[MCQs] DNA and Chromosome Structure, DNA Replication and Repair Quiz (19 test)

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DNA and Chromosome Structure, DNA Replication and Repair (19 test)

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Question 1

A 3-year-old boy is evaluated for symptoms of autism including poor eye contact and severe speech delay. Family history indicates that his mother had trouble in school due to “attention deficit,” that he has an older brother with mental disability, and that another older brother and older sister are normal. The figure below diagrams the results of genetic analysis of the boy (lane 4), his mother (lane 5), and his siblings (lanes 1-3) for fragile X syndrome (MIM*309550). Which of the following best describes the results of this genetic analysis?

A	Immunoblot analysis reveals the boy is not affected with fragile X syndrome but one brother is.
B	Northern blot analysis reveals the boy is affected with fragile X syndrome and that his mother is a carrier.
C	Southern blot analysis reveals the boy is affected with fragile X syndrome as is one brother.
D	Western blot analysis reveals that the boy and his siblings do not have fragile X syndrome.
E	Southern blot analysis depicts normal results for this family.

Question 1 Explanation:

The figure accompanying this question shows DNA fragments displayed by Southern blotting (Dr Ed Southern developed the technique) with lanes 3 and 4 displaying single, diffuse bands derived from expanded DNA fragments from boys with fragile X syndrome. Males have one X chromosome and thus one DNA fragment per X chromosome gene (lane 4 for the patient, lane 3 for his affected brother, lane 1 for his unaffected brother), while women have two X chromosomes and two derived DNA fragments (lane 2 for the patient’s normal sister, lane 5 for his carrier mother). The abnormally large DNA fragments (upper band in lane 5, single bands in lanes 3 and 4) are diffuse because abnormal fragile X alleles are unstable, with variable numbers of amplified triplet repeats among the different cells in one individual. Answers a (immunoblotting), d (Western blotting), or e (reverse transcription) are incorrect in suggesting the boy is unaffected, while Northern blotting in answer b examines RNA rather than DNA fragments. For Southern blotting, the patient’s genomic DNA (from blood, skin biopsy, amniocentesis, etc) is extracted, digested with the appropriate restriction enzymes, size-fractionated on agarose gels, blotted (transferred onto a membrane using capillary pressure), stabilized by heat/drying, hybridized with radioactive sequencespecific DNA probes, and exposed to x-ray film. Those fragments of size-fractionated DNA that are complementary to the probe are highlighted as dark bands on the autoradiogram. Southern analysis has been displaced by rapid DNA sequencing for most single gene diseases but is still used for conditions with amplified triplet repeats (eg, Huntington chorea MIM*143100 or myotonic dystrophy MIM*160900 in addition to fragile X syndrome MIM*309550) because of the dramatic DNA size differences. Analogous blotting techniques were named as puns on the name of Southern, including “Northern” blotting to detect RNA molecules and “Western” blotting to detect proteins. Immunoblotting is one version of the “Western” blot technique where proteins are separated by electrophoresis and transferred to a membrane for probing with antibody to the proteins of interest. Boys with fragile X syndrome have ~400 to 1000 triplet repeats in a region proximal to the fragile X gene coding sequence, inactivating gene expression and causing intellectual disability, lax joints, subtle facial changes, and occasional large testes. Immunoblotting is one version of the “Western” blot technique, where proteins are separated by electrophoresis and transferred to a membrane for probing with antibody to the proteins of interest. Reverse transcription of RNA into single, then double-stranded DNA, followed by polymerase chain reaction amplification (RT-PCR), is a more sensitive method for RNA measurement. RT-PCR could demonstrate altered levels of fragile X mRNA but would not measure triplet repeats because these are outside of the fragile X coding region. Fragile X syndrome (MIM*309550) with amplified CGG repeats in the 5′ region upstream of the coding strand and Huntington chorea (MIM*143100) with similarly amplified CAG repeats exemplify a new category of genetic disease that involves triplet repeat instability. Fragile X syndrome is one of many disorders causing mental disability with autistic symptoms.

Question 2

A 2-year-old Scottish-American girl presents with deeply pigmented and scarred skin and her growth is delayed. Her dermatologist obtains a skin biopsy, suspecting the xeroderma pigmentosum group of diseases (MIM*278730); these have decreased ability to repair thymine-thymine dimers in DNA that are caused by ultraviolet light (sunlight exposure). Which of the following strategies would best measure unscheduled DNA synthesis (DNA repair) in the patient’s skin fibroblasts?

A	Cell synchrony, then incubation with labeled iodine in G phase to complex with newly created hydroxyl groups in deoxyribose residues
B	Incubation with labeled purines to replace newly synthesized bases on the outside of the DNA duplex
C	Cell synchrony, then incubation with labeled deoxyribonucleotides in G phase to measure extension of single DNA strands in the 3′ to 5′ direction
D	Incubation with labeled deoxyribonucleotide triphosphates to measure extension of both strands in the 5′ to 3′ direction
E	Incubation with labeled deoxyribonucleotide triphosphates to measure extension of both strands in the 3′ to 5′ direction

Question 2 Explanation:

DNA replication occurs in the 5′ to 3′ direction on both strands of DNA during the S phase of the cell cycle, as indicated in correct answer d. Measure of unscheduled DNA synthesis would not require cell synchrony as in answers a or c, DNA strands are not extended in the 3′ to 5′ direction (answer e), and purine bases are on the inside of the DNA duplex rather than outside, as indicated in answer b. DNA replication consists of five steps—unwinding by helicases, initiation of primase by short RNAs, 5′ to 3′ addition of deoxyribonucleotides via their triphosphates and DNA polymerase in elongation of the “leading” strand (discontinuously using Okazaki fragments for the “lag” strand), replacement of RNA primers with newly synthesized DNA, and sealing of gaps (on the lag strand) with DNA ligase. DNA repair occurs at all cell cycle phases, and among the types is excision repair that uses endonuclease to nick strands adjacent to thymine dimers, DNA polymerase to degrade and replace damaged nucleotides, and ligase to restore integrity. All types of DNA synthesis will incorporate deoxyribonucleotide triphosphates rather than isolated bases or deoxyribose residues, and the new nucleotides are synthesized in the 5′ to 3′ direction using both strands of the DNA helix as templates.

Question 3

A 55-year-old Caucasian male becomes exhausted after jogging 2 miles despite a prior schedule of jogging 30 miles per week. Evaluation discloses severe anemia with elevated white blood cell count, and bone marrow examination suggests a diagnosis of chronic myeloid leukemia (MIM*608232). Which of the following results would be most diagnostic of chromosome 22 deletion that can be seen in this leukemia?

A	Increased incorporation of labeled acetate into histones of chromatin regions on chromosome 22
B	Altered DNA restriction patterns of chromosome 22 regions with methylation-sensitive endonucleases
C	Altered pattern of small RNAs along leukemic chromatin
D	Altered DNA sequence for at least one chromosome 22 locus
E	Decreased transcription from loci on chromosome 22

Question 3 Explanation:

Chromosome deletion involves a breakage of DNA, with loss of the distal chromosome. The primary change in chronic myelogenous leukemia with chromosome 22 deletion would be a break in the chromosomal DNA that deletes certain sequences, joining DNA before the deletion point with telomeric sequences that foster chromosome replication (answer d). The other answers refer to changes in gene expression that could occur as a result of chromosome alteration, and do occur in various cancers—altered primary transcription (answer e), altered “second” code of epigenesis through histone modification (answer a), altered DNA methylation (answer b), or altered decoration patterns of small RNAs that influence chromatin transcription (answer c). It is recognized now that all cancers involve genetic changes, even though few are hereditary. An early example was the discovery of a Philadelphia chromosome (terminally deleted chromosome 22) in chronic myeloid leukemia that in fact derived from translocation of 22 material onto chromosome 9, joining the Abelson ABL oncogene to the BCR (breakpoint cluster region) to produce a fusion tyrosine kinase product that can be inhibited by imatinib (Gleevec). Specific chromosome translocations have now been documented for several forms of cancer, allowing antibody therapy targeting their fusion oncogenic proteins.

Question 4

A 20-year-old Caucasian male with sensorineural deafness has a mother, grandmother, and other maternal relatives who are also affected. He has been counseled that his family’s pattern fits autosomal dominant inheritance (eg, MIM*124900 among >10 types), conferring a 50% risk for his offspring to be deaf with each pregnancy. His family has been involved with the deaf community and has resisted gene testing, but he and his wife are interested in preconception/prenatal diagnosis options. Which of the following statements accurately conveys DNA testing options for the man and his wife, focusing particularly on restriction fragment length polymorphism (RFLP) analysis?

A	The man and his wife could have RFLP analysis and determine fetal outcome by which paternal RFLP allele was inherited.
B	Survey of RFLPs for an allele cotransmitted with deafness in the man and his affected family members could provide a marker for deafness in his offspring.
C	Finding an RFLP-encoded protein could distinguish mutated and normal genes for deafness.
D	Survey of the man’s DNA for mutations outside of the putative deafness gene would reveal a restriction site that reflected an amino acid coding change.
E	Restriction of the man’s DNA would produce fragments of the same size but with differences in charge.

Question 4 Explanation:

Restriction fragment length polymorphisms (RFLPs), now largely supplanted by single nucleotide polymorphisms (SNPs) detected by DNA sequencing, must be linked (located near) to putative disease genes by study of families with affected and unaffected members (answer b and not a, since the man’s restriction fragments would not be linked with his deaf or normal alleles). The restriction site recognition sequence may vary among different individuals, yielding different DNA fragment sizes after cleavage (not different charges as in answer e or derived proteins as in answer c). Variable nucleotides that produce or eliminate restriction enzyme cleavage are often located outside of coding regions (not answer d). In order for an RFLP to cotransmit with normal or abnormal disease alleles, it must be within a few thousand base pairs of the disease gene. Even closely linked RFLPs or SNPs will require study of 10 or more family members to yield reliable probabilities for linkage and thus of disease diagnosis. The method of RFLP analysis involves restriction of human DNA with endonucleases isolated from bacteria, producing an array of fragment sizes that can be separated by agarose or polyacrylamide electrophoresis. The differently sized DNA fragments migrate at speeds inversely proportional to their weights, and specific gene fragments can be detected amidst background DNA fragments after Southern transfer, hybridization with fluorescent or radiolabeled DNA probes, and visualization by fluorescence or autoradiography. Direct detection of the single nucleotide polymorphisms that alter restriction sites is now possible through DNA sequencing, now yielding a “Hap Map” of DNA markers that can rapidly locate genes for Mendelian or multifactorial diseases by determining which markers on which chromosome regions are coinherited with disease phenotypes.

Question 5

It is well known that DNA polymerases synthesize DNA only in the 5′ to 3′ direction. Yet, at the replication fork, both strands of parental DNA are being replicated with the synthesis of new DNA. An experiment examines incorporation of radio-labeled nucleotide triphosphates in unsynchronized cells and finds a small amount of incorporated label in DNA fragments of small size. Which of the following statements explains why the experiment is compatible with new DNA from both strands being added in the 5′ to 3′ direction?

A	A different DNA polymerase replicates the complementary DNA strand as a series of small fragments.
B	A 3′ to 5′ DNA polymerase is active in cells with high rates of DNA replication.
C	DNA synthesis on complementary strand produces small 5′ to 3′ fragments that are continuously ligated together.
D	DNA replication switch strands every 100 nucleotides, a detail not resolved by microscopy.
E	DNA synthesis on the complementary strand does not use RNA primers.

Question 5 Explanation:

All DNA polymerases and repair enzymes can synthesize DNA only in the 5′ to 3′ direction, posing an apparent contradiction since the DNA strand (Crick) complementary to the coding strand (Watson strand) is oriented 3′ to 5′. The Watson strand of DNA is synthesized continuously in the 5′ to 3′ direction, while a different mechanism for replication on the Crick strand uses RNA primers (answer e is incorrect) and a distinctive enzyme (primase) to produce small Okazaki fragments (different DNA polymerases are not used as incorrectly suggested in answers a and b). Okazaki fragments of about 1000 nucleotides are synthesized in a 5′ to 3′ direction and then joined together by DNA ligase (DNA replication does not switch strands as suggested in answer e). Thus, while the overall direction of growth of the lagging strand that is made up of small fragments is in fact in the 3′ to 5′ direction, the actual polymerization of individual nucleotides is in the 5′ to 3′ direction.

Question 6

A newborn boy has a cleft lip and palate with severe limb defects. The hands emerge directly from the shoulders (phocomelia means literally “flipper” limb) and the legs consist of the roots of the thighs only. A medical student searches the Online Mendelian Inheritance in Man database and suggests the diagnosis of Roberts syndrome (MIM*268300), an autosomal recessive disorder associated with altered chromatin condensation. The attending postulates that the chromatin change may impair DNA replication, and discusses why mammalian genomes are 100 times the size of bacterial genomes, yet are replicated in just a few minutes. Which of the following best explains this fact?

A	Eukaryotic DNA polymerases are extraordinarily fast compared with prokaryotic polymerases.
B	The higher temperature of mammalian cells allows for an exponentially higher replication rate.
C	Hundreds of replication forks work simultaneously on each piece of mammalian genomic DNA.
D	Histones and microRNAs complexed with bacterial DNA speed up the rate of DNA replication.
E	The presence of histones in mammalian chromatin speeds up the rate of DNA replication.

Question 6 Explanation:

Despite the large size of mammalian genomes, DNA is replicated bidirectionally from multiple initiation sites and reduces copying times to minutes. Similar replication rates and cell temperatures among mammals and most bacteria (Archaea in ocean floor vents are exceptions) eliminate answers a and b, while the presence of histones and microRNAs in mammalian chromatin will slow replication and gene expression (incorrect answers d and e). Chromatin is most condensed at metaphase and least at interphase when most genes are active, explaining why chromosome analysis is performed on metaphase cells. Roberts syndrome (MIM*268300) exhibits altered chromosome condensation through special techniques for chromosome analysis, now defined by mutations in a gene stabilizing chromosome pairing during cell division. The resulting alteration in cell division likely impairs critical cell growth in embryonic regions such as the palate and limbs. The hundreds of initiation sites for DNA replication on chromosomes share a consensus sequence called an autonomous replication sequence (ARS). Thus, while the process of DNA replication in mammals is similar to that in bacteria, with DNA polymerases of similar optimal temperatures and speed, the many replication forks allow for a rapid synthesis of chromosomal DNA. Proteins such as histones, which are bound to mammalian chromosomes, inhibit DNA replication or transcription. Dissociation of the protein-DNA complex (chromatin) and unwinding of DNA supercoils (followed by chromatin reassembly) is part of the replication process.

Question 7

A 24-year-old African-American female requests pregnancy counseling because of learning differences and her family history; her DNA analysis for fragile X syndrome, an X-linked recessive disorder, is shown in the figure below. The test examines amplification of a 3 bp (triplet) CGG repeat adjacent to the fragile X mental retardation-1 (FMR-1) gene (MIM*309550) on the X chromosome by determining DNA fragment size after restriction cleavage and electrophoresis (Southern blot). Her analysis (lane 5) is shown after that of her father (lane 1), normal sister (lane 2), and her brothers with mental disability (lanes 3 and 4). Her mother also had mild mental disability. The bands of higher molecular weight (slower mobility by electrophoresis) derive from fragile X gene fragments that contain higher number of CGG repeats. A discrete upper band is seen in the patient’s mother (lane 2), a more diffuse upper band in the patient (lane 5), and very diffuse, slowly moving bands in the patient’s disabled brothers (lanes 3 and 4). Which of the following would be the best interpretation of the results?

A	The patient has one fragile X gene with expanded triplet repeats and a 50% risk for a son with fragile X syndrome
B	The patient has one fragile X gene with expanded triplet repeats and a 25% risk for a son with fragile X syndrome.
C	The patient has a restriction fragment length polymorphism within her fragile X genes with no increased risk for a child with fragile X syndrome.
D	The patient has one fragile X gene with expanded repeats that will produce a string of abnormal amino acids in fragile X protein; she has a 25% risk for a son with fragile X syndrome.
E	The patient has one fragile X gene with expanded triplet repeats and a 100% risk for a son with fragile X syndrome.

Question 7 Explanation:

The female’s upper band, shown in lane 5 of the figure below Question 182, is derived from a fragile X gene fragment with amplified triplet repeats that makes her a carrier with a 25% risk for a son with fragile X syndrome (50% risk for a son ∞ 50% risk for his inheriting her abnormal fragile X allele—answer b and not answer c, suggesting an RFLP). Amplification above approximately 60 triplet repeats produces unstable DNA replication during meiosis as shown by the highly amplified bands in her brothers in lanes 3 and 4. The analysis implies that the female’s deceased mother was also a fragile X carrier. The bands deriving from high numbers of repeats are diffuse because of variable DNA replication (not decreased as in answer a) and thus different sizes (differing numbers of repeats) in fragile X gene fragments from different cells of each individual. Location of these triplet repeat regions in the untranslated region would not produce a string of amino acids (note that CGG would encode proline) as implied by answer d.

Question 8

A 2-year-old child presents with poor weight gain, sparse hair, unusual face with sunken eyes, and skin that reddens markedly in sunlight. A diagnosis of Cockayne syndrome (MIM*216400) is considered and studies of DNA synthesis are undertaken. In vitro assay of labeled thymidine incorporation reveals decreased levels of DNA synthesis compared to controls, but normal-sized labeled DNA fragments. The addition of protein extract from normal cells gently heated to inactivate DNA polymerase restores DNA synthesis in Cockayne cells to normal. Which of the following enzymes used in DNA replication is most likely to be defective in Cockayne syndrome?

A	DNA-directed DNA polymerase
B	Unwinding proteins
C	DNA polymerase I
D	DNA-directed RNA polymerase
E	DNA ligase

Question 8 Explanation:

Before DNA replication can actually begin, unwinding protein must open segments along the DNA double helix. A defective unwinding protein (helicase) slows the overall rate of DNA synthesis, but does not alter the size of replicated DNA fragments. Defects in DNA synthesis or transcription may produce a phenotype of accelerated aging, as in Cockayne syndrome (MIM*216400). After unwinding, DNA-directed RNA polymerase (primase) catalyzes the synthesis of a complementary RNA primer of approximately 50 to 100 bases on each DNA strand. Then, DNA-directed DNA polymerase III adds deoxyribonucleotides to the 3′ end of the primer RNA, which replicates a segment of DNA, the Okazaki fragment. DNA polymerase I then removes the primer RNA and adds deoxyribonucleotides to fill the gaps between adjacent Okazaki fragments. The fragments are finally joined together by DNA ligase to create a continuous DNA chain.

Question 9

A 20-year-old African-American male consults a physician because his father and a paternal aunt developed colon cancer in their mid-20s. Neither relative had multiple colonic polyps, so a diagnosis of hereditary nonpolyposis colon cancer (HNPCC-MIM*114500) is considered. The man’s physician discusses testing for HNPCC, knowing that affected patients have genes with microsatellite instability, many gene regions containing abnormal, small loops of unpaired DNA. These DNA loops are the result of a mutation affecting which of the following?

A	Mismatch repair
B	Chain-break repair
C	Base excision repair
D	Depurination repair
E	Nucleotide excision repair

Question 9 Explanation:

One of the most common types of inherited cancers is hereditary nonpolyposis colon cancer (HNPCC-MIM*114500). Most cases are associated with mutations of either of two genes that encode proteins critical in the surveillance of mismatches. Mismatches are due to copying errors leading to one- to five-base unmatched pieces of DNA. Two- to five-base-long unmatched bases form miniloops. Normally, specific proteins survey newly formed DNA between adenine methylated bases within a GATC sequence. Mismatches are removed and replaced. First, a GATC endonuclease nicks the faulty strand at a site complementary to GATC; then an exonuclease digests the strand from the GATC site beyond the mutation. Finally, the excised faulty DNA is replaced. In HNPCC, the unrecognized mismatches accumulate, leading to malignant growth of colon epithelium. The other forms of DNA repair are important for rectifying damage from ultraviolet light.

Question 10

A recombinant viral DNA containing a 300 bp human gene sequence is replicated in vitro using media containing bromodeoxyuridine triphosphate, completely saturating the DNA with bromodeoxyuridine residues. The bromo-substituted viral DNA is then switched to media with regular deoxyuridine triphosphate and allowed to undergo two rounds of DNA replication. The human DNA fragment is then cleaved by restriction enzyme digestion and subjected to ultracentrifugation in cesium chloride under conditions that will separate bromo-substituted DNA duplexes, hybrid bromo-substituted/normal duplexes, and normal duplexes. Which of the results depicted in the figure below (with their interpretations) should be obtained?

A	One DNA band of low density, meaning all DNA duplexes have two strands with normal uridine
B	One DNA band of intermediate density, meaning all DNA duplexes have one strand with uridine, one with bromouridine
C	One DNA band of high density, meaning all DNA duplexes have two strands with bromouridine
D	One DNA band of high density and one of intermediate density, indicating the corresponding types of DNA duplexes
E	One DNA band of intermediate density and one of low density, indicating the corresponding types of DNA duplexes

Question 10 Explanation:

The replication of double-stranded DNA is semiconservative, meaning that each strand separates and serves as a template for synthesis of a new complementary strand. Bromodeoxyuridine has a bromyl group at the fifth position of the pyrimidine ring, simulating the 5-methyl group of thymine and allowing it to replace thymidine in DNA (uridine is normally incorporated only into RNA). Incorporation of bases with the large bromine atom makes bromo-dU DNA strands more dense and separable from thymine-substituted DNA. The first round of replication of a bromodeoxyuridine (bromo-dU)-labeled DNA helix in a solution with normal dU will yield two daughter duplexes, each with one bromo-dU and one dU strand. The second round of replication will yield four daughter duplexes, two with one bromo-dU and one dU strand (intermediate density, middle of the tube in the figure below Question 185) and two with both strands containing dU (low density, top of the tube). Centrifugation to equilibrium in cesium chloride causes a density gradient to form in the centrifuge tube with banding of DNA duplex molecules at their corresponding density.

Question 11

A 3-year-old boy offspring of first-cousin parents has a falloff in growth despite normal food intake. The child also has skin that is more darkly pigmented than his parents, and his skin also becomes bright red after sunlight exposure and leaves scars after healing. A skin biopsy is obtained and the cultured fibroblasts tested for xeroderma pigmentosum (XP-MIM*278700). The laboratory reports positive testing with assignment to complementation group A. Which of the following is the most likely information in this report?

A	The patient’s fibroblasts had low amounts of thymine dimers after ultraviolet light exposure, which were increased by fusion with other XP cell lines except those in complementation group A.
B	The patient’s fibroblasts had low exogenous nucleotide-excision repair that could not be restored by fusion with any XP cell lines.
C	The patient’s fibroblasts had low levels of RNA polymerase that were restored by fusion with XP cell lines except those from complementation group A
D	The patient’s fibroblast had low exogenous nucleotide-excision repair that was restored to normal levels by fusion with all XP cell lines except those in complementation group A.
E	The fibroblast had high amounts of thymine dimers after x-ray exposure, which were increased by fusion with all XP cell lines.

Question 11 Explanation:

Xeroderma pigmentosum is caused by defects in excision repair, a mechanism involving several different enzymes and thus several different genes. Cells from XP patients should have lower nucleotide-excision repair activity and higher amounts of thymine dimers. Since the individual enzymes contributing to excision repair cannot be assayed, patients must be grouped by complementation testing—fusing patient cells with other XP cell lines to see which complement or correct the defective excision repair and high levels of thymine dimers—the question states that the patient falls into complementation group A and implies all XP cells except those from group A should rectify abnormalities in the patient’s cells (answer d). Incorrect answers postulate either low thymine dimers (answer a), low RNA polymerase that is not an excision-repair enzyme (answer c), or correction by all XP cell lines (answers b and e). Xeroderma pigmentosum (XP-MIM*278700) is an autosomal recessive disorder with enhanced DNA damage from ultraviolet irradiation, manifest by photosensitivity (reddening of skin and eyes from sunlight), pigmentation, blistering and scarring, and a 1000-fold greater risk for skin cancer. Autosomal recessive inheritance implies an enzyme deficiency (heterozygous carriers will be normal with 50% enzyme levels), and the deficient enzyme is part of a complex mediating excision repair. A UV-specific exonuclease nicks the dimer on its 5′ side, DNA polymerase I replicates the damaged sequence, the damaged piece is hydrolyzed by the5′ to 3′ exonuclease activity of DNA polymerase I, and DNA ligase joins the patched and original DNA. Cultured skin cells (fibroblasts) from patients with XP are deficient in nucleotide excision repair, and fusion with cells from different XP patients could restore excision-repair activity (complement the repair deficiency). Systematic testing of various XP fibroblast lines demonstrated there were seven groups of XP cells, implying seven different enzymes involved in the process, each susceptible to mutations that would produce the disease.

Question 12

A 21-year-old female has a history of frequent pneumonias and is hospitalized with fever, severe cough, low blood pressure, and bruises on her lower limbs that are interpreted as hemorrhage (purpura). Streptococcus pneumoniae is cultured from her bloodstream and she improves after antibiotic therapy. She undergoes testing to evaluate white cell responsiveness to bacterial endotoxins that can be altered in the presence of mutated toll-like receptors (TLRs-MIM*603030). White cells from the patient and a control are isolated (buffy coat after centrifugation of heparinized blood) and incubated with endotoxin in the presence of tritiated thymidine to measure stimulation of DNA synthesis. DNA is isolated from the leucocyte samples and size-fractionated, yielding a high-molecular-weight fraction with slight differences and low-molecular-weight fractions with dramatically greater incorporation of tritiated thymidine in the control versus the patient. This result is interpreted as showing the patient has decreased leucocyte response to endotoxin, which is best explained by which of the following?

A	The high-molecular-weight fraction is mostly indicative of DNA repair, while smaller DNA fragments reflect new and discontinuous DNA replication on both DNA strands.
B	The high-molecular-weight fraction arises from 3′ to 5′ DNA replication, while the smaller DNA fragments arise from 5′ to 3′ replication that reflects new DNA synthesis.
C	The high-molecular-weight fraction reflects continuous DNA replication on the leading (Watson) strand; this strand then serves as a template for new and discontinuous DNA synthesis on the lag (Crick) strand.
D	The low-molecular-weight fraction is a direct measure of DNA replication rather than repair, reflecting discontinuous 5′ to 3′ synthesis of DNA fragments on the lag strand using primase, RNA primers, and ligase.
E	The low-molecular-weight fraction is a direct measure of DNA replication rather than repair, reflecting newly synthesized Okazaki fragments directed by both DNA strands.

Question 12 Explanation:

DNA polymerase moves along the DNA in one direction and synthesizes new DNA from both antiparallel strands at one replication fork. Since DNA is always synthesized 5′ to 3′, one of the strands (the leading strand) is synthesized continuously, whereas the other (the lagging strand) is synthesized in short stretches called Okazaki fragments that are made only during DNA replication (as implied by correct answer d). Answers a and e are incorrect because they posit discontinuous synthesis on both strands, b is incorrect because it posits 3′ to 5′ synthesis, and c is incorrect because it violates semiconservative DNA replication. Discontinuous synthesis on the lag strand requires an RNA primer because DNA polymerase I, the major enzyme for DNA replication, can only add bases to an existing strand in the presence of template strand. A unique enzyme called primase synthesizes the short RNA molecules required to prime synthesis of Okazaki fragments and a ligase joins these fragments to elongate the lagging strand. Initiation of DNA replication occurs at specific sites with easily unwound AT-rich sequences characterized as autonomously replicating sequences (ARS) or origin recognition complexes (ORC) in E. coli. Formation and procession of the replication fork requires helicases for DNA unwinding, SSBs (single-strand binding proteins) to prevent premature reannealing, and DNA polymerase I for chain elongation.

Question 13

A 6-month-old Caucasian boy has been followed since birth because of extreme growth failure. His birth weight was 4 lb after a term 40-week uncomplicated pregnancy. Measures of tritiated thymidine incorporation by his stimulated white blood cells show a marked decrease compared to controls. Which of the following is the most likely interpretation of these results?

A	Homozygous nonsense mutation in the gene synthesizing the RNA primer used in DNA replication
B	Homozygous nonsense mutation in DNA polymerase
C	Homozygous mutation in helicase (rep protein) that continuously unwinds duplex DNA at the replication fork during synthesis
D	Homozygous mutation in an exonuclease used in excision repair
E	Homozygous mutation in DNA ligase that repeatedly joins the ends of DNA along the growing strand

Question 13 Explanation:

Homozygous mutations in DNA repair enzymes can lead to primordial dwarfism (short stature beginning with intrauterine growth retardation and low birth weight) or predispose to cancers as with hereditary nonpolyposis colon cancer (HNPCCMIM*114500). Other answers involve mutations in key DNA replication components and would be incompatible with life, presenting as an early embryonic death that might not be recognized as a pregnancy. DNA replication involves continuous synthesis in the 5′ to 3′ direction from the leading strand and discontinuous DNA synthesis, also in the 5′ to 3′ direction, by DNA polymerase III on the lagging strand. RNA primers, primase enzyme, and DNA ligase are used repeatedly to make and join Okazaki DNA fragments along the growing lagging strand. Mutation of both alleles for the major DNA polymerase, DNA polymerase III, the gene synthesizing RNA primer, primase, or the special replication DNA ligase would make DNA replication and cell division impossible and be genetically lethal to the embryo.

Question 14

A 19-year-old female experiences recurrent respiratory infection and night sweats after becoming pregnant, attributing these symptoms to pregnancy until she notes a weight loss rather than expected weight gain. Obstetric evaluation reveals a positive test for tuberculosis and then a low white blood cell count. She admits to multiple sexual partners in the past, and an HIV antibody test is positive. Her doctor places her on a regimen of antituberculous drugs and recommends azidothymidine (AZT) therapy against her AIDS (MIM*609432) because it reduces maternal-to-child AIDS transmission by 30%. Which of the following describes the mechanism of action of AZT?

A	It inhibits viral protein synthesis.
B	It inhibits RNA synthesis.
C	It inhibits viral DNA polymerase.
D	It stimulates DNA provirus production.
E	It inhibits viral reverse transcriptase.

Question 14 Explanation:

The AIDS treatment drug azidothymidine (AZT) inhibits viral reverse transcriptase that is essential for replication of the human immunodeficiency virus (as in answer e). AZT interrupts reverse transcriptase because it lacks a hydroxyl group for polymerization but does not inhibit DNA polymerase, RNA polymerase, or protein synthesis (incorrect answers a-d). Reverse transcriptase is an RNAdirected DNA polymerase that synthesizes the HIV- DNA provirus, which in turn synthesizes new viral RNA. Since AZT (zidovudine) was approved in 1987, nonnucleoside reverse transcriptase inhibitors (eg, nevirapine) and viral protease inhibitors (eg, ritonavir) were developed that allow very effective combination therapies that provide long life spans and low fetal transmission rates for AIDS patients. AZT does have side effects, including muscle aches, hepatomegaly, and even lactic acidosis.

Question 15

Nucleic acids extracted from normal and prostate cancer (MIM*176807) tissue are labeled with differently colored fluorescent probes and compared for their degree of hybridization to an array of DNA segments on a glass slide (DNA chip, technique known as comparative genomic hybridization—CGH—or microarray analysis). The DNA segments are chosen to represent the entire genome, and a segment from chromosome 2 gives a much stronger hybridization signal with the cancer tissue. This DNA segment is then used as a probe to characterize the nucleic acid that is more abundant in prostate cancer tissue. This nucleic acid is stable in acid, degraded by base, and migrates as a 30-base pair species in high or low salt by polyacrylamide gel electrophoresis. Which of the following statements best describes the nucleic acid that is more abundant in prostate cancer?

A	A 60 bp, double-stranded DNA fragment from chromosome 2
B	A 30 bp, single-stranded DNA fragment from chromosome 2
C	A 60 bp, double-stranded RNA fragment encoded by a gene on chromosome 2
D	A 30 bp, single-stranded RNA encoded by a gene on chromosome 2
E	A 30 bp, single-stranded RNA with a high degree of secondary structure, most likely a transfer RNA encoded by chromosome 2

Question 15 Explanation:

DNA is composed of 2′-deoxyribose sugars, while RNA is composed of ribose sugars. The 2′ and 3′ hydroxyls of ribose allow alkaline release of one proton and initiation of phosphate cleavage by oxide, making RNA unstable at basic pH and less stable in general than DNA. The lability of the prostate cancer nucleic acid in base and its similar electrophoretic migration in high or low salt suggests it is a single-stranded RNA with little secondary structure (correct answer d). DNA species would be stable in base (incorrect answers a and b), and high salt would dissociate hydrogen bonds in double-stranded or highly coiled RNAs, reducing their apparent molecular weight estimated by migration on gel electrophoresis (incorrect answers c and e). Comparative genomic hybridization or microarray analysis provides a rapid method for defining nucleic acid excess or deficiency in patient samples, revealing subtle extra or missing chromosome segments in the patient’s genomic DNA (high resolution chromosome analysis) or abnormally abundant/deficient RNAs in cancer tissues. DNA “chips” composed of 40,000 to 1 million segments arrayed on a glass slide are hybridized with fluorescently labeled patient and control samples, and a machine graphs the relative hybridization signals for each DNA segment, ordered by their chromosome location. DNA samples from patients with autism or related disabilities reveal novel microdeletions and microduplications in these disorders, while RNA samples from cancer tissues reveal tumor-specific hybridization patterns (RNA profiles) or particular microRNA markers (implied by this question) that forecast tumor type, prognosis, and optimum therapy.

Question 16

Which of the following is the correct sequence of events in gene repair mechanisms in patients without a mutated repair process?

A	Nicking, excision, replacement, sealing, and recognition
B	Nicking, recognition, excision, sealing, and replacement
C	Nicking, sealing, recognition, excision, and replacement
D	Recognition, nicking, excision, replacement, and sealing
E	Sealing, recognition, nicking, excision, and replacement

Question 16 Explanation:

In all of the forms of DNA repair in normal cells, a common sequence of events occurs: 1. The single or multiple base abnormality is surveyed and detected by a specific protein or proteins. 2. The DNA is nicked on one side of the damaged DNA. 3. A specific enzyme excises the damaged portion (steps 2 and 3 can be combined if an exonuclease cuts on both sides of the damaged DNA). 4. The damaged portion of the strand is replaced by resynthesis catalyzed by DNA polymerase I. 5. A ligase seals the final gap. 6. With some variability, these general principles apply in nucleotide excision repair (segments of about 30 nucleotides), base excision repair of single bases, and mismatch repair of copying errors (1-5 bases).

Question 17

Which of the following indicates the correct sequence of chromosome constituents, going from smaller to larger?

A	Chromatin fibril, nucleosome, histone octamer, and chromosome loop
B	Nucleosome, chromatin fibril, chromosome loop, and histone octamer
C	Chromatin fibril, histone octamer, nucleosome, and chromosome loop
D	Nucleosome, histone octamer, chromosome loop, and chromosome fibril
E	Histone octamer, nucleosome, chromatin fibril, and chromosome loop

Question 17 Explanation:

DNA does not exist naked in eucaryotic cells, but as a complex with histones and nonhistone proteins. The four core histones are present as the dimers H2A-H2B and H3-H4, and these dimers, in turn, pair to produce a histone octamer. The histone octamer associates with DNA and histone H1 to form nucleosomes—arrayed like beads on a string to form a chromatin fibril of 30 nm width. Each histone octamer complexes with 146 base pairs of coiled (superhelical) DNA, and each of the core histones or histone H1 is susceptible to chemical modification by acetylation, methylation, phosphorylation, ribosylation, or ubiquination. Phosphorylation of histone H1 causes chromatin condensation, and the 10 nm fibril can condense to form successive 30 nm fibrils, chromosome loops or domains, and finally the highly condensed metaphase chromosomes that are analyzed in human cytogenetics. Chromatin upstream of active genes is typically less condensed, forming hypersensitive sites to nuclease. The modification of histones, along with methylation of DNA and changes in chromosome condensation, comprises a “second code” for regulation of gene expression. These changes are often distinguished as “epigenetic” incontrast to genetic changes in the primary code of DNA to RNA to protein.

Question 18

A 2-year-old Caucasian girl presents with a 2-week history of fatigue, pallor, and recent chest pain localized to her sternum. Her physical examination is remarkable for general irritability, tenderness over her sternum, and a lemon-yellow hue to her skin. Blood counts show anemia with a hemoglobin of 5 g/dL (normal >12) and extreme leucocytosis (white blood cell count 130,000 cells/cubic millimeter with normal <12,000). A diagnosis of acute lymphocytic leukemia (MIM*187040) is made after further bone marrow studies, mandating a therapy protocol that includes radiation therapy of the central nervous system to destroy leukemic lymphoblasts. Which of the following describes the rationale for radiation therapy of certain cancers?

A	Cross-linking of DNA in rapidly growing cancer cells
B	Demethylation of DNA to prevent progression of precancerous cells
C	Cleavage of DNA double strands necessary for transcription in G phase
D	Disruption of DNA-RNA transcription complexes necessary for G phase
E	Disruption of purine rings in DNA crucial for S phase and rapidly growing cells

Question 18 Explanation:

Chemo- and radiotherapies target deregulated and/or rapidly growing cancer cells that have more active DNA replication (S phase) and transcription compared to quiescent cells in G or stationary phase. Radiation damages cellular DNA by opening purine rings and rupturing phosphodiester bonds, not by other actions suggested by incorrect answers a to d. Chemical agents such as formaldehyde can cross-link DNA, and inhibitors of DNA methylation, such as methotrexate (an inhibitor of folic acid), were the first anticancer drugs. Experimental gene therapies for cancer include the inhibition of oncogene expression and the enhancement of tumor suppressor gene activity. These therapies target particular DNA-RNA transcription complexes or signal transduction cascades that are active in cancer cells.

Question 19

A major obstacle to gene therapy involves the difficulty of homologous gene replacement. Which of the following strategies best addresses this issue?

A	A recombinant vector contains complementary DNA sequences that will facilitate site-specific recombination.
B	A recombinant vector expresses antisense nucleotides that will hybridize with the targeted mRNA.
C	A recombinant vector replaces inessential viral genes with a functional human gene.
D	A recombinant vector transfects patient cells, which are returned to the patient.
E	A recombinant vector contains DNA sequences that target its expressed protein to lysosomes.

Question 19 Explanation:

Challenges for gene therapy include the construction of recombinant viral genomes that can propagate the replacement gene (gene constructs or vectors), delivery of the altered gene to the appropriate tissues (gene targeting), and recombination at the appropriate locus so that replacement of the defective gene is achieved (site-specific recombination). The latter step positions DNA sequences in the vector so that the replacement gene pairs and recombines precisely with homologous DNA in the native gene. Ex vivo transfection (introduction of vector DNA into patient cells outside the body) is an ideal method for gene targeting if the engineered cells can repopulate the tissue/organ in question. Transfection of bone marrow stem cells with a functional adenosine deaminase gene, followed by bone marrow transplantation back to the patient, has been successful in restoring immunity to children with severe deficiency. Even when tissue targeting and precise gene replacement are feasible, mimicking the appropriate patterns of gene expression can be a substantial barrier to gene therapy. Injection of deficient enzymes into plasma (enzyme therapy) has been successful in disorders such as non-neurologic Gaucher disease (MIM*231000)—storage of lipids in the spleen and bone—and takes advantage of cellular pathways that target enzymes to lysosomes.

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1. A 3-year-old boy is evaluated for symptoms of autism including poor eye contact and severe speech delay. Family history indicates that his mother had trouble in school due to “attention deficit,” that he has an older brother with mental disability, and that another older brother and older sister are The figure below diagrams the results of genetic analysis of the boy (lane 4), his mother (lane 5), and his siblings (lanes 1-3) for fragile X syndrome (MIM*309550). Which of the following best describes the results of this genetic analysis?

Immunoblot analysis reveals the boy is not affected with fragile X syndrome but one brother is.
Northern blot analysis reveals the boy is affected with fragile X syndrome and that his mother is a carrier.
Southern blot analysis reveals the boy is affected with fragile X syndrome as is one brother.
Western blot analysis reveals that the boy and his siblings do not have fragile X syndrome.
Southern blot analysis depicts normal results for this family.

2. A 2-year-old Scottish-American girl presents with deeply pigmented and scarred skin and her growth is Her dermatologist obtains a skin biopsy, suspecting the xeroderma pigmentosum group of diseases (MIM*278730); these have decreased ability to repair thymine-thymine dimers in DNA that are caused by ultraviolet light (sunlight exposure). Which of the following strategies would best measure unscheduled DNA synthesis (DNA repair) in the patient’s skin fibroblasts?

Cell synchrony, then incubation with labeled iodine in G phase to complex with newly created hydroxyl groups in deoxyribose residues
Incubation with labeled purines to replace newly synthesized bases on the outside of the DNA duplex
Cell synchrony, then incubation with labeled deoxyribonucleotides in G phase to measure extension of single DNA strands in the 3′ to 5′ direction
Incubation with labeled deoxyribonucleotide triphosphates to measure extension of both strands in the 5′ to 3′ direction
Incubation with labeled deoxyribonucleotide triphosphates to measure extension of both strands in the 3′ to 5′ direction

3. A 55-year-old Caucasian male becomes exhausted after jogging 2 miles despite a prior schedule of jogging 30 miles per Evaluation discloses severe anemia with elevated white blood cell count, and bone marrow examination suggests a diagnosis of chronic myeloid leukemia (MIM*608232). Which of the following results would be most diagnostic of chromosome 22 deletion that can be seen in this leukemia?

Increased incorporation of labeled acetate into histones of chromatin regions on chromosome 22
Altered DNA restriction patterns of chromosome 22 regions with methylation-sensitive endonucleases
Altered pattern of small RNAs along leukemic chromatin
Altered DNA sequence for at least one chromosome 22 locus
Decreased transcription from loci on chromosome 22

4. A 20-year-old Caucasian male with sensorineural deafness has a mother, grandmother, and other maternal relatives who are also affected. He has been counseled that his family’s pattern fits autosomal dominant inheritance (eg, MIM*124900 among >10 types), conferring a 50% risk for his offspring to be deaf with each His family has been involved with the deaf community and has resisted gene testing, but he and his wife are interested in preconception/prenatal diagnosis options. Which of the following statements accurately conveys DNA testing options for the man and his wife, focusing particularly on restriction fragment length polymorphism (RFLP) analysis?

The man and his wife could have RFLP analysis and determine fetal outcome by which paternal RFLP allele was inherited.
Survey of RFLPs for an allele cotransmitted with deafness in the man and his affected family members could provide a marker for deafness in his offspring.
Finding an RFLP-encoded protein could distinguish mutated and normal genes for deafness.
Survey of the man’s DNA for mutations outside of the putative deafness gene would reveal a restriction site that reflected an amino acid coding change.
Restriction of the man’s DNA would produce fragments of the same size but with differences in charge.

5. It is well known that DNA polymerases synthesize DNA only in the 5′ to 3′ direction. Yet, at the replication fork, both strands of parental DNA are being replicated with the synthesis of new An experiment examines incorporation of radio-labeled nucleotide triphosphates in unsynchronized cells and finds a small amount of incorporated label in DNA fragments of small size. Which of the following statements explains why the experiment is compatible with new DNA from both strands being added in the 5′ to 3′ direction?

A different DNA polymerase replicates the complementary DNA strand as a series of small fragments.
A 3′ to 5′ DNA polymerase is active in cells with high rates of DNA replication.
DNA synthesis on complementary strand produces small 5′ to 3′ fragments that are continuously ligated together.
DNA replication switch strands every 100 nucleotides, a detail not resolved by microscopy.
DNA synthesis on the complementary strand does not use RNA primers.

6. A newborn boy has a cleft lip and palate with severe limb The hands emerge directly from the shoulders (phocomelia means literally “flipper” limb) and the legs consist of the roots of the thighs only. A medical student searches the Online Mendelian Inheritance in Man database and suggests the diagnosis of Roberts syndrome (MIM*268300), an autosomal recessive disorder associated with altered chromatin condensation. The attending postulates that the chromatin change may impair DNA replication, and discusses why mammalian genomes are 100 times the size of bacterial genomes, yet are replicated in just a few minutes. Which of the following best explains this fact?

Eukaryotic DNA polymerases are extraordinarily fast compared with prokaryotic polymerases.
The higher temperature of mammalian cells allows for an exponentially higher replication rate.
Hundreds of replication forks work simultaneously on each piece of mammalian genomic DNA.
Histones and microRNAs complexed with bacterial DNA speed up the rate of DNA replication.
The presence of histones in mammalian chromatin speeds up the rate of DNA replication.

7. A 24-year-old African-American female requests pregnancy counseling because of learning differences and her family history; her DNA analysis for fragile X syndrome, an X-linked recessive disorder, is shown in the figure The test examines amplification of a 3 bp (triplet) CGG repeat adjacent to the fragile X mental retardation-1 (FMR-1) gene (MIM*309550) on the X chromosome by determining DNA fragment size after restriction cleavage and electrophoresis (Southern blot). Her analysis (lane 5) is shown after that of her father (lane 1), normal sister (lane 2), and her brothers with mental disability (lanes 3 and 4). Her mother also had mild mental disability. The bands of higher molecular weight (slower mobility by electrophoresis) derive from fragile X gene fragments that contain higher number of CGG repeats. A discrete upper band is seen in the patient’s mother (lane 2), a more diffuse upper band in the patient (lane 5), and very diffuse, slowly moving bands in the patient’s disabled brothers (lanes 3 and 4). Which of the following would be the best interpretation of the results?

The patient has one fragile X gene with expanded triplet repeats and a 50% risk for a son with fragile X syndrome.
The patient has one fragile X gene with expanded triplet repeats and a 25% risk for a son with fragile X syndrome.
The patient has a restriction fragment length polymorphism within her fragile X genes with no increased risk for a child with fragile X syndrome.
The patient has one fragile X gene with expanded repeats that will produce a string of abnormal amino acids in fragile X protein; she has a 25% risk for a son with fragile X syndrome.
The patient has one fragile X gene with expanded triplet repeats and a 100% risk for a son with fragile X syndrome.

8. A 2-year-old child presents with poor weight gain, sparse hair, unusual face with sunken eyes, and skin that reddens markedly in A diagnosis of Cockayne syndrome (MIM*216400) is considered and studies of DNA synthesis are undertaken. In vitro assay of labeled thymidine incorporation reveals decreased levels of DNA synthesis compared to controls, but normal-sized labeled DNA fragments. The addition of protein extract from normal cells gently heated to inactivate DNA polymerase restores DNA synthesis in Cockayne cells to normal. Which of the following enzymes used in DNA replication is most likely to be defective in Cockayne syndrome?

DNA-directed DNA polymerase
Unwinding proteins
DNA polymerase I
DNA-directed RNA polymerase
DNA ligase

9. A 20-year-old African-American male consults a physician because his father and a paternal aunt developed colon cancer in their mid-20s. Neither relative had multiple colonic polyps, so a diagnosis of hereditary nonpolyposis colon cancer (HNPCC-MIM*114500) is The man’s physician discusses testing for HNPCC, knowing that affected patients have genes with microsatellite instability, many gene regions containing abnormal, small loops of unpaired DNA. These DNA loops are the result of a mutation affecting which of the following?

Mismatch repair
Chain-break repair
Base excision repair
Depurination repair
Nucleotide excision repair

10. A recombinant viral DNA containing a 300 bp human gene sequence is replicated in vitro using media containing bromodeoxyuridine triphosphate, completely saturating the DNA with bromodeoxyuridine residues. The bromo-substituted viral DNA is then switched to media with regular deoxyuridine triphosphate and allowed to undergo two rounds of DNA The human DNA fragment is then cleaved by restriction enzyme digestion and subjected to ultracentrifugation in cesium chloride under conditions that will separate bromo-substituted DNA duplexes, hybrid bromo-substituted/normal duplexes, and normal duplexes. Which of the results depicted in the figure below (with their interpretations) should be obtained?

One DNA band of low density, meaning all DNA duplexes have two strands with normal uridine
One DNA band of intermediate density, meaning all DNA duplexes have one strand with uridine, one with bromouridine
One DNA band of high density, meaning all DNA duplexes have two strands with bromouridine
One DNA band of high density and one of intermediate density, indicating the corresponding types of DNA duplexes
One DNA band of intermediate density and one of low density, indicating the corresponding types of DNA duplexes

11. A 3-year-old boy offspring of first-cousin parents has a falloff in growth despite normal food intake. The child also has skin that is more darkly pigmented than his parents, and his skin also becomes bright red after sunlight exposure and leaves scars after A skin biopsy is obtained and the cultured fibroblasts tested for xeroderma pigmentosum (XP-MIM*278700). The laboratory reports positive testing with assignment to complementation group A. Which of the following is the most likely information in this report?

The patient’s fibroblasts had low amounts of thymine dimers after ultraviolet light exposure, which were increased by fusion with other XP cell lines except those in complementation group A.
The patient’s fibroblasts had low exogenous nucleotide-excision repair that could not be restored by fusion with any XP cell lines.
The patient’s fibroblasts had low levels of RNA polymerase that were restored by fusion with XP cell lines except those from complementation group A.
The patient’s fibroblast had low exogenous nucleotide-excision repair that was restored to normal levels by fusion with all XP cell lines except those in complementation group A.
The fibroblast had high amounts of thymine dimers after x-ray exposure, which were increased by fusion with all XP cell lines.

12. A 21-year-old female has a history of frequent pneumonias and is hospitalized with fever, severe cough, low blood pressure, and bruises on her lower limbs that are interpreted as hemorrhage (purpura). Streptococcus pneumoniae is cultured from her bloodstream and she improves after antibiotic She undergoes testing to evaluate white cell responsiveness to bacterial endotoxins that can be altered in the presence of mutated toll-like receptors (TLRs-MIM*603030). White cells from the patient and a control are isolated (buffy coat after centrifugation of heparinized blood) and incubated with endotoxin in the presence of tritiated thymidine to measure stimulation of DNA synthesis. DNA is isolated from the leucocyte samples and size-fractionated, yielding a high-molecular-weight fraction with slight differences and low-molecular-weight fractions with dramatically greater incorporation of tritiated thymidine in the control versus the patient. This result is interpreted as showing the patient has decreased leucocyte response to endotoxin, which is best explained by which of the following?

The high-molecular-weight fraction is mostly indicative of DNA repair, while smaller DNA fragments reflect new and discontinuous DNA replication on both DNA strands.
The high-molecular-weight fraction arises from 3′ to 5′ DNA replication, while the smaller DNA fragments arise from 5′ to 3′ replication that reflects new DNA synthesis.
The high-molecular-weight fraction reflects continuous DNA replication on the leading (Watson) strand; this strand then serves as a template for new and discontinuous DNA synthesis on the lag (Crick) strand.
The low-molecular-weight fraction is a direct measure of DNA replication rather than repair, reflecting discontinuous 5′ to 3′ synthesis of DNA fragments on the lag strand using primase, RNA primers, and ligase.
The low-molecular-weight fraction is a direct measure of DNA replication rather than repair, reflecting newly synthesized Okazaki fragments directed by both DNA strands.

13. A 6-month-old Caucasian boy has been followed since birth because of extreme growth His birth weight was 4 lb after a term 40-week uncomplicated pregnancy. Measures of tritiated thymidine incorporation by his stimulated white blood cells show a marked decrease compared to controls. Which of the following is the most likely interpretation of these results?

Homozygous nonsense mutation in the gene synthesizing the RNA primer used in DNA replication
Homozygous nonsense mutation in DNA polymerase
Homozygous mutation in helicase (rep protein) that continuously unwinds duplex DNA at the replication fork during synthesis
Homozygous mutation in an exonuclease used in excision repair
Homozygous mutation in DNA ligase that repeatedly joins the ends of DNA along the growing strand

14. A 19-year-old female experiences recurrent respiratory infection and night sweats after becoming pregnant, attributing these symptoms to pregnancy until she notes a weight loss rather than expected weight Obstetric evaluation reveals a positive test for tuberculosis and then a low white blood cell count. She admits to multiple sexual partners in the past, and an HIV antibody test is positive. Her doctor places her on a regimen of antituberculous drugs and recommends azidothymidine (AZT) therapy against her AIDS (MIM*609432) because it reduces maternal-to-child AIDS transmission by 30%. Which of the following describes the mechanism of action of AZT?

It inhibits viral protein synthesis.
It inhibits RNA synthesis.
It inhibits viral DNA polymerase.
It stimulates DNA provirus production.
It inhibits viral reverse transcriptase.

15. Nucleic acids extracted from normal and prostate cancer (MIM*176807) tissue are labeled with differently colored fluorescent probes and compared for their degree of hybridization to an array of DNA segments on a glass slide (DNA chip, technique known as comparative genomic hybridization—CGH—or microarray analysis). The DNA segments are chosen to represent the entire genome, and a segment from chromosome 2 gives a much stronger hybridization signal with the cancer This DNA segment is then used as a probe to characterize the nucleic acid that is more abundant in prostate cancer tissue. This nucleic acid is stable in acid, degraded by base, and migrates as a 30-base pair species in high or low salt by polyacrylamide gel electrophoresis. Which of the following statements best describes the nucleic acid that is more abundant in prostate cancer?

A 60 bp, double-stranded DNA fragment from chromosome 2
A 30 bp, single-stranded DNA fragment from chromosome 2
A 60 bp, double-stranded RNA fragment encoded by a gene on chromosome 2
A 30 bp, single-stranded RNA encoded by a gene on chromosome 2
A 30 bp, single-stranded RNA with a high degree of secondary structure, most likely a transfer RNA encoded by chromosome 2

16. Which of the following is the correct sequence of events in gene repair mechanisms in patients without a mutated repair process?

Nicking, excision, replacement, sealing, and recognition
Nicking, recognition, excision, sealing, and replacement
Nicking, sealing, recognition, excision, and replacement
Recognition, nicking, excision, replacement, and sealing
Sealing, recognition, nicking, excision, and replacement

17. Which of the following indicates the correct sequence of chromosome constituents, going from smaller to larger?

Chromatin fibril, nucleosome, histone octamer, and chromosome loop
Nucleosome, chromatin fibril, chromosome loop, and histone octamer
Chromatin fibril, histone octamer, nucleosome, and chromosome loop
Nucleosome, histone octamer, chromosome loop, and chromosome fibril
Histone octamer, nucleosome, chromatin fibril, and chromosome loop

18. A 2-year-old Caucasian girl presents with a 2-week history of fatigue, pallor, and recent chest pain localized to her sternum. Her physical examination is remarkable for general irritability, tenderness over her sternum, and a lemon-yellow hue to her skin. Blood counts show anemia with a hemoglobin of 5 g/dL (normal >12) and extreme leucocytosis (white blood cell count 130,000 cells/cubic millimeter with normal <12,000). A diagnosis of acute lymphocytic leukemia (MIM*187040) is made after further bone marrow studies, mandating a therapy protocol that includes radiation therapy of the central nervous system to destroy leukemic lymphoblasts. Which of the following describes the rationale for radiation therapy of certain cancers?

Cross-linking of DNA in rapidly growing cancer cells
Demethylation of DNA to prevent progression of precancerous cells
Cleavage of DNA double strands necessary for transcription in G phase
Disruption of DNA-RNA transcription complexes necessary for G phase
Disruption of purine rings in DNA crucial for S phase and rapidly growing cells

19. A major obstacle to gene therapy involves the difficulty of homologous gene Which of the following strategies best addresses this issue?

A recombinant vector contains complementary DNA sequences that will facilitate site-specific recombination.
A recombinant vector expresses antisense nucleotides that will hybridize with the targeted mRNA.
A recombinant vector replaces inessential viral genes with a functional human gene.
A recombinant vector transfects patient cells, which are returned to the patient.
A recombinant vector contains DNA sequences that target its expressed protein to lysosomes.

Chromosome mcq DNA mcq

[MCQs] DNA and Chromosome Structure, DNA Replication and Repair Quiz (19 test)

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DNA and Chromosome Structure, DNA Replication and Repair (19 test)

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