Book Qs

Question 1

A nucleotide sequence 5′-TCCGAT-3′ within human genomic DNA underwent spontaneous mutation resulting in a nucleotide substitution, producing the sequence 5′-TTCGAT-3′. Which one of the following chemical reactions could account for the nucleotide substitution above?
A. Conversion of a deoxyribosyl group to a ribosyl group
B. Deamination of a pyrimidine
C. Hydrolysis of the N-glycosidic bond in a purine
D. Hydrolysis of a phosphodiester bond
E. Methylation of a cytosine

Answer 1

B Deamination of cytosine converts the base to a uracil. Uracil will be replaced by thymidine either by repair mechanisms or during the next round of DNA replication.
A Spontaneous conversion of deoxribose to ribose is not a common reaction and could not, in any case, account for the mutation listed in the question.
C Hydrolysis of the N-glycosidic bond of a purine does occur in cells, resulting in abasic sites in the DNA that must be repaired. However, the mutation was a pyrimidine-to-pyrimidine transition and did not involve purine bases.
D Hydrolysis of a phosphodiester bond would result in a single-stranded break in the DNA backbone. It is highly unlikely that this would produce a nucleotide base change.
E Methylation of cytosine does not alter its base pairing and is not mutagenic.

Question 2

Which of the following characteristics of nucleosomes is important for their ability to package genomic DNA into chromatin?
A. The histone constituents of nucleosomes contain a relatively high proportion of lysine and arginine amino acids.
B. Multiple posttranslational modifications can be present on multiple amino acids in the N-terminal regions of histones within nucleosomes.
C. Nucleosomes consist of two copies each of histone H2A, H2B, H3, and H4.
D. The wrapping of DNA around nucleosome particles induces negative supercoils in the DNA molecule.

Answer 2

. A The relatively high percentage of positively charged lysine and arginine residues in histones facilitates wrapping of the negatively charged DNA molecule around the surface of the nucleosome.
B Posttranslational modification of N-terminal tails in histones influences the structure of chromatin and the accessibility of DNA. However, these modifications are not necessary for chromatin to form.
C Although nucleosomes do consist of two copies each of histone H2A, H2B, H3, and H4. However, this fact contributes only to the content of DNA

Question 3

Which of the following observations is evidence for semiconservative DNA replication?
A. DNA synthesis proceeds in the 5′ → 3′ direction.
B. DNA synthesis occurs on both lagging and leading strands in the replication fork.
C. After DNA synthesis, each daughter molecule contains one parental strand and one new strand.
D. During DNA synthesis, replication proceeds in both directions from the origin of replication.
E. DNA synthesis in eukaryotes requires the activity of more than one DNA polymerase enzyme.

Answer 3

C During semiconservative replication, the parental DNA molecule is unwound (by the helicase enzyme), and the two single strands serve as templates for synthesis of new DNA. After synthesis is complete, the resulting two DNA molecules contain one parental strand and one newly synthesized strand; that is, the parental strand is “semiconserved” in the daughter strand.
A DNA synthesis does preoceed in a 5′ → 3′ direction. However, this does not account for the semiconservative nature of DNA synthesis.
B DNA synthesis on leading and lagging strands is evidence of the semidiscontinuous nature of DNA synthesis.
D DNA synthesis is bidirectonal, proceeding in both directions from the origin of replication. However, this does not account for the semiconservative nature of DNA synthesis.
E DNA synthesis in eukaryotes requires the activity of more than one DNA polymerase enzyme. However, this does not account for the semiconservative nature of DNA synthesis.

Question 4

The following DNA sequence is the template for synthesis of newly transcribed RNA: 5′-GGGAAAT-3′
Which of the following sequences represents the RNA synthesized from the template?
A. 5′-AUUUCCC-3′
B. 5′-GGGAAAU-3′
C. 5′-UAAAGGG-3′
D. 5′-CCCUUUA-3′

Answer 4

A RNA synthesis proceeds in the 5′ → 3′ direction. Therefore, the template strand is “read” by the RNA polymerase enzyme in the 3′ → 5′ direction, with the resulting RNA complementary to the DNA template.
B This sequence is identical (except for the substitution of uracil for thymidine) to the template strand. RNA synthesized from the DNA template strand will be complementary; that is, the nucleotides in the RNA will base pair with the DNA template and have the opposite 5′ → 3′ orientation.
C This sequence is reversed but not complementary to the DNA (and has a substitution of uracil for thymidine). RNA synthesized from the DNA template strand will be complementary; that is, the nucleotides in the RNA will base pair with the DNA template and have the opposite 5′ → 3′ orientation.
D This sequence is not complementary to the template strand; its 5′ → 3′ orientation will not allow it to base pair with the template strand.

Question 5

The anticancer drug cytarabine can be metabolically activated into the compound shown below. Which of the following enzymes is inhibited by this compound? (Picture of Pyrimidine-TP)

A. DNA polymerase δ
B. DNA primase
C. Ribosome
D. RNA polymerase II
E. Helicase

Answer 5

A The activated form of cytarabine competes with the normal, cellular substrate deoxycytidine 5′-triphosphate (dCTP), inhibiting the activity of DNA polymerases, including DNA polymerase δ. Inhibition of DNA synthesis is lethal to actively dividing cells.
B DNA primase, part of the DNA polymerase α-DNA primase complex, synthesizes short RNA primers to initiate DNA replication. Primase uses ribonucleotide triphosphates as substrate. The metabolite of cytarabine shown in the question has a hydrogen at the 2′ position, making this a deoxynucleoside triphosphate, a substrate for DNA polymerases.
C Ribosomes polymerize amino acids, synthesizing proteins from an mRNA template. Aminoacyl-tRNAs are the substrate for protein synthesis by ribosomes. Aminoacyl-tRNAs do not resemble the metabolite of cytarabine shown in the question.
D RNA polymerase II polymerizes ribonucleotide triphosphates to make mRNAs. The metabolite of cytarabine shown in the question has a hydrogen at the 2′ position, making this a deoxynucleoside triphosphate, a substrate for DNA polymerases but not RNA polymerases.
E Helicase unwinds the DNA double helix to provide access to single-stranded DNA template by the DNA polymerases. Helicase consumes adenosine triphosphate (ATP) as a source of energy during the DNA unwinding. The metabolite of cytarabine shown in the question contains arabinose rather than ribose and has a hydrogen at the 2′ position, and the base is cytosine, which are major structural differences with ATP.

Question 6

One mechanism of potential mutation is the insertion of an incorrect nucleotide by DNA polymerases δ and ɛ into the newly synthesized DNA molecule. Which of the following enzymes, present at the replication fork, repairs nucleotide misincorporation by DNA polymerase δ during DNA synthesis?
A. DNA ligase
B. DNA helicase
C. DNA topoisomerase
D. DNA polymerase δ
E. DNA polymerase α

Answer 6

D DNA polymerase δ has 3′ → 5′ exonuclease (proofreading) activity that allows the enzyme to backtrack and remove a misincorporated nucleotide and insert the correct nucleotide.
A DNA ligase is important for several DNA repair pathways that operate to remove mutations that arise via multiple mechanisms. However, the vast majority of nucleotide misincorporations are repaired during the DNA synthesis process by the proofreading activity of DNA polymerase δ.
B DNA helicase unwinds the DNA double helix during DNA replication. It is also involved in several DNA repair pathways that operate to remove mutations that arise via multiple mechanisms. However, the vast majority of nucleotide misincorporations are repaired during the DNA synthesis process by the proofreading activity of DNA polymerase δ.
C DNA topoisomerase removes supercoils generated ahead of the replication fork, the result of the unwinding of the DNA double helix during DNA replication. It is not specifically involved in the repair of nucleotide misincorporation during DNA synthesis.
E DNA polymerase α, along with the primase enzymes, synthesizes short primers that are elongated by DNA polymerase δ during DNA synthesis. It is not specifically involved in the repair of nucleotides misincorporated by DNA polymerase δ during DNA synthesis.

Question 7

The PEPCK-C gene is transcribed in the liver under normal conditions, and its transcription is induced significantly by glucocorticoids. It was reported in the literature that a nucleotide substitution, at a position outside the protein coding region of the PEPCK-C gene, resulted in consistently elevated levels of PEPCK-C transcription that were only increased slightly by glucocorticoid treatment. Which of the following regulatory elements is most likely the site of this mutation?
A. Core promoter
B. Enhancer element
C. Silencer element
D. Transcriptional start site
E. Polyadenylation site

Answer 7

C A silencer element reduces or completely suppresses transcription of associated genes when it is bound by specific, repressive transcription factors. A debilitating mutation in a silencer element would result in higher than normal transcription of the associated gene. Because many enhancers and transcriptional activator proteins oppose the actions of silencers, the observation that PEPCKC transcription was only increased slightly by glucocorticoids is consistent with a mutation in a silencer element.
A Because the mutation did not reduce transcription of the PEPCK-C gene, the core promoter elements must be functioning normally. As such, the mutation cannot be in the core promoter sequences.
B Activation of transcription by glucocorticoids is mediated by binding of the glucocorticoid receptor (GR) to a specific nucleotide sequence (the glucocorticoid response element; GRE) that is present in enhancers associated with genes regulated by these hormones. A mutation in the GRE sequence would result in loss of inducibility by glucocorticoids but would not result in consistently elevated levels of PEPCK-C transcription.
D The transcriptional start site is the nucleotide position of a gene where the RNA polymerase initiates polymerization of mRNA. The transcriptional start site is not a specific, conserved nucleotide sequence. Rather, it is determined primarily by the location of core promoter elements (e.g., the TATA box). A nucleotide substitution at the transcriptional start site would likely have no effect on the efficiency of transcription.
E Transcription is terminated when RNA polymerase II cleaves the mRNA transcript at sites located downstream of the polyadenylation signal sequence (AATAAA). Following cleavage, 100 to 200 nontemplate adenosine residues are added to the end of the mRNA by enzymes associated with the RNA polymerase II complex. A mutation in this sequence would not alter transcript levels in the manner outlined in the question.

Question 8

p16 is a member of the INK family of CDK inhibitor proteins (CIP). It functions by binding to CDK4 and CDK6. What effect does the action of p16 have on the cell cycle?
A. p16 promotes the associations between CDK4/6 and cyclin D.
B. p16 triggers the release of E2F from RB.
C. p16 blocks passage of cell cycle through restriction point in late G1.
D. p16 promotes the hyperphosphorylation of RB.

Answer 8

C Binding of p16 to CDK4 and CDK6 prevents them from associating with cyclin D. Without active cyclin D-CDK4/6 complexes, cells cannot pass the restriction point in late G1.
A Binding of p16 to CDK4 and CDK6 prevents (not promotes) the associations between CDK4/6 and cyclin D.
B Binding of p16 to CDK4 and CDK6 (which prevents associations between CDK4/6 and cyclin D) blocks the hyperphosphorylation of RB (by cyclin D-CDK4/6) and thus prevents the release of E2F from RB.
D Binding of p16 to CDK4 and CDK6 (which prevents associations between CDK4/6 and cyclin D) blocks the hyperphosphorylation of RB (by cyclin D-CDK4/6).

Question 9

A strain of bacteria resistant to a specific antibiotic was isolated in a hospital laboratory. Biochemical analysis demonstrated that treatment of the sensitive strain with the antibiotic resulted in a block in bacterial RNA synthesis, whereas in the resistant strain, RNA synthesis was unaffected. The resistant strain is unaffected by which of the following drugs?
A. Chloramphenicol
B. Tetracycline
C. Rifampicin
D. Acyclovir
E. Amoxicillin

Answer 9

C Rifampicin inhibits initiation of RNA synthesis by RNA polymerase holoenzyme in bacteria.
A Chloramphenicol is a broad-spectrum antibiotic that inhibits protein synthesis at the elongation phase by inhibiting peptidyl transferase activity of the bacterial ribosome.
B Tetracyclines bind to the small 30S subunit of bacterial ribosomes and inhibit translational elongation.
D Acyclovir is an inhibitor of viral DNA replication; it is not active against bacteria.
E Amoxicillin is a β-lactam antibiotic that inhibits bacterial cell wall synthesis by blocking the cross-linking of the peptidoglycan polymers.

Question 10

What effect does damage to DNA have on p53 protein?
A. Inactivates p53
B. Stabilizes p53
C. Degrades p53
D. Dephosphorylates p53

Answer 10

B DNA damage triggers the phosphorylation and thus stabilization of p53.
A DNA damage promotes the phosphorylation and release of p53 from MDM2 (a ubiquitin ligase that facilitates the degradation or inactivation of p53).
C DNA damage promotes the phosphorylation and release of p53 from MDM2. Phosphorylated p53 is not degraded.
D DNA damage promotes the phosphorylation of p53.

Question 11

. Which of the following events is characteristic of both the extrinsic and intrinsic pathways of apoptosis?
A. Activation of caspase 8
B. Cleavage of the proapoptotic factor, BID
C. Involvement of Fas-associated death domain protein (FADD)
D. Activation of caspases 3, 6, and 7

Answer 11

D Caspases 3, 6, and 7 are activated by both the extrinsic and intrinsic pathways of apoptosis.
A Activation of caspase 8 is characteristic of the extrinsic pathway of apoptosis.
B Release of cytochrome-c from mitochondria is characteristic of the intrinsic pathway of apoptosis. Cleavage of BID to tBID is catalyzed by caspase 8 (extrinsic pathway) but not by caspase 9 (intrinsic pathway).
C Recruitment of caspase 8 by FADD is characteristic of the extrinsic pathway of apoptosis.

Question 12

. How do genetic mutations in protooncogenes contribute to cancer?
A. They alter the structure or amount of the gene’s protein product.
B. They disrupt the protein product’s ability to repair damaged DNA.
C. They disrupt the protein product’s ability to arrest the cell cycle.
D. They cause a reduction in the amount of the gene’s protein product.

Answer 12

A Mutations in protooncogenes that change their structure or increase their level of expression contribute to the development of cancer.
B Mutations in the genes for tumor suppressors (not protooncogenes) disrupt the encoded protein’s ability to repair damaged DNA.
C Mutations in genes for tumor suppressors (not oncogenes) disrupt the encoded protein’s ability to regulate the cell cycle.
D Mutations in protooncogenes are associated with an increase (not a reduction) in the amount and/or activity of the gene’s protein product.

Question 13

Which of the following DNA sequences (only one strand of the double-stranded sequence is shown) would have the greatest thermal stability?
A. GGAATCCG
B. TTATTCCG
C. AGGATTTC
D. GGCTTTTT
E. CCCCGGGA

Answer 13

E This sequence has the greatest number of G:C base pairs (seven total). G:C base pairs are held together by three hydrogen bonds, compared with two hydrogen bonds for A:T base pairs, and therefore require more energy (heat) to denature.
A Of the sequences listed, this sequence does not have the greatest of G:C base pairs and therefore does not have the greatest thermal stability. G:C base pairs are held together by three hydrogen bonds, compared with two hydrogen bonds for A:T base pairs, and therefore require more energy (heat) to denature.
B Of the sequences listed, this sequence does not have the greatest of G:C base pairs and therefore does not have the greatest thermal stability. G:C base pairs are held together by three hydrogen bonds, compared with two hydrogen bonds for A:T base pairs, and therefore require more energy (heat) to denature.
C Of the sequences listed, this sequence does not have the greatest of G:C base pairs and therefore does not have the greatest thermal stability. G:C base pairs are held together by three hydrogen bonds, compared with two hydrogen bonds for A:T base pairs, and therefore require more energy (heat) to denature.
D Of the sequences listed, this sequence does not have the greatest of G:C base pairs, and therefore does not have the greatest thermal stability. G:C base pairs are held together by three hydrogen bonds, compared with two hydrogen bonds for A:T base pairs, and therefore require more energy (heat) to denature.

Question 14

Valproic acid is a drug used clinically to treat seizures. Biochemically, it is an inhibitor of histone deacetylases. Treating cells with valproic acid will likely alter the charge of which of the following amino acids in histones?
A. Aspartate
B. Glutamate
C. Histidine
D. Arginine
E. Lysine

Answer 14

E Histones are rich in the basic amino acids arginine and lysine. Acetylation occurs only on the positively charged ɛ-amino groups of lysine residues. This neutralizes the basic charge and reduces the affinity of DNA for nucleosomes, making it more accessible to the transcriptional machinery.
A Aspartate (aspartic acid) has a net negative charge. It is not subject to acetylation.
B Glutamate (glutamic acid) has a net negative charge. It is not subject to acetylation.
C At physiological pH, the imidazole ring of histidine has a net positive charge. However, it is not subject to acetylation.
D Arginine has a positive charge and is an important constituent of histones. However, it is not subject to acetylation.

Question 15

Cells from a patient that cannot produce the enzyme GlcNAc phosphotransferase (which phosphorylates mannose on asparagine-linked oligosaccharide chains) would be unable to transport newly synthesized glycoproteins to which of the following locations?
A. Extracellular space
B. Lysosomes
C. Plasma membrane
D. Mitochondria
E. Nucleus

Answer 15

B Proteins destined for lysosomes are processed by the endoplasmic reticulum (ER) and Golgi such that they contain one or more phosphorylated mannose residues.
A The presence of phosphorylated mannose residues on proteins does not play a role in directing them for secretion. Instead, proteins destined for secretion may be processed via standard or signal-regulated pathways and secreted constitutively or in a regulated manner with the aid of a tryptophan-rich domain and the absence of retention motifs.
C The presence of phosphorylated mannose residues on proteins does not play a role in directing them to plasma membranes. Instead, proteins destined for plasma membranes remain embedded in the outer layer of the ER. This is aided by the presence of an apolar region in the N-terminus of the polypeptide that serves as a stop translation sequence.
D The presence of phosphorylated mannose residues on proteins does not play a role in directing them to mitochondria. Instead, proteins destined for mitochondria contain an N-terminal hydrophobic α-helix that interacts with specific auxiliary proteins (chaperones) from the heat shock protein family (e.g., HSP70).

Question 16

The DNA translocation generating the Philadelphia chromosome involves which protooncogene and is diagnostic for which disease?
A. RAS G protein, colon carcinoma
B. RAS G protein, chronic myelogenous leukemia
C. c-MYC transcription factor, Burkitt’s lymphoma
D. ABL tyrosine kinase, chronic myelogenous leukemia
E. p53 transcription factor, chronic myelogenous leukemia

Answer 16

D The Philadelphia chromosome (der (22)) results from the reciprocal translocation of the abl protooncogene from chromosome 9 to chromosome 22 and the BCR region of chromosome 22 to
9. This translocation creates a BCR-ABL fusion gene (and an unregulated tyrosine kinase protein product) and is associated with chronic myelogenous leukemia (CML).
A The Philadelphia chromosome does not involve the RAS G protein, nor is it associated with colon carcinoma. Mutations in the RAS G protein involve point mutations (not translocations). Colon carcinoma is linked to loss of function mutations in the colorectal carcinoma (DCC) gene.
B The Philadelphia chromosome is not linked to mutations involving the RAS G protein but is associated with CML.
C The Philadelphia chromosome does not involve c-MYC. The latter is overexpressed in Burkitt’s lymphoma due to a translocation between chromosomes 8 and 14.
E The Philadelphia chromosome does not involve p53, but it is associated with CML.

Question 17

During the elongation phase of prokaryotic translation, for each amino acid added to the growing peptide chain, two GTP molecules are hydrolyzed to GDP. Which of the following aspects of ribosome function require the hydrolysis of one of these GTP molecules?
A. Peptide bond formation
B. Aminoacyl-tRNA delivery to A site
C. Binding of CAP-binding complex to the 7-methylguanosine cap of mRNA
D. Assembly of the small and large ribosomal subunits

Answer 17

B One GTP molecule is hydrolyzed by the G protein elongation factor-1 (EF-1 in eukaryotes) when an amino acyl tRNA is inserted into the A site. A second is expended during the translocation process. A The energy for peptide bond formation comes from an ATP consumed during charging of the tRNA by aminoacyl tRNA synthetases. The energy from that ATP is stored in the bond between the amino acid and the tRNA.
C The binding of the eIF4F (CAP binding complex) to the 7-methylguanosine cap of mRNA occurs at the initiation of mRNA translation and does not consume GTP.
D The assembly of the small and large ribosomal subunits on the mRNA occurs during the initiation of translation.

Question 18

The mRNA sequence 5′-GCG ACG UCC-3′, is being translated by a ribosome. Which of the following sequences represent the anticodons of the aminoacyl-tRNAs that were used (in order) during translation?
A. 5′-UCC-3′ 5′-ACG-3′ 5′-GCG-3′
B. 5′-GGA-3′ 5′-CGU-3′ 5′-GCG-3′
C. 5′-GCG-3′ 5′-ACG-3′ 5′-UCC-3′
D. 5′-CGC-3′ 5′-CGU-3′ 5′-GGA-3′
E. 5′-CGC-3′ 5′-GCA-3′ 5′-UCC-3′

Answer 18

D Codons and anticodons pair in an antiparallel manner forming a small double helix. If the mRNA has the sequence 5′-GCG ACG UCC-3′, the first aminoacyl-tRNA anticodon would have the sequence 5′-CGC-3′ to base pair with the mRNA. The second aminoacyl-tRNA anti-codon would be 5′-CGU-3′; the third anticodon would be 5′-GGA-3′.
A Codons and anticodons pair in an antiparallel manner, forming a small double helix. This sequence is not complementary to the mRNA; that is, it cannot base pair in an antiparallel fashion with the mRNA.
B Codons and anticodons pair in an antiparallel manner, forming a small double helix. This sequence is not complementary to the mRNA; that is, it cannot base pair in an antiparallel fashion with the mRNA.
C Codons and anticodons pair in an antiparallel manner, forming a small double helix. This sequence is not complementary to the mRNA; that is, it cannot base pair in an antiparallel fashion with the mRNA.
E Codons and anticodons pair in an antiparallel manner, forming a small double helix. This sequence is not complementary to the mRNA; that is, it cannot base pair in an antiparallel fashion with the mRNA.

Question 19

Which one of the following antibiotics will inhibit cytoplasmic protein synthesis in both prokaryotic and eukaryotic organisms?
A. Erythromycin
B. Tetracycline
C. Streptomycin
D. Chloramphenicol
E. Puromycin

Answer 19

E Puromycin acts upon both eukaryotic and prokaryotic peptidyl transferases, releasing small peptides ending in puromycin at their C-terminal ends.
A Erythromycin inhibits protein synthesis at the elongation phase by binding to the prokaryotic 50S subunit and blocking translocation of the ribosome
B Tetracyline is a broad-spectrum antibiotic that binds the 30S ribosomal subunit in bacteria and impedes access to the acceptor (A) site by amino acyl-tRNAs. It does not affect eukaryotic ribosomes.
C Streptomycin inhibits bacterial protein synthesis by binding to the 30S subunit, preventing proper assembly of ribosomes. It does not interfere with eukaryotic ribosomes.
D Chloramphenicol binds the bacterial 50S ribosomal subunit and inhibits peptidyl transferase activity. It does not inhibit cellular eukaryotic ribosomes. However, at high (toxic) doses, it can inhibit mitochondrial protein synthesis. This can lead to severe toxicity in individuals who lack the ability to metabolize chloramphenicol (e.g., neonates).

Question 20

You are cloning BamHI-cut genomic DNA into the plasmid vector pBR322, which has both an ampicillin resistance gene and a tetracycline resistance gene. If the inserts are ligated into the BamHI site in the tetracycline resistance gene of the vector, the entire population of transformed bacteria should first be grown on
A. media containing no antibiotic
B. media containing ampicillin
C. media containing tetracycline
D. media containing both ampicillin and tetracycline

Answer 20

B Because a genomic DNA insert will inactivate the tetracycline resistance gene, the transformants must be grown on ampicillin. This will select against any bacterial cells that were not transformed by a plasmid.
A The absence of any antibiotic in the growth media would allow transformed (contains plasmid) and untransformed (lacks plasmid) to grow.
C Ligation of the inserts into the tetracycline resistance gene inactivates its ability to confer tetracycline resistance to transformed bacteria. Therefore, no bacterial colonies (transformed and un-transformed) will grow in media containing tetracycline.
D Ligation of inserts into the tetracycline resistance gene inactivates its ability to confer tetracycline resistance to transformed bacteria. Although the transformed bacteria are resistant to ampicillin, they (and the untransformed bacteria) will not grow on a plate containing tetracycline.

Question 21

When the ribosome reaches a stop codon on mRNA, release factors act to change the specificity of peptidyl transferase so that the growing peptide chain is transferred to which of the following moieties?
A. Water
B. eIF4F
C. Formyl-methionine
D. UDP-GlcNAc
E. The E site

Answer 21

A Release factors interact with peptidyl transferase, allowing it to transfer the nascent polypeptide chain to water instead of to the α amino group of an incoming aminoacyl tRNA.
B eIF4F is the complex that binds the 7-methyl guanosine cap of mRNA to seed the formation of the preinitiation complex in eukaryotes.
C Formyl-methionine is charged to the initiator tRNA in prokaryotes. It is the first amino acid of all proteins made in prokaryotes, but it does not participate in any way with termination.
D UDP-GlcNAc is the substrate for the initial step of N-linked glycosylation of proteins following their synthesis. It does not participate in translational termination.
E The E (or exit) site is the position on the ribosome where spent tRNA in the P site is transferred during ribosomal translocation.

Question 22

Recombinant human granulocyte-macrophage growth factor (GMCSF), a hematopoietic factor used to treat neutropenia, can be produced in a human cell line engineered to synthesize the protein. Analysis of the glycosylation of recombinant GMCSF (rGMCSF) was performed with PNGase F, an enzyme that catalyzes the complete removal of N-linked oligosaccharides from proteins at the site of attachment. Treatment of rGMCSF released an eight carbohydrate oligosaccharide. Which of the following is true regarding the glycosylated rGMCSF?
A. The oligosaccharide is attached to rGMSCF at an amino acid that can also serve as a site of phosphorylation.
B. The oligosaccharide is attached to rGMSCF via a mechanism that is insensitive to tunicamycin treatment.
C. The oligosaccharide is attached to rGMCSF via a hydroxyl functional group.
D. The oligosaccharide is attached to rGMCSF via an acid amide functional group.
E. The oligosaccharide is attached to rGMCSF via a mechanism that does not require dolichol phosphate.

Answer 22

D N-linked glycosylation of arginine residues is via an acid amide (CONH2) functional group.
A N-linked glycosylation occurs on arginine residues. Phosphorylation is limited to serine, threonine, and tyrosine amino acids.
B Tunicamycin inhibits N-linked glycosylation at the first step of core oligosaccharide synthesis (the transfer of phosphorylated N-acetyl-D-glucosamine [GlcNAc-P] from UDP-GlcNAc to dolichol phosphate).
C O-linked glycosylation is via a hydroxyl functional group, on either serine or threonine amino acids.
E Dolichol phosphate is required for the initial N-linked glycosylation of arginine on the target protein.

Question 23

. If you are screening a library by nucleic acid sequence homology with a radiolabeled DNA probe, how are the bacterial colonies of interest identified?
A. Microscope
B. X-ray film
C. Scintillation counter
D. Culture techniques

Answer 23

B The location of a radioactive label on a filter is detected with X-ray film (autoradiography).
A A microscope is not used to detect the radioactive signals from a radiolabeled DNA probe.
C A scintillation counter is designed to detect and quantify radioactive signals, but it cannot be used to determine which bacterial colonies contain nucleic acids that hybridize the radiolabeled DNA probe.
D Although culturing of transformed bacterial colonies on an agar plate is needed to prepare colonies for the probing experiment, culturing techniques do not allow one to identify the colonies whose nucleic acids hybridize to the radiolabeled DNA probe.

Question 24

Sometimes the gene responsible for a particular genetic disease has not yet been cloned, so researchers must use a probe that is genetically linked to the gene they wish to follow. If a new restriction fragment length polymorphism (RFLP) is detected using a probe linked to the gene, this result may be interpreted to mean
A. that a mutation has not occurred
B. that the activity of the gene’s protein product will not change
C. that a mutation has occurred in the gene or in the region flanking the gene
D. that the recognition sequences for restriction enzymes are unchanged

Answer 24

C Detection of a new RFLP means that the length of the DNA fragment that hybridizes to the probe is different; therefore, a mutation has occurred. However, without additional information, it is not possible to determine the location of the mutation itself.
A Deletion of a new RFLP means that the length of the DNA fragment that hybridizes to the probe is different; therefore, a mutation has occurred.
B Some genetic mutations detected by RFLP have been linked to defective protein products and disease (e.g., sickle cell disease).
D RFLP is detected because of mutations that alter the restriction sites for certain restriction enzymes.

Question 25

A 34-year-old woman whose mother died from breast cancer and whose sister was recently diagnosed with the disease comes to you for genetic testing. The results of the tests indicate she has a homozygous mutation in her BRCA1 genes. To which types of DNA damaging agents will this woman be particularly sensitive?
A. Alkylating agents
B. Intercalating agents
C. Medical X-rays
D. Sunlight
E. Adduct forming agents (e.g., benzo(a)pyrene)

Answer 25

C BRCA1 codes for a protein involved in the homologous recombination repair pathway. This pathway is particularly important in repairing double-stranded breaks in the DNA. Ionizing radiation, such as X-rays, are capable of producing double-stranded breaks (among other types of damage).
A BRCA1 codes for a protein involved in the homologous recombination repair pathway. This pathway is particularly important in repairing double-stranded breaks in the DNA. Alkylating agents induce a wide variety of methylated base changes and even alkylation of phosphodiesters into phosphotriesters. However, they do not particularly cause double-stranded breaks in the DNA.
B BRCA1 codes for a protein involved in the homologous recombination repair pathway. This pathway is particularly important in repairing double-stranded breaks in the DNA. Intercalating agents fit between the stacked bases of double-stranded DNA, distorting the structure of the double helix. They interfere with DNA synthesis and transcription and can be mutagenic, but they do not particularly cause double-stranded breaks.
D BRCA1 codes for a protein involved in the homologous recombination repair pathway. This pathway is particularly important in repairing double-stranded breaks in the DNA. The nonionizing radiation (ultraviolet light) in sunlight primarily results in the formation of pyrimidine dimers and does not particularly cause double-stranded breaks.
E BRCA1 codes for a protein involved in the homologous recombination repair pathway. This pathway is particularly important in repairing double-stranded breaks in the DNA. Adduct-forming agents such as BDPE (a chemical resulting from the metabolic transformation of benzo(a)pyrene) add bulky chemical groups to the DNA molecule but do not particularly cause double-stranded breaks.

Question 26

A 31-year-old man who emigrated from Vietnam and recently returned from a visit to his home country presents in your office with a chief complaint of sore throat, fever, and malaise. Upon physical examination, you observe a gray membrane that covers his tonsils and pharynx. Upon questioning, you find that he has never been vaccinated for diphtheria. In cells exposed to the toxin, which of the following processes related to gene expression are affected as a direct result of the action of the diphtheria toxin?
A. Transcriptional initiation
B. Transcriptional elongation
C. Transcriptional termination
D. Translational initiation
E. Translational elongation

Answer 26

E Diphtheria toxin catalyzes the covalent attachment of ADP to the catalytic unit of eEF-G, irreversibly inactivating it. Translocation, and therefore elongation, is blocked in eukaryotic cells.
A Diphtheria toxin does not affect any transcriptional process.
B Diphtheria toxin does not affect any transcriptional process.
C Diphtheria toxin does not affect any transcriptional process.
D Diphtheria toxin blocks translational elongation but does not have a direct effect on translational initiation.

Question 27

The public health department in your state recently reported the isolation of a tetracycline-resistant strain of Treponema pallidum in your area. Biochemical analysis has determined that the resistance is associated with a change in ribosome structure, which alters the ribosome’s function. Which of the following functions in protein synthesis is most likely to be altered in the tetracycline-resistant strain of T. pallidum?
A. Assembly of the 30S preinitiation complex
B. Entry of the 50S ribosomal subunit to create the initiation complex
C. Binding of aminoacyl-tRNAs to the A site
D. Peptidyl transferase activity of the 70S ribosome
E. Translocation of the 70S ribosome

Answer 27

C Tetracylclines bind to the small 30S subunit of bacteria, blocking access of aminoacyl-tRNAs to the A site. An alteration in ribosome structure, perhaps affecting the A site, could potentially disrupt the binding of tetracycline to the ribosome, generating a resistant phenotype.
A The 30S preinitiation complex consists of initiation factors, the 30S ribosomal subunit, and fMet-tRNAiMet and the mRNA. Tetracycline does not affect the formation of this complex and therefore is unlikely to be altered in a tetracycline-resistant strain of bacteria. Streptomycin, however, does inhibit this step.
B Once the preinitiation complex is formed, the 50S ribosomal subunit joins to create the initiation complex. Tetracycline does not affect the formation of this complex and therefore is unlikely to be altered in a tetracycline-resistant strain of bacteria.
D Formation of a peptide bond between the aminoacyl-tRNA in the A site and the peptidyl-tRNA in the P site is catalyzed by a ribozyme that has peptidyl transferase activity and is located in the large ribosomal subunit. Tetracycline does not affect this activity and therefore is unlikely to be altered in a tetracycline-resistant strain of bacteria. Chloramphenicol, however, is an antibiotic that inhibits peptidyl transferase.
E Once a peptide bond is formed, the hydrolysis of one GTP by an elongation factor is associated with translocation of the ribosome, one codon down the mRNA. Tetracycline does not affect this activity and therefore is unlikely to be altered in a tetracycline-resistant strain of bacteria. Erythromycin, however, is an antibiotic that binds the 50S subunit and inhibits translocation.

Question 28

You must evaluate the genetic status of a family with a history of Huntington disease and wish to determine the number of triplet repeats in each individual. What would be the best way to estimate the number of triplet repeats?
A. Real-time polymerase chain reaction (RT-PCR)
B. cDNA library construction
C. Variable number of tandem repeat (VNTR) analysis
D. Restriction fragment length polymorphism (RFLP) analysis

Answer 28

C Huntington disease is associated with the presence of a variable number of tandem repeats (VNTR) regions in which the sequence CAG is found repeated between 37 and 100 times.
A RT-PCR is used to quantify the amount of a target sequence in a sample (relative to a different sample). RT-PCR cannot specifical be used to estimate the number of CAG repeats.
B Construction of a complementary DNA (cDNA) library will not help estimate the number of CAG repeats in regions of a person’s genome.
D RFLP analysis is unlikely to help estimate the number of CAG repeats in regions of a person’s genome.

Question 29

A couple who had a child affected with cystic fibrosis (CF) requests genetic testing because the mother is now pregnant. They wish to know the genetic status of the fetus. You administer a PCR amplification test for the DF508 mutation responsible for 70% of all cases of CF and obtain the data diagrammed below. What do you tell the family? (no picture, might be a dumb question to put in here…)

A. The fetus does not carry the DF508 mutant allele for the CF gene.
B. The fetus does carry the DF508 mutation.
C. The fetus is heterozygous for the DF508 CF mutation.
D. The fetus is homozygous for the DF508 CF mutation.

Answer 29

A The fetus does not carry the DF508 mutant allele for CF. Its parents are heterozygous, and its brother is homozygous for the DF508 mutant allele.
B The fetus does not carry the DF508 mutant allele for CF.
C The fetus is not heterozygous for the DF508 mutant allele.
D The fetus is not homozygous for the DF508 mutant allele.

Question 30

The following table lists the GC content of the genomes of several pathogenic bacteria: (not sure how the table will look… still no picure/table capabilities)

Organism	% GC content of genome
Mycobacterium tuberculosis	65.5
Clostridium botulinum	28.2
Francisella philomiragia	32.6
Burkholderia mallei	68.5
Escherichia coli O157H7	50.3
Based on the difference in thermal stability of an organism's genomes, which of the following pairs of organisms would be suitable candidates for developing a PCR-based clinical laboratory assay that would allow the amplification of DNA from one pathogen but not the other?
A. M. tuberculosis and B. mallei
B. E. coli and M. tuberculosis
C. B. mallei and E. coli
D. F. philomiragia and B. mallei
E. C. botulinum and F. philomiragia

Answer 30

D The difference in genomic GC content between F. philomiragia and B. mallei is the largest of the pairings provided. The temperature at which the strands of the DNA double helix of the F. philomiragia genome separates would be much lower than the temperature at which the genomic DNA of B. mallei denatures.
A The GC content and hence the thermal stability of these two organisms are very close, which would make exploiting the small difference difficult if not impossible.
B Although there is a modest difference in GC content in the genomes of these two organisms, it is not the most significant of the pairings provided. The relative similarity in GC content and hence thermal stability of these two organisms would make exploiting differences in thermal stability difficult.
C Although there is a modest difference in GC content in the genomes of these two organisms, it is not the most significant of the pairings provided. The relative similarity in GC content and hence thermal stability of these two organisms would make exploiting differences in thermal stability difficult.
E The GC content and hence the thermal stability of these two organisms are very close, which would make exploiting the small difference difficult if not impossible.

Question 31

A 44-year-old woman who emigrated from Poland three years earlier comes to your office with a chief complaint of fatigue of six months’ duration. A previous doctor told her she was anemic and had prescribed oral iron supplements, which she stopped taking because of the taste. Physical examination was unremarkable. During your conversation, she remarks that her mother had died from colon cancer at age 51, as did her maternal grandfather at age 55. Laboratory tests indicate she was indeed anemic, and a fecal occult blood test was positive. On a follow-up visit, colonos-copy and following pathologic examination of biopsied lesions resulted in the diagnosis of colon adenoma with villous features. Genetic testing indicated that the woman has a mutation resulting in a defect of a DNA repair pathway. Which of the following pathways is most likely dysfunctional in this patient?
A. Homologous recombination repair
B. Mismatch excision repair
C. Nonhomologous end-joining repair
D. Nucleotide excision repair
E. Transcription coupled repair

Answer 31

B The woman has a mutation in one of the genes necessary for mismatch excision repair (MER), resulting in hereditary nonpolyposis colorectal cancer. The familial history of colon cancer and the early onset of her disease are consistent with her diagnosis. The absence of MER results in the inability to repair nucleotide misincorporations that are not fixed during replication by the proofreading activity of DNA polymerase δ.
A Genetic defects in homologous recombination repair (BRCA1/2 mutations, Fanconi’s anemia) are typically not associated with hereditary colon cancer. Given the patient’s family history, it is unlikely that this pathway would be affected.
C Genetic defects in nonhomologous end-joining repair are associated with severe, early-onset phenotypes (e.g., severe combined immunodeficiencies) and are typically not associated with hereditary colon cancer. Given the patient’s medical history and her family history, it is unlikely that this pathway would be affected.
D Genetic defects in NER pathways are associated with severe, early-onset phenotypes (e.g., xeroderma pigmentosum) and are typically not associated with hereditary colon cancer. Given the patient’s medical history and her family history, it is unlikely that this pathway would be affected.
E Genetic defects in transcription coupled repair pathways are associated with severe, early-onset phenotypes (e.g., tricothio-dystrophy) and are typically not associated with hereditary colon cancer. Given the patient’s medical history and her family history, it is unlikely that this pathway would be affected.

Question 32

A physician is exploring the best treatment options for his 56-year-old female patient who has been diagnosed with breast cancer. To show that his patient may be a good candidate for Herceptin, he plans to examine a biopsy of her cancerous tissue for expression levels of which of the following proteins?
A. BCR-ABL tyrosine kinase
B. HER2/NEU family of receptors
C. Insulin receptors
D. FAS receptors

Answer 32

B Herceptin (trastuzumab) is a monoclonal antibody directed against HER2/NEU receptors (which are members of the EGF family of receptors).
A The BCR-ABL tyrosine kinase is associated with chronic myelogenous leukemia (CML) and treated with Gleevec (imatinib).
C The insulin receptor is not a target of Herceptin.
D The FAS receptor is not a target of Herceptin. The Fas receptor is linked to the extrinsic pathway of apoptosis.

Question 33

A 2-year-old male patient was brought to the physician by his mother, who stated that her son’s left eye had a glow similar to that of a cat’s eye caught in headlights. She also noticed that his eyes appeared crossed, and he appears to have diminished vision in his left eye. The physician sends the child to an ophthalmologist, who confirms that the child has tumors in his retina. What is the most likely cause of this patient’s condition?
A. A “loss of function” mutation in the p53 gene
B. A chromosomal translocation involving the BCR and ABL genes
C. A “gain of function” mutation in the RAS gene
D. A “loss of function” mutation in the RB gene
E. A point mutation in the N-MYC gene

Answer 33

D This patient has retinoblastoma marked by tumors in the retina. Retinoblastoma is linked to “loss of function” mutations and deletions in the RB gene.
A A “loss of function” mutation in the p53 gene is not linked to retinoblastoma (marked by tumors in the retina).
B A chromosomal translocation involving the BCR and ABL genes is linked to chronic myelogenous leukemia (CML), not retinoblastoma.
C A “gain of function” mutation in the RAS gene is not linked to retinoblastoma.
E A point mutation in the N-MYC gene is linked to neuroblastoma (not retinoblastoma).

Question 34

A novel pharmaceutical agent in development claims to inhibit the activity of CDK2. The effect of this agent on the cell cycle should include
A. inhibition of the G1/S transition
B. maintenance of RB in a hyperphosphorylated state
C. inhibition of the CDK inhibitor proteins p21 and p27
D. promotion of the transcription of genes for cyclin E and cyclin A

Answer 34

A Inhibition of the activity of CDK2 would prevent cells from exiting the G1 phase and thus entering the S phase.
B Because cyclin E/A-CDK2 complexes keep RB in the hyperphosphorylated state, this CDK2 inhibitory agent would do the opposite.
C p21 and p27 are natural inhibitors of CDK2 and are not noted as targets of the novel CDK2 inhibitor.
D This CDK2 inhibitor would promote the hypophosphorylated state of RB and thus decrease the transcription of cyclin E and cyclin A.

Question 35

You carry out an experiment in which human brain RNA is analyzed via hybridization to a DNA probe. Which of the following best describes the hybridization technique used in this experiment?
A. Western blotting
B. Southern blotting
C. Northern blotting
D. DNA microarray

Answer 35

C Northern blotting refers to a process by which RNA (immobilized on a membrane) is probed with a DNA molecule.
A Western blotting refers to a process by which proteins (immobilized on a membrane) are probed with antibodies.
B Southern blotting refers to a process by which DNA (immobilized on a membrane) is probed with a DNA molecule.
D DNA microarray refers to a process by which DNA (immobilized on a solid support) is probed with complementary DNA (cDNA) molecules.

Question 36

. You wish to clone the gene encoding a peptide hormone that will permit medical students to go without sleep while they are studying for their medical board examinations. You plan to express this gene in bacteria so that the hormone can be purified in commercial quantities. You have an antibody that recognizes the peptide hormone. What kind of recombinant DNA library will you screen?
A. Genomic library cloned into a bacterial artificial chromosome (BAC) vector
B. cDNA library cloned into an expression vector
C. cDNA library cloned into the plasmid pBR322
D. Genomic library cloned into a yeast artificial chromosome (YAC) vector

Answer 36

B If you plan to identify a clone using an antibody, you must use an expression library that will produce the protein and then perform western blots on samples from each colony.
A, C, D It is best to screen a cDNA library (which contains only the expressed sequences) whose fragments are cloned into an expression vector.

Question 37

While working in a medical mission in a small village in Honduras, a 7-year-old male child is brought to you by his parents. The parents tell you that the child has always had “very bad skin and been very sick,” and whenever they can afford it, they take the boy to a hospital in the nearest city for treatment of his skin lesions. While the child was still an infant, the doctors told the parents that he cannot ever be exposed to sunlight, a restriction that is difficult to impose as both parents work in the fields all day. Upon examination of the boy, you find significant patches of discolored skin with areas of severe blistering on his face, arms, and neck that are oozing and raw (indicative of xeroderma pigmentosum). Molecular analysis of the DNA from the boy's skin keratinocytes would reveal an abundance of which of the following types of lesions?
A. Double-stranded breaks
B. Pyrimidine dimers
C. Chromosomal translocations
D. Nucleotide deletions
E. Nucleotide substitutions

Answer 37

B The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. The most common form of DNA damage resulting from sunlight exposure is the production of thymidine dimers. Nucleotide excision repair recognizes and removes bulky adducts that alter or distort the normal shape of DNA, which includes thymidine dimers.
A The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. Double strand breaks are often caused by ionizing radiation (e.g., X-rays) an involve repair mechanisms (nonhomologous end-joining and homologous recombination repair) distinct from nucleotide excision repair.
C The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. Chromosomal translocations are swapping of large DNA fragments between chromosomes, usually the result of significant double strand breaks that are repaired by nonhomologous end joining. Nucleotide excision repair does not influence this pathway.
D The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. Nucleotide deletions are primarily the result of errors in replication and are not repaired by nucleotide excision repair pathways.
E Incorrect. The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. Nucleotide substitutions are primarily the result of errors in replication and are not repaired by nucleotide excision repair pathways.

Question 38

A recent report1 identified a G → C transversion in the gene encoding the hematopoietic transcription factor GATA1 in the genome of a patient with the congenital syndrome Diamond-Blackfan anemia. The deletion caused a loss of exon 2 in the transcribed mRNA. Biochemical analysis of cells homozygous for this mutation indicated that no functional GATA1 protein was produce. What is the likely location of the nucleotide substitution in the GATA1 genes of this patient?
A. 5′-UTR
B. 3′-UTR
C. Splice donor site
D. Promoter
E. Enhancer

Answer 38

C Mutations in intron–exon boundaries that affect splicing tend to cause loss of exons in the transcribed mRNA. In this case, the loss of exon 2 leads to the production of GATA1 that lacks the transactivation domain.
A The 5′-UTR can contain sequences that influence translation efficiency. However, such a mutation is extremely unlikely to result in loss of an exon, as was observed in this case.
B The 3′-UTR can contain sequences that influence translation efficiency and mRNA stability, as well as the signal for termination and polyadenylation. However, it is essentially impossible that a mutation in this region would result in the loss of an exon.
D The promoter is required for gene transcription. Because the transcript was detected, the promoter is operating normally.
E Enhancers influence the activity of associated promoters to increase the transcription of genes. Because the transcript was detected, any enhancers must be operating normally.

Question 39

A child born from consanguineous parents has a congenital defect in pituitary development. Genetic analysis revealed a five nucleotide deletion associated with the child's PIT1 genes that was associated with significantly diminished levels of PIT1 mRNA. PIT1 genes activity is required for normal pituitary development. Biochemical analysis of cells biopsied from the child determined that the acetylation of chromatin associated with the PIT1 core promoter was significantly lower than in matched cells from a healthy donor. The five nucleotide deletion is likely in which of the following elements associated with the PIT1 gene?
A. Enhancer
B. Polyadenylation signal sequence
C. Silencer
D. Splice donor site
E. Transcriptional start site

Answer 39

A Enhancers are bound by transcription factors that activate transcription of an associated core promoter. One mechanism for achieving this is the recruitment of histone acetyltransferase (HAT) enzymes by the enhancer-bound transcription factor. Because the histone acetylation levels were diminished in the patient, it is likely that the deletion was in an enhancer that resulted in the absence of HAT enzyme recruitment.
B Transcription is terminated when RNA polymerase II cleaves the mRNA transcript at sites located downstream of the polyadenylation signal sequence (AATAAA). Following cleavage, 100 to 200 nontemplate adenosine residues are added to the end of the mRNA by enzymes associated with the RNA polymerase II complex. A mutation in this sequence would not alter the acetylation of the gene’s core promoter.
C Silencer elements are bound by transcription factors that repress transcription of an associated core promoter. One mechanism for achieving this is the recruitment of histone deacetylase (HDAC) enzymes by the silencer-bound transcription factor. The observation of diminished histone acetylation levels in the patient is the exact opposite of what would be predicted if the silencer element was mutated.
D A splice donor site is the terminal nucleotide within an exon that is joined to the first nucleotide of the subsequent exon (the splice acceptor site). As such, these sequences have no influence on histone acetylation at the promoter of the gene.
E The transcriptional start site is the nucleotide position of a gene where the RNA polymerase initiates polymerization of mRNA. The transcriptional start site is not a specific, conserved nucleotide sequence, although most transcripts do initiate with a purine. Rather, it is determined primarily by the location of core promoter elements (e.g., the TATA box). A nucleotide substitution at the transcriptional start site would likely have no effect on the efficiency of transcription.

Question 40

A 26-year-old woman in generally good health comes to your clinic for a check-up, as she and her husband have decided to start a family. She is currently using a skin patch for transdermal delivery of low-dose synthetic estrogen and progesterone and will discontinue its use immediately. Both of these drugs activate gene transcription through mechanisms similar to another steroid family, glucocorticoids. At genes that are activated by estrogen, what changes in histone modification will likely occur in the short term (hours to days) once the woman discontinues her use of the contraceptive patch?
A. Decreased acetylation
B. Increased acetylation
C. No change in acetylation
D. Variable changes in acetylation, depending on the gene

Answer 40

A The estrogen receptor functions in a manner similar to the glucocorticoid receptor. It is a ligand-activated nuclear transcription factor that binds DNA. The estrogen receptor recruits coactivators with histone acetyltransferase (HAT) activity, to the genes it activates. The HAT enzymes acetylate histones on chromatin, leading to a decompacted structure and subsequent transcription. The loss of estrogen delivery once she discontinues the patch will result in the loss of estrogen receptor binding and activity at these genes. Histone deacetylase enzymes will then remove the acetyl groups on modified histones at these genes, resulting in reduced transcription.
B Increased acetylation is associated with transcriptional activation by the ligand-bound estrogen receptor. Discontinuing the contraceptive estrogen patch will have the opposite effect, resulting in loss of histone acetylation.
C The estrogen receptor (ER) is a ligand-activated nuclear transcription factor that binds DNA, leading to histone acetylation at genes that it activates. Discontinuing the delivery of estrogen will lead to loss of ER binding at target genes and a reduction in histone acetylation.
D Histones within nucleosomes associated with genes activated by estrogen will always have elevated acetylation. As such, discontinuing the delivery of estrogen will lead to a reduction in histone acetylation.