BIOL 112 Study Guide - Midterm Guide: Dna Ligase, Reaction Rate, Genome Size

629 views24 pages

redjellyfish523

9 Feb 2016

School

Department

Course

Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

jakbfousda and 39238 others unlocked

BIOL 112 Full Course Notes

Verified Note

59 documents

Document Summary

Biol 112: unit 1 practice questions for midterm 2. Describe in general terms how enzymes can increase reaction rates. Contrast the progress curve of an enzyme catalyzed reaction and uncatalyzed reaction. Compare competitive, non-competitive, allosteric enzyme inhibitions as a way of regulating enzyme activity. List the factors that can affect enzyme activity. Study questions: the graph presents three activation energy profiles for a chemical reaction (the hydrolysis of sucrose): an uncatalyzed reaction, and the same reaction catalyzed by two different enzymes. On each, draw the curve for the enzyme catalyzed reaction. Biol 112 unit 2 practice questions midterm 2 preparation. Unit 2: genome and information flow from dna to proteins. Discuss the diversity in genome sizes among organisms and if genome size is a good predictor of biological (organismal) complexity. Compare genome structure and arrangement in bacterial cells versus eukaryote cells. Discuss in general terms the biological role of dna as an information molecule for storage and transmitting genetic information.

Related Questions

describe the definitions of substrate, enzyme active site and its general characteristics, and apoand

holo-enzymes.

Â· describe the fact that enzyme catalysis is specific in terms of the type of reaction and the exact

substrate structure.

Â· Explain the difference in the âlock-and-keyâ model and the induced fit model of enzyme

substrate interaction.

Â· describe that reaction rate constant thus reaction rate is dependent on the activation energy.

Â· describe the relationship between the energy of the transition state and the activation energy.

Â· describe the relative binding affinity of a substrate and its transition state to the enzyme.

Â· Explain how enzymes speed up reaction rate, i.e., how do enzymes lower the activation

energy. Note: Factors other than preferential stabilization of transition state also contribute to increased

reaction rate. They include 1) enzymes placing one substrate (reactant) next to the other substrate (reactant)

so that the reaction is no longer dependent on the collision rate of the two substrates; 2) enzymes sequester the

substrate(s), minimizing solvent competition thus speeding up the reaction; 3) in enzyme complexes with each

enzyme catalyzing one step in a multi-step reaction sequence, the product of one reaction is fed to the active site

of another enzyme for the next step without the product diffusing away from the enzyme complex thereby

increasing reaction rate.

Â· describe the Michaelis-Menten equation relating the initial velocity to total substrate

concentration.

Â· describe the assumptions that went into the derivation of the Michaelis-Menten equation. ([E]

is constant, steady state assumption, etc)

Â· describe the relationship between Vmax and kcat and the fact that kcat is a constant for a given

enzyme and substrate at a given temperature and solution condition; whereas Vmax can differ

for a given enzyme and substrate at a given temperature and solution condition, depending

on the enzyme concentration.

Â· describe the definition of Kd (equilibrium dissociation constant for ES into E and S).

Â· describe the relationship between Kd and the affinity of substrate for the enzyme, i.e., the

higher the Kd, the weaker the binding between E and S.

Â· describe the relationship between Kd and Km, i.e., in general, the higher the Kd, the higher theKm.

Â· Use the Michaelis-Menten equation and the relationship between Vmax and kcat to do

calculations.

Â· describe the other names for the catalytic rate constant, kcat (i.e., the turn over number; also

called krelease) and why it is so called (kcat reports on per unit time per enzyme molecule the

number of product molecules formed/released, thus the name turnover number or krelease).

Â· Determine Vmax and Km using both the Michaelis-Menten plot (initial velocity versus [S]) and

the Lineweaver-Burk Plot (the double reciprocal plot), given the data of initial velocity and

[S].

Â· Draw the Michaelis-Menten plot AND the double reciprocal plot (Lineweaver Burk), given

the Vmax and Km values.

Â· describe the fact that the term kcat/Km is the best parameter for measuring the catalytic

capability (efficiency) of an enzyme.

Â· describe the existence of irreversible enzyme inhibition and the general mechanisms of

irreversible inhibition.

Â· describe the different types of reversible inhibitions, namely the competitive, uncompetitive,

and mixed inhibitions (noncompetitive inhibition is a special case in which the inhibitor

binds E and ES with equal affinity therefore Î± is equal to Î±â â Km is not affected).

Â· Describe each type of inhibition in terms of inhibitor binding site (active site or other than

the active site), how the inhibitor affects vmax and Km and the quantitative relationship of

each parameter with and without the inhibitor present and their relationship with [I] and KI

or KI

Â· Show the characteristics of each type of inhibition on a Michaelis-Menten plot (initial

velocity versus total substrate concentration), i.e. draw the curves with and without the

competitive inhibitor.

Â· Show the characteristics of each type of inhibition on a double reciprocal plot (LineweaverBurk

Plot) by drawing the lines with and without the competitive inhibitor.

Â· Given the Michaelis-Menten or double reciprocal plot with and without the inhibitor,

determine the type of inhibitor.

Â· describe the fact that transition state analogs are potent inhibitors.

Â· Explain why transition state analogs are potent inhibitors.

Â· describe that not all enzymes obey Michaelis-Menten kinetics.

Â· describe situations when an enzyme will not obey Michaelis-Menten kinetics (when the enzyme

catalyzes reactions with multiple substrates or when the enzyme is allosteric, i.e. binding of

the substrate at one active site affects the other active site(s) for the same substrate).

Â· Name the four types of catalytic mechanisms. (electrostatic catalysis, general/specific acid

base catalysis, metal ion catalysis, covalent catalysis)

Â· Contrast acid/base catalysis to general acid/base catalysis, i.e., comment on their similarity

and difference.

Â· Name the eight common general acid/base pairs in proteins and write the formula of the

acid and base form for each pair (hint: consider the eight amino acids that have ionizable

side chainsâ¦)

Â· describe the meaning of covalent catalysis.

Â· describe the reaction catalyzed by chymotrypsin and the three catalytic residues in

chymotrypsin.

Â· Draw the interactions among the three catalytic residues (catalytic triad).

Â· State the catalytic role of each of the catalytic residues and how they accomplish that.

Â· describe the steps in the formation of the acyl-enzyme intermediate, and the substrate and

products in this first part of the chymotrypsin mechanism.

Â· Draw the transition state structure for the formation of the acyl-enzyme intermediate and

indicate by drawing how it is preferentially stabilized.

Â· describe the steps in the regeneration of chymotrypsin (deacylation), the substrates and

product of this second part of the chymotrypsin mechanism.

Â· Draw the transition state for deacylation and explain how it is preferentially bound to or

stabilized by the enzyme.

Â· Explain what general acid and base catalysis refers to and what groups act as the general acid

and base in the chymotrypsin mechanism.

Â· Explain what covalent catalysis refers to in the chymotrypsin mechanism.

Â· Explain the basis for the substrate specificity of chymotrypsin (consider the composition of

the active site â chymotrypsin specifically cleaves peptide bonds adjacent to aromatic

residues)

Â· Explain the term âserine proteaseâ. Answer: these are proteases (enzymes that

hydrolyze peptide bonds) that have serine as a catalytic residue (that acts as a

nucleophile, which can attack the carbonyl carbon of the peptide bond).

Â· Key features of Hexokinase, enolase and lysozyme mechanisms that were discussed in class

(specific methods of catalysis utilized by these enzymes; general description of overall

catalytic transformation)

Â· describe the four regulatory strategies of enzyme catalysis

Â· describe the definition of feedback inhibition (mechanism of cellular regulation that regards an

enzyme that catalyzes the production of a given substance that then inhibits the cell when

concentration of that substance has reached an appropriate level â this balances/regulates

amount of product provided with amount needed)

Â· describe the most common type of covalent modification and the types of enzymes

responsible for the addition and removal of the chemical group in this covalent

modification.

Â· Explain the general mechanism of phosphorylation in regulating enzyme activity.

Â· describe the general naming scheme for active and inactive proteases.

Â· describe that protease inhibitors exist to inhibit protease.

Â· describe that enzymes are regulated through covalent and non-covalent processes.

Â· Noncovalent regulation

o Describe noncovalent modification in general terms. Elaborate on the relation

among enzyme activity, allostery, effector molecules (small molecules) and control

proteins.

o Concerning allosteric regulation, some enzymes have multiple active sites for the

same substrate and the binding of the substrate increases the activity of the enzyme

by stabilizing the R (more active) state over the T (relatively inactive) state. (Lecture

13: âNoncovalent Modification: Allosteric Regulatorsâ)

Ã Draw a curve for the initial reaction velocity versus substrate concentration.

Ã describe and explain the specific shape of the v0 vs. [S] plot.

Ã Draw on the same graph above a v0 vs. [S] curve for the enzyme in the

absence of allostery. and how th i.e., in the absence of conformational change

upon substrate binding. Comment on the shape of this curve.

Ã Comment on the benefit of allosteric activation.

Ã When an inhibitor is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

Ã When an activator is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

Ã When an activator is present for such an allosteric enzyme, the v0 versus.

Ã describe the relationship between feedback inhibition and allosteric regulation

and the reason behind the term âfeedback inhibitionâ.

Â· Covalent regulation

o describe the most common covalent modification, the enzymes that catalyze the

modification and the removal of that modification, respectively and the amino acid

residues that tend to be the site of this modification.

o Write the structure of the added group bonded to those residues and the reaction for

this covalent modification.

o Rationalize the effectiveness of phosphorylation on altering the activity of an

enzyme.

o describe the mechanism of chymotrypsinogen activation to chymotrypsin (cleavage of

zymogen by trypsin).

What are the correct answers to the parts of this quiz marked incorrect? (that is my one question. Please take your time. I value accuracy over speed. )

IncorrectQuestion 2

0 / 1 pts
Studies done by Buchner showed that ground-up yeast cells were able to convert sugar to alcohol. The components of the mixture that were responsible for this transformation were

DNA molecules.

enzymes.

lipids.

carbohydrates.

Bloom's Level: 2. Understand
Learning Outcome: 06.01
Section: 06.01
Topic: History of Microbiology

IncorrectQuestion 3
0 / 1 pts
Bacteria that can absorb light energy and convert it into ATP are commonly called

cytochromes.

lysosomes.

chemotrophs.

phototrophs.

cytochromes AND chemotrophs.

Bloom's Level: 1. Remember
Learning Outcome: 06.02
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 4
0 / 1 pts
The readily usable energy currency of cells is

electricity.

the electron transport system.

ATP.

CTP.

the electron transport system AND CTP.

Bloom's Level: 1. Remember
Learning Outcome: 06.02
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 5
0 / 1 pts
NAD in its reduced form is symbolized as

NADH.

NAD.

NAD-.

NADP.

Bloom's Level: 1. Remember
Learning Outcome: 06.03
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 6
0 / 1 pts
The most common starting pathway for the breakdown of sugars is

respiration.

fermentation.

glycolysis.

oxidation.

Bloom's Level: 2. Understand
Learning Outcome: 06.04
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 7
0 / 1 pts
Which of the following processes generates the greatest amount of energy?

fermentation

aerobic respiration

the Entner-Duodoroff pathway

glycolysis

fermentation AND the Entner-Duodoroff pathway

Bloom's Level: 2. Understand
Learning Outcome: 06.05
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 8
0 / 1 pts
Fermentation

uses an inorganic molecule as the final electron acceptor.

uses an organic molecule as the final electron acceptor.

results in the production of a large amount of ATP.

is necessary in some organisms to produce reduced electron carriers.

uses an organic molecule as the final electron acceptor AND is necessary in some organisms to produce reduced electron carriers.

Bloom's Level: 2. Understand
Learning Outcome: 06.05
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 9
0 / 1 pts
The use of the suffix "ase" on a word denotes a(n)

substrate.

product.

enzyme.

electron acceptor.

substrate AND electron acceptor.

Bloom's Level: 1. Remember
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 10
0 / 1 pts
Enzymes speed up reactions by

raising activation energy.

producing heat.

reducing entropy.

lowering activation energy.

raising activation energy AND reducing entropy.

Bloom's Level: 2. Understand
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 11
0 / 1 pts
Most enzymes

are generalists and typically recognize a number of different substrates.

are specialists and typically recognize a single substrate.

are active over a wide pH range.

have several active sites.

are generalists and typically recognize a number of different substrates AND have several active sites.

Bloom's Level: 2. Understand
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 12
0 / 1 pts
Allosteric enzymes

may bind two substrates.

are used to bind to other enzymes.

have an additional binding site that is involved in regulating enzyme activity.

are twice as fast as single site enzymes.

may bind two substrates AND are twice as fast as single site enzymes.

Bloom's Level: 2. Understand
Learning Outcome: 06.09
Section: 06.02
Topic: Microbial Metabolism

Question 13
1 / 1 pts
Feedback inhibition

is a means of regulating the amount of product produced.

often involves the use of allosteric enzymes.

involves inhibiting the last of a series of reactions.

results in raising the activation energy for the reaction.

is a means of regulating the amount of product produced AND often involves the use of allosteric enzymes.

Bloom's Level: 2. Understand
Learning Outcome: 06.10
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 14
0 / 1 pts
Coenzymes are derivatives of

minerals.

proteins.

lipids.

vitamins.

Bloom's Level: 1. Remember
Learning Outcome: 06.07
Section: 06.02
Topic: Microbial Metabolism

Question 15
1 / 1 pts
Environmental factors that may affect enzyme activity include

temperature

pH

salt

age

temperature, pH AND salt

Bloom's Level: 2. Understand
Learning Outcome: 06.08
Section: 06.02
Topic: Microbial Metabolism

Question 16
1 / 1 pts
Most enzymes function best at

acidic pH and high salt concentrations.

basic pH and low salt concentrations.

neutral pH and high salt concentrations.

slightly above pH 7 and low salt concentrations.

Bloom's Level: 2. Understand
Learning Outcome: 06.08
Section: 06.02
Topic: Microbial Metabolism

Question 17
1 / 1 pts
Mercury in mercurochrome inhibits growth

through competitive inhibition.

through a reversible action.

by oxidizing the sulfhydryl groups in cysteine.

by changing the shape of proteins.

by oxidizing the sulfhydryl groups in cysteine AND by changing the shape of proteins.

Bloom's Level: 3. Apply
Learning Outcome: 06.10
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 18
0 / 1 pts
The term precursor metabolite refers to molecules that

activate cell components.

are used in biosynthesis.

result from cell activities.

are present but inactive.

Bloom's Level: 2. Understand
Learning Outcome: 06.11
Section: 06.03
Topic: Microbial Metabolism

IncorrectQuestion 19
0 / 1 pts
Glycolysis

requires oxygen.

produces 3 ATP molecules.

produces 4 molecules of NAD.

may occur under aerobic or anaerobic conditions.

produces 3 ATP molecules AND produces 4 molecules of NAD.

Bloom's Level: 2. Understand
Learning Outcome: 06.12
Section: 06.03
Topic: Microbial Metabolism

IncorrectQuestion 20
0 / 1 pts
The transition step

links glycolysis to the pentose phosphate pathway.

links the pentose phosphate pathway to the Entner-Duodoroff pathway.

links glycolysis to the TCA cycle.

takes place in the matrix of the nucleus.

Bloom's Level: 1. Remember
Learning Outcome: 06.11
Section: 06.03
Topic: Microbial Metabolism

Question 21
1 / 1 pts
The TCA cycle produces

ATP through substrate-level phosphorylation.

NADH.

FADH2.

precursor metabolites.

All of the choices are correct.

Bloom's Level: 1. Remember
Learning Outcome: 06.12
Section: 06.03
Topic: Microbial Metabolism

Question 22
1 / 1 pts
The electron transport system

requires a membrane.

utilizes the nuclear membrane of eukaryotes.

utilizes the mitochondrial membrane of prokaryotes.

generates a concentration gradient of protons.

requires a membrane AND generates a concentration gradient of protons.

Bloom's Level: 2. Understand
Learning Outcome: 06.13
Section: 06.04
Topic: Microbial Metabolism

IncorrectQuestion 23
0 / 1 pts
Which type of phosphorylation does not require a membrane?

Substrate-level phosphorylation

Oxidative phosphorylation

Photophosphorylation

All types of phosphorylation require a membrane.

Bloom's Level: 2. Understand
Learning Outcome: 06.13
Section: 06.04
Topic: Microbial Metabolism

Question 24
1 / 1 pts
Which pairing is incorrect?

A:T

G:C

A:U

A:G

Bloom's Level: 3. Apply
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 25
1 / 1 pts
RNA is characterized by which one of the following features?

Deoxyribose.

Thymine.

Ribose.

Double-stranded.

Bloom's Level: 1. Remember
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 26
1 / 1 pts
DNA is characterized by which of the following feature(s)?

Ribose.

Single-stranded.

Deoxyribose.

Thymine.

Deoxyribose AND thymine.

Bloom's Level: 1. Remember
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 27
1 / 1 pts
Which of the following is/are true of DNA replication?

It starts at the origin of replication.

Nucleotides are added to the 3' end.

It requires an RNA primer to get started.

It utilizes polymerases.

All of the choices are true.

Bloom's Level: 2. Understand
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

Question 28
1 / 1 pts
Which is true about the RNA transcript?

It is formed using the DNA minus strand as a template.

It has the same 5'-3' orientation as the DNA positive strand.

It is made in short fragments that are then stitched together.

The template starts at the promoter region.

It is formed using the DNA minus strand as a template, it has the same 5'-3' orientation as the DNA positive strand AND the template starts at the promoter region.

Bloom's Level: 2. Understand
Learning Outcome: 07.04
Section: 07.03
Topic: Genetics

IncorrectQuestion 29
0 / 1 pts
How many nucleotides are in a codon?

1

2

3

4

5

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

Question 30
1 / 1 pts
Which may be or is an RNA molecule?

AGCCTAC

GGGCCCA

GCCCUUA

AGCCTAC AND GGGCCCA

GGGCCCA AND GCCCUUA

Bloom's Level: 2. Understand
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

IncorrectQuestion 31
0 / 1 pts
Which molecule carries an anticodon?

DNA

mRNA

rRNA

tRNA

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

Question 32
1 / 1 pts
AUG

is only used as the start codon.

codes for methionine.

determines the reading frame.

is one of the stop codons.

codes for methionine AND determines the reading frame.

Bloom's Level: 2. Understand
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 33
0 / 1 pts
The A-site

is found on the RNA polymerase enzyme.

is found on the 30S ribosome.

is found on the 70S ribosome.

is the amino acid site.

is found on the 70S ribosome AND is the amino acid site.

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 34
0 / 1 pts
A stop codon

codes for the stop amino acid s-methyl-methionine.

forms a hairpin loop forcing the ribosome to fall off.

codes for no amino acid.

enhances the binding of the e-polymerase.

codes for the stop amino acid s-methyl-methionine AND forms a hairpin loop forcing the ribosome to fall off.

Bloom's Level: 2. Understand
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 35
0 / 1 pts
Some segments of the precursor mRNA in eukaryotes are non-coding and are called

exons.

introns.

integrans.

uselessans.

Bloom's Level: 1. Remember
Learning Outcome: 07.06
Section: 07.04
Topic: Genetics

Question 36
1 / 1 pts
Gene regulation may entail

turning on genes only when needed.

turning off genes when not needed.

turning on or off entire groups of genes.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

IncorrectQuestion 37
0 / 1 pts
The set of genes in bacteria that are linked together and transcribed as a single unit is referred to as a(n)

operon.

regulon.

operator.

repressor.

Bloom's Level: 1. Remember
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 38
1 / 1 pts
The molecules that bind to a repressor and cause it to no longer bind to the operator are called

activators.

repressors.

introns.

inducers.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 39
1 / 1 pts
Repressors

are involved in negative control.

are involved in positive control.

always bind to the promoter.

bind or do not bind to the operator depending on their shape (conformation).

are involved in negative control AND bind or do not bind to the operator depending on their shape (conformation).

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 40
1 / 1 pts
Activators

are involved in negative control.

are involved in positive control.

always bind to the promoter.

are allosteric proteins.

are involved in positive control AND are allosteric proteins.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 41
1 / 1 pts
Signal transduction

is the relay of information about conditions outside a cell to inside the cell.

often relies on a two component system.

may involve phosphorylation of various molecules.

is used by certain pathogens to sense low magnesium conditions.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.07
Section: 07.05
Topic: Genetics

IncorrectQuestion 42
0 / 1 pts

Quorum sensing

is used by bacteria to sense the density of cells.

is used by bacteria to sense the proximity of eukaryotic cells

is used by bacteria to limit the density of cells.

is used by eukaryotes to sense the presence of bacteria.

does not influence the expression of genes by a bacteria.

Quorum sensingBloom's Level: 2. UnderstandLearning Outcome: 07.07Section: 07.05Topic: Genetics

Question 43
1 / 1 pts
In DNA sequence analysis

the + strand of DNA is used.

the start codon is ATG.

ORFs are searched for.

codon usage is a helpful indicator for protein coding areas.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.12
Section: 07.08
Topic: Genetics

IncorrectQuestion 44
0 / 1 pts
DNA replication is usually unidirectional.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

IncorrectQuestion 45
0 / 1 pts
There are 4 functional types of RNA.

True

False

Bloom's Level: 1. Remember
Learning Outcome: 07.04
Section: 07.03
Topic: Genetics

Question 46
1 / 1 pts
Antisense RNA is the complement of the plus strand and may be useful in inhibiting translation.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 07.11
Section: 07.07
Topic: Genetics

IncorrectQuestion 47
0 / 1 pts
A drug that inhibits the activity of bacterial DNA gyrase ______ be a good antibiotic because _______.

would; it would inhibit all DNA synthesis

would NOT; it would inhibit all DNA synthesis, even our own cell's DNA synthesis

would; it would selectively inhibit bacterial DNA synthesis due to differences between bacterial/human gyrase

would NOT; the DNA replication machinery could still function, even with impaired DNA gyrase enzyme

Bloom's Level: 5. Evaluate
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

IncorrectQuestion 48
0 / 1 pts
The regulatory protein

binds to the promoter region of DNA.

may inhibit or enhance transcription.

may control translation of the operon.

affects the activity of the DNA polymerase.

binds to the promoter region of DNA AND affects the activity of the DNA polymerase.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 49
1 / 1 pts
The source of variation among microorganisms that were once identical is

antibiotic resistance.

virulence factors.

sigma factors.

mutation.

Bloom's Level: 1. Remember
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

Question 50
1 / 1 pts
The characteristics displayed by an organism in any given environment is its

genotype.

archaetype.

mutatotype.

phenotype.

Bloom's Level: 2. Understand
Learning Outcome: 08.01
Section: 08.01
Topic: Genetics

Question 51
1 / 1 pts
Transposons

are informally known as jumping genes.

may cause insertion mutations.

may cause knockout mutations.

were first recognized in plants.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 08.05
Section: 08.02
Topic: Genetics

Question 52
1 / 1 pts
The largest group of chemical mutagens consists of

radiation.

base analogs.

nitrous acid.

alkylating agents.

Bloom's Level: 1. Remember
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

Question 53
1 / 1 pts
Nitrous acid most frequently causes mutations by

substituting oxygen for hydrogen bonds.

converting keto groups to amino groups.

altering the bonding ratios of nitrogen bases.

converting cytosine to uracil.

Bloom's Level: 2. Understand
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

IncorrectQuestion 54
0 / 1 pts
Chemical mutagens that mimic the naturally occurring bases are called

nitrogen mustards.

alkylating agents.

base analogs.

nitrous oxide.

Bloom's Level: 1. Remember
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

IncorrectQuestion 55
0 / 1 pts
Irradiation of cells with ultraviolet light may cause

4 nucleotides to covalently bind together.

thymine dimers.

adenine complementary base pairing with cytosine.

the addition of uracil.

Bloom's Level: 2. Understand
Learning Outcome: 08.08
Section: 08.03
Topic: Genetics

Question 56
1 / 1 pts
Thymine dimers are dealt with by

no repair mechanisms.

photoreactivation repair.

SOS repair.

excision repair.

photoreactivation repair AND excision repair.

Bloom's Level: 2. Understand
Learning Outcome: 08.11
Section: 08.04
Topic: Genetics

IncorrectQuestion 57
0 / 1 pts
X-rays

have no effect on DNA.

cause thymine trimers.

cause single and double strand breaks in DNA molecules.

make the DNA radioactive.

Bloom's Level: 1. Remember
Learning Outcome: 08.08
Section: 08.03
Topic: Genetics

IncorrectQuestion 58
0 / 1 pts
DNA repair mechanisms occur

only in prokaryotes.

only in eukaryotes.

in both eukaryotes and prokaryotes.

in neither eukaryotes or prokaryotes.

Bloom's Level: 1. Remember
Learning Outcome: 08.09
Section: 08.04
Topic: Genetics

IncorrectQuestion 59
0 / 1 pts
Antibiotics

cause mutations to occur.

may act as alkylating mutagens.

provide an environment in which pre-existing mutants survive.

increase the rate of spontaneous mutation.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

IncorrectQuestion 60
0 / 1 pts
Prokaryotic cell mutations can be observed very quickly because the prokaryotic chromosome is

diploid.

polyploid.

haploid.

polysomal.

Bloom's Level: 2. Understand
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

Question 61
1 / 1 pts
The diploid character of eukaryotic cells may mask the appearance of a mutation since

this may be a frame shift.

the mutation is often reversible.

the mutation may be palindromic.

the matching chromosome may carry the dominant gene.

Bloom's Level: 2. Understand
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

IncorrectQuestion 62
0 / 1 pts
A clever technique that streamlines the identification of auxotrophic mutants is

gas chromatography.

replica plating.

direct selection.

reversion.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

IncorrectQuestion 63
0 / 1 pts
To increase the proportion of mutants in a population of bacteria one may use

direct selection.

replica plating.

penicillin enrichment.

individual transfer.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

Question 64
1 / 1 pts
Bacteria that have properties of both the donor and recipient cells are the result of

UV light.

SOS repair.

frame shift mutations.

genetic recombination.

Bloom's Level: 2. Understand
Learning Outcome: 08.13
Section: 08.06
Topic: Genetics

IncorrectQuestion 65
0 / 1 pts
The mechanism by which genes are transferred into bacteria via viruses is called

ellipsis.

replica plating.

transformation.

transduction.

conjugation.

Bloom's Level: 1. Remember
Learning Outcome: 08.14
Section: 08.07
Topic: Genetics

IncorrectQuestion 66
0 / 1 pts
In conjugation the donor cell is recognized by the presence of

an F plasmid.

a Y chromosome.

diploid chromosomes.

an SOS response.

an F plasmid AND diploid chromosomes.

Bloom's Level: 2. Understand
Learning Outcome: 08.15
Section: 08.08
Topic: Genetics

IncorrectQuestion 67
0 / 1 pts
Gene transfer that requires cell-to-cell contact is

transformation.

competency.

conjugation.

functional genomics.

Bloom's Level: 2. Understand
Learning Outcome: 08.15
Section: 08.08
Topic: Genetics

Question 68
1 / 1 pts
The transfer of vancomycin resistance from Enterococcus faecalis to Staphylococcus aureus is thought to have involved

conjugation.

transformation.

transduction.

transposons.

conjugation AND transposons.

Bloom's Level: 2. Understand
Learning Outcome: 08.17
Section: 08.09
Topic: Genetics

IncorrectQuestion 69
0 / 1 pts
Double-stranded DNA enters the recipient cell during transformation.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 08.13
Section: 08.06
Topic: Genetics

Question 70
1 / 1 pts
Plasmids often carry the information for antibiotic resistance.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 08.16
Section: 08.09
Topic: Genetics

Quiz Score: 31 out of 70

BIOL 112 Study Guide - Midterm Guide: Dna Ligase, Reaction Rate, Genome Size

BIOL 112 Full Course Notes

Document Summary

Get access

Related Documents

BIOL 112 Lecture Notes - Lecture 7: Sachet, Dna Polymerase I, Dna Mismatch Repair

BIOL 112 Study Guide - Final Guide: Nucleic Acid Double Helix, Dna Polymerase, Tata Box

BIOL 112 Chapter Notes - Chapter 6: Reaction Rate, Activation Energy, Common Source

Related Questions