[MCQs] Cancer - Oncology Quiz (19 test)

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Oncology (19 test)

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Question 1

A 25-year-old Caucasian female is evaluated because her father, paternal grandfather, and uncle developed colon cancer before age 40. Examination has revealed an unusually dark complexion similar to that of her father and supernumerary (extra) teeth with multiple caries. Her past medical history indicates she had bony cysts noted on her tibia after x-rays for a fracture, and her physician suspects a diagnosis of autosomal dominant Gardner syndrome (MIM*175100) that has dark complexion, tooth changes, bony tumors (osteomas), and multiple colonic polyps with high rates of colon cancer. DNA testing for the causative adenomatous polyposis coli (APC) gene revealed a nucleotide substitution that changed the DNA sequence from CAGAGGT to CAGGGGT. This mutation ablated the splice junction at the border of exon 7 and its adjacent intron. A partial sequencing result is shown in the figure below, and supports which of the following conclusions about the female’s APC genotype and expression?

A	Homozygous for the normal Gardner allele, some smaller mRNA molecules
B	Heterozygous for normal and mutant Gardner allele, some larger mRNA molecules
C	Heterozygous for normal and mutant Gardner allele, some smaller mRNA molecules
D	Homozygous for mutant Gardner allele, no change in mRNA molecules
E	Homozygous for mutant Gardner allele, some larger mRNA molecules

Question 1 Explanation:

The DNA sequencing gel for the patient (figure below Question 482) indicates some bands at heavier density and others at lighter density, compatible with the autosomal Gardner syndrome/adenomatous polyposis coli (APC) alleles having identical sequence (homozygous) for the heavier bands and different sequences for the lighter bands (heterozygous). The gel can be read from top to bottom as CAG A/G GGT, where A/G represents comigrating DNA fragments of lighter density. The patient has one allele with CAGAGGT sequence (normal) and another with CAGGGGT (mutant), making her heterozygous (answer b, not d or e) and implying that she is affected with Gardner syndrome and at increased risk for colon cancer. Interruption of a splice site for one copy of the gene will yield some larger mRNA molecules (answer b, not a or c) with failure of the intron to be removed. DNA sequencing methodology primes DNA synthesis of key gene regions in the presence of the usual deoxynucleotides (dNTPs) plus low concentrations of a labeled deoxynucleotides that will terminate the synthesis, for example, a dideoxynucleotide (ddNTP) that has no free 3′ OH group for addition of the next nucleotide. Four reactions with A, C, T, and G ddNTPs are run, then applied to adjacent lanes on a gel to separate the labeled, synthesized strands by size. The sequence of the DNA is determined by reading the order of the terminated chains going up the gel.

Question 2

A parent is correcting his child’s photograph for red eye and notes that one of the child’s pupils reflects the flash as white rather than red. An ophthalmologist confirms the presence of a tumor in the back of the white-reflecting eye, telling the parents about the possibility of retinoblastoma (MIM* 180200). The parents return to their pediatrician, confused about the relation of retinoblastoma (Rb) and B-cell lymphoma (bcl) genes they saw on the Internet and the possibility that their child’s tumor is inherited. Which of the following is the most appropriate response?

A	Rb and bcl proteins are polymerases that prevent oncogenesis by stringent DNA repair; the parents are therefore carriers for autosomal recessive Rb deficiency with a one in four recurrence risk.
B	Rb protein binds transcription factors needed for cell division and bcl protein (cyclin D1) stimulates it; Rb is a tumor suppressor gene requiring homozygous mutations (two hits) that are likely sporadic in a child with unilateral tumor.
C	Rb and bcl proteins are DNA-binding factors that suppress promoters near oncogenes and act as tumor suppressors; the child represents a new, dominant-acting mutation and the parents have a minimal recurrence risk.
D	Rb protein stimulates cyclins specific for retinal tissue and bcl does the same for lymphatic cyclins; the child represents a new mutation with excess Rb activity.
E	Rb protein forms complexes with bcl protein that promotes cell division in rapidly proliferating tissues; the child represents a new mutation with excess Rb-bcl complex activity.

Question 2 Explanation:

DNA synthesis occurs only in the S phase of the cell cycle, a process regulated by numerous proteins called cyclins. A key cyclin (cyclin D) is phosphorylated as the cell commits from gap 1 (G1) quiescent phase to the S phase of DNA synthesis, and a cascade of secondary phosphorylations by cyclin-dependent kinases (CDK) activates cyclins, transcription factors, and the DNA synthesis machinery. The Rb protein, discovered as the product of the gene causing hereditary retinoblastoma, undergoes phosphorylation early in S phase and releases its inhibition (binding) of E2F transcription factor, promoting gene activation and DNA synthesis. Rb, like the genes for neurofibromatosis (neurofibromin) or breast cancer (BRCA1), acts as a tumor suppressor gene requiring mutation of both Rb alleles to inactivate Rb protein and foster continuous E2F action and cell proliferation. Individuals with germ-line Rb mutations are more likely to incur a “second hit” in several retinal cells and present with retinoblastomas in both eyes. Individuals without germline Rb mutations are much less likely to incur two Rb mutations in the same retinal cell, having unilateral tumors if they occur at all. The bcl protein, discovered by its excess in B-cell ly mphomas, was then identified as a stimulator of cell division called cyclin D1. Bcl is thus an oncogene, requiring one abnormal allele or “one hit” to be genetically activated in a cell and stimulate proliferation (oncogenesis). Several proteins of cancer-causing viruses (v-onc) have been adapted from putative cellular oncogenes (conc), while others inactivate the Rb protein and promote cell proliferation. Despite characterization of many tumor suppressor and oncogene proteins, the tissue specificity and variable ages of onset of different tumors speak for multifactorial, polygenic pathways to cancer.

Question 3

A large family has many individuals with colon cancer, affecting males and females in a manner compatible with autosomal dominant The patients do not have multiple polyps as in Gardner syndrome or adenomatosis polyposis coli (APC-MIM*175100), so that colonoscopy screening has not been helpful. The family is thought most compatible with hereditary nonpolyposis colon cancer (HNPCC-MIM*114500), and the diagnosis is confirmed by skin biopsy and fibroblast studies. Which of the following best describes the diagnostic results?

A	Defect in mismatch repair with accumulation of nucleotide repeats by replication slippage
B	Defect in double-strand break repair with accumulation of multiple small DNA fragments
C	Defect in base-excision repair with accumulation of depurinated DNA
D	Defect in nucleotide excision repair with accumulation of thymine dimers
E	Defect in nucleotide excision repair with defects in an MSH protein that is homologous to an E coli mut gene

Question 3 Explanation:

Hereditary nonpolyposis colon cancer (HNPCC-MIM*114500) is caused by defects in hMLH1 or hMSH2 genes that cause faulty mismatch repair. Mismatch repair corrects errors in newly synthesized DNA that arise from abnormal basepairing or slippage at the replication fork that introduces point mutations or small nucleotide repeats (microsatellites). The mismatch repair apparatus, first characterized as Mut S, Mut C, and Mut H proteins in E coli, match newly synthesized DNA strands with their templates relative to adenine-methylated GATC sequence sites. New strands have unmethylated GATC sites, and mismatched sequences are excised by exonuclease and restored correctly by polymerase and ligase. Human hMLH genes are homologous to E coli mut genes, but the way in which accumulated point mutations and microsatellites in colon cell DNA lead to cancer is not yet clear. It is known that accumulation of abnormal DNA inhibits the cell cycle, so perhaps the continuous demand for high-turnover intestinal cells drives some colon cells to aberrant and unregulated pathways of proliferation.

Question 4

A 25-year-old adopted female uses a private detective to locate her birth hospital and likely biologic mother. She cannot see a family resemblance in this female and becomes concerned because this female had breast cancer at age 31 with several relatives having early-onset breast cancer. Upon questioning, the female recalls that she and her roommate in the hospital were both unwed, delivering on the same day and both giving up their babies for adoption. Which of the following would be the best option for the adopted female to pursue?

A	Compare blood leucocyte DNA for variable patterns of restriction fragments arising from tandemly repeated sequences (microsatellites) between the adoptee and her possible mother.
B	Compare fragment sizes of DNA copied from blood leucocytes using reverse transcriptase between the adoptee and her possible mother.
C	Compare blood leucocyte DNA fragment size distribution as analyzed by gel electrophoresis between the adoptee and her possible mother.
D	Compare HLA gene fragments obtained from blood leucocyte DNA between the adoptee and her mother.
E	Analyze blood leucocyte DNA for breast-ovarian cancer gene mutations (MIM*113705) in the adoptee and her prospective mother.

Question 4 Explanation:

Restriction fragment length polymorphisms (RFLPs) arising from variable numbers of tandem repeats (VNTRs) are visualized by Southern blotting, yielding patterns called DNA fingerprints. If the restriction patterns from several VNTRs are used as in answer a, then differences between prospective parent and child can exclude parentage or support it (nonexclusion) at a probability equal to the number of RFLPs examined and the frequencies of the DNA fragment sizes in that population. Visualization of specific RFLP patterns using appropriate DNA probes after Southern blotting is required, since the gross distribution of DNA sizes (incorrect answer c), including that made by reverse transcriptase (incorrect answer b) will be similar in all individuals. Specific analysis of a gene encoding the highly variable HLA cell surface proteins would only provide one RFLP/gene variant for comparison (incorrect answer d), and the female would be at only 50% risk to inherit a breast-ovarian cancer gene (BRCA1 or 2-MIM*113705) from her mother even if the female were her biological mother and did have such a mutant allele (incorrect answer e). DNA fingerprinting involves (1) isolation of DNA from parent/child or forensic specimens using blood, skin, or semen; (2) PCR amplification and radioactive labeling of DNA from variable regions in each sample; (3) separation of the variable DNA fragments by gel electrophoresis; and (4) comparison of the DNA fragment patterns among samples. Since numbers of arrays of repeats of 2, 3, or 4 bp (microsatellites) may vary from 5 to 100 at a particular chromosome locus, particular alleles may occur in less than 1% of the population. As a result, analysis of three loci, each with two alleles, can produce odds as high as (100)6 that the pattern matches a putative father or suspect as compared to a random person from the general population.

Question 5

A research study examines the nature of telomeres using DNA probes specific for the 6-nucleotide 5′-TTAGGG-3′ repeats present in variable numbers at chromosome ends. The study includes children and adults with normal and abnormal development, including several patients with cancer (many genetic disorders cause both developmental disability and higher risks for cancer). Which of the following would be a likely result of this study?

A	Longer telomeres in older or disabled patients and in cancer tissues.
B	Shorter telomeres in younger patients and in cancer tissues, longer telomeres in some patients with developmental disabilities.
C	Shorter telomeres in older or disabled patients and in cancer tissues.
D	Longer telomeres in cancer tissue and single, specific telomere rearrangements in some patients with developmental disabilities.
E	Shorter telomeres in cancer tissue and single, specific telomere rearrangements in some patients’ developmental disabilities.

Question 5 Explanation:

Special DNA structures at the end of chromosomes, called telomeres, consist of repetitive DNAs including a particular 6-base pair repeat (5′-TTAGGG-3′) at each chromosome terminus. The number of 6-base pair repeats is normally in the thousands, but has been noted to shorten in tumor tissues and in normal tissues as a function of aging. Multicolor, fluorescent DNA probes to regions adjacent to telomeric repeats detect subtle deletions (missing signal) or duplications (extra signal) that are more common in these regions (subtelomere fluorescent in situ hybridization [FISH]). The described research study uses probes to the 6-bp repeats rather than the adjacent regions, and thus will show fewer repeats (shorter telomeres) in older patients and cancer tissues (incorrect answers a and d) without revealing rearrangement of adjacent DNA in select patients with disabilities (incorrect answers b and e). Subtelomere FISH studies, now being supplanted by microarray analysis, can reveal subtle chromosome aberrations in 3% to 6% of individuals with developmental disabilities who have a normal routine karyotype.

Question 6

A 22-year-old single Ashkenazi Jewish female has a strong family history of breast cancer, with her mother, maternal aunt, and grandmother all having breast cancer before menopause. She understands that the family likely exhibits autosomal dominant inheritance for predisposition to breast cancer due to mutations in the BRCA tumor suppressor genes; a pathologic BRCA mutation causes an ∼80% lifetime risk for breast and 45% risk for ovarian cancer. Her mother undergoes testing and a mutation in the BRCA1 gene (MIM*113705) is found, potentially defining her daughter’s cancer risks. Which of the following maternal test results and consequent lifetime cancer risks for the daughter are most accurate?

A	Homozygous mutation in the BRCA1 TATA box, 80% cancer risk for the daughter
B	Heterozygous mutation in the BRCA1 TATA box, 80% cancer risk for the daughter
C	Heterozygous mutation in the BRCA1 coding sequence changing the third nucleotide in a codon for glycine, 80% cancer risk in the daughter
D	Heterozygous frameshift mutation in the BRCA1 coding sequence, 40% cancer risk for the daughter
E	Homozygous frameshift mutation in the BRCA1 coding sequence, 80% cancer risk for the daughter

Question 6 Explanation:

Tumor suppressor genes like BRCA1 and 2 require expression of both alleles to inhibit cell growth and transformation to malignancy. A pathologic BRCA1 mutation would be expected to affect one of the two alleles (incorrect answers a, e) and severely impact protein expression or function (incorrect answer c—“wobble” in the nucleotide code allows for differences in the third position of codons to encodethe same amino acid). Mutations affecting the RNA polymerase-transcription factor IID binding site (TATA box) or frameshifts that garble the amino acid code would be pathologic, that is, would cause disease rather than representing harmless DNA variation (polymorphism). Since pathologic mutations for tumor suppressors will be heterozygous (homozygosity would be extremely rare), the daughter has a 50% chance to inherit the mutant allele with its 80% lifetime risk for breast cancer. This would give a joint probability of 50% to inherit the mutation and 80% lifetime cancer risk if she did = 40% risk for breast cancer (and 22.5% risk for ovarian cancer). If a BRCA1 mutation were found in the mother that was likely to be pathologic (cause her breast cancer), the daughter could have the DNA test and elect prophylactic mastectomy and oophorectomy if positive. Young women can either plan marriage and child-bearing before surgery or have eggs frozen for future use.

Question 7

Cancers may be caused by mutations in two types of genes, those that affect cell proliferation (oncogenes, tumor suppressor genes) and those that control rates of mutation (caretaker genes—ie, those altering DNA repair). A child develops a wide gait and is found to have different leg lengths. On examination, one leg is larger (hemihypertrophy) and a mass is found in the lower abdomen. Surgery reveals a Wilms tumor (MIM*194070), and the IGF2 (insulin-like growth factor 2) gene is known to be specifically overexpressed in Wilms tumor. Restriction analysis of the child’s tumor DNA showed that both IGF2 alleles were undermethylated. It is known that normal adult tissues have one methylated and one undermethylated IGF2 allele, while the single IGF2 allele in sperm DNA is always undermethylated. The results are best summarized as which of the following?

A	A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies IGF2 as a caretaker gene
B	A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies IGF2 as an oncogene
C	A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies it as a tumor suppressor gene
D	A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and causes Wilms tumor in females only
E	A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and causes Wilms tumor in males only

Question 7 Explanation:

The undermethylation of IGF2 alleles in sperm as compared to methylated and undermethylated alleles in diploid adult tissues suggests that the gene is imprinted. During germ cell formation, certain chromosome regions are patterned differently or “imprinted” in males versus females, producing different levels of maternal versus paternal allele expression in tissues of the resulting embryos and adults. These imprinting patterns are associated with DNA methylation at key CpG dinucleotide sites, up- or down-regulating allele transcription and expression. Biparental origin of alleles in diploid tissues is thus necessary for normal levels of gene expression, a balance disrupted by uniparental disomy (both copies of a chromosome and its alleles from one parent) or loss of imprinting (LOI) of one allele. Imprinted genes are often growth factors, perhaps arising to regulate embryo-fetal growth but persisting in diploid tissues as potential sources of cell proliferation and tumors. The specific increased IGF2 expression in Wilms tumor suggests it is a potential regulator of cell proliferation rather than rates of mutation. The fact that one allele can increase expression and cause a tumor suggests IGF2 is an oncogene rather than a tumor suppressor that must have both alleles inactivated to produce a tumor.

Question 8

An abdominal mass is palpated during a well visit for a 15-month-old African American boy with a normal past medical history. Urinalysis demonstrates blood cells and other cells of unusual morphology, suggesting a diagnosis of Wilms tumor (MIM*194070). The physicians suspected erosion of surrounding kidney with invasion of the renal pelvis, causing tumor cells to be excreted in urine. Because a version of the p57 gene is known to be silenced preferentially in Wilms tumor, analysis of p57 gene methylation and expression was performed on the urine tumor cells with the results shown below (normal cells-N, tumor cells-T). Which of the following best summarizes these results?

A	Bi-allelic p57 gene methylation and with normal p57 gene expression making a diagnosis of Wilms tumor unlikely
B	Bi-allelic p57 gene methylation and with increased p57 gene expression making a diagnosis of Wilms tumor unlikely
C	Bi-allelic p57 gene methylation and with decreased p57 gene expression supporting the diagnosis of Wilms tumor
D	Undermethylation of both p57 alleles with increased p57 gene expression supporting the diagnosis of Wilms tumor
E	Methylation of one p57 allele with increased p57 gene expression supporting the diagnosis of Wilms tumor

Question 8 Explanation:

Southern analysis using methylation-sensitive or insensitive endonuclease restriction in the first two panels of the figure suggests both p57 alleles in the tumor are methylated. Only the larger fragment is detected in the patient sample when restricted with methylation-sensitive endonuclease (second panel). The accompanying Northern blot shows markedly decreased amounts of p57 mRNA, supporting a connection between p57 allele methylation and silencing of expression. The maternal allele of p57 is undermethylated and expressed in normal tissues, and its loss of imprinting (methylation) is one of the few situations where epigenetic silencing appears to be the sole carcinogenic event

Question 9

A 50-year-old Caucasian male presents with a markedly enlarged tonsil and recurrent infections with serum immunoglobulin deficiency. He has enlarged lymph nodes along his neck. Chromosome analysis demonstrates a translocation between the immunoglobulin heavy-chain locus on chromosome 14 and an unidentified gene on chromosome 8. Which of the following is the most likely cause of his phenotype?

A	The translocation has deleted constant chain exons on chromosome 14 and prevented heavy-chain class switching.
B	The translocation has deleted the interval containing diversity (D) and joining (J) regions.
C	The translocation has activated a tumor-promoting gene on chromosome 8.
D	The translocation has deleted the heavy-chain constant chain Cμ so that virgin B-cells cannot produce IgM on their membranes.
E	The translocation has deleted an immunoglobulin transcription factor gene on chromosome 8.

Question 9 Explanation:

This case is an example of Burkitt lymphoma, which may affect the tonsils or other lymphoid tissues. The translocation places the myc oncogene on chromosome 8 downstream of the very active heavy-chain locus on chromosome 14, activating myc gene expression in B cells and their derivatives. The translocation is likely an aberrant form of the normal DNA rearrangements that generate unique heavy-chain genes in each B cell. The translocation joins one chromosome 8 to one chromosome 14, leaving their homologs unaffected. The cause for the phenotype must therefore be transacting, since cis-acting effects would pertain only to the translocated loci and not affect the homologous untranslocated loci. Activation of a tumor-promoting gene (oncogene) on chromosome 8 could produce an enlarged tonsil, while underactivity of immunoglobulin production due to one-half expression could decrease immune function but would not completely ablate the processes in choices a, b, d, and e. At the genetic level, transacting events are autosomal dominant in that one of the two homologous loci is abnormal and produces a phenotype. Mutations of cis-acting events must disrupt both homologous loci to produce phenotypes, making them autosomal recessive at the genetic level.

Question 10

The Warburg effect refers to the seemingly paradoxical preference for glycolysis over oxidative phosphorylation in many cancer tissues. Which of the following statements best reflects this paradox?

A	Glycolysis consumes 4 ATP early in the pathway, produces 2 ATP later, and the net of -2 ATP per glucose is less than the 2 ATP produced per oxidation of citrate.
B	Glycolysis consumes 2 ATP early in the pathway, produces 4 ATP later, and the net of 2 ATP per glucose is less than 1/6 of that produced by oxidative phosphorylation.
C	Glycolysis consumes 4 ATP early in the pathway, produces 6 ATP later, and the net of 2 ATP per glucose is less than 1/6 of that produced by oxidative phosphorylation.
D	Glycolysis consumes 2 ATP early in the pathway, produces 4 ATP later, and the net of 2 ATP per glucose is less than the 3 produced by oxidative phosphorylation.
E	Glycolysis consumes 6 ATP early in the pathway, produces 12 ATP later, and the net of 6 ATP per glucose is less than the 8 produced by oxidative phosphorylation.

Question 10 Explanation:

Two ATPs are expended in the first steps of glycolysis and 4 ATPs generated by later steps, yielding a net of 2 ATP molecules generated by glycolysis of one glucose molecule (incorrect answers a, c, and e). The citric acid cycle produces 3 ATP from each of three molecules of NADH produced, two from a molecule of FADH2, and 1 ATP at substrate level from conversion of succinyl-CoA to succinate (12 ATP total— incorrect answer d). The rapid proliferation and high density of cancer cells, along with need for rapid cytoplasmic energy conversion (as in muscle), may explain their preference for glycolysis. ATP hydrolysis occurs during phosphorylation of glucose to glucose 6-phosphate by glucokinase (hexokinase) and phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate by phosphofructokinase. ATP is generated during conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate by phosphoglycerate kinase and in the conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase. Since two molecules of 1,3-bisphosphoglycerate are generated from one glucose molecule (and subsequently two molecules of phosphoenolpyruvate are generated), each of these steps results in generation of 2 ATPs.

Question 11

A 45-year-old Caucasian female seeks evaluation for vague complaints of “tired blood” and “rheumatism.” Her physician first considers chronic fatigue syndrome or depression, but performs a complete blood count (CBC) showing a hemoglobin of 9.0 g/dL (normal mean 14), white blood cell count of 3000/mm3 (normal mean 7.4), and platelet count of 30,000/mm3 (normal 150-350). A diagnosis of aplastic anemia is made and bone marrow chromosome studies initiated with knowledge that aplastic anemia can be a preleukemic state. A previously unreported chromosome translocation is found that suggests the female has a very unusual leukemia. The hematology-oncology consultant has no protocols gleaned from experience with like patients to rely on, but wishes to try a metabolic treatment that will inhibit leukemic cell division without using harsh chemotherapy poisons that could further damage the bone marrow. Which of the following strategies would cause the greatest inhibition of DNA synthesis?

A	Lowering aspartate
B	Lowering carbamoyl phosphate
C	Increasing uric acid
D	Increasing glutamine
E	Increasing orotic acid

Question 11 Explanation:

Glutamine and aspartate participate in both purine and pyrimidine synthesis, so increasing glutamine would not inhibit DNA synthesis (incorrect answer d). Carbamoyl phosphate synthase II (as opposed to CPS I in the urea cycle) initiates pyrimidine synthesis by producing carbamoyl phosphate that adds to aspartic acid and eventually forms orotic acid (incorrect answer e). Purine synthesis begins with ribose 5- phosphate and phosphoribosyl pyrophosphate (PRPP) via PRPP synthase, so decreasing CPS II would not effect purine synthesis (incorrect answer b). Uric acid is formed from purines through xanthine and xanthine oxidase, and its excess might produce some feedback inhibition for PRPP synthase that is inhibited by both purines and pyrimidines to help coordinate their synthesis and equivalent incorporation into DNA. However, the main effect of uric acid excess is deposit of the chemical in joint spaces to cause the severe pain of gout (incorrect answer c). During purine ring biosynthesis, the amino acid glycine is completely incorporated to provide C4, C5, and N7. Glutamine contributes N3 and N9, aspartate provides N1, and derivatives of tetrahydrofolate furnish C2 and C8. Carbon dioxide is the source of C6. In pyrimidine ring synthesis, C2 and N3 are derived from carbamoyl phosphate, while N1, C4, C5, and C6 come from aspartate.

Question 12

One of the first chemotherapeutic agents was methotrexate, a compound that was effective in killing rapidly dividing cells like those of This compound would be expected to elevate concentrations of which of the following compounds?

A	Homocysteine and dUMP
B	Thymine and choline
C	Serine and methionine
D	Glycine and methionine
E	Homocysteine and thymine

Question 12 Explanation:

The vitamin folic acid is provided commercially and pharmaceutically as the stable 5- formyltetrahydrofolate known as folinic acid or its synthetic analog leucovorin. Addition of folate to foods (bread) and encouragement of preconceptional vitamins with folate was prompted by its ability to lower the incidence of neural tube defects by two- to threefold. Folic acid is biologically active as the interconvertible forms tetrahydrofolate (THF), methyl THF, and other methylated forms (methylene THF, N5, N10-methylene THF) that are important for one-carbon (methyl) transfers and interconversions (glycine-serine, formate-formylmethionine, formate-CO2 homocysteine-methionine, and uracil-thymine). THF is required in two steps of purine synthesis and thus required in the de novo synthesis of ATP and GTP. Although de novo synthesis of the pyrimidine ring does not require tetrahydrofolate, the methylation of deoxyuridine monophosphate (dUMP) to form thymine from uracil does. In this thymidylate synthetase reaction, methylene THF donates a methyl group and is converted to dihydrofolate, which requires action of dihydrofolate reductase to regenerate THF. Methotrexate inhibits dihydrofolate reductase, depletes THF pools, and thus would elevate substrates of enzymes dependent on this cofactor like dUMP. THF is also a cofactor for methionine synthase that coverts homocysteine to methionine, an enzyme deficient in one form of homocystinuria (MIM*236200).

Question 13

A 35-year-old Caucasian female notes a lump in her breast and undergoes biopsy that shows breast cancer. The biopsy tissue is sectioned and shown to react with an antibody to the epidermal growth factor (EGF) receptor-2, also called Her-2. Which of the following describes the most likely treatment strategy for this female?

A	Intravenous infusion of EGF to activate signal transduction in normal breast tissue
B	Intravenous infusion of EGF antibody to block signal transduction in breast cancer tissue
C	Oral dosing of EGF to activate signal transduction in normal and breast cancer tissue
D	Oral dosing of EGFR2 antibody (Herceptin) to block signal transduction in normal breast tissue
E	Intravenous infusion of EGFR2 antibody (Herceptin) to block signal transduction in breast cancer tissue

Question 13 Explanation:

Epidermal growth factor (EGF) can interact with four receptors including EGF receptor-2 (Her2) as shown in Fig. 20 of the High-Yield Facts. The resulting signal transduction cascade promotes cell survival and growth, so provision of EGF (intravenously since oral proteins will be degraded in the intestine) would promote growth of normal and cancer cells (incorrect answers a-d). Cancer cells are now tested for protein or microRNA markers that can forecast prognosis and responses to therapy. Some marker proteins like the Her-2 receptor in breast cancer become targets for chemotherapy, illustrated by Herceptin in breast cancer and Gleevec inhibitor of the fusion oncogenic protein in chronic myelogenous leukemia (MIM*608232).

Question 14

Gardner syndrome or adenomatous polyposis of the colon (MIM*175100) is an autosomal dominant condition characterized by retinal changes, extra teeth, skin cysts, and multiple colonic polyps with high rates of malignant transformation. A family is encountered in which a great-grandfather, grandmother, and father are affected with Gardner syndrome and develop intestinal cancer in their thirties. The father brags that none of his four children have inherited Gardner syndrome because they lack skin cysts and have not had cancer. The chance that at least one child has inherited the Gardner syndrome allele and the reason the children have not manifested cancer are which of the following?

A	¼, ascertainment bias
B	½, variable cancer predisposition
C	¾, early-onset disease manifestation
D	1/16, incomplete medical evaluation
E	15/16, later-onset disease manifestation

Question 14 Explanation:

The father is affected with Gardner syndrome (MIM*175100), an autosomal dominant disease. Therefore, each of his four children has a half chance of receiving the allele that causes Gardner syndrome and a half chance of receiving the normal allele. The probability that none of his four children received the allele for Gardner syndrome is thus the joint probability of four independent events, computed by the product ½ × ½ × ½ × ½ = 1/16. The probability that at least one child has received the abnormal Gardner syndrome allele is thus 1 – 1/16 = 15/16. Gardner syndrome is one of many genetic disorders that may not be obvious in early childhood. Intestinal cancer in particular has a later onset, with 50% of patients being affected by age 30 to 35. More extensive evaluation of the children for internal signs of disease (eg, the bony tumors) is required before the father can conclude that he has not transmitted the gene. The evaluations would be important; many patients choose colonectomy to avoid lethal colonic cancers. A recent option for the family would be DNA diagnosis for mutations in the adenomatous polyposis coli (APC) gene. Presymptomatic DNA diagnosis offers a new approach to genetic diseases of later onset, but is ethically controversial when minors are involved.

Question 15

Studies of the eye tumor retinoblastoma have revealed an Rb locus on the long arm of chromosome 13 that influences retinoblastoma occurrence. Patients with retinoblastoma 13q-deletions often develop bilateral tumors (both sides), in contrast to more common forms of retinoblastoma that occur at one site. Which of the following phrases best explains this phenomenon?

A	Rb is an oncogene.
B	Rb is a tumor suppressor gene.
C	Rb mutations ablate a promoter sequence.
D	Rb mutations ablate an enhancer sequence.
E	Rb mutations must always involve chromosome abnormalities.

Question 15 Explanation:

The two-hit hypothesis was developed by Knudsen to explain why patients with hereditary retinoblastoma (germ-line mutations [MIM*180200]) have multiple, bilateral tumors while those with sporadic tumors (no family history) have single tumors. A germ-line mutation (first hit) alters one Rb allele and confers enhanced susceptibility to retinoblastoma. A somatic mutation (second hit) inactivating the other homologous Rb allele can then occur in any tissue. If it occurs in the retina, a tumor is born. The multiple tumors thus represent the sites at which somatic mutations have occurred in the retina. In sporadic cases, two somatic mutational events must take place. Since these somatic mutations are relatively rare events, it is extremely uncommon for more than one tumor to develop. It is curious that, although retinoblastoma susceptibility is inherited in a dominant fashion, tumor development is a recessive event, requiring the inactivation of both alleles. Genes such as Rb are called tumor suppressor genes, in contrast to oncogenes, in which only one of the two homologous alleles must be altered to initiate malignant transformation. Alteration of an enhancer or promoter site on one Rb allele would thus not be sufficient to cause cancer, since the other Rb allele would not be affected. Obvious chromosome changes such as 13q—are rare compared to other mutations that alter Rb function.

Question 16

In Burkitt lymphoma, there is increased expression of a hybrid protein with an amino-terminus similar to immunoglobulin (Ig) heavy chain and an unknown carboxy-terminus. Which of the following best explains this phenomenon?

A	Chromosome translocation that brings together an Ig heavy chain with an oncogene
B	Chromosome duplication involving a segment with an oncogene
C	Chromosome translocation involving a segment with Ig heavy chains
D	Chromosome deletion removing an oncogene
E	Chromosome deletion removing a tumor suppressor gene

Question 16 Explanation:

Chromosome translocations may often promote tumors in somatic cells by placing regulatory genes next to promoters that aberrantly increase their expression. Burkitt lymphoma, a B-cell lymphoma that usually occurs in childhood, often involves reciprocal translocation of chromosomes 8 and 14. The result of this is to place the c-myc proto-oncogene from 8q24 into the immunoglobulin heavychain locus at 14q32. Because immunoglobulin genes are actively transcribed, this move alters the normal regulatory control of c-myc. Another example is the Philadelphia chromosome, a shortened chromosome 22 caused by translocation t(9:22)(q34:q11). This translocation is seen in almost all patients with chronic myelogenous leukemia (CML) and in a percentage of patients with acute lymphoblastic leukemia (ALL). The Philadelphia chromosome is seen with increased frequency in individuals with Down syndrome.

Question 17

A couple request genetic counseling because the wife has contracted early-onset breast cancer at age The husband has a benign family history, but the wife has several relatives who developed cancers at relatively early ages. Affected relatives include a sister (colon cancer, age 42), a brother (colon cancer, age 46), mother (breast cancer, age 56), maternal aunt (leukemia, age 45), maternal uncle (muscle sarcoma, age 49), and a nephew through the brother with colon cancer (leukemia, age 8). Which of the following is an accurate conclusion from the family history?

A	No genetic predisposition to cancer, since most individuals have different types of cancer
B	Possible autosomal dominant inheritance or multifactorial inheritance of cancer predisposition
C	Germ-line mutations in an oncogene, with somatic mutations that suppress the oncogene
D	Germ-line mutations in a tumor suppressor gene, with neoplasia from chemical exposure
E	Mitochondrial inheritance of tumor predisposition evidenced by the affected maternal relatives

Question 17 Explanation:

Genetic predisposition to cancer is best understood by the Knudsen hypothesis, where two independent mutations or “hits” are required to produce neoplasia of a somatic tissue. In many hereditary cancers, the “first hit” is a germ-line mutation that is transmitted in families. Individuals who inherit this mutation are much more likely to develop cancer through a “second hit” in their somatic cells. The second hit can be any mutation that removes the homologous allele (loss of heterozygosity); mechanisms include missense mutation, chromosome deletion, and chromosome nondisjunction. For tumor suppressor genes like those responsible for neurofibromatosis 1 (MIM*162200) or the Li-Fraumeni syndrome (MIM*114480), the first hit removes one suppressor allele and the second hit removes the homologous suppressor allele. The family in the question is an example of a “cancer family” that exhibits the bone, breast, colon, and blood cancers that are typical of Li-Fraumeni syndrome. The mechanism involves mutations in the src tumor suppressor gene.

Question 18

A normal 6-year-old girl has a strong family history of cancer, including several relatives with Li-Fraumeni syndrome, an autosomal dominant condition that predisposes to breast and colon cancer. Her parents request that she have genetic testing for a possible cancer gene. Which of the following is the major ethical concern about such testing?

A	Nonmaleficence
B	Beneficence
C	Autonomy
D	Informed consent
E	Confidentiality

Question 18 Explanation:

Presymptomatic DNA testing of individuals in cancer families is increasingly available. However, testing of minors is controversial because they may not be old or mature enough to understand the personal, medical, and financial implications. They, therefore, cannot give truly informed consent. Beneficence is the ethical imperative to do good for patients, while nonmaleficence is the imperative to do no harm. Autonomy refers to a patient’s right to make decisions regarding his or her health care, and confidentiality to the privilege of doctor-patient communication.

Question 19

Deletions of 11p13 may result in Wilm tumor, aniridia, genitourinary malformations, and mental retardation (WAGR syndrome). In some patients, however, not all features are seen. Additionally, individual features of this syndrome may be inherited separately in a Mendelian fashion. Limited features may also be seen in patients without visible chromosomal deletions. Which of the following is the most likely mechanism for this finding?

A	Mitochondrial inheritance
B	Imprinting
C	Germ-line mosaicism
D	Uniparental disomy
E	Contiguous gene syndrome

Question 19 Explanation:

Contiguous gene syndromes, also known as microdeletion syndromes, occur when deletions result in the loss of several different closely linked loci. Depending on the size of the deletion, different phenotypes may result. Mutations in the individual genes may result in isolated features that may be inherited in a Mendelian fashion.

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1. A 25-year-old Caucasian female is evaluated because her father, paternal grandfather, and uncle developed colon cancer before age Examination has revealed an unusually dark complexion similar to that of her father and supernumerary (extra) teeth with multiple caries. Her past medical history indicates she had bony cysts noted on her tibia after x-rays for a fracture, and her physician suspects a diagnosis of autosomal dominant Gardner syndrome (MIM*175100) that has dark complexion, tooth changes, bony tumors (osteomas), and multiple colonic polyps with high rates of colon cancer. DNA testing for the causative adenomatous polyposis coli (APC) gene revealed a nucleotide substitution that changed the DNA sequence from CAGAGGT to CAGGGGT. This mutation ablated the splice junction at the border of exon 7 and its adjacent intron. A partial sequencing result is shown in the figure below, and supports which of the following conclusions about the female’s APC genotype and expression?

Homozygous for the normal Gardner allele, some smaller mRNA molecules
Heterozygous for normal and mutant Gardner allele, some larger mRNA molecules
Heterozygous for normal and mutant Gardner allele, some smaller mRNA molecules
Homozygous for mutant Gardner allele, no change in mRNA molecules
Homozygous for mutant Gardner allele, some larger mRNA molecules

2. A parent is correcting his child’s photograph for red eye and notes that one of the child’s pupils reflects the flash as white rather than red. An ophthalmologist confirms the presence of a tumor in the back of the white-reflecting eye, telling the parents about the possibility of retinoblastoma (MIM* 180200). The parents return to their pediatrician, confused about the relation of retinoblastoma (Rb) and B-cell lymphoma (bcl) genes they saw on the Internet and the possibility that their child’s tumor is Which of the following is the most appropriate response?

Rb and bcl proteins are polymerases that prevent oncogenesis by stringent DNA repair; the parents are therefore carriers for autosomal recessive Rb deficiency with a one in four recurrence
Rb protein binds transcription factors needed for cell division and bcl protein (cyclin D1) stimulates it; Rb is a tumor suppressor gene requiring homozygous mutations (two hits) that are likely sporadic in a child with unilateral
Rb and bcl proteins are DNA-binding factors that suppress promoters near oncogenes and act as tumor suppressors; the child represents a new, dominant-acting mutation and the parents have a minimal recurrence
Rb protein stimulates cyclins specific for retinal tissue and bcl does the same for lymphatic cyclins; the child represents a new mutation with excess Rb
Rb protein forms complexes with bcl protein that promotes cell division in rapidly proliferating tissues; the child represents a new mutation with excess Rb-bcl complex

3. A large family has many individuals with colon cancer, affecting males and females in a manner compatible with autosomal dominant The patients do not have multiple polyps as in Gardner syndrome or adenomatosis polyposis coli (APC-MIM*175100), so that colonoscopy screening has not been helpful. The family is thought most compatible with hereditary nonpolyposis colon cancer (HNPCC-MIM*114500), and the diagnosis is confirmed by skin biopsy and fibroblast studies. Which of the following best describes the diagnostic results?

Defect in mismatch repair with accumulation of nucleotide repeats by replication slippage
Defect in double-strand break repair with accumulation of multiple small DNA fragments
Defect in base-excision repair with accumulation of depurinated DNA
Defect in nucleotide excision repair with accumulation of thymine dimers
Defect in nucleotide excision repair with defects in an MSH protein that is homologous to an E coli mut gene

4. A 25-year-old adopted female uses a private detective to locate her birth hospital and likely biologic She cannot see a family resemblance in this female and becomes concerned because this female had breast cancer at age 31 with several relatives having early-onset breast cancer. Upon questioning, the female recalls that she and her roommate in the hospital were both unwed, delivering on the same day and both giving up their babies for adoption. Which of the following would be the best option for the adopted female to pursue?

Compare blood leucocyte DNA for variable patterns of restriction fragments arising from tandemly repeated sequences (microsatellites) between the adoptee and her possible
Compare fragment sizes of DNA copied from blood leucocytes using reverse transcriptase between the adoptee and her possible
Compare blood leucocyte DNA fragment size distribution as analyzed by gel electrophoresis between the adoptee and her possible
Compare HLA gene fragments obtained from blood leucocyte DNA between the adoptee and her
Analyze blood leucocyte DNA for breast-ovarian cancer gene mutations (MIM*113705) in the adoptee and her prospective

5. A research study examines the nature of telomeres using DNA probes specific for the 6-nucleotide 5′-TTAGGG-3′ repeats present in variable numbers at chromosome ends. The study includes children and adults with normal and abnormal development, including several patients with cancer (many genetic disorders cause both developmental disability and higher risks for cancer). Which of the following would be a likely result of this study?

Longer telomeres in older or disabled patients and in cancer
Shorter telomeres in younger patients and in cancer tissues, longer telomeres in some patients with developmental
Shorter telomeres in older or disabled patients and in cancer
Longer telomeres in cancer tissue and single, specific telomere rearrangements in some patients with developmental
Shorter telomeres in cancer tissue and single, specific telomere rearrangements in some patients’ developmental

6. A 22-year-old single Ashkenazi Jewish female has a strong family history of breast cancer, with her mother, maternal aunt, and grandmother all having breast cancer before menopause. She understands that the family likely exhibits autosomal dominant inheritance for predisposition to breast cancer due to mutations in the BRCA tumor suppressor genes; a pathologic BRCA mutation causes an ∼80% lifetime risk for breast and 45% risk for ovarian cancer. Her mother undergoes testing and a mutation in the BRCA1 gene (MIM*113705) is found, potentially defining her daughter’s cancer Which of the following maternal test results and consequent lifetime cancer risks for the daughter are most accurate?

Homozygous mutation in the BRCA1 TATA box, 80% cancer risk for the daughter
Heterozygous mutation in the BRCA1 TATA box, 80% cancer risk for the daughter
Heterozygous mutation in the BRCA1 coding sequence changing the third nucleotide in a codon for glycine, 80% cancer risk in the daughter
Heterozygous frameshift mutation in the BRCA1 coding sequence, 40% cancer risk for the daughter
Homozygous frameshift mutation in the BRCA1 coding sequence, 80% cancer risk for the daughter

7. Cancers may be caused by mutations in two types of genes, those that affect cell proliferation (oncogenes, tumor suppressor genes) and those that control rates of mutation (caretaker genes—ie, those altering DNA repair). A child develops a wide gait and is found to have different leg On examination, one leg is larger (hemihypertrophy) and a mass is found in the lower abdomen. Surgery reveals a Wilms tumor (MIM*194070), and the IGF2 (insulin-like growth factor 2) gene is known to be specifically overexpressed in Wilms tumor. Restriction analysis of the child’s tumor DNA showed that both IGF2 alleles were undermethylated. It is known that normal adult tissues have one methylated and one undermethylated IGF2 allele, while the single IGF2 allele in sperm DNA is always undermethylated. The results are best summarized as which of the following?

A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies IGF2 as a caretaker gene
A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies IGF2 as an oncogene
A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and identifies it as a tumor suppressor gene
A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and causes Wilms tumor in females only
A loss of imprinting (LOI) mutation that increased tumor IGF2 expression and causes Wilms tumor in males only

8. An abdominal mass is palpated during a well visit for a 15-month-old African American boy with a normal past medical Urinalysis demonstrates blood cells and other cells of unusual morphology, suggesting a diagnosis of Wilms tumor (MIM*194070). The physicians suspected erosion of surrounding kidney with invasion of the renal pelvis, causing tumor cells to be excreted in urine. Because a version of the p57 gene is known to be silenced preferentially in Wilms tumor, analysis of p57 gene methylation and expression was performed on the urine tumor cells with the results shown below (normal cells-N, tumor cells-T). Which of the following best summarizes these results?

Bi-allelic p57 gene methylation and with normal p57 gene expression making a diagnosis of Wilms tumor unlikely
Bi-allelic p57 gene methylation and with increased p57 gene expression making a diagnosis of Wilms tumor unlikely
Bi-allelic p57 gene methylation and with decreased p57 gene expression supporting the diagnosis of Wilms tumor
Undermethylation of both p57 alleles with increased p57 gene expression supporting the diagnosis of Wilms tumor
Methylation of one p57 allele with increased p57 gene expression supporting the diagnosis of Wilms tumor

9. A 50-year-old Caucasian male presents with a markedly enlarged tonsil and recurrent infections with serum immunoglobulin He has enlarged lymph nodes along his neck. Chromosome analysis demonstrates a translocation between the immunoglobulin heavy-chain locus on chromosome 14 and an unidentified gene on chromosome 8. Which of the following is the most likely cause of his phenotype?

The translocation has deleted constant chain exons on chromosome 14 and prevented heavy-chain class
The translocation has deleted the interval containing diversity (D) and joining (J)
The translocation has activated a tumor-promoting gene on chromosome
The translocation has deleted the heavy-chain constant chain Cμ so that virgin B-cells cannot produce IgM on their
The translocation has deleted an immunoglobulin transcription factor gene on chromosome

10. The Warburg effect refers to the seemingly paradoxical preference for glycolysis over oxidative phosphorylation in many cancer Which of the following statements best reflects this paradox?

Glycolysis consumes 4 ATP early in the pathway, produces 2 ATP later, and the net of -2 ATP per glucose is less than the 2 ATP produced per oxidation of citrate.
Glycolysis consumes 2 ATP early in the pathway, produces 4 ATP later, and the net of 2 ATP per glucose is less than 1/6 of that produced by oxidative
Glycolysis consumes 4 ATP early in the pathway, produces 6 ATP later, and the net of 2 ATP per glucose is less than 1/6 of that produced by oxidative
Glycolysis consumes 2 ATP early in the pathway, produces 4 ATP later, and the net of 2 ATP per glucose is less than the 3 produced by oxidative
Glycolysis consumes 6 ATP early in the pathway, produces 12 ATP later, and the net of 6 ATP per glucose is less than the 8 produced by oxidative

11. A 45-year-old Caucasian female seeks evaluation for vague complaints of “tired blood” and “rheumatism.” Her physician first considers chronic fatigue syndrome or depression, but performs a complete blood count (CBC) showing a hemoglobin of 9.0 g/dL (normal mean 14), white blood cell count of 3000/mm3 (normal mean 4), and platelet count of 30,000/mm3 (normal 150-350). A diagnosis of aplastic anemia is made and bone marrow chromosome studies initiated with knowledge that aplastic anemia can be a preleukemic state. A previously unreported chromosome translocation is found that suggests the female has a very unusual leukemia. The hematology-oncology consultant has no protocols gleaned from experience with like patients to rely on, but wishes to try a metabolic treatment that will inhibit leukemic cell division without using harsh chemotherapy poisons that could further damage the bone marrow. Which of the following strategies would cause the greatest inhibition of DNA synthesis?

Lowering aspartate
Lowering carbamoyl phosphate
Increasing uric acid
Increasing glutamine
Increasing orotic acid

12. One of the first chemotherapeutic agents was methotrexate, a compound that was effective in killing rapidly dividing cells like those of This compound would be expected to elevate concentrations of which of the following compounds?

Homocysteine and dUMP
Thymine and choline
Serine and methionine
Glycine and methionine
Homocysteine and thymine

13. A 35-year-old Caucasian female notes a lump in her breast and undergoes biopsy that shows breast The biopsy tissue is sectioned and shown to react with an antibody to the epidermal growth factor (EGF) receptor-2, also called Her-2. Which of the following describes the most likely treatment strategy for this female?

Intravenous infusion of EGF to activate signal transduction in normal breast tissue
Intravenous infusion of EGF antibody to block signal transduction in breast cancer tissue
Oral dosing of EGF to activate signal transduction in normal and breast cancer tissue
Oral dosing of EGFR2 antibody (Herceptin) to block signal transduction in normal breast tissue
Intravenous infusion of EGFR2 antibody (Herceptin) to block signal transduction in breast cancer tissue

14. Gardner syndrome or adenomatous polyposis of the colon (MIM*175100) is an autosomal dominant condition characterized by retinal changes, extra teeth, skin cysts, and multiple colonic polyps with high rates of malignant A family is encountered in which a great-grandfather, grandmother, and father are affected with Gardner syndrome and develop intestinal cancer in their thirties. The father brags that none of his four children have inherited Gardner syndrome because they lack skin cysts and have not had cancer. The chance that at least one child has inherited the Gardner syndrome allele and the reason the children have not manifested cancer are which of the following?

¼, ascertainment bias
½, variable cancer predisposition
¾, early-onset disease manifestation
1/16, incomplete medical evaluation
15/16, later-onset disease manifestation

15. Studies of the eye tumor retinoblastoma have revealed an Rb locus on the long arm of chromosome 13 that influences retinoblastoma Patients with retinoblastoma 13q-deletions often develop bilateral tumors (both sides), in contrast to more common forms of retinoblastoma that occur at one site. Which of the following phrases best explains this phenomenon?

Rb is an
Rb is a tumor suppressor
Rb mutations ablate a promoter
Rb mutations ablate an enhancer
Rb mutations must always involve chromosome

16. In Burkitt lymphoma, there is increased expression of a hybrid protein with an amino-terminus similar to immunoglobulin (Ig) heavy chain and an unknown carboxy-terminus. Which of the following best explains this phenomenon?

Chromosome translocation that brings together an Ig heavy chain with an oncogene
Chromosome duplication involving a segment with an oncogene
Chromosome translocation involving a segment with Ig heavy chains
Chromosome deletion removing an oncogene
Chromosome deletion removing a tumor suppressor gene

17. A couple request genetic counseling because the wife has contracted early-onset breast cancer at age The husband has a benign family history, but the wife has several relatives who developed cancers at relatively early ages. Affected relatives include a sister (colon cancer, age 42), a brother (colon cancer, age 46), mother (breast cancer, age 56), maternal aunt (leukemia, age 45), maternal uncle (muscle sarcoma, age 49), and a nephew through the brother with colon cancer (leukemia, age 8). Which of the following is an accurate conclusion from the family history?

No genetic predisposition to cancer, since most individuals have different types of cancer
Possible autosomal dominant inheritance or multifactorial inheritance of cancer predisposition
Germ-line mutations in an oncogene, with somatic mutations that suppress the oncogene
Germ-line mutations in a tumor suppressor gene, with neoplasia from chemical exposure
Mitochondrial inheritance of tumor predisposition evidenced by the affected maternal relatives

18. A normal 6-year-old girl has a strong family history of cancer, including several relatives with Li-Fraumeni syndrome, an autosomal dominant condition that predisposes to breast and colon cancer. Her parents request that she have genetic testing for a possible cancer Which of the following is the major ethical concern about such testing?

Nonmaleficence
Beneficence
Autonomy
Informed consent
Confidentiality

19. Deletions of 11p13 may result in Wilm tumor, aniridia, genitourinary malformations, and mental retardation (WAGR syndrome). In some patients, however, not all features are seen. Additionally, individual features of this syndrome may be inherited separately in a Mendelian fashion. Limited features may also be seen in patients without visible chromosomal deletions. Which of the following is the most likely mechanism for this finding?

Mitochondrial inheritance
Imprinting
Germ-line mosaicism
Uniparental disomy
Contiguous gene syndrome

[MCQs] Cancer – Oncology Quiz (19 test)

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Oncology (19 test)

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