[MCQs] Gene Expression and Regulation Quiz – Part 2 (19 test)

I. Start the exam by click the “Start” button

Gene Expression and Regulation - Part 2 (19 test)

Please wait while the activity loads.
If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If loading fails, click here to try again

Select the ONE answer that is BEST in each question!

Start

Congratulations - you have completed Gene Expression and Regulation - Part 2 (19 test). You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1

Aminoacyl-tRNA synthetases must be capable of recognizing which of the following?

A	A specific tRNA and a specific amino acid
B	A specific rRNA and a specific amino acid
C	A specific tRNA and the 40S ribosomal subunit
D	A specific amino acid and the 40S ribosomal subunit
E	A specific amino acid and the 60S ribosomal subunit

Question 1 Explanation:

Aminoacyl-tRNA synthetases are responsible for charging a tRNA with the appropriate amino acid for translation. Charging a tRNA is a two-step reaction. In the first step, the enzyme forms an aminoacyl-AMP enzyme complex in a reaction that requires one ATP. In the second step, the activated amino acid is attached to the appropriate tRNA, and the enzyme and AMP are released.

Question 2

Ribosomes similar to those of bacteria are found in which of the following?

A	Cardiac muscle cytoplasm
B	Liver endoplasmic reticulum
C	Neuronal cytoplasm
D	Pancreatic mitochondria
E	Plant nuclei

Question 2 Explanation:

Prokaryotic ribosomes have a sedimentation coefficient of 70S, and are composed of 50S and 30S subunits (odd numbers). Eukaryotic cytoplasmic ribosomes, either free or bound to the endoplasmic reticulum, are larger—60S and 40S subunits that associate to an 80S ribosome (even numbers). Nuclear ribosomes are attached to the endoplasmic reticulum of the nuclear membrane. Ribosomes in chloroplasts and mitochondria of eucaryotic cells are more similar to prokaryotic ribosomes than to eukaryotic cytosolic ribosomes. Like bacterial ribosomes, chloroplast and mitochondrial ribosomes use a formylated tRNA and are sensitive to bacterial protein synthesis inhibitors. Antibiotics that target bacterial protein synthesis may interfere with human mitochondrial function at high doses.

Question 3

Endogenous insulin is ineffective in type 2 diabetes mellitus (MIM*222100), leading to high blood glucose levels (hyperglycemia) with gradual effects on blood vessels in eye, kidney, and skin. Pregnant women with diabetes mellitus present high glucose loads to their fetus; this stimulates production of fetal insulin (hyperinsulinemia) and causes rapid fetal growth with large birth weight (macrosomia). The stimulation of fetal growth by insulin may be correlated with its effect on gene expression, which is which of the following?

A	Stimulation of mRNA production by enhancing 5′ capping
B	Acceleration of protein synthesis by phosphorylating initiation inhibitors
C	Acceleration of protein synthesis by phosphorylating the 40S ribosomal subunit
D	Acceleration of protein synthesis by phosphorylating proteinase K
E	Stimulation of mRNA production by enhancing RNA splicing

Question 3 Explanation:

The initiation of protein synthesis is a multistep process that includes several protein factors—eIF-4E and eIF-4G proteins that bind the mRNA cap, eIF-4A and eIF-4B proteins that bind to the 5′ end of mRNA and reduce its secondary structure. The eIF-3 proteins bind to the prior factors, linking them to the 40S ribosomal subunit and scanning for suitable AUG codons that will initiate peptide bond formation. This is usually the 5′-most AUG, but others may be selected by virtue of Kozak consensus sequences that surround it (GCC-AUG-UGG). Other protein factors regulate the rate of protein synthesis initiation by acting on eIF-4E, including those such as BP1 that bind and inactivate eIF-4E unless it is phosphorylated. Insulin and other growth factors act to phosphorylate BP-1 at several sites, releasing eIF-4E and stimulating initiation and protein synthesis. Although stimulation ofprotein synthesis by insulin is one plausible factor in increased fetal growth, other factors undoubtedly contribute to this complex process. Once the EF factors, 40S subunit, and relaxed mRNA are associated (as a 43S preinitiation complex), it binds the 60S ribosomal subunit and its peptidyl transferase that is a component of the 28S rRNA—an example of a ribozyme. In translation, the P site of the ribosome is occupied by the peptidyl-tRNA, which has the growing peptide chain attached. The appropriate aminoacyl-tRNA enters the A site and the carboxyl group of the peptidyl tRNA undergoes nucleophilic attack by the α- amino group of the aminoacyl-tRNA, as catalyzed by peptidyl transferase. The peptide chain is transferred to the tRNA in the A site, which is subsequently displaced to the P site, freeing the A site to bind the next charged tRNA.

Question 4

Which of the following eukaryotic promoter/regulatory elements has the most variable position with respect to the start site of transcription?

A	Downstream promoter element (DPE)
B	Enhancer
C	Initiator sequence
D	Operator
E	TATA box

Question 4 Explanation:

Eukaryotic promoters are much more complex than bacterial promoters. The most common bacterial promoter elements are the –10 and –35 sequences, which are the recognition sites for RNA polymerase binding. In addition, bacterial promoters may have operators (repressor binding sites) and activator binding sites. Eukaryotic promoters have a core region that may contain a TATA box (which is bound by TBP, the TATA binding protein), an initiator sequence, and downstream promoter elements. Many promoters lack one or more of these elements. In addition, eukaryotic promoters have upstream elements such as GC boxes and CCAAT boxes that bind specific transcription factors (Sp1 and CTF, respectively). Enhancers are elements that increase or enhance the rate of transcription initiation from a promoter. Enhancers can be upstream or downstream of the transcription start site and can exert their effect from hundreds or thousands of bases away. In addition, enhancers are orientation-independent (ie, whether their recognition sequence is on the Watson or Crick strand).

Question 5

Which of the following is required for certain types of eucaryotic protein synthesis, but not for prokaryotic protein synthesis?

A	GTP
B	Messenger RNA
C	Ribosomal RNA
D	Peptidyl transferase
E	Signal recognition particle

Question 5 Explanation:

Signal recognition particles (SRPs) recognize the signal sequence on the N-terminal end of proteins destined for the lumen of the endoplasmic reticulum (ER). SRP binding arrests translation and an SRP receptor facilitates import of the nascent protein into the ER lumen. A signal peptidase removes the signal sequence from the protein, which may remain in the membrane or be routed for secretion. Common to both eukaryotic and prokaryotic protein synthesis is the requirement for ATP to activate amino acids. The activated aminoacyl-tRNAs then interact with ribosomes carrying mRNA. Peptidyl transferase catalyzes the formation of peptide bonds between the free amino group of activated aminoacyl-tRNA on the A site of the ribosome and the esterified carboxyl group of the peptidyl-rRNA on the P site; the liberated rRNA remains on the P site.

Question 6

A 26-year-old black African American suffers vision loss after hitting his head while playing handball. Ophthalmologic evaluation reveals detachment of the retina that seems unusual for such mild trauma, and further medical history notes that the male has lax joints, tall and thin build, and chronic joint pain after exercise. The family history is further revealing in that several individuals have had arthritis and detached retina. The symptoms and history suggest a diagnosis of Stickler syndrome (MIM*108300), a collagen disease that results in lax joints, arthritis, and retinal detachments. The locus for Stickler syndrome has been mapped near that for type II collagen on chromosome 12, and mutations in the COL2A1 gene have been described in Stickler syndrome. The family became interested in molecular diagnosis to distinguish normal from mildly affected individuals. Which of the following results would be expected in an individual with a promoter mutation at one COL2A1 gene locus?

A	Western blotting detects no type II collagen chains.
B	Southern blotting using intronic restriction sites yields normal restriction fragment sizes.
C	Reverse transcriptase-polymerase chain reaction (RT-PCR) detects one-half normal amounts of COL2A1 mRNA in affected individuals.
D	Fluorescent in situ hybridization (FISH) analysis using a COL2A1 probe detects signals on only one chromosome 12.
E	DNA sequencing reveals a single nucleotide difference between homologous COL2A1 exons.

Question 6 Explanation:

After the locus responsible for a genetic disease is mapped to a particular chromosome region, “candidate” genes can be examined for molecular abnormalities in affected individuals. The connective tissue abnormalities in Stickler syndrome (MIM*108300) make the COL2A1 collagen locus an attractive candidate for disease mutations, prompting analysis of COL2A1 gene structure and expression. Western blotting detects gene alterations that interfere with protein expression, while use of the reverse transcriptase-polymerase chain reaction (RT-PCR) detects alterations in mRNA levels. Each analysis should detect one-half the respective amounts of COL2A1 protein or mRNA in the case of a promoter mutation that abolishes transcription of one COL2A1 allele. Southern blotting detects nucleotide changes that alter DNA restriction sites, but this is relatively insensitive unless large portions of the gene are deleted. Fluorescent in situ hybridization (FISH) analysis using DNA probes from the COL2A1 locus is a sensitive method for detecting deletions of the entire locus, and DNA sequencing of the entire gene provides the gold standard for detecting any alteration in the regulatory or coding sequences. Nucleotide sequence changes are still subject to interpretation, since they may represent polymorphisms that do not alter gene function. Population studies and/or in vitro studies of gene expression are often needed to discriminate DNA polymorphisms from mutations that disrupt gene function. For any autosomal locus, the interpretation of molecular analyses is complicated by the presence of two homologous copies of the gene.

Question 7

Gyrate atrophy (MIM*258870) is a rare autosomal recessive genetic disorder caused by a deficiency of ornithine aminotransferase. Affected individuals experience progressive chorioretinal degeneration with vision and neurologic defects. The gene for ornithine aminotransferase has been cloned, its structure has been determined, and mutations in affected individuals have been extensively studied. Which of the following mutations best fits with test results showing normal Southern blots with probes from all ornithine aminotransferase exons but absent enzymatic activity?

A	Duplication of entire gene
B	Two-kb deletion in coding region of gene
C	Two-kb insertion in coding region of gene
D	Deletion of entire gene
E	Missense mutation

Question 7 Explanation:

Missense mutations, which cause the substitution of one amino acid for another, may significantly alter the function of the resultant protein without altering the size of DNA restriction fragments detected by Southern blotting. In this case, Northern blot results would most likely also be normal. Single-base changes may also result in nonsense mutations. Large insertions or deletions in the exon or coding regions of the gene alter the Southern blot pattern and usually ablate the activity of one gene copy. In the case of an autosomal locus like that for ornithine aminotransferase, the homologous allele remains active and gives 50% enzyme activity (heterozygote or carrier range with a normal phenotype). Similar effects on enzyme activity would be predicted from complete gene deletions at one locus, while duplication might produce 150% or 50% of normal enzyme activity depending on the status of promoter sites.

Question 8

A 1-year-old Caucasian girl who has been treated for scoliosis and clubfoot is noted to be developmentally delayed with no walking and no speech. Examination reveals increased subcutaneous tissue in her face that causes a coarse appearance, enlarged liver and spleen, scoliosis with a posterior spinal prominence called a gibbus, and increased body hair (hirsutism). A diagnosis of Hurler syndrome (MIM*252800) is considered, and analysis of her white blood cells shows the characteristic deficiency of L-iduronidase. Exogenous L-iduronidase can be taken up by deficient cells via a targeting signal that directs the enzyme to its normal lysosomal location. Which of the following therapeutic strategies is the most realistic and efficient mode of therapy?

A	Germ-line gene therapy
B	Heterologous bone marrow transplant
C	Infection with a disabled adenovirus vector that carries the L-iduronidase gene
D	Injection with L-iduronidase purified from human liver
E	Autologous bone marrow transplant after transfection with a virus carrying the L-iduronidase gene

Question 8 Explanation:

All of the modes of therapy are theoretically possible, and enzyme therapy (ie, injection of purified enzyme) has been successful in several lysosomal deficiencies, particularly those in which the central nervous system is not affected (ie, nonneurologic Gaucher disease [MIM*231000]). Unfortunately, antibodies frequently develop to the injected enzyme and limit the term of successful enzyme delivery. Heterologous bone marrow transplant, preferably from a related donor, offers the most realistic and effective therapy since the graft provides a permanent source of enzyme. Bone marrow transplants do have a 10% mortality, however, and the enzyme diffuses poorly into the central nervous system. Somatic gene therapy (ie, delivery of enzyme to somatic cells via viral vectors or transfected tissue) is now possible; however, targeting of the gene product to appropriate tissues and organelles is still a problem. Transfected autologous bone marrow transplant (ie, marrow from the patient) has been used in a few cases of adenosine deaminase deficiency, an immune disorder affecting lymphocytes. Germ-line gene therapy requires the insertion of functional genes into gametes or blastomeres of early embryos prior to birth. The potential for embryonic damage, lack of knowledge regarding developmental gene control, and ethical controversies regarding selective breeding or embryo experimentation make germ-line therapy unrealistic at present.

Question 9

It is now recognized that all cancers involve genetic changes, even though few are hereditary. An early example was the discovery of a Philadelphia chromosome (terminal deletion of chromosome 22) in leucocytes from patients with chronic myeloid leukemia (MIM*608232). Later work demonstrated that the deleted material was in fact translocated to the terminus of chromosome 9, being an example of a balanced translocation between 9 and 22. Of the following results, which would be the most diagnostic of the 9;22 translocation in chronic myeloid leukemia?

A	All chromosome 22 genes show increased incorporation of labeled acetate into histones of chromatin regions.
B	A particular gene near the terminus of chromosome 22 shows altered DNA restriction endonuclease patterns.
C	The pattern of small RNA “decoration” along leukemic chromatin is changed.
D	Increased transcription of mRNA from genes near the terminus for chromosome 9.
E	Decreased transcription of mRNA from all loci on the chromosome 22 long arm.

Question 9 Explanation:

Because chromosomes are large chains of DNA comprising several thousand genes (links of chain), even small deletions will remove tens to hundreds of genes. Deletions involving the autosomes (chromosomes 1-22) will reduce genes within the deletion from diploid to haploid dosage, decreasing the amount of mRNA produced from these genes (answer e). The other answers concern changes in epigenesis— histone acetylation, DNA methylation, small RNA decoration—or a change in gene sequence that may not occur if the deletion cuts between genes. Changes in epigenesis undoubtedly occur with chromosome rearrangements (eg, translocations, duplicated segments, and interstitial deletions between chromosome ends) that make new links between noncontiguous chromosome segments. Epigenetic changes likely contribute to the diverse phenotypic effects of trisomies, where an extra dose of normal chromosomal DNA sequence is added to cells (eg, trisomy or translocation 21, causing Down syndrome).

Question 10

A 14-year-old Caucasian adolescent of Ashkenazi Jewish descent is evaluated for fatigue and aching legs at night that had been attributed to “growing pains.” Examination revealed a palpable spleen (splenomegaly because the spleen is not normally palpated at this age). His white blood cells showed the typical foamy appearance of cytoplasm due to lipid accumulation in Gaucher disease (MIM*231000). Gaucher disease was considered, and blood obtained for DNA testing, examining the gene for glucocerebrosidase enzyme that, when deficient, causes accumulation of complex lipids in white cells, brain, liver, and spleen. A homozygous glucocerebrosidase gene mutation was found that changed the mRNA sequence of codon 93 from UAC to the UAA depicted in the table below. Which of the following best describes this result?

A	Missense mutation that may interfere with glucocerebrosidase function and could be diagnostic of Gaucher disease
B	Silent mutation that should not interfere with glucocerebrosidase function and is not diagnostic of Gaucher disease
C	Nonsense mutation that produces a truncated, nonfunctional glucocerebrosidase polypeptide and is diagnostic of Gaucher disease
D	Suppressor mutation that interferes with glucocerebrosidase mRNA transcription and is diagnostic of Gaucher disease
E	Frameshift mutation that produces a nonfunctional glucocerebrosidase and is diagnostic of Gaucher disease

Question 10 Explanation:

The replacement of the codon UAC with UAA would be a nonsense mutation, since UAC encodes tyrosine and UAA is a “stop” signal. Missense mutations substitute one amino acid for another, silent mutations do not change the amino acid code, suppressor mutations are rare changes in tRNA that allow normal reading of a stop codon, and frameshift mutations insert or delete a nucleotide so as to change the reading frame (incorrect answers a, b, d, and e).Nonsense or terminating codons in mRNA signal the ribosome to stop translation. Mutation of an internal codon to a termination codon results in a shortened peptide that in the case of enzymes would not function as a catalyst. Homozygous nonsense mutation affecting both glucocerebrosidase alleles would result in no enzyme produced and no ability for the cells to break down stored lipids.

Question 11

A 3-month-old Caucasian boy fails to grow and has very low muscle tone with strange “kinky” hair. A diagnosis of Menkes kinky hair syndrome (MIM*309400) is made by showing deficient copper in fibroblasts. A likely candidate gene encoding an ATP-dependent copper transporter protein is known, and antibodies to this protein are available. Experimental analysis of the patient’s fibroblast DNA is performed to see if a pathologic mutation can be found, thus confirming the copper transport gene as the cause of Menkes disease. The table below shows a partial DNA sequence from the transporter coding strand as amplified by PCR; the amplified sequence is from a region where a TATA box has been located. Which of the following experimental results would be the most compatible with a pathologic mutation in the transporter gene

A	Usual band of usual intensity on Northern and Western blots
B	Absent bands on Northern and Western blots
C	Usual band plus a second band of lower size on Northern blot, usual band on Western blot
D	Usual band on Northern blot, band of unusual size and reduced intensity on Western blot
E	Two unusual bands on Northern blot, usual bands on Western blot

Question 11 Explanation:

The change in the patient’s DNA coding strand sequence from TAC to GAC would change AUG to CUG in mRNA, substituting the amino acid leucine for methionine, but also changing a likely AUG translation initiation signal, since this AUG is near a TATA signal mediating transcription initiation. The mutation would logically affect the conformation of the translated protein and alter its mobility on electrophoresis while also ablating an AUG translation signal and reducing the amount of protein detected on Western blot (correct answer d, incorrect answers a-c and e). Transcription of mRNA should not be affected, also making answers b, c, and e that suggest an abnormal signal by Northern blotting incorrect. Note that RNA is transcribed 5′ to 3′ from the coding strand of DNA that is 3′ to 5′ it is the 5′ to 3′ complement of this coding strand that is represented in genome databases so the gene sequence can be easily visualized in terms of its product RNA. When researchers examine a cloned DNA sequence for potential coding activity, they must consider reads from both complimentary strands to see if either gives a sensible peptide translation. Sequences encoding RNA can come from either DNA strand, indicating that the asymmetry of leading and lagging strands during DNA replication is not preserved for transcription. Another facet of the patient’s mutation is its change from T (thymine, a pyrimidine) to G (guanine, a purine). This is a transversion with change in base class rather than a transition that substitutes one purine or pyrimidine for another. Transversions will usually have more severe consequences than transitions, and both will be less severe than frameshift or nonsense mutations that garble or truncate the protein product.

Question 12

Lipoproteins transport lipids from intestine and liver, where they are absorbed and metabolized to peripheral tissues via the bloodstream. They contain core proteins (apo proteins) that bind various types of lipids, thus producing lipoprotein complexes of varying densities—very low density lipoproteins (VLDL), low- density lipoproteins (LDL), and high-density lipoproteins (HDL) after plasma lipoprotein electrophoresis. A disorder called abetalipoproteinemia (MIM*200100) was distinguished by its deficiency of the bands for very low density lipoproteins (VLDL) and low-density lipoproteins (LDL) after plasma electrophoresis. Affected patients have ataxia (wide gait, incoordination), retinopathy (retinal degeneration with blindness), and myopathy (muscle weakness). VLDL and LDL contain the same apoB core protein, their different densities produced by association with different lipids, and apoB became good candidate for mutation in the disorder. Initial characterization of the apoB gene together with its mRNA and protein products demonstrated identical mRNA sizes in all tissues. Western blotting using antibody specific for the amino-terminus of apoB protein showed a 100-kDa species in liver and a 48-kDa species in intestine, as did antibody specific for the C-terminus of the 48-kDa intestinal protein. Molecular study of a patient with abetalipoproteinemia showed normal mRNA sizes, but 100-kDa peptide species were identified in both liver and intestine using the amino- or carboxy-terminus antibody probes. Which of the following processes is most likely deficient in this patient with apobetalipoproteinemia?

A	RNA splicing
B	DNA amplification
C	Transcription initiation
D	Transcription factor phosphorylation
E	RNA editing

Question 12 Explanation:

The apoB gene, which encodes apolipoprotein B, is transcribed in the liver normally and the mRNA is translated into a 100-kDa protein (apoB100). In the intestine, however, the apoB mRNA is edited by the enzyme cysteine deaminase, which alters a CAA codon (encoding a glutamine) to UAA (a stop codon). This change results in synthesis of a 48-kDa form of apolipoprotein B (apoB48) that lacks the carboxy-terminal end. The apoB100 and apoB48 proteins serve different functions in the liver and the intestine. A mutation that ablates apoB mRNA editing and its resulting apoB48 protein product in intestine could interfere with lipid absorption and production of VLDL and LDL in plasma. Other answers besides e indicate processes that would affect mRNA transcription, not protein size. Although RNA editing is not widespread, there are a growing number of other examples. Alternative RNA splicing is a much more common means of generating alternative forms of a protein. In splicing, intron sequences are removed and exon (protein coding) sequences are joined together. Use of alternative splice sites can include or exclude certain exons, resulting in different primary sequence, conformation, and size of a protein. RNA splicing will usually preserve the carboxy-terminal exon, since it is needed for mRNA polyadenylation and its transport from nucleus to cytoplasm.

Question 13

During the analyses described in Question 226, another patient with abetalipoproteinemia was found to have negligible amounts of apoB protein in either liver or intestine. Analysis of the patient’s DNA and RNA would most likely yield which of the following results?

A	Substitution of the last base of a codon near the middle of the first apoB exon
B	Insertion of two bases near the apoB amino-terminus
C	Transversion producing a nonsense mutation within the first apoB intron
D	Transition affecting the 30th base of the first apoB intron
E	Base substitution changing glycine for alanine near the apoB carboxyl-terminus

Question 13 Explanation:

The structures of glycine and alanine are quite similar, with the –H side group of glycine being replaced with the –CH3 side group of alanine. Consequently, a mutation causing such a change, particularly near the carboxy-terminus is unlikely to produce a dysfunctional protein. Similarly, the third position of mRNA codons is least crucial for amino acid coding due to “wobble”—similar or identical amino acids are often produced regardless of the third codon nucleotide (see High-Yield Facts, Table 8). In contrast, a mutation inserting two bases into DNA and mRNA will change the reading frame of translation, producing a truncated (misread termination codon) or garbled amino acid sequence with apoB dysfunction. Introns are spliced out of RNA and do not encode segments of the protein product unless splicing is defective; intronic mutations should therefore not disrupt protein function unless they affect the splice sites near exon-intron junctions.

Question 14

Part of the triplet genetic code involving mRNA codon triplets that start with U is shown below. Using the portion of the genetic code shown, which of the following mutations in the 3′ to 5′ DNA template segments corresponds to a nonsense mutation?

A	ACGACGACG to ACAAACACG
B	AGGAATATG to AGGAATATT
C	AGAATAACA to AAAATAACA
D	AAAATGAGC to AAAATAAGC
E	AACAACAAC to AACAAGAAC

Question 14 Explanation:

The following mutations were shown in the DNA changes: a. Missense (Cys to Leu) b. Nonsense (Tyr to stop) c. Missense (Ser to Phe) d. Harmless (Tyr to Tyr) e. Missense (Leu to Phe) Most of the mutations shown result in a missense effect, with a different amino acid being incorporated into the same site in a protein. This may or may not have an effect depending on its location. Some single-base mutations are harmless because of the degeneracy of the genetic code, whereby more than one triplet code exists for all amino acids except tryptophan and methionine. Choice a contains two mutations, one degenerate, and the other missense. DNA coding: 3′-ACGACGACG-5′ to 3′-ACAAACACG-5′ mRNA: 5′-UGCUGCUGC-3′ to 5′-UGUUUGUGC-3 Protein: Cys-Cys-Cys to Cys-Leu-Cys Nonsense mutations occur when the reading of the normal termination signal is changed. This can occur by mutation to a stop signal as in choice b, by deletions near a stop codon, or by insertions.

Question 15

The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following?

A	Increased concentrations of glucose and cyclic AMP (cAMP)
B	Decreased concentrations of glucose and cAMP
C	Increased concentrations of glucose, decreased concentration of cAMP
D	Decreased concentrations of glucose, increased concentration of cAMP
E	Increased concentrations of glucose and adenosine triphosphate (ATP)

Question 15 Explanation:

Several operons in Escherichia coli, including the lac operon, are subject to catabolite repression. In the presence of glucose, there is decreased manufacture of cyclic AMP (cAMP) by adenylate cyclase. Low glucose levels increase production of cAMP, which binds to the catabolite activator protein (CAP). The cAMP-CAP complex binds to the promoters of several responsive operons at catabolite activator protein (CAP) binding sites, greatly enhancing transcription of operon RNA. This positive control stimulates use of more exotic metabolites when glucose is not available and conserves energy when glucose is plentiful. High levels of glucose lower cAMP levels and direct metabolism toward constitutive glucose pathways such as glycolysis.

Question 16

Which of the following regulators are said to act in “cis?”

A	The lac repressor and the lac operator
B	The lac operator and mammalian transcription factors
C	The lac operator and mammalian enhancers
D	Mammalian transcription factors and enhancers
E	The lac repressor and mammalian transcription factors

Question 16 Explanation:

Certain regulatory elements act on genes on the same chromosome (“cis”), while others can regulate genes on the opposite chromosome (“trans”). The terminology makes analogy to carbon-carbon double bonds, where two modifying groups may both be above or below the bond (cis) or opposite it (trans). Cis regulatory elements such as the lac operator and promoter or mammalian enhancers are usually DNA sequences (regulatory sequences) adjoining or within the regulated gene. Transregulatory elements such as the lac repressor protein or mammalian transcription factors are usually diffusible proteins (regulatory factors) that can interact with adjoining target genes or with target genes on other chromosomes. Classification of bacterialelements as cis or trans requires mating experiments where portions of a second chromosome are introduced by transduction (with bacteriophage) or conjugation (with other bacteria). The distinction between cis and trans is fundamental for understanding how regulators work.

Question 17

The proopiomelanocortin (POMC-MIM*176830) gene encodes several regulatory proteins that affect pituitary function. Children with severe brain defects such as holoprosencephaly (MIM*157170) often have abnormalities in the hypothalamic-pituitary axis. In different brain regions, proteins encoded by this gene have different carboxy-terminal peptides. Which of the following best explains the regulatory mechanism?

A	POMC transcription is regulated by different factors in different brain regions.
B	POMC translation elongation is regulated by different factors in different brain regions.
C	POMC translation elongation is regulated by different factors in different brain regions.
D	POMC protein undergoes different protein processing in different brain regions.
E	POMC protein forms different allosteric complexes in different brain regions.

Question 17 Explanation:

The POMC gene provides a mammalian example in which several proteins are derived from the same RNA transcript. Unlike the polycistronic mRNA of the bacterial lactose operon, mammalian cells generate several mRNAs or proteins from the same gene by variable protein processing or by alternative splicing. Variable protein processing preserves the peptide products of some gene regions but degrades those from others. Alternative splicing would often produce proteins composed of different exon combinations with the same terminal exon and carboxy-terminal peptide, but could remove the terminal exon in some proteins and produce different C-terminal peptides. Different transcription factors or enhancers in different brain regions could regulate the total amounts of POMC gene transcript, but not the types of protein produced. Elongation of protein synthesis involves GTP cleavage, but is not differentially regulated in mammalian tissues. The POMC gene may overexpress in primary adrenal deficiency and increase melanin deposition with hyperpigmentation.

Question 18

A 3-year-old Chinese boy has severe anemia with prominence of the forehead (frontal bossing) and cheeks. The red cell hemoglobin concentration is dramatically decreased, and it contains only β-globin chains with virtual deficiency of α-globin chains. Which of the following mechanisms is the most likely explanation?

A	A transcription factor regulating the α-globin gene is mutated.
B	A regulatory sequence element has been mutated adjacent to an α-globin gene.
C	A transcription factor regulating the β-globin gene is mutated.
D	A transcription factor regulating the α- and β-globin genes is deficient.
E	A deletion has occurred surrounding an α-globin gene.

Question 18 Explanation:

Imbalance of globin chain synthesis occurs in the thalassemias. Deficiency of α-globin chains α-thalassemia—MIM*141900) is common in Asian populations and may be associated with abnormal hemoglobins composed of four β-globin chains (hemoglobin H) or (in fetuses and newborns) of four γ-globin chains (hemoglobin Bart). Mutation in a transcription factor necessary for expression of α-globin could ablate α-globin expression, since the same factor could act in trans on all four copies of the α-globin genes (two α-globin loci). Mutation of a regulatory sequence element that acts in cis would inactivate only one α-globin gene, leaving others to produce α-globin in reduced amounts (mild α-thalassemia). Deletions of one α-globin would produce a similar mild phenotype, and deficiencies of transcription factors regulating α- and β-globin genes would not produce chain imbalance.

Question 19

A mutation that results in a valine replacement for glutamic acid at position 6 of the β chain of hemoglobin S hinders normal hemoglobin function and results in sickle cell anemia (MIM*602903) when the patient is homozygous for this mutation. This is an example of which of the following types of mutation?

A	Deletion
B	Frameshift
C	Insertion
D	Missense
E	Nonsense

Question 19 Explanation:

Missense mutations are those in which a single-base change (point mutation) results in a codon that encodes for a different amino acid residue. The effects of these types of mutations can range from very minor or even undetectable to major, depending on the importance of the altered residue to protein folding and function. Nonsense mutations are also point mutations in which the affected codon is altered to a stop (nonsense) codon, resulting in a truncated protein. Frameshift mutations are due to one or two base pair insertions or deletions such that the reading frame is altered. These mutations generally lead to truncated proteins as well, since in most protein coding regions the unused reading frames contain numerous stop codons.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect. Get Results

There are 19 questions to complete.

←

List

→

Return

Shaded items are complete.

1	2	3	4	5
6	7	8	9	10
11	12	13	14	15
16	17	18	19	End

Return

See all quizzes of the Gene Expression and Regulation at here:

Part 1 (20 test) | Part 2 (19 test – end)

II. Preview all questions below

1. Aminoacyl-tRNA synthetases must be capable of recognizing which of the following?

A specific tRNA and a specific amino acid
A specific rRNA and a specific amino acid
A specific tRNA and the 40S ribosomal subunit
A specific amino acid and the 40S ribosomal subunit
A specific amino acid and the 60S ribosomal subunit

2. Ribosomes similar to those of bacteria are found in which of the following?

Cardiac muscle cytoplasm
Liver endoplasmic reticulum
Neuronal cytoplasm
Pancreatic mitochondria
Plant nuclei

3. Endogenous insulin is ineffective in type 2 diabetes mellitus (MIM*222100), leading to high blood glucose levels (hyperglycemia) with gradual effects on blood vessels in eye, kidney, and skin. Pregnant women with diabetes mellitus present high glucose loads to their fetus; this stimulates production of fetal insulin (hyperinsulinemia) and causes rapid fetal growth with large birth weight (macrosomia). The stimulation of fetal growth by insulin may be correlated with its effect on gene expression, which is which of the following?

Stimulation of mRNA production by enhancing 5′ capping
Acceleration of protein synthesis by phosphorylating initiation inhibitors
Acceleration of protein synthesis by phosphorylating the 40S ribosomal subunit
Acceleration of protein synthesis by phosphorylating proteinase K
Stimulation of mRNA production by enhancing RNA splicing

4. Which of the following eukaryotic promoter/regulatory elements has the most variable position with respect to the start site of transcription?

Downstream promoter element (DPE)
Enhancer
Initiator sequence
Operator
TATA box

5. Which of the following is required for certain types of eucaryotic protein synthesis, but not for prokaryotic protein synthesis?

GTP
Messenger RNA
Ribosomal RNA
Peptidyl transferase
Signal recognition particle

6. A 26-year-old black African American suffers vision loss after hitting his head while playing Ophthalmologic evaluation reveals detachment of the retina that seems unusual for such mild trauma, and further medical history notes that the male has lax joints, tall and thin build, and chronic joint pain after exercise. The family history is further revealing in that several individuals have had arthritis and detached retina. The symptoms and history suggest a diagnosis of Stickler syndrome (MIM*108300), a collagen disease that results in lax joints, arthritis, and retinal detachments. The locus for Stickler syndrome has been mapped near that for type II collagen on chromosome 12, and mutations in the COL2A1 gene have been described in Stickler syndrome. The family became interested in molecular diagnosis to distinguish normal from mildly affected individuals. Which of the following results would be expected in an individual with a promoter mutation at one COL2A1 gene locus?

Western blotting detects no type II collagen chains.
Southern blotting using intronic restriction sites yields normal restriction fragment sizes.
Reverse transcriptase-polymerase chain reaction (RT-PCR) detects one-half normal amounts of COL2A1 mRNA in affected individuals.
Fluorescent in situ hybridization (FISH) analysis using a COL2A1 probe detects signals on only one chromosome 12.
DNA sequencing reveals a single nucleotide difference between homologous COL2A1 exons.

7. Gyrate atrophy (MIM*258870) is a rare autosomal recessive genetic disorder caused by a deficiency of ornithine aminotransferase. Affected individuals experience progressive chorioretinal degeneration with vision and neurologic The gene for ornithine aminotransferase has been cloned, its structure has been determined, and mutations in affected individuals have been extensively studied. Which of the following mutations best fits with test results showing normal Southern blots with probes from all ornithine aminotransferase exons but absent enzymatic activity?

Duplication of entire gene
Two-kb deletion in coding region of gene
Two-kb insertion in coding region of gene
Deletion of entire gene
Missense mutation

8. A 1-year-old Caucasian girl who has been treated for scoliosis and clubfoot is noted to be developmentally delayed with no walking and no speech. Examination reveals increased subcutaneous tissue in her face that causes a coarse appearance, enlarged liver and spleen, scoliosis with a posterior spinal prominence called a gibbus, and increased body hair (hirsutism). A diagnosis of Hurler syndrome (MIM*252800) is considered, and analysis of her white blood cells shows the characteristic deficiency of L-iduronidase. Exogenous L-iduronidase can be taken up by deficient cells via a targeting signal that directs the enzyme to its normal lysosomal location. Which of the following therapeutic strategies is the most realistic and efficient mode of therapy?

Germ-line gene therapy
Heterologous bone marrow transplant
Infection with a disabled adenovirus vector that carries the L-iduronidase gene
Injection with L-iduronidase purified from human liver
Autologous bone marrow transplant after transfection with a virus carrying the L-iduronidase gene

9. It is now recognized that all cancers involve genetic changes, even though few are hereditary. An early example was the discovery of a Philadelphia chromosome (terminal deletion of chromosome 22) in leucocytes from patients with chronic myeloid leukemia (MIM*608232). Later work demonstrated that the deleted material was in fact translocated to the terminus of chromosome 9, being an example of a balanced translocation between 9 and Of the following results, which would be the most diagnostic of the 9;22 translocation in chronic myeloid leukemia?

All chromosome 22 genes show increased incorporation of labeled acetate into histones of chromatin regions.
A particular gene near the terminus of chromosome 22 shows altered DNA restriction endonuclease patterns.
The pattern of small RNA “decoration” along leukemic chromatin is changed.
Increased transcription of mRNA from genes near the terminus for chromosome 9.
Decreased transcription of mRNA from all loci on the chromosome 22 long arm.

10. A 14-year-old Caucasian adolescent of Ashkenazi Jewish descent is evaluated for fatigue and aching legs at night that had been attributed to “growing pains.” Examination revealed a palpable spleen (splenomegaly because the spleen is not normally palpated at this age). His white blood cells showed the typical foamy appearance of cytoplasm due to lipid accumulation in Gaucher disease (MIM*231000). Gaucher disease was considered, and blood obtained for DNA testing, examining the gene for glucocerebrosidase enzyme that, when deficient, causes accumulation of complex lipids in white cells, brain, liver, and spleen. A homozygous glucocerebrosidase gene mutation was found that changed the mRNA sequence of codon 93 from UAC to the UAA depicted in the table below. Which of the following best describes this result?

Missense mutation that may interfere with glucocerebrosidase function and could be diagnostic of Gaucher disease
Silent mutation that should not interfere with glucocerebrosidase function and is not diagnostic of Gaucher disease
Nonsense mutation that produces a truncated, nonfunctional glucocerebrosidase polypeptide and is diagnostic of Gaucher disease
Suppressor mutation that interferes with glucocerebrosidase mRNA transcription and is diagnostic of Gaucher disease
Frameshift mutation that produces a nonfunctional glucocerebrosidase and is diagnostic of Gaucher disease

11. A 3-month-old Caucasian boy fails to grow and has very low muscle tone with strange “kinky” A diagnosis of Menkes kinky hair syndrome (MIM*309400) is made by showing deficient copper in fibroblasts. A likely candidate gene encoding an ATP-dependent copper transporter protein is known, and antibodies to this protein are available. Experimental analysis of the patient’s fibroblast DNA is performed to see if a pathologic mutation can be found, thus confirming the copper transport gene as the cause of Menkes disease. The table below shows a partial DNA sequence from the transporter coding strand as amplified by PCR; the amplified sequence is from a region where a TATA box has been located.

Which of the following experimental results would be the most compatible with a pathologic mutation in the transporter gene

Usual band of usual intensity on Northern and Western blots
Absent bands on Northern and Western blots
Usual band plus a second band of lower size on Northern blot, usual band on Western blot
Usual band on Northern blot, band of unusual size and reduced intensity on Western blot
Two unusual bands on Northern blot, usual bands on Western blot

12. Lipoproteins transport lipids from intestine and liver, where they are absorbed and metabolized to peripheral tissues via the They contain core proteins (apo proteins) that bind various types of lipids, thus producing lipoprotein complexes of varying densities—very low density lipoproteins (VLDL), low- density lipoproteins (LDL), and high-density lipoproteins (HDL) after plasma lipoprotein electrophoresis. A disorder called abetalipoproteinemia (MIM*200100) was distinguished by its deficiency of the bands for very low density lipoproteins (VLDL) and low-density lipoproteins (LDL) after plasma electrophoresis. Affected patients have ataxia (wide gait, incoordination), retinopathy (retinal degeneration with blindness), and myopathy (muscle weakness). VLDL and LDL contain the same apoB core protein, their different densities produced by association with different lipids, and apoB became good candidate for mutation in the disorder. Initial characterization of the apoB gene together with its mRNA and protein products demonstrated identical mRNA sizes in all tissues. Western blotting using antibody specific for the amino-terminus of apoB protein showed a 100-kDa species in liver and a 48-kDa species in intestine, as did antibody specific for the C-terminus of the 48-kDa intestinal protein. Molecular study of a patient with abetalipoproteinemia showed normal mRNA sizes, but 100-kDa peptide species were identified in both liver and intestine using the amino- or carboxy-terminus antibody probes. Which of the following processes is most likely deficient in this patient with apobetalipoproteinemia?

RNA splicing
DNA amplification
Transcription initiation
Transcription factor phosphorylation
RNA editing

13. During the analyses described in Question 226, another patient with abetalipoproteinemia was found to have negligible amounts of apoB protein in either liver or intestine. Analysis of the patient’s DNA and RNA would most likely yield which of the following results?

Substitution of the last base of a codon near the middle of the first apoB exon
Insertion of two bases near the apoB amino-terminus
Transversion producing a nonsense mutation within the first apoB intron
Transition affecting the 30th base of the first apoB intron
Base substitution changing glycine for alanine near the apoB carboxyl-terminus

14. Part of the triplet genetic code involving mRNA codon triplets that start with U is shown below. Using the portion of the genetic code shown, which of the following mutations in the 3′ to 5′ DNA template segments corresponds to a nonsense mutation?

ACGACGACG to ACAAACACG
AGGAATATG to AGGAATATT
AGAATAACA to AAAATAACA
AAAATGAGC to AAAATAAGC
AACAACAAC to AACAAGAAC

15. The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following?

Increased concentrations of glucose and cyclic AMP (cAMP)
Decreased concentrations of glucose and cAMP
Increased concentrations of glucose, decreased concentration of cAMP
Decreased concentrations of glucose, increased concentration of cAMP
Increased concentrations of glucose and adenosine triphosphate (ATP)

16. Which of the following regulators are said to act in “cis?”

The lac repressor and the lac operator
The lac operator and mammalian transcription factors
The lac operator and mammalian enhancers
Mammalian transcription factors and enhancers
The lac repressor and mammalian transcription factors

17. The proopiomelanocortin (POMC-MIM*176830) gene encodes several regulatory proteins that affect pituitary function. Children with severe brain defects such as holoprosencephaly (MIM*157170) often have abnormalities in the hypothalamic-pituitary In different brain regions, proteins encoded by this gene have different carboxy-terminal peptides. Which of the following best explains the regulatory mechanism?

POMC transcription is regulated by different factors in different brain regions.
POMC translation elongation is regulated by different factors in different brain regions.
POMC transcription has different enhancers in different brain regions.
POMC protein undergoes different protein processing in different brain regions.
POMC protein forms different allosteric complexes in different brain regions.

18. A 3-year-old Chinese boy has severe anemia with prominence of the forehead (frontal bossing) and The red cell hemoglobin concentration is dramatically decreased, and it contains only β-globin chains with virtual deficiency of α-globin chains. Which of the following mechanisms is the most likely explanation?

A transcription factor regulating the α-globin gene is mutated.
A regulatory sequence element has been mutated adjacent to an α-globin gene.
A transcription factor regulating the β-globin gene is mutated.
A transcription factor regulating the α- and β-globin genes is deficient.
A deletion has occurred surrounding an α-globin gene.

19. A mutation that results in a valine replacement for glutamic acid at position 6 of the β chain of hemoglobin S hinders normal hemoglobin function and results in sickle cell anemia (MIM*602903) when the patient is homozygous for this mutation. This is an example of which of the following types of mutation?

Deletion
Frameshift
Insertion
Missense
Nonsense

[MCQs] Gene Expression and Regulation Quiz – Part 2 (19 test)

I. Start the exam by click the “Start” button

Gene Expression and Regulation - Part 2 (19 test)

II. Preview all questions below

Share this:

Related

[MCQs] Gene Expression and Regulation Quiz – Part 1 (20 test)

[MCQs] Molecular Basis of Disease — Pathology, Laboratory Medicine Quiz – Part 1 (20 test)

You may also like

Leave a Comment Cancel Reply