - Lecture 33: The Sequence, Exonuclease, Permease
purplebeaver856
BIOLCHEM 415 Lecture Notes - Lecture 30: Histone H2B, Semiconservative Replication, Cisplatin
purplebeaver856
BIOLCHEM 415 Lecture Notes - Lecture 38: Cloning, Phosphodiester Bond, Dideoxynucleotide
purplebeaver856

Related Questions

As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.Drag each phrase to the appropriate bin depending on whether it describes the synthesis of the leading strand, the synthesis of the lagging strand, or the synthesis of both strands.a. multiple primers needed b. daughter strand elongates toward replication fork c. only one primer neededd. synthesized 5' to 3'e. daughter strand elongates away from replication fork f. made continuously g. made in segments

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.

1. _________________________ is the name of the process that produces proteins using RNA molecules.

a.Transcription

b Translation

c. Replication

d. Mutagenesis

2. The peptide bonding site on the ribosome:

a. A site

b. P site

c. E site

d. D site

3. The name given to the bond formed between two amino acids.

a. Hydrogen bond

b. Double bond

c. Peptide bond

d. Negative bond

4. __________________________ is the movement of the mRNA molecule on the ribosome to open up the A-site for the new incoming amino acid.

a. Transformation

b. Mutation

c. Translocation

d. Initiation

5. What does the term semi-conservative mean.

a. Replication of RNA is semi-conservative, each new strand consists of one old strand of RNA and one new strand of RNA.

b. Replication of DNA is semi-conservative, each new strand consists of one old strand of DNA and one new strand of DNA.

c. Replication of DNA is semi-conservative, each new strand consists of new strands of nucleotides.

d. Transcription of RNA is semi conservative.

6. This molecule functions during protein synthesis to recognize the codon and bring the proper amino acid back to the ribosome for translation.

a. mRNA

b. tRNA

c. rRNA

d. DNA polymerase

7. What is meant by the base-pairing rule.

a. Bases on a single strand of DNA are always found as pairs

b. Adenine always base pairs with guanine; cytosine with thymine

c. Adenine always base pairs with thymine; cytosine with guanine

d. Bases on a single strand of RNA are found in pair.

8. In this step of protein synthesis, amino acids are added to a growing polypeptide chain.

a. Initiation.

b. Elongation.

c. Termination.

d. Transpiration.

9. Three complementary bases found in a mRNA are called ____________________.

a. Anti-codon.

b. Triplet.

c. Doublet.

d. Codons.

10. Which statement best describes the job or function of the messenger RNA.

a. Carries the map of the gene sequence in the form of codons to the cytoplasm

for translation of the protein.

b. It transfers amino acids to the ribosome for protein synthesis.

c. It forms the ribosomes.

d. It helps replicate DNA.