Essential Biology | Comprehensive Study Guide, Exam Questions & Answers
(11/12 SW) Which is the most abundant phospholipid in phosphatidylcholine. ex) The most abundant lipids in cell membranes are the
animal cell membranes?Choose one: phospholipids, which have a phosphate-containing, hydrophilic head linked to a pair
phosphatidylinositol of hydrophobic, hydrocarbon tails. One of the most abundant phospholipids in the
glycolipid membranes of animals and plants, phosphatidylcholine, has the small molecule
phosphatidylcholine choline attached to a phosphate group as its hydrophilic head. Phosphatidylinositol
cholesterol is a minor component of the plasma membrane that is concentrated in the cytosolic
triacylglycerol half of the lipid bilayer, where it has a special role in relaying signals from the cell
surface to the cell interior. Glycolipids are also a fairly minor component of cell
membranes, where they are concentrated in the outer half of the bilayer. Glycolipids
are not phospholipids, as they have sugars as part of their hydrophilic heads.
Cholesterol is indeed abundant in the membranes of animal cells, but it is not a
phospholipid. And triacylglycerols are not phospholipids and are not found in cell
membranes. They are hydrophobic fats found in lipid droplets in the cell cytosol.
8 SW) Researchers have created plasmids that only allow A, D. Strong promoter, operator recognized by Lac repressor protein. ex) Effective
expression of inserted genes in response to a metabolite. transcription relies on strong binding of RNA polymerase to promoters. This can
Researchers can add these plasmids to E. coli cells and occur if the promoter itself is strong because it is a good match for the consensus
increase the expression of the inserted gene by adding the DNA sequences recognized by RNA polymerase or if RNA polymerase is recruited
appropriate metabolite to the culture media. Plasmids to a weak promoter by an activator binding to a regulatory site. Normally, robust
containing which combination of operator and promoter transcription from the Lac Operon requires both the presence of lactose (to remove
allow activation of gene expression in response to an repressor protein) and the binding of the CAP activator protein. The requirement for
added metabolite? the CAP activator can be overridden with a strong promoter.
Choose one or more:
A.operator recognized by Lac repressor protein
B.operator recognized by Trp repressor protein
C.weak promoter
D.strong promoter
SW) Which of the following statements most accurately The gene for the tryptophan repressor is expressed constitutively, which means it is
describes the expression of the repressor protein of the a gene that is transcribed at a relatively constant level regardless of the cell
tryptophan operon? Choose one: environmental conditions. Explain)The Trp operon contains the genes that code for
The gene for the tryptophan repressor is turned on in the necessary enzymes to synthesize the amino acid tryptophan from precursors. If
response to high levels of tryptophan in the cell. tryptophan is low in a cell, the operon is expressed so that the cell can synthesize
The gene for the tryptophan repressor is turned off in the tryptophan it needs. If tryptophan in the cell is high, then the operon is turned
response to high levels of tryptophan in the cell. off. The Trp repressor is a molecular "switch" that controls gene expression at the
The gene for the tryptophan repressor is turned on in Trp operon. If tryptophan binds to the Trp repressor, then the repressor adopts a
response to low levels of tryptophan in the cell. conformation that allows it to bind to the promoter region of the Trp operon and
The gene for the tryptophan repressor is turned off in prevent transcription. In the absence of tryptophan, the Trp repressor is inactivated
response to low levels of tryptophan in the cell. and it can no longer bind to the promoter, allowing transcription to take place. For
The gene for the tryptophan repressor is expressed the Trp repressor to be able to function as the molecular switch, it must always be
constitutively. present in the cell so that it can quickly respond to changes in the concentration of
tryptophan in the cell.
SW) Based on your understanding of DNA replication, E: Both daughter helices will have a gap at only one end. EX)After one round of
which statement best describes the daughter helices in a replication in the absence of telomerase, the daughter helices in a eukaryotic cell
eukaryotic cell after one round of replication has taken will have a gap at only one end. When the final primer at the end of the newly
place in the absence of telomerase? Choose one: synthesized lagging strand is removed, it cannot be replaced by DNA because there
A. A repair polymerase will replace all missing DNA, so is no already-synthesized DNA for polymerase to extend from. This leaves a gap on
both daughter helices will be the same size as the parent one end of each daughter helix. Repair polymerases also require primers, so this
helix. mechanism of DNA lesion repair cannot fix the "end replication problem." Without a
B. One daughter helix will have a gap at one end, while the special mechanism to replicate the ends of linear double helices, DNA will be lost
other daughter helix will be normal. during each round of cell division. Some cells use the activity of telomerase to
C. Both daughter helices will have gaps at both ends. extend the length of DNA so as to avoid sequence erosion due to the end
D. One daughter helix will have gaps at both ends, while replication problem. The image below shows one side of DNA replication. You can
the other daughter helix will be normal. mirror this information to reveal the other side, but be sure to maintain awareness of
E. Both daughter helices will have a gap at only one end. 5' and 3', leading and lagging.
Which of the following statements is/are true of long Two choices: They are involved in X chromosome inactivation AND they can
noncoding RNAs? Choose one or more: regulate the translation and stability of mRNAs. EX) Long noncoding RNAs are
They can silence genes by promoting the formation of involved in X chromosome inactivation and can regulate the translation and stability
euchromatin. of mRNAs. For example, Xist is a long noncoding RNA that remains nearby the X
They can trigger the activity of histone acetyltransferases. chromosome after transcription and recruits chromatin-remodeling enzymes that
They are involved in X chromosome inactivation. condenses one of the X chromosomes in females. Alternatively, long noncoding
They can regulate the translation and stability of mRNAs. RNAs can act as scaffolds by virtue of their ability fold upon themselves and
assume unique conformations. Indeed, the RNA in telomerase is believed to
function as a scaffold for protein subunits.
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, Essential Biology | Comprehensive Study Guide, Exam Questions & Answers
In bacterial cells, the tryptophan operon encodes the It activates the tryptophan repressor, which shuts down expression of the
genes needed to synthesize tryptophan. What happens tryptophan operon. EX) The tryptophan repressor is always present in the bacterial
when the concentration of tryptophan inside a cell is high? cell and its conformation depends on the presence or absence of the amino acid
Choose one: tryptophan. If tryptophan is present, it will bind to the repressor and this
It inactivates the tryptophan repressor, which shuts down conformation is able to bind to the operator sequence of the Trp operon to turn off
the tryptophan operon. transcription. If tryptophan is absent, the repressor protein is no longer able to bind
It activates the tryptophan repressor, which breaks down to the operator, which allows transcription to proceed. In this way, the concentration
excess tryptophan. of tryptophan provides feedback inhibition of the Trp operon, using the tryptophan
It activates the tryptophan repressor, which shuts down repressor as the on/off switch. The figure provides an overview of this regulation.
expression of the tryptophan operon.
It inactivates the tryptophan repressor, allowing
transcription of the tryptophan operon.
It activates the tryptophan repressor, allowing transcription
of the tryptophan operon.
Which of the following statements is not true about the A. EX) Specialized cells, such as liver and kidney cells, do contain the same genes,
differences between liver cells and kidney cells in the but they just express them differently. This is referred to as differential gene
same organism? Choose one: expression. We know that all cells in a human are the descendants of an initial cell
A. They contain different genes. that was created at fertilization, and so they must be genetically identical to one
B. They contain the entire set of instructions needed to another. Because most specialized cell types in a multicellular organism contain all
form the whole organism. the same genes that were present in the fertilized egg that gave rise to them, they
C. They express different genes. retain the instructions necessary to form the whole organism. Thus, cell
D. They contain different proteins. differentiation is instead achieved by changes in gene expression.
Using powerful new sequencing technologies, Gene B Only. Ex) Because gene B is expressed in both the liver and the brain, it
investigators can now catalog every RNA molecule made could encode a housekeeping protein, which is the description of a gene
by a cell and determine at what quantities these RNAs are product/protein that is found in all cells of a multicellular organism. Gene A, on the
present. In an experiment, researchers measured the other hand, is expressed in the brain but not in the liver, which is consistent with a
relative quantities of two different mRNAs—one pattern of differential expression. However, an examination of additional cell types
transcribed from gene A, the other from gene B—in two would be necessary to more conclusively support the prediction that gene B codes
different cell types. Gene B is expressed in both the liver for a housekeeping protein. RNA polymerase, which is necessary to synthesize
and the brain whereas gene A is expressed in the brain but RNA from a DNA template, is an example of a housekeeping protein, as all cells
not in the liver. Which most likely encodes a housekeeping utilize transcription to express genes.
protein? Choose one:
A. gene B. only
B. both genes
C. gene A only
D. neither gene
Which of the following statements concerning leucine A. Ex) Leucine zipper proteins dimerize through interactions between the leucine
zipper protein dimerization and DNA binding is correct? zipper domain α helices on two subunits. The helices contain hydrophobic side
Choose one: chains, including many leucine side chains, which pack tightly together. The α
A. Leucine zipper proteins function as a dimer with both helices extend around both sides of the DNA molecule. The polar side chains in
subunits making contact with the sequence-specific DNA these regions of the α helices make hydrogen bonds with specific bases in the DNA
site. major groove, leading to sequence-specific binding.
B. Leucine zipper proteins contain many leucine amino
acids in the DNA-binding region that facilitate sequence-
specific DNA binding.
C. Leucine zipper proteins use ionic bonds to bind with the
sequence-specific DNA site.
D. Leucine zipper protein dimerization is facilitated by polar
amino acids in the dimerization domains.
Which of the following mutations would be least likely to C. mutation of a leucine to valine in the dimerization domain of the protein. EX)
disrupt the function of the leucine zipper protein? Hint: Dimerization between the two subunits requires hydrophobic amino acids like
Consider which amino acids are hydrophobic. Choose one: leucine to pack tightly. Mutation of the hydrophobic leucine to another hydrophobic
A. mutation of a leucine to aspartic acid in the dimerization amino acid would be less disruptive than the other listed mutations. Changing the
domain of the protein hydrophobic leucine to a charged amino acid in the dimerization domain will very
B. mutation of the DNA-binding arginine to alanine in the likely disrupt dimerization, which is required for protein function. Changing the polar,
DNA-binding domain of the protein charged arginine to a hydrophobic alanine will likely disrupt the ability of the protein
C. mutation of a leucine to valine in the dimerization to form hydrogen bonds and bind to the DNA, again disrupting protein function.
domain of the protein Finally, altering the sequence of the binding site in the DNA will alter the bonds that
D. mutation of the guanine nucleotide to a thymine can be formed and disrupt the sequence-specific interaction between the protein
nucleotide in the binding site on the DNA and DNA, disrupting function.
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, Essential Biology | Comprehensive Study Guide, Exam Questions & Answers
In the first reaction of glycolysis (the pathway that begins A. -30.5 kJ/mole EX) The first reaction of glycolysis can be written: glucose + ATP
the oxidative breakdown of sugars), the enzyme → glucose 6-phosphate + ADPThe "reverse" reaction, which is a hydrolysis, can be
hexokinase uses ATP to catalyze the phosphorylation of written: glucose 6-phosphate + H2O → glucose + PiWhen these two reactions are
glucose to glucose 6-phosphate and ADP. The ΔG° of this "added" together, they produce: glucose + ATP + glucose 6-phosphate + H2O
reaction is a favorable -16.7 kJ/mole.Another sometimes →glucose 6-phosphate + ADP + glucose + PiBecause glucose and glucose 6-
active enzyme, called glucose 6-phosphatase, effectively phosphate are present on both sides of this equation, they cancel out, leaving:ATP
"reverses" this reaction, hydrolyzing glucose 6-phosphate + H2O → ADP + PiSo, the ΔG° of ATP hydrolysis can be calculated by simply
back to glucose and releasing a phosphate. The ΔG° of adding together the ΔG° values for the two reactions given:-16.7 kJ/mole + -13.8
this reaction is -13.8 kJ/mole.Based on these values, what kJ/mole = -30.5 kJ/mole.When canceling out the carbohydrate, you are in essence
is the ΔG° for the hydrolysis of ATP: ATP + H2O → ADP + adding the reactions together to achieve the net equation. When achieving the net
Pi? equation, you must add together the associated energetic values as well.
Choose one:
A. -30.5 kJ/mole
B. +30.5 kJ/mole
C. -2.9 kJ/mole
D. +61 kJ/mole
E. +2.9 kJ/mole
Adenine, a purine, has a N and NH2 to hydro bond with The integrity of the interaction would decrease because one of the two hydrogen
Tyrosine's NH and O-. Guanine, the other purine, would bonds would not be able to form. Ex) If guanine replaced adenine, the integrity of
have an NH and O- to bond with. What would happen to the protein-DNA interaction would decrease because one of the two hydrogen
the helix-3 interaction (a homeodomain with 3 overlapping bonds would not be able to form. This is because adenine contains an amino group
a-helixes) with DNA if a mutation occurred that altered this hydrogen atom that interacts with an asparagine carbonyl oxygen atom; however,
adenine to guanine? this interaction would not form if guanine were present in place of adenine. If
Choose one: guanine were present, there would actually be a repulsion between the partially
A. The integrity of the interaction would increase because negatively charged carbonyl group oxygen of the asparagine residue and the
the hydrogen bonds would still be able to form and would partially negatively charged carbonyl group oxygen of the guanine nitrogenous
be stronger. base. Hydrogen bonds form when there is a favorable interaction between a partial
B. The integrity of the interaction would remain the same negative charge (from electronegative atoms such as O or N) and a partial positive
because the same two hydrogen bonds would still be able charge, such as a hydrogen atom covalently bonded to a more electronegative
to form. atom.
C. The integrity of the interaction would decrease because
one of the two hydrogen bonds would not be able to form.
The maltose operon contains genes that code for proteins C and E. Ex) There are different ways to accomplish operon induction: (1) the sugar
that catabolize the disaccharide maltose. Similar to the Lac inducer can remove a repressor from an operon with an efficient promoter
operon, which is only efficiently transcribed in the presence (releasing the brakes, in a sense), or (2) the inducer can cause an activator to bind
of lactose, the maltose operon is only efficiently to an operon with an inefficient promoter (stepping on the gas pedal). Cells use both
transcribed in the presence of maltose. How might of these mechanisms; for example, lactose removes a repressor. While in theory the
induction of the maltose operon in response to maltose be maltose operon could be regulated by either method, in reality maltose causes an
achieved?Choose one or more: activator to bind, which increases the affinity for RNA polymerase at the otherwise
A.Maltose removes an activator from an operon with an inefficient promoter.
inefficient promoter.
B.Maltose causes an activator to bind an operon with an
efficient promoter.
C.Maltose causes an activator to bind an operon with an
inefficient promoter.
D.Maltose causes a repressor to bind an operon with an
efficient promoter.
E.Maltose removes a repressor from an operon with an
efficient promoter.
For which of the following is Lac operon on? D. (Simple Ex) Glucose is preferred, so it's not going to switch to lactose unless
Q. + Glucose + Lactose glucose is gone AND lactose is present. More complicated ex) When lactose is
B. - Glucose - Lactose absent, the Lac repressor binds to the Lac operator and shuts off expression of the
C. + Glucose - Lactose operon. Addition of lactose increases the intracellular concentration of a related
D. - glucose, + lactose, compound, allolactose; allolactose binds to the Lac repressor, causing it to undergo
a conformational change that releases its grip on the operator DNA. When glucose
is absent, cyclic AMP (red circle) is produced by the cell, and CAP binds to DNA.
For the operon to be transcribed, glucose must be absent (allowing the CAP
activator to bind) and lactose must be present (releasing the Lac repressor). LacZ,
the first gene of the operon, encodes the enzyme β-galactosidase, which breaks
down lactose to galactose and glucose
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, Essential Biology | Comprehensive Study Guide, Exam Questions & Answers
Which of the following describes the Lac operon in E. coli E. Ex) The expression of the Lac operon is tightly regulated by the type of
when lactose, but not glucose, is present in the culture carbohydrate present in the environment. Glucose is the preferred carbohydrate
medium?Choose one: source and only in its absence will the necessary small molecule, cAMP, be found.
A. The Lac repressor, but not CAP, is bound to the Lac cAMP binding to the CAP activator is necessary but not sufficient for the expression
operon's regulatory DNA, and the Lac operon is not of the Lac operon. In addition to low glucose levels, there must also be lactose
expressed. present in the environment. If this is the case, allolactose, a derivative of lactose,
B. Neither CAP nor the Lac repressor is bound to the Lac can bind to the Lac repressor and cause it to release from the promoter. Therefore,
operon's regulatory DNA, and the Lac operon is as outlined in the figure below, the Lac operon is only expressed when glucose is
expressed. absent and lactose is present because the CAP activator is bound to the promoter,
C. CAP and the Lac repressor are both bound to the Lac but the Lac repressor is not. This ensures that the genes in the Lac operon are
operon's regulatory DNA, and the Lac operon is not expressed only in the conditions where lactose is the only carbohydrate source
expressed. available.
D. Neither CAP nor the Lac repressor is bound to the Lac
operon's regulatory DNA, and the Lac operon is not
expressed.
E. CAP, but not the Lac repressor, is bound to the Lac
operon's regulatory DNA, and the Lac operon is
expressed.
c-Met is an oncogene that contributes to the development C. Ex) In their 2009 article (MicroRNA-1/206 Targets c-Met and Inhibits
of certain cancers by triggering cell division and tumor Rhabdomyosarcoma Development), Yan and colleagues found an inverse
growth. In a 2009 article, Yan and colleagues found correlation between microRNA-1/206 and c-Met protein levels. MicroRNAs are
regions in the 3' untranslated region of c-Met mRNA noncoding regulatory RNAs that down-regulate gene expression by binding
complementary to microRNA-1/206. In addition, higher complementary mRNAs and destroying them via the RNA-induced silencing
levels of microRNA-1/206 were associated with slower cell complex (RISC). Researchers use double-stranded RNAs to prevent the expression
proliferation. What is a likely explanation for the inverse of selected genes, a technique known as RNA interference.
correlation between microRNA-1/206 and cell
proliferation?Choose one:
A. MicroRNA-1/206 codes for a protein that directly binds
to and inhibits c-Met protein.
B. MicroRNA-1/206 stabilizes c-Met mRNA, leading to
enhanced translation.
C. MicroRNA-1/206 targets c-Met mRNA for destruction via
RISC.
D. MicroRNA-1/206 codes for a tumor suppressor protein
that directly inhibits cell proliferation.
Researchers assayed the activity of enzyme F in three B, C, AND D. Ex) There are seven steps from DNA to active protein and gene
different types of tissue from the same mouse by expression can be controlled at any of these. C, 'Transcriptional Control', is quite
determining the amount of enzyme product produced per common and is the most energy efficient, but it takes the longest amount of time to
milligram of tissue per unit time. As shown in the graph produce an active protein. In contrast, D, 'Protein Activity Control', can quickly lead
below (Graph depicts much higher activity in the liver, but to active protein but is more energy intensive. B, is 'Translation Control'
not the kidney or muscle cells.) results indicate more
product generation in the liver compared to the kidney and
muscle samples.Which of the following factors might
explain the different results among the three tissues?
Choose one or more:
A.differences in the DNA content among the tissue types
B.differences in the post-translational modifications of the
enzyme among the tissue types
C.differences in the transcription of the gene encoding the
enzyme among the tissue types
D.differences in the translation of the mRNA encoding the
protein among the tissue types
What problem with replication of linear chromosomes does C. Ex) Although the leading strand can be replicated all the way to the chromosome
telomerase address?Choose one: tip, the lagging strand cannot. When the final RNA primer on the lagging strand is
A. The leading strand continues farther during replication, removed, there is no enzyme that can replace it with dna. But that a strategy to deal
so chromosomes would lengthen in each replication cycle with this problem a lagging strand would become shorter with each round of DNA
without telomerase. replication and, after repeated cell divisions, the chromosomes themselves would
B. The leading strand stops short of the 5' end during shrink eventually losing valuable genetic information. A bacteria avoid this end
replication, so chromosomes would shorten in each replication problem by having circular DNA molecules. Eukaryotes get around it by
replication cycle without telomerase. adding long repetitive nucleotide sequences to the ends of every chromosome.
C. The lagging strand stops short of the 3' end during These sequences which are incorporated into structures called telomeres attract an
replication, so chromosomes would shorten in each enzyme called telemerase to the chromosomes. Telomerase carries its own RNA
replication cycle without telomerase. template but it uses to add multiple copies of the same repetitive DNA sequence to
D. The lagging strand continues farther during replication, the lagging strand template.
so chromosomes would lengthen in each replication cycle
without telomerase.
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(11/12 SW) Which is the most abundant phospholipid in phosphatidylcholine. ex) The most abundant lipids in cell membranes are the
animal cell membranes?Choose one: phospholipids, which have a phosphate-containing, hydrophilic head linked to a pair
phosphatidylinositol of hydrophobic, hydrocarbon tails. One of the most abundant phospholipids in the
glycolipid membranes of animals and plants, phosphatidylcholine, has the small molecule
phosphatidylcholine choline attached to a phosphate group as its hydrophilic head. Phosphatidylinositol
cholesterol is a minor component of the plasma membrane that is concentrated in the cytosolic
triacylglycerol half of the lipid bilayer, where it has a special role in relaying signals from the cell
surface to the cell interior. Glycolipids are also a fairly minor component of cell
membranes, where they are concentrated in the outer half of the bilayer. Glycolipids
are not phospholipids, as they have sugars as part of their hydrophilic heads.
Cholesterol is indeed abundant in the membranes of animal cells, but it is not a
phospholipid. And triacylglycerols are not phospholipids and are not found in cell
membranes. They are hydrophobic fats found in lipid droplets in the cell cytosol.
8 SW) Researchers have created plasmids that only allow A, D. Strong promoter, operator recognized by Lac repressor protein. ex) Effective
expression of inserted genes in response to a metabolite. transcription relies on strong binding of RNA polymerase to promoters. This can
Researchers can add these plasmids to E. coli cells and occur if the promoter itself is strong because it is a good match for the consensus
increase the expression of the inserted gene by adding the DNA sequences recognized by RNA polymerase or if RNA polymerase is recruited
appropriate metabolite to the culture media. Plasmids to a weak promoter by an activator binding to a regulatory site. Normally, robust
containing which combination of operator and promoter transcription from the Lac Operon requires both the presence of lactose (to remove
allow activation of gene expression in response to an repressor protein) and the binding of the CAP activator protein. The requirement for
added metabolite? the CAP activator can be overridden with a strong promoter.
Choose one or more:
A.operator recognized by Lac repressor protein
B.operator recognized by Trp repressor protein
C.weak promoter
D.strong promoter
SW) Which of the following statements most accurately The gene for the tryptophan repressor is expressed constitutively, which means it is
describes the expression of the repressor protein of the a gene that is transcribed at a relatively constant level regardless of the cell
tryptophan operon? Choose one: environmental conditions. Explain)The Trp operon contains the genes that code for
The gene for the tryptophan repressor is turned on in the necessary enzymes to synthesize the amino acid tryptophan from precursors. If
response to high levels of tryptophan in the cell. tryptophan is low in a cell, the operon is expressed so that the cell can synthesize
The gene for the tryptophan repressor is turned off in the tryptophan it needs. If tryptophan in the cell is high, then the operon is turned
response to high levels of tryptophan in the cell. off. The Trp repressor is a molecular "switch" that controls gene expression at the
The gene for the tryptophan repressor is turned on in Trp operon. If tryptophan binds to the Trp repressor, then the repressor adopts a
response to low levels of tryptophan in the cell. conformation that allows it to bind to the promoter region of the Trp operon and
The gene for the tryptophan repressor is turned off in prevent transcription. In the absence of tryptophan, the Trp repressor is inactivated
response to low levels of tryptophan in the cell. and it can no longer bind to the promoter, allowing transcription to take place. For
The gene for the tryptophan repressor is expressed the Trp repressor to be able to function as the molecular switch, it must always be
constitutively. present in the cell so that it can quickly respond to changes in the concentration of
tryptophan in the cell.
SW) Based on your understanding of DNA replication, E: Both daughter helices will have a gap at only one end. EX)After one round of
which statement best describes the daughter helices in a replication in the absence of telomerase, the daughter helices in a eukaryotic cell
eukaryotic cell after one round of replication has taken will have a gap at only one end. When the final primer at the end of the newly
place in the absence of telomerase? Choose one: synthesized lagging strand is removed, it cannot be replaced by DNA because there
A. A repair polymerase will replace all missing DNA, so is no already-synthesized DNA for polymerase to extend from. This leaves a gap on
both daughter helices will be the same size as the parent one end of each daughter helix. Repair polymerases also require primers, so this
helix. mechanism of DNA lesion repair cannot fix the "end replication problem." Without a
B. One daughter helix will have a gap at one end, while the special mechanism to replicate the ends of linear double helices, DNA will be lost
other daughter helix will be normal. during each round of cell division. Some cells use the activity of telomerase to
C. Both daughter helices will have gaps at both ends. extend the length of DNA so as to avoid sequence erosion due to the end
D. One daughter helix will have gaps at both ends, while replication problem. The image below shows one side of DNA replication. You can
the other daughter helix will be normal. mirror this information to reveal the other side, but be sure to maintain awareness of
E. Both daughter helices will have a gap at only one end. 5' and 3', leading and lagging.
Which of the following statements is/are true of long Two choices: They are involved in X chromosome inactivation AND they can
noncoding RNAs? Choose one or more: regulate the translation and stability of mRNAs. EX) Long noncoding RNAs are
They can silence genes by promoting the formation of involved in X chromosome inactivation and can regulate the translation and stability
euchromatin. of mRNAs. For example, Xist is a long noncoding RNA that remains nearby the X
They can trigger the activity of histone acetyltransferases. chromosome after transcription and recruits chromatin-remodeling enzymes that
They are involved in X chromosome inactivation. condenses one of the X chromosomes in females. Alternatively, long noncoding
They can regulate the translation and stability of mRNAs. RNAs can act as scaffolds by virtue of their ability fold upon themselves and
assume unique conformations. Indeed, the RNA in telomerase is believed to
function as a scaffold for protein subunits.
Stuvia 2026-2027
, Essential Biology | Comprehensive Study Guide, Exam Questions & Answers
In bacterial cells, the tryptophan operon encodes the It activates the tryptophan repressor, which shuts down expression of the
genes needed to synthesize tryptophan. What happens tryptophan operon. EX) The tryptophan repressor is always present in the bacterial
when the concentration of tryptophan inside a cell is high? cell and its conformation depends on the presence or absence of the amino acid
Choose one: tryptophan. If tryptophan is present, it will bind to the repressor and this
It inactivates the tryptophan repressor, which shuts down conformation is able to bind to the operator sequence of the Trp operon to turn off
the tryptophan operon. transcription. If tryptophan is absent, the repressor protein is no longer able to bind
It activates the tryptophan repressor, which breaks down to the operator, which allows transcription to proceed. In this way, the concentration
excess tryptophan. of tryptophan provides feedback inhibition of the Trp operon, using the tryptophan
It activates the tryptophan repressor, which shuts down repressor as the on/off switch. The figure provides an overview of this regulation.
expression of the tryptophan operon.
It inactivates the tryptophan repressor, allowing
transcription of the tryptophan operon.
It activates the tryptophan repressor, allowing transcription
of the tryptophan operon.
Which of the following statements is not true about the A. EX) Specialized cells, such as liver and kidney cells, do contain the same genes,
differences between liver cells and kidney cells in the but they just express them differently. This is referred to as differential gene
same organism? Choose one: expression. We know that all cells in a human are the descendants of an initial cell
A. They contain different genes. that was created at fertilization, and so they must be genetically identical to one
B. They contain the entire set of instructions needed to another. Because most specialized cell types in a multicellular organism contain all
form the whole organism. the same genes that were present in the fertilized egg that gave rise to them, they
C. They express different genes. retain the instructions necessary to form the whole organism. Thus, cell
D. They contain different proteins. differentiation is instead achieved by changes in gene expression.
Using powerful new sequencing technologies, Gene B Only. Ex) Because gene B is expressed in both the liver and the brain, it
investigators can now catalog every RNA molecule made could encode a housekeeping protein, which is the description of a gene
by a cell and determine at what quantities these RNAs are product/protein that is found in all cells of a multicellular organism. Gene A, on the
present. In an experiment, researchers measured the other hand, is expressed in the brain but not in the liver, which is consistent with a
relative quantities of two different mRNAs—one pattern of differential expression. However, an examination of additional cell types
transcribed from gene A, the other from gene B—in two would be necessary to more conclusively support the prediction that gene B codes
different cell types. Gene B is expressed in both the liver for a housekeeping protein. RNA polymerase, which is necessary to synthesize
and the brain whereas gene A is expressed in the brain but RNA from a DNA template, is an example of a housekeeping protein, as all cells
not in the liver. Which most likely encodes a housekeeping utilize transcription to express genes.
protein? Choose one:
A. gene B. only
B. both genes
C. gene A only
D. neither gene
Which of the following statements concerning leucine A. Ex) Leucine zipper proteins dimerize through interactions between the leucine
zipper protein dimerization and DNA binding is correct? zipper domain α helices on two subunits. The helices contain hydrophobic side
Choose one: chains, including many leucine side chains, which pack tightly together. The α
A. Leucine zipper proteins function as a dimer with both helices extend around both sides of the DNA molecule. The polar side chains in
subunits making contact with the sequence-specific DNA these regions of the α helices make hydrogen bonds with specific bases in the DNA
site. major groove, leading to sequence-specific binding.
B. Leucine zipper proteins contain many leucine amino
acids in the DNA-binding region that facilitate sequence-
specific DNA binding.
C. Leucine zipper proteins use ionic bonds to bind with the
sequence-specific DNA site.
D. Leucine zipper protein dimerization is facilitated by polar
amino acids in the dimerization domains.
Which of the following mutations would be least likely to C. mutation of a leucine to valine in the dimerization domain of the protein. EX)
disrupt the function of the leucine zipper protein? Hint: Dimerization between the two subunits requires hydrophobic amino acids like
Consider which amino acids are hydrophobic. Choose one: leucine to pack tightly. Mutation of the hydrophobic leucine to another hydrophobic
A. mutation of a leucine to aspartic acid in the dimerization amino acid would be less disruptive than the other listed mutations. Changing the
domain of the protein hydrophobic leucine to a charged amino acid in the dimerization domain will very
B. mutation of the DNA-binding arginine to alanine in the likely disrupt dimerization, which is required for protein function. Changing the polar,
DNA-binding domain of the protein charged arginine to a hydrophobic alanine will likely disrupt the ability of the protein
C. mutation of a leucine to valine in the dimerization to form hydrogen bonds and bind to the DNA, again disrupting protein function.
domain of the protein Finally, altering the sequence of the binding site in the DNA will alter the bonds that
D. mutation of the guanine nucleotide to a thymine can be formed and disrupt the sequence-specific interaction between the protein
nucleotide in the binding site on the DNA and DNA, disrupting function.
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In the first reaction of glycolysis (the pathway that begins A. -30.5 kJ/mole EX) The first reaction of glycolysis can be written: glucose + ATP
the oxidative breakdown of sugars), the enzyme → glucose 6-phosphate + ADPThe "reverse" reaction, which is a hydrolysis, can be
hexokinase uses ATP to catalyze the phosphorylation of written: glucose 6-phosphate + H2O → glucose + PiWhen these two reactions are
glucose to glucose 6-phosphate and ADP. The ΔG° of this "added" together, they produce: glucose + ATP + glucose 6-phosphate + H2O
reaction is a favorable -16.7 kJ/mole.Another sometimes →glucose 6-phosphate + ADP + glucose + PiBecause glucose and glucose 6-
active enzyme, called glucose 6-phosphatase, effectively phosphate are present on both sides of this equation, they cancel out, leaving:ATP
"reverses" this reaction, hydrolyzing glucose 6-phosphate + H2O → ADP + PiSo, the ΔG° of ATP hydrolysis can be calculated by simply
back to glucose and releasing a phosphate. The ΔG° of adding together the ΔG° values for the two reactions given:-16.7 kJ/mole + -13.8
this reaction is -13.8 kJ/mole.Based on these values, what kJ/mole = -30.5 kJ/mole.When canceling out the carbohydrate, you are in essence
is the ΔG° for the hydrolysis of ATP: ATP + H2O → ADP + adding the reactions together to achieve the net equation. When achieving the net
Pi? equation, you must add together the associated energetic values as well.
Choose one:
A. -30.5 kJ/mole
B. +30.5 kJ/mole
C. -2.9 kJ/mole
D. +61 kJ/mole
E. +2.9 kJ/mole
Adenine, a purine, has a N and NH2 to hydro bond with The integrity of the interaction would decrease because one of the two hydrogen
Tyrosine's NH and O-. Guanine, the other purine, would bonds would not be able to form. Ex) If guanine replaced adenine, the integrity of
have an NH and O- to bond with. What would happen to the protein-DNA interaction would decrease because one of the two hydrogen
the helix-3 interaction (a homeodomain with 3 overlapping bonds would not be able to form. This is because adenine contains an amino group
a-helixes) with DNA if a mutation occurred that altered this hydrogen atom that interacts with an asparagine carbonyl oxygen atom; however,
adenine to guanine? this interaction would not form if guanine were present in place of adenine. If
Choose one: guanine were present, there would actually be a repulsion between the partially
A. The integrity of the interaction would increase because negatively charged carbonyl group oxygen of the asparagine residue and the
the hydrogen bonds would still be able to form and would partially negatively charged carbonyl group oxygen of the guanine nitrogenous
be stronger. base. Hydrogen bonds form when there is a favorable interaction between a partial
B. The integrity of the interaction would remain the same negative charge (from electronegative atoms such as O or N) and a partial positive
because the same two hydrogen bonds would still be able charge, such as a hydrogen atom covalently bonded to a more electronegative
to form. atom.
C. The integrity of the interaction would decrease because
one of the two hydrogen bonds would not be able to form.
The maltose operon contains genes that code for proteins C and E. Ex) There are different ways to accomplish operon induction: (1) the sugar
that catabolize the disaccharide maltose. Similar to the Lac inducer can remove a repressor from an operon with an efficient promoter
operon, which is only efficiently transcribed in the presence (releasing the brakes, in a sense), or (2) the inducer can cause an activator to bind
of lactose, the maltose operon is only efficiently to an operon with an inefficient promoter (stepping on the gas pedal). Cells use both
transcribed in the presence of maltose. How might of these mechanisms; for example, lactose removes a repressor. While in theory the
induction of the maltose operon in response to maltose be maltose operon could be regulated by either method, in reality maltose causes an
achieved?Choose one or more: activator to bind, which increases the affinity for RNA polymerase at the otherwise
A.Maltose removes an activator from an operon with an inefficient promoter.
inefficient promoter.
B.Maltose causes an activator to bind an operon with an
efficient promoter.
C.Maltose causes an activator to bind an operon with an
inefficient promoter.
D.Maltose causes a repressor to bind an operon with an
efficient promoter.
E.Maltose removes a repressor from an operon with an
efficient promoter.
For which of the following is Lac operon on? D. (Simple Ex) Glucose is preferred, so it's not going to switch to lactose unless
Q. + Glucose + Lactose glucose is gone AND lactose is present. More complicated ex) When lactose is
B. - Glucose - Lactose absent, the Lac repressor binds to the Lac operator and shuts off expression of the
C. + Glucose - Lactose operon. Addition of lactose increases the intracellular concentration of a related
D. - glucose, + lactose, compound, allolactose; allolactose binds to the Lac repressor, causing it to undergo
a conformational change that releases its grip on the operator DNA. When glucose
is absent, cyclic AMP (red circle) is produced by the cell, and CAP binds to DNA.
For the operon to be transcribed, glucose must be absent (allowing the CAP
activator to bind) and lactose must be present (releasing the Lac repressor). LacZ,
the first gene of the operon, encodes the enzyme β-galactosidase, which breaks
down lactose to galactose and glucose
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Which of the following describes the Lac operon in E. coli E. Ex) The expression of the Lac operon is tightly regulated by the type of
when lactose, but not glucose, is present in the culture carbohydrate present in the environment. Glucose is the preferred carbohydrate
medium?Choose one: source and only in its absence will the necessary small molecule, cAMP, be found.
A. The Lac repressor, but not CAP, is bound to the Lac cAMP binding to the CAP activator is necessary but not sufficient for the expression
operon's regulatory DNA, and the Lac operon is not of the Lac operon. In addition to low glucose levels, there must also be lactose
expressed. present in the environment. If this is the case, allolactose, a derivative of lactose,
B. Neither CAP nor the Lac repressor is bound to the Lac can bind to the Lac repressor and cause it to release from the promoter. Therefore,
operon's regulatory DNA, and the Lac operon is as outlined in the figure below, the Lac operon is only expressed when glucose is
expressed. absent and lactose is present because the CAP activator is bound to the promoter,
C. CAP and the Lac repressor are both bound to the Lac but the Lac repressor is not. This ensures that the genes in the Lac operon are
operon's regulatory DNA, and the Lac operon is not expressed only in the conditions where lactose is the only carbohydrate source
expressed. available.
D. Neither CAP nor the Lac repressor is bound to the Lac
operon's regulatory DNA, and the Lac operon is not
expressed.
E. CAP, but not the Lac repressor, is bound to the Lac
operon's regulatory DNA, and the Lac operon is
expressed.
c-Met is an oncogene that contributes to the development C. Ex) In their 2009 article (MicroRNA-1/206 Targets c-Met and Inhibits
of certain cancers by triggering cell division and tumor Rhabdomyosarcoma Development), Yan and colleagues found an inverse
growth. In a 2009 article, Yan and colleagues found correlation between microRNA-1/206 and c-Met protein levels. MicroRNAs are
regions in the 3' untranslated region of c-Met mRNA noncoding regulatory RNAs that down-regulate gene expression by binding
complementary to microRNA-1/206. In addition, higher complementary mRNAs and destroying them via the RNA-induced silencing
levels of microRNA-1/206 were associated with slower cell complex (RISC). Researchers use double-stranded RNAs to prevent the expression
proliferation. What is a likely explanation for the inverse of selected genes, a technique known as RNA interference.
correlation between microRNA-1/206 and cell
proliferation?Choose one:
A. MicroRNA-1/206 codes for a protein that directly binds
to and inhibits c-Met protein.
B. MicroRNA-1/206 stabilizes c-Met mRNA, leading to
enhanced translation.
C. MicroRNA-1/206 targets c-Met mRNA for destruction via
RISC.
D. MicroRNA-1/206 codes for a tumor suppressor protein
that directly inhibits cell proliferation.
Researchers assayed the activity of enzyme F in three B, C, AND D. Ex) There are seven steps from DNA to active protein and gene
different types of tissue from the same mouse by expression can be controlled at any of these. C, 'Transcriptional Control', is quite
determining the amount of enzyme product produced per common and is the most energy efficient, but it takes the longest amount of time to
milligram of tissue per unit time. As shown in the graph produce an active protein. In contrast, D, 'Protein Activity Control', can quickly lead
below (Graph depicts much higher activity in the liver, but to active protein but is more energy intensive. B, is 'Translation Control'
not the kidney or muscle cells.) results indicate more
product generation in the liver compared to the kidney and
muscle samples.Which of the following factors might
explain the different results among the three tissues?
Choose one or more:
A.differences in the DNA content among the tissue types
B.differences in the post-translational modifications of the
enzyme among the tissue types
C.differences in the transcription of the gene encoding the
enzyme among the tissue types
D.differences in the translation of the mRNA encoding the
protein among the tissue types
What problem with replication of linear chromosomes does C. Ex) Although the leading strand can be replicated all the way to the chromosome
telomerase address?Choose one: tip, the lagging strand cannot. When the final RNA primer on the lagging strand is
A. The leading strand continues farther during replication, removed, there is no enzyme that can replace it with dna. But that a strategy to deal
so chromosomes would lengthen in each replication cycle with this problem a lagging strand would become shorter with each round of DNA
without telomerase. replication and, after repeated cell divisions, the chromosomes themselves would
B. The leading strand stops short of the 5' end during shrink eventually losing valuable genetic information. A bacteria avoid this end
replication, so chromosomes would shorten in each replication problem by having circular DNA molecules. Eukaryotes get around it by
replication cycle without telomerase. adding long repetitive nucleotide sequences to the ends of every chromosome.
C. The lagging strand stops short of the 3' end during These sequences which are incorporated into structures called telomeres attract an
replication, so chromosomes would shorten in each enzyme called telemerase to the chromosomes. Telomerase carries its own RNA
replication cycle without telomerase. template but it uses to add multiple copies of the same repetitive DNA sequence to
D. The lagging strand continues farther during replication, the lagging strand template.
so chromosomes would lengthen in each replication cycle
without telomerase.
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