BI 203 Exam 2

The growing polypeptide chain coming off the ribosomal complex is fairly unstable. It has a tendency to fold back on itself and can aggregate with adjacent polypeptides. If these processes are allowed to occur, degraded, improperly folded, or large aggregates of nonfunctional proteins would result. How does the cell prevent this?

A class of proteins called chaperones prevents these problems. As polypeptides come off the ribosomes, chaperones quickly bind to and stabilize the growing chain. They prevent improper or premature folding until the entire chain is synthesized and the completed protein can fold properly.

Which molecule has a critical role in transcription elongation? A. P-TEFb B. HAT C. miRNA D. lncRNA

A. P-TEFb

In prokaryotes, sigma factors are necessary for ... A. specific binding to certain promoter elements. B. low-affinity binding upstream from promoters. C. elongation of the RNA strand to its end. D. faster transcription

A. specific binding to certain promoter elements.

Which process is not associated with translational regulation? A. Dephosphorylation of 4E-BPs that bind eIF4E and prevent its interaction with eIF4G B. Autolytic degradation of the mRNA by folding back on itself C. Cleavage of mRNA by miRNA/RISC complexes D. Phosphorylation of eIF2, which inhibits GDP/GTP exchange

B. Autolytic degradation of the mRNA by folding back on itself

The half-lives of proteins in the cell vary widely, ranging from A. milliseconds to seconds. B. three to seven minutes. C. minutes to days. D. days to weeks.

C. minutes to days.

A proteasome is a A. vesicle containing proteolytic enzymes. B. precursor to lysosomes. C. complex of a proteolytic enzyme and the protein that is being degraded. D. multisubunit protease complex that degrades proteins marked for destruction.

D. multisubunit protease complex that degrades proteins marked for destruction.

Deamination of cytosine results in the conversion of cytosine to A. hypoxanthine. B. adenine. C. 7-methyl guanosine. D. uracil.

D. uracil.

How do DNA polymerases differ from RNA polymerases in terms of primer requirement?

DNA polymerases require a primer, whereas RNA polymerases do not.

One unique feature of prokaryotes is that their mRNAs are often polycistronic. What does this mean?

Prokaryotic bacteria are polycistronic, meaning that a single mRNA can encode multiple proteins, each with its own start and stop sites.

Explain both the closed- and open-promoter complex that is formed during the initiation of transcription in E. coli.

RNA polymerase initially binds nonspecifically to DNA and moves along until it binds to the -10 and -35 promoter sequences. This forms the closed-promoter complex. The polymerase unwinds the DNA at the initiation site to form the open-promoter complex, and transcription is initiated.

Mammals also regulate gene expression with mRNA lifetime (ranging from <30 minutes to ~20 hours). One would expect the shorter-lived mRNAs to code for (select all that apply):

Regulatory proteins Transcription factors

Suppose a culture of E. coli is growing in media containing only glucose. The cells eventually deplete this glucose supply. Just as the last bit of glucose is used up, lactose and other sugars become available. How does both the loss of glucose and the addition of lactose affect expression of the lac operon in these cells?

The lac operon is repressed when lactose is absent and glucose is present. When lactose becomes available, it forms a metabolite that binds to the repressor, causing its removal from the operator site on the DNA. At the same time, glucose levels drop, which induces CAP to activate RNA polymerase. The combined loss of the repressor and activation of RNA polymerase causes rapid transcription of the lac operon genes.

A single mRNA can code for more than one polypeptide chain.

True

Suppose an E. coli cell is in an environment that is rich in glucose and poor in lactose. Which statement best explains gene expression of enzymes used in lactose breakdown? A. A repressor binds to the operator site of the lac operon and inhibits transcription of the genes involved in lactose breakdown. B. The high levels of glucose cause a cascade that ends with CAP binding to RNA polymerase, which activates transcription of the genes involved in lactose breakdown. C. Glucose binds to a repressor protein, which binds to the operator site and blocks RNA polymerase from transcribing the genes involved in lactose breakdown. D. Glucose binds to the operator site of the lac operon, repressing transcription of the genes involved in lactose breakdown.

A. A repressor binds to the operator site of the lac operon and inhibits transcription of the genes involved in lactose breakdown.

How does nucleotide-excision repair differ from base-excision repair? A. Base-excision repair recognizes and removes single damaged bases, whereas nucleotide-excision repair is more general, recognizing many different kinds of lesions that distort the DNA molecule. B. Nucleotide-excision repair reverses the chemical reaction that caused the lesion, whereas base-excision repair removes the damaged bases and replaces them with normal ones. C. Only the base is removed in base-excision repair, whereas the entire nucleotide is removed in nucleotide-excision repair. D. Base-excision repair requires no protein components and can occur by simple absorption of UV light, whereas nucleotide-excision repair requires several enzymes.

A. Base-excision repair recognizes and removes single damaged bases, whereas nucleotide-excision repair is more general, recognizing many different kinds of lesions that distort the DNA molecule.

Which statement about translational initiation is false? A. In prokaryotes, ribosomes often bind the mRNA and can scan 5ʹ or 3ʹ until recognizing a Shine-Dalgarno sequence. B. Viral mRNAs contain internal ribosome entry sites that allow ribosomes to bind to an internal site of the mRNA. C. Initiation codons in prokaryotic cells are preceded by Shine-Dalgarno sequences. D. 5ʹ 7-methylguanosine caps serve as the point of recognition and binding for ribosomes in eukaryotic cells.

A. In prokaryotes, ribosomes often bind the mRNA and can scan 5ʹ or 3ʹ until recognizing a Shine-Dalgarno sequence.

Which rationale best describes the ability of an enhancer to mediate transcription from very distant sites? A. Looping of the DNA can occur, allowing the transcription factor to come into proximity of the RNA polymerase. B. When needed, enhancers are spliced into a region closer to the promoter. C. They are recognized by RNA polymerase, which binds to the enhancers and then slides down the promoter toward the gene. D. All of the above

A. Looping of the DNA can occur, allowing the transcription factor to come into proximity of the RNA polymerase.

Which statement about translational regulation of ferritin is false? A. The iron response element is a unique sequence of amino acids near the amino terminus of the growing polypeptide. B. In the absence of iron, an iron regulatory protein binds the IRE, preventing translation. C. In the presence of iron, the ferritin protein is translated. D. The iron binding protein must bind the mRNA within about 70 bases of the 5ʹ mRNA cap.

A. The iron response element is a unique sequence of amino acids near the amino terminus of the growing polypeptide.

Which statement is true of all known DNA polymerases? A. They synthesize DNA in the 5ʹ to 3ʹ direction, and they require a preformed primer hydrogen-bonded to the template. B. They synthesize DNA in the 5ʹ to 3ʹ direction, and they possess primase activity. C. They require a preformed primer, and they possess helicase activity. D. They synthesize DNA in the 3ʹ to 5ʹ direction, and they possess exonuclease activity.

A. They synthesize DNA in the 5ʹ to 3ʹ direction, and they require a preformed primer hydrogen-bonded to the template.

Glycosylphosphatidylinositol (or GPI) anchors are found A. at the carboxy terminus of membrane proteins. B. at the amino terminus of membrane proteins. C. on the cytoplasmic side of membrane channel proteins. D. on the amino terminus of the 28S of certain ribosomal complexes associated with secretion.

A. at the carboxy terminus of membrane proteins.

The function of aminoacyl tRNA synthetases is to A. covalently attach amino acids to their corresponding tRNA molecules. B. synthesize tRNA molecules. C. catalyze the formation of the aminoacyl ATP intermediate during amino acid attachment to tRNAs. D. catalyze the formation of a peptide bond between amino acids.

A. covalently attach amino acids to their corresponding tRNA molecules.

A reporter gene is used to ... A. identify regulatory sequences from the upstream regions of other genes. B. determine if a protein binds to a given sequence element. C. determine if a gene contains introns. D. determine the stability of a protein.

A. identify regulatory sequences from the upstream regions of other genes.

A ribozyme is defined as an enzyme A. in which an RNA molecule is responsible for the catalytic activity. B. that catalyzes the cleavage of RNA. C. that is involved in translation. D. that catalyzes the addition of ribose moieties to RNA.

A. in which an RNA molecule is responsible for the catalytic activity.

The genes that encode the immunoglobulin light chain consist of regions that are called ... A. variable (or V); constant (or C); joining (or J) B. variable ( or V); diversity (or D); joining (or J) C. constant (or C); diversity (or D); joining (or J) D. variable ( or V); diversity (or D); constant (or C)

A. variable (or V); constant (or C); joining (or J)

What are three ways in which an enhancer's function differs from that of a promoter?

An enhancer can (1) act at great distance from the initiation site it regulates; (2) be either upstream or downstream from the initiation site; and (3) function in either orientation (forward or backward).

Suppose that a researcher wishes to determine the effect of shutting off production of protein Z in the cell. How could production of protein Z be shut off without altering any of Z's transcriptional control?

An excellent example is ferritin, an iron-storage protein, that is regulated by iron at the translational level. When iron is scarce, a protein binds to a sequence at the 5ʹ end of the mRNA, called the iron response element (IRE), and inhibits translation. In the presence of iron, the IRE-binding protein is released from the mRNA and translation resumes. This regulatory mechanism can be exploited for controlling protein Z's expression by inserting an IRE in the 5ʹ region (within 70 nucleotides of the cap) of the gene. Growing the cells in media containing low levels of iron should effectively inhibit X translation, creating observable effects.

According to the central dogma of molecular biology, transcription of genetic information occurs via A. DNA-dependent DNA synthesis. B. DNA-dependent RNA synthesis. C. RNA-dependent DNA synthesis. D. RNA-dependent protein synthesis.

B. DNA-dependent RNA synthesis.

Cells regulate gene expression to conserve which resource? A. Water B. Energy C. Nucleotides D. Amino acids

B. Energy

In Drosophila, genes encoding heat shock proteins are normally not expressed but are arrested at an early transcription phase. However, if these organisms are exposed to sudden bursts of heat, transcription elongation is stimulated and the heat shock proteins are expressed. What is the most likely explanation that could account for this heat shock effect? A. Heat causes repression of HAT expression. B. Heat stimulates PTEFb expression. C. Heat leads to dephosphorylation of the C terminal domain of RNA polymerase. D. Heat causes expression of an inhibitor of PTEFb.

B. Heat stimulates PTEFb expression.

A research group working with E. coli isolated several lac operon mutants. The group measured b-galactosidase activity under different environmental conditions, as shown in the table below. Which statement draws a plausible conclusion about one of these mutants? A. Mutant I has an operator region that no longer binds repressor. B. Mutant II has a repressor that no longer binds to the operator. C. Mutant III has a promoter site that no longer binds RNA polymerase. D. Mutant IV has a mutation in the b-galactosidase gene, rendering its gene product inactive.

B. Mutant II has a repressor that no longer binds to the operator.

In eukaryotic gene expression, how do promoters differ from enhancers? A. Promoters are regulatory sequences that are usually far upstream from the coding region they control, while enhancers are usually within a few hundred base pairs of a coding region. B. Promoters are often active at low levels, allowing for a minimal level of gene expression, while enhancers are utilized only to up-regulate the level of gene expression. C. Enhancers are strictly on/off regulatory elements, while promoters determine the level of transcription at any given time. D. Promoters are typically able to bind only specific transcription factors, while a single enhancer may up-regulate several different genes.

B. Promoters are often active at low levels, allowing for a minimal level of gene expression, while enhancers are utilized only to up-regulate the level of gene expression.

Which statement about the attachment of amino acids to tRNAs is false? A. The amino acid is first joined to AMP, forming an aminoacyl AMP intermediate. B. Two molecules of ATP are required for the process, one at each step. C. Aminoacyl tRNA synthetases catalyze the reaction. D. The amino acid is transferred to the 3ʹ end of the tRNA.

B. Two molecules of ATP are required for the process, one at each step.

A scientist used a mammalian cell extract to see whether it contained any chromatin remodeling factors. She incubated the cell extract with and without ATP in the presence of chromatin containing promoters X, Y, and Z and tested whether RNA polymerase would bind to any of the promoters. The table below summarizes the results. Do the results suggest that a chromatin remodeling factor was present in the cell extract? A. Yes, because Promoter Y binds RNA polymerase independent of the presence of ATP. B. Yes, because Promoter X binds RNA polymerase only when ATP is present. C. No, because Promoter Z cannot bind RNA polymerase even when ATP is absent. D. No, because there is not enough data to tell whether a chromatin remodeling factor may be present.

B. Yes, because Promoter X binds RNA polymerase only when ATP is present.

Protein phosphatases A. catalyze the addition of phosphate residues to proteins. B. catalyze the removal of phosphate residues from proteins. C. catalyze the addition of glycosylphosphatidylinositol to proteins. D. are proteins that specifically bind phosphorylated proteins.

B. catalyze the removal of phosphate residues from proteins.

Ferritin expression is stimulated by iron because iron A. stimulates a protein to bind the ferritin mRNA and inhibit its degradation. B. stimulates the dissociation of a translational inhibitor from the ferritin mRNA. C. binding stabilizes the ferritin protein. D. stimulates the extension of poly-A tails of ferritin mRNAs.

B. stimulates the dissociation of a translational inhibitor from the ferritin mRNA.

The twisting of the parental DNA strands around each other ahead of a replication fork is relieved by enzymes called A. DNA helicases. B. topoisomerases. C. DNA ligases. D. DNA polymerases.

B. topoisomerases.

What is the main difference between base-excision repair and nucleotide-excision repair?

Base excision removes one nucleotide; nucleotide excision replaces an oligonucleotide.

Suppose that gene X is induced in E. coli cells when they are treated with high levels of ethanol. A mutation in region 1 in the upstream region of gene X results in high-level expression under basal conditions, and a mutation in region 2 results in an uninducible gene. What would you expect from a strain in which both regions 1 and 2 are mutated?

Because a mutation in region 1 causes high-level expression even in the absence of ethanol, region 1 must bind a transcriptional repressor and regulation of gene X must occur by relief of repression. Since a mutation in region 2 blocks inducibility of the gene, region 2 must bind a transcriptional activator: In the absence of an activator, the repressor becomes irrelevant, and the gene is inactive, regardless of conditions. Mutations in both regions 1 and 2 would have the same effect as the single mutation in region 2, because in the absence of a bound transcriptional activator, the repressor is irrelevant, and the gene is inactive regardless of the presence or absence of ethanol.

Explain how a prokaryotic chromosome can replicate and divide normally with no telomeres and only one origin of replication.

Because a prokaryotic chromosome is circular, it has no ends, so no telomeres are required. Because prokaryote genomes are a single small circle and the rate of prokaryotic chromosome replication is faster than that of eukaryotes, a single origin of replication is sufficient.

What function does covalent modification of proteins serve in the cell that proteolysis cannot?

Both mechanisms regulate enzymes, but covalent modification permits recycling of proteins; proteolysis is irreversible.

Suppose that a gene has three exons and two introns in the following order: 5ʹ - exon 1 - intron 1 - exon 2 - intron 2 - exon 3 - 3ʹ Which of the following could not result from alternative splicing in this gene? A. 5ʹ - exon 1 - exon 2 - exon 3 - 3ʹ B. 5ʹ - exon 2 - exon 3 - 3ʹ C. 5ʹ - exon 2 - exon 1 - 3ʹ D. 5ʹ - exon 1 - exon 3 - 3ʹ

C. 5ʹ - exon 2 - exon 1 - 3ʹ

Which process both stabilizes and increases the efficiency of translation of an mRNA? A. Editing B. Splicing C. Addition of a 7-methylguanosine cap D. Addition of the CCA sequence to the 3ʹ end

C. Addition of a 7-methylguanosine cap

mRNAs are degraded in the cytoplasm by A. shortening of the poly-A tail and degradation from the 3′ end by nucleases. B. removal of the 5′ cap and degradation from the 5′ end by nucleases. C. Both a and b D. Neither a nor b

C. Both a and b

A research group characterized regions of DNA within chromatin from mouse cells. They also characterized the histones flanking these regions. The table below summarizes some of their results. Which statement provides a plausible hypothesis based on these data? A. Regions I and IV are promoters. B. Region III is an enhancer. C. Region II is a promoter. D. Regions I, II, and IV are genes under active transcription.

C. Region II is a promoter.

Which of the following statements regarding somatic hypermutation is false? A. The enzyme activation-induced deaminase (AID) is a key player in somatic hypermutation. B. Somatic hypermutation is thought to be the result of a high frequency of errors during DNA repair. C. Somatic hypermutation is thought to control the proliferation of B lymphocytes by rendering their genome irreplicable. D. Somatic hypermutation substantially increases affinity for antigen.

C. Somatic hypermutation is thought to control the proliferation of B lymphocytes by rendering their genome irreplicable. Rationale: C is correct. Somatic hypermutation does not affect proliferation of B lymphocytes. The other statements are all correct.

A student made the following table to summarize what he had learned about bacterial transcriptional control mechanisms. A fellow classmate pointed out that the table is not accurate. Which statement describes the problem? A. The terms listed for the types of regulatory proteins for both positive and negative control systems are incorrect. B. The wild-type regulatory protein in positive control systems causes transcription of the operon to be turned off. C. The mutated nonfunctional regulatory protein in negative control systems causes transcription of the operon to be turned on. D. Both the mutated nonfunctional regulatory protein in positive and negative control systems cause transcription of the operon to be turned on.

C. The mutated nonfunctional regulatory protein in negative control systems causes transcription of the operon to be turned on.

Which of the following statements about transcriptional termination in prokaryotes is false? A. Termination is signaled by a GC-rich interval that forms a stem-loop structure in the RNA. B. Transcription terminates when the RNA polymerase dissociates from its DNA template. C. The ribosome comes to a UAA, UAG, or UGA stop codon and transcription ceases. D. A segment of RNA froms a stable stem-loop structure by complementary base pairing.

C. The ribosome comes to a UAA, UAG, or UGA stop codon and transcription ceases. Rationale: C is correct. The ribosome is involved in translation, not transcription. The three other choices describe processes involved in transcription termination.

Some bacterial promoter regions are regulated by _______ that stimulate RNA polymerase binding and by _______ that block RNA polymerase binding. A. inducers; suppressors B. triggers; arrestors C. activators; repressors D. promoters; operators

C. activators; repressors

T cell receptors bind to _______ on the surface of other cells during immune responses. A. antibodies B. B cells C. antigens D. T cell proteins

C. antigens

The primary function of rRNAs in the ribosome is to A. serve as a scaffold for the ribosomal proteins. B. assist in the proper positioning of tRNAs along the mRNA template. C. catalyze peptide bond formation. D. assist in the proper folding of ribosomal proteins.

C. catalyze peptide bond formation.

The proteins shaped like a "double chamber" that are involved in protein folding are called A. Hsp70 proteins. B. protein disulfide isomerases. C. chaperonin proteins. D. peptidyl prolyl isomerases.

C. chaperonin proteins.

Eukaryotic transcriptional activators have activation domains that interact with A. DNA binding domains. B. enhancers and promoters. C. mediator proteins and transcription factors. D. histones.

C. mediator proteins and transcription factors.

During mismatch repair in E. coli, the parental strand is recognized by A. single-stranded breaks. B. glycosylated adenines. C. methylated adenines. D. methylation of the O6 position of guanine residues.

C. methylated adenines.

The initiator codon in prokaryotes is A. the first codon located at the 5ʹ end of the mRNA. B. recognized by scanning of the ribosome downstream of the 5ʹ 7-methylguanosine cap. C. recognized via the Shine-Dalgarno sequence. D. the first 5ʹ AUG of the mRNA.

C. recognized via the Shine-Dalgarno sequence.

Which of the following RNAs is RNA polymerase II not responsible for transcribing? A. mRNA B. miRNA C. tRNA D. lncRNA

C. tRNA

Which statement regarding tRNAs is false? A. tRNAs are approximately 70-80 bases long and form a cloverleaf structure. B. All tRNAs have a CCA sequence at their 3ʹ terminus. C. tRNAs differ in sequence only at the anticodon. D. There are several modified bases present in mature tRNAs.

C. tRNAs differ in sequence only at the anticodon.

Which of the following enzymes serves an important role in both somatic hypermutation and class switch recombination? A. Helicase B. Exonuclease C. DNA glycosylase D. Activation-induced deaminase E. Serine-threonine kinase

D. Activation-induced deaminase

Estimates of mutation rates for a variety of genes indicate that the frequency of errors during replication is much lower than would be predicted on the basis of complementary base pairing. What accounts for the higher degree of fidelity? A. Conformational changes in DNA polymerase B. 3ʹ to 5ʹ exonuclease activity of DNA polymerase C. Requirement of a primer for DNA synthesis by DNA polymerase D. All of the above

D. All of the above

Which statement about cis-acting elements is true? A. They are specific DNA sequences that control the transcription of adjacent genes. B. Various proteins specifically recognize and bind to these cis-acting sequences. C. They may be directly adjacent to the gene they control or far away. D. All of the above

D. All of the above

Which statement concerning elongation of DNA at the replication fork is false? A. The leading strand is synthesized continuously in the direction of replication fork movement. B. The lagging strand is synthesized in Okazaki fragments backward from the overall direction of replication. C. The Okazaki fragments are joined by the action of DNA ligase. D. Both strands are synthesized continuously at the replication fork.

D. Both strands are synthesized continuously at the replication fork.

Which is a mechanism of histone change that is mediated by chromatin remodeling factors? A. Proteolytic breakdown of histone proteins B. Acetylation of histone proteins C. Ubiquitination of histone proteins D. Conformational change of histone tertiary structure

D. Conformational change of histone tertiary structure

Which process is not a common lipid modification to proteins? A. N-myristylation B. Prenylation C. GPI anchor addition D. Glycosylation

D. Glycosylation

What is the major difference between histone modification and chromosome remodeling? A. Histone modification enzymes excise nucleosomes from the chromatin while chromatin remodeling enzymes add phosphate, acetyl, or methyl groups to the entire nucleosome. B. The histone modification enzymes remove methyl and acetyl groups from nucleosomes, and chromosome remodeling enzymes replace them. C. Histone modification enzymes recruit repressors and activators, while chromosome remodeling enzymes recruit corepressors and coactivators. D. Histone modification enzymes add side groups to the nucleosome proteins, while chromosome remodeling enzymes move, alter, or remove entire nucleosomes.

D. Histone modification enzymes add side groups to the nucleosome proteins, while chromosome remodeling enzymes move, alter, or remove entire nucleosomes.

The lac operon is regulated by both a positive transcriptional control system and a negative transcriptional control system. Under what conditions will both systems operate to stimulate transcription of the lac operon? A. High glucose and high allolactose B. Low glucose and low allolactose C. High glucose and low allolactose D. Low glucose and high allolactose

D. Low glucose and high allolactose

Which of the following is not an example of posttranslational modification? A. Glycosylation B. Proteolysis C. Palmitoylation D. Self-splicing

D. Self-splicing

Which statement about pyrimidine dimers is false? A. They are lesions in DNA caused by UV radiation. B. They are formed between adjacent pyrimidines on a DNA strand. C. Their formation blocks DNA replication and transcription. D. They can be repaired by photoreactivation in human cells.

D. They can be repaired by photoreactivation in human cells.

A eukaryotic regulatory protein controls expression of a gene. Some data regarding this protein are listed in the table below. Which hypothesis about the mechanism of this protein can be ruled out? A. This protein blocks an activating transcription factor from binding to the promoter. B. This protein prevents RNA polymerase from binding to the promoter. C. This protein competes with a transcription activator for binding to an enhancer. D. This protein binds to Mediator proteins in the transcription initiation complex.

D. This protein binds to Mediator proteins in the transcription initiation complex.

cAMP activates cAMP-dependent protein kinase by A. stimulating its phosphorylation. B. stimulating the dimerization of kinase subunits. C. stimulating the release of a translational inhibitory protein bound to its mRNA. D. binding regulatory subunits and inducing their release from the catalytic subunits.

D. binding regulatory subunits and inducing their release from the catalytic subunits.

The 5′ end of messenger RNA is modified by A. polyadenylation. B. polyribothymidine addition. C. capping with methylation. D. capping with 7-methylguanosine.

D. capping with 7-methylguanosine.

The lac operon in E. coli is regulated by lactose, which ... A. activates an activator of transcription. B. inactivates an activator of transcription. C. activates a repressor of transcription. D. inactivates a repressor of transcription.

D. inactivates a repressor of transcription.

The first amino acid of eukaryotic polypeptides is A. the amino acid encoded by the first 5ʹ codon. B. valine. C. N-formylmethionine. D. methionine.

D. methionine.

The nuclear localization signal (NLS) is not removed (cleaved off) after entry of the protein into the nucleus. In contrast, the targeting sequences of proteins destined for organelles such as the mitochondria are located at the N-terminus of the protein and are cleaved off once the proteins reach the lumen of the organelle. What are two likely reasons that the nuclear localization signal (NLS) is not cleaved off for nuclear proteins?

During mitosis, the nuclear envelope disintegrates, and nuclear proteins are released into the cytoplasm. Once the envelope re-forms around the daughter nuclei, the nuclear proteins need to be re-transported into the nuclei, explaining a need for retention of NLS. In some organisms, such as yeast, the nuclear envelope remains intact during mitosis. Thus, there are likely other reasons for retention of NLS that are of equal or greater importance. A second possible reason is that nuclear localization signals are internal to the polypeptide. Cleaved targeting sequences are typically located at the N-terminus of the protein. Cleavage of an internal sequence has much more consequence for the overall structure of the protein than removal of an N-terminal sequence. In addition, many nuclear proteins shuttle in and out of the nucleus, which would be a third reason for retaining the NLS.

You know that there is a protein in a particular cell line that binds within the first 200 bp of a gene's promoter to activate its transcription. You have another cell line in which the same gene is not expressed, and you suspect that this is due to the lack of binding to this protein. How could electrophoretic-mobility shift assays (EMSA) help test your hypothesis?

EMSA is an excellent method for determining if cellular proteins interact on a particular piece of DNA. First, the 200 bp piece of DNA would be radiolabeled. It would then be incubated with nuclear extracts from the two cell lines and a no-extract control. The reaction with no-extract or probe-only control would yield an unbound probe at the bottom of the gel. If proteins were present in the nuclear extracts that bound the probe, the protein bound to the probe would slow or retard the migration of the probe through the gel, resulting in a "shift" of the probe to a point higher in the gel than the unbound free probe. If the hypothesis is correct, the extract from the cell line with the gene that was not expressed would result in a pattern that looks very similar to that of the probe-only control. The extract from the cell line that does express the gene would have a band that is shifted up in the gel.

Chromatin remodeling factors are responsible for shuffling and rearranging DNA sequences in chromatin to make them either more or less accessible for transcription.

False

Somatic hypermutation is focused in the diversity region of the immunoglobulin gene.

False

True or False: The promoter sequence is transcribed first, followed by the DNA of the gene.

False

Explain the two steps involved in pre-mRNA splicing.

First, the pre-mRNA is cleaved at the 5′ splice site, and the 5′ end of the intron is joined to an adenine nucleotide within the intron. Second, there is simultaneous cleavage at the 3′ splice site and ligation of the two exons.

In vitro experiments have shown that formation of the preinitiation transcription complex on RNA polymerase II promoters requires the dephosphorylated form of the RNA polymerase II large subunit, whereas the phosphorylated form is associated with elongating complexes. Describe how the timing of phosphorylation might influence gene transcription.

If phosphorylation occurred before the formation of the preinitiation complex, then RNA polymerase II would not bind the promoter, and transcription would be inhibited. In contrast, phosphorylation after the formation of the preinitiation complex would favor the elongating form of the complex, and transcription would be stimulated. Thus, the timing of activation of kinases and phosphatases can be a factor in the regulation of gene transcription.

Unlike eukaryotic cells, prokaryotic cells do not have a nucleus. What are some possible benefits of the evolution of a nucleus in eukaryotes?

In contrast to prokaryotes, eukaryotic RNA is extensively modified by processes such as splicing and RNA editing, which are critical to the accurate transfer of genetic information from gene to protein. Sequestering the chromosomes in a separate compartment allows RNA-processing events to be more efficient, because the enzymes involved are sequestered in the nucleus and hence have less interference from unrelated enzymes. In addition, because RNA is retained in the nucleus until the processing events are complete, inappropriate translation of unprocessed RNAs does not occur. Furthermore, an additional level of control comes from the entry of regulated transcription factor into the nucleus. This would not be possible if the cytosol were not separated from the nucleus and chromosomes.

How can prokaryotes and eukaryotes tell which is the new strand in mismatch repair?

In prokaryotes, template strands are methylated; newly synthesized strands are not. This allows repair enzymes to differentiate between the template and the new strand. IN eukaryotes, the mismatch is recognized by an enzyme that nicks the new strand. Repair enzymes then excise a portion of the nicked strand, and DNA polymerase replaces it with the proper strand.

Why does DNA polymerase synthesize DNA only in the 5ʹ to 3ʹ direction?

In the 5ʹ to 3ʹ direction, the energy for polymerization comes from hydrolysis of the 5ʹ triphosphate of the free dNTP as it is added to the 3ʹ hydroxyl group of the existing chain. If the direction were 3ʹ to 5ʹ, proofreading repair could not be performed because the removal of the terminal nucleotide would leave no 5ʹ triphosphate to provide the energy to bond to the 3ʹ hydroxyl group of the incoming nucleotide triphosphate.

Explain why accuracy is more important for DNA replication than for transcription.

Mistakes made during DNA replication are heritable and thus are passed on to generations of cells in a lineage. The accumulation of such mistakes would eventually lead to a breakdown of essential processes in the cell and cell death. In contrast, mistakes made in RNA transcription are not heritable, and the cell can afford a few aberrant RNA transcripts as long as the majority encode the wild-type protein. Hence, an elaborate system of proofreading has evolved for DNA replication but not for RNA transcription.

A research lab working with mouse cells characterized three mutant cell lines summarized in the attached tableShortanswer Nr9.png . Which of these is most likely to contain a mutated form of P-TEFb? Explain.

Mutant cell line X most likely carries a mutated form of P-TEFb because this protein is responsible for releasing paused RNA polymerase from its paused state, thereby allowing it to continue elongating messenger RNA transcripts.

Regulation of protein function is critical to cell physiology. Phosphorylation is frequently used to regulate protein function and is often referred to as an on/off switch for proteins. How are proteins phosphorylated, and what is a unique characteristic of phosphorylation that makes it such an ideal molecular switch?

Proteins are phosphorylated by protein kinases that covalently transfer a phosphate group to a protein. The phosphate group is generally supplied by ATP, which provides the energy necessary for the reaction. Proteins are phosphorylated on either serine/threonine or tyrosine amino acids by serine/threonine kinases or tyrosine kinases, respectively. Phosphorylation of proteins may activate or inactivate the function of a particular protein. The unique characteristic of phosphorylation is that it is readily reversible by protein phosphatases, so that a protein can easily be switched on or off simply by phosphorylation/dephosphorylation with no significant impact on the rest of the protein.

What are the expected phenotypic consequences of a mutation in Ran GAP that reduces its affinity for cytoplasmic filaments of the nuclear pore?

Ran is important to both protein import into and export out of the nucleus. As part of importin recycling to the cytoplasm, Ran in the GTP-bound form complexes with importin inside the nucleus. Formation of this complex results in the release of importin from cargo and the importin/Ran-GTP complex is then recycled back to the cytoplasm. During nuclear protein export in general, GTP-bound Ran binds to exportin inside the nucleus to give rise to a Ran-GFP/exportin/cargo complex that is exported as a unit through the nuclear pore into the cytoplasm. In either case, Ran functions as a GTP-dependent molecular switch. Ran GAP bound to the cytoplasmic filaments of the nuclear pore complex activates the GTPase activity of Ran in the cytoplasm and bound GTP is cleaved to GDP. If Ran GAP were not bound to the cytoplasmic filaments, its effective local concentration would be diluted, and the efficiency of Ran activation would be low. Low affinity would cause a distinct decrease in the Ran GTP gradient across the nuclear pore. This would reduce the efficiency of both import into the nucleus, which is dependent on importin recycling, and export from the nucleus. Phenotypically, cell growth rates would slow and the defect might be so severe as to be lethal.

Prepare a spider map of the four mechanisms that contribute to antibody diversity. Provide a brief explanation of each mechanism

Site-specific recombinations, combining different light chains and heavy chains, class switch recombination, somatic hypermutation

Explain what TFIIH is, including its function in initiating transcription.

TFIIH is a multisubunit factor that has two roles: The first is to act as a helicase to unwind DNA around the initiation site. The second role is to act as a protein kinase that phosphorylates repeated sequences on the CTD of RNA polymerase II.

Suppose that a single protein is significantly overexpressed in one of two different cell lines. The proteins are identical, but Northern blot analysis indicates that the mRNA for the protein in the overexpressing cell line is about 200 base pairs shorter. Subsequent reverse transcriptase PCR has revealed that the truncation occurs in the 3ʹ untranslated region. How might this lead to overexpression of the protein?

The 3ʹ UTR can serve several roles in regulating mRNA translation. It is likely that there is a translational repressor that recognizes a specific sequence and interacts with eIF-4E and prevents initiation in the full-length mRNA. In the cell line with the truncated mRNA, that overexpresses the protein, the repressor is lost; the loss of repression allows the protein to be expressed continuously, as long as the mRNA is present.

What would happen in a bacterium in which the Shine-Dalgarno sequence was mutated?

The Shine-Dalgarno sequence is required for the 16S ribosomal RNA to base pair match with and direct the initiation of translation. A mutation in this sequence would prevent 16S ribosomal RNA attachment and block or greatly impair protein synthesis.

Okazaki fragments are a consequence of

The ability of DNA polymerase to synthesize only in the 5' to 3' direction.

Sixty-four different codons translate into only 20 amino acids. How are tRNAs adapted to address this?

The genetic code is redundant, meaning that most amino acids are specified by more than one codon. Three of these codons signal translation termination, which leaves 61 triplets to code for amino acids. Although some amino acids are linked to more than one tRNA, a type of base pairing called "wobble" allows a single tRNA anticodon to recognize more than one codon. Thus G, in addition to pairing with its favored partner, C, can also pair with U; while inosine (I), present in several tRNA anticodons, can pair with C, U, or A. Thus, for example in E. coli, wobble reduces the number of necessary tRNAs from 61 to 40.

UV irradiation often induces DNA damage in the form of pyrimidine dimers. Briefly describe the two mechanisms that cells have developed to repair this damage. Which mechanism is lacking in humans?

The most direct process is photoreactivation, which uses energy from visible light to break the cyclobutane ring structure formed by the two adjacent bases. Photoreactivation is not observed in humans. The other process is nucleotide excision repair, in which DNA on either side of the dimer is cleaved by 3ʹ and 5ʹ nucleases and the short fragment containing the dimer is excised out. The resulting gap is filled in by DNA polymerase.

Protein transport into the endoplasmic reticulum (ER) or mitochondria requires unfolding of the protein and threading the amino acid chain through a channel or pore into the organelle. In contrast, protein transport into the nucleus does not require unfolding of the protein. Why is unfolding unnecessary for nuclear protein import?

The nuclear pore complex is a huge structure composed of about 30 different pore proteins, and its central channel has a diameter of approximately 10-40 nm, which is large enough for even very large protein complexes. For this reason, it is unnecessary to unfold the protein, which can pass through the nuclear pore in its native state.

Why would chromosomes shorten with each replication cycle in the absence of telomerase?

The synthesis of the lagging strand involves the continual synthesis of RNA primers, which are then elongated by DNA polymerase, as the replication fork moves along a DNA molecule. The enzyme that synthesizes the primers, primase, is template-directed, meaning that it cannot synthesize a primer de novo but instead needs a template. This is problematic once the fork reaches the end of the chromosome because the template runs out. This problem is solved by telomerase, which extends the template strand by a repeat unit, thus providing a template for primase and, consequently, replication of the lagging strand to its full length.

What would be the likely result of a mutation in the enzyme protein disulfide isomerase (PDI)?

There would be significant problems with protein folding. PDI is responsible for the formation of disulfide bonds between cysteine residues, important in stabilizing the folded structures of many proteins.

Why are topoisomerases required in replication?

Topoisomerases are needed because as DNA unwinds, or as the two strands of the double helix get pulled apart, the helix ahead of the unwinding helix can get stuck (much like the strands of a twisted rope get stuck as one tries to separate them). Topoisomerases catalyze reactions that break the strand, allowing the helix to relax and then rejoin the strand. This is repeated over and over and prevents the torsional strain that would prevent unwinding.

MicroRNAs can regulate translation by targeting proteins that can either cleave an mRNA or repress its translation.

True

The activity of transcription factors is frequently regulated by controlling their localization within eukaryotic cells, such that only in response to a given stimulus is the transcription factor transported into the nucleus, where it can activate its target genes. For a transcription factor that has a nuclear localization signal (NLS), describe two mechanisms by which regulated transport can occur.

Two possible mechanisms involve masking of the NLS such that it is not recognized by importin. This masking can be: (1) intermolecular, whereby another protein binds and obscures the NLS, or (2) intramolecular, whereby the protein itself folds in such a way that the signal is masked. Upon stimulation, the masking protein dissociates or the protein structure is altered such that the NLS is exposed, and transport to the nucleus can occur. Other mechanisms are also possible. For example, the transcription factor may be tethered to another protein that is attached to a cytoplasmic structure, such as a membrane protein or a cytoskeletal protein. In this case, the nuclear localization signal is exposed, but the protein is physically retained in the cytoplasm until the appropriate stimulus liberates it, and it is free to make its way into the nucleus.

In eukaryotic cells, different mRNAs have different half-lives, and therefore are degraded at different rates. What is the usefulness of this wide range of half-lives?

Unstable mRNAs frequently code for regulatory proteins, and the levels of these proteins within the cell vary rapidly in response to environmental stimuli. mRNAs encoding metabolic enzymes or structural proteins generally have longer half-lives, and the levels of these proteins in the cell do not fluctuate frequently.

What is allosteric regulation of a protein?

When a protein is regulated allosterically, it is regulated by the binding of a molecule at a site other than the active site. The binding induces a conformational change in the protein that alters the active site, either increasing or decreasing the enzyme's activity.