Biology 1001A Lecture Notes - Lecture 6: Okazaki Fragments, Nucleic Acid Double Helix, Alpha Helix

44 views4 pages

carminewalrus319

28 Mar 2016

School

Department

Course

Professor

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

Document Summary

Independent study outcomes: purine and pyrimidine base-pairing in dna/rna. Pair by complementary base pairing: a-t, c-g. Semi-conservative: in dna replication, a double stranded dna splits into two single strands that are templates for the synthesis of a new strand, then becoming double stranded dna (from one we make two) Semi-discontinuous: also known as discontinuous replication, is the replication in which a dna strand is formed in short lengths that are synthesized in the direction opposite of dna unwinding: these short lengths are called okazaki fragments. Leading strand: dna strand synthesized in the direction of dna unwinding (at origin) Lagging strand: dna synthesized discontinuously in opposite direction of dna unwinding: general action of proteins in fig. Single stranded binding proteins (ssb"s): stabilize single stranded dna and prevent the two strands at the replication fork from reforming double stranded dna: topoisomerase: avoids twisting of dna ahead of replication fork by cutting the dna, turning the.

Related Questions

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.

ceruleangnat663

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

skyhamster121

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.

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

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

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.

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

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl 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 3â² hydroxyl group (âOH). 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 90Â°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.

Biology 1001A Lecture Notes - Lecture 6: Okazaki Fragments, Nucleic Acid Double Helix, Alpha Helix

Document Summary

Get access

Related Documents

Biology 1001A Lecture Notes - Lecture 9: Okazaki Fragments, Dna Clamp, Dna Replication

Biology 1001A Lecture Notes - Lecture 7: Telomere, S Phase, Telomerase

Biology 1001A Lecture Notes - Semiconservative Replication, Polynucleotide, Pyrimidine

Related Questions