Question 1

You are studying lipoprotein processing in a strain of mutant mice. You find that these mice synthesize the same apolipoprotein composition contained in lipoproteins made in the liver and the intestines, whereas, in wild-type mice there are 2 different sized apolipoproteins in these same tissues. Examination of the mRNA processing in the mutant mice would most likely show a defect in which of the following processes?
A) alternative polyadenylation
B) alternative splicing
C) differential capping
D) posttranscriptional modification converting a U to a T residue
E) RNA editing

Accepted Answer

RNA editing is a molecular process through which some cells can make discrete changes to specific nucleotide sequences within an mRNA molecule after it has been generated by RNA polymerase. There are 2 major types of RNA editing with 1 being a C-to-U change catalyzed by cytidine deaminase that deaminates a cytidine base into a uridine base. An example of C-to-U editing is with the apolipoprotein B gene in humans. ApoB-100 is expressed in the liver and apoB-48 is expressed in the intestines. The B-100 form has a CAA sequence that is edited to UAA, a stop codon, in the intestines

Question 2

You are studying the processes of RNA transcription in a cell-free system. You find that although the RNA polymerase II in your system initiates transcription as expected, it does not complete the process. Which of the following modifications to the enzyme is most likely not taking place, thereby, explaining the lack of complete transcription?
A) 5′ capping
B) acetylation
C) methylation
D) phosphorylation
E) ubiquitination

Accepted Answer

Phosphorylation of the RNA polymerase II C-terminal domain (CTD) is crucial for the transition from transcription initiation to elongation. Without this modification, the enzyme may not proceed to complete transcription.  In order for RNA pol II to initiate, as well as to continue, transcription it must be phosphorylated. The transcription factor complex TFIIH contains a kinase that phosphorylates serine residues in the C-terminal domain (CTD) of the large subunit of RNA pol II. This phosphorylation event allows transcription initiation to commence. The 5′ end of all eukaryotic mRNAs are capped and this process occurs after the newly synthesizing mRNAs are around 20-30 bases long. At this point, RNA pol II pauses and a different kinase phosphorylates additional residues in the CTD. This pausing and regulatory phosphorylation event allows for the potential of attenuation in the rate of transcription. Lack of phosphorylation would, therefore, prevent continued transcription.

Question 3

Which of the following represents a characteristic feature of the cap structure on the 5′ end of most eukaryotic mRNAs?
A) it is a methylated adenine
B) it is a methylated cytidine
C) it is a methylated guanine
D) it is an adenine found attached to a cytidine dinucleotide in the context A-C-C
E) it is a guanine found attached to a cytidine dinucleotide in the context G-C-C

Accepted Answer

The cap structure on the 5′ end of most eukaryotic mRNAs is a 7-methylguanylate (methylated guanine) linked to the mRNA via a 5′ to 5′ triphosphate bridge.  The 5′ end of all eukaryotic mRNAs are capped with a unique 5′→5′ linkage to a 7-methylguanosine residue. The capped end of the mRNA is thus protected from exonucleases and more importantly is recognized by specific proteins of the translational machinery.

Question 4

You are comparing the process of RNA splicing in an mRNA from normal and mutant cells of the same tissue. You find that the mutant mRNA is longer than the mRNA from the normal cells. Examination of the sequences of the gene encoding this mRNA reveals that there are nucleotide changes at the boundary of one of the exons in the mutant cell gene. Which of the following nucleotides would be expected to be found at the 3′ end of upstream donor exons and the 5′ end of downstream acceptor exons in the normal gene but not in the mutant gene?
A) AG-G
B) AG-AG
C) CA-GA
D) CG-G
E) G-AG

Accepted Answer

The correct answer is A because AG-G is the consensus sequence for the 3' end of upstream donor exons and the 5' end of downstream acceptor exons in the normal gene. In the mutant gene, there are nucleotide changes at the boundary of one of the exons, which would disrupt the consensus sequence and prevent proper splicing.  Analysis of the human genome and the encoded mRNA have revealed highly conserved consensus sequences at the 5′ and 3′ ends of essentially all mRNA exons and introns. The 3′ end of all upstream exons contains AG with the G being the donor nucleotide attached to the 5′ end of the downstream exon. The 5′ end of the acceptor exon is always a G residue.

Question 5

Which of the following relates to the transcription of a eukaryotic mRNA gene?
A) begins only when repressors are removed from nucleosomes in which the gene lies
B) occurs when the gene is in heterochromatin
C) requires binding of RNA polymerase II to the CCAAT box
D) requires binding of TFIID to the TATA box
E) requires eIF-2

Accepted Answer

Transcription of a eukaryotic mRNA gene requires the binding of TFIID to the TATA box, which is a crucial step in the formation of the transcription initiation complex.  There are numerous protein factors controlling transcription initiation. Two basal promoter elements that are found in essentially all eukaryotic mRNA genes are the TATA box and the CAAT box. Many of the basal transcription factors are identified by the fact that they control the activity of RNA pol II. Thus, the nomenclature of these proteins is TFII, for transcription factor of RNA pol II. TFIID is the factor that binds to the TATA-box.

Question 6

You are examining the process of RNA synthesis in a cell line derived from a pancreatic tumor. You discover that the cells do not contain any U1 small nuclear RNA (snRNA). The lack of this RNA species in these cells is most likely to have which of the following effects on overall RNA synthesis?
A) defective polyA tail addition
B) failure of capping mRNAs
C) loss of intron splicing
D) no synthesis of any mRNAs
E) synthesis of only rRNAs

Accepted Answer

U1 snRNA is a component of the spliceosome, which is essential for the splicing of introns from pre-mRNA. Without U1 snRNA, intron splicing cannot occur properly.  Nuclear mRNA introns undergo a splicing reaction catalyzed by specialized RNA-protein complexes called small nuclear ribonucleoprotein particles (snRNPs, pronounced snurps). The RNAs found in snRNPs are identified as U1, U2, U4, U5, and U6. Lack of the U snRNPs would lead to a failure to properly splice mRNAs.

Question 7

The process of RNA editing leads to sequence changes that are not reflective of the genetic information contained within the gene for an edited RNA. RNA editing occurs through the action of specific nucleotide deaminases that change either adenine to inosine or cytidine to uridine. These nucleotide changes can affect the coding or splicing capacity of the RNA. Which of the following proteins is an important example of the result of RNA editing?
A) apo-B48
B) calcitonin gene related peptide (CGRP)
C) ferritin
D) glutamate dehydrogenase
E) tissue factor (factor III)

Accepted Answer

Apo-B48 is an important example of RNA editing, where a cytidine to uridine change creates a stop codon, resulting in a truncated protein compared to apo-B100. RNA editing in mammals involves the deamination of adenine (A) to produce inosine (I) in pre-mRNAs. The process of A to I editing is catalyzed by ADARs (ADAs acting on RNA). In some pre-mRNAs this alters splicing; in some it alters codon(s). RNA editing also occurs to convert C to U. A C-to-U edit in the gene for apolipoprotein B leads to tissue-specific generation of apoB- 48 in intestines, whereas, the nonedited mRNA makes apoB-100 in liver.

Question 8

If the DNA strand shown below is used as a template for RNA polymerase, what would be the sequence of the resultant mRNA following transcription?
5′-CATTCCATAGCATGT-3′
A) 5′-ACAUGCUAUGGAAUG-3′
B) 5′-CAUUCCAUAGCAUGU-3′
C) 5′-GUAAGGUAUCGUACA-3′
D) 5′-UGUACGAUACCUUAC-3′

Accepted Answer

The mRNA sequence is complementary to the DNA template strand and is synthesized in the 5' to 3' direction. The DNA template 3'-5' strand is 3'-GTACGATATGGAATG-5', so the mRNA is 5'-ACAUGCUAUGGAAUG-3'. DNA template strands are used by RNA polymerases to direct the synthesis of an RNA. The nucleotides incorporated are ribonucleotides and they will be complimentary to the sequences of the template strand and oriented in the antiparallel direction. Whenever RNA polymerase encounters an adenosine, it will incorporate a uracil into the RNA molecule.

Question 9

A mutation that resulted in the loss of histone deacetylation activity in a eukaryotic cell line would be expected to have what effect on transcription?
A) conversion of activators into repressors
B) conversion of repressors into activators
C) enhanced rate of transcription
D) no effect
E) reduced rate of transcription

Accepted Answer

Histone deacetylation typically represses transcription by making chromatin more compact. Loss of this activity would lead to a more open chromatin structure, enhancing transcription.  Histone deacetylases are responsible for changing the state of histone acetylation. The consequences of reduced histone acetylation are altered chromatin structure, making it difficult for the transcriptional machinery to engage the DNA. Thus, histone deacetylation is most often associated with reduced capacity for transcription initiation.

Question 10

Which of the following best describes the role of proteins that contain a basic helix-loop-helix (bHLH) domain, such as in the MyoD protein involved in the determination of muscle lineages?
A) binds to specific regions of DNA
B) interacts with the splicesome
C) participates in the polyadenylation of mRNAs
D) participates in the splicing of mRNAs
E) promotes the transnuclear transport of processed mRNAs

Accepted Answer

Proteins with a basic helix-loop-helix (bHLH) domain, like MyoD, are transcription factors that bind to specific DNA sequences to regulate gene expression. The helix-loop-helix domain is a protein interaction domain characteristic of numerous transcription factors. The basic region is required for the recognition of, and binding to, specific DNA sequences.

Question 11

Which of the following is the sequence of the DNA strand that serves as the template strand for transcription of the following strand of RNA?
RNA: 5′-AAGCGUUUGGG-3′
A) 3′-AAGCGTTTGGG-5′
B) 5′-AAGCGTTTGGG-3′
C) 5′-CCCAAACGCTT-3′
D) 5′-GGGTTTGCGAA-3′
E) 5′-TTCGCAAACCC-3′

Accepted Answer

DNA template strands are used by RNA polymerases to direct the synthesis of an RNA. The nucleotides incorporated are ribonucleotides and they will be complimentary to the sequences of the template strand and oriented in the antiparallel direction. Whenever RNA polymerase encounters an adenosine, it will incorporate a uracil into the RNA molecule.

Question 12

You are studying RNA processing in a cell-free system. You find that incomplete RNA processing is occurring in your system and that the failure is most likely due to incomplete assembly of the splicesome complex. Which of the following RNA-processing events is most likely defective in your assay system?
A) addition of poly(A)
B) guiding ribosomes to the endoplasmic reticulum
C) modification of the 5′ end of hnRNA
D) removal of introns in hnRNA
E) transport of mRNA to the cytoplasm

Accepted Answer

The spliceosome is responsible for the removal of introns from hnRNA (heterogeneous nuclear RNA) during RNA processing. If the spliceosome assembly is incomplete, this splicing process would be defective.  A spliceosome is a complex of small nuclear RNAs (snRNAs) and proteins, which is responsible for the catalytic removal of introns from a transcribed pre-mRNA in the process referred to as RNA splicing

Question 13

You have developed a strain of transgenic mice that express a human β-globin gene. Your transgenic construct includes all 3 exons of the human β-globin gene plus 500 nucleotides of both 5′- and 3′-flanking sequences. Examination of these transgenic mice shows that the human β-globin construct is expressed at levels about 10-fold lower than the endogenous mouse β-globin gene. Which of the following regulatory elements is most likely missing from your transgenic construct, thereby explaining the reduced level of expression?
A) enhancer elements
B) mRNA splicing signals
C) polyadenylation signals
D) promoter elements
E) silencer elements

Accepted Answer

Enhancer elements are regulatory DNA sequences that, when bound by specific proteins, can significantly increase the transcription of a gene. The absence of these elements in the transgenic construct could lead to lower expression levels of the human β-globin gene compared to the endogenous mouse gene.  Enhancer elements are cis-acting- DNA sequences that can increase transcription of genes. Enhancers can usually function in either orientation and at various distances from a promoter. The β-globin construct used to create the transgenic mice most likely did not include the appropriate enhancer element(s) that are present in the endogenous mouse β-globin gene, thereby accounting for the reduced expression of the transgene.

Question 14

The completion of the human genome project demonstrated that the human genome codes for approximately 25,000 genes. However, these limited numbers of genes actually encode several hundreds of thousands of different proteins. Which of the following processes provides the majority of the additional diversity of the protein products from these genes?
A) DNA transposition
B) mRNA splicing
C) mutagenesis
D) rRNA processing
E) tRNA modification

Accepted Answer

mRNA splicing, particularly alternative splicing, allows a single gene to produce multiple protein variants by including or excluding certain exons during the processing of the pre-mRNA.  The precise splicing outcome of a transcribed gene is controlled by complex interactions between cis-regulatory splicing signals and trans-acting regulators. Alternative splicing is a prevalent mechanism for generating transcriptome and proteome diversity. It can modulate gene function, affect organismal phenotype, and cause disease.

Question 15

Many members of the nuclear family of receptors are prevented from activity through protein-protein interaction. The glucocorticoid receptor, for example, is released from hsp90 (a chaperone protein) following binding by the appropriate steroid hormone. Which of the following represents the next event for the localization of the receptor-hormone complex?
A) cytoplasmic surface of the plasma membrane
B) Golgi cisterna
C) specific DNA sequences
D) specific mRNA transcripts
E) 50S ribosomal subunit

Accepted Answer

The glucocorticoid receptor-hormone complex translocates to the nucleus where it binds to specific DNA sequences to regulate gene expression. The steroid/thyroid hormone receptor superfamily is a class of proteins that reside in the cytoplasm and bind their lipophilic hormone ligands in this compartment as these hormones are capable of freely penetrating the hydrophobic plasma membrane. Upon binding ligand, the hormone-receptor complex translocates to the nucleus and binds to specific DNA sequences termed hormone response elements (HREs). The binding of the complex to an HRE results in altered transcription rates of the associated gene.

Question 16

You are characterizing the expression of a gene suspected of causing a form of ovarian cancer. You have isolated the gene and found that it is approximately 35 kilobases (kb) in length. Transcriptional analysis reveals that there are several discrete mRNAs produced by this gene ranging in length from 2.0 to 5.5 kb in normal and cancerous cells. Which of the following processes is the most likely cause of the difference in size of the mRNAs produced by this gene?
A) poly(A) addition
B) posttranslational modification
C) recombination
D) RNA editing
E) splicing

Accepted Answer

The difference in size of the mRNAs is most likely due to splicing, where introns are removed and exons are joined in different combinations, resulting in mRNAs of varying lengths.  The precise splicing outcome of a transcribed gene is controlled by complex interactions between cis-regulatory splicing signals and trans-acting regulators. Alternative splicing is a prevalent mechanism for generating transcriptome and proteome diversity. Alternative splicing can modulate gene function, affect organismal phenotype, and cause disease.

Question 17

You are studying the activities of the apolipoproteins, apoB-48 and apoB-100. Both of these proteins are encoded by the same gene; yet apoB-48 is primarily expressed in the intestine, while apoB-100 expression is restricted to the liver. Which of the following is the most likely molecular process accounting for this difference in gene expression?
A) different initiation codons
B) DNA methylation
C) gene duplication
D) reiterated sequences
E) RNA editing

Accepted Answer

RNA editing is the process that results in the production of apoB-48 in the intestine and apoB-100 in the liver from the same gene. In the intestine, a specific cytidine to uridine change in the mRNA results in a premature stop codon, leading to the shorter apoB-48 protein.  RNA editing in mammals involves the deamination of adenine (A) to produce inosine (I) in pre-mRNAs. The process of A-to-I editing is catalyzed by ADARs (ADAs acting on RNA). In some pre-mRNAs this alters splicing; in some it alters codon(s). RNA editing also occurs to convert C to U. A C-to-U edit in the gene for apolipoprotein B leads to tissue-specific generation of apoB- 48 in intestines, whereas, the nonedited mRNA makes apoB-100 in liver.

Question 18

You are isolating and characterizing the nucleic acids from a variety of different cell types. You fractionate and hydrolyze a specific sample of the nucleic acids. You find that the sample being analyzed contains pseudouridine. Which of the following is the most likely nucleic acid you have analyzed?
A) DNA
B) mtDNA
C) mRNA
D) rRNA
E) tRNA

Accepted Answer

Pseudouridine is a modified nucleoside commonly found in tRNA.  More than 300 different tRNAs have been characterized and shown to have extensive similarities. They vary in length from 60-95 nucleotides (18- 28 kD) and exhibit a cloverleaf-like secondary structure that is composed of an anticodon loop, a D loop, and a TΨC loop. The D loop contains the modified nucleotide, dihydrouridine, and the TΨC loop contains pseudouridine, hence the derivation of the names for these loops.

Question 19

You are carrying out experiments designed to characterize the structure-function relationships of tRNAs. You isolate an enzyme with tRNA nucleotidyltransferase activity and show that it adds a CCA sequence to the 3′ end of human tRNAs. Which of the following characteristics best distinguishes this enzyme from other RNA polymerases?
A) requires only a GG template
B) synthesizes a specific sequence without a template
C) uses both mRNA and tRNA as substrates
D) uses nucleoside monophosphate as substrates
E) works in a 3′→5′ direction

Accepted Answer

tRNA nucleotidyltransferase adds a CCA sequence to the 3' end of tRNAs without using a template, distinguishing it from other RNA polymerases that require a template to synthesize RNA.  Since all human tRNAs contain variable nucleotide sequence yet also have the CCA sequence added posttranscriptionally, there is no template-specific control for the activity of tRNA nucleotidyltransferase. The growing 3′ terminus of the tRNA progressively refolds to allow the solitary active site of the enzyme to reuse a single CTP-binding site. The ATP-binding site is then created collaboratively by the refolded CC terminus and the enzyme, and nucleotide addition to any tRNA ceases when the nucleotide-binding pocket of the enzyme is full. The template for CCA addition is, therefore, a dynamic ribonucleoprotein structure and not a typical nucleotide template as for other RNA polymerases.

Question 20

You are studying the rRNA synthesis and processing in a mutant cell line. You find that these cells synthesize a correct precursor rRNA, but do not correctly process it into the expected 3 rRNAs that are found in normal cells. Which of the following organelles is likely to be missing or defective in these cells, thus accounting for the failed rRNA processing?
A) endoplasmic reticulum
B) Golgi complex
C) nuclear matrix
D) nucleolus
E) ribosome

Accepted Answer

The nucleolus is responsible for the synthesis and processing of rRNA, so a defect or absence in the nucleolus would lead to improper rRNA processing.  During or immediately following transcription of pre-rRNA from rDNA in the nucleolus, the ribosomal RNA precursor (pre-rRNA) is modified and associates with a few ribosomal proteins. In order to generate the mature 18S, 5.8S, and 28S rRNAs from the precursor 45S rRNA, the precursor rRNA goes through a series of cleavages in order to remove the external and internal spacers.

Quiz 36: Transcription: RNA Synthesis, Processing, and Regulation

Which of the following represents a characteristic feature of the cap structure on the 5′ end of most eukaryotic mRNAs?

Which of the following relates to the transcription of a eukaryotic mRNA gene?

You are examining the process of RNA synthesis in a cell line derived from a pancreatic tumor. You discover that the cells do not contain any U1 small nuclear RNA (snRNA) . The lack of this RNA species in these cells is most likely to have which of the following effects on overall RNA synthesis?

If the DNA strand shown below is used as a template for RNA polymerase, what would be the sequence of the resultant mRNA following transcription? 5′-CATTCCATAGCATGT-3′

A mutation that resulted in the loss of histone deacetylation activity in a eukaryotic cell line would be expected to have what effect on transcription?

Which of the following best describes the role of proteins that contain a basic helix-loop-helix (bHLH) domain, such as in the MyoD protein involved in the determination of muscle lineages?

Which of the following is the sequence of the DNA strand that serves as the template strand for transcription of the following strand of RNA? RNA: 5′-AAGCGUUUGGG-3′