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16.3: Gene Regulation in Prokaryotes - Biology

16.3: Gene Regulation in Prokaryotes - Biology


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Many prokaryotic genes are organized in operons, linked genes transcribed into a single mRNA encoding two or more proteins. Regulating the activity of an operon (rather than multiple single genes encoding single proteins) allows better coordination of the synthesis of several proteins at once. In E. coli, the regulated lac operon encodes three enzymes involved in the metabolism of lactose (an alternative nutrient to glucose). Regulation of an operon (or of a single gene for that matter) can be by repression or by induction. When a small metabolite in a cell binds to a regulatory repressor or inducer protein, the protein undergoes an allosteric change that allows it to bind to a regulatory DNA sequence…, or to un-bind from the DNA. We will see examples of such regulation in the lac and trp operons. Lac operon gene regulation is an example of gene repression as well as induction. Trp (tryptophan) operon regulation is by gene repression. In both the operons, changes in levels of intracellular metabolites reflect the metabolic status of the cell and elicit appropriate changes in gene transcription. We will look at the regulation of both operons.

The mRNA transcribed from the lac operon is simultaneously translated into those three enzymes, as shown below.

A. Mechanisms of Control of the Lac Operon

In the animal digestive tract (including ours), genes of the E. coli lac operon regulate the use of lactose as an alternative nutrient to glucose. Think cheese instead of chocolate! The operon consists of lacZ, lacY, and lacA genes that were called structural genes. By definition, structural genes encode proteins that participate in cell structure and metabolic function. As already noted, the lac operon is transcribed into an mRNA encoding the Z, Y and A proteins.

Let’s take a closer look at the structure of the lac operon and the function of the Y, Z and A proteins (below).

The lacZ gene encodes β-galactosidase, the enzyme that breaks lactose (a disaccharide) into galactose and glucose. The lacY gene encodes lactose permease, a membrane protein that facilitates lactose entry into the cells. The role of the lacA gene (a transacetylase) in lactose energy metabolism is not well understood. The I gene to the left of the lac Z gene is a regulatory gene (to distinguish it from structural genes). Regulatory genes encode proteins that interact with regulatory DNA sequences associated with a gene to control transcription. The operator sequence separating the I and Z genes is a transcription regulatory DNA sequence.

The E. coli lac operon is usually silent (repressed) because these cells prefer glucose as an energy and carbon source. In the presence of sufficient glucose, a repressor protein (the I gene product) is bound to the operator, blocking transcription of the lac operon. Even if lactose is available, cells will not be use it as an alternative energy and carbon source when glucose levels adequate. However, when glucose levels drop, the lac operon is active and the three enzyme products are translated. We will see how limiting glucose levels induce maximal lac operon transcription by both derepression and direct induction, leading to maximal transcription of the lac genes only when necessary (i.e., in the presence of lactose and absence of glucose). Let’s look at some of the classic experiments that led to our understanding of E. coli gene regulation in general, and of the lac operon in particular.

In the late 1950s and early 1960s, Francois Jacob and Jacques Monod were studying the use of different sugars as carbon sources by E. coli. They knew that wild type E. coli would not make the (eta )-galactosidase, (eta )-galactoside permease or (eta )-galactoside transacetylase proteins when grown on glucose. Of course, they also knew that the cells would switch to lactose for growth and reproduction if they were deprived of glucose! They then searched for and isolated different E. coli mutants that could not grow on lactose, even when there was no glucose in the growth medium. Here are some of the mutants they studied:

  1. One mutant failed to make active (eta )-galactosidase enzyme but made permease.
  2. One mutant failed to make active permease but made normal amounts of (eta )-galactosidase.
  3. Another mutant failed to make transacetylase..., but could still metabolize lactose in the absence of glucose. Hence the uncertainty of its role in lactose metabolism.
  4. Curiosly, one mutant strain failed to make any of the three enzymes!

Since double mutants are very rare and triple mutants even rarer, Jacob and Monod inferred that the activation of all three genes in the presence of lactose were controlled together in some way. In fact, it was this discovery that defined the operon as a set of genes transcribed as a single mRNA, whose expression could therefore be effectively coordinated. They later characterized the repressor protein produced by the lacI gene. Jacob, Monod and Andre Lwoff shared the Nobel Prize in Medicine in 1965 for their work on bacterial gene regulation. We now know that negative and positive regulation of the lac operon (described below) depend on two regulatory proteins that together, control the rate of lactose metabolism.

1. Negative Regulation of the Lac Operon by Lactose

Refer to the illustration below to identify the players in lac operon derepression.

The repressor protein product of the I gene is always made and present in E. coli cells. I gene expression is not regulated! In the absence of lactose in the growth medium, the repressor protein binds tightly to the operator DNA. While RNA polymerase is bound to the promoter and ready to transcribe the operon, the presence of the repressor bound to the operator sequence close to the Z gene physically blocks its forward movement. Under these conditions, little or no transcript is made. If cells are grown in the presence of lactose, the lactose entering the cells is converted to allolactose. Allolactose binds to the repressor sitting on the operator DNA to form a 2-part complex, as shown below.

The allosterically altered repressor dissociates from the operator and RNA polymerase can transcribe the lac operon genes as illustrated below

2. Positive Regulation of the Lac Operon; Induction by Catabolite Activation

The second control mechanism regulating lac operon expression is mediated by CAP (cAMP-bound catabolite activator protein or cAMP receptor protein). When glucose is available, cellular levels of cAMP are low in the cells and CAP is in an inactive conformation. On the other hand, if glucose levels are low, cAMP levels rise and bind to the CAP, activating it. If lactose levels are also low, the cAMP-bound CAP will have no effect. If lactose is present and glucose levels are low, then allolactose binds the lac repressor causing it to dissociate from the operator region. Under these conditions, the cAMP-bound CAP can bind to the operator in lieu of the repressor protein. In this case, rather than blocking the RNA polymerase, the activated Camp-bound CAP induces even more efficient lac operon transcription. The result is synthesis of higher levels of lac enzymes that facilitate efficient cellular use of lactose as an alternative to glucose as an energy source. Maximal activation of the lac operon in high lactose and low glucose is shown below.

cAMP-bound CAP is an inducer of transcription. It does this by forcing the DNA in the promoter-operator region to bend. And since bending the double helix loosens H-bonds, it becomes easier for RNA polymerase to find and bind the promoter on the DNA strand to be transcribed…, and for transcription to begin. cAMP-CAPinduced bending of DNA is illustrated below.

3. Lac Operon Regulation by Inducer Exclusion and Multiple Operators

In recent years, additional layers of lac operon regulation have been uncovered. In one case, the ability of lac permease to transport lactose across the cell membrane is regulated. In another, additional operator sequences have been discovered to interact with a multimeric repressor to control lac gene expression.

A) Regulation of Lactose use by Inducer Exclusion

When glucose levels are high (even in the presence of lactose), phosphate is consumed to phosphorylate glycolytic intermediates, keeping cytoplasmic phosphate levels low. Under these conditions, unphosphorylated EIIAGlc binds to the lactose permease enzyme in the cell membrane, preventing it from bringing lactose into the cell.

The role of phosphorylated and unphosphorylated EIIAGlc in regulating the lac operon are shown below.

High glucose levels block lactose entry into the cells, effectively preventing allolactose formation and the derepression of the lac operon. Inducer exclusion is thus a logical way for the cells to handle an abundance of glucose, whether or not lactose is present. On the other hand, if glucose levels are low in the growth medium, phosphate concentrations in the cells rise sufficiently for a specific kinase to phosphorylate the EIIAGlc. Phosphorylated EIIAGlc then undergoes an allosteric change and dissociates from the lactose permease, making it active so that more lactose can enter the cell. In other words, the inducer is not “excluded” under these conditions!

The kinase that phosphorylates EIIAGlc is part of a phosphoenolpyruvate (PEP)- dependent phosphotransferase system (PTS) cascade. When extracellular glucose levels are low, the cell activates the PTS system in an effort to bring whatever glucose is around into the cell. But the last enzyme in the PTS phosphorylation cascade is the kinase that phosphorylates EIIAGlc. Phosphorylated EIIAGlc dissociates from the lactose permease, re-activating it, bringing available lactose into the cell from the medium.

B) Repressor Protein Structure and Additional Operator Sequences

The lac repressor is a tetramer of identical subunits (below).

Each subunit contains a helix-turn-helix motif capable of binding to DNA. However, the operator DNA sequence downstream of the promoter in the operon consists of a pair of inverted repeats spaced apart in such a way that they can only interact two of the repressor subunits, leaving the function of the other two subunits unknown… that is, until recently!

Two more operator regions were recently characterized in the lac operon. One, called O2, is within the lac z gene itself and the other, called O3, lies near the end of, but within the lac I gene. Apart from their unusual location within actual genes, these operators, which interact with the remaining two repressor subunits, went undetected at first because mutations in the O2 or the O3 region individually do not contribute substantially to the effect of lactose in derepressing the lac operon. Only mutating both regions at the same time results in a substantial reduction in binding of the repressor to the operon.

B. Mechanism of Control of the Tryptophan Operon

If ample tryptophan (trp) is available, the tryptophan synthesis pathway can be inhibited in two ways. First, recall how feedback inhibition by excess trp can allosterically inhibit the trp synthesis pathway. A rapid response occurs when tryptophan is present in excess, resulting in rapid feedback inhibition by blocking the first of five enzymes in the trp synthesis pathway. The trp operon encodes polypeptides that make up two of these enzymes.

Enzyme 1 is a multimeric protein, made from polypeptides encoded by the trp5 and trp4 genes. The trp1 and trp2 gene products make up Enzyme 3. If cellular tryptophan levels drop because the amino acid is rapidly consumed (e.g., due to demands for proteins during rapid growth), E. coli cells will continue to synthesize the amino acid, as illustrated below.

On the other hand, if tryptophan consumption slows down, tryptophan accumulates in the cytoplasm. Excess tryptophan will bind to the trp repressor. The trp-bound repressor then binds to the trp operator, blocking RNA polymerase from transcribing the operon. The repression of the trp operon by trp is shown below.

In this scenario, tryptophan is a co-repressor. The function of a co-repressor is to bind to a repressor protein and change its conformation so that it can bind to the operator.


16.2 Prokaryotic Gene Regulation

The DNA of prokaryotes is organized into a circular chromosome supercoiled in the nucleoid region of the cell cytoplasm. Proteins that are needed for a specific function, or that are involved in the same biochemical pathway, are encoded together in blocks called operons. For example, all of the genes needed to use lactose as an energy source are coded next to each other in the lactose (or lac) operon.

In prokaryotic cells, there are three types of regulatory molecules that can affect the expression of operons: repressors, activators, and inducers. Repressors are proteins that suppress transcription of a gene in response to an external stimulus, whereas activators are proteins that increase the transcription of a gene in response to an external stimulus. Finally, inducers are small molecules that either activate or repress transcription depending on the needs of the cell and the availability of substrate.


Prokaryotic Gene Regulation

The DNA of prokaryotes is organized into a circular chromosome, supercoiled within the nucleoid region of the cell cytoplasm. Proteins that are needed for a specific function, or that are involved in the same biochemical pathway, are encoded together in blocks called operons. For example, all of the genes needed to use lactose as an energy source are coded next to each other in the lactose (or lac) operon, and transcribed into a single mRNA.

In prokaryotic cells, there are three types of regulatory molecules that can affect the expression of operons: repressors, activators, and inducers. Repressors and activators are proteins produced in the cell. Both repressors and activators regulate gene expression by binding to specific DNA sites adjacent to the genes they control. In general, activators bind to the promoter site, while repressors bind to operator regions. Repressors prevent transcription of a gene in response to an external stimulus, whereas activators increase the transcription of a gene in response to an external stimulus. Inducers are small molecules that may be produced by the cell or that are in the cell’s environment. Inducers either activate or repress transcription depending on the needs of the cell and the availability of substrate.


Post-transcriptional Regulation

Once a eukaryotic gene is transcribed, it has to be processed by removal of the introns. In larger genes with many exons, several different versions of the protein coding sequence may be produced by alternative splicing, the selection of different sets of exons to form the mRNA. In addition, the mRNA is protected from degradation in the cytoplasm by the addition of nucleotides at both the 5′ and 3′ ends of the mRNA. At the 5′ end is a methyl-guanosine cap, which helps identify the mRNA to the ribosome. At the 3′ end is a string of adenine nucleotides, the poly-A tail, which helps to stabilize the mRNA as it moves to the ribosome.

The translation of the mRNA may then be either enhanced or repressed by proteins that bind to the untranslated 5′ and 3′ regions of the mRNA. These proteins may be themselves modified in response to various factors in the cell’s environment. In addition to these regulatory proteins, regulatory RNAs may bind to a mRNA to control its translation.

After translation, the protein may be further modified. In many cases, the initiating methionine, the amino acid encoded by the start codon AUG, is removed. Some proteins, like trypsin and other digestive enzymes, are synthesized as an inactive precursor protein that has to be activated by removing part of the protein. The protein precursor of insulin is a larger protein cut into several pieces, two of which are then linked together as the A and B chains of active insulin. The alpha and beta chains of the protein hemoglobin are the products of two different genes.

Generally, transcriptional regulators select which genes are transcribed in a given cell, while post-transcriptional factors come into play to determine whether or not the encoded protein product is made.

The next lesson is on hormone production.


17.1.1 Prokaryotic versus Eukaryotic Gene Expression

Since prokaryotic organisms are single-celled organisms that lack a cell nucleus, their DNA floats freely in the cell’s cytoplasm. When a particular protein is needed, the gene that codes for it is transcribed in mRNA, which is simultaneously translated into protein. When the protein is no longer needed, transcription stops. As a result, the primary method to control how much of each protein is expressed in a prokaryotic cell is the regulation of transcription.

Eukaryotic cells, in contrast, have intracellular organelles that add to their complexity. In eukaryotic cells, the DNA is contained inside the cell’s nucleus, where it is transcribed into mRNA. The newly synthesized mRNA is then modified and transported out of the nucleus into the cytoplasm, where ribosomes translate the mRNA into protein. The processes of transcription and translation are physically separated by the nuclear membrane transcription occurs only within the nucleus, and translation occurs only in the cytoplasm. The regulation of gene expression in eukaryotes can occur at all stages of the process (Figure 17.2).

Figure 17.2 Prokaryotic transcription and translation occur simultaneously in the cytoplasm, and regulation occurs at the level of transcription. In eukaryotes, transcription and translation are physically separated, and gene expression is regulated at many different levels.

Some of the differences in the regulation of gene expression between prokaryotes and eukaryotes are summarized in Table 17.1.

Table 17.1 Differences in prokaryotic and eukaryotic gene regulation.

Prokaryotic organisms

Eukaryotic organisms

DNA is found in the cytoplasm

Transcription and translation occur almost simultaneously

Transcription occurs in the nucleus prior to translation, which occurs in the cytoplasm.

Gene expression is regulated primarily at the transcriptional level

Gene expression is regulated at many levels: epigenetic, transcriptional, nuclear shuttling, post-transcriptional, translational, and post-translational


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Gene Expression Steps

Gene expression steps, as already mentioned, can be found in more detail on the protein synthesis page. Any gene that codes for an amino acid sequence that produces a polypeptide chain or a protein is called a structural gene.

The below image shows a codon wheel. Messenger RNA codons are comprised of a combination of three of the following nucleobases:

A broad set of combinations give codes for one or more amino acids, as well as for start and stop signals implemented during the translation process. For example, AAA – starting from the center of the codon wheel and moving outward – codes for the amino acid lysine (Lys).

Structural genes have various components:

  • Start site: the first part of a gene that tells messenger RNA when and where to begin the transcription process.
  • Promoter: not part of the mRNA transcript but a part that assists in its formation.
  • Enhancers: catalysts that speed up the transcription rate.
  • Silencers: decelerators of the transcription rate. Some proteins are produced at certain times such as during puberty or fetal development silencer sequences stop these proteins from being produced when they are not required.
  • Exons: the part of the gene that codes for amino acid sequences.
  • Introns: non-coding parts of the gene that are not transcribed by messenger RNA but spliced out before mRNA leaves the nucleus. An intron is a regulatory and protective part of a structural gene.

Gene expression is specific to the transcription and translation of DNA gene sequences in eukaryotes and prokaryotes. While eukaryotic gene expression happens inside and outside of the cell nucleus in two distinct stages, prokaryotic gene expression occurs nearly simultaneously in free-floating DNA within the cell cytoplasm.

The following four transcription steps describe eukaryotic processes.

  • Initiation: a double strand of DNA splits so that an enzyme (RNA polymerase) can recognize the start site and attach to the promoter.
  • Elongation: RNA polymerase moves along the open strand of DNA to produce a growing strand of pre-mRNA.
  • Termination: the finished strand of pre-mRNA detaches from the DNA.
  • Processing: introns are spliced (by spliceosomes) and the exons joined to produce mature mRNA that codes for a single protein.

Translation is the follow-on step from transcription it is also composed of four similarly-named gene expression steps:

  • Initiation: the mRNA strand leaves the cell nucleus and binds to a ribosome. Ribosomes can be compared to assembly-line machines in a factory that produce the final product – proteins or polypeptide chains. A first transfer RNA molecule attaches to the ribosome in the form of a start codon. Each transfer RNA (tRNA) carries a single amino acid that fits according to each mRNA codon. A start codon carries the amino acid methionine .
  • Elongation: more tRNA molecules bring specific amino acids to the appropriate section of mRNA. This mRNA runs through the ribosome a little like a ribbon through an old typewriter. Amino acids are bound to each other by an enzyme called peptidyl transferase.
  • Termination: after reaching a stop signal called the stop codon, the translation phase ends.
  • Post-translation processing: the finished protein or polypeptide is used either inside the cell or sent outside of the cell to carry out its required function.


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A: Leprosy is a bacterial disorder which affects the skin, respiratory tract, and nerves. It is a conta.

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A: Several bacteria and fungus are present in the environment, which may harm the human population. The.

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A: Reproductive health is the state of complete physical, mental and social well-being and not merely t.

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Q: How many major types of cells gastric glands have? Explain.

A: The inner membrane of the stomach contains some glands, known as gastric glands. The gastric glands .

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Biology for AP Courses

Control of gene expression in eukaryotic cells occurs at which level(s)?
a. only the transcriptional level
b. epigenetic and transcriptional levels
c. epigenetic and transcriptional and translational levels
d. epigenetic and transcriptional, translational, and post-translational levels

Problem 2

What do figures X and Y in the graphic illustrate?
a. Transcription and translation in a eukaryotic cell (figure X) and a prokaryotic cell (figure Y).
b. Transcription and translation in a prokaryotic cell (figure X) and a eukaryotic cell (figure Y).
c. Transcription in a eukaryotic cell (figure X) and translation in a prokaryotic cell (figure Y).
d. Transcription in a prokaryotic cell (figure X) and translation in a eukaryotic cell (figure Y)

Problem 3

If glucose is absent but lactose is present, the lac operon will be:
a. activated
b. repressed
c. partially activated
d. mutated

Problem 4

.What would happen if the operator sequence of the lac operon contained a mutation that prevented the repressor protein from binding the operator?
a. In the presence of lactose, the lac operon will not be transcribed.
b. In the absence of lactose, the lac operon will be transcribed.
c. The cAMP-CAP complex will not increase RNA synthesis.
d. The RNA polymerase will not bind the promoter.

Problem 5

What would happen if the operator sequence of the trp operon contained a mutation that prevented the repressor protein from binding to the operator?
a. In the absence of tryptophan, the genes trpA-E will not be transcribed.
b. In the absence of tryptophan, only genes trpE and trpD will be transcribed.
c. In the presence of tryptophan, the genes trpA-E will be transcribed.
d. In the presence of tryptophan, the trpE gene will not be transcribed.

Problem 6

What are epigenetic modifications?
a. the addition of reversible changes to histone proteins and DNA
b. the removal of nucleosomes from the DNA
c. the addition of more nucleosomes to the DNA
d. mutation of the DNA sequence

Problem 7

Which of the following statements about epigenetic regulation is false?
a. Histone protein charge becomes more positive when acetyl groups are added.
b. DNA molecules are modified within CpG islands.
c. Methylation of DNA and histones causes nucleosomes to pack tightly together.
d. Histone acetylation results in the loose packing of nucleosomes.

Problem 8

Which of the following is true of epigenetic changes?
a. They only allow gene expression.
b. They allow movement of histones.
c. They change the DNA sequence.
d. They are always heritable.

Problem 9

The binding of what is required for transcription start?
a. a protein
b. DNA polymerase
c. RNA polymerase
d. a transcription factor

Problem 10

What would be the outcome of a mutation that prevented DNA binding proteins from being produced?

a. decreased transcription because transcription factors would not bind to transcription binding sites
b. decreased transcription because enhancers would not be able to bind to transcription factors
c. increased transcription because repressors would not be able to bind to promoter regions
d. increased transcription because RNA polymerase would be able to increase binding to promoter regions

Problem 11

What will result from the binding of a transcription factor to an enhancer region?
a. decreased transcription of an adjacent gene
b. increased transcription of a distant gene
c. alteration of the translation of an adjacent gene
d. initiation of the recruitment of RNA polymerase

Problem 12

Which of the following are involved in posttranscriptional control?
a. control of RNA splicing
b. ubiquitination
c. proteolytic cleavage
d. phosphorylation

Problem 13

Gene A is thought to be associated with color blindness. The protein corresponding to gene A is isolated. Analysis of the protein recovered shows there are actually two different proteins that differ in molecular weight that correspond to gene A. What is one reason why there may be two proteins corresponding to the gene?
a. One protein had a 5’ cap and a poly-A tail in its mRNA, and the other protein did not.
b. One protein had a 5’ UTR and a 3’ UTR in its RNA, and the other protein did not.
c. The gene was alternatively spliced.
d. The gene produced mRNA molecules with differing stability.

Problem 14

Binding of an RNA binding protein will change the stability of the RNA molecule in what way?
a. increase
b. decrease
c. neither increase nor decrease
d. either increase or decrease

Problem 15

A mutation in the 5’UTR that prevents any proteins from binding to the region will:
a. increase or decrease the stability of the RNA molecule
b. prevent translation of the RNA molecule
c. prevent splicing of the RNA molecule
d. increase or decrease the length of the poly-A tail

Problem 16

Post-translational modifications of proteins can affect which of the following?
a. mRNA splicing
b. 5’capping
c. 3’polyadenylation
d. chemical modifications

Problem 17

A mutation is found in eIF-2 that impairs the initiation of translation. The mutation could affect all but one of the following functions of eIF-2. Which one would not be affected?
a. The mutation prevents eIF-2 from binding to RNA.
b. The mutation prevents eIF-2 from being phosphorylated.
c. The mutation prevents eIF-2 from binding to GTP.
d. The mutation prevents eIF-2 from binding to the 40S ribosomal subunit

Problem 18

The addition of a ubiquitin group to a protein does what?
a. increases the stability of the protein
b. decreases translation of the protein
c. increases translation of the protein
d. marks the protein for degradation

Problem 19

What are cancer-causing genes called?
a. transformation genes
b. tumor suppressor genes
c. oncogenes
d. protooncogenes

Problem 20

Targeted therapies are used in patients with a certain gene expression pattern. A targeted therapy that prevents the activation of the estrogen receptor in breast cancer would be beneficial to what type of patient?
a. patients who express the EGFR receptor in normal cells
b. patients with a mutation that inactivates the estrogen receptor
c. patients with over-expression of ER alpha in their tumor cells
d. patients with over-expression of VEGF, which helps in tumor angiogenesis

Problem 21

In a new cancer treatment, a cold virus is genetically modified so that it binds to, enters, and is replicated in cells, causing them to burst. The modified cold virus cannot replicate when wildtype p53 protein is present in the cell. How does this treatment treat cancer without harming healthy cells?
a. The modified virus only infects and enters cancer cells.
b. The modified virus replicates in normal and cancer cells.
c. The modified virus only infects and enters normal cells.
d. The modified virus replicates only in cancer cells.

Problem 22

A drug designed to switch silenced genes back on in cancer cells would result in what?
a. prevent methylation of DNA and deacetylation of histones
b. prevent methylation of DNA and acetylation of histones
c. prevent deacetylation of DNA and methylation of histones
d. prevent acetylation of DNA and demethylation of histones

Problem 23

What are positive cell-cycle regulators that can cause cancer when mutated called?
a. transformation genes
b. tumor suppressor genes
c. oncogenes
d. mutated genes

Problem 24

Which best distinguishes prokaryotic and eukaryotic cells?
a. Prokaryotes possess a nucleus whereas eukaryotes do not, but eukaryotes show greater
compartmentalization that allows for greater regulation of gene expression.
b. Eukaryotic cells contain a nucleus whereas prokaryotes do not, and eukaryotes show greater
compartmentalization that allows for greater regulation of gene expression.
c. Prokaryotic cells are less complex and perform highly-regulated gene expression whereas
eukaryotes perform less-regulated gene expression.
d. Eukaryotic cells are more complex and perform less-regulated gene expression whereas
prokaryotic cells perform highly-regulated gene expression.

Problem 25

Which statement is correct regarding the distinction between prokaryotic and eukaryotic gene expression?
a. Prokaryotes regulate gene expression at the level of transcription whereas eukaryotes regulate at
multiple levels including epigenetic, transcriptional and translational.
b. Prokaryotes regulate gene expression at the level of translation whereas eukaryotes regulate at the level of transcription to manipulate protein levels.
c. Prokaryotes regulate gene expression with the help of repressors and activators whereas eukaryotes regulate expression by degrading mRNA transcripts, thereby controlling protein
levels.
d. Prokaryotes control protein levels using epigenetic modifications whereas eukaryotes control protein levels by regulating the rate of transcription and translation.

Problem 26

All the cells of one organisms share the genome. However, during development, some cells develop into
skin cells while others develop into muscle cells. How can the same genetic instructions result in two different cell types in the same organism? Thoroughly explain your answer.

Problem 27

Which of the following statements describes prokaryotic transcription of the lac operon?
a. When lactose and glucose are present in the medium, transcription of the lac operon is
induced.
b. When lactose is present but glucose is absent, the lac operon is repressed.
c. Lactose acts as an inducer of the lac operon when glucose is absent.
d. Lactose acts as an inducer of the lac operon when glucose is present.

Problem 28

The lac operon consists of regulatory regions such as the promoter as well as the structural genes lacZ, lacY, and lacA, which code for proteins involved in lactose metabolism. What would be the outcome of a mutation in one of the structural genes of the lac operon?
a. Mutation in structural genes will stop transcription.
b. Mutated lacY will produce an abnormal $eta$ galactosidase protein.
c. Mutated lacA will produce a protein that will transfer an acetyl group to $eta$ galactosidase.
d. Transcription will continue but lactose will not be metabolized properly.

Problem 29

In some diseases, alteration to epigenetic modifications turns off genes that are normally expressed.
Hypothetically, how could you reverse this process to turn these genes back on?

Problem 30

Flowering Locus C (FLC) is a gene that is responsible for flowering in certain plants. FLC is expressed in new seedlings, which prevents flowering. Upon exposure to cold temperatures, FLC expression decreases and the plant flowers. FLC is regulated through epigenetic modifications. What type of epigenetic modifications are present in new seedlings and after cold exposure?
a. In new seedlings, histone acetylations are present upon cold exposure, methylation occurs.
b. In new seedlings, histone deacetylations are present upon cold exposure, methylation occurs.
c. In new seedlings, histone methylations are present upon cold exposure, acetylation occurs.
d. In new seedlings, histone methylations are present upon cold exposure, deacetylation occurs

Problem 31

A mutation within the promoter region can alter gene transcription. Describe how this can happen.
a. Mutated promoters decrease the rate of transcription by altering the binding site for the transcription factor.
b. Mutated promoters increase the rate of transcription by altering the binding site for the transcription factor.
c. Mutated promoters alter the binding site for transcription factors to increase or decrease the
rate of transcription.
d. Mutated promoters alter the binding site for transcription factors and thereby cease transcription of the adjacent gene.

Problem 32

What could happen if a cell had too much of an activating transcription factor present?
a. The transcription rate would increase, altering cell function.
b. The transcription rate would decrease, inhibiting cell functions.
c. The transcription rate decreases due to clogging of the transcription factors.
d. The transcription rate increases due to clogging of the transcription factors.

Problem 33

3. The $wnt$ transcription pathway is responsible for key changes during animal development. Based on the transcription pathway shown below. In this diagram, arrows indicate the transformation of one substance into another. Square lines, or the lines with no arrowheads, indicate inhibition of the product below the line. Based on this, how would increased $wnt$ gene expression affect the expression of Bar-1?
FIGURE 16.16

Problem 34

Describe how RBPs can prevent miRNAs from degrading an RNA molecule.
a. RBPs can bind first to the RNA, thus preventing the binding of miRNA, which degrades RNA.
b. RBPs bind the miRNA, thereby protecting the mRNA from degradation.
c. RBPs methylate miRNA to inhibit its function and thus stop mRNA degradation.
d. RBPs direct miRNA degradation with the help of a DICER protein complex.

Problem 35

How can external stimuli alter post-transcriptional control of gene expression?
a. UV rays can alter methylation and acetylation of proteins.
b. RNA binding proteins are modified through phosphorylation.
c. External stimuli can cause deacetylation and demethylation of the transcript.
d. UV rays can cause dimerization of the RNA binding proteins.

Problem 36

Protein modifications can alter gene expression in many ways. Describe how phosphorylation of proteins can alter gene expression.
a. Phosphorylation of proteins can alter translation, RNA shuttling, RNA stability or post transcriptional modification.
b. Phosphorylation of proteins can alter DNA replication, cell division, pathogen recognition
and RNA stability.
c. Phosphorylated proteins affect only translation and can cause cancer by altering the p53
function.
d. Phosphorylated proteins affect only RNA shuttling, RNA stability, and post-translational
modifications.

Problem 37

Changes in epigenetic modifications alter the accessibility and transcription of DNA. Describe how
environmental stimuli, such as ultraviolet light exposure, could modify gene expression.
a. UV rays could cause methylation and deacetylation of the genes that could alter the accessibility and transcription of DNA.
b. The UV rays could cause phosphorylation and acetylation of the DNA and histones which could alter the transcriptional capabilities of the DNA.
c. UV rays could cause methylation and phosphorylation of the DNA bases which could become dimerized rendering no accessibility of DNA.
d. The UV rays can cause methylation and acetylation of histones making the DNA more tightly packed and leading to inaccessibility.

Problem 38

New drugs are being developed that decrease DNA methylation and prevent the removal of acetyl groups from histone proteins. Explain how these drugs could affect gene expression to help kill tumor cells.
a. These drugs maintain the demethylated and the acetylated forms of the DNA to keep transcription of necessary genes “on”.
b. The demethylated and the acetylated forms of the DNA are reversed when the silenced gene is expressed.
c. The drug methylates and acetylates the silenced genes to turn them back “on”.
d. Drugs maintain DNA methylation and acetylation to silence unimportant genes in cancer cells.

Problem 39

How can understanding the gene expression pattern in a cancer cell tell you something about that specific form of cancer?
a. Understanding gene expression patterns in cancer cells will identify the faulty genes, which is helpful in providing the relevant drug treatment.
b. Understanding gene expression will help diagnose tumor cells for antigen therapy.
c. Gene profiling would identify the target genes of the cancer-causing pathogens.
d. Breast cancer patients who do not express EGFR can respond to anti-EGFR therapy.

Problem 40

Explain what personalized medicine is and how it can be used to treat cancer.
a. Personalized medicines would vary based on the type of mutations and the gene’s expression pattern.
b. The medicines are given based on the type of tumor found in the body of an individual.
c. The personalized medicines are provided based only on the symptoms of the patient.
d. The medicines tend to vary depending on the severity and the stage of the cancer.

Problem 41

Which of the following is found in both prokaryotes and eukaryotes?
a. 3’ poly-A tails
b. 5’ caps
c. promoters
d. introns

Problem 42

The enzyme ployadenylate polymerase catalyzes the addition of adenosine monophosphate to the 3’ ends of mRNAs to form a poly-A tail. If the enzyme were blocked so that it could not function, the result would be:
a. increased mRNA stability in eukaryotes, and decreased mRNA stability in prokaryotes
b. decreased mRNA stability in eukaryotes, and no effect in prokaryotes
c. no effect in eukaryotes, and increased mRNA stability in prokaryotes
d. no effect in eukaryotes, and decreased mRNA stability in prokaryotes

Problem 43

Describe two ways in which gene regulation differs and two ways in which it is similar in prokaryotes and eukaryotes.
a. Prokaryotes show co-transcriptional translation whereas eukaryotes perform transcription prior
to translation in both cell types, regulation occurs through the binding of transcription factors, activators, and repressors.
b. Prokaryotes perform transcription prior to translation whereas eukaryotes show cotranscriptional translation (the processes occur in the same organelle).
c. Prokaryotes show co-transcriptional translation that is regulated prior to translation whereas eukaryotes perform transcription prior to translation that is regulated only at the level of transcription. In both domains, transcription factors, activators, and repressors provide regulation.
d. Prokaryotes show co-transcriptional translation that occurs in the nucleus whereas eukaryotes
show transcription prior to translation. In both cell types, regulation occurs using transcription factors, activators, and repressors.

Problem 44

Lactose digestion in $E$ . coli begins with its hydrolysis by the enzyme $eta$ -galactosidase. The gene encoding $eta$ -galactosidase, lacZ, is part of a coordinately regulated operon containing other genes required for lactose utilization. Which of the following figures correctly depicts the interactions at the $lac$ operon when lactose is not being utilized?

Problem 45

What would be the result of a mutation in the repressor protein that prevented it from binding lactose?
a. The repressor will bind to lactose when it is removed from the operator.
b. The repressor will bind the operator in the presence of lactose.
c. The repressor will not bind the operator in the presence of lactose.
d. The repressor will not bind the operator in the absence of lactose.

Problem 46

What type of modification might be observed in the GR gene in all newborn rats?
a. The DNA will have many methyl molecules.
b. The DNA will have many acetyl molecules.
c. The DNA will have few methyl groups.
d. The histones will have many acetyl groups.

Problem 47

What type of modification will be observed in the GR gene in the highly nurtured rats?
a. The DNA will have many methyl molecules.
b. The DNA will have many acetyl molecules.
c. The DNA will have few methyl groups.
d. The histones will have few acetyl groups.

Problem 48

The level of transcription of a gene is tested by creating deletions in the gene as shown in the illustration. These modified genes are tested for their level of transcription: $(++)$ normal transcription levels $(+)$ low transcription levels $(+++)$ high transcription levels. Which deletion is in an enhancer involved in regulating the gene?
a. deletion 1
b. deletion 2
c. deletion 3
d. deletion 4

Problem 49

Which deletion is in a repressor involved in regulating the gene?
a. deletion 1
b. deletion 2
c. deletion 3
d. deletion 4

Problem 50

The diagram provided shows different regions (1-5) of a pre-mRNA molecule, a mature-mRNA molecule, and the protein corresponding to the mRNA. A mutation in which region is most likely to be damaging to the cell?
a. 1
b. 2
c. 3
d. 5

Problem 51

What do regions 1 and 5 correspond to?
a. exons
b. introns
c. promoters
d. untranslated regions

Problem 52

What are regions 1 through 5 in the diagram?
a. 1, 3, and 5 are exons 2 and 4 are introns.
b. 2 and 4 are exons 1,3, and 5 are introns.
c. 1 and 5 are exons 2, 3, and 4 are introns.
d. 2, 3, and 4 are exons 1 and 5 are introns.

Problem 53

A mutation results in the formation of the mutated maturemRNA as indicated in the diagram. Describe what type of mutation occurred and what the likely outcome of the mutation is.
a. Mutation in the GU-AG sites of introns produced a non-functional protein.
b. A transversion mutation in the introns led to alternative splicing, producing a functional protein.
c. A transversion mutation in the GU-AG site mutated this mRNA, producing a non-functional protein.
d. Transition mutations in the introns could produce a functional protein.

Problem 54

The diagram illustrates the role of p53 in response to UV exposure. What would be the result of a mutation in the p53 gene that inactivates it?
a. Skin will peel in response to UV exposure.
b. Apoptosis will occur in response to UV exposure.
c. No DNA damage will occur in response to UV exposure.
d. No peeling of skin will occur in response to UV exposure.

Problem 55

Which of the following will not occur in response to UV exposure if a p53 mutation inactivates the p53
protein?
1. Damage to DNA
2. p53 activation
3. p21 activation
4. Apoptosis
a. 1, 2, and 3
b. 3 and 4
c. 3
d. 2, 3, and 4

Problem 56

What happens when tryptophan is present?
a. The repressor binds to the operator, and RNA synthesis is blocked.
b. RNA polymerase binds to the operator, and RNA synthesis is blocked.
c. Tryptophan binds to the repressor, and RNA synthesis proceeds.
d. Tryptophan binds to RNA polymerase, and RNA synthesis proceeds.

Problem 57

What happens in the absence of tryptophan?
a. RNA polymerase binds to the repressor
b. the repressor binds to the promoter
c. the repressor dissociates from the operator
d. RNA polymerase dissociates from the promoter

Problem 58

Anabaena is a simple multicellular photosynthetic cyanobacterium. In the absence of fixed nitrogen, certain newly developing cells along a filament express genes that code for nitrogen-fixing enzymes and become nonphotosynthetic heterocysts. The specialization is advantageous because some nitrogen-fixing enzymes function best in the absence of oxygen. Heterocysts do not carry out photosynthesis but instead provides adjacent cells with fixed nitrogen and receives fixed carbon and
reduced energy carriers in return. As shown in the diagram above, when there is low fixed nitrogen in the environment, an increase in the concentration of free calcium ions and 2-oxyglutarate stimulates the expression of genes that produce two transcription factors (NtcA and HetR) that promote the expression of genes responsible for heterocyst development. HetR also causes production of a
signal, PatS, that prevents adjacent cells from developing as heterocysts. Based on your understanding of the ways in which signal transmission mediates cell function, which of the following predictions is most consistent with the information given above?
a. In an environment with low fixed nitrogen, treating the Anabaena cells with a calciumbinding compound should prevent heterocyst differentiation.
b. A strain that overexpresses the patS gene should develop many more heterocysts in a low nitrogen environment.
c. In an environment with abundant fixed nitrogen, free calcium levels should be high in all cells,
preventing heterocysts from developing.
d. In environments with abundant fixed nitrogen, loss of the hetR gene should induce heterocyst
development.

Problem 59

Which of the following statements about Anabaena is false?
a. Decreasing the concentration of free calcium ions will prevent heterocyst development.
b. In the presence of fixed nitrogen, NtcA will not be expressed.
c. Low fixed nitrogen levels result in increased PatS levels.
d. A mutation in NtcA that makes it nonfunctional will also allow adjacent cells to develop as heterocysts.

Problem 60

The operon model describes expression in prokaryotes. Describe this model and the essential difference in the way in which expression is regulated in eukaryotes.


What is Regulation of Gene Expression?

It is the process which enables cells to control when and how to regulate gene expression. However, this regulation is quite complicated, and any sort of malfunctioning can prove to be detrimental for cells leading to the occurrence of several diseases, including cancer.

Typically, the regulation of gene expression helps to conserve space and energy. Also, through it, living organisms adapt to the changes in their surroundings.

Furthermore, it is normal for each cell to have different active genes which are responsible for facilitating distinct functions. For example, liver cells are responsible for removing toxins from the bloodstream, while the neurons are responsible for transmitting signals.