Regulation of Gene Expression

Transcription factors can be activated in several ways as well as inhibited under certain conditions. Which one of the following describes a common theme in the structure of DNA-binding proteins? A. The presence of a specific helix that lies across the major or minor groove of DNA B. The ability to recognize RNA molecules with the same sequence C. The ability to form multiple hydrogen bonds between the protein peptide backbone and the DNA phosphodiester backbone D. The presence of zinc E. The ability to form dimers with disulfide linkages

A. The presence of a specific helix that lies across the major or minor groove of DNA All DNA-binding proteins contain an α-helix that binds to the major or minor groove in DNA. These proteins do not recognize RNA molecules (thus, B is incorrect), nor do they form bonds between the peptide backbone and the DNA backbone (thus, C is incorrect; if this were correct, how could there be any specificity in protein binding to DNA?). Only zinc fingers contain zinc, and dimers are formed by hydrogen bonding—not by disulfide linkages.

X chromosome inactivation is a process by which one of two X chromosomes in mammalian females is condensed and inactivated to prevent overexpression of X-linked genes. What would most likely be true about the degree of DNA methylation and histone acetylation on the inactivated X chromosome?

Cytosines in CpG islands would be hypermethylated, and histone proteins would be deacetylated. Both conditions are associated with decreased gene expression, and both are important in maintaining X inactivation.

Many transcription factors, which act as dimers, bind to palindromic sequences in their target DNA. Which one of the following double-stranded DNA sequences shows perfect dyad symmetry (the same sequence of bases on both strands)? A. GAACTGCTAGTCGC B. GGCATCGCGATGCC C. TAATCGGAACCAAT D. GCAGATTTTAGACG E. TGACCGGTGACCGG

B. GGCATCGCGATGCC The sequence, if read in the 5′-to-3′ direction, is identical to the complementary sequence read 5′-to-3′. None of the other sequences fits this pattern.

E. coli will only express genes for lactose metabolism when lactose is present in the growth medium. In E. coli, under high-lactose, high-glucose conditions, which one of the following could lead to maximal transcription activation of the lac operon? A. A mutation in the lac I gene (which encodes the repressor) B. A mutation in the CRP-binding site leading to enhanced binding C. A mutation in the operator sequence D. A mutation leading to enhanced cAMP levels E. A mutation leading to lower binding of repressor

D. A Mutation leading to lover binding of repressor In order to transcribe the lac operon, the repressor protein (lac I gene product) must bind allolactose and leave the operator region, and the cAMP-CRP complex must bind to the promoter in order for RNA polymerase to bind. Of the choices offered, only raising cAMP levels can allow transcription of the operon when both lactose and glucose are high. Raising cAMP, even though glucose is present, will allow the cAMP-CRP complex to bind and recruit RNA polymerase. Answers that call for mutations in the repressor (answers A and E) will not affect binding of cAMP-CRP. Mutations in the DNA (answers B and C) do not allow CRP binding in the absence of cAMP.

Expression of the lactose operon in E. coli can be quite complex. A mutation in the lac I (repressor) gene of a "noninducible" strain of E. coli resulted in an inability to synthesize any of the proteins of the lac operon. Which one of the following provides a rational explanation? A. The repressor has lost its affinity for the inducer. B. The repressor has lost its affinity for the operator. C. A trans-acting factor can no longer bind to the promoter. D. The CAP protein is no longer being made. E. Lactose feedback inhibition becomes constitutive.

A. The repressor has lost its affinity for the inducer The repressor will bind to the operator and block transcription of all genes in the operon unless prevented by the inducer allolactose. If the repressor has lost its affinity for the inducer, it cannot dissociate from the operator and the genes in the operon will not be expressed (thus, E is incorrect). If the repressor has lost its affinity for the operator (answer B), then the operon would be expressed constitutively. Because the question states that there is a mutation in the I (repressor) gene, answer D is incorrect, and mutations in the I gene do not affect trans-acting factors from binding to the promoter, although the only other one for the lac operon is the CRP.

Which of the following mutations is most likely to result in reduced expression of the lac operon? A. cya− (no adenylyl cyclase made) B. i− (no repressor protein made) C. Oc (operator cannot bind repressor protein) D. One resulting in impaired glucose uptake

A. cya- (no adenylyl cyclase made) In the absence of glucose, adenylyl cyclase makes cyclic adenosine monophosphate (cAMP), which forms a complex with the catabolite activator protein (CAP). The cAMP-CAP complex binds the CAP site on the DNA, causing RNA polymerase to bind more efficiently to the lac operon promoter, thereby increasing expression of the operon. With cya− mutations, adenylyl cyclase is not made, and so the operon is unable to be maximally expressed even when glucose is absent and lactose is present. The absence of a repressor protein or decreased ability of the repressor to bind the operator results in constitutive (essentially constant) expression of the lac operon.

A patient presents with a β-thalassemia. Such a disorder could result from a mutation located in which of the following? Choose the one best answer.

B. A β-thalassemia refers to a disorder in which the number of α-globin chains exceeds that of the β-globin genes. This can occur because of a stop codon being introduced into an exon of the β-globin gene, or loss of a splice site in the β-globin gene (which could occur in an intron or exon). An imbalance in chain synthesis could also occur because of a mutation in the β-globin gene promoter, or in the α-globin gene promoter that enhanced α-globin synthesis relative to β-globin gene synthesis. A mutation in an intron of the α-globin gene would not lead to more α-globin protein than β-globin protein

Which of the following is best described as cis-acting? A. Cyclic adenosine monophosphate response element-binding protein B. Operator C. Repressor protein D. Thyroid hormone nuclear receptor

B. Operator The operator is part of the DNA itself, and so is cis-acting. The cyclic adenosine monophosphate response element-binding protein, repressor protein, and thyroid hormone nuclear receptor protein are molecules that diffuse (transit) to the DNA, bind, and affect the expression of that DNA and so are trans-acting

Altered eukaryotic DNA can lead to mutations; however, the alteration in DNA does not necessarily have to be within an exon. Which one of the following best represents an epigenetic alteration in DNA which could lead to altered gene regulation? A. Deamination of C to U in DNA B. Deamination of A to I in DNA C. Methylation of C residues in DNA D. Substitution of an A for a G in DNA E. A simple base deletion in the DNA

C. Methylation of C residues in DNA Epigenetic events include histone acetylation and DNA methylation—alterations to the DNA which do not involve altering the base-pairing characteristics of the DNA (or causing insertions or deletions within the DNA). Deamination of C or A residues (to U or I, respectively) will lead to altered base pairing when the DNA is replicated (methylation of C does not alter the base-pairing properties of the C). Substitution of one base for another also leads to an alteration in base-pairing properties

A eukaryotic cell line grows normally at 30°C, but at 42°C, its growth rate is reduced owing to iron toxicity. At the elevated temperature, the cell displays elevated free intracellular iron levels, coupled with high levels of transferrin receptor. Ferritin levels within the cell are extremely low at the elevated temperature. These results can be explained by a single-nucleotide mutation in which one of the following proteins? A. Transferrin B. Ferritin C. Transferrin receptor D. IRE-BP E. RNA polymerase

D. IRE-BP Ferritin is the cellular storage protein for iron. Transferrin is the transport protein for iron in plasma. The transferrin receptor binds the iron-transferrin complex for transport of iron into the cell. The synthesis of both the transferrin receptor and ferritin are controlled by the IRE-BP. At the low temperature, the IRE-BP binds to the 3′-end of the transferrin receptor mRNA, stabilizing the mRNA so that it can be translated to produced transferrin receptor proteins. When intracellular iron levels increase, the iron binds to IRE-BP, displacing the protein from the mRNA, which leads to degradation of the mRNA and reduced synthesis of the transferrin receptor. In a similar fashion, the IRE-BP binds to the 5′-end of the ferritin mRNA, blocking ferritin synthesis. When intracellular iron levels increase, the IRE-BP falls off the mRNA, and ferritin is synthesized to bind the intracellular iron and prevent free iron toxicity in the cell. In this cell line, the IRE-BP is mutated such that at the elevated temperature it cannot bind iron, meaning that the IRE-BP remains bound to the transferrin receptor and ferritin mRNA molecules. This leads to a lack of ferritin and to overexpression of transferrin receptor in the membrane. The cell will accumulate iron but will not have adequate ferritin for the iron to bind to, leading to elevated free iron levels in the cell

Which of the following is the basis for the intestine-specific expression of apolipoprotein B-48? A. DNA rearrangement and loss B. DNA transposition C. RNA alternative splicing D. RNA editing E. RNA interference

D. RNA Editing The production of apolipoprotein (apo) B-48 in the intestine and apo B-100 in liver is the result of RNA editing in the intestine, where a sense codon is changed to a nonsense codon by posttranscriptional deamination of cytosine to uracil. DNA rearrangement and transposition, as well as RNA interference and alternative splicing, do alter gene expression but are not the basis of apo B-48 tissue-specific production.

The ZYA region of the lac operon will be maximally expressed if: A. cyclic adenosine monophosphate levels are low. B. glucose and lactose are both available. C. the attenuation stem-loop is able to form. D. the CAP site is occupied.

D. The CAP site is occupied It is only when glucose is gone, cyclic adenosine monophosphate (cAMP) levels are increased, the cAMP-catabolite activator protein (CAP) complex is bound to the CAP site, and lactose is available that the operon is maximally expressed (induced). If glucose is present, the operon is off as a result of catabolite repression. The lac operon is not regulated by attenuation, a mechanism for stopping transcription in some operons such as the trp operon

In response to foreign organisms, humans produce a variety of antibodies that can bind to the organism. In the production of human antibodies, which one of the following can cause the production of different proteins from a single gene? A. Pretranscription processing of hnRNA B. Removal of introns from hnRNA C. Addition of a cap to the 5′-end of hnRNA D. Addition of a cap to the 3′-end of hnRNA E. Alternative sites for synthesizing the poly(A) tail

E. Alternative sites for synthesizing the poly(A) tail After the gene is transcribed (posttranscription), the use of alternative splice sites or sites for addition of the poly(A) tail can result in different mRNAs (and therefore different proteins) from a single hnRNA. Introns are inert (noncoding) and would have no effect. Alternative splicing would remove exons, but all introns are removed from the hnRNA. The cap (on the 5′-end) is required for translation but would not alter the reading frame of the protein

Which of the following is most likely to be true in hemochromatosis, a disease of iron accumulation? A. The messenger RNA for the transferrin receptor is stabilized by the binding of iron regulatory proteins to its 3′-iron-responsive elements. B. The messenger RNA for the transferrin receptor is not bound by iron regulatory proteins and is degraded. C. The messenger RNA for ferritin is not bound by iron regulatory proteins at its 5′-iron-responsive element and is translated. D. The messenger RNA for ferritin is bound by iron regulatory proteins and is not translated. E. Both B and C are correct.

E. Both B and C When iron levels in the body are high, as is seen with hemochromatosis, there is increased synthesis of the iron-storage molecule, ferritin, and decreased synthesis of the transferrin receptor (TfR) that mediates iron uptake by cells. These effects are the result of cis-acting iron-responsive elements not being bound by trans-acting iron regulatory proteins, resulting in degradation of the messenger RNA (mRNA) for TfR and increased translation of the mRNA for ferritin.

Patients with estrogen receptor-positive (hormone responsive) breast cancer may be treated with the drug tamoxifen, which binds the estrogen nuclear receptor without activating it. Which of the following is the most logical outcome of tamoxifen use? A. Increased acetylation of estrogen-responsive genes B. Increased growth of estrogen receptor-positive breast cancer cells C. Increased production of cyclic adenosine monophosphate D. Inhibition of the estrogen operon E. Inhibition of transcription of estrogen-responsive genes

E. Inhibition of transcription of estrogen-responsive genes. Tamoxifen competes with estrogen for binding to the estrogen nuclear receptor. Tamoxifen fails to activate the receptor, preventing its binding to DNA sequences that upregulate expression of estrogen-responsive genes. Tamoxifen, then, blocks the growth-promoting effects of these genes and results in growth inhibition of estrogen-dependent breast cancer cells. Acetylation increases transcription by relaxing the nucleosome. Cyclic adenosine monophosphate is a regulatory signal mediated by cell-surface rather than nuclear receptors. Mammalian cells do not have operons

Bacteria can coordinately express several genes simultaneously. Which one of the following explains why several different proteins can be synthesized from a typical prokaryotic mRNA? A. Any of the three reading frames can be used. B. There is redundancy in the choice of codon/tRNA interactions. C. The gene contains several operator sequences from which to initiate translation. D. Alternative splicing events are commonly found. E. Many RNAs are organized in a series of consecutive translational cistrons.

E. Many RNAs are organized in a series of consecutive translational cistrons. Many prokaryotic genes are organized into operons, in which one polycistronic mRNA contains the translational start and stop sites for several related genes. Although each gene within the mRNA can be read from a different reading frame, the reading frame is always consistent within each gene (thus, A is incorrect). Redundancy in codon/tRNA interactions has nothing to do with multiple cistrons within an mRNA (thus, B is incorrect). Operator sequences are in DNA and initiate transcription, not translation (thus, C is incorrect). Alternative splicing occurs only in eukaryotes (which have introns), not in prokaryotes (thus, D is incorrect).

An altered response to hormones can occur if the receptor contains a mutation. A nuclear receptor has a mutation in its transactivation domain, such that it can no longer bind to other transcription factors. Which one of the following is most likely to occur when this receptor binds its cognate ligand? A. Inability to bind to DNA B. Enhanced ability to bind to DNA C. Enhanced transcription of hormone-responsive genes D. Enhanced dimerization of hormone receptors E. Reduced transcription of hormone-responsive genes

E. Reduced transcription of hormone-responsive genes The transactivation domain of the receptor is required to recruit other positive acting factors to the promoter region of the gene in order to enhance transcription. Lack of this domain, and reduced recruitment of coactivators, would lead to reduced transcription. The receptor would still be able to bind to DNA (i.e., through a different site on the receptor, the DNA-binding domain), although its affinity for DNA is not enhanced by lack of the transactivation domain. The transactivation domain is not related to the dimerization domain of hormone receptors.