Log in
HomeClass Notes1,200,000US530,000

BSC 310 Lecture 36: 6.1-6.2 - DNA Binding Proteins, Negative Control

11 views3 pages
School
University of Alabama
Department
Biological Sciences
Course
BSC 310
Professor
Julie Olson

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

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers

Related Documents

BSC 310 Lecture Notes - Lecture 29: Chemotaxis, Cell Membrane, Histidine
graysardine594
BSC 310 Lecture Notes - Lecture 81: Camp Receptor Protein, Catabolite Repression, Lac Operon
silvercicada472
BSC 310 Lecture Notes - Lecture 83: Fluorescence Microscope, Green Fluorescent Protein, Dna Replication
silvercicada472

Related Questions

Part A:

In prokaryotes and eukaryotes cell reproduction is similar in that it requires some kind of signal to proceed, DNA replication, careful separation of the DNA into two equal portions, and division of the cytoplasm. How is cell reproduction different in prokaryotes and eukaryotes?

Select one:

a. In a eukaryotic organism, the signals for cell division are usually not related to the environment but to the function of the entire organism.

b. Eukaryotic cells constantly divide whenever environmental conditions are adequate and the cells are large enough.

c. The linear chromosomes in prokaryotes are constantly being updated.

d. Prokaryotic cell division requires cytokinesis before mitosis because of their small size.

e. The sexual phase of most prokaryotes (bacteria) is essential to the rapid cell division required for infection.

Part B:

What does the name Cyclin Dependent Kinase stand for?

Select one:

a. A protein important to the cell cycle that is dependent on DNA synthesis.

b. An enzyme that cuts DNA so that chromosomes can segregate.

c. An enzyme that phosphorylates other proteins using ATP only when the protein Cyclin is bound to the enzyme.

d. An enzyme that rolls along microtubules dragging a vesicle.

e. An enzyme that connects chromosomes to the spindle fiber.

Part C:

The Cell Cycle is divided into Mitosis, Cytokinesis, and Interphase. Mitosis can be ordered into the following phases:

Select one:

a. Prophase/Prometaphase/Metaphase/Anaphase/Telophase

b. Anaphase/Prometaphase/Metaphase/Prophase/Telophase

c. Metaphase/Anaphase/Telophase/Prophase/Prometaphase

d. Telophase/Prophase/Prometaphase/Metaphase/Anaphase

e. Anaphase/Telophase/Prophase/Prometaphase/Metaphase

Part D:

The Cell Cycle is divided into Mitosis, Cytokinesis, and Interphase. Interphase can be ordered into the following phases:

Select one:

a. S/G1/G2

b. G2/G1/S

c. G1/S/G2

d. G2/S/G1

e. G1/G2/S

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

The mutation changes the number of domains in the enzyme, which makes it work more efficiently

The mutation changes the amino acid sequence of the lactase protein

The mutation increases the number of copies of the lactase gene that will be found in your genome

The mutation changes whether the lactase sequence is found in an intron or exon

The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

The substrate will increase the reaction rate by binding to the allosteric site

The substrate will increase the reaction rate by competing with the inhibitor for the active site

The reaction rate will not change unless the inhibitor can be removed

The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind

The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

Transcription regulators bind to the 5' UTR region of a gene

Regulators bind via complementary base-pairing to certain DNA molecules

Covalent bonds form between the transcription regulator and the atoms of the DNA backbone

Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene

Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

DNA -> RNA -> protein

All cells arise from pre-existing cells

DNA provides the complete instructions to create a cell

The identity of a cell is determined through gene expression patterns

All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment

Enzymes to digest lactose are only made when energy is low and lactose is available

The activator can override the inhibition of the lac operon by the repressor

The repressor can control the enhancer, while the activator can control the promoter

When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

Phosphorylation and acetylation of DNA affect its ability to be compacted

Changes to the sequence of DNA change whether DNA will wrap around histone proteins

Covalent modifications of histones affect the ability of the transcription initiation complex to form

Histones provide the codon sequences needed for translation to occur

The histone code affects which amino acids will get added to a polypeptide

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.