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After the first round of replication, Meselson and Stahl saw only one band in between the size predicted for DNA containing only 15N or 14N. After this observation, which hypothesis for DNA replication could be eliminated?
It breaks hydrogen bonds between the two DNA strands.
The primer is RNA-synthesized by RNA polymerase, which does not require a 3’ OH group.
Mismatch repair requires the ability to distinguish between template and newly synthesized DNA strands. How can E. coli distinguish between these two strands?
Template DNA is methylated.
Eukaryotic DNA contains introns.
Why is it that low-fidelity eukaryotic DNA polymerases are able to replicate DNA that contains abnormal bases, distorted structures, or bulky lesions, whereas high-fidelity DNA polymerases stall at these areas?
Low-fidelity DNA polymerases have large active sites that can bind to these regions.
You identify a mutant whose chromosomes shorten after each round of replication. A mutation in which gene would explain this observation?
Understanding DNA recombination is important in understanding ___________.
a. DNA repair
b. gene linkage traits
c. genetic variation
. List the different proteins and enzymes taking part in bacterial replication. Give thefunction of each in the replication process.
DNA polymerase III is the primary replication polymerase. It elongates a new
nucleotide strand from the 3'–OH of the primer.
DNA polymerase I removes the RNA nucleotides of the primers and replaces them
with DNA nucleotides.
DNA ligase connects Okazaki fragments by sealing nicks in the sugar phosphate
DNA primase synthesizes the RNA primers that provide the 3'–OH group needed
for DNA polymerase III to initiate DNA synthesis.
DNA helicase unwinds the double helix by breaking the hydrogen bonding between
the two strands at the replication fork.
DNA gyrase reduces DNA supercoiling and torsional strain that is created ahead of
the replication fork by making double-stranded breaks in the DNA and passing
another segment of the helix through the break before resealing it. Gyrase is also
called topoisomerase II.
Initiator proteins bind to the replication origin and unwind short regions of DNA.
Single-stranded binding protein (SSB protein) stabilizes single-stranded DNA prior
to replication by binding to it, thus preventing the DNA from pairing with
. DNA polymerases always synthesize new DNA by adding nucleotides on to the 5′ phosphate.
The proofreading function of DNA polymerases involves 5′ à 3′ exonuclease activity.
Okazaki fragments are involved in both lagging and leading DNA strand synthesis.
. DNA primase requires only a DNA template to initiate RNA primer synthesis.
RNA has a hydroxyl group on the 2’-carbon atom of its sugar component....
the template strand
promoter, RNA coding sequence, terminator
What would the result be if a specific sigma subunit were mutated?
The bacterial holoenzyme binds to which part of the promoter?
–10 and –35 consensus sequence
In rho-dependent transcription termination, the rho factor binds to ___________.
Which is NOT one of the DNA sequences known to regulate gene transcription?
basal transcription apparatus
a. RNA polymerase I requires a termination factor (like rho factor).
b. RNA polymerase II synthesizes hundreds of bases beyond what is needed for the mRNA, which is degraded by the protein Rat1.
c. RNA polymerase III terminates transcription after a termination sequence is transcribed.
d. All of the above statements are correct.......
How is the structure of RNA similar to that of DNA? How is it different?
RNA and DNA are polymers of nucleotides that are held together by phosphodiester
bonds. An RNA nucleotide contains a ribose sugar, whereas a DNA nucleotide
contains a deoxyribose sugar. Also, the pyrimidine base uracil is found in RNA but
thymine is not. DNA, however, contains thymine but not uracil. Finally, an RNA
polynucleotide is typically single-stranded even though RNA molecules can pair with
other complementary sequences. DNA molecules are almost always double-stranded.
Describe the structure of bacterial RNA polymerase.
. Bacterial RNA polymerase consists of several polypeptides. The RNA polymerase
core enzyme is composed of four polypeptide subunits: two copies of the alpha
subunit, the beta subunit, and the beta prime subunit. The addition of a sigma factor to
the core enzyme forms the RNA polymerase holoenzyme.
What are the two basic types of terminators found in bacterial cells? Describe the
structure of each.
Rho-independent terminators consist of inverted repeats that can form a hairpin structure. Immediately following the inverted repeats is a string of six adenine nucleotides.
Rho-dependent terminators require the interaction of the protein rho with RNA polymerase. Two features are typical for rho-dependent
termination: (1) variable DNA sequences that cause RNA polymerase to pause during
transcription and (2) upstream from variable region lies DNA sequence that encodes
RNA devoid of secondary structure, which also serves as the rho binding site.
How is the process of transcription in eukaryotic cells different from that in bacterial
Eukaryotic transcription requires the action of three RNA polymerases. Each type of
polymerase recognizes and transcribes from different types of promoters. Binding to
the promoter and initiation from the promoter requires the action of many protein
factors; different promoters require different sets of protein factors. The RNA
molecule produced by transcription in eukaryotic cells usually requires extensive
processing, such as the addition of a 5' cap, a 3' poly(A) tail, and the removal of
introns prior to becoming functional. Bacterial promoters tend to be more uniform in
composition, and only one RNA polymerase does transcription. Bacterial RNAs are
typically functional once transcription has taken place.
An RNA molecule has the following percentages of bases: A = 23%, U = 42%,
C = 21%, G = 14%.
a. Is this RNA single-stranded or double-stranded? How can you tell?
The RNA molecule is likely to be single-stranded. If the molecule was doublestranded,
we would expect nearly equal percentages of adenine and uracil, as well
as equal percentages of guanine and cytosine. In this RNA molecule, the
percentages of these potential base pairs are not equal, so the molecule is singlestranded.
. RNA polymerases carry out transcription at a much slower rate than DNA
polymerases carry out replication. Why is speed more important in replication than in
. DNA polymerases are required to replicate much larger regions of DNA, such as
entire chromosomes. Speed is essential to complete the replication process in a timely
manner. RNA polymerases typically transcribe only small areas of the chromosomes.
The speed required for replication by DNA polymerases is not needed by the RNA
polymerases to transcribe these smaller regions.
List at least five properties that DNA polymerases and RNA polymerases have in
common. List at least three differences.
Similarities: (1) Both use DNA templates, (2) DNA templates are read in the 3' to 5'
direction, (3) the complementary strand is synthesized in a 5' to 3' direction that is
antiparallel to the template, (4) both use triphosphates as substrates, and (5) their
actions are enhanced by accessory proteins.
Differences: (1) RNA polymerases use ribonucleoside triphosphates as substrates,
whereas DNA polymerases use deoxyribonucleoside triphophates; (2) DNA
polymerases require a primer that provides an available 3'–OH group where
synthesis begins, whereas RNA polymerases do not require primers to begin
synthesis; and (3) RNA polymerases synthesize a copy off only one of the DNAstrands, whereas DNA polymerases can synthesize copies off both strands
Suppose that the string of A’s following the inverted repeat in a rho-independent
terminator was deleted, but the inverted repeat was left intact. How would this
deletion affect termination? What would happen when RNA polymerase reached this
Termination would not take place if the string of A’s following the inverted repeat
were deleted. Although RNA polymerase may stall briefly at the hairpin, the presence
of the A–U base pairs is needed to destabilize the DNA-RNA interaction and to end
transcription. If the string of A’s is not present, then transcription by RNApolymerase would continue
The following diagram represents a transcription unit in a hypothetical DNA
On the basis of the information given, is this DNA from a bacterium or from a
The DNA is from a bacterium as evidenced by the TATAAT sequence that
corresponds to the –10 consensus sequence of a bacterial promoter and the
TTGACA sequence that is identical to the –35 consensus sequence of a bacterial
Sigma subunits dissociate from RNA polymerase after transcription has terminated.
Initiation of transcription does not require a primer.
Which mRNA processing event adds stability to the mRNA?
-addition of a 5’ cap
-addition of a poly(A) tail to the 3’ end
What kind of RNA functions in splicing and is associated with the spliceosome?
small nuclear RNA (snRNA)
5’ splice site
alternative 3’ cleavage sites
They are made up of only ribosomal RNA molecules.
genes from which they were transcribed
What is the concept of colinearity? In what way is this concept fulfilled in bacterial and
Colinearity is the concept that the sequence of codons in the DNA of a gene is the same
as the sequence of amino acids in the protein. If we examine the location of three codons
for three amino acids in a protein, the order of the codons in the gene is always the same
as the order of the amino acids in the protein. The presence of large regions of non
coding DNA in introns means that DNA sequences that code for adjacent amino acids
may be separated by many base pairs of DNA, but introns to not alter the order of codonsin the gene.
What are the three principal elements in mRNA sequences in bacterial cells?
(1) The 5' untranslated region, which contains the Shine-Dalgarno sequence
(2) The protein-encoding region
(3) The 3' untranslated region
. What are some of the modifications in tRNA that occur through processing?
(1) The precursor RNA may be cleaved into smaller molecules.
(2) Nucleotides at both the 5' and 3' ends of the tRNAs may be removed or trimmed.
(3) Standard bases can be altered by base-modifying enzymes.
Describe the basic structure of ribosomes in bacterial and eukaryotic cells.
Ribosomes in both eukaryotes and bacteria consist of a complex of protein and RNA
molecules. A functional ribosome is composed of a large and a small subunit. The
bacterial 70S ribosome consists of a 30S small subunit and a 50S large subunit. Within
the small subunit are a single 16S RNA molecule and 21 proteins. The 23S and the 5S
RNA molecules, along with 31 proteins, are found in the large bacterial subunit. The
eukaryotic 80S ribosome is comprised of a 60S large subunit and a 40S small subunit.
Three RNA molecules, the 28S RNA, the 5.8S RNA, and the 5S RNA, are located in the
large subunit as well as 49 proteins. The eukaryotic small subunit contains only a single
18S RNA molecule and 33 proteins.
What is the origin of siRNAs and microRNAs? What do these RNA molecules do in the
. The siRNAs originate from the cleavage of mRNAs, RNA transposons, and RNA
viruses by the enzyme Dicer. Dicer may produce multiple siRNAs from a single doublestranded
RNA molecule. The double-stranded RNA molecule may occur due to the
formation of hairpins or by duplexes between different RNA molecules. The miRNAs
arise from the cleavage of individual RNA molecules that are distinct from other genes.
The enzyme Dicer cleaves these RNA molecules that have formed small hairpins. A
single miRNA is produced from a single RNA molecule.
Both siRNAs and microRNAs silence gene expression through a process called RNA
interference. Both function by shutting off gene expression of a cell’s own genes or to
shut off expression of genes from the invading foreign genes of viruses or tranposons.
The microRNAs typically silence genes that are different from those from which the
microRNAs are transcribed. However, the siRNAs usually silence genes from which
they are transcribed.
Consider Duchenne muscular dystrophy and the dystrophin gene. The gene causing Duchenne muscular dystrophy encompasses more than 2 million nucleotides, but less than 1% of the gene encodes the
protein dystrophin. On the basis of what you now know about gene structure and RNA
processing in eukaryotic cells, provide a possible explanation for the large size of thedystrophin gene
. The large size of the dystrophin gene is likely due to the presence of many intervening
sequences or introns within the coding region of the gene. Excision of the introns through
RNA splicing yields the mature mRNA that encodes the dystrophin protein.
How do the mRNA of bacterial cells and the pre-mRNA of eukaryotic cells differ? How
do the mature mRNAs of bacterial and eukaryotic cells differ?
Bacterial mRNA is translated immediately upon being transcribed. Eukaryotic pre-mRNA
must be processed. Bacterial mRNA and eukaryotic pre-mRNA have similarities in
structure. Each has a 5' untranslated region as well as a 3' untranslated region. Both also
have protein-coding regions. However, the protein-coding region of the pre-mRNA is
disrupted by introns. The eukaryotic pre-mRNA must be processed to produce the mature
mRNA. Eukaryotic mRNA has a 5' cap and a poly(A) tail, unlike bacterial mRNAs.
Bacterial mRNA also contains the Shine-Dalgarno consensus sequence. Eukaryotic
mRNA does not have the equivalent.
. In the early 1990s, Carolyn Napoli and her colleagues were working on petunias,
attempting to genetically engineer a variety with dark purple petals by introducing
numerous copies of a gene that codes for purple petals (C. Napoli, C. Lemieux, and R.
RNA Jorgensen. 1990. Plant Cell 2:279–289). Their thinking was that extra copies of the gene
would cause more purple pigment to be produced and would result in a petunia with an
even darker hue of purple. However, much to their surprise, many of the plants carrying
extra copies of the purple gene were completely white or had only patches of color.
Molecular analysis revealed that the level of the mRNA produced by the purple gene was
reduced 50-fold in the engineered plants compared with levels of mRNA in wild-type
plants. Somehow, the introduction of extra copies of the purple gene silenced both the
introduced copies and the plant’s own purple genes. Explain why the introduction of
numerous copies of the purple gene silenced all copies of the purple gene.
The overexpression of the purple gene mRNA led potentially to the formation of doublestranded
regions by these RNA molecules because of areas of homology within the
mRNAs being produced. These double-stranded molecules stimulated RNA silencing
mechanisms or the RNA-Induced-Silencing Complex (RISC) leading to rapid
degradation of the mRNA molecules. The result would be a reduction in translation of
the protein needed for the production of purple petals and the phenotypic loss of
Identify the following items and, for each item, give a
brief description of its function.
a. 5' untranslated region
c. AAUAAA consensus sequence
d. Transcription start site
e. 3' untranslated region
h. Poly(A) tail
i. 5' cap
. The 5' untranslated region lies upstream of the translation start site. In bacteria, the
ribosome binding site or Shine-Dalgarno sequence is found within the 5' untranslated
region. However, eukaryotic mRNA does not have the equivalent sequence, and a
eukaryotic ribosome binds at the 5' cap of the mRNA molecule.
b. The promoter is the DNA sequence that the transcription apparatus recognizes and
binds to initiate transcription.
c. The AAUAAA consensus sequence lies downstream of the coding region of the
gene. It determines the location of the 3' cleavage site in the pre-mRNA molecule.
d. The transcription start site begins the coding region of the gene and is located 25 to 30
nucleotides downstream of the TATA box.
e. The 3' untranslated region is a sequence of nucleotides at the 3' end of the mRNA that
is not translated into proteins. However, it does affect the translation of the mRNA
molecule as well as the stability of the mRNA.
f. Introns are noncoding sequences of DNA that intervene within coding regions of a
g. Exons are transcribed regions that are not removed in intron processing. They include
the 5' UTR, coding regions that are translated into amino acid sequences, and the
h. A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.
i. The 5' cap functions in the initiation of translation and mRNA stability.
Multiple protein products are often produced from single eukaryotic genes.
. The 5′ and 3′ untranslated regions (UTRs) of processed mRNA molecules are derived from introns.
Introns are degraded in the cytoplasm.
Transcription and translation take place simultaneously in bacterial cells.
Y, X, Z
. If instead of 20 amino acids there were 200 amino acids, then how many nucleotides would you predict to be in a codon?
The wobble rules account for non–Watson and Crick nucleotide pairing at which position of the codon and anticodon, respectively?
Any given DNA sequence has ______ possible reading frames, and the correct one is set by a(n) ___________________.
The formation of peptide bonds during elongation is catalyzed by an enzyme called the ribozyme. The ribozyme is made of what macromolecule?
. Simultaneous transcription and translation does NOT occur in which of the following locations?
What is the one gene, one enzyme hypothesis? Why was this hypothesis an important advance in our understanding of genetics?
The one gene, one enzyme hypothesis proposed by Beadle and Tatum states that each gene encodes a single, separate protein. Now that we know more about the nature of enzymes and genes, it has been modified to the one gene, one polypeptide hypothesis because many enzymes consist of multiple polypeptides. The original hypothesis helped establish a linear link between genes (DNA) and proteins.
Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but accept the same amino acids.
How are tRNAs linked to their corresponding amino acids?
. Each of the 20 different amino acids that are commonly found in proteins has a
corresponding aminoacyl-tRNA synthetase that covalently links the amino acid to the correct tRNA molecule.
Arrange the following components of translation in the approximate order in which
they would appear or be used during protein synthesis.
-70S initiation complex
-elongation factor Tu
-peptidyl transferase -fMet-tRNAfMet -30S
-initiation complex -initiation factor 3
-elongation factor G -release factor 1
. In what ways are spliceosomes and ribosomes similar? In what ways are they different?
. Spliceosomes and ribosomes are both large complexes that are composed of several different RNA and protein molecules. In essence, the spliceosome and ribosome are RNA-based enzymes or ribozymes. The RNA molecules in both structures are necessary for catalysis. The 23S rRNA molecule in the ribosome catalyzes the formation of peptide bonds between amino acids. Other rRNAs are also important for protein synthesis. In spliceosomes, the snRNA molecules catalyze the cutting and splicing of pre-mRNA molecules to produce mature mRNA. The catalytic RNAs of the ribosome and the spliceosome may have originated during the RNA world when RNA molecules served to store information and to catalyze reactions that sustained life.
The primary sequence of the protein determines the quaternary structure of a protein
During translation initiation in prokaryotes, one rRNA base pairs with a sequence in mRNA to position a ribosome at the start codon.
The codon for methionine appears only at the beginning of the mRNA for a protein, not in the middle or in the end.
. The first three bases at the 5' end of an mRNA are the AUG at which translation begins.
. Ribosomes move along an mRNA in the 5' to 3' direction.
A special tRNA that does not have an attached amino acid binds to stop codons to terminate translation.
DNA not methylated
Some have acetyltransferase activity.
They block the role of enhancers.
5’ cap and 3’ poly(A) tail
stabilizing translation machinery
List some important differences between bacterial and eukaryotic cells that affect
the way in which the genes are regulated.
a. Bacterial genes are frequently organized into operons with coordinate
regulation, and genes with operons can be transcribed as on a single long
mRNA. Eukaryotic genes are not organized into operons and are singly
transcribed from its own promoter.
b. In eukaryotic cells, DNA must be unwound from histone proteins prior to
transcription occurring. Essentially, the chromatin must assume a more open
configuration state, allowing for access by transcription associated factors.
c. Activator and repressor molecules function in both eukaryotic and bacterial
cells. However, in eukaryotic cells activators appear to be more common than
in bacterial cells.
d. In bacteria, transcription and translation can occur concurrently. In eukaryotes,
the nuclear membrane separates transcription from translation both physically
and temporally. This separation results in a greater diversity of regulatory
mechanisms that can occur at different points during gene expression.
Where are DNase I hypersensitivity sites found and what do they indicate about the
nature of chromatin?
. DNase I hypersensitivity sites are typically found approximately 1000 nucleotides
upstream of a transcription start site. This sensitivity to DNase I suggests a more
relaxed or open configuration state for the chromatin, allowing for access to the
DNA of transcriptional regulatory proteins as well as access by DNase I.
Essentially the DNA is more exposed due to the lack of nuclesome structure.
Briefly explain how transcriptional activator proteins and repressors affect the level
of transcription of eukaryotic genes.
Transcriptional activator proteins stimulate transcription by binding DNA at
specific base sequences such as an enhancer or regulatory promoter and attracting
or stabilizing the basal transcriptional factor apparatus. Repressor proteins bind
to silencer sequences or promoter regulatory sequences. These proteins may inhibit
transcription by either blocking access to the enhancer sequence by the activator
protein, preventing the activator from interacting with the basal transcriptionapparatus, or preventing the basal transcription factors from being assembled
Outline the role of alternative splicing in the control of sex differentiation in
Sex development in fruit flies depends on alternative splicing as well as a cascade
of genetic regulation. Early in the development of female fruit flies, a femalespecific
promoter is activated stimulating transcription at the sex-lethal (Sxl) gene.
Splicing of the pre-mRNA of the transformer (tra) gene is regulated by the Sxl
protein. The mature mRNA produces the Tra protein. In conjunction with another
protein, the Tra protein stimulates splicing of the pre-mRNA from the doublesex
(Dsx) gene. The resulting Dsx protein is required for the embryo to develop female
characteristics. Male fruit flies do not produce the Sxl protein, which results in the
Tra pre-mRNA in male fruit flies being spliced at an alternate location. The
alternate Tra protein is not functional, resulting in the Dsx pre-mRNA splicing at a
different location as well. Protein synthesis from this mRNA produces a malespecific
doublesex protein, which causes development of male-specific traits.
Briefly list some of the ways in which siRNAs and miRNAs regulate genes.
. (1) Through cleavage of mRNA sequences through “slicer activity”: The binding
of RISCs containing either siRNA or miRNA to complementary sequences in
mRNA molecules stimulate cleavage of the mRNA through “slicer activity.”
This is followed by further degradation of the cleaved mRNA.
(2) Through binding of complementary regions with the mRNA molecule by
miRNAs to prevent translation: The miRNAs as part of RISC bind to
complementary mRNA sequences preventing either translation initiation or
elongation, which results in premature termination.
(3) Through transcriptional silencing due to methylation of either histone proteins
or DNA sequences: The siRNA bind to complementary DNA sequences within
the nucleus and stimulate methylation of histone proteins. Methylated
histones bind DNA more tightly preventing transcriptional factors from
binding the DNA. The miRNA molecules bind to complementary DNA
sequences and stimulate DNA methylases to directly methylate the DNA
sequences, which results in transcriptional silencing.
(4) Through slicer-independent mRNA degradation stimulated by miRNA binding
to complementary regions in the 3' UTR of the mRNA: A miRNA binds to the
AU rich element in the 3' UTR of the mRNA stimulating degradation using
RISC and dicer.
. What are some of the characteristics of Arabidopsis thaliana that make it a good
model genetic organism?
. Useful characteristics of Arabidopsis thaliana
Is an angiosperm and thus shares characteristics and life cycle similarities
with other flowering plants.
Is capable of self-fertilization or cross-fertilization.
Is small in size, which is useful in laboratory environments.
Can grow under low illumination levels, which also is useful in laboratory
Is a prolific reproducer with each plant capable of producing 10,000 to
Has a high germination frequency from its seeds.
Has a small genome (~125 million base pairs) that has been completely
Has a number of ecotypes or variants that are available, which differ in
genotypes and phenotypic characteristics.
Can uptake genes from other organisms by means of a Ti plasmid.
X31b is an experimental compound that is taken up by rapidly dividing cells.
Research has shown that X31b stimulates the methylation of DNA. Some cancer
researchers are interested in testing X31b as a possible drug for treating prostate
cancer. Offer a possible explanation for why X31b might be an effective anticancer
Cancer cells are typically rapidly dividing cells. DNA methylation particularly in
regions with many CpG sequences (CpG islands) is associated with transcriptional
repression. If the X31b molecules can be uptaken by the rapidly dividing cancers
cells and then stimulate methylation of DNA sequences in the cancer cells,
transcriptional repression of genes in the cancer cells would be expected. The
repression of transcription could affect the growth of the cancer cells and
potentially cause a loss of viability of these cells.
. How do repressors that bind to silencers in eukaryotes differ from repressors that
bind to operators in bacteria?
. In bacteria, repressors that bind to the operator block RNA polymerase from
binding to the promoter. Repressors that bind to silencers in eukaryotes block
transcriptional activator proteins from binding at an activator site, thus eliminating
. What will be the effect on sexual development in newly fertilized Drosophila
embryos if the following genes are deleted?
A. If the Sxl gene is absent, then female specific splicing of the tra pre-mRNA will
not occur resulting in the production of a nonfunctional Tra protein. Thus, the
fruit flies will develop male specific characteristics.
If the tra gene is absent, then a male-specific Dsx protein is synthesized allowing
for the development of male specific characteristics.
If the Dsx gene is deleted, then no doublesex protein will be produced, resulting
in flies with intersex characteristics. Female flies lacking the female specific
doublesex protein will have both male and female characteristics. Male flies
lacking the male-specific doublesex protein will also have both male and female
. In the fungus Neurospora, about 2−3% of cytosine bases are methylated. A recent
study isolated those DNA sequences in Neurospora that contained 5-methylcytosine
and found that almost all methylated sequences were located in the relict copies of
transposable genetic elements. Based on these observations, propose a possible
explanation for why Neurospora methylates its DNA and why DNA methylation in
this species is associated with transposable genetic elements.
Heavy DNA methylation of cytosine to yield 5-methylcytosine containing
nucleotides is associated with transcriptional repression in other organisms such as
vertebrate animals and plants. Because almost all of the methylated sequences were
associated with relic copies of transposons, this suggests that the methylation may
be part of a process for inactivating these transposable elements. The Neurospora
methylation could be a mechanism for inactivating the expression of genes found in
these invading nucleic acid sequences and thus preventing the transposon
sequences from spreading within the Neurospora genome.
. A common feature of many eukaryotic mRNAs is the presence of a rather long 3'
UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an
enzyme important in cellular metabolism. Certain cells—termed U937D cells—
have lots of CK-B mRNA, but no CK-B enzyme is present. In these cells, the 5'
end of the CK-B mRNA is bound to ribosomes, but the mRNA is apparently not
translated. Something inhibits the translation of the CK-B mRNA in these cells.
Researchers introduced numerous short segments of RNA containing only 3' UTR
sequences into U937D cells. As a result, the U937D cells began to synthesize the
CK-B enzyme, but the total amount of CK-B mRNA did not increase. The
introduction of short segments of other RNA sequences did not stimulate the
synthesis of CK-B; only the 3' UTR sequences turned on the translation of the
On the basis of these results, propose a mechanism for how CK-B translation is
inhibited in the U937D cells. Explain how the introduction of short segments of
RNA containing the 3' UTR sequences might remove the inhibition.
. From the above experimental data, translation of the CK-B protein is inhibited in
the U937D cells—the CK-B mRNA is present and bound to the ribosome, but no
protein is synthesized. A possible mechanism for the inhibition of translation could
be the binding of translational repressors to the 3' UTR region of the CK-B mRNA.
The action of soluble proteins inhibiting translation seems to be suggested by the
response of the U937 cells to the short RNA sequences containing the 3' UTR.
When these sequences are introduced to the U937D cells, the synthesis of CK-B
occurs. Possibly, exogenously applied 3' UTR sequences bind to the translational
repressor proteins, making them unavailable to bind to the CK-B mRNA. If these
factors are not present on the CK-B mRNA, then synthesis of the CK-B protein can
Separation of transcription and translation by the nuclear membrane allowed the evolution of additional mechanisms of gene regulation.
In eukaryotes, most structural genes are found within operons.
In a general sense, highly condensed DNA bound with histone proteins represses gene expression.
When regions of chromatin become actively transcribed, they become more sensitive to digestion by DNase I.
. Acetylation involves the addition of acetyl groups to histone proteins and usually represses transcription.
Eukaryotic enhancers are also known as insulators.
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