* Genetics - Extra Questions

*The answer is B.* This child has many of the characteristic features of Down syndrome (DS). The abnormalities found in DS result from extra genetic material from chromosome 21. Three cytogenetic abnormalities can lead to DS: 1. Meiotic nondisjunction accounts for nearly 95% of DS cases. Meiotic nondisjunction (failure of homologous chromosomes to separate during meiosis) of chromosome 21 occurs in the ovum, resulting in the inheritance of 3 copies in one daughter cell (trisomy) and 1 copy in the other daughter coil (monosomy). Nondisjunction is almost always of maternal origin, and increased maternal age is a risk factor. 2. Unbalanced Robertsonian translocations account for 2%-3% of DS cases. These individuals have 46 chromosomes, but an extra arm of chromosome 21 is attached to another chromosome (translocation). Approximately one-third of these cases are due to a balanced translocation in one parent. These balanced translocations are associated with high recurrence risk. Genetic counseling for the parents is indicated if a translocation is identified in the infant. 3. Mosaicism accounts for <2% of cases. Affected individuals have 2 cell: lines 1 with a normal genotype, and 1 with trisomy 21. The proportion of affected cells determines the severity of DS features. *Educational Objective:* Common findings in Down syndrome include cognitive impairment, facial dysmorphism. and cardiac defects; 95% of cases are caused by the presence of an extra chromosome 21 (trisomy) resulting from nondisjunction. Unbalanced Robertsonian translocations or mosaicism are less common causes.

A 1-hour-old girl born to a 40-year-old woman is brought to the nursery for evaluation. The pregnancy and delivery were uncomplicated. Physical examination shows mid-face hyperplasia with a flat mud bridge, up-slenting palpebral fissures, a small mouth, and a single palmer crease bilaterally. Cardiac auscultation reveals a blowing holosystolic murmur heard best along the eternal border. Which of the following mechanisms is the most likely cause of this infant's clinical findings? A Genomic imprinting B. Mosaicism C. Partial deletion D. Triplet expansion E. Uniparental disomy

*The answer is B.* Using the Hardy-Weinberg equilibrium, q2 (the disease frequency) is equal to 1 in 2,500, so q equals 1 in 50. The heterozygote (carrier) frequency is 2pq (q is 1 in 50, and p is very close to 1), or 1 in 25. The disease the child is exhibiting is CF, and the population frequencies are for those individuals of northern European heritage.

A 1-year-old boy has been hospitalized twice for lung infections, was often wheezing and out of breath, and was lagging in both height and weight in his growth charts, despite a very good appetite. A subsequent test found that his sweat contained elevated levels of chloride ions. If the frequency of affected individuals for this disease is 1 in 2,500 in a specific population, what is the frequency of carriers within that population? (A) 1 in 10 (B) 1 in 25 (C) 1 in 50 (D) 1 in 100 (E) 1 in 1,250

*The answer is B.* The boy has the classic symptoms of CF, an autosomal recessive disorder. The prevalence of CF in the northern European population is 1 in 2,500, with a carrier frequency of 1 in 25. The mutated protein is the CFTR, which regulates chloride transport across membranes. The drying of the pancreatic duct leads to a reduction of secretions from the pancreas reaching the intestine, which leads to the digestive problems exhibited by patients with CF.

A 1-year-old boy has been hospitalized twice for lung infections, was often wheezing and out of breath, and was lagging in both height and weight in his growth charts, despite a very good appetite. A subsequent test found that his sweat contained elevated levels of chloride ions. The inheritance pattern for this disorder is which one of the following? (A) Autosomal dominant (B) Autosomal recessive (C) X-linked recessive (D) X-linked dominant (E) Mitochondrial (F) Multifactorial

*The answer is F.* The patient likely has Becker's muscular dystrophy (BMD), a mildler~ more slowly developing dystrophinopathy than Duchenne's muscular dystrophy, BMD demonst rates X- linked recessive inheritance in that affected male individuals inherit a defective copy of the X chromosome from heterozygous (asymptomatic) mothers. There is no male-to-male transmission . Heterozygous females may be affected but usually not as severely as males.

A 14-year-old boy presents to his family physician with complaints of muscle weakness that has been affecting his ability to play soccer. Physical examination reveals gross enlargement of his calf muscles, and a biopsy reveals active fiber degeneration and regeneration with significant replacement by fat and connective tissue. The boy's family history is positive for similar symptoms in a maternal uncle, whose condition progressed to a wheelchair-bound state by the age of 35 years. Which of the following is the mode of genetic inheritance of this disorder? A. Autosomal dominant B. Autosomal recessive C. Mitochondrial inheritance D. Somat ic mutation E. X-linked dominant F. X-linked recessive

*The answer is A.* Because both chromosome 7's in the daughter cells came from one parent, the mother, the daughter cells have uniparental disomy.

A conception is trisomic for chromosome 7. During the first cell division, the paternal chromosome 7 is lost, leaving two maternal chromosome 7's in the daughter cells. The daughter cells will thus have which one of the following? (A) uniparental disomy (B) monosomy 7 (C) double trisomy (D) a deletion

*The answer is A.* Neurofibromatosis 1 (NF1) is an autosomal dominant disease with variable expressivity. The family history and the clinical findings in the patient confirm the diagnosis of NF1. The fact that the patient's brother had it means that it probably was not due to a new mutation. One of the parents would probably be found to have some mild manifestation of the disease upon examination, as it is fully penetrant.

A 15-year-old boy is referred to a genetics clinic to rule out neurofibromatosis 1. He reports having 25 café-au-lait spots and has started getting lumps and bumps on his skin since he hit puberty. During the family history, he describes his brother as being born with bowed legs and reports that he died at age 12 from a tumor in his neck that had been there since birth. He remembers that his brother had some birthmarks, but not nearly as many as he has. He does not recall his parents having any birthmarks, but they are not with him at the appointment. What inheritance pattern for the disease is occurring in this family? (A) autosomal dominant (B) autosomal recessive (C) X-lined dominant (D) X-linked recessive (E) multifactorial

*The answer is E.* Klinefelter syndrome is most commonly caused by a meiotic nondisjunction event during parental gametogenesis that results in a 47,XXY karyotype. Variants include 46,XY/47,XXY mosaicism and 48,XXXY. In general, patients with higher numbers of X chromosomes are more likely to have more severe manifestations. The disorder is usually not diagnosed until puberty when the characteristic physical signs begin to develop. The major features are as follows: 1. Klinefelter syndrome causes primary testicular failure due to hyalinization and fibrosis of the seminiferous tubules. This resume in small, firm testes and azoospermia (infertility). Leydig cell dysfunction also occurs and leads to testosterone deficiency. Gonadotropin (FSH, LH) levels are increased secondary to gonadal failure . 2. Testosterone deficiency results in development of a eunuchoid body habitus. Patients have tall stature and gynecomastia. Facial and body hair is sparse or absent and muscle mass is decreased. 3. Mild intellectual disability is seen in some patients. although the majority have normal intelligence. Psychosocial abnormalities (eg, lack of insight, poor judgment) are also common. (Choice A) Arachnodactyly, scoliosis and aortic root dilation are signs of Marfan syndrome, which occurs due to an inherited defect of the extracellular matrix protein fibriilin. (Choice B) Macroorchidism, large jaw, and intellectual disability are soon in patients with fragile X syndrome, an X-linked disorder caused by mutations in the fragile X mental retardation 1 gene. (Choice C) In females, loss of an X chromosome (45,XO karyotype) results in Tumor syndrome, which presents with short stature, broad chest, and primary amenorrhea. (Choice D) Preder-Willi syndrome is characterized by short stature, hypotonia, intellectual disability, and obesity. The most common cause is a microdeletion affecting the paternal chromosome 15q11 ~13 critical region. *Educational Objective:* 47,XXY is the most common genotype causing Klinefelter syndrome. Patients present with tall stature, small, firm testes azoospermia; and gynecomastia. Mild intellectual disability is seen in some patients, and the severity generally increases with each additional X chromosome.

A 15-year-old patient is referred to the physician by a teacher who is concerned about the patient's Iearning abilities and behavior. The patients reading and writing skills are significantly impaired compared to other classmates, and the patient often misbehaves in class despite receiving numerous detentions. Neuropsychological assessment shows mild intelectual disability. Cytogenetic studies show a karyotype containing 47 chromosomes. Which of the following findings are most likely to be present on further evaluation? A Arachnodactyly, scoliosis, aortic root dilation B. Macroorchidism. large jaw and ears C. Short stature, broad chest, amenorrhea D. Short stature, hypotonia, obesity E. Tall stature, gynecomastia, azoospermia

*The answer is E.* This patient has the classic phenotype of Prader-Willi syndrome that is caused by the inheritance of a partial deletion (microdeletion) of chromosome 15 from the father. The corresponding maternal alleles in the deleted regions have been imprinted, and some inactivated, leading to a loss of gene expression and the phenotype observed in the patient.

A 16-year-old female has a constant sense of hunger, obesity, almond shaped eyes, strabismus, mental retardation, and retarded puberty development. Which of the following best describes the reason for this chromosomal abnormality? (A) Trinucleotide repeat (B) Nondisjunction (C) Point mutation (D) Translocation (E) Deletion

*The answer is A.* The patient is displaying the signs of maturity onset diabetes of the young (MODY), which can be due to a mutation in the pancreatic glucokinase gene, such that its Km is increased. The increase in the Km for glucokinase would lead to glucose only being metabolized at higher-than-normal levels. Once glucose is metabolized in the β cells of the pancreas, and ATP levels increase, then insulin can be released. The glucokinase mutation causes insulin release to occur at higher-than-normal glucose levels. The mother also expresses the mutant glucokinase gene. During pregnancy, the effect of placental hormones tends to inhibit insulin's action, and in a mother with MODY, in which insulin is not being released appropriately owing to the glucokinase mutation, blood glucose levels rise significantly during the pregnancy, leading to gestational diabetes. MODY, in terms of the glucokinase mutation, is transmitted in an autosomal dominant manner.

A 19-year-old male, at a routine physical exam for sports activities (long distance running) at his college, is noticed to have elevated fasting blood glucose levels (about 7.5 mM). Measurements of C-peptide and insulin levels were close to normal under fasting conditions. After eating, blood glucose levels are only slightly elevated above the normal fasting levels before stabilizing at the fasting levels. The student indicates that he is not drinking or urinating excessively, but that he remembers that his mother had gestational diabetes when pregnant with him. This alteration in glucose homeostasis is best typified by which one of the following types of inheritance? (A) Autosomal dominant (B) Autosomal recessive (C) Sex-linked (D) Mitochondrial (E) Multifactorial

*The answer is C.* Since the father has two mutated hemoglobin genes (both of his β-globin genes contain the E6V mutation) and his partner has two normal β-globin genes, all offspring will have one normal and one sickle gene and therefore will be carriers (the children will have sickle trait). Since this is an autosomal recessive trait and not a sex-linked trait, the sex of the offspring is irrelevant.

A 20-year-old African-American male presents to the ER with severe abdominal, low back, and rib pain. He has had similar episodes all his life. Blood work reveals severe microcytic, hypochromic anemia. A peripheral smear shows abnormally shaped red blood cells and a hemoglobin electrophoresis confirms his abnormality. If the above patient has children with a partner with normal hemoglobin genes, which one of the following patterns will occur? (A) All sons will have the same disease as the father, while all daughters will be carriers. (B) All children will have the same disease as the father. (C) All children will be carriers for the disease (contain the mutated gene, but do not express the disease phenotype). (D) All children will have normal hemoglobin, like their mother. (E) All daughters will have the same disease as their father, while all sons will be carriers.

*The answer is B.* Sickle cell anemia is due to a point mutation on the β globin chain of the hemoglobin where glutamic acid is replaced by valine in the sixth position (E6V). Under deoxygenated conditions, the valine in this position can form hydrophobic interactions with another deoxygenated hemoglobin molecule, leading to the polymerization of the hemoglobin within the cell. The long hemoglobin polymer alters the shape of the red cell, and leads to a loss of red cell elasticity, hemolysis, and "sludging." The altered, or sickle, shape of the cells prevents them from entering capillaries, leading to vaso-occlusive crises and the typical symptoms observed by individuals in a sickle cell crisis. This is an autosomal recessive disease, and the probability that two carriers will have a child affected with sickle cell disease is 25%.

A 20-year-old African-American male presents to the ER with severe abdominal, low back, and rib pain. He has had similar episodes all his life. Blood work reveals severe microcytic, hypochromic anemia. A peripheral smear shows abnormally shaped red blood cells and a hemoglobin electrophoresis confirms his abnormality. Which of the following best describes this genetic abnormality? (A) Trinucleotide repeat (B) Point mutation (C) Trisomy (D) Deletion (E) Translocation

*The answer is A.*Because both chromosome 7's in the daughter cells came from one parent, the mother, the daughter cells have uniparental disomy.

A conception is trisomic for chromosome 7. During the first cell division, the paternal chromosome 7 is lost, leaving two maternal chromosome 7's in the daughter cells. The daughter cells will thus have which one of the following? (A) uniparental disomy (B) monosomy 7 (C) double trisomy (D) a deletion

*The answer is D.* Monosomy X (Turner syndrome) can be due to either maternal nondisjunction (an egg is created lacking an X chromosome, whereas another egg has two X chromosomes) or paternal nondisjunction (a sperm lacks the X chromosome, whereas another sperm contains both the X and Y chromosomes). Turner syndrome is not caused by translocation events, either reciprocal or Robertsonian (recall that Robertsonian translocations only occur amongst acrocentric chromosomes, and the X chromosome if not acrocentric).

A 20-year-old female presents for an infertility workup. She has never had a menstrual period. She is short with a broad chest, webbed neck, and low-set ears. It is demonstrated that she has an abnormal karyotype. The cause of the woman's abnormal karyotype is which one of the following? (A) Maternal nondisjunction (B) Paternal nondisjunction (C) Both maternal and paternal nondisjunction (D) Either maternal or paternal nondisjunction (E) A reciprocal translocation (F) A Robertsonian translocation

*The answer is B.* This patient has Turner syndrome or 45 XO. She is missing a second sex (X) chromosome. She would not have a Barr body and would be infertile. She is phenotypically female.

A 20-year-old female presents for an infertility workup. She has never had a menstrual period. She is short with a broad chest, webbed neck, and low-set ears. It is demonstrated that she has an abnormal karyotype. Which one of the following best describes the cause of this genetic abnormality? (A) Trisomy (B) Monosomy (C) Trinucleotide repeat (D) Translocation (E) Point mutation

*The answer is C.* Because multiple family members are affected and because mutations at the retinoblastoma gene are known to be sometimes nonpenetrant, the man in question is most likely an obligate carrier of the mutation who did not experience a second mutation in this gene during his fetal development.

A 20-year-old man has had no retinoblastomas but has produced two offspring with multiple retinoblastomas. In addition, his father had two retinoblastomas as a young child, and one of his siblings has had 3 retinoblastomas. What is the most likely explanation for the absence of retinoblastomas in this individual? A. A new mutation in the unaffected individual, which has corrected the disease-causing mutation B. Highly variable expression of the disorder C. Incomplete penetrance D. Multiple new mutations in other family members E. Pleiotropy

*The answer is E.* The neuromuscular lesions, ragged red skeletal muscle fibers, and lactic acidosis in these family members together suggest mitochondria! encephalomyopathy. Mitochondrial disorders follow a maternal inheritance pattern, as an embryo's mitochondria are inherited from the ovum only. Mitochondria are responsible for ATP production via oxidative phosphorylation, which is why mitochondrial defects tend to cause lactic acidosis and primarily affect tissues with the highest metabolic rates (e.g., neural tissue, muscular tissue). Though many mitochondrial proteins are coded for in the nuclear genome. mitochondria also contain their own genome, which is also vulnerable to mutations. Defects in the mitochondrial genome may occur in any number of the mitochondria within a cell, and the severity of mitochondrial diseases is often related to the proportion of abnormal to normal mitochondria within a patients cells Heteroplasmy describes the condition of having different organellar genomes (eg mutated and wild-type) within a single cell. For mitochondrial diseases, patients with more severe disease are those with a higher proportion of defective mitochondrial genomes within their cells. *Educational Objective:* The presence of lactic acidosis and ragged red skeletal muscle fibers histologically suggest a mitochondrial myopathy. There may be variable clinical expression of mitochondrial DNA defects in different affected family members due to heteroplasmy, which is the coexistence of both mutated and wild-type versions of mitochondrial genomes in an individual cell.

A 21-year-old man comes to the physician complaining of recent onset of vision loss. Neuroimaging studies reveal several small infarcts in the occipital lobes bilaterally. Skeletal muscle biopsy reveals ragged appearing muscle fibers. The patient's 56-year-old mother has chronic intermittent muscle weakness and an elevated serum lactate level. The patient's family history is also significant for a maternal uncle who developed hemiplegia at the age of 35. Assuming each of these family members has the same condition, the variability in their clinical findings is best explained by: A. Variable penetrance B. Mosaicism C. Uniparental disomy D. Anticipation E. Heteroplasmy

*The answer is B.* Fragile X syndrome, the disorder displayed by the patient, exhibits variable expressivity depending upon the number of repeats inherited. Some males may be asymptomatic, but a grandson of a male who is asymptomatic may have full expression of the disease (remember, males cannot transmit an X chromosome to their sons, only to their daughters). Though the abnormality may be present in individuals, they may show no features of the disease. With full expression, everyone] with genetic abnormalities would show the full spectrum of the disease process (100% penetrance and expressivity). Homoplasmy and heteroplasmy are terms used in mitochondrial inheritance. Heteroplasmy refers to the fact that some mitochondria contain normal genomes and other mutated genomes. Homoplasmy refers to all mitochondria containing the same genome. Gene dosage refers to the expression of too many genes such as in trisomies, which is not the case in triplet repeat disorders.

A 22-year-old male has a long face, large forehead, ears and jaw, large testes, autism, speech and language delays, and hand flapping. His only sister is normal except for extreme shyness. The phenotypic pattern of this man's disorder is best described as which one of the following? (A) Full expressivity (B) Variable expressivity (C) Gene dosage effect (D) Homoplasmy (E) Heteroplasmy

*The answer is D.* The patient has fragile X syndrome, the most common inherited mental retardation disorder in males. There is a trinucleotide repeat of the FMR1 gene (in the 5'-untranslated region) on the X chromosome. The more repeats, the more severe the signs and symptoms. Males are more severely affected than females (only one X chromosome), and many women with fragile X appear asymptomatic except for excessive shyness (awkward social interactions). Fragile X syndrome was named for chromosome breaks in this region of the chromosome when cells are cultured in a folate-deficient medium.

A 22-year-old male has a long face, large forehead, ears and jaw, large testes, autism, speech and language delays, and hand flapping. His only sister is normal except for extreme shyness. Which one of the following best describes the basis of this chromosomal abnormality? (A) Autosomal dominant (B) Codominant (C) Autosomal recessive (D) Trinucleotide repeat (E) Mitochondrial inheritance

*The answer is C.* Cystathionine beta-synthase deficiency is the enzyme defect present in classic homocystinuria. Homocystinuria is characterized clinically by ectopia lentis, mental retardation, marfanoid habitus and osteoporosis in addition to vascular problems. Pleiotropy is the occurrence of multiple phenotypic manifestations, often in different organ systems, as a result of a single genetic defect. *Educational Objective:* Pleiotropy describes instances where multiple phenotypic manifestations result from a single genetic mutation. Most syndromic genetic illnesses exhibit pleiotropy.

A single missense mutation in the gene coding for cystathionine beta synthase causes a variety of phenotypic manifestations including skeletal deformities, mental retardation and vascular thromboses. This phenomenon is referred to as: A. Polyploidy B. Genetic linkage C. Pleiotropy D. Variable penetrate E. Segregation F. Imprinting

*The answer is B.* B. The disease is a mitochondrial disorder, LHON. Mitochondrial diseases are maternally inherited, as all of the mitochondria in a developing embryo are derived from the egg, and none from the sperm. Since the patient is a male, none of his mutant mitochondria will enter the egg, and none of his children will express the disease, nor will they be carriers of the disease.

A 22-year-old male reports to his eye doctor that he lost the center portion of the vision in his right eye over a week's time and 2 weeks later, the same thing happened in his left eye. He has two older brothers who had the same thing happen to them at age 21 and 23, and one older sister who had the same eye problems starting at age 24. If the patient has children, which one of the following patterns will occur? (A) All of his children will have the disease. (B) None of his children will have the disease. (C) Only his sons will have the disease. (D) Only his daughters will have the disease. (E) All of his children will be carriers of the disease.

*The answer is A.* As the patient has a mitochondrial disorder (which he inherited from his mother), the patient's sister also has the same mitochondrial disorder. As a female, the woman will pass on mutant mitochondria to all of her children, who will express the disease.

A 22-year-old male reports to his eye doctor that he lost the center portion of the vision in his right eye over a week's time and 2 weeks later, the same thing happened in his left eye. He has two older brothers who had the same thing happen to them at age 21 and 23, and one older sister who had the same eye problems starting at age 24. If the patient's sister has children, which of the following patterns will occur? (A) All of her children will have the disease. (B) None of her children will have the disease. (C) Only her sons will have the disease. (D) Only her daughters will have the disease. (E) All of her children will be carriers of the disease.

*The answer is D.* LHON is due to a mutation in the mitochondrial genome, so it is classified as mitochondrial inheritance. All mitochondria is inherited from the mother since the mitochondria associated with sperm does not enter the egg. All of the offspring of an affected mother will have the disease (100% penetrance), although the expressivity is quite variable depending on the degree of heteroplasmy exhibited by each child. Males cannot transmit mitochondrial diseases.

A 22-year-old male reports to his eye doctor that he lost the center portion of the vision in his right eye over a week's time and 2 weeks later, the same thing happened in his left eye. He has two older brothers who had the same thing happen to them at age 21 and 23, and one older sister who had the same eye problems starting at age 24. Which one of the following typifies this type of genetic abnormality? (A) Autosomal dominant (B) Autosomal recessive (C) Sex-linked (D) Mitochondrial (E) Trinucelotide repeat

*The answer is E.* Red-green color blindness is an X-linked recessive disorder. The male who has the disease will transmit his X chromosome (with the mutation) to his daughter, who will then be a carrier of the disease (since the X chromosome inherited from the mother contains a normal allele). The father transmits his Y chromosome to his son, who inherits his X chromosome from his mother, so the son does not inherit the mutation and will be normal in terms of red-green color discrimination.

A 22-year-old male's lifelong dream was to be a fighter pilot for the US Air Force. He passed his vision test without a problem, but failed the color portion of the standard Snellen chart. If the above patient marries a normal female and has one son and one daughter, which one of the following inheritance patterns will occur? (A) The son will be color-blind, but the daughter will be normal. (B) The daughter will be color-blind, but the son will be normal. (C) Both children will be color-blind. (D) The son will be a carrier, but the daughter will be normal. (E) The daughter will be a carrier, but the son will be normal.

*The answer is D.* Red and green are the standard colors on a Snellen eye chart, and are used to screen for the most common "color blindness" in boys, red-green. This is an X-linked recessive trait. Females can be carriers with no color-perception problems, but can have red-green color blindness if they are homozygous for the mutation. Half of a carrier female's sons will be normal and half will be red-green color-blind. None of an affected male's sons will have the problem, but all of his daughters will be carriers. The mother of the person in question is a carrier for red-green color blindness, and she would have a 50% chance of passing the mutated allele on one of her X chromosomes to the patient's brother.

A 22-year-old male's lifelong dream was to be a fighter pilot for the US Air Force. He passed his vision test without a problem, but failed the color portion of the standard Snellen chart. What is the probability that the subject's brother would have the same problem? (A) 100% (B) 75% (C) 67% (D) 50% (E) 33% (F) 25% (G) 0%

*The answer is D.* Sickle cell anemia Is an autosomal recessive hemoglobinopathy. Affected patients must inherit 2 mutant genes for hemoglobin S (HbS), one from each parent. In order for an offspring of 2 parents without sickle cell anemia to be affected, both parents must carry the sickle cell trait. Any offspring resulting from such pairings will have a 1 in 4 chance of having sickle cell disease (SCD). This patient is a carrier of the sick cell trait as she has a child with sickle cell anemia. Hemoglobin electrophoresis, a type of gel electrophoresis, is used to determine if the patient's new husband also has sickle cell trait. Abnormal hemoglobin (eg, HbS), moves at a slower speed than normal hemoglobin due to the replacement of glutamic acid by valine. The husband's test resits combined with the maternal family history will determine the risk that future offspring will inherit sickle cell anemia. *Educational Objective:* Sickle cell anemia is an autosomal recessive hemoglobinopathy. In order for a child to have sickle cell disease, both parents must be carriers. Hemoglobin electrophoresis can be used to determine the carrier status of a prospective parent who has no history of sickle cell anemia.

A 24-year-old African American woman comes to the office with her husband for prenatal counseling. She has a 3-year-old child with sickle cell anemia from a previous marriage, and the child's father died in a car accident. The patient remarried last year and is interested in having more children. She and her new husband do not have sickle cell anemia, and the patient's husband has no other children. However, the patient and her husband are worried that their future children could have sickle cell anemia. A urine pregnancy test is negative. Which of the following is the best initial test that can be offered to this couple? A Chorionic villous sampling during future pregnancy B. Maternal hemoglobin electrophoresis C. Northern blot analysis of paternal blood sample D. Paternal hemoglobin electrophoresis E. Paternal karyotypa analysis

*The answer is B.* Because of genomic imprinting, both maternal and paternal haploid sets of chromosomes are required for normal development. When there are two paternal haploid sets of chromosomes in a conceptus, a placenta will develop but not an embryo.

A 24-year-old woman is diagnosed as having a complete molar pregnancy with enlargement of the chorionic villi and absence of an embryo. Cytogenetic analysis of the products of conception revealed a 46,XX karyotype. The molar pregnancy was caused by which one of the following? (A) preeclampsia (B) two haploid sets of paternal chromosomes (C) trophoblastic neoplasia (D) elevated hCG levels (E) enlarged uterus

*The answer is D.* The pedigree shows that only males are affected by the drug intolerance. Specifically, male offspring of unaffected parents are affected. There is no evidence of male-to-male transmission. This pattern is most consistent with X-linked recessive inheritance from an asymptomatic carrier female in the first generation. In X-linked recessive inheritance: 1 . Affected males will always produce unaffected sons and carrie daughters. 2. Carrier females have a 50% chance of producing an affected son or carrier daughter. G6PD deficiency, which causes acute hemolytic anemia on exposure to oxidant drugs, follows an X-linked recessive paper of inheritance *Educational Objective:* In X~linked recessive inheritance 1) affected mains will always produce unaffected sons and carrier daughters, and 2) carrier females have a 50% chance of producing affected sons and carrier daughters. G6PD deficiency follows this inheritance pattern and causes acute hemolytic anemia in response to oxidant drugs.

A 25-year-old man experiences severe intolerance to certain medications. On 2 occasions, his reactions to various drugs have necessitated hospital admission. His family pedigree with respect to this condition is shown below, with the red arrow indicating his position within the family. Assume that this condition demonstrates complete penetrance and is rare in the general population. This condition most likely exhibits which of the following inheritance patterns? A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive E. Mitochondrial

*The answer is A.* The most likely explanation for mild expression in a heterozygous carrier is that when X inactivation occurred in the affected individual, the random process happened to inactivate most of the X chromosomes that carried the normal version of the factor VIII gene. Thus, most of the active X chromosomes in this individual would carry the mutation and would not produce factor VIII, leading to a clinically expressed deficiency.

A 25-year-old woman has mild expression of hemophilia A. A genetic diagnosis reveals that she is a heterozygous carrier of a mutation in the X linked factor VIII gene. What is the most likely explanation for mild expression of the dis- ease in this individual? A. A high proportion of the X chromosomes carrying the mutation are active in this woman B. Her father is affected, and her mother is a heterozygous carrier C. Nonsense mutation causing truncated protein D. One of her X chromosomes carries the SRY gene E. X inactivation does not affect the entire chromosome

*The answer is B.* This patient has Klinefelter syndrome or 47 XXY. This is caused by a nondisjunction of the X chromosomes in either the egg, which is then combined with a Y chromosome during fertilization, or the sperm, in which case an XY sperm combines with an egg carrying one X chromosome. Even though one of the X chromosomes will become a Barr body, men with the karyotype 47 XXY have reduced testosterone levels as compared to men who are 46 XY. While symptoms are not seen in all individuals, the ones described in this question are the classic symptoms for Klinefelter syndrome. Trinucelotide repeats, point mutations, translocations, or deletions will not give rise to the 47 XXY genotype. That can only occur through a nondisjunction event in either the formation of the mother's eggs or father's sperm. The nondisjunction event that leads to 47 XXY appears to occur evenly between the mother and father.

A 27-year-old male underwent a workup for what appeared to be a feminizing disorder since he was tall with sparse facial hair, small testes, and gynecomastia. He also had poor coordination and language and reading difficulties. During the workup, it was discovered that he had an abnormal karyotype. Which of the following best describes the reason for his abnormal karyotype? (A) Trinucleotide repeat (B) Nondisjunction (C) Point mutation (D) Translocation (E) Deletion

*The answer is B.* The patient has Factor V Leiden, which is the most common hereditary hypercoagulable disorder in the United States. Individuals who inherit one copy of this mutation (heterozygote) are at an increased risk for clotting, but that risk is less than someone who is homozygous for the mutation. Thus, the mutated allele is not completely dominant (since the homozygous state increases the risks of clots), and is termed incomplete dominance. Not everyone who inherits just one mutated allele will develop a clotting problem. Codominant would imply two mutations of the gene (either or both of which would produce the disease process). The mutated gene for Factor V Leiden is not on the X or Y chromosome, so it is not sex-linked. Since a person inheriting just one mutated allele can express disease symptoms, the inheritance process is not autosomal recessive.

A 32-year-old female has had multiple deep venous thrombosis and pulmonary emboli, especially during her pregnancies. She is on chronic warfarin therapy. Her father died of a pulmonary embolus after a retinal detachment operation. Neither of her three siblings nor her two children have had any clotting problems. Which one of the following inheritance patterns/descriptions best typifies this genetic problem? (A) Autosomal dominant (B) Autosomal incomplete dominant (C) Codominant (D) Autosomal recessive (E) Sex-linked

*The answer is C.* Acute myelogenous leukemia (AML) is characterized by failure of immature myeloid precursors (myeloblasts) to differentiate into mature granulocytes. It is divided into eight types. M0 through M7. The M3 variant of AML, acute promyelocytic leukemia is associated with the cytogenetic abnormality t(15,17). Here, the gene for retinoic acid receptor alpha from chromosome 17 is transferred to chromosome 15 in a location adjacent to the PML (promyelocyte leukemia) gene. Fusion of these two genes produces a chimeric gene product PML/RARα. which codes for an abnormal retinoic acid receptor. This abnormal fusion gene product inhibits myeloblast differentiation, producing acute promyelocytic leukemia. The clinical manifestations of AML, including anemia (fatigue, pallor), thrombocytopenia (petechiae, hemorrhages) and neutropenia (fever, opportunistic infections), result from marrow replacement by leukemic cells. *Educational Objective:* The cytogenetic defect t(1517) is associated with acute promyelocytic leukemia (AML type M3). Translocation of the gene for the retinoic acid receptor alpha from chromosome 17 to chromosome 15 leads to formation of the fusion gene PML/RARα This abnormal fusion gene product inhibits differentiation of myeloblasts and triggers the development of acute promyelocytic leukemia.

A 32-year-old male presents to the ER with recent onset of severe fatigue, exertional dyspnea and fever. Cytogenetic studies of this patient's blood cells demonstrate a 15;17 chromosomal translocation. Which of the following proteins is most likely malfunctioning in the affected cells? A. Epidermal growth factor receptor B. Platelet derived growth factor receptor C. Retinoic acid receptor D. GTP-binding protein E. Retinoblastoma gene product

*The answer is C.* The most common type of hair loss in males and females is known as androgenetic alopecia (male pattern baldness). The pattern and severity of the baldness varies between males and females, and circulating androgen levels along with the degree of genetic predisposition are thought to play a prominent role in determining clinical manifestations. Androgenetic alopecia demonstrates polygenic inheritance with variable penetrance. Key sites of genetic influence have been identified on the X and Y chromosomes and also on the short arm of chromosome 20. Recent research has demonstrated the importance of specific androgen receptor gene variations in the development of androgenetic alopecia. The androgen receptor gene is located on the X chromosome, and the variations of this gene that are associated with androgenetic alopecia are inherited in an X-linked recessive manner. *Educational Objective:* Androgenetic alopecia is the most common cause of hair loss in both males and females, and demonstrates polygenic inheritance with variable penetrance. The pattern and severity of the baldness varies between males and females, and circulating androgen levels along with the degree of genetic predisposition are thought to play a prominent role in detemining clinical manifestations.

A 32-year-old woman comes to the physician because she is worried that she will go bald. Both her father and paternal grandmother suffered from early-onset of baldness, but no one on the maternal side of her family is bald. Physical examination reveals a normal appearing hairline without evidence of hair thinning. After reassuring the woman, you explain the genetics underlying the most common form of hair loss in both men and women. The most likely inheritance pattern of this condition is A. Autosomal dominant B. Autosomal recessive C. Polygenic D. Mitochondrial E. Sporadic

*The answer is D.* This patient has a marfanoid habit's - tail and slender build with disproportionately long arms, legs, and fingers. The flesh-colored nodules on his lips and tongue are likely mucosal neuromas which are unencapsulated, thickened proliferation: of neural tissue. This combination of clinical findings is consistent with multiple endocrine neoplasia type 2B (MEN2B), which is due to an inherited mutation in the RET proto-oncogene. This patient also has a history of total thyroidectomy, which was likely due to medullary thyroid cancer (MTC) (benign thyroid masses are usually treated medically or with partial thyroidectomy). Early recognition of MEN2B is important as almost all patients will develop MTC and prophylactic thyroidectomy can be lifesaving. Other possible manifestations of MENZB include pheochromocytoma and intestinal ganglioneuromas (often causing associated constipation). *Educational Objective:* Multiple endocrine neoplasm type 2B (MEN2B) is characterized by medullar carcinoma of the thyroid, pheochromocytoma, marfanoid habitus, and oral and intestinal mucosal neuromas.

A 34-year-old man comes to the office due to oral and perioral nodules. The nodules appeared several months ago, and the patient reports that they are similar to other lesions that were removed 10 years earlier. He has a history of total thyroidectomy 5 years ago following the discovery of a palpable thyroid mass. On examination. the patient is tall and slender with disproportionately long arms and legs. His fingers are also long and thin. Oral inspection shows several small, flesh-colored nodules on his lips and tongue. This patient most likely suffers from which of the following conditions? A. Ehlers-Danlos syndrome B. Marfarn syndrome C. Multiple endocrine neoplasia type 1 D. Multiple endocrine neodasia type 2B E. Neurofibromatosis type 1 F. Neurofibromatosis type 2

*The answer is C.* The politician has the phenotype of an individual with Marfan syndrome, which is inherited in an autosomal dominant pattern. The mutation is in the protein fibrillin, which is a glycoprotein found in elastic fibers in connective tissue. Since this is an autosomal dominant disorder, the man has a one-in-two chance (50%) of passing the mutation on to one of his children.

A 35-year-old male woodcutter and politician from Illinois is very tall and thin with arachnidactyly, pectus excavatum, a high arched palate, lens dislocation, and aortic insufficiency. His wife is of normal height and weight, and does not exhibit any of the same symptoms as the politician. What is the probability that one of their children will exhibit features similar to the father's? (A) 100% (B) 75% (C) 50% (D) 25% (E) 0%

*The answer is C.* Alpha-1 antitrypsin (A1AT) deficiency is a codominant process. Codominance means that multiple versions of the gene may be active or expressed, and the genetic trait is due to the effects of both the expressed alleles. The SERPINA1 gene on chromosome 14 codes for A1AT, which is a protein that protects tissues (especially the lungs) from neutrophil elastase. A1AT is synthesized in the liver and secreted into the circulation. Under normal conditions, neutrophils in the lung engulf and destroy particulate matter in the air we breathe. At times, the protease elastase escapes from the neutrophils, and is inactivated by A1AT. In the absence of functional A1AT, the elastase destroys the lung cells, and chronic obstructive pulmonary disease (COPD) will occur very early in life. If the patient smokes, the condition is greatly exacerbated. A common mutation in A1AT leads to misfolding and accumulation of the misfolded form of the protein in the liver. The accumulation of this inactive, misfolded protein can lead to cirrhosis of the liver, and eventual liver failure.

A 35-year-old nonsmoking male has been diagnosed with emphysema. His father died of emphysema at age 30, but he smoked. His father also had cirrhosis and recurrent pancreatitis but did not drink alcohol. Which one of the following inheritance patterns typifies this disease process? (A) Autosomal dominant (B) Incomplete dominance (C) Codominant (D) Autosomal recessive (E) Sex-linked

*The answer is A.* The baby will be born a phenotypically normal female as two of the X chromosomes will be inactivated per cell, and become Barr bodies. In this manner, there are no gene dosage effects occurring during the development of the fetus. There is no Y chromosome, so this baby will not be male. Trisomies for autosomal chromosomes are lethal except for trisomy 21, trisomy 18, and trisomy 13. However, polysomies of the X chromosome can be asymptomatic.

A 40-year-old pregnant woman is very concerned about chromosomal abnormalities in her fetus. She undergoes amniocentesis, and the fetus is demonstrated to have the karyotype 47 XXX. Which of the following is the most likely outcome of this pregnancy? (A) A phenotypically normal female. (B) A phenotypically normal male. (C) Since trisomies are almost always lethal, there will be a spontaneous abortion. (D) Since trisomies are almost always lethal, this will end in a stillbirth. (E) Since trisomies are almost always lethal, the baby will die shortly after birth.

*The answer is C.* The condition described is Huntington disease (HD). It usually manifests in patients between 30 and 50 years of age. It is associated with an increased number of CAG tandem repeats on chromosome 4 as a trinucleotide repeat disorder. The mutation for HD is transmitted in an autosomal dominant fashion. HD demonstrates anticipation, meaning that with each successive affected generation, the disease manifests earlier and gets worse faster, as it did in this vignette. Three classic clinical signs of HD are a gradual onset of change in mood, chorea (also known as St. Vitus dance), and eventual dementia. Although forgetfulness and mood instability are early symptoms of HD, full-blown dementia is a late sign. Choreiform movements are fluid but involuntary and purposeless motions of the limbs. As the caudate and putamen atrophy, the lateral ventricles enlarge. This ventricular enlargement, however, is more typical of late rather than early HD. These symptoms can be treated with tetrabenazine, a vesicular monoamine transporter (VMAT) inhibitor~ which works to decrease dopamine levels.

A 45-year-old man comes to a physician accompanied by his wife. They report that the patient has been struggling with depression for over a year now. He had no past history of psychiatric illness prior to the onset of his current depressive symptoms. In addition, he has become more irritable lately. They are also concerned because recently he has been restless and constantly moving his arms in a rapid, nonrythmic manner that he cannot suppress. The patient's wife notes that her father-in-law developed similar symptoms in his 60s and later developed dementia. What is the biologic basis of this patient 's likely condition? A. α-Synuclein concretions B. Aggregated tau proteins C. CAG repeats D. Organ-specific amyloidosis E. SOD1 defect

*The answer is D.* In light of his relative youth and the frequency of cancer in his close relatives, this patient most likely has hereditary non polyposis colon cancer (HNDCC) . The process of mismatch repair is defective in pat ients with hereditary nonpolyposis colorectal cancer. Mismatch repair proteins such as MSH2, MSH6, and MLH1 normally detect the mismatched bases so that they can removed and replaced. In HNDCC, also known as Lynch syndrome, these proteins are mutated, leading to defective mismatch detection and repair. The gross specimen in the vignette shows a circumferential tumor with central ulceration that is narrowing the colonic lumen, which explains the patient's presentation with abdominal distent ion.

A 46-year-old man presents to the physician with abdominal distention. An X-ray is followed by a contrast enema, which yields distinctive findings that lead his physician to recommend surgery. A portion of the patient's large intestine is removed. The image shows the gross specimen. Questioning of the patient reveals that 80% of his relatives have had colon cancer most often in their forties. This patient likely has a defect in which of the following forms of DNA repair? A. Apurinic/apyrimidinic site repair B. Base excision repair C. Double-strand break repair D. Mismatch repair E. Pyrimidine dimer repair

*The answer is D.* This child and his uncle appear to have Friedreich ataxia, a trinucleotide repeat disease (GAA). Like other trinucleotide repeat diseases, illness occurs because the unstable microsatellite regions on certain chromosomes have triplet codons that expand, typically worsening from generation to generation (and often making the age at onset earlier for each successive generation, termed "anticipation"). These regions of massively expanded triplet repeats within the introns of the allele (most commonly between 600 and 1200 in Friedreich ataxia) cause a decrease in the product of a gene, frataxin, at the transcription stage. This gene silencing occurs because the size of the trinucleotide repeat leads to a conformation change to heterochromatin. His abnormally high arch is known as pes cavus and is seen in Friedreich ataxia.

A 6-year-old boy cannot play soccer or participate in gym class. He staggers when he walks and falls frequently. In addition to this ataxic gait, the pediatrician notes the abnormally high arch of his feet and discovers a dysrhythmia on further work-up. The family reports that the boy's uncle also has this condition, but his symptoms did not appear until he was 12 years of age. What is the molecular mechanism of this disease? A. Unstable repeats affect protein folding B. Unstable repeats affect protein splicing C. Unstable repeats cause an amino acid substitution D. Unstable repeats impede gene transcription E. Unstable repeats result in a truncated protein

*The answer is B.* The patient has classic symptoms of Prader-Willi syndrome: intellectual disability, short stature, hypotonia, hyperphagia, obesity, small hands and feet, and hypogonadism. The images illustrate some of the typical facial features (narrow forehead, downward corners of the mouth). Both Prader-Willi syndrome and Angelman syndrome have been localized to the 15q12 band, although they are very different entities. This difference is because of sex-specific imprinting of genes at that locus. Some genes at 15q12 are imprinted (turned off via histone and DNA modifications) when inherited from the mother, and others are imprinted when inherited by the father. In Prader-Willi syndrome, the genes that would normally only be active on the chromosome inherited from the father (because they are maternally imprinted) have been deleted in the paternal chromosome, thus the patient has no functional copies of these genes.

A 6-year-old boy is brought to the clinic by his mother, who thinks he is "eating too much" because he is gaining weight. She says that as a baby he had hypotonia and a weak cry, and he sta1ted walking at age 2 years. His mother also says he has always been "slower" and shorter than other kids. Which of the following is the most likely cause of the patient's condition? A. Deletion of q12 on the maternally derived chromosome 15 B. Deletion of q12 on the paternally derived chromosome 15 C. Maternal alcohol use during pregnancy D. Trinucleotide CAG repeats E. Trinucleotide CGG repeats

*The answer is C.* Deletions involving the long arm of chromosome 22 can result in facial, cardiac and immunological abnormalities. DiGeorge syndrome is one such manifestation. DiGeorge syndrome is defined by thymic aplasia and failure of parathyroid formation, due to defective embryonic development of the third and fourth pharyngeal pouches. Patients typically present with hypocalcemic tetany and recurrent viral and fungal infections due to T-cell deficiency. Cardiac defects associated with this syndrome are Tetralogy of Fallot and interrupted aortic arch. Chromosome 22q11.2 deletion is found in 90% of cases of DiGeorge syndrome. *Educational Objective:* A variety of genetic disorders can result in facial and/or palatal malformations, including deletions of the long arm of chromosome 22. However, deletions involving the long arm of chromosome 22 are also associated with DiGeorge syndrome (congenital thymic and parathyroid aplasia, congenital cardiovascular anomalies).

A Caucasian newborn with facial dysmorphia and cleft palate in found to have a deletion involving the long arm of chromosome 22. These findings are most consistent with: A. Kartagener's syndrome B. Tuberous sclerosis C. DiGeorge syndrome D. Friedreich's ataxia E. Marfan syndrome F. Down syndrome G. Turner's syndrome.

*The answer is E.* This is an example of mitochondrial inheritance in that all transmission is from the female (all children of an affected female display the trait, with variable expressivity), and an energy-intensive organ is most severely affected. The family history indicates that male transmission of the disorder does not occur (the mother's brother), and that variable expressivity is evident due to the degree of heteroplasmy inherited by each child.

A couple in their mid-30s has had three children, two boys and one girl. One boy and one girl have severe hearing loss, but the other boy has only a mild hearing loss. The father has normal hearing, but the mother does wear a hearing aid on her right ear. The mother's brother is also hard of hearing, but he has three children who have no hearing loss. A likely mode of inheritance for this disorder is which one of the following? (A) Autosomal recessive (B) Autosomal dominant (C) X-linked dominant (D) X-linked recessive (E) Mitochondrial (F) Triplet repeat expansion

*The answer is C.* One of the potential parents most likely has a translocation, which is causing the formation of abnormal gametes, in terms of chromosome number, during meiosis. FISH analysis may miss translocations and would not be able to detect this problem. A chromosomal translocation in either parent (either reciprocal or Robertsonian) can lead to the problems in conception the couple is experiencing.

A couple is having trouble bringing a pregnancy to term. They have experienced three miscarriages in the past 2 years and want to understand what the problem may be. An initial test that should be run on the couple is which one of the following? (A) Karyotype analysis of the female (B) Karyotype analysis of the male (C) Karyotype analysis of both the potential parents (D) FISH analysis of the mother using chromosome- specific probes (E) FISH analysis of the father using chromosome- specific probes (F) FISH analysis of both potential parents using chromosome-specific probes

*The answer is D.* The translocations would cause problems during meiosis, leading to gametes either lacking a chromosome (or a portion of a chromosome) or gained an additional chromosome (or portion thereof). This would lead to autosomal monosomies or trisomies after fertilization, the majority of which are incompatible with life and lead to early termination of the pregnancy. Multiple X chromosomes are tolerated, and would not lead to pregnancy termination. Trisomy 21 leads to Down syndrome, which is compatible with life and does not lead to miscarriage. Lack of the Y chromosome would lead to females, which does not lead to pregnancy termination.

A couple is having trouble bringing a pregnancy to term. They have experienced three miscarriages in the past 2 years and want to understand what the problem may be. The miscarriages were most likely caused by which one of the following? (A) Multiple X chromosomes (B) Autosomal monosomies (C) Autosomal trisomies (D) Either autosomal monosomies or trisomies (E) Trisomy 21 (F) Lack of the Y chromosome

*The answer is C.* The probability of having a child affected by an autosomal recessive disorder is given by the product of the following individual probabilities: P(affected child given carrier parents) x P(carrier mother) x P(carrier father) In this example: P(affected child given carrier parents) = 1/4. as the child of 2 carriers of a recessive condition has a 1/4 chance of being affected P(carrier mother) = 1, as the patient must be a carrier given that her first son has Pompe disease P(carrier father) can be calculated using Hardy-Weinberg analysis Hardy-Weinberg analysis can be used to relate gene/allele, disease. and carrier frequencies if 1 of these values is known: Gene/allele frequency: By convention, p = frequency of normal (dominant) allele. and q = frequency of mutant (recessive) allele in the population of interest Disease frequency: As homozygous recessive individuals (with the disease) must have 2 copies of the recessive (mutant) allele, the frequency of homozygous recessive individuals (disease frequency) = q x q = q^2 Carrier frequency: Heterozygous individuals (disease carriers) have only 1 mutant allele (genotype is either pq or qp), so in general, cancer frequency = 2pq. For rare autosomal recessive disorders, p = 1, therefore, the probability of being a carrier approximates to 2x frequency of the mutant allele or 2q. In this example, we are given q= disease frequency = 1/40.000. Therefore, q = square root (1/ 40,000) = 1/200 and P(carrier father) as 2q = 2 x (1/200) = 1/100. In sum, the probability of this patient having a second affected child is: (1/4 x 1 x 1/100) = 1/400.

A healthy 31-year-old woman comes to the office because she and her husband desire a second child. The husband has Klinefelter syndrome and is infertile, and the patient's son, who was conceived via donor insemination, was recently diagnosed with glycogen storage disease type II (Pompe disease). This rare autosomal recessive disease is known to affect 1 in 40,000 of the general population. What is the probability that the patient would have a second affected child with a new, healthy sperm donor? A. 1/4 B. 1/240 C. 1/400 D. 1/800 E. 1/40,000 F. 1/160,000

*The answer is A.* The pat ient suffers from Duchenne's muscular dystrophy, an X- linked inherited disease characterized by muscular pseudohypertrophy (via fibrofatty replacement of muscle fibers) and muscle weakness.

A mother brings her 5-year-old boy, who has a 2-year history of progressive muscle weakness, to the pediatrician. She notices that the boy has trouble moving himself from a sitting position to a standing position and that he has to push off of the floor or pull himself up in order to stand. Physical examination reveals a happy boy with normal height and weight, normal reflexes, and large calves. A biopsy of his gastrocnemius would most likely show which of the following? A. Fibrofatty replacement of muscle fibers B. Increased proportion of slow-twitch red muscle fibers C. Massive necrosis of all muscle fibers D. Unusually high capillary density E. Hypertrophy of fast-twitch white muscle fibers

*The answer is C.* If the woman is expressing three Barr bodies, then she has four X chromosomes per cell, three of which have been inactivated. This would give her a total of 48 chromosomes, and four of those would be X chromosomes, for a karyotype of 48 XXXX. Any karyotype with a Y chromosome would be a male.

A phenotypically normal woman underwent a karyotype analysis for difficulties in conceiving. She was found to contain three Barr bodies, but no translocations or large deletions. Her karyotype would be best represented by which one of the following? (A) 48 XXXXY (B) 46 XX (C) 48 XXXX (D) 48 XXXY (E) 48 XXYY

*The answer is A.* The mother is homozygous for a CF mutation (aa) so she can only pass along a mutated gene (a). The father is presumably homozygous for the normal gene (AA), so he can only pass on a normal gene (A). Therefore, all their children will be heterozygotes (Aa), or carriers of a CF mutation.

What is the risk that the child of a mother with cystic fibrosis will be a carrier of the disease? (A) 100% (B) 75% (C) 50% (D) 25%

*The answer is C.* Two allele loci are said to be in linkage disequilibrium when o pair of alleles from two loci are inherited together in the same gamete (haplotype) more or less often than would be expected by random chance alone given their corresponding allele frequencies. However, it is important to realize that linkage disequilibrium does not always imply physical proximity between the allelic loci. Although linkage disequilibrium can be the result of physical linkage of genes (on the same chromosome). it can also occur even of the genes are on different chromosomes due to mutations, genetic drift., migration, selection pressure, and non-random mating. To estimate the probability of two alleles appearing together, multiply their occurrence rates. Note that the Hardy-Weinberg principle (2pq) is not applicable since we are comparing allelic frequency at two distinct loci. DQA1*0501-DQB1*0201 haplotype = [Frequency of DQA1*0501] * [Frequency of DQB1'0201] = 0.3 * 0.2 = 0.06 In this example, the observed frequency (given in the question) is 0.20 which is greater than the expected frequency of 0.06. Hence, the population is said to be at linkage disequilibrium. *Educational Objective:* Two allele loci are said to be in linkage disequilibrium when a pair of alleles are inherited together in the same gamete (haplotype) more or less often than would be expected given random chance. It is important to understand that this can occur even if the genes are on different chromosomes.

A study is undertaken to map the HLA-DQ loci in a population with a high incidence of celiac sprue. High-resolution HLA typing of the DQA1 and DQB1 loci is performed using polymerase chain reaction sequencing. The frequency of the DQA1'0501-DQB1"0201 haplotype, which has been strongly implicated in autoimmunity, is found to be 0.20. However, in the same population, the frequency of the DQA 1 '0501 allele is 0.3 and the frequency of the DQB1'0201 allele is 0.2. Which of the following terms best explains the observed DQA1'0501-DQB1"0201 haplotype frequency in this population? A. Heteroplasmy B. Increased penetrance C. Linkage disequilibrium D. Pleiotropy E. Segregation

*The answer is A.* Alan is Barbara's father and his sister Alice is Blaine's mother. Alan is Blaine's uncle and his daughter Barbara is Blaine's first cousin.

Alan has hemophilia A. His sister, Alice, has one son, Blaine. Blaine also has hemophilia A. Alan and his wife Annette have 2 children, Bart and Barbara. Barbara has a daughter, Cassie, and a son Chip. Cassie and Blaine are married and have a son, Daniel, with hemophilia A. They are now expecting fraternal twins, a boy and a girl. How are Barbara and Blaine related? (A) first cousins (B) first cousins once removed (C) second cousins (D) second cousins once removed

*The answer is E.* Because Barbara's daughter Cassie has a son with hemophilia A, Cassie must have received the mutation from her mother. The mutation could not have come from Blaine because he cannot pass on his X chromosome to his son. Both Barbara and Cassie are obligate carriers.

Alan has hemophilia A. His sister, Alice, has one son, Blaine. Blaine also has hemophilia A. Alan and his wife Annette have 2 children, Bart and Barbara. Barbara has a daughter, Cassie, and a son Chip. Cassie and Blaine are married and have a son, Daniel, with hemophilia A. They are now expecting fraternal twins, a boy and a girl. What is Barbara's risk to be a carrier of hemophilia A? (A) 0% (B) 25% (C) 50% (D) 75% (E) 100%

*The answer is C.* Blaine can only pass on a Y chromosome to sons. Cassie can either pass on the X chromosome with the mutation and her son will have hemophilia A, or pass on the normal X chromosome, in which case her son would be normal and not affected with hemophilia A. The risk of being affected with hemophilia A is thus 50%.

Alan has hemophilia A. His sister, Alice, has one son, Blaine. Blaine also has hemophilia A. Alan and his wife Annette have 2 children, Bart and Barbara. Barbara has a daughter, Cassie, and a son Chip. Cassie and Blaine are married and have a son, Daniel, with hemophilia A. They are now expecting fraternal twins, a boy and a girl. What is Cassie and Blain's son's (the fraternal twin) risk to have hemophilia A? (A) 0% (B) 25% (C) 50% (D) 75% (E) 100%

*The answer is C.* Because Blaine has hemophilia A, he can only pass on an X chromosome with the mutation. Cassie is an obligate carrier with one X chromosome carrying the mutation and a normal X chromosome. The female fraternal twin can either receive a mutated X chromosome from both parents and have hemophilia A, or receive the mutated X from Blaine and a normal X from Cassie and be a carrier. The risk of being affected with hemophilia A is thus 50%.

Alan has hemophilia A. His sister, Alice, has one son, Blaine. Blaine also has hemophilia A. Alan and his wife Annette have 2 children, Bart and Barbara. Barbara has a daughter, Cassie, and a son Chip. Cassie and Blaine are married and have a son, Daniel, with hemophilia A. They are now expecting fraternal twins, a boy and a girl. What is Cassie and Blaine's daughter's (the fraternal twin) risk to be affected with hemophilia A? (A) 0% (B) 25% (C) 50% (D) 75% (E) 100%

*The answer is C.* The glucose-6-phosphate dehydrogenase alleles allow one to trace X chromosomes throughout the pedigree, and to determine the probabilities that someone has inherited the X allele that leads to the disease. Analyzing individual III-3, who has the disease, one can determine that an "A" polymorphic form of glucose-6-phosphate dehydrogenase travels with the disease locus. III-1 inherited his "A" allele from his mother (II-1), who contains two "A" alleles, one on each X chromosome. Since we do not know which X chromosome in II-1 contains the mutated allele, II-1 has a 50% chance of passing on the X chromosome with the mutated allele to her daughter, III-1 (the "B" allele in III-1 came from her father). When III-1 and III-2 have their child, IV-1, the "A" allele in IV-1 had to have come from the mother, as the father passed the Y chromosome to IV-1, and not his X chromosome. This is the same chromosome that has a 50% chance of carrying the mutation, so there is a 50% chance that IV-1 will express the disease.

An X-linked recessive disorder is found in a particular family. Using the glucose-6-phosphate dehydrogenase allele as a marker, which contains two polymorphic forms, A and B, all family members of the pedigree were genotyped for the presence of either the A, or B, or both alleles. Considering the pedigree shown, what is the probability that individual IV-1 will express this disease? (A) 100% (B) 75% (C) 50% (D) 25% (E) 0%

*The answer B.* Trisomy 18 (Edwards syndrome) is the second most common autosomal trisomy identified in liveborn infants, with trisomy 21 the most common. Ninety percent of Edwards syndrome cases result from nondisjunction, with cytogenetic studies demonstrating a full trisomy (47, +18). The clinical manifestations of Edwards syndrome include the abnormalities categorized by system below. 1. Face; micrognathia, microstomia, eye defects (microphthalmia, cataracts, coloboma). low-set and malformed ears. prominent occiput 2. CNS: microcephaly, neural tube defects (kg, meningocele, anencephaly), holoprosencephaly, Arnold-Chiari malformation, severe mental retardation, delayed psychomotor development 3. Musculoskeletal: clenched hands with overlapping fingers (index finger overrides the middle finger and fifth finger overrides the fourth finger), rocker-bottorn feet, short sternum, hypertonia 4. Cardiac ventricular septal defect, patent duct arteriosus 5. Gastrointeatinal: Meckel'a diverticulum, malrotation Trisomy 18 can be detected prenatally with the use of ultrasound. Suggestive findings include intrauterine growth restriction and polyhydramnios, especially in e fetus with abnormal hand arrangement.

An infant born prematurely to a 42-year-old Caucasian female in small for gestational age. Physical examination reveals microcephaly, low-set ears, prominent occiput and small mandible. The infant's fists are clenched and the fingers overlap. A bilateral fool deformity is observed. Which of the following is the most likely karyotype abnormality in this infant? A. Trisomy 21 B. Trisomy 18 C. Trisomy 13 D. 47,XXX E. 47,XXY

*The answer is A.* Achondroplasia is an autosomal dominant disease. There is no family history because achondroplasia is often caused by a new mutation.

Baby John was diagnosed with achondroplasia shortly after birth. What inheritance pattern should be discussed with the parents? (A) autosomal dominant (B) autosomal recessive (C) X-linked dominant (D) X-linked recessive (E) multifactorial

*The answer is C.* G6PD deficiency is X-linked. The risk that Britney received the mutation from her mother Lynne, an obligate carrier, is 50% or 0.5. The chance that the fetus will be a girl is 50% or 1⁄2. The chance that the girl will be a carrier is 50% if the mother is a carrier. So, 1⁄2 x 0.5 x 1⁄2 = 1⁄8.

Britney and Kevin have two healthy sons, Preston and Jaden. Britney has a full brother, Brian, with G6PD deficiency. Britney's mom, Lynne, has two brothers with G6PD deficiency. Britney is currently 10 weeks pregnant by her new partner, Isaa. What is the risk the current fetus has G6PD deficiency? (A) 1/2 (B) 1/4 (C) 1/8 (D) 1/16 (E) 1/32

*The answer is D.* The probability that individual II-2 transmitted the mutated allele to III-1 is 100%, and the probability that III-1 transmitted the mutated allele to IV-1 is also 50%. The probability that IV-1 has transmitted the mutated allele to V-1 is 50%, so the overall probability of the disease allele in II-2 being transmitted to V-1 is 1/4. The same is true for the chances of II-4 transmitting to III-2, to IV-2, and to V-1. For V-1 to express the disease, all of these events have to occur, such that there is a 1/4 chance that V-1 will express the disease (1/4 x 1/4) = 1/16.

Consider the pedigree shown in which the affected individuals are expressing a rare autosomal recessive disease. What is the probability that the unborn child of the marriage will express the disease? (A) 1/2 (B) 1/4 (C) 1/8 (D) 1/16 (E) 1/32 (F) 1/64 (G) 1/128

*The answer is C.* If sufficient numbers of normal X chromosomes are inactivated, there may not be enough of the normal gene product present for proper functioning. In these cases, there may be a partial or complete disease phenotype due to the fact that the majority of the gene product produced will be defective or nonfunctional.

Female carriers of X-linked recessive diseases sometimes exhibit some symptoms of the disease. The cause of this is which of the following? (A) variable expressivity of the X-linked gene (B) mitochondrial inheritance (C) skewed X chromosome inactivation (D) incomplete penetrance of the X-linked gene

*The answer is B.* In Fragile X syndrome the triplet repeat expansion, CGG, must reach a certain number of repeats before there is clinical manifestation of the disease. The repeat expands with succeeding generations and eventually will reach the critical number. That is why males without the disease can pass it on to subsequent generations where it appears because the threshold number of repeats has been reached. Females with a high number of repeats may also express some manifestations of the disease because of skewed X inactivation.

Fragile X syndrome is one of the most common causes of mental retardation in humans. It generally acts like an X-linked recessive disease, but some males do not have the disease yet they can pass it on, and some females are affected. The cause of the disease explains these observations. Fragile X syndrome is caused by which one of the following mechanisms? (A) a deletion of the Prader-Willi/Angelman gene on the father's X chromosome (B) a triplet repeat expansion (C) chromosome breakage (D) having two X chromosomes

*The answer is C.* Transposons can cause mutations in their former site when they relocate, alter gene expression at sites where they integrate, inactivate a gene by integrating somewhere in its sequence, and move pieces of nontransposon DNA to a new location in the genome. Although much of the time this is a detrimental event, sometimes the changes are beneficial and spread through the population.

Genetic variability in an organism (including humans) is significantly affected by which one of the following? (A) microsatellite DNA (B) satellite DNA (C) transposons (D) heterochromatin

*The answer is C.* Although protein-coding genes account for much of what are recognized as heritable traits, RNA-coding genes and epigenetic control are also important in gene expression, both normal and abnormal.

Heritable traits, both normal and disease producing, are determined by which of the following? (A) introns and exons of protein-coding genes with epigenetic control (B) RNA-coding genes under epigenetic control (C) protein-coding genes, RNA-coding genes, and epigenetic control (D) protein-coding genes, processed pseudogenes and retrogenes, and epigenetic control

*The answer is B.* Because there are no maternal copies of the Prader Willi/Angelman genes present, this is equivalent to a deletion of that area of the maternal chromosome 15 and Angelman syndrome will result.

If a sperm from the father with two copies of chromosome 15 fertilizes an egg from the mother that has no copies of chromosome 15, the conceptus would have the normal disomic complement of chromosome 15. Assuming that the conception goes to term, which one of the following phenotypes would the child have? (A) normal (B) Angelman syndrome (C) Prader-Willi syndrome (D) Beckwith-Wiedemann syndrome

*The answer is D.* Pleiotropy is when a gene mutation produces diverse phenotypic events. Marfan syndrome is one of the best examples of pleiotropy.

In Marfan syndrome, the affected protein, Fibrillin-1, is active in three parts of the body: the aorta, the suspensory ligaments of the lens, and the periosteum or connective tissue. This is an example of which of the following?

*The answer is D.* In lethal disorders, all the mutated genes are lost in each generation and these represent a third of the alleles for that mutated gene. In a population at equilibrium, the number of new mutations equals the number of genes lost, so that number of new mutations replacing those lost is one-third, or 33%.

In X-linked recessive lethal disorders, the mutant gene is not always inherited from a carrier female (Haldane's rule). What approximate percentage of affected males is attributable to a new mutation? (A) 100% (B) 75% (C) 66% (D) 33%

*The answer is C.* Both copies of the X chromosome in females are active only for a short time early in development.

In humans, the female is functionally hemizygous due to X chromosome inactivation. The inactivated chromosome is thus composed of which of the following? (A) satellite 1 DNA (B) beta-satellite DNA (C) facultative heterochromatin (D) constitutive heterochromatin (E) euchromatin

*The answer is A.* Myotonic dystrophy is caused by a triplet repeat expansion that expands with each succeeding generation. The larger the repeat, the earlier the onset and the more severe the disease is. This phenomenon is called anticipation and differs from incomplete penetrance and variable expressivity in that once the critical repeat threshold is reached, the disease is manifested with severity depending on the number of repeats.

In myotonic dystrophy, the severity of the disease increases with each succeeding generation. This phenomenon is called: (A) anticipation (B) incomplete penetrance (C) genomic imprinting (D) variable expressivity

*The answer is B.* Because Joe is unaffected, he can be a carrier or not be a carrier. He has a 1 in 3 chance of not being a carrier and a 2 in 3 chance of being a carrier.

Joe's brother has cystic fibrosis. What is the risk that Joe is a carrier? (A) 1/3 (B) 2/3 (C) 1/4 (D) 1/2

*The answer is A.* Because different genes (loci) can be involved in the development of leukemia, there is locus heterogeneity.

Mutations in different autosomal recessively inherited genes may result in the development of leukemia in Fanconi anemia patients. Which of the following best describes why this can happen? (A) locus heterogeneity (B) allelic heterogeneity (C) genotype-phenotype correlation (D) de novo mutations (E) variable expressivity

*The answer is B.* The gene from one parent that is methylated is not expressed, but the gene on the chromosome from the opposite sex is expressed.

One of the mechanisms by which genes are imprinted is which of the following? (A) change in DNA sequence (B) methylation of genes (C) heteroplasmy (D) trisomy rescue

*The answer is B.* Because Sally's father does not have hemophilia B, he does not have the X chromosome with the mutated gene to pass on to Sally. Therefore, the risk for Sally to have a child with hemophilia B is near 0.

Sally has a paternal uncle with hemophilia B, an X-linked recessive disease. Her risk of having a child with hemophilia B is best described as which of the following? (A) near 100% (B) near 0% (C) 50% with all male children (D) 50% for all children

*The answer is A.* If the normal gene is inactivated in a large enough number of cells, then there would be more defective gene products present than normal gene products and disease symptoms will be the result.

Some female carriers of hemophilia B (an X-linked recessive disease) have symptoms of the disease. Which of the following is the most likely explanation for how this occurs? (A) The X chromosome for the normal gene is inactivated in a majority of cells in the body. (B) Triplet repeat expansion. (C) Incomplete penetrance. (D) Variable expressivity.

*The answer is B.* Hypervariable minisatellite DNA is found near the telomere and at other chromosome locations. Satellite 1 and alpha satellite DNA are found at the centromeres. Microsatellite DNA is dispersed throughout all the chromosomes.

Which noncoding DNA is found near the telomeres of the chromosomes? (A) microsatellite DNA (B) hypervariable minisatellite DNA (C) satellite 1 DNA (D) alpha satellite DNA

*The answer is C.* A sibling of an affected individual, in whom the disease is due to autosomal recessive transmission, has a two-in-three chance of being a carrier. The allele distribution from the parents leads to four possibilities: having the disease, being homozygous for the wild- type allele, and two possible ways of being a carrier (either inherit the variant allele from the mother or father). Since the sister does not express the disease, she has two chances to be a carrier, and one chance to be homozygous normal, or a two-in-three chance of carrying the mutated allele.

The 1-year-old boy has a sister who is phenotypically normal. What is the probability that the sister is a carrier of the disease? (A) 100% (B) 75% (C) 67% (D) 50% (E) 33% (F) 25% (G) 0%

*The answer is E.* The main component of ribosomes is ribosomal RNA or rRNA. The other RNAs participate in the processes of transcription and translation but are not components of the ribosomes.

The central dogma of molecular biology is that DNA is transcribed into RNA, which is then translated into a protein. The translation takes place on the ribosomes. Which of the following RNAs are the main components of the ribosomes? (A) tRNA (B) snoRNA (C) snRNA (D) mRNA (E) rRNA

*The answer is B.* If the disease frequency for this autosomal recessive disorder (sickle cell anemia) is 1 in 400, then q2 5 1/400, and q 5 1/20. The heterozygote frequency, according to the Hardy-Weinberg equilibria, is 2pq, or 2 x 1 x 1/20, or 1 in 10. For the purposes of this calculation, we are considering p, which is really 19/20, to be functionally equivalent to 1.

The disease frequency for sickle cell anemia in the African-American population is 1 in 400. What is the carrier frequency in this population? (A) 1 in 5 (B) 1 in 10 (C) 1 in 20 (D) 1 in 40 (E) 1 in 50 (F) 1 in 100

*The answer is E.* Individual II-3 does not contribute to this calculation, as he is from outside the family, and the disease is rare in the population, so his risk of being a carrier is very low. The probability that individual II-2 is a carrier is 2/3, as she is an unaffected sibling of an affected individual. There is a 50% chance that II-2 will transmit the mutated allele to her son (III-1), such that the probability that III-1 will inherit this allele is 2/3 3 1/2, or 1/3 (33%).

The family in the pedigree shown has one family member with an autosomal recessive disease. Assuming that this is a rare disorder, what is the probability that individual III-1 is a carrier of the mutated allele? (A) 100% (B) 75% (C) 67% (D) 50% (E) 33% (F) 25% (G) 0%

*The answer is D.* If the disease frequency in the population is 1 in 1,000, and the disorder is X-linked, that means that out of 1,000 men, one would have the disease, as each man contains one X chromosome. Since women contain two X chromosomes, the carrier frequency is 1 in 500, as 500 women would contain 1,000 X chromosomes, one of which would contain the mutated allele.

The gene frequency for an X-linked recessive disease is 1 in 1,000 in the general population. What is the frequency of affected males in this population? (A) 1 in 10 (B) 1 in 100 (C) 1 in 500 (D) 1 in 1,000 (E) 1 in 2,000

*The answer is A.* What has been determined so far is that the amount of noncoding DNA corresponds with the biological complexity of an organism. The human genome is composed of 98% noncoding DNA and no other organism studied to date has this amount of noncoding DNA. Humans have 30,000 genes compared to 19,100 for the roundworm Caenorhabditis elegans. The number of chromosomes has no correspondence with biological complexity. For example, carp (a fish) have 100 chromosomes and humans have 46.

The genomes of a number of organisms, including humans, have now been characterized and compared. Which of the following describes one of the findings of these endeavors? (A) There is a correspondence between the biological complexity of an organism and the amount of noncoding DNA. (B) There is a correspondence between the biological complexity of an organism and the amount of coding DNA. (C) There is a correspondence between the biological complexity of an organism and the number of chromosomes. (D) There is no correspondence between the biological complexity of an organism and the amount of coding DNA, noncoding DNA, or the number of chromosomes.

*The answer is A.* Noncoding DNA such as introns, pseudogenes, and repetitive elements (such as satellite DNA) make up the rest of the genome.

The human genome codes for 30,000 genes that make up 2% of the DNA in the human nuclear genome. The remaining nuclear genome consists of which of the following DNA elements? (A) noncoding DNA (B) repetitive DNA (C) intron DNA (D) pseudogenes (E) satellite DNA

*The answer is E. Double helix DNA is coiled around histones that are organized into nucleosomes, which form the extended chromatin that compacts into a metaphase chromosome.

The levels of DNA packaging are depicted in which of the following sequences? (A) alpha satellite DNA, heterochromatin, centromere, euchromatin, metaphase chromosome (B) constitutive heterochromatin, euchromatin, facultative heterochromatin, metaphase chromosome (C) purines, pyrimidines, phosphates, nucleosome, 30 nm chromatin fiber, metaphase chromosome (D) double helix DNA, nucleosome, 30 nm chromatin fiber, extended chromatin, metaphase chromosome (E) double helix DNA, histones, nucleosomes, extended chromatin, metaphase chromosome

*The answer is A.* Methylation of DNA is one of the primary ways that a gene can be "turned off". Methylation plays a crucial role in genomic imprinting.

The modification of DNA that can make transcription of a DNA segment unlikely and thus "silence" a gene containing that segment is which one of the following? (A) methylation of cytosine nucleotides (B) acetylation of histones (C) retrotransposition (D) transcription

*The answer is D.* Alu repeats are located in the GC rich, R-band positive areas of the chromosome that contain many genes. There are many copies of the other sequences in the human genome, but they are not as abundant as the Alu sequences.

The most abundant sequence in the human genome is which one of the following? (A) rRNA tandem repeats (B) microsatellite DNA (C) satellite DNA (D) Alu repeats

*The answer is D.* A DNA nucleotide consists of one of the nitrogenous bases adenine, thymine, cytosine or guanine, the sugar deoxyribose, and a phosphate group. Uracil and the sugar ribose are components of RNA nucleotides.

The nitrogenous bases that make up the nucleotides of DNA are listed in which one of the following? (A) deoxyribose and ribose (B) deoxyribose, ribose, and phosphate (C) adenine, thymine, cytosine, uracil (D) adenine, thymine, cytosine, guanine

*The answer is D.* The full mutation threshold for Fragile X is reached when there are 200 repeats.

The repeat size of each individual's FMR-1 gene or genes is noted on the pedigree below. Which person has a full mutation in the FMR-1 gene and would have Fragile X syndrome? (A) person A (B) person B (C) person C (D) person D

*The answer is C.* The CGG repeat in Fragile X is outside the FMR1 gene on the X chromosome. The threshold for the manifestation of Fragile X syndrome is 200 repeats.

The second leading cause of inherited mental retardation is caused by highly expanded repeats outside of a gene on the X chromosome, which may result in hypermethylation of the gene so that it is not expressed. The syndrome that results from these highly expanded repeats is which one of the following? (A) Friedreich ataxia (B) Myotonic dystrophy (C) Fragile X (D) Spinocerebellar ataxia type 3

*The answer is C.* Because her mother is an obligate carrier of Alport syndrome, there is a 50% chance that she passed on the X chromosome with the mutation and a 50% chance that she passed on the normal X chromosome.

What is III-1's risk to be a carrier of Alport syndrome, an X-linked recessive condition? (A) 0 (B) 25% (C) 50% (D) 100%

*The answer is D.* This is an example of an X-linked recessive disease. Only males express the disease, and they obtained the mutated allele from their mothers, who do not express the disease (so it cannot be an X-linked dominant disorder). It is also observed that certain generations can be skipped, but that the gene is still passed through the pedigree via the female members of the family.

What is the most likely mode of transmission of this disease? (A) Autosomal recessive (B) Autosomal dominant (C) Mitochondrial (D) X-linked recessive (E) X-linked dominant

*The answer is D.* Since individuals IV-1 and III-4 are expressing the disease, their mothers (III-2 and II-4) must be carriers of the disease. III-2 also must have acquired the disease allele from her mother, II-2. II-3 is the sister of both II-2 and II-4, and would have had a 50% chance of inheriting the mutated allele from her mother, I-2.

What is the probability that individual II-3 is a carrier of the disease? (A) 0% (B) 25% (C) 33% (D) 50% (E) 67% (F) 75% (G) 100%

*The answer is A.* As this is an X-linked recessive disorder, any male who has the mutated allele will express the disease, and would not be a carrier for the disease. Since individual III-5 does not express the disease, he does not carry the mutated allele, and cannot pass the mutated allele on to his daughters.

What is the probability that individual III-5 is a carrier of the disease? (A) 0% (B) 25% (C) 33% (D) 50% (E) 67% (F) 75% (G) 100%

*The answer is G.* Individual IV-3's father has the disease, and he has passed on his X chromosome (which carries the mutated allele) to his daughters. The daughter does not express the disease because she also carries a normal allele that was inherited from her mother. All daughters of fathers with X-linked disorders will be carriers of the disease.

What is the probability that individual IV-3 is a carrier of the disease? (A) 0% (B) 25% (C) 33% (D) 50% (E) 67% (F) 75% (G) 100%

*The answer is E.* Because the cause of achondroplasia in John is a new mutation, it is extremely unlikely to happen again so the risk is ~0.

What is the recurrence risk for the couple to have another child with achondroplasia? (A) 50% (B) 25% (C) 3%-5% (D) 1%-2% (E) 0%

*The answer is E.* Imprinting does not change the DNA sequence of a gene. The maternal and paternal copies of genes are mostly active or silent at the same time. The end result of a deletion of chromosome 15 or UPD is that there are no paternal copies of the gene(s) involved in the syndrome, so there is no difference in phenotype.

Which of the following is a characteristic of genomic imprinting? (A) Most genes must bear the parent of origin imprint for proper expression. (B) The parent of origin copy to be imprinted differs from gene to gene, and most genes require an imprint. (C) The phenotype of a child with Prader Willi syndrome is different depending on whether the child has a deletion on chromosome 15 or UPD for the chromosome. (D) During gamete formation, the imprint is removed from the genes and replaced with an imprint of the opposite sex. (E) Imprinting does not disturb the primary DNA sequence.

*The answer is B.* Duchenne muscular dystrophy is X-linked recessive.

Which of the following is an X linked recessive disease? A. Acute intermittent porphyria B. Duchenne muscular dystrophy C. Huntington disease D. Marfan syndrome

*The answer is D.* If the full sibling's status was unknown, he would have a 1 in 4 risk of being unaffected and not carrying a CF mutation gene, a 2 in 4 risk of being unaffected but a carrier of a CF mutation and a 1 in 4 risk of having CF. Because he is unaffected, there are 3 possible independent outcomes. He now has a 1 in 3 chance of not carrying a mutated CF gene, but a 2 in 3 chance of being a carrier of a CF mutation.

Which of the following is the risk that an unaffected full sibling of a patient with cystic fibrosis (CF) carries a mutated CF gene? (A) 1 in 2 (B) 1 in 4 (C) 3 in 4 (D) 2 in 3

*The answer is C.* The 171 base pair repeat unit of alpha satellite DNA makes up much of the centromeric DNA.

Which one of the following is a major component of centromeric DNA? (A) the Barr body (B) the XY body (C) alpha satellite DNA (D) Z-DNA

*The answer is C.* DNA is chemically modified by methylation and is less likely to be transcribed into RNA.

Which one of the following is the mechanism responsible for genomic imprinting? (A) acetylation (B) phosphorylation (C) methylation (D) transposition

*The answer is A.* In DNA, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C). In RNA, adenine pairs with uracil (U).

Which one of the following is the way bases are paired in a double helix of DNA? (A) A-T, G-C (B) A-U, G-C (C) A-G, C-T (D) A-C, G-T

*The answer is D.* Sex-linked traits are passed via the X chromosome. Males transmit their Y chromosome to their sons, and their X chromosomes to their daughters. Thus, an affected male cannot transmit a mutated X allele to his son, so the presence of male-to-male transmission in a pedigree categorically eliminates X-linked transmission as the genetic pattern of inheritance.

Which one of the following observations would rule out a sex-linked trait in an extended family pedigree? (A) Males expressing the disease (B) Females expressing the disease (C) Female-to-male transmission (D) Male-to-male transmission (E) The disease-skipping generations in the pedigree

*The answer is C.* In Pedigree C, the condition appears in every generation in both sexes. Pedigree A is a possibility, but only males are affected and Pedigree B the condition "skips" a generation and only males are affected.

Which pedigree best represents X-linked dominant inheritance for a nonlethal condition? (A) pedigree A (B) pedigree B (C) pedigree C