Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank A+

$35.00
Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank A+

Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank A+

$35.00
Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank A+

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

CHAPTER 6: HOW CELLS READ THE GENOME: FROM DNA TO PROTEIN

© Garland Science 2015

1.1. DNA and RNA polymerase differ in all of the following EXCEPT…

  1. the nucleotide substrates they incorporate.
  2. their requirement for a primer.
  3. their error rate.
  4. the type of chemical reaction they catalyze.
  5. their processivity.

1.2. What enzyme is depicted in the following schematic drawing?

  1. DNA polymerase
  2. RNA polymerase
  3. Ribosome
  4. Reverse transcriptase
  5. Topoisomerase

1.3. The sequence of a region of DNA around the 5′ end of a gene in Escherichia coli is shown below. The –10 hexamer and the transcription start site are highlighted. What would be the sequence of the first 10 nucleotides of the mRNA transcribed from this gene? Write down the sequence from 5′ to 3′, e.g. CGGAUAAACT.

5′…GCGCTTGGTATAATCGCTGGGGGTCAAAGAT…3′

1.4. Due to their high transcription rate, active ribosomal RNA (rRNA) genes can be easily distinguished in electron micrographs of chromatin spreads. They have a characteristic “Christmas tree” appearance, where the DNA template is the “trunk” of the tree and the nascent RNA transcripts form closely packed “branches.” At the base of each branch is an RNA polymerase extending that branch, while RNA processing complexes at the tip of the branch form terminal “ornaments.” The top of the tree represents the … of the rRNA gene, and the “ornaments” are at the … end of the nascent rRNA molecules.

  1. end; 3′
  2. end; 5′
  3. beginning; either 3′ or 5′
  4. beginning; 3′
  5. beginning; 5′

1.5. Which of the following types of noncoding RNA chiefly functions in the processing and chemical modification of ribosomal RNAs (rRNAs)?

  1. Small nuclear RNAs (snRNAs)
  2. Small nucleolar RNAs (snoRNAs)
  3. Small interfering RNAs (siRNAs)
  4. Transfer RNAs (tRNAs)
  5. MicroRNAs (miRNAs)

1.6. For the bacterial transcription machinery, which of the following mRNA sequences would you expect to constitute a potent transcriptional termination signal? Note that the two underlined regions in each sequence are complementary to each other.

  1. 5’… UGGCCCAGUCGGAAGACUGGGCCUUUUGUUUU…3′
  2. 5’… UGGCCCAGUCGGAAGACUGGGCCCGCGGAGCU…3′
  3. 5’… UUUUGUUUUAGGCCCAGUCGGAAGACUGGGCCA…3′
  4. 5’… CGCGGAGCUAGGCCCAGUCGGAAGACUGGGCCA…3′

1.7. The transcript for which of the following noncoding RNA in our cells is expected to undergo 5′ cap addition after transcription?

  1. 5S rRNA
  2. miR-21 (a microRNA)
  3. tRNAPhe
  4. 5.8S rRNA
  5. 18S rRNA

1.8. This large and complex general transcription factor has a DNA helicase activity that exposes the template for RNA polymerase II transcription. It also has a kinase activity that phosphorylates the C-terminal domain of the polymerase on Ser5 leading to promoter clearance. It is…

  1. TFIIB
  2. TFIID
  3. TFIIE
  4. TFIIF
  5. TFIIH

1.9. This general transcription factor recognizes the TATA box in RNA polymerase II promoters. It is…

  1. the only single-subunit general transcription factor.
  2. able to introduce a rather sharp kink in the double helix upon binding to DNA.
  3. responsible for the phosphorylation of the RNA polymerase CTD during transcription initiation.
  4. also responsible for the recognition of the BRE element in the promoter.
  5. All of the above.

1.10. Of the following proteins or protein complexes, which one does NOT typically interact with an elongating RNA polymerase II?

  1. Histone-modifying enzymes
  2. Capping enzymes
  3. Chromatin remodeling complexes
  4. Mediator complex
  5. Histone chaperones

1.11. How does a eukaryotic cell deal with the superhelical tension in its genomic DNA resulting from the activity of RNA polymerases?

  1. DNA gyrase introduces negative supercoils, keeping the DNA under constant tension.
  2. The RNA polymerases are allowed to rotate freely around their templates during transcription, leading to the relaxation of the tension.
  3. DNA topoisomerases rapidly remove the superhelical tension caused by transcription.
  4. The nucleosomes adjust the tension by binding to positively supercoiled regions behind a moving RNA polymerase.
  5. All of the above.

1.12. Comparing mRNA molecules from human and Escherichia coli cells, which of the following is typically NOT true?

  1. A human mRNA has a special 5′ cap, while a bacterial mRNA does not.
  2. A human mRNA has a poly-A tail, while a bacterial mRNA does not.
  3. A human mRNA undergoes alternative splicing, while a bacterial mRNA does not.
  4. A human mRNA contains noncoding sequences, while a bacterial mRNA does not.
  5. A typical human mRNA encodes one protein, while many bacterial mRNAs encode several different proteins.

1.13. Eukaryotic pre-mRNAs undergo a number of modifications such as capping at the 5′ end. A 5′ cap…

  1. consists of a modified terminal adenine nucleotide.
  2. has a 3′-to-5′ linkage between the terminal nucleotide and the 5′ end of the pre-mRNA.
  3. contains a triphosphate bridge between the terminal base and the 5′ end of the pre-mRNA.
  4. carries a negative charge in the terminal base due to methylation.
  5. is identical for all mRNAs that are transcribed by RNA polymerase II.

1.14. After the first and before the second chemical step of RNA splicing, the intron of the pre-mRNA…

  1. is still covalently connected to the 3′ exon and has an internal branch in the shape of a lariat.
  2. is still covalently connected to the 3′ exon and is linear.
  3. is still covalently connected to the 5′ exon and has an internal branch in the shape of a lariat.
  4. is still covalently connected to the 5′ exon and is linear.
  5. is still covalently connected to both of its flanking exons and is linear.

1.15. In the following qualitative histogram, which two curves better correspond to human exon and intron length distributions, respectively?

  1. Curves 1 and 2
  2. Curves 2 and 1
  3. Curves 2 and 3
  4. Curves 3 and 2
  5. Curves 3 and 1

1.16. To ensure the fidelity of splicing, the spliceosome…

  1. hydrolyzes ATP to undergo complex rearrangements.
  2. examines the splicing signals on the pre-mRNA several times.
  3. assembles on the pre-mRNA co-transcriptionally.
  4. takes advantage of “exon definition.”
  5. All of the above.

1.17. A primary mRNA transcript with three exons is depicted below. Which of the following mature mRNA products of this transcript is a result of exon skipping?

1.18. The enzyme poly-A polymerase is responsible for adding 3′ poly-A tails to eukaryotic mRNAs. This enzyme…

  1. cuts the mRNA after recognition of the cleavage/polyadenylation signal by CstF and CPSF proteins.
  2. polymerizes the tail using an RNA template that is part of the enzyme.
  3. is extremely processive.
  4. normally adds about 1000 A nucleotides to the mRNA.
  5. uses ATP as a substrate.

1.19. Several mechanisms contribute to the diversity of the mRNAs and proteins encoded by a single gene in our genome. Which of the following is normally NOT one of them?

  1. Alternative choice of polyadenylation sites
  2. Alternative choice of translation initiation sites
  3. Alternative choice of transcription initiation sites
  4. Alternative choice of the reading frames
  5. Alternative choice of splice sites

1.20. Which of the following better describes a typical, actively translated mRNA in its journey from the nucleus to the cytosol?

  1. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (5′ end first), and remains linear.
  2. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (3′ end first), and remains linear.
  3. Initially linear, the mRNA enters the cytosol (5′ end first), and adopts a circular conformation.
  4. Initially linear, the mRNA enters the cytosol (3′ end first), and adopts a circular conformation.
  5. Initially linear, the mRNA enters the cytosol (3′ end first), and remains linear.

1.21. As an mRNA molecule is processed in the nucleus, it loses some proteins and binds to new ones, some of which are used in mRNA surveillance pathways. The presence of which of the following molecules on an mRNA is a signal that the mRNA is still NOT ready for nuclear export?

  1. Cap-binding complex
  2. Exon junction complex
  3. snRNPs used in splicing
  4. poly-A-binding proteins
  5. SR proteins

1.22. The nucleolus is a dynamic subcompartment within the nucleus and its size varies depending on the circumstances. In which of the following cells would you NOT expect to see the nucleoli?

  1. A yeast cell undergoing DNA replication
  2. A human neuron in a quiescent (G0) state
  3. A human macrophage active in phagocytosis
  4. A fruit fly embryonic nucleus in the G2 phase of the cell cycle
  5. A mouse embryonic cell in the metaphase stage of mitosis

1.23. Cajal bodies in the eukaryotic cell nucleus…

  1. are stockpiles of fully mature snRNPs and other RNA processing components.
  2. can only be observed by electron microscopy.
  3. are absolutely required in all cell types.
  4. speed up the maturation and assembly of snRNPs and snoRNPs.
  5. are the main sites of pre-mRNA splicing.

1.24. Which of the following processes takes place in the nucleoli within the eukaryotic nucleus?

  1. Ribosome assembly
  2. rRNA gene transcription
  3. Telomerase assembly
  4. tRNA processing
  5. All of the above

1.25. Which of the following is true about the molecule shown in the following drawing?

  1. Its gene is transcribed by RNA polymerase I.
  2. There are about 50 genes in our genome encoding this type of molecule.
  3. This molecule normally undergoes various covalent modifications.
  4. It is normally composed of about 20 monomers.
  5. Its gene transcript normally undergoes alternative splicing.

1.26. How is tRNA splicing different from mRNA splicing in eukaryotic cells?

  1. tRNA splicing does not proceed via transesterification reactions.
  2. tRNA splicing is carried out by proteins only.
  3. tRNA splicing does not create a lariat intermediate.
  4. tRNA splicing involves RNA endonuclease and RNA ligase activities.
  5. All of the above.

1.27. The modified nucleotide shown below is normally found in mature…

  1. tRNAs.
  2. ribosomes.
  3. spliceosomes.
  4. snoRNAs.
  5. All of the above.

1.28. Each aminoacyl-tRNA synthetase activates a certain amino acid by attaching it to its corresponding tRNA(s) through a high-energy linkage…

  1. between the amino group of the amino acid and the 3′ hydroxyl group of the tRNA.
  2. between the amino group of the amino acid and the 5′ hydroxyl group of the tRNA.
  3. between the carboxyl group of the amino acid and the 3′ or 2′ hydroxyl group of the tRNA.
  4. between the carboxyl group of the amino acid and the 5′ hydroxyl group of the tRNA.
  5. between the amino group of the amino acid and the 3′ or 2′ hydroxyl group of the tRNA.

1.29. Activation of an amino acid by an aminoacyl-tRNA synthetase involves the covalent linkage of…

  1. AMP to the amino acid.
  2. ADP to the amino acid.
  3. AMP to the tRNA.
  4. ADP to the tRNA.
  5. ADP to the enzyme.

1.30. On the ribosome, the mRNA is read from …, and the polypeptide chain is synthesized from…

  1. 5′ to 3′; C- to N-terminus.
  2. 5′ to 3′; N- to C-terminus.
  3. 3′ to 5′; C- to N-terminus.
  4. 3′ to 5′; N- to C-terminus.

1.31. An elongating ribosome is bound to appropriate tRNAs in both the A and the P sites and is ready for peptidyl transfer. What happens next?

  1. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site tRNA.
  2. The amino end of the polypeptide chain is released from the P-site tRNA and joined to the free carboxyl group of the amino acid linked to the A-site tRNA.
  3. The carboxyl end of the amino acid is released from the A-site tRNA and joined to the free amino group of the polypeptide chain linked to the P-site tRNA.
  4. The amino end of the amino acid is released from the A-site tRNA and joined to the free carboxyl group of the polypeptide chain linked to the P-site tRNA.

1.32. Which of the following nucleotides is hydrolyzed in both transcription and in translation elongation?

  1. ATP
  2. GTP
  3. TTP
  4. UTP
  5. CTP

1.33. How fast does a bacterial ribosome move on an mRNA?

  1. At about 2 nucleotides per second, significantly lower than the speed of the RNA polymerase.
  2. At about 5 nucleotides per second, comparable to the speed of the RNA polymerase.
  3. At about 10 nucleotides per second, significantly lower than the speed of the RNA polymerase.
  4. At about 20 nucleotides per second, comparable to the speed of the RNA polymerase.
  5. At about 60 nucleotides per second, comparable to the speed of the RNA polymerase.

1.34. Which of the following features is common between the bacterial and eukaryotic ribosomes in translation initiation?

  1. They both use an initiator tRNA that carries formylmethionine.
  2. They both bind to the 5′ end of the mRNA and move forward to find the start codon.
  3. They both recognize the start codon by interacting with the Shine–Dalgarno sequence.
  4. They both interact with various translation initiation factors.
  5. All of the above.

1.35. The following mRNA sequence is taken from the middle of exon 3 in a mature mRNA that has 12 exons. Knowing that this mRNA does not undergo nonsense-mediated decay, which of the reading frames shown is the correct one for this mRNA? Write down 1, 2, or 3 as your answer.

1.36. Polysomes…

  1. are large cytoplasmic assemblies made of several ribosomes each translating their exclusive mRNA.
  2. are only found in the eukaryotic cytoplasm.
  3. can take advantage of the circularization of eukaryotic mRNA (by interactions between the 5′ and 3′ ends of the mRNA) to further speed up the rate of protein synthesis.
  4. are mostly translationally inactive and are normally used by the cell to store the ribosomes and their associated mRNAs for future use.
  5. All of the above.

1.37. The specific transfer RNA used for the incorporation of selenocysteine in proteins recognizes the UGA codon, which is normally a translation stop codon. What prevents this tRNA from reading through all the other “legitimate” UGA stop codons in the cells and causing a massive, disastrous recoding?

  1. The cells that have selenocysteine in their proteins do not normally use the UGA stop codon and always use the other two stop codons (UAA and UAG) instead.
  2. This tRNA is made in such a small amount that its side effects are negligible.
  3. This tRNA only interacts with the UGA codon in the P site of the ribosome, and therefore does not interfere with the normal function of the codon in translation termination, which takes place in the A site.
  4. The mRNAs encoding the selenocysteine-containing proteins also contain additional sequences required for the recoding event.

1.38. The structural formulas for a few antibiotics that target the ribosome are shown below. Which one do you think is that of puromycin? Hint: puromycin mimics the structure of an amino acid linked to a tRNA molecule.

Source: Wikipedia.com License: Public domain.

1.39. In a gene that normally contains three exons, which of the following changes probably will NOT activate the nonsense-mediated mRNA decay pathway? The sizes of exons 1 to 3 are 100, 150, and 200 nucleotide pairs, respectively.

  1. A nonsense mutation in exon 1.
  2. A nonsense mutation in exon 2.
  3. A mutation in the first intron resulting in the inclusion of a large intronic fragment in the mature mRNA.
  4. A frameshift mutation in exon 1.
  5. A mutation in exon 2 leading to its loss through exon skipping.

1.40. This complex uses ATP and forms an “isolation chamber” for its misfolded protein substrates, allowing them to fold to their native conformation. It is called…

  1. a proteasome.
  2. a ribosome.
  3. Mediator.
  4. a chaperonin.
  5. an exosome.

1.41. What is the function of the 19S cap in the proteasome complex?

  1. Recognizing the proteins that will be degraded
  2. Unfolding the proteins that will be degraded and threading them into the central 20S cylinder for digestion
  3. Removing the polyubiquitin chain from the proteins that will be degraded
  4. Hydrolyzing ATP to make the digestion process highly efficient
  5. All of the above

1.42. How many ATP molecules must be hydrolyzed by the proteasome for the digestion of a protein that has been tagged for degradation with a polyubiquitin chain?

  1. None; the digestion does not require ATP hydrolysis and is exergonic
  2. One
  3. Two
  4. Ten
  5. Many; the number depends on the specific protein

1.43. In humans, nearly 80% of proteins are acetylated on their N-terminal residue, a modification known to be recognized by a specific E3 enzyme, which directs the ubiquitylation of the protein for rapid degradation. Does this mean that all of these proteins are actively degraded at any given time?

  1. Yes; this high turnover rate ensures that their activity is under tight control.
  2. Yes; but they are not fully degraded in this way and can still function as protein fragments.
  3. No; the destruction signal can be buried in the interior of the protein or bound to other proteins.
  4. No; the E3 enzyme recognizing this mark is inactive most of the time.

1.44. In the following representation of a single-stranded RNA molecule, the nucleotide residues are drawn as gray circles, while base-pairing interactions between them are indicated by dashed lines. The structure of this RNA molecule has…

  1. one hairpin loop and one pseudoknot.
  2. one pseudoknot only.
  3. three hairpin loops and a four-stem junction.
  4. two hairpin loops and a three-nucleotide bulge.
  5. three hairpin loops.

1.45. In the following diagram showing the overall flow of genetic information in all living cells, what is the name of the process indicated with a question mark?

1.46. Sort the following events in the order that they occur during transcription initiation in Escherichia coli. Your answer would be a four-letter string composed of letters A to D only; e.g. DCAB.

(A) Abortive initiation trials

(B) σ Factor release from the RNA polymerase holoenzyme

(C) Binding of the holoenzyme to the promoter in the “closed” complex

(D) Formation of the transcription bubble

1.47. Your friend who is studying the sequence conservation around the branch-point site on a subset of pre-mRNAs in Macaca mulatto (rhesus macaque) has sent you the following sequence logo of the region of interest, but has forgotten to indicate the position of the branch-point nucleotide itself within this region. Where do you think it is? Write down the position number (1 to 8) as your answer.

1.48. You have identified an RNA-binding protein that can bind directly (with varying affinities) to mRNA molecules that bear the following sequences near the 3′ splice sites of their exons. Based on these results, what is the five-nucleotide consensus sequence for the binding site of this protein? Your answer would be a five-letter string composed of letters A, U, C, or G only; e.g. UUUAG.

5’…AACGG…3′
5’…AUCGG…3′
5’…GACGC…3′
5’…AACCC…3′
5’…AUCGC…3′
5’…AACAC…3′
5’…AACCC…3′

1.49. In the following schematic drawing, a transcribing RNA polymerase is moving toward the right on the DNA template as indicated. If the polymerase is allowed to rotate freely while the DNA is kept from doing so, would the enzyme rotate in a right-handed (R) or left-handed (L) fashion around the template? Write down R or L as your answer.

1.50. Indicate true (T) and false (F) statements below regarding mRNA splicing in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

( ) Introns must be removed in the order in which they occur along the pre-mRNA.

( ) Nucleosomes tend to be positioned over exons (on average, about one nucleosome per exon).

( ) Exon size tends to be much more variable than intron size.

( ) The exon junction complexes mark the sites on the mRNA where splicing has not been successful.

1.51. No matter where translation begins, only one polypeptide sequence pattern can be obtained from the translation of an mRNA chain with the sequence (AC)n; that is, an mRNA with the sequence 5’…ACACACACAC…3′. In contrast, an mRNA chain with the sequence (ACG)n can be translated to polypeptide sequences of three distinct patterns, depending on the choice of the reading frame. Using similar judgment, how many different polypeptide sequence patterns can be obtained from an mRNA chain with the sequence (ACGU)n? Write down your answer in digits, e.g. 12.

1.52. Three consecutive nucleotides in RNA (such as AUC or GUA) constitute a triplet called a …, which can specify an amino acid or a stop signal for translation.

1.53. The anticodon sequence of a phenylalanine-specific tRNA is 5′-GAA-3′. Given the wobble base-pairing rules presented in the following table, how many different codons can be “read” by this tRNA? Write down your answer in digits, e.g. 12.

Wobble base in the codonPossible anticodon bases
UA, G, or I
CG or I
AU
GC

1.54. Indicate true (T) and false (F) statements below regarding the accuracy of translation by the ribosome. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

( ) The difference in the binding affinity of a codon to a correct versus incorrect anticodon CANNOT by itself account for the high accuracy of translation.

( ) GTP hydrolysis by EF-Tu both speeds up translation and increases its accuracy.

( ) The ribosome can detect correct Watson–Crick base-pairing between the codon and anticodon in the A site, and consequently trigger GTP hydrolysis by EF-Tu.

( ) Even after GTP hydrolysis by EF-Tu, the ribosome can prevent wrong amino acid incorporation by kinetic proofreading.

1.55. Indicate whether each of the following is located (or takes place) in the large (L) or the small (S) ribosomal subunit. Your answer would be a four-letter string composed of letters L and S only, e.g. LLLS.

( ) Peptide bond formation

( ) Codon–anticodon interaction

( ) The path of the mRNA

( ) The polypeptide exit tunnel

1.56. Indicate true (T) and false (F) statements below regarding the folding of proteins upon their synthesis. Your answer would be a five-letter string composed of letters T and F only, e.g. TFFFF.

( ) A globular protein can fold to its native form while still within the ribosome.

( ) A protein can begin to fold and to bind to other proteins while still being synthesized on the ribosome.

( ) A protein may undergo several cycles of hsp70 binding and release while still being synthesized on the ribosome.

( ) A molten globule can form only after the synthesis of the protein is complete.

( ) A protein typically undergoes hsp60-assisted folding while still being synthesized.

1.57. Fill in the blank. These RNA molecules can catalyze a wide variety of biochemical reactions, including peptide bond formation and RNA splicing. They can be found in nature or selected in vitro for a desired function. Such a molecule is called a(n) …

1.58. Indicate true (T) and false (F) statements below regarding the RNA world hypothesis. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

( ) According to this hypothesis, RNA in primitive cells was responsible for storing genetic information, while proteins were responsible for the catalysis of chemical reactions.

( ) The existence of natural ribozymes supports this hypothesis.

( ) In support of this hypothesis, all peptides in present-day cells are made by the ribosome.

( ) All present-day cells use DNA as their hereditary material.

Answers:

  1. Answer: D

Difficulty: 2

Section: From DNA to RNA

Feedback: Despite the lack of homology, the reactions catalyzed by these two polymerases are analogous, even though the enzymes work on different substrates. DNA polymerases require a primer and have a much lower error rate compared to RNA polymerases. They are also not as processive on their own.

  1. Answer: B

Difficulty: 2

Section: From DNA to RNA

Feedback: The enzyme synthesizes RNA using one of the DNA strands as template.

  1. Answer: GGUCAAAGAU

Difficulty: 3

Section: From DNA to RNA

Feedback: The mRNA sequence starts from the transcription start site and is the same as the sequence of the coding strand of the DNA (and complementary to the “template strand”) except that T is replaced with U.

  1. Answer: E

Difficulty: 3

Section: From DNA to RNA

Feedback: The shortest rRNA “branches” at the “top” of the tree emanate from the beginning of the gene since their transcription has just started. All branches extend at their 3′ end at their base while their 5′ end starts to associate with RNA processing complexes.

  1. Answer: B

Difficulty: 1

Section: From DNA to RNA

Feedback: In the nucleolus, the snoRNAs help process and chemically modify the rRNAs in various ways.

  1. Answer: A

Difficulty: 3

Section: From DNA to RNA

Feedback: Many bacterial transcription terminators consist of a stable hairpin followed by a string of U nucleotides (that form weak A-U base pairs with the template), which together help disengage the mRNA transcript from RNA polymerase.

  1. Answer: B

Difficulty: 2

Section: From DNA to RNA

Feedback: MicroRNAs are mainly transcribed by the RNA polymerase II machinery and, similar to mRNAs, their primary transcripts normally undergo capping and polyadenylation.

  1. Answer: E

Difficulty: 1

Section: From DNA to RNA

Feedback: TFIIH is a large complex with several key functions in transcription initiation.

  1. Answer: B

Difficulty: 2

Section: From DNA to RNA

Feedback: It is the multisubunit TFIID, containing a TATA-binding protein, which, upon binding to the TATA box, creates a sharp kink in the double helix that serves as a landmark to attract other transcription factors.

  1. Answer: D

Difficulty: 2

Section: From DNA to RNA

Feedback: In the initiation phase of transcription, Mediator mediates the interaction of RNA polymerase with the transcriptional activators and is required for the activation of transcription by enhancers.

  1. Answer: C

Difficulty: 2

Section: From DNA to RNA

Feedback: Unlike bacteria, which maintain a fairly constant level of negative supercoiling in their circular DNA, eukaryotic topoisomerases rapidly remove the superhelical tension caused by DNA translocating machines such as the RNA polymerases.

  1. Answer: D

Difficulty: 2

Section: From DNA to RNA

Feedback: Noncoding regions are also found in E. coli mRNAs, although these regions are typically short compared to those in human cells.

  1. Answer: C

Difficulty: 1

Section: From DNA to RNA

Feedback: The cap is a 7-methyl G residue linked to the 5′ end of the transcript via a 5′-to-5′ triphosphate bridge. Methyl transferase enzymes methylate the terminal guanine base (giving it a positive charge) and sometimes other nucleotides, creating different 5′ caps.

  1. Answer: A

Difficulty: 2

Section: From DNA to RNA

Feedback: Due to the first transesterification step, the branch point in the intron is covalently bound to the 5′ end of the intron, and the intron is no longer continuous with the upstream (5′) exon.

  1. Answer: A

Difficulty: 3

Section: From DNA to RNA

Feedback: Our exons have a narrow distribution of size around the mean value of about 150 nucleotide pairs, whereas introns can vary greatly in size from about 10 to over 100,000 nucleotide pairs.

  1. Answer: E

Difficulty: 1

Section: From DNA to RNA

Feedback: The spliceosome uses ATP hydrolysis to produce a series of rearrangements, allowing the splicing signals to be checked and rechecked to ensure the accuracy of the process. The fact that splicing is coupled to transcription provides an advantage and helps the cell keep track of the exons, as does exon definition by SR proteins and other factors that mark the splice sites around each exon.

  1. Answer: B

Difficulty: 2

Section: From DNA to RNA

Feedback: Splicing “mistakes” include exon skipping and intron retention. Additionally, selection of cryptic splice sites can lead to the partial inclusion of exons or introns.

  1. Answer: E

Difficulty: 2

Section: From DNA to RNA

Feedback: After the mRNA is cleaved, this enzyme adds about 200 A nucleotides (from ATP) one by one without using a template.

  1. Answer: D

Difficulty: 2

Section: From DNA to RNA

Feedback: Eukaryotic cells take advantage of various mechanisms such as alternative splicing and alternative choices of polyadenylation sites, transcription start sites, and translation start sites to create a highly complex pool of gene products in the cell. Normally, however, a cellular gene is interpreted in only one of the possible reading frames.

  1. Answer: C

Difficulty: 2

Section: From DNA to RNA

Feedback: Normally, the export of an mRNA to the cytosol occurs with the 5′ cap proceeding first. The mRNA can then circularize through the interaction of the cytosolic cap-binding proteins with the cytosolic poly-A-binding proteins to facilitate rapid translation.

  1. Answer: C

Difficulty: 2

Section: From DNA to RNA

Feedback: The snRNPs remain associated with the pre-mRNA until it is fully spliced. Their presence indicates that the mRNA is still being processed.

  1. Answer: E

Difficulty: 2

Section: From DNA to RNA

Feedback: During the M phase, the chromosomes condense and the nucleoli disappear. In telophase, the chromosomal regions contributing their rRNA genes to the subcompartment coalesce again and their transcription resumes.

  1. Answer: D

Difficulty: 1

Section: From DNA to RNA

Feedback: In Cajal bodies, snRNPs and snoRNPs undergo maturation and assembly at enhanced rates compared to the rates estimated for cells in which these aggregates are disrupted. However, this enhancement is not crucial for all cells. Pre-mRNA splicing is thought to happen in mRNA production “factories” dispersed throughout the nucleus.

  1. Answer: E

Difficulty: 1

Section: From DNA to RNA

Feedback: The nucleolus is a factory for the processing of various noncoding RNAs and their assembly into larger complexes.

  1. Answer: C

Difficulty: 2

Section: From RNA to Protein

Feedback: It is a transfer RNA (tRNA) molecule, which is normally composed of nearly 80 nucleotides, is heavily modified, and is transcribed from one of the ~500 tRNA genes in our genome by RNA polymerase III.

  1. Answer: E

Difficulty: 2

Section: From RNA to Protein

Feedback: tRNA splicing appears to be simpler compared to the splicing of pre-mRNAs.

  1. Answer: E

Difficulty: 2

Section: From RNA to Protein

Feedback: The modified nucleotide is pseudouridine, one of the most abundant modified nucleotides in cellular RNA, and it is found in tRNAs, rRNAs, snRNAs, and snoRNAs. The “guide” snoRNAs in the nucleolus are themselves involved in the pseudouridylation of the other RNAs.

  1. Answer: C

Difficulty: 2

Section: From RNA to Protein

Feedback: The amino group of the amino acid is free, and will serve as the attacking group in the formation of the peptide bond on the ribosome. The carboxyl group, on the other hand, forms a high-energy and susceptible ester bond with the tRNA in preparation for that same reaction.

  1. Answer: A

Difficulty: 1

Section: From RNA to Protein

Feedback: The amino acid is initially activated by an ester linkage, through its carboxyl group, to an AMP molecule resulting from the hydrolysis of an ATP molecule. The amino acid is then transferred onto the appropriate tRNA molecule.

  1. Answer: B

Difficulty: 2

Section: From RNA to Protein

Feedback: During translation elongation, the ribosome translocates on the mRNA from the 5′ end to the 3′ end in consecutive sets of three nucleotides. The polypeptide is concomitantly elongated at its C-terminus by addition of individual amino acids.

  1. Answer: A

Difficulty: 2

Section: From RNA to Protein

Feedback: In the fundamental reaction of protein synthesis, a peptide bond is formed between the carboxyl group at the end of a growing polypeptide chain and a free amino group on an incoming aminoacyl-tRNA.

  1. Answer: B

Difficulty: 1

Section: From RNA to Protein

Feedback: In transcription, GTP is a polymerization substrate along with ATP, UTP, and CTP. Additionally, GTP is hydrolyzed by several factors involved in translation, including the eukaryotic elongation factors eEF1 and eEF2.

  1. Answer: E

Difficulty: 3

Section: From RNA to Protein

Feedback: The polypeptide chain grows on the ribosome by about 20 amino acids per second, which is equivalent to an mRNA translocation rate of 60 nucleotides per second. This is comparable to the rate of transcription by the bacterial RNA polymerase (about 50 nucleotides per second). Indeed, a bacterial ribosome can translate mRNAs co-transcriptionally, by closely following an RNA polymerase and translating the same mRNA molecule that is being produced by the polymerase.

  1. Answer: D

Difficulty: 2

Section: From RNA to Protein

Feedback: The initiator tRNA in eukaryotes carries methionine. Unlike the cap-binding and the 5′-to-3′ “scanning” of the mRNA by the eukaryotic translation initiation machinery, bacterial ribosomes find the start codon by directly interacting with the ribosome-binding sites in the mRNA. Initiation factors in both systems assist the ribosomes in initiating with high efficiency and speed.

  1. Answer: 1

Difficulty: 3

Section: From DNA to RNA

Feedback: Frames 2 and 3 contain a stop codon (UAG at the sixth codon, UGA at the first codon, respectively), which is not consistent with the sequence coding for a protein.

  1. Answer: C

Difficulty: 2

Section: From RNA to Protein

Feedback: Polysomes form as a result of frequent initiation events leading to several ribosomes bound to and translating each mRNA simultaneously. If the mRNA is circularized, the ribosomes that have finished the synthesis of the protein can reinitiate immediately for another round.

  1. Answer: D

Difficulty: 2

Section: From RNA to Protein

Feedback: The presence of the UGA codon is only one of the signals required for the recoding event. Other sequences (that can form stem-loop structures) in the vicinity of the UGA codon on the mRNA collaborate with a specific protein factor to incorporate the amino acid into the growing protein chain only for particular mRNAs and not any mRNA with a UGA codon.

  1. Answer: C

Difficulty: 3

Section: From RNA to Protein

Feedback: Part of the antibiotic puromycin has a striking resemblance to the 3′ end of an aminoacyl-tRNA. As a result, the ribosome mistakes it for an authentic amino acid substrate and covalently incorporates it into the growing polypeptide chain, thus causing premature termination.

  1. Answer: E

Difficulty: 3

Section: From RNA to Protein

Feedback: Nonsense mutations in the upstream exons can directly trigger nonsense-mediated decay (NMD). Frameshift mutations in these exons are likely to introduce premature termination codons and activate NMD as well. Similarly, retention of intronic sequences carries a high risk of premature stop codons. In some cases, exon skipping can lead to NMD if the size of the skipped exon is not a multiple of 3 and therefore its absence changes the reading frame of the downstream exon.

  1. Answer: D

Difficulty: 1

Section: From RNA to Protein

Feedback: The hsp60-like proteins form a large barrel-shaped complex, sometimes called a chaperonin, that forms an “isolation chamber” for the folding process of a protein after it has been fully synthesized.

  1. Answer: E

Difficulty: 1

Section: From RNA to Protein

Feedback: The 19S cap in the proteasome is a large complex with various functions including substrate recognition, de-ubiquitylation, unfolding and feeding into the 20S core, as well as regulatory roles.

  1. Answer: E

Difficulty: 1

Section: From RNA to Protein

Feedback: The AAA proteins in the cap hydrolyze tens or hundreds of ATP molecules to unfold and thread each protein into the central 20S digestive core.

  1. Answer: C

Difficulty: 3

Section: From RNA to Protein

Feedback: The modification is not necessarily exposed for the E3 enzymes to recognize. Exposure of the modification can occur if, for example, the protein becomes misfolded.

  1. Answer: D

Difficulty: 3

Section: The RNA World and the Origins of Life

Feedback: The RNA molecule folds into a structure (shown below) with two hairpins; one with a five-nucleotide loop, and the other with a four-nucleotide loop. The former hairpin also has a three-nucleotide bulge in its 5′ stem. A three-way junction is formed at the base of the two hairpins.

  1. Answer: Translation

Difficulty: 1

Section: From DNA to RNA

Feedback: The diagram shows the pathway from DNA to protein.

  1. Answer: CDAB

Difficulty: 2

Section: From DNA to RNA

Feedback: In bacterial transcription initiation, binding of the RNA polymerase holoenzyme to the promoter region in the closed complex is followed by the formation of the open complex involving the formation of the transcription bubble. After a number of abortive initiations, RNA polymerase clears the promoter and σ factor is released.

  1. Answer: 7

Difficulty: 3

Section: From DNA to RNA

Feedback: The branch-point nucleotide is often an A residue surrounded by other intronic sequences that are required for splicing.

  1. Answer: AACGC

Difficulty: 3

Section: From DNA to RNA

Feedback: The most common nucleotide at each position of the alignment is chosen for the consensus sequence. Note that none of the examined mRNAs has the consensus sequence.

  1. Answer: R

Difficulty: 3

Section: From DNA to RNA

Feedback: When the polymerase is moving to the right, it induces positive supercoiling which hinders helix opening. If the enzyme is allowed to rotate freely (which is normally not the case in vivo), right-handed rotation will eliminate the superhelical tension, as if the polymerase is twisting to follow along the right-handed double helix.

  1. Answer: FTFF

Difficulty: 2

Section: From DNA to RNA

Feedback: Exon size tends to be much more uniform than intron size and averages at about 150 nucleotide pairs, close to the length of DNA wrapped by one nucleosome. The bound nucleosome is suggested to act as a “speed bump” to allow the assembly of splicing proteins involved in exon definition.

  1. Answer: 1

Difficulty: 3

Section: From RNA to Protein

Feedback: Even though the polypeptide sequence pattern for the (ACGU)n RNA polymer is more complicated than for the other two examples, there is only one such pattern regardless of the choice of the reading frame: (Thr.Tyr.Val.Arg)n. For the (AC)n RNA, also only one polypeptide sequence pattern is possible: (Thr.His)n.

  1. Answer: codon

Difficulty: 1

Section: From RNA to Protein

Feedback: Three consecutive nucleotides in a protein-coding sequence of mRNA constitute a codon. Each codon specifies either one amino acid or a stop in the translation process.

  1. Answer: 2

Difficulty: 3

Section: From RNA to Protein

Feedback: The Phe codons UUC and UUU can be recognized by this tRNA, according to the wobble pairing rules.

  1. Answer: TTTT

Difficulty: 2

Section: From RNA to Protein

Feedback: The difference in affinity between correct and incorrect codon–anticodon matches cannot by itself account for the high accuracy of translation. The ribosome further assesses the correctness of the match; in the case of a correct match, EF-Tu hydrolyzes its bound GTP. This contributes to the speed and fidelity of translation. Even after EF-Tu release, there is a second proofreading opportunity for the ribosome to prevent an incorrect amino acid from being added to the growing chain.

  1. Answer: LSSL

Difficulty: 2

Section: From RNA to Protein

Feedback: The RNA components of the ribosome in the large and small subunit are responsible for peptide bond formation and the verification of the codon–anticodon match, respectively. mRNA is threaded through the small subunit, while the nascent protein emerges from the large subunit.

  1. Answer: FTTFF

Difficulty: 2

Section: From RNA to Protein

Feedback: Proteins can start folding as soon as they emerge from the ribosome (or even within the exit tunnel of the ribosome, but only for small helical segments) and one domain can be fully folded and functional before an entire multidomain protein is completely synthesized. However, folding within the hsp60 chaperonin complex typically happens only post-translationally.

  1. Answer: ribozyme

Difficulty: 1

Section: The RNA World and the Origins of Life

Feedback: Ribozymes are RNAs capable of catalyzing many different biochemical reactions. The ribosome provides an example in which the rRNA of the large subunit bears peptidyl transferase activity. Many other functions not found naturally can be screened for in the laboratory.

  1. Answer: FTFT

Difficulty: 2

Section: The RNA World and the Origins of Life

Feedback: The hereditary material in all present-day cells is DNA, but according to the RNA world hypothesis, RNA preceded DNA (as genetic material) as well as proteins (as catalysts). Some short peptides are synthesized without the ribosome, an observation that provides clues as to how the early transition from an RNA world to a protein-dominated world might have initiated.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

Chapter 7: CONTROL OF GENE EXPRESSION

© Garland Science 2015

1.59. Indicate true (T) and false (F) statements below regarding gene control. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.

( ) Development of multicellular organisms from a fertilized egg only rarely involves DNA rearrangements in specialized cells.

( ) A typical human cell expresses less than 1% of its approximately 30,000 genes at any given time.

( ) Genes that are expressed in all cell types usually vary in their level of expression in different cell types.

( ) Many differentiated plant cells can be fully de-differentiated and give rise to an entire plant.

1.60. In the following schematic graph of a hypothetical set of RNA-seq data, the number of reads is plotted for a region of chromosome containing two genes, from samples obtained from two different tissues. Which gene (X or Y) do you think is more likely a “housekeeping” gene? Which region (1 or 2) within gene Y most likely corresponds to an exon?

  1. Gene X; region 1
  2. Gene X; region 2
  3. Both genes; region 1
  4. Gene Y; region 1
  5. Gene Y; region 2

1.61. Studying the expression of a transcription regulatory protein in two cell types, you have performed experiments showing that the mRNA encoding the protein is present at comparable levels in the cytosol of both cell types. However, based on the expression of its target genes, you suspect that the protein activity might be significantly different in the two cell types. Which of the following steps in expression of the gene encoding this protein is more likely to be differentially controlled in these cell types?

  1. Transcription
  2. Translation
  3. mRNA transport
  4. mRNA degradation

1.62. In analysis using two-dimensional gel electrophoresis of the proteins expressed in different cell types, the number of spots that are different in different cells usually exceeds the number of common spots, and even the common spots can still have different intensities. The spots representing which of the following proteins would you expect to be among the common spots when compared across several cell types?

  1. RPL10 (a ribosomal protein)
  2. HBA1 (a hemoglobin subunit)
  3. Insulin (a hormone)
  4. Tyrosine aminotransferase (a metabolic enzyme)
  5. All of the above

1.63. What determines the time and place that a certain gene is transcribed in the cell?

  1. The type of cis-regulatory sequences associated with it
  2. The relative position of cis-regulatory sequences associated with it
  3. The arrangement of various cis-regulatory sequences associated with it
  4. The specific combination of transcription regulators present in the nucleus
  5. All of the above

1.64. The majority of transcription regulators make sequence-specific contacts with DNA in the major groove. In the two diagrams below, where are the contact surfaces that are exposed in the major groove?

  1. 1, 3
  2. 1, 4
  3. 2, 3
  4. 2, 4

1.65. Considering the diagrams below that show hydrogen-bond donors and acceptors as well as hydrophobic groups in four DNA base pairs, which of the following do you think is the most difficult to accomplish by DNA-binding proteins?

  1. Distinguishing between A-T and T-A in the major groove
  2. Distinguishing between A-T and C-G in the major groove
  3. Distinguishing between C-G and G-C in the major groove
  4. Distinguishing between C-G and G-C in the minor groove
  5. Distinguishing between C-G and T-A in the minor groove

1.66. Protein subunits that interact specifically with DNA sequences …

  1. typically recognize sequences of two to three nucleotide pairs in length.
  2. do so via hydrogen bonds, ionic bonds, and hydrophobic interactions.
  3. typically form about five weak interactions at the protein–DNA interface.
  4. often bind loosely to DNA.
  5. All of the above.

1.67. Which of the following DNA-binding motifs uses β sheets to recognize DNA bases?

  1. The helix–turn–helix motif
  2. The leucine zipper
  3. The zinc finger motif
  4. The helix–loop–helix motif
  5. None of the above

1.68. The following sequence logo represents the preferred cis-regulatory sequences of an imaginary transcription regulator that functions as a dimer. Would you expect this sequence to be recognized by a homodimer (M) or a heterodimer (T)? Write down M or T as your answer.

1.69. A DNA-binding protein recognizes a specific eight-nucelotide sequence in DNA. Assuming that its binding is perfectly specific, how many binding sites are expected to exist for it in human genomic DNA, which is composed of about 6 × 109 nucleotide pairs? What about if the protein forms a homodimer? Note that the target sequence can be oriented either way in the double-stranded DNA.

  1. About 200,000; about 30,000
  2. About 2000; about 300
  3. About 200; about 30
  4. About 200,000; about 300
  5. About 200,000; about 3

1.70. A DNA-binding protein binds cooperatively to DNA by fairly weak homodimerization. The binding of this protein to DNA …

  1. shows a sigmoidal binding curve.
  2. occurs in more of an all-or-none manner, compared to a protein that is a constitutive dimer.
  3. occurs despite the fact that the protein molecules are predominantly monomers in solution.
  4. All of the above.

1.71. Indicate true (T) and false (F) statements below regarding transcription regulators. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

( ) About 10% of the protein-coding genes in most organisms encode transcription regulators.

( ) The packaging of eukaryotic DNA into chromatin dampens the effect of cooperative binding of transcription regulators.

( ) Some DNA-binding proteins can induce bends or kinks in their target DNA.

( ) Dimerization of transcription regulators increases the chance of binding to nonspecific sequences.

1.72. The Trp operon in Escherichia coli encodes the components necessary for tryptophan biosynthesis. In the presence of the amino acid in a bacterium, …

  1. the tryptophan operator is bound to the tryptophan repressor.
  2. the tryptophan repressor is bound to bacterial RNA polymerase.
  3. the expression of the tryptophan repressor is shut off.
  4. the operon genes are expressed.
  5. All of the above.

1.73. Under which of the following conditions is the Lac operon in Escherichia coli fully turned on?

  1. Low glucose and lactose levels
  2. Low glucose but high lactose levels
  3. High glucose but low lactose levels
  4. High glucose and lactose levels
  5. Low cAMP and high glucose levels

1.74. Consider a cis-regulatory enhancer sequence in the Escherichia coli chromosome that is located thousands of nucleotide pairs upstream of the gene that it regulates. If the regulatory sequence is mutated to become nonfunctional, the introduction of the wild-type enhancer on a plasmid fails to regulate the gene. This implies that …

  1. the regulatory sequence encodes a regulatory protein that binds near the promoter of the target gene and controls RNA polymerase binding.
  2. the regulation of the target gene involves looping out of the intervening DNA, and the promoter of the cis-regulatory sequence must be on the same chromosome.
  3. the regulatory sequence can bind directly to the RNA polymerase.
  4. the regulatory sequence cannot bind to a protein.

1.75. Transcription regulation has similarities and differences in bacteria and in eukaryotes. Which of the following is correct in this regard?

  1. Most bacterial genes are regulated individually, whereas most eukaryotic genes are regulated in clusters.
  2. The rate of transcription for a eukaryotic gene can vary in a much wider range than for a bacterial gene (which is, at most, only about 1000-fold).
  3. DNA looping for gene regulation is the rule in bacteria but the exception in eukaryotes.
  4. Transcription regulators in both bacteria and eukaryotes usually bind directly to RNA polymerase.
  5. The default state of both bacterial and eukaryotic genomes is transcriptionally active.

1.76. Which of the following is directed by transcription activators in eukaryotic cells in order to provide a more accessible DNA for the transcription machinery?

  1. Nucleosome remodeling
  2. Histone removal
  3. Histone replacement
  4. Histone modifications
  5. All of the above

1.77. The following schematic drawing represents the activation of transcription for a eukaryotic gene (gene X). Indicate what component (A to D) in the drawing corresponds to each of the following. Your answer would be a four-letter string composed of letters A to D only, e.g. ADBC.

( ) Mediator

( ) cis-Regulatory sequence

( ) A general transcription factor

( ) “Spacer” DNA that may encode lncRNAs

1.78. To prevent spurious transcription from a gene, acetylation of histones—which is carried out by histone acetyl transferases ahead of a moving RNA polymerase II—is quickly reversed by histone deacetylases and histone methyl transferases in the wake of the polymerase, leaving a trail of specific methylated histones. Which of the following curves do you think better represents the distribution of this specific histone methylation mark with respect to a gene?

1.79. Two transcription activators cooperate to recruit a coactivator to a cis-regulatory sequence and activate transcription of a nearby gene. If each of the activators increases the affinity of the coactivator for the reaction site (and therefore the rate of transcription) by 100-fold, how much would you expect the affinity to increase when both activators are bound to DNA compared to when none is bound?

  1. 10-fold
  2. 100-fold
  3. 200-fold
  4. 10,000-fold

1.80. In the following schematic diagram of a region of a mammalian genome, genes A to D (triangles) are located in between a number of insulator elements (white squares) and barrier sequences (black squares). If the cis-regulatory sequence (oval) is bound by an abundant repressor protein, which gene would you expect to be expressed at a higher rate in this cell?

.

Reference: Even-skipped Regulatory Module (Questions 23-26)

Expression of the Even-skipped (Eve) gene in early Drosophila embryos is under the control of several transcription regulators. In one example, one of the Eve stripes is positioned near the anterior region of the embryo, and its regulatory module contains binding sites for Bicoid and Hunchback (activators) as well as Giant and Krüppel (inhibitors) such that the gene is expressed only in the region where concentrations of the two activators are high and the concentrations of the two inhibitors are low. A reporter gene can be placed under the control of this module, and it can be shown to form a stripe in the same place in the embryo as the corresponding stripe of Eve. Answer the following question(s) based on these findings.

1.81. In the following schematic diagrams of an early Drosophila embryo, in which region would you expect to find the reporter protein put under the control of the regulatory module mentioned above?

1.82. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Giant?

  1. It would be expressed in all seven stripes.
  2. It would be expressed in stripe 5 only.
  3. It would expand to cover a broad anterior region of the embryo.
  4. It would fail to express efficiently in the stripe 2 region.
  5. It would be expressed throughout the whole embryo.

1.83. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Bicoid?

  1. It would be expressed in all seven stripes.
  2. It would be expressed in stripe 5 only.
  3. It would expand to cover a broad anterior region of the embryo.
  4. It would fail to express efficiently in the stripe 2 region.
  5. It would be expressed throughout the whole embryo.

1.84. In the Eve regulatory modules, the binding sites for Giant and Krüppel are usually close to, or even overlapping with, those of Bicoid and Hunchback. This implies that Eve expression is regulated by … between the activators and regulators.

  1. Feedback regulation
  2. Cooperation
  3. Competition
  4. Feed-forward regulation

1.85. In the following schematic drawings showing four different network motifs in gene transcription circuits, indicate whether each of the motifs 1 to 4 corresponds to a feed-forward loop (E), flip-flop device (F), negative feedback loop (N), or positive feedback loop (P). Your answer would be a four-letter string composed of letters E, F, N, or P only, e.g. PNEF.

Answer: PENF Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: The circuits 1 and 3 include indirect feedback loops (with positive and negative feedback, respectively). Circuit 4 has two interconnected flip-flop switches.

Reference: Feed-forward Loops (Questions 28-30)

Consider the two feed-forward loops below containing three transcription regulators A, B, and C, where A receives the input signal and C generates the output. In the so-called coherent loop (left), A activates C both directly and indirectly, whereas in an incoherent loop (right), A activates C via one route and inactivates it via the other. Answer the following question(s) based on these network motifs.

1.86. Considering the two particular feed-forward designs shown, would you expect the coherent (C) or the incoherent (I) loop to have a stable output and respond best to input signals that are above a certain threshold? Write down C or I as your answer.

1.87. Considering coherent and incoherent feed-forward loops, eight different designs are possible using three components and involving activation (positive) and inhibition (negative) regulation only. In all such loops, A would regulate both B and C, and B would regulate only C. How many of these designs constitute a coherent loop? Write down your answer as a number, e.g. 7.

1.88. Would you expect a coherent (C) or an incoherent (I) loop to generate the following response pattern? In this example, A is stimulated by the input, and the transcription of C is measured as the output. Write down C or I as your answer.

1.89. Indicate true (T) and false (F) statements below regarding DNA methylation in humans. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

( ) Methylation of adenines is the most common DNA methylation in humans.

( ) Methylated cytosine can be accidentally deaminated to produce thymine, leading to a C-to-T transition.

( ) Cytosine methylation often occurs within a 5′-CG-3′ sequence.

( ) Shortly after fertilization, a genome-wide wave of demethylation takes place.

1.90. Indicate true (T) and false (F) statements below regarding DNA methylation in humans. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

( ) DNA methylation makes repression of gene expression “leakier.”

( ) DNA methylation at the promoter region is usually an indication of active transcription.

( ) CG islands are nonpermissive for RNA polymerase assembly.

( ) The dinucleotide 5′-GC-3′ is much more common than 5′-CG-3′ in the human genome.

1.91. Indicate which of the following enzymatic activities corresponds to A, B, or C in the following schematic drawing of DNA methylation events in the human genome. Your answer would be a three-letter string composed of letters A, B, and C only, e.g. CBA.

( ) DNA demethylating enzyme

( ) Maintenance methyl transferase enzyme

( ) De novo methyl transferase enzyme

1.92. Beckwith-Wiedemann syndrome in humans is characterized by “overgrowth” in childhood, sometimes leading to unusually large parts of the body. An imprinted gene cluster on chromosome 11 is associated with this disease. The cluster contains several genes including Igf2 and H19. Igf2 encodes an insulin-like growth factor that is maternally imprinted, i.e. the maternal copy is not expressed. However, the DNA methylation pattern of this locus is not different between the two homologous chromosomes. On the other hand, H19 is also imprinted and its methylation pattern does differ between the two parental chromosomes. H19 is transcribed into a noncoding RNA that appears to silence the transcription of the Igf2 gene in cis. Would you expect the H19 locus to be hypermethylated in the maternally inherited chromosome (M) or paternally inherited chromosome (P)? Write down M or P as your answer.

1.93. In the following pedigree, females and males are indicated by circles and squares, respectively, and the presence of a rare disease caused by a loss-of-function mutation in an imprinted autosomal gene is indicated by black color. Is the gene maternally imprinted (M; i.e. not expressed from the maternally inherited chromosome) or paternally imprinted (P)? Write down M or P as your answer.

1.94. Which of the following is true regarding genomic imprinting?

  1. It is an epigenetic phenomenon.
  2. It occurs in most animals.
  3. It always involves inactivation of genes through direct DNA methylation.
  4. It can “unmask” recessive alleles but cannot “mask” dominant ones.
  5. All of the above.

1.95. You have engineered the X chromosomes in female mice such that one X chromosome expresses green fluorescent protein when active, while the other expresses red fluorescent protein. You have used these mice to study cancer in females. You know that each tumor is a clonal cellular proliferation, meaning all of its proliferating cells are descendants of a single original cancer-causing cell. It follows that, unless X-chromosome inactivation is perturbed in tumors, …

  1. all tumor cells in one mouse should express the same fluorescent protein (either red or green), but tumor cells from different mice can show either red or green fluorescence.
  2. the cells in any tumor should all express the same fluorescent protein (either red or green), but independently derived tumors in the same mouse can show either green or red fluorescence.
  3. different cells within each tumor can express different fluorescent proteins, and the tumors would therefore show yellow fluorescence, but each cell shows either red or green fluorescence.
  4. each cell can express both fluorescent proteins and would therefore emit yellow fluorescence, and the tumors would glow in yellow as well.
  5. different tumors would show red, yellow, or green fluorescence.

1.96. Fill in the blank in the following paragraph regarding X-chromosome inactivation. Do not use abbreviations.

“The presence of two copies of the X chromosome in females but only one in males poses a potential problem for regulation of gene expression: everything else being equal, the higher copy number of the X-linked genes can lead to an imbalanced overproduction of their products in females compared to males. Organisms have evolved different … mechanisms to solve this problem. In mammals, for example, the almost complete inactivation of one of the X chromosomes provides the solution to the dosage problem.”

1.97. Two copies of the gene A exist in a diploid mammal. However, only one copy is expressed in every somatic cell. Different somatic cells in the body of each organism have inactivated one or the other allele in a seemingly random fashion, but when they divide, the daughter cells inherit the choice of the inactive allele faithfully. This is an example of …

  1. X-inactivation
  2. Genomic imprinting
  3. Loss of heterozygosity
  4. Monoallelic gene expression
  5. Alternative gene splicing

1.98. Indicate true (T) and false (F) statements below regarding riboswitches. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

( ) A riboswitch permits transcription elongation only when bound to its small-molecule ligand.

( ) Riboswitches appear to be a modern evolutionary invention.

( ) Riboswitches are often located near the 3′ end of mRNAs, and therefore fold after the rest of the mRNA.

( ) Riboswitches are exclusively found in prokaryotes.

1.99. In the following schematic diagram of five pairs of alternative RNA splicing patterns, which pair depicts exon skipping only? In exon skipping, an exon is spliced out of the RNA transcript along with its flanking introns. In the diagram, exons are colored dark blue and introns are colored yellow. Light blue indicates possible exons.

1.100. Imagine a gene encoding a pre-mRNA with twelve exons, ten of which can be alternatively spliced in vivo. Assuming that alternative splicing for this RNA only occurs through exon skipping, how many different proteins can possibly be made from this pre-mRNA as a result of alternative splicing?

  1. About 10
  2. About 20
  3. About 100
  4. About 200
  5. About 1000

1.101. The alternative splicing of a certain transcript can result in the production of two mRNA isoforms, one predominant in muscle cells and the other in neurons. The gene contains an exon that is skipped in muscle cells but retained in neurons. You create a mutant version of the gene in which the 3′ splice site near this exon is deleted. However, when you introduce this into a culture of neural cells, an even longer pre-mRNA is produced, consistent with the activation of a secondary splice site located near the deleted one. Is the secondary splice site within the exon (E) or its neighboring intron (I)? Write down E or I as your answer.

1.102. Stimulated B lymphocytes switch from the synthesis of membrane-bound to secreted antibody molecules by increasing the concentration of a subunit of the trimeric CstF complex that cleaves and polyadenylates mRNAs. How does this up-regulation of CstF bring about the production of soluble antibodies?

  1. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted.
  2. It activates a weak polyadenylation site in the immunoglobulin primary transcript and prevents splicing, creating a shorter molecule that is secreted.
  3. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a shorter antibody molecule that is secreted.
  4. It activates a weak polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted.
  5. It favors a strong polyadenylation site in the immunoglobulin primary transcript, aborting translation and creating a shorter antibody molecule that is secreted.

1.103. Indicate true (T) and false (F) statements below regarding RNA editing. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

( ) RNA editing in animal cells mainly takes place through adenine or cytosine deamination.

( ) RNA editing can change the pattern of pre-mRNA splicing.

( ) RNA editing in coding regions can result in changes in the protein sequence.

( ) Retroviruses such as HIV encode RNA-editing components to confront the host defense mechanisms.

1.104. You have transfected HeLa cells with a gene encoding the human immunodeficiency virus (HIV) protein Rev and have induced the expression of the protein. You incubate the cell culture in the presence or absence of leptomycin B and later measure the localization of Rev inside the cells by immunofluorescence microscopy. Leptomycin B specifically inhibits the cellular Crm1 protein, a nuclear transport receptor that is essential for the normal function of Rev in the HIV life cycle. Your results are tabulated below. Which condition (1 or 2) do you think corresponds to the presence of leptomycin B?

ConditionCytoplasmic Rev (%)Nuclear Rev (%)
13070
2397

1.105. When unfolded proteins accumulate in the lumen of the endoplasmic reticulum due to inefficient folding conditions, a membrane-bound protein kinase is activated and phosphorylates a subunit of the translation initiation factor eIF2. Indicate which of the following events would (Y) or would not (N) occur as a result. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY.

( ) eIF2 binds tightly to the ribosome.

( ) Translation initiation is stimulated.

( ) The GDP-bound form of eIF2 accumulates.

( ) eIF2B is activated.

1.106. A researcher studying the human immunodeficiency virus (HIV) has infected human T cells with wild-type or Rev-deficient viruses. She extracts cytoplasmic RNA from the cells and separates each isolated RNA mixture by agarose-gel electrophoresis. She then uses northern analysis using HIV-specific probes to identify those bands that contain viral RNA. The results are presented in the following schematic drawing. Which lane (1 or 2) corresponds to cells infected with the Rev­-deficient virus? What is the size of the full-length HIV genome?

  1. Lane 1; about 9000 nucleotides
  2. Lane 1; about 4000 nucleotides
  3. Lane 1; about 2000 nucleotides
  4. Lane 2; about 9000 nucleotides
  5. Lane 2; about 2000 nucleotides

1.107. Some viruses encode a protease that cleaves the translation initiation factor eIF4G, rendering it unable to bind to eIF4E. What is the consequence of this cleavage?

  1. It shuts down most of the cellular translation machinery, which causes the release of virus particles.
  2. It favors viral protein synthesis because IRES-dependent translation initiation is inhibited.
  3. It shuts down most host-cell protein synthesis and diverts the translation machinery to IRES-dependent initiation, thus favoring viral protein synthesis.
  4. It favors viral protein synthesis by shutting down translation from uORFs.
  5. It shuts down IRES-dependent translation, forcing the virus into latency.

1.108. Gcn4 is a protein kinase that can phosphorylate and inactivate the initiation factor eIF2. The Gcn4 mRNA contains several short upstream open reading frames (uORFs) that negatively regulate its translation initiation. The phosphorylation of eIF2 by Gcn4, therefore, can control Gcn4 expression …

  1. in a negative feedback loop.
  2. in a positive feedback loop.
  3. in a feed-forward loop.

1.109. Indicate true (T) and false (F) statements below regarding mRNA stability and degradation. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.

( ) Any factor that affects the translation efficiency of an mRNA tends to have the opposite effect on its degradation.

( ) Decapping of mRNA normally occurs after poly-A tail shortening removes the poly-A tail and starts chewing into the 3′-UTR.

( ) Most mRNA decay in the cells generally proceeds via endonucleolytic cleavage.

( ) As a general rule, eukaryotic mRNAs have a shorter half-life compared to bacterial mRNAs.

1.110. Under iron-starvation conditions in the cell,

  1. cytosolic aconitase binds to the 3′ UTR of the mRNA encoding the transferrin receptor.
  2. cytosolic aconitase binds to the 3′ UTR of the mRNA encoding ferritin.
  3. cytosolic aconitase blocks translation of the mRNA encoding the transferrin receptor.
  4. the mRNA encoding the transferrin receptor is cleaved by an endonuclease.
  5. All of the above.

1.111. Indicate true (T) and false (F) statements below regarding P-bodies. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.

( ) P-bodies are membrane-enclosed organelles scattered throughout the cytoplasm.

( ) mRNAs that are to be degraded are transferred from P-bodies to stress granules containing decapping and exonuclease enzymes.

( ) When translation initiation is blocked, stress granules typically shrink.

( ) mRNAs associated with stress granules are primed for translation.

1.112. Which of the following classes of noncoding RNAs is NOT directly involved in RNA interference?

  1. miRNA
  2. snoRNA
  3. piRNA
  4. siRNA

1.113. Indicate whether each of the following descriptions better matches miRNAs (M) or siRNAs (S) in animal cells. Your answer would be a four-letter string composed of letters M and S only, e.g. MMMM.

( ) They usually cleave their target RNAs using the slicing activity of the Argonautes.

( ) They typically show perfect complementarity with their target RNA.

( ) They can bind to the RITS as well as the RISC complex.

( ) They seem to be the more ancient form of small noncoding RNAs.

1.114. Which strand in an miRNA precursor will serve as the guide strand in the RNA-induced silencing complex (RISC)? Analysis of miRNA sequences has revealed an asymmetry in the stability of the double-stranded RNA precursor. The strand showing lower thermodynamic stability near its 5′ end (nucleotides 2 to 6 in the mature guide strand) is normally selected as the guide strand, and the other strand is usually degraded. In the following miRNA precursor, which strand (1 or 2) do you think will be incorporated into the active RISC? Write down 1 or 2 as your answer.

1.115. The schematic diagram below shows the processing of a class of small RNAs that are involved in RNA interference (RNAi). Which of the following is true regarding these RNAs?

1.116. What is the function of RNA-dependent RNA polymerases in RNAi?

  1. They prevent the spread of the RNAi pathway by replicating the target RNAs.
  2. They help amplify the RNAi response by replicating the target RNAs.
  3. They produce additional copies of the siRNAs to ensure that the RNAi response is sustained and spread.
  4. They are viral proteins that prevent the spread of RNAi by preferentially replicating siRNA sponges.

1.117. A certain region of a mammalian genome is transcribed at low but sustained levels using two flanking promoters. The RNA products are annealed together, processed, and used to recruit the RITS complex to this region, as well as chromatin modifying enzymes, RNA-dependent RNA polymerase, and a Dicer enzyme. Indicate whether this pathway is more closely related to that seen in miRNA (M), piRNA (P), or siRNA (S) interference pathways. Write down M, P, or S as your answer.

1.118. In many animals, siRNAs can only function within the cell in which the siRNA is introduced. In contrast, the worm Caenorhabditis elegans can be fed with bacteria that synthesize double-stranded RNAs, and the RNAi spreads throughout the animal. A genetic screen to identify genes involved in systemic RNAi led to the discovery of Sid-1, a transmembrane protein expressed in most tissues in the adult worm. Loss-of-function mutations in the gene encoding Sid-1 restrict the RNAi activity to the cells surrounding the digestive tract after siRNA feeding. Ectopic expression of Sid-1 in Drosophila melanogaster cells that normally lack systemic RNAi and are unable to take up siRNAs from the medium, enables these cells to passively take up siRNA . These findings suggest that …

  1. Sid-1 is necessary and sufficient for systemic RNAi and siRNA uptake, respectively.
  2. Sid-1 is necessary but not sufficient for systemic RNAi and siRNA uptake, respectively.
  3. Sid-1 is not necessary but is sufficient for systemic RNAi and siRNA uptake, respectively.
  4. Sid-1 is not necessary and not sufficient for systemic RNAi and siRNA uptake, respectively.

1.119. Comparing the bacterial CRISPR system and the eukaryotic RNAi mediated by siRNAs, indicate whether each of the following CRISPR components is most analogous to Argonaute (A), siRNA (S), or target mRNA (T) in the RNAi pathway. Your answer would be a three-letter string composed of letter A, S, and T only.

( ) Cas

( ) Viral DNA

( ) crRNA

1.120. Consider the descriptions below for three different lncRNAs, and determine whether they function in cis (C) or in trans (T). Your answer would be a three-letter string composed of letters C and T only, e.g. CCT.

( ) The tsiX RNA is transcribed from the X-inactivation center in the X chromosome that will not be inactivated. It is antisense with respect to Xist, repressing its transcription and function.

( ) The ANRIL lncRNA is an antisense transcript in a gene cluster expressing inhibitors of cell-cycle proteins. It silences the entire cluster by recruiting heterochromatin proteins of the Polycomb group.

( ) The PTENP1 lncRNA is similar to an mRNA that encodes the protein PTEN, and contains many of the miRNA response elements present in PTEN mRNA; it therefore protects the PTEN mRNA from miRNA-mediated silencing by acting as a decoy.

Answers:

  1. Answer: TFTT

Difficulty: 1

Section: An Overview of Gene Control

Feedback: The changes in gene expression that underlie the development of multicellular organisms do not generally involve changes in the DNA sequence of the genome. At any one time, a typical human cell expresses 30–60% of its approximately 30,000 genes at some level. But even those genes that are expressed in all cell types usually vary in their level of expression from one cell type to the next. Under the right conditions, differentiated plant cells can regenerate an entire adult plant.

  1. Answer: D

Difficulty: 3

Section: An Overview of Gene Control

Feedback: A housekeeping gene is expected to be expressed in nearly all cells, including brain and liver cells. This is the case for gene Y, for which the mRNA levels are significant in both cell types, according to the RNA-seq data. The regions with higher numbers of reads in the RNA-seq results correspond to exons.

  1. Answer: B

Difficulty: 2

Section: An Overview of Gene Control

Feedback: Since the mRNA levels are similar in both cell types, difference in protein activity can be attributed to differences in translation or in post-translational control.

  1. Answer: A

Difficulty: 2

Section: An Overview of Gene Control

Feedback: Ubiquitously expressed gene products include ribosomal proteins, DNA and RNA polymerases, and DNA repair enzymes, among others.

  1. Answer: E

Difficulty: 2

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: The expression of a gene can be determined by its cis-regulatory sequences and transcription regulators that function in trans.

  1. Answer: B

Difficulty: 2

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: Nearly all transcription regulators make the majority of their contacts with the major groove (1 and 4 in the drawings) because the major groove is wider and displays more molecular features than does the minor groove (2 and 3).

  1. Answer: D

Difficulty: 2

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: In the minor groove, precise recognition of A-T versus T-A (or C-G versus G-C) is not readily feasible. However, an A-T can still be distinguished from C-G, for example.

  1. Answer: B

Difficulty: 1

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: DNA-binding proteins typically recognize specific sequences that are 5 to 10 nucleotide pairs in length. This involves forming 20 or so weak interactions at the interface, including hydrogen-bonding, ionic bonding, and hydrophobic interactions. The sum of all protein–DNA interactions can result in high affinity and specificity.

  1. Answer: E

Difficulty: 1

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: All of the examples presented in A to D use α helices to interact with DNA. A limited number of DNA-binding motifs use β sheets to recognize DNA bases.

  1. Answer: T

Difficulty: 3

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: Homodimeric transcription regulators typically have a symmetrical arrangement which is reflected in their target DNA sequence in the form of an inverted repeat. The sequence logo shown is not in agreement with an inverted repeat, or even a direct repeat; the DNA is therefore expected to be recognized by a heterodimer.

  1. Answer: E

Difficulty: 3

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: The expected number of target 8-mer sequences is calculated as:

(6 × 109 8-mer/genome) × (2 orientations per double strand) / (48 8-mers/target sequence)

= ~200,000 target sequences/genome.

Similarly, the expected number of target 16-mer sequences is calculated as:

(6 × 109 16-mer/genome) × (2 orientations per double strand) / (416 16-mers/target sequence)

= ~3 target sequences/genome.

  1. Answer: D

Difficulty: 1

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: Cooperative binding creates a sigmoidal binding curve with two main states: at low protein concentrations, the protein is mostly a monomer and the binding site on DNA is mostly unoccupied; at high protein concentrations, the great majority of the DNA sites are occupied by the protein dimer.

  1. Answer: TFTF

Difficulty: 1

Section: Control of Transcription by Sequence-specific DNA-binding Proteins

Feedback: Nucleosome packaging of DNA and dimerization of transcription regulators promote cooperative DNA binding by these proteins. Some proteins that bind to DNA can significantly bend the DNA molecule. About 10% of protein-coding genes in most cells encode transcription regulators, making them one of the largest classes of proteins in the cell.

  1. Answer: A

Difficulty: 1

Section: Transcription Regulators Switch Genes On and Off

Feedback: When tryptophan is present, it binds to the repressor which then binds to the operator to turn off the operon.

  1. Answer: B

Difficulty: 1

Section: Transcription Regulators Switch Genes On and Off

Feedback: The Lac operon is fully turned on when glucose is absent AND lactose is present.

  1. Answer: B

Difficulty: 2

Section: Transcription Regulators Switch Genes On and Off

Feedback: DNA looping can occur during bacterial gene regulation, where the intervening DNA acts as a tether to enormously increase the probability that the proteins bound near the promoter and those bound to the cis-regulatory sequences will collide with each other. This effect requires that both sequences be on the same DNA molecule, or be somehow linked physically.

  1. Answer: B

Difficulty: 2

Section: Transcription Regulators Switch Genes On and Off

Feedback: Variation in transcription rates across the genome is much greater in eukaryotic cells (about one million-fold) compared to prokaryotic cells (about one thousand-fold). In eukaryotes, genes are often transcribed and regulated individually, and DNA looping occurs in the regulation of nearly every gene. In these cells, transcription regulators often assemble in groups and do not directly contact RNA polymerases. The default state of most eukaryotic DNA packaged into nucleosomes is off.

  1. Answer: E

Difficulty: 1

Section: Transcription Regulators Switch Genes On and Off

Feedback: Eukaryotic transcription activator proteins can direct local alterations in chromatin structure. This can be achieved through covalent histone modifications, nucleosome remodeling or removal, and histone replacement.

  1. Answer: DBAC

Difficulty: 2

Section: Transcription Regulators Switch Genes On and Off

Feedback: Please refer to Figure 7–17.

  1. Answer: B

Difficulty: 3

Section: Transcription Regulators Switch Genes On and Off

Feedback: The methylation mark would be found in the body of actively transcribed genes (but not near their promoter, which requires a more open nucleosome packaging).

  1. Answer: D

Difficulty: 2

Section: Transcription Regulators Switch Genes On and Off

Feedback: The combined effect of two transcription activators binding to the same regulatory sequence in DNA is synergistic, not simply additive.

  1. Answer: C

Difficulty: 3

Section: Transcription Regulators Switch Genes On and Off

Feedback: Gene D is not protected from heterochromatin expansion by any barrier sequence, while genes A and B are repressed by the cis-regulatory element bound to a repressor. Gene C, however, is not under these repressive effects, thanks to its flanking insulator elements and barrier sequences.

  1. Answer: B

Difficulty: 1

Refer to: Even-skipped Regulatory Module

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: This module is normally activated in the second stripe, where inhibition is weak and activation is strong.

  1. Answer: C

Difficulty: 2

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: As expected, lack of Giant or Krüppel results in a broad expression pattern for the second stripe.

  1. Answer: D

Difficulty: 2

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: As expected, lack of Bicoid or Hunchback results in an inefficient expression of the second stripe.

  1. Answer: C

Difficulty: 2

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: This helps sharpen the boundaries flanking the Eve stripes.

  1. Answer: PENF

Difficulty: 2

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: The circuits 1 and 3 include indirect feedback loops (with positive and negative feedback, respectively). Circuit 4 has two interconnected flip-flop switches.

  1. Answer: C

Difficulty: 3

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: This also depends on the particular parameters of the system. However, coherent feed-forward motifs are generally more stable and more robust to perturbations.

  1. Answer: 4

Difficulty: 2

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: Half of the possible designs yield coherent loops.

  1. Answer: I

Difficulty: 3

Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: The two-phase response reflects the incoherent nature of this feed-forward motif.

  1. Answer: FTTT

Difficulty: 1

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: The most common DNA methylation in vertebrate DNA occurs on cytosine (C) nucleotides largely in the sequence 5′-CG-3′. The accidental deamination of the methylated cytosine can then potentially lead to a mutation. Shortly after fertilization in mammals, a genome-wide wave of cytosine demethylation takes place and the vast majority of methyl groups are lost from the DNA.

  1. Answer: FFFT

Difficulty: 2

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: DNA methylation helps to make repression of gene expression less leaky. RNA polymerase is often found bound to promoters within unmethylated CG islands, even when the associated gene is not being actively transcribed. Except in CG islands, 5′-CG-3′ dinucleotides are rare in vertebrate genomes.

  1. Answer: CAB

Difficulty: 2

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Arrow A represents the maintenance of DNA methylation after DNA replication, while arrows B and C represent de novo DNA methylation and DNA demethylation, respectively.

  1. Answer: P

Difficulty: 3

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: On the maternally inherited chromosome, H19 is transcribed and silences Igf2. This does not occur on the paternal chromosome. As a result, Igf2 is maternally imprinted with no maternal-specific methylation.

  1. Answer: M

Difficulty: 3

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: If a disease-causing gene is maternally imprinted, carrier females silently pass it to the next generation; their sons may have affected children, but their daughters would not.

  1. Answer: A

Difficulty: 2

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Genomic imprinting, observed in placental mammals and some flowering plants, is an epigenetic form of inheritance. It is mediated through DNA methylation in mammals, but it does not always involve gene inactivation by direct methylation. It can unmask recessive alleles or mask dominant alleles.

  1. Answer: B

Difficulty: 2

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Since different cells in the female body have randomly inactivated one or the other X chromosome, tumors derived from them will inherit the inactivation pattern and can therefore express one or the other fluorescent protein. However, all cells within each tumor are expected to express the same fluorescent protein.

  1. Answer: dosage compensation

Difficulty: 2

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Dosage compensation allows mammals to equalize the dosage of X-chromosome gene products between males and females.

  1. Answer: D

Difficulty: 1

Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Monoallelic expression can result from genomic imprinting, but not all examples of monoallelic expression are due to imprinting.

  1. Answer: FFFF

Difficulty: 2

Section: Post-transcriptional Controls

Feedback: Riboswitches are short RNA sequences that can change conformation upon ligand binding and regulate transcription. These ancient regulators are common in bacteria as well as in eukaryotes. They are often found near the 5′ end of mRNAs and fold co-transcriptionally.

  1. Answer: A

Difficulty: 2

Section: Post-transcriptional Controls

Feedback: The alternative splicing event shown in A is exon skipping: the third exon is missing (together with its flanking introns) in one of the two mRNA variants. Shown in B is the retention of a single intron (together with its flanking exons) in one of the mRNA variants.

  1. Answer: E

Difficulty: 2

Section: Post-transcriptional Controls

Feedback: Each alternative exon is either skipped or retained, possibly creating 210 (= ~1000) different mature mRNAs.

  1. Answer: I

Difficulty: 2

Section: Post-transcriptional Controls

Feedback: The cryptic splice site resides within the upstream intron, resulting in the production of a longer mRNA when this splice site is chosen.

  1. Answer: B

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: The suboptimal cleavage/poly-A addition site in the transcript is encountered first but is usually skipped in unstimulated B lymphocytes, leading to production of a longer antibody molecule that is anchored in the plasma membrane. When activated to produce antibodies, the B lymphocyte increases its CstF concentration; as a result, cleavage now occurs at the suboptimal site, and the shorter transcript is produced.

  1. Answer: TTTF

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Two principal types of mRNA editing in animals are the deamination of adenine to produce inosine and, less frequently, the deamination of cytosine to produce uracil. Both types can have profound effects on the meaning of the RNA message, including changing the pattern of pre-mRNA splicing or changing the amino acid sequence of the protein coded by the mRNA. RNA from some retroviruses, including the human immunodeficiency virus (HIV), is extensively edited by the host cell, presumably as part of a defense mechanism to hold the virus in check.

  1. Answer: 2

Difficulty: 2

Section: Post-transcriptional Controls

Feedback: By inhibiting Crm1, leptomycin B inhibits the nuclear export of Rev, resulting in its nuclear accumulation.

  1. Answer: NNYN

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Under stress conditions, eIF2 phosphorylation prevents its release from the guanine nucleotide exchange factor eIF2B, sequestering the latter in a nonfunctional form, and consequently preventing further eIF2 guanine nucleotide exchange and activation. As a result, translation initiation is largely inhibited.

  1. Answer: D

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: In the absence of a functional Rev protein (lane 2), only the fully spliced viral RNAs are exported from the nucleus to the cytosol. Rev is required for the export of the full-length viral RNA.

  1. Answer: C

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Translation of internal ribosome entry site (IRES)-containing mRNAs does not rely on the common cap-dependent initiation mechanism.

  1. Answer: B

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Since phosphorylated eIF2 promotes the skipping of the uORFs in favor of the translation of the main Gcn4 ORF, the kinase activity of Gcn4 provides a possible positive feedback control mechanism.

  1. Answer: TFFF

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Eukaryotic mRNAs are generally more stable than bacterial mRNA. Eukaryotic mRNA decay is mediated by 5′ exonucleases (after decapping) or 3′ exonucleases, usually after poly-A tail shortening leaves only about 25 nucleotides of the tail. Since the poly-A shortening and the translation machinery compete for the same components, factors that affect one of them also affect the other one in an opposite direction. Some mRNAs are destroyed by specific endonucleases.

  1. Answer: A

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: The iron-sensitive RNA-binding protein aconitase can bind to the 3′ untranslated region (UTR) of the transferrin receptor mRNA and increase production of the receptor by blocking endonucleolytic cleavage of the mRNA. The addition of iron to cells results in the release of aconitase from the mRNA and thus decreases the stability of the mRNA.

  1. Answer: FFFT

Difficulty: 1

Section: Post-transcriptional Controls

Feedback: Even though they somehow function as “organelles,” P-bodies are not membrane-enclosed. Many mRNAs are degraded in P-bodies, but some mRNAs move from P-bodies to stress granules, which contain mRNA-binding proteins that prime the mRNAs for translation. Translation itself, however, does not occur in stress granules. When translation initiation is blocked, stress granules enlarge as more and more nontranslated mRNAs are stored in them.

  1. Answer: B

Difficulty: 1

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: The small nucleolar RNAs do not function in RNA interference.

  1. Answer: SSSS

Difficulty: 2

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: The small interfering RNAs (siRNAs) appear to be the most ancient form of RNA interference. Having an exact match to their target RNA molecules, they direct the cleavage of the target. siRNAs can also control transcription of target genes by binding to the RNA-induced transcriptional silencing (RITS) complex.

  1. Answer: 1

Difficulty: 2

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: Note the presence of mismatches and even a bulge along the duplex. Strand 1 contains mostly A-U base pairs, as well as a mismatch, near its 5′ end (left). Strand 2 contains mostly C-G pairs near its 5′ end (right).

  1. Answer: C

Difficulty: 2

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: These are miRNAs

  1. Answer: C

Difficulty: 1

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: The RNA interference (RNAi) response can be amplified in some organisms using RNA-dependent RNA polymerases. In these cases, RNA-dependent RNA polymerases use siRNAs as primers to produce additional copies of double-stranded RNAs that are then cleaved into siRNAs.

  1. Answer: S

Difficulty: 1

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: RNA interference can direct heterochromatin formation, altering the packaging of nuclear chromatin and expression of genes.

  1. Answer: A

Difficulty: 2

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: According to these results, Sid-1 is both necessary for systemic RNAi in the nematode and sufficient for siRNA uptake in fly cells: without it, systemic RNAi in C. elegans is lost; additionally, when ectopically expressed, it confers the ability to take up siRNA in D. melanogaster cells.

  1. Answer: ATS

Difficulty: 1

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: Despite lacking homology, these two pathways share similarities: Cas can be guided by crRNAs to target invading viral DNA; similarly, an Argonaute or Piwi protein uses siRNAs to seek and cleave target mRNAs.

  1. Answer: CCT

Difficulty: 2

Section: Regulation of Gene Expression by Noncoding RNAs

Feedback: The first two lncRNAs act in cis; i.e. they affect only the chromosome from which they are transcribed. The third example, however, acts in trans and regulates the function of other genes after it diffuses from its site of synthesis

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