molecular bio exams 2-4 textbook questions

Ch 11. 2

A DNA polymerase cannot synthesize a DNA strand de novo fro1n a mixture of deoxynucleotide triphosphates. Explain the role of the primer and template in the reaction promoted by this enzyme.

Ch 11. 2

A DNA polymerase cannot synthesize a DNA strand de novo fro1n a mixture of deoxynucleotide triphosphates. Explain the role of the primer and template in the reaction promoted by this enzyme.

The primer isa preexisting strand of RNA or DNA to which new nucleotides are added at the 3' end. The template is a
longer strand, paired with the primer, that determines the identity of each new nucleotide added via base pairing.

Ch 11. 3

Arthur Kornberg and his colleagues incubated an extract of E. coli with a mixture of dATP, dTTP, dGTP, and dCTP. Only the dTTP was labeled with 32P in the ex-phosphate group. After incubation, the mixture was treated with trichloroacetic acid to precipitate DNA, but not the dNTPs. The precipitate was collected, and the extent of [32P]dTMP incorporated into DNA was determined.
(a) If any one of the dNTPs were omitted fro1n the incubation mixture, would radioactivity be found in the precipitate? Explain.
(b) Would radioactivity be found in the precipitate if 32P labeled the 13-or 'Y-phosphate position? Explain.

(a) No. In the absence of any one dNTP, the polymerase would stop incorporating the other three dNTPs as soon as it encountered a template residue that should pair with the missing dNTP, and incorporation of 32P ,would be undetectable.
(b) No. DNA synthesis releases the Beta and Gamma phosphates of dNTPs as pyrophosphate.

Ch 11. 9

Five DNA polymerases have been identified in E. coli. Two of them- DNA polymerases IV and V- lack a 3'~ 5' exonuclease activity. What is the function of the 3'~ 5' exonuclease activity present in many DNA polymerases? How does the absence of such an activity in DNA polymerases IV and V make sense in the context of the cellular function of these enzymes?

The 3'-5' exonuclease is a proofreading function, eliminating mismatched nucleotides incorrectly inserted into the growing DNA strand. DNA polymerases IV and V function in translesion DNA synthesis during DNA repair, a situation where the accuracy of DNA polymerization becomes less important (as described in Chapter 12).

Ch 11. 11

Eukaryotic origins are tightly regulated so as not to fire more than once during S phase. Explain the key points of regulation of this process in eukaryotic cells.

The preRC forms only in G , not in other phases of the 1
cell cycle. Cyclin kinases produced only in S phase are needed to assemble the remaining proteins to produce active replication forks. Origins do not fire a second time, because new preRC complexes cannot form until the cell completes its cycle and returns to G •

Ch 11. 12c

How many T subunits are actually present in an active Pol III replisome acting at a replication fork in vivo? Explain the function of these 'T subunits.

Three. Two of the three 7 subunits, and their accompanying polymerase core subunits, act on the lagging strand and may permit more efficient DNA synthesis on the lagging strand.

Ch 11. 15

The enzyme telomerase is considered a reverse transcriptase, synthesizing DNA on an RNA template. What is the source of the RNA used in the synthesis of telomeres, where is it located, and how many nucleotides of the RNA are actually used as template for DNA synthesis?

The RNA template for telomere synthesis is telomerase RNA (RT) which is bound tightly to the telomerase reverse transcriptase (TERT) to form the active telomerase holoenzyme. A very short segment of the TR, equivalent to an organism's telomere repeat sequence (typically 6 or 7 nucleotides), is used over and over again as a template in telomere DNA synthesis.

Ch 12. 1

The base-pairing atoms of thymine are not directly involved in the cyclobutane ring of a pyrimidine dime1; formed by UV irradiation. Why does a pyrimidine dimer stall the replicative DNA polymerase?

The cross-linked pyrimidine dimer causes a distortion in the DNA that prevents base pairing in the active site of the DNA polymerase.

Ch 12. 2

For the nucleotide sequence AAC(06-meG)TGCAC, with a damaged (methylated) G residue, what would be the sequence of each strand of the double-stranded DNA in the following situations?
(a) Replication occurs before repair
(b) The DNA is acted upon by a glycosylase and then repaired, but only after replication has occurred.
(c) Two rounds of replication occur, followed by repair.

a) AACOTGCAC
TTGTACGTG

b) AACGTGCAC
TTGTACGTG

c) figure

Ch 12. 3

Name three of the common ways in which DNA lesions are incurred. What is required for these DNA lesions to result in a mutation?

Any three of the following: defects in repair; UV light; TLS error-prone bypass; oxidative damage; spontaneous hydro- lysis. For these lesions to become 1nutations, replication must occur (before repair), producing a naturally occurring base pair that is different from the original base pair

Ch 12. 7
What type of mutation is most likely to result from the
following lesions (if left unrepaired)?
a) Deamination of cytosine: G=C to ___
b) Formation of 8-oxoguanine: G=C to ___
c) Deamination of adenine: A= T to ___

a) G=C to A=T.
(b) G=C to T=A.
(c) A=T to G=C

Ch 12. 8
The human disease known as xerodenna pigmentosum
(XP) arises from mutations in at least seven different
genes. The resulting deficiencies are generally in enzymes
involved in some part of the pathway for nucleotide excision repair. The various types of XP are denoted A through
G (XPA, XPB, etc.), with a few additional variants lumped
together under the label XP-V Cultures of fibroblasts from
healtl1y individuals and from patients with XPG are irradiated with UV light. The DNA is isolated and denatured,
and the resulting single-stranded DNA is exa1nined by
analytical ultracentrifugation.

(a) Samples from the normal fibroblasts show a significant reduction in the average molecular weight of the
single-stranded DNA after irradiation, but samples
from the XPG fibroblasts show no such reduction.
Why might this be?

XP mutant cells usually contain a mutation in a gene
required for NER, the main repair pathway for UV lesions in humans. Single-strand DNA breaks are produced
during NER, accounting for the short ssDNA fragments
observed in normal cells after irradiation. However, a defective NER system in the XPG mutant cells prevents formation of single-strand breaks.

(b) If you assume that an NER system is operative in
fibroblasts, which step might be defective in the cells
of patients with XPG? Explain.

XPG cells are defective at a step preceding the first strand incision of NER, which is needed to produce the fragmented ssDNA

Ch 12. 9
Describe the most critical difference between global nucleotide excision repair and transcription-coupled repair.

The initiation step. In global NER, the XPC protein recognizes the lesion. In TCR, RNA polymerase recognizes the lesion, by stalling at the lesion.

Ch 12. 12
A gene is found that has a sequence of 11 contiguous A
residues in one strand. Mutations occur at an elevated
frequency in this gene, mostly in the region with the repeated A residues. Most of these mutations result in inactivation of the encoded protein, with many amino acids either missing or altered. What type of mutations would account for these observations, and how might they occur?

Frameshift mutations would lead to alteration of many
amino acid residues in the protein product and could be
caused by template slippage of DNA polymerase in the region with the repeated A residues.

Ch 12. 14
In humans, a fetus lacking at least one good copy of a gene
encoding any one of dozens of key DNA repair enzymes is
usually nonviable. If the fetus has one good copy of the gene
and one mutant (inactive) copy, the individual will have a
fully functional DNA repair system, but a higher than normal probability of acquiring cancer in middle age. Explain.

With only one good copy of a key DNA repair gene, there
is a good chance that the gene will be inactivated in one or
more cells by normal mutagenic processes such as unrepaired oxidative damage. Whenever this happens, a DNA
repair system is no longer present in that cell, and the mutation rate increases substantially. Eventually, mutations
occur in- and inactivate- one or more genes that normally
control cell division, and a cancer cell is born.

Ch 12. 16
If an oxidative lesion occurs spontaneously in a single-
stranded DNA fragment generated on the lagging strand
during replication, it is not readily repaired by nucleotide
excision repair or base excision repair. Explain why.

NER and BER occur only in double-stranded DNA. Both
processes excise the damaged base or bases from the dam-
aged strand, leaving a gap that can be filled in- but only if
an undamaged complementary strand is present.

Ch 12. 17
O6-Methylguanine lesions are repaired directly by transfer
of the methyl group to O6-methylguanine methyltransferase. A very high level of metabolic energy is invested in
this simple methyl transfer reaction. Describe this energy
investment.

Each molecule of O6-methylguanine methyltransferase is
used only once and is degraded after the repair reaction.
Thus, the reaction expends all the energy needed to synthesize the protein, along with all the energy used to mark the
protein for degradation and to carry out that degradation.

Ch 6. 7
Why does DNA predominantly contain thymine rather
than uracil as one of its pyrimidine bases?

The presence of T rather than U as one of the primary pyrimidine nucleotides in RNA is a likely mechanism by which
cells monitor mutations in DNA Uracil is regularly produced in DNA largely by the slow, nonenzymatic hydrolytic
deamination of cytosine; having thymine as the base in DNA
allows the efficient detection and repair of C-to-U mutations.

Ch 13. 10
Unlike recombination, the repair of double-strand breaks by
nonhomologous end joining creates mutations. Explain why.

NHEJ involves degradation by nucleases and processing of
the DNA ends, leading to some loss of base pairs.

Ch. 15.1
The sequences of promoters tend to be rich in A and T residues. Suggest why this is so.

The promoter is a site for RNA polymerase loading and initiation of transcription. To accomplish this, the RNA polymerase must form an open complex in which the two DNA strands are separated over a short distance. Due to the effects of AT-rich versus GC-rich DNA on the overall thermal stability of DNA, the strand separation is more readily accomplished in sequences that are AT-rich (see Chapter 4).

Ch 15.2
The -10 and -35 sequences in bacterial promoters are separated by about two turns of the DNA double helix. How would transcription be affected if a deletion were introduced in the promoter region that moved the -35 sequence to the -29 position?

The deletion would move the - 35 sequence closer to the - 10 sequence by half a helical turn of the DNA, putting the two elements on opposite faces of the DNA duplex.
This would dramatically reduce binding of sigma factor to the promoter, thereby decreasing transcription efficiency.

Ch 15.3
The gene encoding the E. coli enzyme B-galactosidase begins with the sequence ATGACCATGATTACG. What is the sequence of the mRNA transcript specified by this part of the gene?

5'-AUGACCAUGAUUACG. The sequence reported for a gene is, by convention, that of the coding strand, and sequences are always written in the 5'~3' direction.

Ch 15.5
The sequence of the consensus -10 region is TATAAT. If no genes, tesA and tesB, have identical promoter sequences except in the -10 region, where the tesA sequence is TAATAT and the tesB sequence is TGTCGA, ·which gene do you expect to be 1nore efficiently transcribed, and why?

Assuming the two genes use similar transcription initiation modes, tesA will be more efficiently transcribed. The tesB - 10 sequence deviates more from the consensus sequence, and its higher G= C content will be more difficult to melt.

Ch 15.12
Gene A encodes protein A. A genetic engineer excises a pro1noter sequence for gene A from the DNA and reinserts it at the other end of gene A, oriented so that an RNA polymerase binding at the pro1noter will transcribe across gene A. Will the 1nRNA synthesized by the RNA polymerase still possess a sequence that produces a functional protein A? Why or why not?

No. The two strands of a DNA molecule are antiparallel
and complementary (not identical). When the promoter is inverted, it will direct RNA synthesis that uses what was
originally the coding strand as the template strand. The mRNA sequence, derived from a different DNA strand and synthesized in the opposite direction, would be very different from the mRNA produced by the original gene and might not even contain an open reading frame.

Ch 15.13
In most organisms, specialized DNA repair systems are closely linked to transcription. Suggest a biological rationale for this close relationship.

Errors in the genetic information in actively transcribed genes are of greater immediate importance to the cell than errors in silent genes. Transcription-coupled DNA re- pair focuses repair on those DNA sequences most heavily utilized by the cell

Ch 15.14
In bacteria, there are many examples of two (or even more) genes being transcribed from one pro1noter- for example, the promoter is followed by gene A and then gene B, with both genes transcribed into a single mRNA. In some cases, the first gene in the linear sequence is transcribed at much higher levels than the second gene (i.e., many but not all of the mRNAs do not include gene B). What kind of DNA sequences might be present between the first and second genes to account for the lower level of transcription of geneB?

There may be a palindromic sequence that allows forma- tion of a hairpin structure during transcription, creating a p-independent terminator that is not perfectly efficient, or an imperfect rut sequence to guide the loading of the p protein. In both cases, the sequences would have to be imperfect so that some transcripts could be elongated through geneB.

Ch 19.2
Activator proteins A and B are required to express geneX. Analysis of the DNA upstream from the gene X promoter identified an 18 bp sequence ·with near twofold symmetry that is required for activation. Purification of the gene A and gene B products sho'\>ved that both proteins form homodimers, but neither the A nor the B homodimer binds the 18 bp site. What are the possible functions of the A and B activators with respect to the 18 bp site? Propose a test of one of your ideas.

Because the 18 bp site is a near palindrome, the activator probably functions as a dimer. Given that both protein A and protein B are required for activating gene X, they may form a heterodimer that has specificity for the binding site. This could be tested in an electrophoretic mobility shift assay or any other assay that measures DNA binding. Alternatively, the 18 bp site might be bound by a third, unidentified protein, and proteins A and B might bind different DNA sites. To test this, the site could be used in a functional DNA-binding assay to follow the binding protein during purification, allowing identification of the correct protein. Footprinting could be used as an assay (see Chapter 20), and the DNA sequence could be used to make an affinity chromatography resin to aid purification (see Chapter 7). A final possibility is that protein A and/or protein B interact with a different protein to bind the 18 bp site. This could be tested by purification using a functional assay such as footprinting. The purified active protein would reveal the additional protein.

Ch 19.3
Most proteins that regulate gene expression bind at specific DNA sequences, recognizing those sequences primarily through protein-DNA interactions within in the major groove of the DNA. Why is the major groove used for sequence recognition more often than the minor groove or the phosphoribose backbone?

There are more hydrogen-bond donor and acceptor groups on the nucleotide bases in the major groove than on those in the minor groove, providing much better discrimination between bases.

Ch 19.4
Briefly describe the relationship between chromatin structure and transcription in eukaryotes.

Heterochromatin is highly condensed and transcriptionally inert, because the histone proteins make promoters inaccessible. The less-condensed euchromatin has undergone a structural remodeling, allowing some regions to be transcribed. The alterations include covalent modification (such as acetylation) of histones and displacement of nucleosomes, creating exposed regions of DNA that are probably binding sites for regulatory proteins.

Ch 19.9
A repressor protein effectively blocks transcription from bacterial gene X. A mutant form of the repressor is en- gineered with an altered DNA-binding site in the helix-turn-helix motif. This mutant repressor does not repress transcription from gene X. When the mutant repressor is expressed at high levels on a plasmid that is introduced into the bacterial cell, transcription of X is increased even though the wild-type repressor (capable of binding its normal DNA binding site and shutting down transcription) is present in the same cell. Explain.

Most repressors with helix-turn-helix motifs function as oligomers, many as homodimers. When the plasmid-encoded mutant repressor is synthesized at high levels in the cell, most of the "1-wild-type repressor molecules synthesized are incorporated into less functional heterodi1ners with a mutant repressor subunit.

Ch 19.13
Steroid hormone receptors are located in the cytoplasm, where they can interact with incoming hormones. However, steroid hormones act by regulating gene function, and genes are in the nucleus. How is this regulation achieved?

On binding a hormone molecule, the steroid hormone receptor dimerizes and the hormone-receptor complex is transported into the nucleus.

Ch 19.14
Expression of the CRP transcription activator in E. coli readily leads to transcription of the lactose metabolism genes when lactose is present and glucose is not. If a particular eukaryotic activator is expressed in the appropriate eukaryotic cell, introduced on an engineered virus or plasmid, it often does not trigger transcription of its target gene. Explain.

There are several possible explanations. Many eukaryotic genes are regulated by more than one activator protein, and another activator may be needed. Many eukaryotic genes are encapsulated (and silenced) in heterochromatin, and remodeling of the chromatin may be required in the region where the gene is located to allow, activation. The protein may need to be modified, and the modifying enzyme (e.g., kinase) may not be present in the cell. Finally, perhaps the activator cannot be transported into the nucleus.

Ch 21.3
A histone acetyltransferase (HAT) is activated, transferring acetyl groups to histones in a particular region
of the genome. What amino acid residues in histones
are generally modified by HATs? What is the likely
effect of the modifications on the transcription levels
of genes in that region? What enzymes reverse the
effects of HATs?

HATs generally modify Lys residues in the C-terminal tails of histones. Histone acetylation reduces the affinity of nucleosomes for DNA and can increase the transcriptional activity of a chromosomal region. The acetyl groups are removed by histone deacetylases (HDACs).

Ch 21.6
Perhaps 3,000 or more transcription factors participate in
the activation of human genes. However, this is far fewer
than the number of genes in the human genome (- 20,000
to 25,000). Explain how specific gene activation is achieved
When genes outnumber gene activators by 10:1.

First, most genes are regulated by multiple transcription factors (activators), and different (often unique) combinations of factors are used at different genes. Second, families of activators form heterodimers, such that a family of four related proteins can make a total of 10 different dimeric species that can recognize 10 different DNA sequences

Ch 21.9
Housekeeping genes are those that must be expressed at
all times, providing a protein or RNA that is essential for
general cellular metabolism. They are often expressed
at a low but constant level. If an essential housekeeping
gene were experimentally moved from euchromatin to
a region of heterochromatin, what would be the likely
effect on the cell?

Gene expression is generally silenced in regions of heterochromatin. If the gene were essential, as most housekeeping genes are, moving it into heterochromatin would be lethal for the cell.

Ch 21.11a
A scientist is studying the function of a type of nuclear
steroid receptor protein in mouse cells. She introduces
various mutations into the gene encoding the receptor
protein and transfers the genes into mice. If mutations
are introduced that (a) eliminate the nuclear import
signal in the receptor protein

With elimination of the nuclear import signal, the receptor could bind the hormone in the cytoplasm, but the complex would not be imported into the nucleus to activate gene expression.

Ch 10.4
Within the histone structure, do protein modifications
occur primarily near the N-terminus or near the C-
terminus? What features distinguish the structure ·where
modifications occur?

Most histone modifications occur near the N-terminus.
This part of the histone molecule is a relatively unstructured tail that extends out from the nucleosome core.

Ch 10.10
Haw does epigenetic inheritance differ from Mendelian
inheritance?

"Epigenetic inheritance" refers to chromatin modifications (particularly histone modifications) that are retained in the chromatin after cell division and affect gene
transcription. Such modifications are not encoded in the
DNA and thus are not subject to Mendelian inheritance.

Ch 16.1
What would be the likely cellular effects of a large deletion
in the gene encoding the polymerase responsible for adding 3' poly(A) tails to eukaryotic mRNAs?

Inactivation of the polymerase would lead to incomplete pre-mRNA processing, including 3' end formation, splicing, editing, and transport- and would certainly be lethal.

Ch 16.2
What is the minimum number of transesterification re-
actions required to splice an intron from a precursor
transcript?

Two; one to cleave the 5' exon-intron junction and one to join the two exons, with release of the intron.

Ch 16.4
Self splicing introns do not require an energy source, such
as ATP or GTP, to catalyze splicing. How does self-splicing
proceed with a reasonable yield of products?

Self-splicing catalyzes phosphodiester exchange reactions with no net loss or gain of energy. Bonds are broken and re-formed with different nucleotides, and there is no change in the number of phosphodiester bonds, just in the covalent bonding partners. Thus, there is no net change in free energy from reactants to products.

Ch 16.7
Is it correct to call an RNA molecule that catalyzes a reaction on itself an enzyme? Explain your answer.

Sometimes. By definition, enzymes remain unchanged after catalysis, which is not the case for self-splicing and self-cleaving RNAs. They are catalysts because they enhance the rate of bond cleavage and/or joining, but in their natural form they have only one reaction turnover. However, some of these RNAs can be engineered to bind to separate substrate molecules and catalyze reactions on multiple such molecules, so they are inherently capable of functioning as enzymes. Some, such as RNase P, do so in their natural state.

Ch 16.12
All living systems share key properties (see Chapter 1).
Why did the discovery of RNA catalysis trigger the development of an RNA World hypothesis as a stage in the evolution of life?

The most important properties of living systems are the capacity to catalyze reactions and the capacity to store information that defines the structure of the catalysts. RNA has both of these properties.

Ch 16.17
RNA enzymes (ribozymes) have been discovered in
cells ranging from bacteria to humans. What kinds of
reactions are catalyzed by these naturally occurring
ribozymes?

Many naturally occurring ribozymes catalyze reactions involving other RNAs, mainly cleavage or splicing reactions. Ribozymes also catalyze the formation of peptide bonds on ribosomes.

Ch 16.18b (1)
The discovery of RNA catalysis is sometimes cited as a classic case of serendipity, but it required a thoroughly prepared
mind In the late 1970s and early 1980s, RNA splicing was
still a new concept. Thomas Cech and his colleagues set out
to investigate the splicing of an intron in rRNA of the pro-
tozoan Tetrahymena thermophila. They selected this RNA
for several reasons: rRNAs are much 1nore abundant than
most mRNAs, this rRNA has an intron, and Tetrahymena is a
single-celled eukaryote that can be grown in large quantities.
To carry out their early studies, Cech and colleagues
devised a method to produce unspliced precursor rRNAs.

Some very stable or tightly bound protein may have remained and catalyzed the intron-splicing reaction

Ch 16.8b 2
As described in their 1981 paper, they isolated nuclei from
Tetrahymena cells, lysed them, and added buffer, labeled
rNTPs, and a-amanitin, then incubated the reaction mix
under conditions in which the cellular RNA poly1nerases
would synthesize RNA. They extracted the labeled RNA
by a method that used phenol and chloroform, treatments
that normally remove most protein. The result demonstrated that the splicing reaction was working, but it did not demonstrate RNA catalysis. Why?

Some very stable or tightly bound protein may have remained and catalyzed the intron-splicing reaction

Ch 16.18d
Does this experiment demonstrate self-splicing (RNA
catalysis) of the intron? If so, why didn't the experiment described in the first paper do so?

This experiment is a compelling demonstration of self-splicing and RNA catalysis. There are no introns in bacterial rRNA. Expressing the rRl'~A seg1nent with the use of bacterial enzymes eliminated the possibility that contaminating intron-splicing enzymes from Tetrahymena were present. The Tetrahy1nena rRNA gene is the only Tetrahymena macromolecule in the reaction mixture and was deproteinized before the experiment.

Ch 22.3
As a researcher, you wish to scan the sequences of all the
genes of the mouse genome to determine how many genes
have multiple sites for 3 '-end cleavage. What sequence
would you scan for? Would a scan for a single sequence
find all of the sites?

The RNA-binding site for proteins that carry out 3' -end
cleavage and polyadenylation is defined most reliably
by the sequence AAUAAA. In the DNA, the sequence is
AATAAA, located do,,vnstream from the final gene exon.
Because AATAAA is a consensus sequence, scanning only
for this sequence would not find all the 3' -end cleavage
sites; some will vary slightly from the consensus.

Ch 22.6
As organisms become more co1nplex, so do the nu1nbers
and structures of introns. Introns in vertebrates range up
to 100,000 nucleotides in length. These long introns often
include canonical splicing signals, but the signals are not
recognized by the cellular spliceosome. How are these
nonproductive splicing sites suppressed?

Sequences called intronic splicing silencers, or ISSs, are
bound by proteins that suppress splicing at these sites.

Ch 6.1
In the 1980s, Tom Cech and Sidney Altman discovered
that RNA could function as an enzyme. List two properties of enzymes that must hold true for this characterization to be correct

An enzyme (RNA or protein) must (1) increase the rate of
a chemical reaction and (2) remain unchanged on completion of a catalytic cycle.

Ch 6.12
The cells of many eukaryotic organisms have highly specialized systems that repair G-T mismatches in DNA. The
mismatch is always converted to a G= C, never to an A T
base pair: This G-T mismatch repair system occurs in addition to a more general repair system that fixes virtually
all types of mismatches. Suggest why cells might require
a specialized G-T mismatch repair system, and why cells
would specifically convert the G-T to a G= C.

Cytosine deamination to form uracil is a slow but constant reaction in all cells. In many eukaryotes, hundreds of C residues are converted to U residues every day, in every
cell, creating G- U base pairs that are "seen" by the repair
system as G- T pairs. Because the G is correct and the
U is the damaged base, repairing G- T to G= C restores the
correct genetic information. Repair to A= T would cause
a mutation.

18.2

Name the type of chemical bonds that link (a) adjacent amino acids in a protein; (b) an amino acid to tRNA; (c) adjacent nucleotides in RNA; (d) a codon in mRNA to an anti codon in tRNA; (e) the two subunits of a ribosome.

(a) Amide or peptide. (b) Ester. (c) Phosphodiester.

(d) Hydrogen bonds. (e) Noncovalent bonds, including hydrogen bonds and van der Waals forces.

18.4

Discuss the advantages to the cell of having multiple ribosomes translating a single mRNA molecule.

Using polysomes, the cell can produce several protein molecules on a single mRNA molecule. Because mRNAs have an average lifetime of just a few minutes in the cell, polysomes maximize the number of proteins that can be made per unit time.

18.7

The isoleucyl-tRNA synthetase has a proof-reading function that improves the fidelity of the aminoacylation reaction, , whereas the histidyl-tRNA synthetase lacks such a proofreading function. Explain why.

Isoleucyl-tRNA synthetase sometimes catalyzes the addition of valine to tRNA^Ile; valine is similar to but smaller than isoleucine and can readily fit into the synthetase active site. Histidine has no close structural analogs among the amino acids, greatly lowering the chance that tRNA^His will be charged with the incorrect amino acid.

19.5

MicroRNAs known as small temporal RNAs (stRNAs) have been discovered in higher eukaryotes. Describe their characteristics and general function.

The primary transcript of an miRNA that is an stRNA is about 70 nucleotides long, with self-complementary internal sequences that form hairpin structures. The precursor is cleaved into 20- to 25-nucleotide partial duplexes, one strand of which can bind complementary stretches in cellular mRNAs. This binding can inhibit gene expression by blocking translation or facilitating mRNA degradation.