Chapter 10: Human Genetics

*The answer is C.* For multifactorial disorders, the risk of recurrence for first-degree relatives is approximately the square root of its incidence.

In Ireland, the incidence of neural tube defects is 1 in 200. What is the approximate recurrence risk for first-degree relatives? (A) 1 in 100 (B) 1 in 37 (C) 1 in 14 (D) 1 in 5

*The answer is C.* The bleeding disorder most likely to be asymptomatic is Type 1 von Willebrand disease because there is von Willebrand factor present but in reduced amounts.

The bleeding disorder that is most likely to be asymptomatic is which one of the following? (A) hemophilia A (B) hemophilia B (C) Type 1 von Willebrand disease (D) Type 3 von Willebrand disease

*The answer is C.* Assortative mating, where "like mates with like" violates the assumption that there is random mating in the population. Assortative mating would cause an increase in the frequency of the gene (or genes) responsible for whatever characteristic was associated with the assortative mating, like tall with tall, short with short, etc.

Which one of the following violates the assumptions upon which the Hardy-Weinberg Law is based? (A) There is no selection against any allele in the population. (B) There is a constant mutation rate where lethal genes are replaced by new mutations. (C) There is a large population with assortative mating. (D) There is no migration into the population.

*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 C.* Mucopolysaccharidosis is due to lysosomal enzyme deficiencies that result in the lack of complete degradation of glycosaminoglycans, resulting in the buildup of incomplete degradation products in cells.

Which of the following is a genetic metabolic disease involving degradation pathways? (A) glycogen storage disease (B) maple syrup urine disease (C) mucopolysaccharidosis (D) galactosemia

*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 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 C.* A missense mutation results in the change of only a single amino acid.

The severe form of alpha-1 antitrypsin deficiency is the result of a single nucleotide substitution that produces a single amino acid substitution. This is best described as a A. Frameshift mutation B. In-frame mutation C. Missense mutation D. Nonsense mutation E. Splice-site mutation

*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.* 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.* In genetically normal (46, XX) females. one X chromosome is normally randomly deactivated in each cell during early embryonic development. X-inactlvatlon lyonization) is maintained across cell division, resulting in clusters of cells throughout the body that express either the maternal or paternal X chromosome. This mosaic pattern of X-chromosome expression generally prevents X-linked recessive conditions from manifesting in female carriers. However, in rare cases, skewed lyonization (uneven inactivation of maternal/paternal X chromosome due to chance alone) may result in females developing an X-linked recessive condition (eg, classic hemophilia). The process of lyonization converts the inactive X chromosome into condensed heterochromatin, which can be identified on microscopy as a compact body at the periphery of the nucleus (Barr body). Heterochromatin consists of heavily methylated DNA (eg, cytosine converted to methylcytosine) and deacetylated histones. which cause it to have a low level of transcriptional activity (Choice C). A small proportion of genes remain transcriptionally active on the inactivated X chromosome. For this reason, inheritance of an abnormal number of X chromosomes can cause clinical manifestations due to a gene-dosage effect, as seen with Tumor (45, XO) and Klinefelter (47, XXY) syndromes.

A 24-year-old woman comes to the office after discovering a new mole on her right leg. She is worried that it might be skin cancer as she has used tanning beds several times s year since age 18. Physical examination shows s 5-mm brown, oval macule on her anterior thigh with a homogeneous coloration and discrete borders. The lesion appears darker than her other moles. A biopsy of the lesion shows normal-appearing news cells clustered in the epidermis, end she is diagnosed with a benign acquired melanocytic; nevus. During histologist analysis, her epithelial cells are each found to contain a condensed body composed of heavily methylated DNA at the periphery of the nucleus. This region is most likely associated with which of the following genetic findings? A. DNA supercoil accumulation B. Double-strand DNA break repair C. History acetylation D. Impaired mismatch repair E. Low transcription activity

*The answer is A.* The presence of an expanded trinucleotide repeat in the 5′ untranslated region of the gene is an accurate test for fragile X syndrome.

A 6-year-old boy has a family history of mental retardation and has developmental delay and some unusual facial features. He is being evaluated for possible fragile X syndrome. Which of the following would be most useful in helping establish the diagnosis? A. Genetic test for a trinucleotide repeat expansion in the fragile X gene B. IQ test C. Karyotype of the child's chromosomes D. Karyotype of the father's chromosomes E. Measurement of testicular volume

*The answer is A.* An F8 intron gene inversion, called a "flip" inversion, is responsible for the majority cases of hemophilia A.

A gene inversion, called a "flip" inversion, is the most common mutation in which one of the following? (A) hemophilia A (B) hemophilia B (C) von Willebrand disease (D) Christmas disease

*The answer is D.* Because there is one normal chromosome 22 and one deleted chromosome 22, there is a 50% chance of passing one or the other on with each pregnancy.

A woman comes to clinic because of her family history of tetralogy of Fallot (a conotruncal heart defect). Her father was born with a heart defect, has immunity problems, and schizophrenia. Her brother has cleft palate and a heart defect as well. The patient was studied cytogenetically and found to have a microdeletion of 22q11.2 by FISH. What is the best estimate of her recurrence risk for a future pregnancy? (A) 2%-3% (B) 5%-6% (C) 10% (D) 50% (E) 100%

*The answer is A.* Only one mature oocyte results from each primary oocyte. At each of the two cell divisions in meiosis, a polar body is formed, which usually degenerates.

At the completion of oogenesis, how mature oocytes are formed from each primary oocyte? (A) one (B) two (C) three (D) four

*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 A.* In an autosomal recessive disease, those individuals who are represented by "2pq" are heterozygotes who carry a copy of a mutated disease-causing gene.

In an autosomal recessive disease, the people represented by the "2pq" figure in the equation "p2 + 2pq + q2" are: (A) carriers of the disease (B) affected by the disease (C) those who have a new mutation (D) those without the mutant gene

*The answer is A.* Imprinting refers to the differential transcriptional activity of genes inherited from the father versus the mother.

In studying a large number of families with a small deletion in a specific chromosome region, it is noted that the disease phenotype is distinctly different when the deletion is inherited from the mother as opposed to the father. What is the most likely explanation? A. Imprinting B. Mitochondrial inheritance C. Sex-dependent penetrance D. X-linked dominant inheritance E. X-linked recessive inheritance

*The answer is D.* The hallmarks of children with chromosomal anomalies are mental retardation and multiple congenital anomalies. In this case, the individual has learning problems that have not yet been recognized as mental retardation, and he has minor anomalies rather than major birth defects that cause cosmetic or surgical problems. The physician was astute to suspect a chromosomal anomaly even when the developmental disability and alterations in physical appearance were subtle. Other indications for a karyotype include a couple with multiple miscarriages, an individual at risk for inheriting or transmitting a chromosomal rearrangement, a child with ambiguous external genitalia, or an individual with characteristics of a chromosomal syndrome such as Down's, Turner's, or Klinefelter's syndrome. Chromosome translocations are characteristic of many types of cancer, but these occur in somatic cancer cells rather than in the patient's germ line.

A 10-year-old boy is referred to the physician because of learning problems and a lack of motivation in school. His family history is unremarkable. Physical examination is normal except for single palmar creases of the hands and curved fifth fingers (clinodactyly). The physician decides to order a karyotype. Which of the following indications for obtaining a karyotype would best explain the physician's decision in this case? a. A couple with multiple miscarriages, or a person who is at risk for an inherited chromosome rearrangement b. A child with ambiguous genitalia who needs genetic sex assignment c. A child with an appearance suggestive of Down's syndrome or other chromosomal disorder d. A child with mental retardation and/or multiple congenital anomalies e. A child who is at risk for cancer

*The answer is C.* The disease-producing allele of the gene is associated with the presence of the HindII site. Notice that both affected males show two smaller bands (75 and 40 bp). II-3, a carrier female, also has these two smaller bands in her pattern, in addition to a larger PCR product (115 bp), representing the absence of the HindII site on her normal chromosome. III-2 has only the larger PCR product (notice the density because both chromosomes yielded this product). She is homozygous for the normal allele. Choice A, carrier, would be correct if her pattern had looked like those of II-3 and III-1. All the males shown are hemizygous (choice B) for the dystrophin gene because they have only one copy. II-1 and III-3 are hemizygous for the disease-producing allele, and II-2 is hemizygous for the normal allele. No one in the family is homozygous for the disease producing allele (choice D). In an X-linked pattern, this would be characteristic of a female with two copies of the disease-producing allele and is very rarely seen. III-2 is not a manifesting heterozygote (choice E) because she has no symptoms and is not a heterozygote.

A 14-year-old boy has Becker muscular dystrophy (BMD), an X-linked recessive disease. A maternal uncle is also affected. His sisters, aged 20 and 18, wish to know their genetic status with respect to the BMD. Neither the boy nor his affected uncle has any of the known mutations in the dystrophin gene associated with BMD. Family members are typed for a HindII restriction site polymorphism that maps to the 5′ end of intron 12 of the dystrophin gene. The region around the restriction site is amplified with a PCR. The amplified product is treated with the restriction enzyme HindII and the fragments separated by agarose gel electrophoresis. The results are shown below. What is the most likely status of individual III-2? A. Carrier of the disease-producing allele B. Hemizygous for the disease-producing allele C. Homozygous for the normal allele D. Homozygous for the disease-producing allele E. Manifesting heterozygote

*The answer is B.* This patient's short stature, short and thick neck, broad chest, and shortened fourth metacarpals are characteristic of Tumor syndrome (Ts). Bicuspid aortic valve is the most common congenital cardiac malformation in patients with TS. It is usually an isolated abnormality, however, it may occur in combination with other anomalies, particularly aortic coarctation. A nonstenotic bicuspid aortic valve manifests with an aortic ejection sound, which an early systolic, high-frequency click heard over the right second interspace. As the valve calcifies or may result in progressive valvular dysfunction, manifesting with aortic stenosis or regurgitation and associated murmurs. Patients with a bicuspid aortic valve are susceptible to infectious endocarditic due to abnormal leaflet structure and turbulent flow.

A 14-year-old girl is brought to the office for a routine physical examination. The patient will be starting her freshman year of high school soon. She plays the clarinet in the school band but does not play any sports. The patient says that she is not sexually active and does not use tobacco, alcohol, or illicit drugs. Height is at the 5th percentile, and weight is at the 25th percentile. Her temperature is 36.7 C (98 F), blood pressure is 120/80 mm Hg, pulse is 88/min, and respirations are 16/min. Physical examination shows a short and thick neck, a broad chest, and shortened fourth metacarpels bilaterally. A murmur is heard on cardiac auscultation. Which of the following would most likely be seen in an echocerdiogram? A. Atrial sepal defect B. Bicuspnd aortic valve C. Mitral stenosis D. Mitral valve prolapse E. Patent ductus arteriosus F. Ventricular septa defect

*The answer is E.* Children with chromosome abnormalities often exhibit poor growth (failure to thrive) and developmental delay with an abnormal facial appearance. This baby is too young for developmental assessment, but the catlike cry should provoke suspicion of cri-du-chat syndrome. Cri-du-chat syndrome is caused by deletion of the terminal short arm of chromosome 5 [46,XX,del(5p), also abbreviated as 5p−] as depicted in panel e. When a partial deletion or duplication like this one is found, the parents must be karyotyped to determine if one carries a balanced reciprocal translocation. The other karyotypes show (a) deletion of the short arm of chromosome 4 [46,XY,del(4p) or 4p−]; (b) XYY syndrome (47,XYY); (c) deletion of the long arm of chromosome 13 [46,XX,del(13q) or 13q−]; (d) Klinefelter's syndrome (47,XXY). Most disorders involving excess or deficient chromosome material produce a characteristic and recognizable phenotype (e.g., Down's, cri-du-chat, or Turner's syndrome). The deletion of 4p− (panel A) produces a pattern of abnormalities (syndrome) known as Wolf-Hirschhorn syndrome; deletion of 13q− produces a 13q− syndrome (no eponym). The mechanism(s) by which imbalanced chromosome material produces a distinctive phenotype is completely unknown.

A 2-week-old baby is hospitalized for inadequate feeding and poor growth. The parents are concerned by the child's weak cry. An experienced grandmother accompanies them, saying she thought the cry sounded like a cat's meow. The grandmother also states that the baby doesn't look much like either parent. The physician orders a karyotype after noting a small head size (microcephaly) and subtle abnormalities of the face. Which of the results pictured below is most likely? a. Result A b. Result B c. Result C d. Result D e. Result E

*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 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 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 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 D.* As a translocation carrier, it is possible that she can transmit the translocated chromosome, containing the long arms of both 14 and 21, to each of her offspring. If she also transmits her normal copy of chromosome 21, then she will effectively transmit two copies of chromosome 21. When this egg cell is fertilized by a sperm cell carrying another copy of chromosome 21, the zygote will receive 3 copies of the long arm of chromosome 21. The miscarriages may represent fetuses that inherited 3 copies of the long arm and were spontaneously aborted during pregnancy.

A 26-year-old woman has produced two children with Down syndrome, and she has also had two miscarriages. Which of the following would be the best explanation? A. Her first cousin has Down syndrome. B. Her husband is 62 years old. C. She carries a reciprocal translocation involving chromosomes 14 and 18. D. She carries a Robertsonian translocation involving chromosomes 14 and 21. E. She was exposed to multiple x-rays as a child.

*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 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 C.* Chromosomal abnormalities are responsible for about 50% of first trimester spontaneous abortions, and of these the most common cause is trisomy (52%). The most common trisomy in spontaneous abortion is trisomy 16. Polyploidy (choice A) is seen in 22% and monosomy (choice B) in 19%.

A 37-year-old woman is brought to emergency department because of crampy abdominal pain and vaginal bleeding for 3 hours. She is 11 weeks pregnant. This is her first pregnancy. Her pregnancy has been unremarkable until this episode. Her temperature is 36.8 C (98.2 F), pulse is 106/min, blood pressure is 125/70 mm Hg, and respiration rate is 22/min. Speculum examination shows the presence of blood in the vagina and cervical dilatation. Inevitable spontaneous abortion is suspected. After discussing the condition with the patient, she gave her consent for dilatation and curettage. What is the most common cause of spontaneous abortions? A. Chromosomal abnormality, polyploidy B. Chromosomal abnormality, monosomy X C. Chromosomal abnormality, trisomy D. Effects of environmental chemicals E. Immunologic rejection F. Infection G. Maternal endocrinopathies H. Physical stresses I. Teratogenic drugs

*The answer is A.* The fetus has unbalanced chromosomal material (additional chromosomal material on one copy of chromosome 18). One of the parents is likely to be a carrier of a reciprocal translocation involving chromosome 18 and one other chromosome (un-specified in stem).

A 38-year-old woman in her 15th week of pregnancy undergoes ultrasonography that reveals an increased area of nuchal transparency. Amniocentesis is recommended and performed at 16 weeks' gestation. The amniotic karyotype is 46,XYadd(18)(p.11.2), indicating additional chromosomal material on the short arm of one chromosome 18 at band 11.2. All other chromosomes are normal. What is the most likely cause of this fetal karyotype? A. A balanced reciprocal translocation in one of the parents B. A balanced Robertsonian translocation in one of the parents C. An isochromosome 18i(p) in one of the parents D. Nondisjunction during meiosis 1 in one of the parents E. Nondisjunction during meiosis 2 in one of the parents

*The answer is D.* The fact that the mother of the affected child has an affected brother and an affected nephew through her sister suggests X-linked recessive inheritance. This is made more likely because the symptoms suggest a mucopolysaccharidosis (storage of glycosaminoglycans) and because one type exhibits X-linked recessive inheritance (Hunter's syndrome or MPS type II). When evaluating the possibility of an X-linked disorder, it is important to remember the pattern of inheritance of the X chromosome. Females have two X chromosomes, which are passed along in a random fashion. They pass any given X chromosome to 50% of their sons and 50% of their daughters. For an X-linked recessive condition, those daughters who inherit the affected allele are heterozygous carriers of the disorder but are not affected. Since males have only one X chromosome, those who inherit the affected allele are affected with the disorder. Given X-linked recessive inheritance, the mother must have the abnormal allele on one of her X chromosomes (she is an obligate carrier) in order for her son and brother to be affected. The fetus thus has a 1/2 chance of being a boy and a 1/2 chance of being affected given male sex, resulting in a 1/4 (25%) overall risk of being affected.

A 4-year-old boy presents to the physician's office with coarse facies, short stature, stiffening of the joints, and mental retardation. Both parents, a 10-year-old sister, and an 8-year-old brother all appear unaffected. The patient's mother is pregnant. She had a brother who died at 15 years of age with similar findings that seemed to worsen with age. She also has a nephew (her sister's son) who exhibits similar features. Based on the probable mode of inheritance, the risk that her fetus is affected is a. 100% b. 67% c. 50% d. 25% e. Virtually 0

*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 A.* The patient suffered from Huntington disease. The gene that is responsible for this autosomal dominant disorder, which is localized to chromosome 4, has an excess of trinucleotide repeats and codes the protein huntingtin, which results in atrophy of the caudate nucleus. The exact mechanism whereby the mutant huntingtin causes brain injury remains unclear. Patients begin developing a progressive cognitive decline and choreoathetosis between 20 and 50 years of age, and death often follows within the next 20 years. Depression is a common comorbidity. other diseases involving expansion mutations of trinucleotide repeats include myotonic dystrophy, fragile X syndrome, and Friedreich ataxia. The trinucleotide repeat disorders demonstrate a phenomenon known as genetic anticipation, in which symptom onset occurs at a progressively earlier age in successive generations.

A 42-year-old man presents to the clinic with complaints of uncontrollable movements. During the interview, he seems depressed and states that he has "known this was coming for a long time now." He states his father suffered from a movement disorder and eventually cognitive decline. You note that he exhibits some mild involuntary writhing movements in his limbs. Which of the following is the likely cause of this disease? A. Expansion of CAG trinucleotide repeats on chromosome 4 B. Expansion of CCG trinucleotide repeats on chromosome 11 C. Expansion of CGG trinucleotide repeats on chromosome X D. Expansion of CTG trinucleotide repeats on chromosome 19 E. Expansion of GAA trinucleotide repeats on chromosome 9

*The answer is C.* This patient has recurrent infections (possibly reflecting neutropenia), pallor (anemia). and ecchymoses (thrombocytopenia), with a peripheral blood smear that shows several abnormal myelocyte precursors containing coarse rod-shaped intracytoplasmic granules called Auer rods. This presentation is characteristic of acute myelogenous leukemia (AML). The M3 variant of AML, acute promyelocytic leukemia (APML). characterized by the presence of promyelocytes on smear, is associated with disseminated intravascular coagulation (bleeding, thrombocytopenia, prolonged prothrombin and activated thromboplastin time). Affected eels exhibit the cytogenetic abnormality t(15,17) This cytogenetic change represents a translocation between the retinoic acid receptor alpha (RARα) gene on chromosome 17 and promyelocytic leukemia (PML) gene on chromosome 15. Fusion of these 2 genes produces a chimeric gene product, PML/RARα. which codes for an abnormal retinoic acid receptor. This abnormal fusion gene inhibits promyelocyte differentiation and triggers the development of APML. Management is with all-trans retinoic acid.

A 43-year-old man comes to the hospital due to recurrent episodes of fever and sore throat despite multiple courses of antibiotic therapy. For the past several months, he has also felt "run down' and fatigued all the time. His wife adds that he bruises easily and has had bleeding gums on several occasions. His temperature is 37.8 C (100.2 F). On examination. he has mucosal pelter, pharyngeal erythema, and multiple ecchymoses on his extremities. His peripheral blood smear is shown in the image below. Which of the following chromosomal abnormalities is most likely present in the affected cells? A t(8;14) B. t(9;22) C. t(11;14) D. t(15;17) E. 13q-

*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 C.* This clinical vignette describes the typical presentation of xeroderma pigmentosum, which literally means pigmented dry skin. This autosomal recessive (AR) condition occurs due to decreased ability to repair DNA following damage by UV light The skin of affected individuals is normal at birth. The disease manifests during the first year of life with erythema, scaling and subsequent hyperpigmentation and lentigo formation on light-exposed areas (especially the face). Later, the skin of affected areas shows atrophy, telangiectasias, and intermingling areas of hypo-and hyperpigmentation. Skin malignancies, including squamous cell carcinoma, basal cell carcinoma, and malignant melanoma. develop as early as at 5-6 years of life. Normally the regions of DNA damaged by UV radiation are excised and replaced. In xeroderma pigmentosum, the genes that code for various DNA repair enzymes are abnormal. Impairment of DNA repair results due to defects in excision of abnormal nucleotides or defects in replacement of nucleotides following excision.

A 5-year-old girl is brought to the physician by her mother. The mother tells you that the girl's skin becomes red and scaling with only minimal sun exposure. She began to notice this phenomenon when the child was 7 months old. Now the girl's skin is thin and hyperpigmented. The patient has a few nevi on her hands that have been rapidly enlarging. The defective gene in this patient is responsible for A. Regulation of cell cycle B. Signal transduction C. DNA excision repair D. DNA mismatch repair E. Prevention of microdeletions F. Regulation of apoptosis

*The answer is C.* Base excision repair is responsible for repairing various non-bulky DNA base alterations, including depurination, alkylation, oxidation, and deamination Excessive consumption of dietary nitrites can promote the deamination of cytosine, adenine, and guanine to form uracil, hypoxanthine, and xanthine, respectively. If these abnormal bases are not removed and replaced with the correct base, DNA mutations and carcinogenesis may result. Base excision repair (not to be confused with nucleotide excision repair or mismatch repair) begins with the recognition of abnormal bases by specific glycosylases. These cleave the altered DNA bases from the parent DNA molecule, leaving an empty sugar-phosphate site called an apurinic/apyrimidinic site (AP). An endonuclease then cleaves the 5' end of the AP site before a lyase (or phosphodiesterase) enzyme subsequently completes extraction of the AP site from the DNA molecule by removing the remaining sugar-phosphate group. DNA polymerase then fits the gap with the correct sugar-phosphate base, and the final nick is sealed by Ilgase.

A 58-year-old man comes to the office due to difficulty swallowing for the past several months. He has the most trouble with sold foods and says, "They seem to get stuck in my throat if I don't chew a lot" The patient has no chest pain or heartburn and has lost 4.5 kg (10 lb) in the last 3 months. He has been an avid hunter for many years and frequently cures the meat he eats with sodium nitrite Physical examination is unremarkable. Endoscopy shows an ulcerated mass in the distal third of the esophagus, and biopsy samples ere obtained from the mass and adjacent normal mucosa. Analysis of the samples shows accelerated cytosine deamination of chromosomal DNA in both normal and malignant epithelial cells. This damage is most likely to be repaired through which of the following enzymatic sequences? A. Endonuclease. polymerase. glycosylase, lyase, ligaee B. Endonuclease, polymerase, lyase, glycosylase, Iigase C. Glycosylase, endonuclease, lyase, polymerase, ligase D. Glycosylase, ligase, lease, endonuclease, polymerase E. Lyase, endonuclease, glycosylase, polymerase, ligase

*The answer is A.* The affected grandfather has marker alleles DS2 and DS3. There is no information about which one is in linkage phase with his disease-producing huntingtin allele. On the basis of the pedigree alone, the daughter has a 25% change of inheriting the grandfather's disease- producing huntingtin allele (choice B); however, she would like more information. Because her father (II-1) does not wish to be tested or know about his genetic status with respect to Huntington's, it is unethical to test the daughter for the triplet repeat expansion. The results would necessarily reveal the status of her father also. By doing an indirect genetic test, one can see the daughter has inherited one of her marker alleles (DS2) from the grandfather via her father. This means she has a 50% chance of developing Huntington's because there is a 50% chance that DS2 is a marker for the disease producing huntingtin allele in the grandfather and a 50% chance it is not (and DS3 is). Notice the result does not reveal additional information about her father (II-1). Before her testing, he had a 50% chance of having the disease-producing huntingtin allele. His risk is still 50% with the information from the daughter's test. However, if the father (II-1) does develop Huntington's in the future, that will mean that the daughter has a 100% chance of having the disease also (choice D).

A 66-year-old man (I-2) has recently been diagnosed with Huntington disease, a late-onset, autosomal dominant condition. His grand-daughter (III-1) wishes to know whether she has inherited the disease-producing allele, but her 48-year-old father (II-1) does not wish to be tested or to have his status known. The grandfather, his unaffected wife, the granddaughter, and her mother (II-2) are tested for alleles of a marker closely linked to the huntingtin gene on 4p16.3. The pedigree and the results of testing are shown below. What is the best information that can be given to the granddaughter (III-1) about her risk for developing Huntington disease? A. 50% B. 25% C. Marker is not informative D. Nearly 100% E. Nearly 0%

*The answer is F.* This patient has fragile X syndrome, which is the most common cause of inherited mental retardation and, after trisomy 21, is the second most common cause of genetically associated mental deficiencies. The genetic basis of this disease is a triplet (CGG) repeat expansion in the 5′ untranslated region of a gene (FMR-1) on the X chromosome. The standard diagnostic testing for fragile X syndrome uses molecular genetic techniques. The exact number of CGG triplet repeats can be determined by Southern blotting or by amplification of the repeat with a polymerase chain reaction and gel electrophoresis. Southern blot analysis provides a more accurate estimation of the number of CGG triplet repeats if a full mutation is present (with a large CGG expansion). Males with a permutation (moderate expansion but not sufficient to cause classic symptoms of fragile X) may have a fragile X tremor/ataxia syndrome (FXTAS) that presents later in life, usually after 50 years of age. Fragile X is also seen in females where learning disabilities and mild mental retardation characterize the syndrome.

A 9-year-old boy is referred to a pediatric clinic by his school psychologist because of poor academic performance, cognitive delay, and hyperkinetic behavior. Family history is significant for early dementia and ataxia in the maternal father. Physical examination reveals that the boy has a long thin face with prominent ears, some facial asymmetry, and a prominent forehead and jaw. His vital signs are normal, his lungs are clear to auscultation, and heart sounds are normal. His abdomen is soft, nontender, and nondistended. Examination of the extremities showed hyperextensible finger joints. The examining physician suspects a possible genetic disorder. What would be the best test to diagnose this disease? A. Brain MRI B. Cytogenetic testing for fragile X C. Developmental evaluation by a speech/language therapist D. EEG E. Measurement of testicular volume F. Southern blot analysis

*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.* The presence of consanguinity (double line in the figure) is a red flag for autosomal recessive inheritance because, although disease-causing alleles are rare, the probability of a homozygous individual escalates dramatically when the same rare allele descends through two branches of a family. Using a lowercase r to denote the retinitis pigmentosa allele, the affected male (individual II-2 in the pedigree) has a genotype of rr. His prospective mate has a very low risk to be a carrier for this rare disease, making her genotype RR. Their children will all have genotypes Rr, making them carriers but not affected. Retinitis pigmentosa is another disease manifesting genetic heterogeneity, with autosomal dominant, autosomal recessive, and X-linked recessive forms. Carriers of autosomal recessive diseases are heterozygotes with one normal and one abnormal allele. Many autosomal recessive diseases involve enzyme deficiencies, indicating that 50% levels of enzymes found in heterozygotes are sufficient for normal function. The probability that an affected individual will encounter a mate who is a carrier is approximately twice the square root of the disease incidence. This figure derives from the Hardy-Weinberg law. Since most recessive diseases have incidences lower than 1/10,000, the risk for unrelated mates to be carriers is less than 1/50, and the chance of having an affected child is less than 1/50 × 1/4 = less than 1/200. Disorders that are fairly common in certain ethnic groups, such as cystic fibrosis, are exceptions to this very low risk.

A child is evaluated by an ophthalmologist and is found to have retinitis pigmentosa, a disorder characterized by pigmentary granules in the retina and progressive vision loss. The pedigree below is obtained and the family comes in for counseling. What is the risk for individual II-2 of having an affected child if he mates with an unrelated woman? a. 100% b. 75% c. 50% d. 25% e. Virtually 0

*The answer is D.* The family history and the likelihood that the boys have a metabolic disease suggest autosomal recessive inheritance. Autosomal recessive conditions tend to have a horizontal pattern in the pedigree. Although there may be multiple affected individuals within a sibship, parents, offspring, and other relatives are generally not affected. Most autosomal recessive conditions are rare; however, consanguinity greatly increases the likelihood that two individuals will inherit the same mutant allele and pass it along to their offspring. The recurrence risk for the fetus will be that for an autosomal recessive condition with carrier parents—1/4 or 25%. This risk is not affected by the sex of the fetus. The disease caused by cystathionine-β-synthase (CS) deficiency is homocystinuria (236300). S-adenosylmethionine accepts methyl groups and is converted to S-adenosylhomocysteine, which yields homocysteine; homocysteine is converted to cystathionine by CS. Methionine and homocysteine (dimerized to homocystine) accumulate, and homocystine is excreted in urine. Pyridoxine is a cofactor for CS and is beneficial in some forms of homocystinuria. Other causes of homocystinuria include defective cobalamin (vitamin B12) metabolism.

A couple comes to the physician's office after having had two sons affected with a similar disease. The first-born son is tall and thin and has dislocated lenses and an IQ of 70. He has also experienced several episodes of deep vein thromboses. The chart mentions deficiency of the enzyme cystathionine-β-synthase, but a diagnosis is not given. The second son was treated from an early age with pyridoxine (vitamin B6) and is less severely affected. No other family members are affected. While taking a family history, the physician discovers that the parents are first cousins. The 38-year-old mother is pregnant, and amniocentesis has demonstrated that the fetus has a 46,XY karyotype. The risk that the fetus will be affected with the same disease is a. 100% b. 67% c. 50% d. 25% e. Virtually 0

*The answer is D.* The fetal karyotype shows a partial trisomy and a partial monosomy. The fetus has 46 chromosomes, indicating euploidy (a multiple of 23), not aneu-ploidy (choice A). The father is the reciprocal translocation carrier, not the fetus (choice B). Nondisjunction during meiosis (choice C) produces full trisomies and chromosomes have normal structure. Although the father is a translocation carrier, his genetic material is balanced, not unbalanced (choice E). He is diploid for all loci.

A couple has one son, who is age 7. Multiple attempts to have a second child have ended in miscarriages and spontaneous abortions. Karyotypes of the mother, the father, and the most recently aborted fetus are represented schematically below. What is the most likely explanation for the most recent pregnancy loss? A. Aneuploidy in the fetus B. Fetus identified as a reciprocal translocation carrier C. Nondisjunction during oogenesis in the mother D. Partial monosomy and trisomy in the fetus E. Unbalanced chromosomal material in the father

*The answer is F.* It is important that the pedigree be an accurate reflection of the family history and that information not be recorded unless specifically mentioned. Pedigree B in the figure omits the double line needed to indicate consanguinity, and pedigree C assumes that the father's affected cousin is the offspring of his uncle rather than being unspecified. Pedigree F correctly illustrates the birth order (third) of the affected female (indicated by arrow) and the consanguinity (double line) represented by the first-cousin marriage.

A couple has three girls, the last of whom is affected with cystic fibrosis. The first-born daughter marries her first cousin—that is, the son of her mother's sister—and they have a son with cystic fibrosis. The father has a female cousin with cystic fibrosis on his mother's side. Select the pedigree that best represents this family history from the diagrams below. a. Diagram A b. Diagram B c. Diagram C d. Diagram D e. Diagram E f. Diagram F g. Diagram G h. Diagram H

*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 D.* The figure shows the correctly drawn pedigree with generations indicated by Roman numerals and individuals by Arabic numbers. As the McKusick numbers indicate, achondroplasia is autosomal dominant, cystic fibrosis autosomal recessive. Since neither parent is affected with achondroplasia, the risk for their next child to be affected is virtually zero (rare chances for germ-line mosaicism or incomplete penetrance are ignored). The person who prompted genetic concern is the proband (III-1). George has a brother with cystic fibrosis, making his parents (I-3, I-4) obligate carriers. He has a 1/4 chance of being normal, a 2/4 chance of being a carrier, and a 1/4 chance of being affected with cystic fibrosis. Since George's possibility of being affected is eliminated by circumstance (he is normal), his odds of being a carrier are 2/3. George's wife is definitely a carrier, giving their next child a 1/6 chance to have cystic fibrosis (2/3 chance George is a carrier × 1/4 chance the child is affected if both are carriers). Although the ΔF508 (three-base pair deletion of phenylalanine codon at position 508 in the cystic fibrosis transmembrane regulator gene) accounts for 70% of cystic fibrosis mutations in whites, George's family may have a different mutation than was detected by DNA analysis in his wife. Their child may therefore have a risk of being a compound heterozygote (two different abnormal cystic fibrosis alleles) but will still be affected.

A couple presents for genetic counseling after their first child is born with achondroplasia, a dwarfing syndrome. The physician obtains the following family history: the husband (George) is the first-born of four male children, and George's next-oldest brother has cystic fibrosis. The wife is an only child, but she had DNA screening because a second cousin had cystic fibrosis and she knows that she is a carrier. There are no other medical problems in the couple or their families. The physician should now draw the pedigree with the female member of any couple on the left. The generations are numbered with Roman numerals and individuals with Arabic numerals; individuals affected with achondroplasia or cystic fibrosis are indicated. Which of the following risk figures applies to the next child born to George and his wife? a. Achondroplasia 1/2, cystic fibrosis 1/4 b. Achondroplasia 1/2, cystic fibrosis 1/8 c. Achondroplasia virtually 0, cystic fibrosis 1/4 d. Achondroplasia virtually 0, cystic fibrosis 1/6 e. Achondroplasia virtually 0, cystic fibrosis 1/8

*The answer is D.* X-linked recessive inheritance is characterized by a predominance of affected males and an oblique pattern. Transmission must be through females with no evidence of male to-male transmission. The lack of affected females would make autosomal dominant inheritance less likely, and the sex ratio plus transmission through three generations would eliminate autosomal recessive inheritance. Polygenic inheritance usually exhibits less frequent transmission, although it is certainly not ruled out in this pedigree. The many normal offspring of affected females rule out maternal inheritance.Individual II-4 in the pedigree that accompanies the question is an obligate carrier because she has an affected brother and affected son. This means that her daughter (III-3) has a 1/2 chance of inheriting the X chromosome with an abnormal allele and 1/2 chance of inheriting the X chromosome with the normal allele. If individual III-3 is a carrier, she has a 1/2 chance of transmitting her abnormal allele to her son. The risk that her son will be affected is thus 1/2 × 1/2 = 1/4, or 25%. Since the daughters of individual III-3 might be carriers (1/2 chance) but will not be affected, individual III-3 has a 1/8 chance of having an affected child.

A different family with retinitis pigmentosa is encountered, and the pedigree shown below is documented. What is the risk that a son born to individual III-3 would be affected? a. 100% b. 75% c. 50% d. 25% e. Virtually 0

*The answer is C.* Autosomal dominant inheritance is suggested by the pedigree because of the vertical pattern of affected individuals and the affliction of both sexes. Autosomal recessive inheritance is ruled out by transmission through three generations, and X-linked recessive inheritance is made unlikely by the presence of affected females. Maternal inheritance should demonstrate transmission to all or most offspring of affected mothers. Polygenic or multifactorial inheritance is not associated with such a high frequency of transmission. Note that X-linked dominant inheritance would also be an explanation for the pedigree. Because the most likely mechanism responsible for the pedigree is autosomal or X-linked dominant inheritance, individual III-3 is affected with the disorder, and she has a 50% risk of transmitting the disease. Discrimination between autosomal and X-linked dominant inheritance could be made by noting the offspring of affected males, such as individual III-4. If X-linked dominant inheritance were operative, affected males would have normal sons and affected daughters. The likely diagnosis is an autosomal dominant form of Charcot-Marie- Tooth disease. Charcot-Marie-Tooth disease exhibits genetic heterogeneity and can exhibit autosomal dominant, autosomal recessive, and X-linked inheritance. Note that the physician could provide counseling based on knowledge of genetics even though the disease is unfamiliar.

A family presents with an unusual type of footdrop and lower leg atrophy that is unfamiliar to their physician. The pedigree below is obtained. Based on the pedigree, what is the risk of individual III-3 having an affected child? a. 100% b. 75% c. 50% d. 25% e. Virtually 0

*The answer is A.* In this pedigree, the disease allele is consistently transmitted with the 1 allele. There is no case in this small number of individuals where recombination between these two loci has occurred. Therefore, in Generation III, there is no recombination seen in any of the 4 individuals. Receiving the one allele always goes together with receiving the disease gene. Linked markers can be "uninformative" (choice E) in some pedigrees if, for example, the same alleles are expressed in all family members. In such a case, it would be impossible to determine any recombination frequency.

A family with an autosomal dominant disorder is typed for a 2 allele marker, which is closely linked to the disease locus. Based on the individuals in Generation III, what is the recombination rate between the disease locus and the marker locus? A. 0 B. 0.25 C. 0.50 D. 0.75 E. 1.0 F. The marker is uninformative

*The answer is C.* When the affected child in the family is of the less-affected sex, then the recurrence risk for an opposite-sex sibling is higher but the risk to a same-sex sibling is also increased. This is because if the less-affected sex is affected by a disorder then there are more of the environmental and genetic factors that cause the disorder present, thus lowering the threshold of liability and increasing the recurrence risk.

A female child is born with pyloric stenosis, which is more common in males than in females. Which of the following best describes the recurrence risk to future siblings? (A) The recurrence risk for a future brother will increase. (B) The recurrence risk for a future sister will increase. (C) The recurrence risk for both a future brother and future sister will increase. (D) The recurrence risk for both a future brother and future sister will not increase.

*The answer is B.* Cytogenetic notation provides the chromosome number (e.g., 46), the sex chromosomes, and a shorthand description of anomalies. Examples include the following: 45,X indicates a female with monosomy X or Turner's syndrome; 47,XX+21 indicates a female with trisomy 21 or Down's syndrome; 46,XX,t(14;21) indicates a female with translocation Down's syndrome; 45,XX−21 indicates a female with monosomy 21. Note the absence of spaces between symbols, and the use of 47,XXX for sex chromosomal aneuploidy ("triple X" syndrome) rather than the more awkward 47,XX+X. (Note also that 45,X is sufficient for X chromosome monosomy, since absence of an X is indicated by the convention of listing sex chromosomes). Translocations that join two chromosomes with minuscule short arms (acrocentric chromosomes—13, 14, 15, 21, and 22) are called Robertsonian translocations. The joined acrocentric chromosomes in a Robertsonian translocation have a single centromere between them and are counted as one chromosome. A normal person who "carries" a Robertsonian translocation therefore has a chromosome number of 45, as in 45,XX,t(14;21). This female has a 5 to 20% risk of transmitting the Robertsonian 14:21 translocation to her offspring and having a child with Down's syndrome—e.g., 46,XX,t(14;21).

A female with Turner's syndrome is denoted by which of the following cytogenetic notations? a. 47,XX,+21 b. 45,X c. 47,XXX d. 46,XX,t(14;21) e. 45,XX,−21

*The answer is C.* Splicing is a component of post-transcriptional processing of RNA transcripts. In splicing, introns species they and exons are expressed. Alternative splicing allows multiple proteins to be translated from single transcript. Though exons are usually expressed, in alternative splicing some exons may be discarded in E controlled fashion. Since multiple proteins can be translated from a single transcript, the proteome (i.e„ entire set of proteins expressed by a genome at Eny given time), is larger than the genome.

A group of cell biologists are researching the genetics of several eukaryotic species. In particular, they are looking into the relative sizes of the species' genomes and the number of proteins expressed by that genome at any given time. During their investigation, they discover that every eukaryotic species they are studying has a larger proteome than genome. Which of the following best explains this phenomenon? A. Heteroplasmy B. Allelic heterogeneity C. Alternative splicing D. Locus heterogeneity E. Pleiotropy

*The answer is A.*DNA methylation by DNA methyltransferase is process by which methyl groups are added to DNA strand If methylation occurs in the promoter region of DNA (i.e., at CPG "islands"), it suppresses gene transcription. If methylation is removed, gene transcription is uninhibited, and constitutive gene transcription and cellular growth occurs. This will contribute to the uninhibited growth pattern of malignant cell

A group of cellular biology researchers is studying the genetic expression of leukemia cells. Specifically, they are looking at the expression of certain proteins and their relative activity levels. They hypothesize that malignant cells will have different characteristics within these areas from normal somatic cells, contributing to their uninhibited growth profile. Which of the following is most likely to be observed in a malignant leukemia cell? A. Decreased activity of DNA methyltransferases B. Decreased ratio of euchromatin:heterochromatin C. Decreased activity of telomerases D. Decreased activity of histone acetyltransferases E. Decreased activity of histone methyltransferases

*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 C.* This is a pattern expected of mitochondrial inheritance because only females transmit mitochondrial DNA to their offspring. Thus, an affected female can transmit the mutation to her offspring of both sexes, but an affected male cannot transmit it.

A large, 3 generation family in whom multiple members are affected with a rare, undiagnosed disease is being studied. Affected males never produce affected children, but affected females do produce affected children of both sexes when they mate with unaffected males. What is the most likely mode of inheritance? A. Autosomal dominant, with expression limited to females B. Y-linked C. Mitochondrial D. X-linked dominant E. X-linked recessive

*The answer is C.* If both parents are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the disease-causing gene (i.e., a 50% chance that the offspring will receive one copy and a 25% chance that the offspring will receive two copies). With 80% penetrance, the probability that the offspring will be affected is 0.75 × 0.8, or 0.6 (60%). The probability that the offspring will be phenotypically normal is 1 - 0.60, or 0.40 (40%).

A man and woman are both affected by an autosomal dominant disorder that has 80% penetrance. They are both heterozygotes for the disease-causing mutation. What is the probability that they will produce phenotypically normal offspring? A. 20% B. 25% C. 40% D. 60% E. 80%

*The answer is C.* One must first determine the probability that the man's mate will also be a heterozygous carrier. If the frequency of affected homozygotes (q²) is 1/40,000, then the allele frequency, q, is 1/200. The carrier frequency in the population (approximately 2q) is 1/100. Three independent events must happen for their child to be homozygous for the mutation. The mate must be a carrier (probability 1/100), the mate must pass along the mutant allele (probability 1/2), and the man must also pass along the mutant allele (probability 1/2). Multiplying the 3 probabilities to determine the probability of their joint occurrence gives 1/100 × 1/2 × 1/2 = 1/400.

A man is a known heterozygous carrier of a mutation causing hyperprolinemia, an autosomal recessive condition. Phenotypic expression is variable and ranges from high urinary excretion of proline to neurologic manifestations including seizures. Suppose that 0.0025% (1/40,000) of the population is homozygous for the mutation causing this condition. If the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is homozygous for the mutation involved? A. 1% (1/100) B. 0.5% (1/200) C. 0.25% (1/400) D. 0.1% (1/1,000) E. 0.05% (1/2,000)

*The answer is A.* A child will inherit a gene for alkaptonuria from the father and the normal allele of this gene from the mother. Conversely, the child will inherit a gene for hereditary sucrose intolerance from the mother and a normal allele of this gene from the father. The child will therefore be a carrier for each disease but will not be affected with either one.

A man who has alkaptonuria marries a woman who has hereditary sucrose intolerance. Both are autosomal recessive diseases and both map to 3q with a distance of 10 cM separating the two loci. What is the chance they will have a child with alkaptonuria and sucrose intolerance? A. 0% B. 12.5% C. 25% D. 50% E. 100%

*The answer is D.* Because the man transmits his X chromosome to all of his daughters, all of the daughters must carry at least one copy of the mutation. The mother will transmit a mutation-carrying X chromosome half the time and a normal X chromosome half the time. Thus, half of the daughters will be heterozygous carriers, and half will be affected homozygotes, having received a mutation from both parents.

A man who is affected with hemophilia A (X-linked recessive) mates with a woman who is a heterozygous carrier of this disorder. What proportion of this couple's daughters will be affected, and what proportion of the daughters will be heterozygous carriers? A. 0%; 50% B. 100%; 0% C. 0%; 100% D. 50%; 50% E. 2/3; 1/3

*The answer is E.* The husband's short stature and morphologic features are suggestive of achondroplasia. Achondroplasia is the most common form of short-limbed dwarfism and is caused by a mutation that results in constitutive activation of fibroblast growth factor receptor 3 (FGFR3). Achondroplasia occurs as a sporadic mutation (due to advanced paternal age) in 85% of cases and as an inherited autosomal dominant (AD) trait in the remaining 15%. Only 1 mutant copy of the FGFR3 gene is sufficient to cause the disorder; 2 copies of the mutant gene (ie, homozygosity) usually result in death shortly after birth. As a result, the husband must be heterozygous for the achondroplasia mutation. A heterozygous parent has a 50% chance of transmitting the mutation. Therefore, the unborn child has a 50% chance of inheriting achondroplasia (Choice A). Because achondroplasia is a rare condition. the chance of the unborn child having a sporadic mutation does not significantly add to the 50% risk of inheriting the disease.

A married couple comes to the physician for routine prenatal counseling. The husband is 120 cm (3 ft 11 in) tall with disproportionately short upper and lower extremities a large head, and a prominent forehead. He is unable to provide a biological family history as he was adopted. His spouse is of average height with normal constitutional features, and her family history is insignificant. They are concerned about their unborn child's height. Which of the following is the best response to their concerns? A. The condition is not inheritable B. The risk depends on the child's gender C. The risk depends on the mother's carrier status D. The risk for the child to be short is about 25% E. The risk for the child to be short is about 50%

*The answer is B.* In myoclonic epilepsy with ragged red fibers (MERRF) syndrome, the abnormal mitochondria in the skeletal muscle cells give them an irregular shape and a blotchy red appearance so that the muscle fibers look ragged.

A mitochondrial disorder characterized by myoclonus, seizures, cerebellar ataxia, dementia, and mitochondrial myopathy in which the skeletal muscle cells have an irregular shape and blotchy red appearance is which one of the following? (A) MELAS (B) MERRF (C) LHON (D) KS

*The answer is B.* It is thought that alleles linked to the HLA-DR3 and HLA-DR4 genes alter the immune response such that an immune response to an environmental antigen like a virus gets out of control and destroys the pancreatic beta cells that make insulin. In Type 2 diabetes, there is almost always some insulin production, but individuals with this condition develop insulin resistance. The genetic component of this disease is not well understood.

A multifactorial disorder for which the genetic component is thought to be involvement of the HLA (human leukocyte antigen) type II genes and the environmental component is hypothesized to be infection with a virus is which one of the following? (A) cancer (B) diabetes Type 1 (C) diabetes Type 2 (D) hypertension

*The answer is A.* The infant described in this clinical vignette has some typical features of Down syndrome. Characteristic appearance of these patients includes flat facies, epicanthal folds, oblique palpebral fissures and a single palmer crease. Congenital heart defects are seen in 50% of patients with Down syndrome, and endocardial cushion defects are most commonly observed. The majority of patients with Down syndrome have trisomy 21 that occurs as a result of chromosomal nondisjunction (failure of chromosomes to separate) during the first meiotic division of the ovum. The incidence of this abnormality increases with maternal age.

A neonate born to 41-year-old woman in her 39th week of gestation has a flattened face and epicanthal folds. The child's echocardiography reveals an endocardial cushion defect. Which of the following most likely occurred prior to conception? A. Meiotic non-disjunction B. Robertsonian translocation C. Expansion of trinucleotide repeats D. Formation of fragile site E. Inactivation of one chromosome F. Deletion of chromosomal part

*The answer is C.* Panel c demonstrates normal X and Y sex chromosomes, but one pair of autosomes is not homologous (arrow). Given the family history of Down's syndrome, the appearance of extra material on the short arm of chromosome 14 (arrow) can be interpreted as material from chromosome 21. Together with the two normal chromosomes 21, this extra 21 material would give three doses of chromosome 21 and result in Down's syndrome. The abnormal chromosome 14 can thus be interpreted as a Robertsonian 14;21 translocation that was inherited by this child and by the previous child with Down's syndrome [i.e., karyotypes of 46,XY,t(14;21) causing Down's syndrome]. Karyotyping the parents would then be important to determine which was the carrier of the 14;21 translocation 45,XX,t(14;21) or 45,XY,t(14;21). Genetic counseling using the appropriate recurrence risk (5 to 10% for male carriers, 10 to 20% for female carriers) could include the option of prenatal diagnosis (fetal karyotyping) for future pregnancies.

A newborn boy feeds poorly, turning blue and choking after breastfeeding. He is also very floppy (hypotonic), has a loud heart murmur, and has some unusual physical findings. These include a flat occiput (brachycephaly), folds over the inner corners of the eyes (epicanthal folds), single creases on the palms (single palmar creases), and a broad space between the first and second toes. Significant in the family history is that one of the parents' three prior children had Down's syndrome. After obtaining a chromosome analysis, which of the results pictured is most likely? a. Result A b. Result B c. Result C d. Result D

*The answer is A.* In most cases of Turner's syndrome there is a lack of one X chromosome, as in panel A, which shows one X (arrow) and no Y chromosome. Other cases involve mosaicism (45,X/46,XX or 45,X/46,XY) or isochromosomes (e.g., 46,X,isoXq). Correlation of karyotypes and phenotypic features of girls with Turner's syndrome has demonstrated that haploinsufficiency (partial monosomy) of the short arm (Xp) is what generates the characteristic manifestations (web neck, shield chest, puffy feet, coarctation). Women with Turner's syndrome also have short stature and infertility due to maldevelopment of the ovaries (streak gonads).

A newborn girl is found to have marked swelling of the dorsal areas of her feet along with a broad (webbed) neck, a broad chest, and a heart murmur that is due to coarctation of the aorta. Her physician suspects a chromosomal disorder and orders a karyotype. Which of the results pictured below is most likely? a. Result A b. Result B c. Result C d. Result D

*The answer is C.* In nondisjunction at meiosis II one of the two daughter cells resulting from cell division in meiosis I would proceed to divide normally during meiosis II, resulting in two normal daughter cells. The nondisjunction of the paired chromosome 21s in the other meiosis I daughter cell would during meiosis II would lead to both chromosome 21's going to one daughter cell and no chromosome 21s going to the other daughter cell.

A nondisjunction of chromosome 21 in meiosis II in a male would yield which combination of the following gametes? (A) one sperm with two chromosome 21s, the rest with one (B) one sperm with no 21s, three with two 21s (C) one sperm with two chromosome 21s, one with no chromosome 21s and two sperm with one chromosome 21 (D) all of the sperm would have one chromosome 21

*The answer is C.* Males always transmit their single X chromosome to their daughters. Therefore, a daughter of a male affected with an X-linked disorder is an obligate carrier for that disorder. When the condition is X-linked recessive, as with most forms of color-blindness, the daughter is unlikely to show any phenotypic evidence that she is carrying this abnormal gene. Offspring of female carriers are of four types: (1) female carrier with one normal and one mutant allele, (2) normal female with two normal alleles, (3) affected male with a single mutant allele, and (4) normal male with a single normal allele. The chance of having an affected child is thus 1/4 or 25%. If the obligate carrier female gives birth to a son, the chance of the son being color-blind is 50%.

A patient presents to the physician's office to ask questions about color blindness. The patient is color-blind, as is one of his brothers. His maternal grandfather was color-blind, but his mother, father, daughter, and another brother are not. His daughter is now pregnant. The risk that her child will be color-blind is a. 100% b. 50% c. 25% d. 12.5% e. Virtually 0

*The answer is B.* The recurrence risk for aneuploidies caused by meiotic nondisjunction is about 1% in addition to the maternal age-related risk. It is not known why the risk for aneuploidy increases slightly after an affected child is born, but parental karyotypes are almost always normal. Surveys of penal institutions have revealed an increased incidence of 47,XYY individuals, but other conditions with mental disability are increased as well. As with other chromosomal syndromes, the phenotype of 47,XYY is variable and can be found coincidentally in normal males. It would therefore be inappropriate to label a child as abnormal in school unless there have been previous concerns about a medical disorder. Males with Klinefelter's syndrome (47,XXY), rather than those with 47,XYY syndrome, are often sterile and may require supplementation with male hormones.

A physician makes the diagnosis of 47,XYY in a 16-year old boy. Which of the following options is most appropriate for the physician during the counseling session that follows the chromosome result? a. Recommend karyotyping of the parents b. Explain that the recurrence risk for such chromosomal aberrations is about 1% c. Urge that the school receive a copy of the karyotype since these boys often have behavior problems d. Recommend testosterone supplementation when the boy reaches puberty e. Inform the parents that their child will be sterile

*The answer is B.* The denominator of the gene frequency is 100, which is obtained by adding the number of genotyped individuals (50) and multiplying by 2 (because each individual has two alleles at the locus). The numerator is obtained by counting the number of alleles of each type: the 4 homozygotes with the 1,1 genotype contribute 8 copies of allele 1; the 1,3 heterozygotes contribute another 8 alleles; and the 1,4 heterozygotes contribute 3 alleles. Adding these together, we obtain 19 copies of allele 1. Dividing by 100, this yields a gene frequency of 0.19 for allele 1. For allele 2, there are two classes of heterozygotes that have a copy of the allele: those with the 2,3 and 2,4 genotypes. These 2 genotypes yield 5 and 9 copies of allele 2, respectively, for a frequency of 14/100 = 0.14.

A population has been assayed for a 4-allele polymorphism, and the following genotype counts have been obtained: On the basis of these genotype counts, what are the gene frequencies of alleles 1 and 2? A. 0.38, 0.28 B. 0.19, 0.14 C. 0.095, 0.07 D. 0.25, 0.25 E. 0.38, 0.20

*The answer is A.* Flat facial features and excessive skin at the nape of the neck are two of numerous phenotypic features associated with Down syndrome (trisomy 21). Other classic phenotypic findings include slanted palpebral fissures and a single transverse palmar crease. Visceral anomalies are common. Cardiac defects are found In approximately half of all infants with Down syndrome, with the endocardial cushion defect (atrioventricular septal defect) and ventricular septal defect most often seen. Gastrointestinal tract abnormalities are also identified in 10-16% of this patient population and can include duodenal atresia, Hirschsprung's disease, and tracheoesophageal fistula. Down syndrome incidence increases with mammal age. The vast majority of cases arise due to chromosomal nondisjunction during maternal meiosis I, which results in a full trisomy 21 present at conception.

A stillborn fetus delivered at the 2801 week of gestation has flat facial features and excessive skin at the posterior neck. Autopsy findings include a ventricular septal defect and duodenal atresia. Which of the following is the most likely karyotype abnormality in this fetus? A. Trisomy 21 B. Trisomy 18 C. Trisomy 13 D. 47, XXX E. 47, XXY F. 47, XYY G. 45, XO

*The answer is C.* Because 60% of the women are homozygous for the normal dominant hemoglobin allele, that means that p2 = 0.60 and the square root, or "p" is equal to 0.77. That means that 1 - p, or "q", the frequency of sickle cell trait, is equal to 0.23. The percentage of heterozygotes, or carriers of the sickle cell trait, is thus 2pq, or 2 x 0.77 x 0.23, which is equal to 0.35 or 35%.

A study was undertaken of pregnant women in North Carolina to determine the ratios of women with sickle cell disease and sickle cell trait. The study found that 60% of the African American women in North Carolina are homozygous for the normal dominant hemoglobin allele. What percentage of the women would be carriers of sickle cell trait? (A) 5% (B) 25% (C) 35% (D) 40%

*The answer is A.* The daughter most likely has Turner syndrome, monosomy X, or 45,X. It should be high on the differential diagnosis list for a female adolescent of short stature who presents with primary amenorrhea. Women with Turner syndrome have streak gonads, and the absence of ovarian function is responsible for failure to develop many secondary sex characteristics.

A woman brings her 16-year-old daughter to a physician because she has not yet begun menstruating. Although her parents are both 1.75 meters, the patient is 1.5 meters and has always been below the 50th percentile in height. Physical examination reveals no breast development. She has no problems in school and is of normal intelligence. What is the most likely underlying basis for her condition? A. A 45,X karyotype B. A balanced reciprocal translocation C. A balanced Robertsonian translocation D. Two Barr bodies E. Deletion of an imprinted locus

*The answer is A.* The condition described is fetal alcohol syndrome, which is marked by a smooth philtrum and a small head. Later in life, affected children can demonstrate delayed growth and at tent ion-deficit hyperactivity disorder. Other abnormalities, such as shortened palpebral fissures and lowset ears, may also be seen, as may cardiac and lung abnormalities. This child has an atrial septal defect (ASD), resulting in a fixed and split-second heart sound, regardless of breathing. The ASD increases left-to-right shunting, causing delayed closure of the pulmonic valve.

A woman presents to t he hospital in labor. She states that she has not had any prenatal care and has not changed any of her habits while pregnant. She delivers a child of normal weight, but his head is small, and his philtrum is smooth. On auscultation of the heart, a split second heart sound is noted both on inspiration and expiration. Aside from a genetic or infectious origin, which of the following habits of the mother would most likely produce this clinical picture? A. Drinking alcohol B. Smoking cigarettes C. Strenuous exercise D. Taking thalidomide E. Using cocaine

*The answer is B.* The grandmother has cystic fibrosis, so her children are obligate carriers. Each cousin therefore has a 1/2 chance of being a carrier. The woman's risk is 1/2 × 1/2 × 1/4 = 1/16 chance of having an affected child. This illustrates the effects of consanguinity.

A woman who married her first cousin wants to know the risk of having a child with cystic fibrosis because her grandmother, who is also her husband's grandmother, died of cystic fibrosis. Her risk is a. 1/8 b. 1/16 c. 1/60 d. 1/120 e. 1/256

*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 B.* Many patients with Turner syndrome are mosaics, that is, they have two or more cell lines with different karyotypes. Although there is a normal, 46,XX cell line present, the majority of cells have the 45,X Turner syndrome karyotype and thus some phenotypic features of Turner syndrome can be expected.

Amniocentesis is performed on a patient at 16 weeks' gestation because of her age (she is 36). The final report to the physician says that the fetus has a 45X/46,XX karyotype, with the 45,X cell line making up 90% of the cells examined. The fetus will most likely have phenotypic features of which of the following syndromes? (A) Fragile X syndrome (B) Turner syndrome (C) Down syndrome (D) Angelman syndrome

*The answer is E.* The African American man and woman each have a 1/8 chance of having sickle trait. They have a 1/64 × 1/4 = 1/256 chance of having a child with sickle cell anemia. There is also a 1/64 × 1/2 = 1/128 chance that their child will have sickle trait.

An African American couple with a normal family history wants to know their chance of having a child with sickle cell anemia. The incidence of sickle cell trait is 1 in 8 for African Americans. The risk in this case is a. 1/8 b. 1/16 c. 1/60 d. 1/120 e. 1/256

*The answer is C.* Trisomy 13 (Patau syndrome) is the third most common autosomal trisomy identified in liveborn infants, and the most severe. Affected children typically die within the first week of life, with only 6% surviving the first six months. In the majority of trisomy 13 infants, cytogenetic studies demonstrate nondisjunction (47, XX +13); this chromosomal abnormality arises during maternal meiosis I and is associated with advanced maternal age. The prominent phenotypic features of trisomy 13 are associated with an early defect in prechordal mesoderm development. As a result, the midface, eye, and forebrain are most markedly affected. The clinical manifestations of Patau syndrome include the abnormalities categorized by system below. 1. Head and neck: severe cleft lip and/or palate, microphthalmia or anophthalmia, coloboma, Cyclops, malformed or absent nose, deafness, scalp defects (aplasia cutis) 2. CNS: severe mental retardation, microcephaly, holoprosencephaly (failure of brain to divide into hemispheres), absent olfactory nerve or bulb, neural tube defects 3. Extremities polydactyl, rocker-bottom feet 4. Cardiac PDA, arid septa defect, ventricular septa defect 5. Renal polycystic kidney disease 6. Gastrointestinal: abdominal defects associated with omphalocela or umbilical hernia. pyloric stenosis

An infant born prematurely to a 38-year-old Caucasian female is small for gestational age. Physical examination reveals a bilateral cleft lip, microcephaly, and microphthalmos. Viscera protrude from an umbilical opening in the child's abdominal wall. Which of the following karyotypes is most likely in this casa? A. Trisomy 21 B. Trisomy 18 C. Trisomy 13 D. 47, XXX E. 47, XXY F. 47, XYY

*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 C.* Down syndrome iS the most common autosomal trisomy identified in liveborn infants. As many as 95% of Down syndrome cases arise duo to chromosomal nondisjunction during maternal meiosis I (47,XX. +21 ), an abnormality that positively correlates with increasing maternal age.

An infant born to a 36-year-old Caucasian female demonstrates some dysmorphic facial features and a dysmorphic murmur at the left stall border. Karyotype analysis is consistent with trisomy 21. Which of the following additional findings would be most expected in this infant? A. Cleft palate B. Polydactyly C. Single palmar crease D. Rocker-bottom feet E. Macrocephaly F. Macroorchidism

*The answer is E.* During meiotic segregation, each parental gamete receives one allele from every genetic locus. The probability of a parental allele being transmitted to offspring is thus 1/2, and the probability of a given genotype appearing in offspring is thus a joint probability. For a maternal Aa versus paternal aa mating, the probability of maternal alleles A or a being transmitted is 1/2, and the probability of transmission of the paternal allele a is 1. The joint probability for an Aa genotype in offspring is thus 1/2 × 1 = 1/2, the same as for an aa genotype. For a maternal Aa versus paternal Aa mating, the probabilities for AA, Aa, or aa genotypes in offspring are all 1/2 × 1/2 = 1/4, but the Aa genotype can occur in two ways (A from mother, a from father, or vice versa).

Assuming that all alleles derive from a single locus, match the mating of an Aa father with an aa mother and their probabilities for genotypes in offspring. a. 1 AA b. 1⁄2 AA, 1⁄2 aa c. 1⁄4 AA, 1⁄2 Aa, 1⁄4 aa d. 1⁄2 AA, 1⁄2 Aa e. 1⁄2 Aa, 1⁄2 aa

*The answer is A.* Autosomal recessive conditions tend to have a horizontal pattern in the pedigree. Men and women are affected with equal frequency and severity. It is the pattern of inheritance most often seen in cases of deficient enzyme activity (inborn errors of metabolism). Autosomal recessive conditions tend to be more severe than dominant conditions and are less variable than dominant phenotypes. Both alleles are defective but do not necessarily contain the exact same mutation. All individuals carry 6 to 12 mutant recessive alleles. Fortunately, most matings involve persons who have mutations at different loci. Since related persons are more likely to inherit the same mutant gene, consanguinity increases the possibility of homozygous affected offspring.

Autosomal recessive conditions are correctly characterized by which of the following statements? a. They are often associated with deficient enzyme activity b. Both alleles contain the same mutation c. They are more variable than autosomal dominant conditions d. Most persons do not carry any abnormal recessive genes e. Affected individuals are likely to have affected offspring

*The answer is C.* The 47,XYY karyotype is an example of sex chromosome aneuploidy, as are Klinefelter's syndrome (47,XXY), Turner's syndrome (45,X), and triple X syndrome (47,XXX). Sex chromosome mixoploidy implies mosaicism, such as 45,X/46,XX with two cell lines in one individual. Autosomal trisomies include Down's syndrome [47,XX+21 (trisomy 21)], Patau's syndrome [47,XX+13 (trisomy 13)], and Edwards' syndrome [47,XY+18 (trisomy 18)].

Chromosomal analysis reveals a 47,XYY karyotype. Which of the following descriptions best fits this abnormality? a. Autosomal trisomy b. A male with Klinefelter's syndrome c. Sex chromosome aneuploidy d. A female with Turner's syndrome e. Sex chromosome triploidy

*The answer is A.* Heritability increases as the concordance between monozygotic twins for a trait increases versus the concordance in dizygotic twins. Because the difference in concordance for cleft palate between monozygotic and dizygotic twins is not very large, heritability is low and there is not a very large genetic component.

Concordance for cleft palate is 25% for monozygotic twins versus 10% for dizygotic twins. Which of the following is indicated by these figures? (A) Heritability for cleft palate is low. (B) There is a large genetic component to cleft palate. (C) Heritability for cleft palate is high. (D) There is a small environmental component to cleft palate.

*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 B.* Crossing-over and random segregation of the maternal and paternal chromosomes occur during meiosis.

Crossing over and random segregation produce much of the genetic variation in human populations. These events occur during which of the following? (A) mitosis (B) meiosis (C) fertilization (D) transcription

*The answer is A.* Incomplete penetrance applies to a normal individual who is known from the pedigree to have an allele responsible for an autosomal dominant trait. Variable expressivity refers to family members who exhibit signs of the autosomal dominant disorder that vary in severity. When this severity seems to worsen with progressive generations, it is called anticipation. A new mutation in the grandson would be extremely unlikely given the affected grandfather. The father could be an example of somatic mosaicism if a back-mutation occurred to allow normal limb development, but there is no reason to suspect mosaicism of his germ cells (germinal mosaicism).

Ectrodactyly is an autosomal dominant trait that causes missing middle fingers (lobster claw malformation). A grandfather and grandson both have ectrodactyly, but the intervening father has normal hands by x-ray. Which of the following terms applies to this family? a. Incomplete penetrance b. New mutation c. Variable expressivity d. Germinal mosaicism e. Anticipation

*The answer is B.* Familial hypercholesterolemia is caused by autosomal dominant inheritance of a mutation in the low density lipoprotein receptor (LDLR) gene. Because it is an autosomal dominant disease, both heterozygous and homozygous individuals are affected. Most cases of hypercholesterolemia in the population are multifactorial in origin.

Familial hypercholesterolemia is caused by which one of the following? (A) autosomal recessive inheritance of a mutation in the LDL receptor gene (B) homozygosity or heterozygosity for a mutation in the LDL receptor gene (C) a mutation in the mitochondria (D) multifactorial inheritance

*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 A.* Females have two alleles for each locus on the X chromosome because of their 46,XX karyotype. One normal allele is by definition sufficient for normal function in X-linked recessive disorders, so that females with one abnormal allele are carriers instead of affected individuals. Only when the companion normal allele is disrupted or missing does the abnormal allele cause disease. The Lyon hypothesis predicts that X inactivation is early, irreversible, and random, but some females inactivate only the X chromosome carrying the normal allele. X autosome translocations may disrupt an X chromosome locus and cause disease because the translocated autosome must remain active to avert embryonic death; nonrandom inactivation of the normal X chromosome thus ablates expression of its normal allele. Females with Turner's syndrome, like males with 46,XY karyotypes, have only one X chromosome and can be affected with X-linked recessive diseases. Conversely, females with triple X or trisomy X syndrome have three alleles at each X chromosome locus and are not affected with X-linked recessive disorders. Since choices c, d, and e each require two genetic changes, they are less common than choice a.

Females occasionally have symptoms of X-linked recessive diseases such as Duchenne's muscular dystrophy, hemophilia, or color blindness. The most common explanation is a. Lyonization b. X chromosome trisomy (47,XXX) c. X autosome-balanced translocation that disrupts the particular X chromosome locus d. Turner's syndrome (45,X) e. 46,XY karyotype in a female

*The answer is E.* The father is affected with Gardner's syndrome, an autosomal dominant disease. Therefore, each of his four children has a 1/2 chance of receiving the allele that causes Gardner's syndrome and a 1/2 chance of receiving the normal allele. The probability that none of his four children received the allele for Gardner's syndrome is thus the joint probability of four independent events, computed by the product 1/2 × 1/2 × 1/2 × 1/2 = 1/16. The probability that at least one child has received the abnormal Gardner's syndrome allele is thus 1 − 1/16 = 15/16. Gardner's syndrome is one of many genetic disorders that may not be obvious in early childhood. Intestinal cancer in particular has a later onset, with 50% of patients being affected by age 30 to 35. More extensive evaluation of the children for internal signs of disease (e.g., the bony tumors) is required before the father can conclude that he has not transmitted the gene. Late-onset disorders are an important category of adult genetic disease, and presymptomatic testing for these diseases is a novel application of DNA diagnosis.

Gardner's syndrome is an autosomal dominant condition characterized by multiple polyps of the intestines, bony tumors, skin cysts, and a high risk of intestinal cancer. A family is encountered in which a greatgrandfather, grandmother, and father are affected with Gardner's syndrome and develop intestinal cancer in their thirties. The father brags that none of his four children have inherited Gardner's syndrome because they lack skin cysts and have not had cancer. The chance that at least one child has inherited the Gardner's syndrome allele, and the reason the children have not manifested cancer, are a. 1/4, ascertainment bias b. 1/2, variable cancer predisposition c. 3/4, early-onset disease manifestation d. 13/16, incomplete medical evaluation e. 15/16, later-onset disease manifestation

*The answer is D.* The frequency of those with the disease, 1 in 500, is equal to 0.02, which is q2. The frequency of the disease gene "q" is the square root of 0.02, or 0.0447. The carrier frequency is 2pq, and "p" is close to 1, but with the disease being so common, a more accurate carrier frequency can be obtained by using the true value of "p", which is 1 - q = 1 - 0.0447 = 0.9553. So 2pq = 2(0.9553)(0.0447) = 0.0849 which is about 8.5%.

Hemochromatosis is an autosomal recessive disease that is relatively common in the population (1 in 500). The disease can be treated successfully by periodic removal of blood (serial phlebotomy) but failure to recognize it can lead to a number of serious conditions. Population screening for carriers is being considered. What is the expected carrier frequency in the population? (A) 0.10% (B) 0.20% (C) 2.25% (D) 8.5% (E) 15%

*The answer D.* Human papilloma viruses have been implicated as a factor in the development of cervical carcinoma. A vaccine has been developed to stop infection with the virus and help prevent the development of cervical cancer.

Human papilloma virus is an infectious environmental factor in which one of the following multifactorial diseases? (A) colorectal cancer (B) hepatocellular cancer (C) lung cancer (D) cervical cancer

*The answer is A.* X-linked recessive diseases should be expressed much more commonly in males than in females because males are hemizygous for the X chromosome (they have only one copy). In the pedigree shown, Individuals I-2 and II-2 are obligate carriers of the trait and have a 1 in 2 chance of transmitting the disease gene to their offspring. Male offspring who receive the X chromosome with the disease-causing allele will develop the disease (Individuals II-3 and III-1), and female offspring who receive the X chromosome with the disease-causing allele will be carriers of the trait. In most cases, the presence of a second, normal X chromosome in these female heterozygotes will prevent the expression of the disease. In some cases, however, inactivation of the normal X-chromosome may occur in an unusually high percentage of her cells. If this happens, most cells will have the X-chromosome with the mutation, and even though she is heterozygous she may manifest symptoms (manifesting heterozygote). That this is the case with III-2 is confirmed by finding lower than expected activity of the enzyme. One would expect a heterozygote to have approximately 50% normal enzyme activity. This woman has only 10%.

Hunter disease is an X-linked recessive condition in which a failure of mucopolysaccharide breakdown results in progressive mental retardation, deafness, skeletal abnormalities, and hepatosplenomegaly. In the family pedigree shown, all affected individuals were diagnosed biochemically by assaying activity of iduronate 2-sulfatase, the enzyme encoded by the gene involved in Hunter syndrome. Activity of the enzyme relative to the normal range is displayed below the symbol for selected individuals in the pedigree. What is the most likely explanation for the presence of the syndrome in individual III-2? A. She is a manifesting heterozygote. B. She is homozygous for the disease-producing allele. C. She is not the daughter of II-1. D. The trait has incomplete penetrance. E. The trait has variable expression.

*The answer is D.* Because the frequency p of the disease gene is 0.08, then the homozygote frequency would be p2 or 0.0064.

Huntington disease is an autosomal dominant disease in which those who are homozygous for the disease gene have a clinical course that is no different from that of heterozygotes. In a population where the frequency of the Huntington gene is 0.08, what is the frequency of those who are homozygous for the gene? (A) 0.064 (B) 0.016 (C) 0.08 (D) 0.0064

*The answer is D.* A lethal dominant gene appears in the population as the result of new mutations. There are no "carriers" who are unaffected; it is lethal to all who inherit it.

If a dominant gene is lethal (fitness = 0), why do you continue to see it in a population? (A) some individuals are carriers (B) it is "protected" in homozygotes (C) it is "protected" in heterozygotes (D) it continues to arise as a new mutation

*The answer is E.* An increasing recurrence risk according to the number of relatives affected is characteristic of polygenic inheritance. The more affected relatives there are, the more evidence there is that an individual's genetic background is shifted toward the threshold for a particular trait; for example, the expectation for tall parents with tall grandparents is to have tall children. Inheritance risks for Mendelian disorders are unaffected by outcomes in prior offspring.

If parents with three affected children have a higher recurrence risk than parents with two affected children, the disease in question is likely to exhibit a. Autosomal dominant inheritance b. Autosomal recessive inheritance c. X-linked recessive inheritance d. X-linked dominant inheritance e. Multifactorial determination inheritance

*The answer is D.* If a disorder is entirely due to environmental factors, then H = 0. If the disorder is entirely genetic, then H 1. A multifactorial disorder would have a value for H somewhere between these two figures because there are both environmental and genetic factors involved in developing the disorder.

If the heritability (H) of a disorder is entirely due to genetic factors, then H would be equal to which one of the following? (A) 0 (B) 0.01 (C) 0.10 (D) 1.0

*The answer is A.* This answer is obtained by taking the square root of the incidence (i.e., the frequency of affected homozygotes) to get a gene frequency for the disease-causing mutation (q) of 1/50 (0.02). The carrier frequency is given by 2pq, or approximately 2q, or 1/25.

If the incidence of cystic fibrosis is 1/2,500 among a population of Europeans, what is the predicted incidence of heterozygous carriers of a cystic fibrosis mutation in this population? A. 1/25 B. 1/50 C. 2/2,500 D. 1/2,500 E. (1/2,500)²

*The answer is B.* In a small population, a greater proportion of "founding" individuals may carry a gene than in the larger population. The small group of Dutch settlers in South Africa reproduced only with members of the group. If the autosomal dominant mutant gene that causes variegate porphyria was overrepresented in that small population, then a larger proportion of individuals with the gene would be born into the group, and the frequency of the gene would increase.

In South Africa, variegate porphyria is found in white South Africans at a higher frequency than would be expected if the population was in Hardy-Weinberg equilibrium. This population originated from a small group of Dutch settlers. The most likely explanation for the high frequency of variegate porphyria in this population is: (A) selection for heterozygotes (B) the founder effect (C) immigration into the population (D) selection against heterozygotes

*The answer is C.*Heterozygotes for the sickle cell gene (S) have resistance to malaria. West Africa is a malaria prone area and having resistance to malaria would enhance an individual's chances of living to reproduce. Thus, the sickle cell gene would be "selected for" in the population and its frequency would gradually increase because those with the gene would have more reproductive success.

In West Africa, the sickle cell gene (S) frequency is 0.15, while in the United States the frequency of (S) is about 0.04. The most likely explanation for this is which one of the following? (A) selection for homozygotes (B) selection against homozygotes (C) selection for heterozygotes (D) selection against heterozygotes

*The answer is C.* In this pedigree, the disease locus alleles are segregating with the ABO blood locus alleles. In each case, individuals who receive the A allele also receive the disease allele. The MN locus is not linked to the AO locus because individuals III-4, -5, and -7 are each recombinants between these loci. The MN locus is not linked to the disease allele because individuals III-4, -5, and -7 are each recombinants between these loci.

In a family study following an autosomal dominant trait through 3 generations, two loci are compared for their potential linkage to the disease locus. In the following 3 generation pedigree, shaded symbols indicate the presence of the disease phenotype, and the expression of ABO blood type and MN alleles are shown beneath each individual symbol. Which of the following conclusions can be made about the linkage of the disease allele, ABO blood group locus, and MN locus? A. The ABO and MN alleles are linked, but assort independently from the disease allele B. The ABO, MN, and disease alleles all assort independently C. The disease allele is linked to the ABO locus D. The disease allele is linked to the ABO and MN loci E. The disease allele is linked to the MN locus

*The answer is C.* Because the couple shares common ancestors (i.e., one set of grandparents), they are more likely to be heterozygous carriers of the same autosomal recessive disease-causing mutations. Thus, their risk of producing a child with an autosomal recessive disease is elevated above that of the general population.

In a genetic counseling session, a healthy couple has revealed that they are first cousins and that they are concerned about health risks for their offspring. Which of the following best characterizes these risks? A. Because the couple shares approximately half of their genes, most of the offspring are likely to be affected with some type of genetic disorder. B. The couple has an increased risk of producing a child with an autosomal dominant disease. C. The couple has an increased risk of producing a child with an autosomal recessive disease. D. The couple has an increased risk of producing a child with Down syndrome. E. There is no known increase in risk for the offspring.

*The answer is E.* Pleiotropy refers to the multiple effects exerted by a single mutation and thus describes the two features observed in this patient.

In assessing a patient with osteogenesis imperfecta, a history of bone fractures, as well as blue sclerae, are noted. These findings are an example of A. allelic heterogeneity B. gain-of-function mutation C. locus heterogeneity D. multiple mutations E. pleiotropy

*The answer is D.* Heterozygotes for either C282Y or H63D are normal, but when both mutations are present in the heterozygous state in an individual that individual is affected. This is called compound heterozygosity.

In hereditary hemochromatosis, some affected individuals are not homozygous for either of the two common mutations, C282Y and H63D, which cause the disease. Which one of the following causes the disease in these individuals? (A) high hepatic concentration of copper (B) heterozygosity for C282Y (C) heterozygosity for H63D (D) compound heterozygosity for the two common mutations

*The answer is B.* Because most metabolic diseases are inherited in an autosomal recessive fashion, heterozygotes have a normal allele and usually produce enough enzymes for normal metabolic function.

In metabolic genetic diseases, heterozygotes are generally normal. Which of the following is the most likely explanation for this observation? (A) Heterozygotes have compound mutations. (B) Heterozygotes produce enough enzymes for normal metabolic function. (C) Heterozygotes produce defective enzymes that are repaired by the normal allele. (D) Heterozygotes produce defective enzymes that are spliced to normal enzymes.

*The answer is C.* In some mitochondrial diseases, like MERRF, there is a mixture of mitochondria, some of which carry the disease-causing mutation and some that are normal. This is called heteroplasmy. If, by chance, an individual inherits more of the normal mitochondria than abnormal, then they may exhibit a milder form of the disease or have no symptoms at all.

In myoclonic epilepsy with ragged red fibers syndrome (MERRF), expression of the disease is highly variable, with some family members being severely affected while others are not affected at all. Which of the following explains the variable expression of this disease? (A) variable expressivity (B) incomplete penetrance (C) heteroplasmy (D) allelic heterogeneity

*The answer is C.* The abl proto-oncogene on chromosome 9 and the bcr proto-oncogene on chromosome 22 are fused by the 9;22 translocation. Deletions of these genes are not associated with CML.

In the 9;22 translocation characteristic of chronic myeloid leukemia (CML), the Philadelphia chromosome has which of the following characteristics? (A) It is deleted for the abl proto-oncogene. (B) It is deleted for the bcr proto-oncogene. (C) It has the abl/bcr fusion gene generated by the 9;22 translocation. (D) It is deleted for the abl/bcr fusion gene.

*The answer is B.* Sickle cell anemia is the classic example of a disorder with a high frequency in a specific population because of heterozygote advantage. Persons who are heterozygous for this mutant allele (hemoglobin AS) have increased resistance to malaria and are therefore at an advantage in areas where malaria is endemic. Founder effect is a special type of genetic drift. In these cases, the founder or original ancestor of a population has a certain mutant allele. Because of genetic isolation and inbreeding in populations such as the Pennsylvania Amish, certain disorders such as maple syrup urine disease are maintained at a relatively high frequency. Fitness is a measure of the ability to reproduce. A genetic lethal implies that affected individuals cannot reproduce and, therefore, cannot pass on their mutant alleles. Natural selection is a theory introduced by Charles Darwin, which postulates that the fittest individuals have a selective advantage for survival.

Increased resistance to malaria is seen in persons with hemoglobin AS, where A is the normal allele and S is the allele for sickle hemoglobin. Which of the following terms applies to this situation? a. Founder effect b. Heterozygote advantage c. Genetic lethal d. Fitness e. Natural selection

*The answer is A.* If an abnormal allele is as likely to be transmitted to the next generation as its corresponding normal allele, it is said to have a fitness of 1. Loss of fitness (decrease in allele frequency after one generation) is also referred to as negative selection. The decreased fitness of achondroplast alleles that are eliminated by negative selection must be balanced by new mutations if the disorder has not disappeared or declined in incidence. Thus, the mutation rate of achondroplasia would be expected to be high relative to those of more benign dominant diseases.

Individuals with achondroplastic dwarfism have about 80% fewer viable offspring than do normal persons, but the incidence of achondroplasia seems to have remained constant for generations. These observations imply a. Decreased fitness, negative selection, and relatively high mutation rates b. Increased fitness, negative selection, and relatively high mutation rates c. Decreased fitness, positive selection, and relatively low mutation rates d. Increased fitness, positive selection, and relatively low mutation rates e. Decreased fitness, positive selection, and relatively high mutation rates

*The answer is E.* Cleft lip with or without cleft palate [CL(P)] is one of the most common congenital malformations. Because of the genetic component of this trait, it tends to be more common in certain families. The more family members affected and the more severe the cleft, the higher the recurrence risk. In addition, CL(P) is more common in males and in certain ethnic groups (i.e., Asians > whites > African Americans).

Isolated cleft lip and palate is a multifactorial trait. The recurrence risk of isolated cleft lip and palate is a. The same in all families b. Not dependent upon the number of affected family members c. The same in all ethnic groups d. The same in males and females e. Affected by the severity of the cleft

*The answer is B.* Because isolated clubfoot is more common in males than in females, the fact that the daughter III-1 has the condition raises the risk to subsequent siblings. Male siblings will be at a greater risk than female siblings, but the risk is elevated for both. Thus, the male fetus III-2 is at the highest risk for isolated clubfoot. None of the other pregnancies in the family are first-degree relatives so the risk is lower for them.

Isolated clubfoot is more common in males than females and follows multifactorial inheritance. Who has the highest recurrence risk for clubfoot? (A) II-4 (B) III-2 (C) III-3 (D) III-4 (E) III-5

*The answer is A.* If heterozygosity for the CF gene conferred protection against cholera, then heterozygotes would probably survive to reproduce. Those without the CF gene would most likely die before producing any more children or die before they reached reproductive age. Those who were homozygous for the gene had CF and died without reproducing. Carriers would survive to reproduce, the CF gene would be passed on, and the gene frequency would increase due to the positive selective pressure on the gene.

It is believed that the cystic fibrosis (CF) gene conferred protection to carriers during the cholera epidemics of the Middle Ages. The CF gene frequency would be expected to do what in those populations? (A) increase (B) decrease (C) stay the same (D) become sex-linked recessive

*The answer is A.* Because mitochondrial diseases are maternally inherited, Jack cannot pass on the disease.

Jack has been diagnosed with a mitochondrial disease. His wife Sally does not have the disease and currently is pregnant. What is the risk that Jack and Sally's child will have the disease? (A) 0% (B) 25% (C) 50% (D) 100%

*The answer is D.* Both parents should be studied because either parent could be a carrier. If neither parent is a carrier, this would mean that there was little risk of having another child with a chromosome abnormality. Carriers of Robertsonian translocations can have normal children, children who are balanced carriers like themselves or children with chromosome abnormalities. In the case of a 13;21 Robertsonian translocation, the risk of having an abnormal child would be 5% if the father is a carrier, and 15% if the mother is a carrier.

Jane and her husband Charlie have a phenotypically normal female child with a balanced Robertsonian translocation between chromosomes 13 and 21. Jane and Charlie wish to have more children. What should their physician recommend as their next course of action? (A) Recommend that they have no more children because of the risk of having an abnormal child. (B) Recommend that Jane be studied to determine if she is a carrier of the Robertsonian translocation. (C) Recommend that Charlie be studied to determine if he is a carrier of the Robertsonian translocation. (D) Recommend that both Jane and Charlie be studied to determine if one of them is a carrier of the Robertsonian translocation

*The answer is C.* Because a Robertsonian translocation leads to the fusion of two chromosomes, in this case chromosomes 13 and 21, there is one less chromosome in the karyotype as a result. The chromosome number thus goes from 46 to 45.

Jane and her husband Charlie have a phenotypically normal female child with a balanced Robertsonian translocation between chromosomes 13 and 21. How many chromosomes does the child have? (A) 46 (B) 47 (C) 45 (D) 48

*The answer is B.* Many common diseases are caused by a combination of environmental and genetic factors, and are described as multifactorial diseases. Examples include diabetes mellitus, schizophrenia, alcoholism, and many common birth defects such as cleft palate or congenital dislocation of the hip. The proportion of genetically identical monozygous twins who share a trait such as diabetes mellitus provides a measure of the genetic contribution to etiology (hereditability). Mendelian disorders are completely determined by the genotype of an individual, and exhibit 100% concordance in identical twins. Sporadic disorders have no genetic predisposition and do not cluster in families except by chance or through similar environmental exposure. Congenital disorders are present at birth, in contrast to juvenile diabetes mellitus, which usually presents during childhood. Sex-limited disorders occur predominantly in males or females, in contrast to the approximately equal sex distribution of juvenile diabetes mellitus.

Juvenile diabetes mellitus is a disorder of carbohydrate metabolism caused by insulin deficiency. The disease often follows a viral infection with inflammation of the pancreatic β cells, but also exhibits genetic predisposition with a 40 to 50% concordance rate in monozygous twins and clustering in families. Juvenile diabetes mellitus is best described as a a. Congenital disorder b. Multifactorial disorder c. Mendelian disorder d. Sporadic disorder e. Sex-limited disorder

*The answer is C.* . Prometaphase and metaphase of mitosis is when the chromosomes are condensed enough to visualize for cytogenetic analysis.

Karyotype analysis can be conducted on cells that have entered which one of the following stages of cell division? (A) meiosis I (B) meiosis II (C) metaphase of mitosis (D) anaphase of mitosis

*The answer is B.* Many common disorders tend to run in families but are not single-gene or chromosomal disorders. These disorders are multifactorial traits, which are caused by multiple genetic and environmental factors. For quantitative traits like height or blood pressure, it is easy to visualize how the alleles at multiple loci plus environmental factors might make additive contributions toward a final phenotype such as a height of 6 ft. For qualitative traits such as cleft lip/palate and other congenital anomalies, a threshold is envisioned that divides normal from abnormal phenotypes. Individuals with more clefting alleles, in combination with an unfavorable intrauterine environment, can cross the threshold and manifest the anomaly. The likelihood of inheriting clefting alleles is increased if there are relatives in the family with cleft lip/palate. Recurrence risks for cleft lip/palate and other multifactorial disorders are thus modified according to the family history.

Many disorders that present in adult life, such as coronary artery disease and hypertension, are multifactorial traits. A multifactorial trait results from a. The interaction between the environment and a single gene b. The interaction between the environment and multiple genes c. Multiple postnatal environmental factors d. Multiple pre- and postnatal environmental factors e. Multiple genes independent of environmental factors

*The answer is D.* Metabolic reactions within biochemical pathways require enzymes to facilitate the reactions. Most metabolic diseases are caused by mutations in the genes that encode the enzymes in a biochemical pathway.

Most genetic metabolic diseases are caused by mutations in which one of the following? (A) mitochondrial genes (B) DNA repair genes (C) genes that code for structural proteins (D) genes that code for enzymes

*The answer is C.* In an autosomal dominant pedigree, there is a vertical pattern of inheritance. Assuming the disorder is not the result of a new mutation, every affected person has an affected parent. The same is true of X-linked dominant pedigrees. However, male-to-male transmission, as seen in this family, excludes the possibility of an X-linked disorder. A person with an autosomal dominant phenotype has one mutant allele and one normal allele. These people randomly pass one or the other of these alleles to their offspring, giving a child a 50% chance of inheriting the mutant allele and therefore being affected with the disorder. This risk is unaffected by the genotypes of the previous offspring.

Mr. Smith is affected with Crouzon's syndrome and has craniosynostosis (i.e., premature closure of the skull sutures) along with unusual facies that includes proptosis secondary to shallow orbits, hypoplasia of the maxilla, and a prominent nose. His son and brother are also affected, although two daughters and his wife are not. Mr. and Mrs. Smith are considering having another child. Their physician counsels them that the risk that the child will be affected with Crouzon's syndrome is a. 100% b. 67% c. 50% d. 25% e. Virtually 0

*The answer is D.* The other time in the cell cycle when there is a response to DNA damage is at the G1 checkpoint before the S (synthesis) phase begins.

One of the two places in the cell cycle where a response to DNA damage occurs is which one of the following? (A) G0 phase (B) metaphase (C) m (synthesis) phase (D) G2 checkpoint

*The answer is E.* If two individuals in a sibship are affected with an autosomal dominant disease, then the usual implication is that one of the parents has the abnormal allele. Parents with one normal and one abnormal allele have a 50% chance of transmitting the abnormal allele with each pregnancy. Complicating the recognition of autosomal dominant inheritance are incomplete penetrance, where there are no signs of the disease phenotype after all relevant medical evaluations, and variable expressivity, where a parent may have more subtle disease than the offspring. Incomplete penetrance applies to this family because the parents have no signs or symptoms of disease. If a mutation occurs in the primordial germ cells, then these cells may have abnormal alleles despite the lack of these alleles in the rest of the body tissues (germinal mosaicism). Germinal mosaicism was thought to be very rare until testing for type I collagen gene mutations in osteogenesis imperfecta allowed verification of germinal mosaicism in this condition. Germinal mosaicism explained why autosomal recessive inheritance had been incorrectly postulated for families with normal parents and multiple affected children. Once a child has received the abnormal allele through the gamete of the mosaic parent, the child has the abnormal allele in all cells, with the usual 50% risk of transmission.

Osteogenesis imperfecta is an autosomal dominant disorder that causes thin, bluish scleras (whites of the eyes), deafness, and multiple bone fractures. Parents have two children with osteogenesis imperfecta, but themselves exhibit no signs of the disease. Which of the following genetic mechanisms is the most likely explanation for two offspring of normal parents to have an autosomal dominant disease? a. Variable expressivity b. Uniparental disomy c. New mutations d. Germinal mosaicism in one parent e. Incomplete penetrance

*The answer is B.* If the abnormal allele is represented as p and the normal as P, an infant affected with phenylketonuria (PKU) has the genotype pp. Parents must be heterozygotes or carriers (Pp) for the child to inherit the p allele from both the mother and father (assuming correct paternity and the absence of unusual chromosomal segregation). Subsequent children have a 1/2 chance of inheriting allele p from the mother and a 1/2 chance of inheriting allele p from the father; the chance that both events will occur to give genotype pp is thus 1/2 × 1/2 = 1/4, or 25%. A normal sibling may be genotype PP (1/4 probability) or Pp (1/2 probability since two different combinations of parental alleles give this genotype). The ratio of these probabilities results in a 2/3 chance (67%) of genotype Pp. Note that genotype pp is excluded because a normal sibling (the first child) is specified.

Phenylketonuria (PKU) is an autosomal recessive disease that causes severe mental retardation if it is undetected. Two normal parents are told by their state neonatal screening program that their third child has PKU. Assuming that the initial screening is accurate, what is the risk that their first child is a carrier for PKU? a. 100% b. 67% c. 50% d. 25% e. Virtually 0

*The answer is D.* Because the trait in this case is 5 times more common in males in females, it means that males are found lower on the liability curve. Fewer factors are needed to cause the disease phenotype in the male. Therefore, a female with the disease is higher on the liability curve and has a larger number of factors promoting disease. The highest risk population in this model of multifactorial inheritance would be the sons (the higher risk group) of affected mothers (the lower risk group). The affected mother had an accumulation of more disease-promoting liabilities, so she is likely to transmit these to her sons, who need fewer liabilities to develop the syndrome.

Pyloric stenosis is 5 times more common in males than in females in certain Japanese populations. The liability curve for the development of this condition in that population is shown below: Within this population, which of the following is most at risk for the development of disease? A. The daughters of affected fathers B. The daughters of affected mothers C. The sons of affected fathers D. The sons of affected mothers

*The answer is A.* When the proband is of the less affected sex, more of the genetic and environmental factors that contribute to the condition are present and subsequent siblings are at a higher risk for the condition because of that. There is also a greater risk to a sibling if the proband is more severely affected, if there are multiple family members affected, or if the affected family member is a first-degree relative.

Recurrence risk for a multifactorially inherited birth defect is higher in which of the following circumstances? (A) the proband is of the less commonly affected sex (B) the proband is less severely affected (C) only one person in the family is affected (D) the proband is a first cousin

*The answer is D.* Hemophilia A is caused by a mutation in the F8 gene for coagulation factor VIII.

Reduced factor VIII clotting activity with normal von Willebrand factor is a finding in which one of the following? (A) von Willebrand disease (B) hemophilia B (C) Christmas disease (D) hemophilia A

*The answer is C.* Parents of children with autosomal recessive disorders are obligate carriers if nonpaternity and rare examples of uniparental disomy (inheritance of chromosomal homologues from the same parent) are excluded. Normal siblings have a 2/3 chance of being carriers because they cannot be homozygous for the abnormal allele. Grandparents have a 1/2 chance of being carriers because one or the other must have transmitted the abnormal allele to the obligate carrier parent. First cousins share a set of grandparents of whom one must be a carrier. There is a 1/2 chance for the aunt or uncle to be a carrier and a 1/4 chance for the first cousin. Half-siblings share an obligate carrier parent and have a 1/2 chance of being carriers. These calculations assume a lack of mutations (Tay-Sachs is rare) and a lack of coincidental alleles (no consanguinity).

Tay-Sachs disease causes cherry red spots in the eye, "startle" responses in infancy, neurodegeneration, and death. Heterozygotes with an abnormal Tay-Sachs allele are termed carriers. What is the risk that the grandmother of an affected child is a carrier? a. 100% b. 67% c. 50% d. 25% e. Virtually 0

*The answer is C.* The 46 doubled chromosomes separate during anaphase, resulting in 92 chromosomes, and if cytokinesis (cell division) fails, one cell with a tetraploid number of chromosomes, 92, is the result instead of two daughter cells with the normal diploid number of 46 in each.

Tetraploid cells are the result of the failure of which one of the following processes? (A) anaphase of mitosis (B) S (synthesis) phase of the cell cycle (C) cytokinesis of mitosis (D) G1 phase of the cell cycle

*The answer is E.* The term familial indicates that a trait or disorder tends to cluster in families. A genetic disorder is one in which there is evidence that a gene or chromosome is involved in the susceptibility to the disease. Evidence for vertical transmission (e.g., father to daughter) is necessary for a disorder to be labeled inherited. Sporadic indicates that evidence for vertical transmission or familial clustering is lacking. Congenital simply means present at birth. Note that many congenital diseases (e.g., congenital AIDS) are not genetic, that adult-onset diseases may be genetic without being congenital, and that diseases may be familial (e.g., chickenpox) without being inherited or genetic. The eugenics movement was based on a fallacy about genetics, as it proposed breeding restrictions based on the assumption that all genetic traits (e.g., Down's syndrome) have a high risk of transmission.

The age of onset of a degenerative neurologic disease is 35. Epidemiologic study of affected persons indicates that most cases occur in the spring, are isolated (i.e., no neighbors or relatives are affected), and occur equally in men and women. However, a subset of cases consists of two affected siblings in a family. The best description of this disease is a. Inherited b. Genetic c. Sporadic d. Congenital e. Familial

*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 C.* The sex chromosomes with differently named homologues allow easy visualization of chromosome sorting during meiosis. Female meiosis only involves X chromosomes; thus, Y chromosomal abnormalities must arise during paternal meiosis or occur spontaneously in offspring. Nondisjunction at paternal meiosis I produces XY secondary spermatocytes and a 24,XY gamete. Fertilization with a 23,X ovum yields a 47,XXY individual (Klinefelter's syndrome). Only nondisjunction at paternal meiosis II produces a 24,YY gamete that yields a 47,XYY individual after fertilization.

The error in meiosis that produces a 47,XYY karyotype is best described by a. Meiosis division I of paternal spermatogenesis b. Meiosis division I of maternal oogenesis c. Meiosis division II of paternal spermatogenesis d. Meiosis division II of maternal oogenesis e. Meiosis division II in either parent

*The answer is A.* The frequency of phenylketonuria, 1/10,000 is q2. So "q" is the square root of 1/10,000 or 1/100. Because "p" is close to 1, the carrier frequency, 2pq is 2(1)(1/100) = 2/100 = 1/50, so 1 in 50 is the carrier frequency.

The frequency of the autosomal recessive disease phenylketonuria is 1 in 10,000. What is the carrier frequency for this disease? (A) 1/50 (B) 1/100 (C) 1/500 (D) 1/1,000

*The answer is D.* Diploid persons have two alleles per autosomal locus, with one being transmitted to each gamete (Mendel's law of segregation). The key to blood group problems is to recognize that a blood type is ambiguous regarding possible alleles—type A persons may have AA or AO genotypes. Once the possible genotypes are deduced from the blood types, potential offspring will represent all combinations of parental alleles. Parents with AB and OO genotypes can only have offspring with genotypes AO (type A) or BO (type B).

The major blood group locus in humans produces types A (genotypes AA or AO), B (genotypes BB or BO), AB (genotype AB), or O (genotype OO). For parents who are type AB and type O, what are the possible blood types of their offspring? a. Type AB child b. Type B child c. Type O child d. Type A or B child e. Type B or AB child

*The answer is E.* Although II-3 has an RFLP pattern consistent with heterozygosity for the PKU allele, she has PKU. The best explanation offered is that recombination has occurred, and although she is heterozygous for the restriction site generating the RFLP pattern, she is homozygous for the mutation causing PKU. The restriction site is 10 million bp upstream from the phenylalanine hydroxylase gene so there is a minimum chance of recombination of 10%. Although this is small, it is the most likely of the options listed. The phenylalanine hydroxylase gene is not on the X chromosome (choice A). Heteroplasmy (choice B) is associated with mitochondrial pedigrees, and the phenylalanine hydroxylase gene is a nuclear one. The RFLP pattern is quite consistent with I-2 being the biologic father (choice C), and he is a known carrier of the PKU mutation because he has another affected child (II-1). If II-3's RFLP pattern showed homozygosity for the marker (identical to II-1), and she had no symptoms, incomplete penetrance (choice D) would be a good choice.

The pedigree below represents a family in which phenylketonuria (PKU), an autosomal recessive disease, is segregating. Southern blots for each family member are also shown for an RFLP that maps 10 million bp upstream from the phenylalanine hydroxylase gene. What is the most likely explanation for the phenotype of II-3? A. A large percentage of her cells have the paternal X chromosome carrying the PKU allele active B. Heteroplasmy C. Male I-2 is not the biologic father D. PKU shows incomplete penetrance E. Recombination has occurred

*The answer is A.* The female II-1 in this family is heterozygous for the marker (from the gel) and also has an unaffected father. Her mother is a carrier and the bottom band in the mother's pattern is associated with the disease-producing allele of the factor VIII gene. All observations are consistent with II-1 being heterozygous (Xx) for the factor VIII gene. She has no symptoms, so she is not a manifesting heterozygote (choice E). She cannot be homozygous for the disease-producing allele (choice B) because her father is unaffected. Homozygosity for the normal allele (choice C) is inconsistent with the results shown on the gel. She has inherited the chromosome from her mother (bottom band) that carries the mutant factor VIII allele, but from her father she has received a chromosome carrying the normal allele. Note that her father is not affected, and the bottom band in his pattern is in linkage phase with the normal allele of the gene. This is a case where linkage phase is different in the mother and the father. Incomplete penetrance (choice D) is not a good choice because the female (II-1) does not have the disease-producing genotype. She is heterozygous for the recessive and (dominant) normal allele. One would expect from her genotype that she would be unaffected.

The pedigree below shows a family in which hemophilia A, an X-linked disorder, is segregating. PCR products for each member of the family are also shown for a short tandem repeat polymorphism located within an intron of the factor VIII gene. What is the best explanation for the phenotype of individual II-1? A. Heterozygous for the disease-producing allele B. Homozygous for the disease-producing allele C. Homozygous for the normal allele D. Incomplete penetrance E. Manifesting heterozygote

*The answer is D.* Mosaicism occurs when a chromosomal anomaly affects one of several precursor cells of an embryo or tissue. The two or more karyotypes that characterize the mosaic cells are separated by a slash in cytogenetic notation. The notation 47,XX,+21 denotes a cell line typical of a female with trisomy 21 (Down's syndrome), while 46,XX is the karyotype expected for a normal female.

The proper cytogenetic notation for a female with Down's syndrome mosaicism is a. 46,XX,+21/46,XY b. 47,XY,+21 c. 47,XXX/46,XX d. 47,XX,+21/46,XX e. 47,XX,+21(46,XX)

*The answer is A.* The recombinant percent of the two loci is very large, meaning that there is a lot of recombination between them. The further apart two loci are, the higher the recombination percent, so the two loci are far apart and not very tightly linked.

The recombinant % of two loci is 66%. Which one of the following is the explanation for this figure? (A) The two loci are far apart and there is low linkage between them. (B) The two loci are close together and there is high linkage between them. (C) The recombinant fraction between the loci is low. (D) The LOD score is high.

*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 D.* Linkage disequilibrium describes an association between a particular polymorphic allele and a trait. Many autoimmune diseases exhibit association with particular human leukocyte antigen (HLA) alleles (i.e., HLA-B27 and ankylosing spondylitis). The association is not necessarily cause and effect (e.g., when viral infections that trigger a disease preferentially infect certain HLA genotypes). Founder effects represent a special case of genetic drift in which rare alleles are introduced into a small population by the migration of ancestors. Genetic linkage implies physical proximity of the allele locus to the gene causing the disease. Linkage differs from allele association in that either allele A or a may be linked in a given family, depending on which allele is present together with the offending gene. Neither assortative mating (preferential mating by genotype) or selection (advantageous alleles) applies to the examples in the questions.

The ship Hopewell arrived on a small island several hundred years ago, carrying numerous pilgrims with diabetes insipidus. This disease is now known to be caused by mutant allele A, and residents of the island have 10 times the frequency of this allele as do those on the mainland. Which of the following terms describes this phenomenon? a. Selection for allele A b. Linkage disequilibrium with allele A c. Linkage to allele A d. Founder effect for allele A e. Assortative mating for allele A

*The answer is C.* The standard karyotype is an arrangement of chromosomes from one cell that is undergoing division at metaphase. At other mitotic stages, the chromosomes are not sufficiently condensed or are too dispersed to allow counting and comparison of pairs under the microscope. After growth, metaphase arrest, separation, hypotonic treatment, and fixing of white blood cells, smearing on a slide yields only about 3% cells that can be analyzed (metaphase spreads). In high-resolution chromosome analysis, less condensed chromosomes in late prophase may be analyzed (prometaphase analysis); however, this process is extremely time-consuming and usually requires focus on a particular chromosome region (e.g., chromosome 15 in a patient suspected of Prader-Willi syndrome, a condition marked by obesity and mental retardation).

The standard karyotype is performed by photomicroscopy of cells at which mitotic stage? a. Interphase b. Prophase c. Metaphase d. Anaphase e. Telophase

*The answer is B.* Albinism is one of many genetic diseases that exhibit locus heterogeneity, which means that mutations at several different loci can produce identical phenotypes.T is more than one locus for albinism, both causing autosomal recessive disease. Each parent must be homozygous for a mutant allele from one albinism locus but heterozygous or homozygous normal at the other locus. Their child would then be an obligate carrier for each type of albinism. A new mutation in the child is also possible, converting one of the parental mutant alleles to normal, but this would be very rare. Autosomal dominant disorders often vary in severity within families (variable expressivity) but occasionally are clinically silent in a person known to carry the abnormal allele (incomplete penetrance).

Two parents are both affected with two different types of albinism, but have a normal child. Which of the following terms best applies to this situation? a. Allelic heterogeneity b. Locus heterogeneity c. Variable expressivity d. Incomplete penetrance e. New mutation

*The answer is E.* The blot indicates that both parents are heterozygous for the mutant allele. Because both are phenotypically normal, the disease must be autosomal recessive. If it had been X-linked recessive, the man would be hemizygous. Thus, the chance they will have an affected child is 25% (0.25).

Two phenotypically normal second cousins marry and would like to have a child. They are aware that one ancestor (great-grandfather) had PKU and are concerned about having an affected offspring. They request ASO testing and get the following results. What is the probability that their child will be affected? A. 1.0 B. 0.75 C. 0.67 D. 0.50 E. 0.25

*The answer is C.* For each pregnancy, the probability that the child will be affected is 1/2. Therefore, the probability that all three children will be affected is the product of the three independent events—that is, 1/2 × 1/2 × 1/2 = 1/8. The probability that all three children will be unaffected is the same. When evaluating the probability that one of the three children will be affected, it must be noted that there are three of eight possible birth orders that have one affected child (Www, wWw, wwW). For two of three children to be affected, there are also three of eight possible birth orders (WWw, WwW, wWW).

Waardenburg syndrome is an autosomal dominant condition that accounts for 1.4% of cases of congenital deafness. In addition to deafness, patients with this condition have atypical facies, including lateral displacement of the inner canthi and partial albinism. A mother has Waardenburg syndrome, her husband is unaffected, and they plan to have a family with three children. What is the probability that one of the three children will be affected? a. 1/8 b. 1/4 c. 3/8 d. 1/3 e. 1/2

*The answer is E.* Pleiotropy refers to the appearance of apparently unrelated characteristics resulting from a single genetic defect. It is often the result of the presence of a single altered molecule in multiple locations in the body, so that the single mutation has effects in multiple organ systems. In Marfan syndrome, for example, a defect in the fibrillin gene causes manifestations of the disease in the eye, aorta, and joints.

Waardenburg syndrome is an autosomal dominant disorder in which patients may exhibit a variety of clinical features, including patches of prematurely grey hair, white eyelashes, a broad nasal root, and moderate to severe hearing impairment. Occasionally, affected individuals display two eyes of different colors and a cleft lip and/or palate. Patients who possess a mutation in the PAX3 gene on chromosome 2 can present with all of these disparate signs and symptoms. Which of the following characteristics of genetic traits is illustrated by this example? A. Anticipation B. Imprinting C. Incomplete penetrance D. Locus heterogeneity E. Pleiotropy

*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 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 A.* Reciprocal translocations (diagram A) involve the exchange of segments between two chromosomes. Robertsonian translocations (diagram B) involve the joining of two acrocentric chromosomes by breakage and reunion of their short arms. Translocations that produce no duplication or deficiency are called balanced. Individuals who have balanced translocations are called "carriers"; they have normal phenotypes unless the translocation alters the expression of an important gene at the breakpoint region. Isochromosomes involve duplication of short (diagram C) or long (diagram E) arms, which produces perfectly metacentric chromosomes deficient in long-or short-arm material, respectively. Paracentric inversions (diagram D) alter the banding pattern but not the shape of the chromosome because they do not involve the centromere.

Which of the diagrams below depicts a reciprocal translocation? a. Diagram A b. Diagram B c. Diagram C d. Diagram D e. Diagram E

*The answer is E.* The incidence of affected homozygotes permits the estimation of the frequency of the recessive mutation in the population. Using the Hardy-Weinberg equilibrium relationship between gene frequency and genotype frequency, the gene frequency can then be used to estimate the frequency of the heterozygous genotype in the population.

Which of the following best characterizes Hardy-Weinberg equilibrium? A. Consanguinity has no effect on Hardy-Weinberg equilibrium. B. Genotype frequencies can be estimated from allele frequencies, but the reverse is not true. C. Natural selection has no effect on Hardy-Weinberg equilibrium. D. Once a population deviates from Hardy-Weinberg equilibrium, it takes many generations to return to equilibrium. E. The frequency of heterozygous carriers of an autosomal recessive mutation can be estimated if one knows the incidence of affected homozygotes in the population.

*The answer is A.* Homologous chromosomes pair during meiosis but not during mitosis. The number of chromosomes is reduced by half, from 46 to 23 during meiosis and the daughter cells are genetically different, but during mitosis, the chromosome number of 46 is maintained and the daughter cells are genetically identical.

Which of the following describes the main difference between meiosis and mitosis? (A) homologous chromosomes pair during meiosis (B) the number of chromosomes is reduced by half during mitosis (C) after meiosis is complete, there are 46 chromosomes in each cell (D) after mitosis is complete there are 23 chromosomes in each cell.

*The answer is A.* Because all of the mitochondria in a conceptus comes from the ovum, with none coming from the sperm, all mitochondria in an individual are maternally inherited. Therefore, all of the children of a mother with a mitochondrial disease will be affected with the disease to some degree.

Which of the following describes what happens in a family where one of the parents has a mitochondrial disease? (A) All of the children of an affected mother will inherit the disease. (B) All of the children of an affected father will inherit the disease. (C) Only males will be affected if the mother has the disease. (D) Only females will be affected if the father has the disease.

*The answer is D.* Allele frequencies may differ among populations when there has been geographic isolation, founder effect, or selection for certain alleles based on different environments. Although African Americans have intermixed with whites in the United States for over 400 years, they retain a higher frequency of sickle cell alleles, which are thought to protect individuals from malarial infection. Each ethnic group has frequencies of polymorphic alleles that reflect its origin; for example, Ashkenazi Jews have a higher frequency of Tay-Sachs alleles; Greeks and other Mediterranean peoples of thalassemia alleles; and whites of cystic fibrosis alleles. Down's syndrome, a chromosomal disorder, has virtually the same frequency of 1 in 600 births in all ethnic groups. The preservation of genetic differences after migration allows the use of highly polymorphic mitochondrial genes to trace relationships among ancient and modern human populations.

Which of the following genetic disorders has a similar incidence in different ethnic groups? a. Cystic fibrosis b. Thalassemias c. Tay-Sachs disease d. Down's syndrome e. Sickle cell anemia

*The answer is A.* Mitochondrial DNA is arranged as a circular piece of double stranded DNA with no introns, only exons, with 37 genes, which make up most of the genome. DNA replication occurs in the mitochondrial matrix and is catalyzed by a different DNA polymerase than nuclear DNA.

Which of the following is characteristic of the mitochondrial genome? (A) There are 37 genes, which make up 93% of the genome. (B) DNA replication is the same as in the nuclear genome. (C) The DNA in the genome is found on 23 chromosomes. (D) Every exon has an intron.

*The answer is B.* The most common cause of Prader-Willi syndrome is a microdeletion in the area of the long arm of chromosome 15 between bands q11 and q13. This area of chromosome 15 is genomically imprinted, so the parent of origin for the chromosome determines what syndrome will occur as a result of the deletion. If the deletion is on the chromosome 15 that came from the father, then Prader-Willi syndrome will result. Angelman syndrome occurs if the microdeletion is on the maternal chromosome 15.

Which of the following is one of the most common causes of Prader-Willi syndrome? (A) a microdeletion on the maternal chromosome 15 (B) a microdeletion on the paternal chromosome 15 (C) a microdeletion on the maternal chromosome 22 (D) a microdeletion on the paternal chromosome 22

*The answer is D.* In meiosis, a 21;21 Robertsonian translocation chromosome will go into one of the daughter cells during meiosis I and the other daughter cell will not receive anything. Thus, a 21;21 Robertsonian translation carrier can only produce gametes that are disomic for chromosome 21 or nullisomic for chromosome 21. When an ovum from a carrier female is fertilized with a normal sperm, the union of the sperm with its one copy of chromosome 21 and the ovum with its two copies of chromosome 21 contained in the 21;21 translocation chromosome will result in three copies of chromosome 21 being present in the conceptus and Down syndrome will be the result. The fertilization of a nullisomic ovum, with no copy of chromosome 21, by a normal sperm with its one copy of chromosome 21, will result in a conceptus that is monosomic for chromosome 21 and this is lethal. Thus, there is almost a 100% chance that any child resulting from this union will have Down syndrome.

Which of the following is the best estimate of the chance that a child produced by the union of a female carrier of a 21;21 Robertsonian translocation carrier and a karyotypically normal male will have Down syndrome? (A) 0% (B) 5% (C) 15% (D) 100%

*The answer is B.* A family history of unexplained miscarriages and mental retardation may indicate that a structural chromosome rearrangement is segregating in the family and the miscarriages and mental retardation are the result of inheriting unbalanced segregants. Cytogenetic testing is not indicated for the other choices.

Which one of the following is an indication that you should offer a patient cytogenetic studies? (A) family history of Huntington disease (B) family history of unexplained miscarriages and mental retardation (C) family history of tall stature (D) family history of cystic fibrosis

*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

*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 A.* This karyotype shows trisomy 21 (47, XY, +21 ), which is diagnostic for Down syndrome, the most common genetic cause of congenital mental retardation. In most cases, Down syndrome results from meiotic nondisjunction in the ovum the parents themselves are usually genetically normal. Advanced maternal age is a risk factor for having a child with Down syndrome. Individuals with Down syndrome have a 10- to 20-fold increased risk of developing acute lymphoblastic leukemia, and their risk for developing acute myelogenous leukemia is also increased.

A 35-year-old woman and her husband have been trying to conceive for more than a year and are being followed by an infertility specialist. The woman is found to have significant scarring and fibrosis involving her fallopian tubes secondary to pelvic inflammatory disease that she had at a young age. After a long struggle. the woman finally becomes pregnant She gives birth to a boy who is evaluated by a pediatrician and found to have a flat nasal bridge. smallmouth, and low-set ears. The pediatrician orders a karyotype analysis on the infant, which is shown below. The infant is most likely to suffer from which of the following conditions? A. Acute lymphoblastic leukemia B. Chronic myelogenous leukemia C. Immotile cilia D. Macroorchidism E. Red blood cell sickling F. Rickets

*The answer is C.*Of the couple's four children, two of the children had a recombinant chromosome resulting from crossing over in the inversion loop during paternal meiosis. The recombinant fraction is the number of recombinant chromosomes divided by the number of children. There were 2 recombinant chromosomes in the 4 children, which is 2/4 or 1/2, which is 0.50.

A couple has had four children, two boys and two girls. The father is a carrier of a pericentric inversion of chromosome 8. Two of the couple's children were born with recombinant chromosomes and died shortly after birth. What is the recombinant fraction for the inversion? (A) 0.10 (B) 0.25 (C) 0.50 (D) 0.75

*The answer is D.* The case described represents one of the more common chromosomal causes of reproductive failure, Turner mosaicism. Turner's syndrome represents a pattern of anomalies including short stature, heart defects, and infertility. Turner's syndrome is often associated with a 45,X karyotype (monosomy X) in females, but mosaicism (i.e., two or more cell lines with different karyotypes in the same individual) is common. However, chimerism (i.e., two cell lines in an individual arising from different zygotes, such as fraternal twins who do not separate) is extremely rare. Trisomy refers to three copies of one chromosome, euploidy to a normal chromosome number, and monoploidy to one set of chromosomes (haploidy in humans).

A couple is referred to the physician because their first three pregnancies have ended in spontaneous abortion. Chromosomal analysis reveals that the wife has two cell lines in her blood, one with a missing X chromosome (45,X) and the other normal (46,XX). Her chromosomal constitution can be described as a. Chimeric b. Monoploid c. Trisomic d. Mosaic e. Euploid

*The answer is B.* The deletion is on the "p" or short arm of chromosome 5 at band 15.31.

A cytogenetics laboratory report states that a patient has a deletion of a chromosome distal to 5p15.31. Which of the following best describes what this means? (A) There is a deletion of a portion of the long arm of chromosome 5 with the breakpoint at band p15.31. (B) There is a deletion of a portion of the short arm of chromosome 5 with the breakpoint at band p15.31. (C) There is a deletion of a portion of the long arm of chromosome 15 at band 5p31. (D) There is a deletion of a portion of the short arm of chromosome 15 at band 5p31.

*The answer is E.* Intellectual disability, gait or posture abnormality, eczema, and a musty body odor are signs of phenytketonuria (PKU). PKU is an autosomal recessive disease caused by mutation of the gene that codes for phenylalanine hydroxylasa. In the United States, phenylalanine levels are measured in all neonates to screen for PKU. Because PKU is inherited in an autosomal recessive fashion, both of the healthy parents must be heterozygous carriers of the mutation. The probability that their next child wil inherit the disease is: p1 = probability that the mother transmits the mutant allele = 1/2 p2 = probability that the father transmits the mutant allele = 1/2 The probability that a child will inherit amutant allele from each carrier parent is equal to p1 x p2 = 1/4, as these are independent events.

A healthy couple, who recency emigrated from Eastern Europe. bring their 3-year-old son to the office for evaluation of an eczematous rash. On examination, the child also shows signs of intellectual disability and gait abnormality and has a musty body odor. What is the likelihood that this couple's next child will be affected with the same disease? A. Same as the general population B. 1/32 C. 1/16 D. 1/8 E. 1/4 F. 1/2

*The answer is C.* Each of the affected individuals on this pedigree has inherited the disorder from asymptomatic parents. consistent with an autosomal recessive inheritance pattern. Classical galactosemia is an autosomal recessive disease. Patients with galactosemia are homozygous for a defective galactose-1-phosphate uridyltransferase gene In general, most enzyme deficiency conditions follow an autosomal recessive inheritance pattern, whereas diseases due to defective non-catalytic proteins tend to follow an autosomal dominant pattern

A patient is suspected of having an inherited disorder. Pedigree analysis shows the following pattern (see the slide below). This patient is most likely to be suffering from which of the following conditions? A. Hemophilia B B. Huntington's disease C. Classic galactosemia D. Lesch-Nyhan syndrome E. Labor hereditary optic neuropathy

*The answer is E.* The frequency of sickle cell disease is elevated in many African populations because heterozygous carriers of the sickle cell mutation are resistant to malarial infection but do not develop sickle cell disease, which is autosomal recessive. Thus, there is a selective advantage for the mutation in heterozygous carriers, elevating its frequency in the population.

An African American couple has produced two children with sickle cell disease. They have asked why this disease seems to be more common in the African American population than in other U.S. populations. Which of the following factors provides the best explanation? A. Consanguinity B. Genetic drift C. Increased gene flow in this population D. Increased mutation rate in this population E. Natural selection

*The answer is B.* It is important to remember that individuals with blood type A can have either genotype AA or AO, and individuals with blood type B can have either genotype BB or BO. Therefore, the frequency of blood type A is the frequency of homozygotes—that is, 0.3 × 0.3—plus the frequency of heterozygotes—that is, 2 (0.3) × 0.6—for a total of 0.45. Thefrequency of blood type B is 0.1 × 0.1 + 2 (0.1) × 0.6 for a total of 0.13. The frequency of individuals with blood type O is simply the frequency of homozygotes—that is, 0.6 × 0.6 = 0.36.

Assume that frequencies for the different blood group alleles are as follows: A = 0.3; B = 0.1; O = 0.6. What is the expected percentage of individuals with blood type B? a. 7% b. 13% c. 27% d. 36% e. 45%

*The answer is B.* More than 90% of patients with Edwards syndrome have some form of congenital heart malformation, the most common being ventricular septal defect with pulmonary and aortic valve defects. Other types of cardiac malformations can also occur.

At birth, a baby exhibits micrognathia, malformed ears, ocular hypertelorism, syndactyly, and clenched hands, with the fifth finger overriding the fourth finger and the index finger overriding the middle finger. A karyotype shows three copies of chromosome 18. Which of the following symptoms is also associated with this disorder? A. Anencephaly B. Congenital cardiac malformations C. Limb atresia D. Pulmonary hamartomas E. Spots on the iris

*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 B.* The gene frequency in X-linked recessive diseases is the same as the disease frequency. In this case, the gene frequency is 1/3,500. The carrier frequency is 2pq, since p is almost equal to 1 the carrier frequency is 2 x 1 x 1/3,500 = 2/3,500 or 1/1,750.

Duchenne muscular dystrophy is a lethal X-linked recessive disease, which affects 1 in 3,500 boys. What is the carrier frequency of this gene mutation in females? (A) 1/3,500 (B) 1/1,750 (C) 1/59 (D) 3/50 (E) 1/25

*The answer is C.* In autosomal recessive diseases, carriers of the disease gene are generally asymptomatic, so it is not obvious who does and who does not carry the gene. Since those who are carrying the gene cannot be identified, the gene is "protected" from removal from the population. Even if that could be accomplished, new mutations would occur that would keep the gene in the population.

Early in the 20th century, eugenics proponents thought that if those with genetic diseases were prevented from having children, then the diseases would disappear. They were erroneous in their reasoning, however. Why wouldn't this strategy work for autosomal recessive diseases? (A) because only females are carriers (B) because only males are carriers (C) because the genes would be "protected" in carriers (D) because 1/4 of all children would be normal

*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 C.* Because the frequency of the normal allele "q" is 0.99, the frequency of the mutated allele "p" is 1 - q or 1 - 0.99, which is 0.01. The frequency of "p" is almost equal to 1, so the heterozygote frequency 2pq is thus 2 x 1 x 0.01 or 0.02.

In achondroplastic dwarfism, an autosomal dominant disease, the gene is lethal in homozygotes. If the frequency of the normal allele is 0.99, what is the heterozygote frequency? (A) 0.98 (B) 0.01 (C) 0.02 (D) 0.99

*The answer is B.* The genotype of each dwarf can be represented as Aa, with the uppercase A representing the achondroplasia allele. The Punnett square below demonstrates that 3/4 possible gamete combinations yield individuals with at least one A allele. Homozygous AA achondroplasia is a severe disease that is usually lethal in the newborn period. The increased likelihood of individuals with achondroplasia marrying each other because of their similar phenotypes is an example of assortative mating.

Little People of America (LPA) is a support group for individuals with short stature that conducts many workshops and social activities. Two individuals with achondroplasia, a common form of dwarfism, meet at an LPA convention and decide to marry and have children. What is their risk of having a child with dwarfism? a. 100% b. 75% c. 50% d. 25% e. Virtually 0

*The answer is B.* During meiosis I, the 23 paired, doubled homologs randomly separate, resulting in two daughter cells with 23 chromosomes each.

The "reduction division" in which the number of chromosomes in a germ cell is reduced from 46 to 23 chromosomes occurs during which of the following? (A) mitosis (B) meiosis I (C) meiosis II (D) synapsis

*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 D.* During meiosis I, the nondisjunction of the paired and doubled X and Y would cause them to go into one of the daughter cells with no X or Y chromosomes going to the other daughter cell. During meiosis II, the doubled X chromosome and the doubled Y chromosome would go to separate daughter cells and the cell with no sex chromosomes would give rise to two daughter cells with no sex chromosomes. Because the X and Y chromosomes are doubled, the daughter cell receiving the X chromosomes will have two copies and the daughter cell receiving the Y chromosomes will have two copies.

The X and Y chromosomes pair in meiosis at the pseudoautosomal regions. A nondisjunction of the X and Y chromosomes in a male during meiosis I would produce which of the following combinations of gametes? (A) one sperm with two X's, and three sperm with Y's (B) two sperm with two X's and two sperm with two Y's (C) one sperm with no X's, and three sperm with an X and a Y (D) a sperm with two X's, a sperm with two Y's, and two sperm with no sex chromosomes

*The answer is B.* In-frame deletions or insertions typically produce an altered protein product (dystrophin), but the alteration is mild enough so that Becker muscular dystrophy results. Frame-shifts usually produce a truncated protein because a stop codon is eventually encountered. The truncated protein is degraded, resulting in an absence of dystrophin and a more severe disease phenotype.

The clinical progression of Becker muscular dystrophy is typically much slower than that of Duchenne muscular dystrophy. This is usually the result of A. gain-of-function mutations in the Duchenne form; loss-of-function mutations in the Becker form B. in-frame deletions or insertions in the Becker form; frameshift deletions or insertions in the Duchenne form C. missense mutations in the Becker form; nonsense mutations in the Duchenne form D. mutations at two distinct loci for these two forms of muscular dystrophy E. nonsense mutations in the Becker form; missense mutations in the Duchenne form

*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 B.* Carriers of some balanced translocations have a significant risk of having a child with a chromosome abnormality. Because the couple has already had an abnormal child, this indicates that viable, abnormal outcomes are possible with this particular balanced translocation and the couple has a significantly elevated risk of having it happen again.

The greatest risk of having a child with a chromosome abnormality will occur with which one of the following? (A) a couple who had a child with a de novo (spontaneously occurring) unbalanced translocation between chromosomes 2 and 5 (B) a couple who had a child with an unbalanced translocation between chromosomes 2 and 5, and the father was identified as a carrier of a balanced 2;5 translocation (C) a couple who had a child with Down syndrome (D) a couple who have had no children

*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 A.* Because males have only a single X chromosome, each affected male has one copy of the disease-causing recessive mutation. Thus, the incidence of an X-linked recessive disease in the male portion of a population is a direct estimate of the gene frequency in the population.

The incidence of Duchenne muscular dystrophy in North America is about 1/3,000 males. On the basis for this figure, what is the gene frequency of this X-linked recessive mutation? A. 1/3,000 B. 2/3,000 C. (1/3,000)² D. 1/6,000 E. 1/9,000

*The answer is C.* The most common inherited bleeding disorder is von Willebrand disease, but only a small number of patients come to medical attention because of symptoms.

The most common bleeding disorder is which one of the following? (A) hemophilia A (B) hemophilia B (C) von Willebrand disease (D) Christmas disease

*The answer is C.* Having a first-degree relative increases recurrence risk for heart disease and having a female first-degree relative (least-affected sex) confers an even higher risk.

There are a number of risk factors in heart disease, but the in the family history the greatest risk would be due to which one of the following? (A) having a maternal uncle with heart disease (B) having a paternal uncle with heart disease (C) having a mother with heart disease (D) having a father with heart disease

*The answer is B.* Inheritance in mitochondrial disorders in exclusively maternal and there is a wide degree of severity among affected individuals because of heteroplasmy. Cell types that have a high ATP requirement are more seriously affected by mitochondrial disorders. If an individual does not a lot of mutated mitochondria, then the threshold where clinical symptoms appear will not be reached. When a sufficient number of mutated mitochondria are present in an individual, the threshold will be reached and the disease manifested.

Which of the following is a characteristic of mitochondrial disorders? (A) They are inherited in a Mendelian fashion. (B) There is a threshold level of mutated mitochondria that must be reached before clinical symptoms appear. (C) All cell types in the body are equally affected by mitochondrial disorders. (D) The degree of severity is the same for most affected individuals.

*The answer is B.* During meiosis, the inverted chromosome must pair with its homolog in a way that forms a loop. Crossing-over within the inversion loop can result in duplications or deletions of parts of the chromosomes. These duplicated and deleted chromosomes are thus in the gamete resulting from the meiosis and when this unbalanced gamete and a normal gamete fuse, the conceptus will have an unbalanced chromosome

Which of the following is the main risk to children of inversion carriers? (A) Down syndrome (B) duplications or deletions (C) chronic myelogenous leukemia (D) Robertsonian translocations

*The answer is C.* A 37-year-old woman has about a 1% risk to have a child with a chromosome abnormality and should be offered amniocentesis to detect chromosome abnormalities in the fetus. Trisomy 21 and Turner syndrome occur spontaneously so cytogenetic studies of the parents would not provide any information on future risk. Parents of a normal child have the population risk for having a child with a chromosome abnormality so there is no indication for offering the test.

Which of the following patients should be offered cytogenetic studies? (A) parents of a child with trisomy 21 (B) parents of a child with Turner syndrome (C) a 37-year-old woman who is pregnant (D) parents of a normal child

*The answer is C.* Carriers of a 13;21 Robertsonian translocation are at risk for having a child with Robertsonian Down syndrome or Robertsonian trisomy 13. All the other Robertsonian translocation carriers have Robertsonian translocations that are lethal when trisomy occurs and most of these conceptions are not even recognized pregnancies. There may be an increased risk of infertility connected with these Robertsonian translocations, but no increased risk of having abnormal children.

Which one of the following Robertsonian translocation carriers has the greatest risk of having an abnormal child? (A) 45,XX,t(14;15) (B) 45,XY,t(15;22) (C) 45,XX,t(13;21) (D) 45,XY,t(14;22)

*The answer is D.* The gene frequency for X-linked recessive diseases is the same as the frequency of affected males, which is the disease frequency because females are generally not affected.

Which one of the following best describes the gene frequency for an X-linked recessive disease? (A) it is equal to 1 (B) it is equal to 2pq - 1 (C) it is equal to its frequency in female carriers (D) it is equal to the disease frequency

*The answer is B.* A secondary spermatocyte results from meiosis I, the reduction division, in a primary spermatocyte and the chromosome number is reduced from 46 to 23.

Which one of the following has a haploid number of chromosomes? (A) primary spermatocyte (B) secondary spermatocyte (C) spermatogonia (D) oogonia

*The answer is D.* The mitochondrial genome has a mutation rate 10 times that of the nuclear DNA, is not protected by histones, consists only of exons, and replication occurs in the mitochondrial matrix.

Which one of the following is a characteristic of the mitochondrial genome? (A) It has a very low mutation rate. (B) The DNA is bound to histones. (C) Replication of the DNA occurs in the nucleus. (D) There are no introns present.

*The answer is D.* Tissues preserved in formalin and frozen tissues that have not been properly cryopreserved do not contain live cells, so they cannot be grown in culture.

Which one of the following is a suitable specimen for cytogenetic analysis? (A) placenta in formalin (B) frozen (not cryopreserved) blood plasma (C) frozen (not cryopreserved) amniotic fluid (D) peripheral blood

*The answer is C.* Meiotic chromosomes are not suitable for routine cytogenetic analysis. Metaphase chromosomes are suitable for cytogenetic analysis in general, but mitotic prometaphase chromosomes are more extended and allow for detailed, high-resolution cytogenetic analysis.

Which one of the following is often the preferred stage for more detailed cytogenetic analysis? (A) meiotic prometaphase (B) meiotic metaphase (C) mitotic prometaphase (D) mitotic metaphase

*The answer is D.* For an autosomal dominant condition, the first occurrence in a family is usually the result of a new mutation that occurred in one of the gametes transmitted by a parent of the affected individual.

A 10-year-old girl is diagnosed with Marfan syndrome, an autosomal dominant condition. An extensive review of her pedigree indicates no previous family history of this disorder. The most likely explanation for this pattern is A. highly variable expression of the disease phenotype B. incomplete penetrance C. mitochondrial compensation in the mother D. new mutation transmitted by one of the parents to the affected girl E. pleiotropy

*The answer is D.* According to the Hardy-Weinberg equilibrium, the frequency of heterozygotes (2pq) is twice the square root of the rare homozygote frequency (q2). The man in the question has a 2/3 chance of being a carrier and a 1/20 chance that his wife is a carrier. His risk for an affected child is 2/3 × 1/20 × 1/4 = 1/120.

A man whose brother has cystic fibrosis wants to know his risk of having an affected child. The prevalence of cystic fibrosis is 1 in 1600 individuals. The risk in this case is a. 1/8 b. 1/16 c. 1/60 d. 1/120 e. 1/256

*The answer is A.* Neurofibromatosis type 2 is caused by a mutation in the NF2 gene on chromosome 22; this mutation is inherited in an autosomal-dominant manner. Autosomal-dominant inheritance shows disease in many generations, with both males and females affected. It is possible for a male or a female to transmit the defective gene to t heir offspri ng.

A 13-year-old boy is diagnosed with neurofibromatosis after an MRI reveals bilateral acoustic neuromas as the cause of his progressive decline in hearing. By which of the following modes is this disease manifested? A. Autosomal dominant B. Autosomal recessive C. Mitochondrial inheritance D. Somatic mutation E. X-linked recessive

*The answer is D.* This patient's hemoglobin electrophoresis from her newborn screen is most consistent with sickle cell trait At birth, infants who are heterozygous for sickle cell trait typically have the greatest amount of fetal hemoglobin (Hb F), followed by hemoglobin A (Hb A), and the smallest amount of hemoglobin S (Hb S). Hb A continues to be higher than Hb S throughout the lifetime of these patients as Hb F naturally declines, offering protection from sickle cell anemia, aplastic crises, and splenic sequestration. Patients with sickle cell trait are usually asymptomatic with normal hemoglobin level, reticulocyte count, and red blood cell (RBC) indices. However, they may develop hematuria, priapism and increased incidence of urinary tract infections. Splenic infarction at high altitudes has also been reported. Patients with sickle cell trait have relative protection from Plasmodium falciparum (malaria), resulting in lower rates of severe malaria and hospitalization than seen in the general population. Possible mechanisms include increased sickling of parasitized sickle cell trait RBCs and accelerated remove of these cells by the splenic monocle-macrophage system. These patients are not immune to malaria. however, and those visiting malaria-endemic areas should still receive prophylaxis

A 2-week-old girl is brought to her primary care provider for a routine visit. The patient was bom by normal spontaneous vaginal delivery at 39 weeks gestation. mother is breastfeeding exclusively, and the infant has regained her birth weight Newborn screening results from hemoglobin electrophoresis are as follows: Hemoglobin F 70% Hemoglobin A 20% Hemoglobin S 10% The patients mother has sickle celll trait, and a maternal cousin has sickle cell anemia. Examination shows a well-appearing infant with no pallor of splenomegaly. Which of the following is most likely true about this patient? A. Life expectancy will be shorter than average B. Mean corpuscular volume will be decreased C. Reticulocyte count wil be elevated D. She has relative protection from Plasmodium falciparum E. She will likely develop pain crises

*The answer is B.* Klinefelter syndrome, which is the result of a 47, XXY chromosome constitution, is characterized, among other things, by tall stature, gynecomastia, and small testes. This combination of features is not seen in the other choices.

A tall male with gynecomastia and small testes should have a cytogenetic study to rule out which of the following? (A) XYY syndrome (B) Klinefelter syndrome (C) Fragile X syndrome (D) Turner syndrome

*The answer is C.* The blot shows the top band in the patterns of I-1 and II-2 (the proband) is associated with the disease-producing allele. Because the fetus has inherited this marker allele from the mother (II-2) and Marfan disease is dominant, the fetus will develop Marfan disease. Choices A and B are recurrence risks associated with the pedigree data. With no blot to examine, choice B, 50% risk would be correct. Choice D would be correct if the blot from the fetal DNA showed both the bottom band (must be from mother) and the top band (from the unaffected father). Choice E is incorrect because Marfan is a dominant disease with no "carrier" status.

A 22-year-old woman with Marfan syndrome, a dominant genetic disorder, is referred to a prenatal genetics clinic during her tenth week of pregnancy. Her family pedigree is shown below (the arrow indicates the pregnant woman). PCR amplification of a short tandem repeat (STR) located in an intron of the fibrillin gene is carried out on DNA from each family member. What is the best conclusion about the fetus (III-1)? A. Has a 25% change of having Marfan syndrome B. Has a 50% chance of having Marfan syndrome C. Will develop Marfan syndrome D. Will not develop Marfan syndrome E. Will not develop Marfan syndrome, but will be a carrier of the disease allele

*The answer is C.* This child has Patau 's syndrome, which is caused by trisomy 13. Patau's syndrome is characterized by cleft lip, micropht halmia, mental retardation, polydactyly, congenital heart disease, and renal defects.

A 25-year-old mother recently gave birth to a child with several birth defects, including a cleft lip, extra digits, and extremely small eyes. What is the most likely etiology of these findings? A. Deletion of 11pl3 B. Deletion of 5p C. Trisomy 13 D. Trisomy 18 E. Trisomy 21

*The answer is C.* Because two of the man's siblings had Tay-Sachs disease, his parents must both be carriers. This clearly elevates his risk above the general population and excludes choice E. He is not affected, so this excludes choice A, which is the probability of inheriting two copies of the disease allele. His risk of inheriting one copy of the disease gene at conception is 1/2 (choice B). However, the fact that he is phenotypically normal at age 30 means that he cannot have inherited copies of the disease gene from both parents. Only 3 possibilities remain: Either he inherited no copies of the mutation, he inherited a copy from his father, or he inherited a copy from his mother. Each of these possibilities is equally likely, and two of them lead to heterozygosity. Thus, the risk that he is a carrier is 2/3.

A 30-year-old man is phenotypically normal, but two of his siblings died from infantile Tay-Sachs disease, an autosomal recessive condition that is lethal by the age of 5. What is the risk that this man is a heterozygous carrier of the disease-causing mutation? A. 1/4 B. 1/2 C. 2/3 D. 3/4 E. Not elevated above that of the general population

*The answer is E.* Chromosomal abnormalities may involve changes in number (i.e., polyploidy and aneuploidy) or changes in structure (i.e., rearrangements such as translocations, rings, and inversions). Extra material (i.e., extra chromatin) seen on chromosome 5 implies recombination of chromosome 5 DNA with that of another chromosome to produce a rearranged chromosome. Since this rearranged chromosome 5 takes the place of a normal chromosome 5, there is no change in number of the autosomes (nonsex chromosomes) or sex chromosomes (X and Y chromosomes). The question implies that all cells karyotyped from the patient (usually 11 to 25 cells) have the same chromosomal constitution, ruling out mosaicism. The patient's clinical findings are similar to those occurring in trisomy 13, suggesting that the extra material on chromosome 5 is derived from chromosome 13, producing an unbalanced karyotype called dup(13) or partial trisomy 13.

A child with cleft palate, a heart defect, and extra fifth fingers is found to have 46 chromosomes with extra material on one homologue of the chromosome 5 pair. This chromosomal abnormality is best described by which of the following terms? a. Polyploidy b. Balanced rearrangement c. Ring formation d. Mosaicism e. Unbalanced rearrangement

*The answer is C.* The disease in the question stem is Duchenne muscular dystrophy. This is caused by an X- linked frameshift mutation and results in accelerated muscle breakdown. Calf muscles become enlarged due to fatty replacement of muscle, called calf pseudohypertrophy. Hemophilia A and Bare inherited in the same manner.

A concerned mother brings her 5-year-old son to the pediatrician's office. She explains that the child has been healthy since birth, but has recently begun to have difficulty standing up. In the office, the physician notices that when the child wants to stand up, he uses his hands and arms to push himself to a standing position, as shown in the image. He is diagnosed with a genetic condition. Which other disorder shares the same inheritance pattern as this patient's genetic disease? A. Cystic fibrosis B. Hemochromatosis C. Hemophilia A D. Marfan syndrome E. Neurofibromatosis type 1

*The answer is D.* Based on the activity of glucocerebrosidase, both of the parents are heterozygous (Aa). Therefore, there is a 25% chance that the child will be affected (aa) and a 50% chance that the child will be a carrier (Aa). Therefore, there is a 75% chance that the child may carry one or more allele of the mutation.

A married couple is screened to assess the risk for Gaucher disease in their children. The activities of glucocerebrosidase in the sera of the mother and father are 45% and 55%, respectively, of the reference value. The couple has one child. Which of the following is the probability of the child possessing one or more alleles of the Gaucher mutation? (A) 0 (B) 0.25 (C) 0.5 (D) 0.75 (E) 1.0

*The answer is C.* Although all individuals, other than identical twins, are genetically unique, we all share some genes in common with our relatives. The more closely we are related, the more genes we have in common. First-degree relatives, such as siblings, parents, and children, share one-half of their genes. Second-degree relatives share one-fourth, and third-degree relatives share one-eighth.

What proportion of genes do a brother and half-sister have in common? a. One b. One-half c. One-fourth d. One-eighth e. One-sixteenth

*The answer is C.* Von Willebrand disease is an autosomal dominant disorder.

Which one of the following inherited bleeding disorders is autosomal dominant? (A) hemophilia A (B) hemophilia B (C) von Willebrand disease (D) Christmas disease

*The answer is A.* Peripheral blood is easily obtained and gives high quality cytogenetic preparations. A skin sample involves minor surgery. A bone marrow biopsy is painful and generally does not yield high quality cytogenetic preparations. Cheek cells are more appropriate for DNA studies because it would be difficult to obtain sufficient numbers of them for tissue culture and they would probably be too contaminated with bacteria to be grown successfully.

Which one of the following is the appropriate specimen for cytogenetic analysis where the patient is a child with dysmorphic features and unexplained mental retardation? (A) peripheral blood (B) skin (C) bone marrow (D) cheek cells

*The answer is C.* In Leber's hereditary optic neuropathy (LHON), heteroplasmy is rare and disease expression is relatively uniform. Heteroplasmy is common in the other mitochondrial diseases listed.

Which one of the following mitochondrial diseases shows the typical mitochondrial inheritance pattern? (A) MELAS (B) MERRF (C) LHON (D) KS

*The answer is D.* Consanguinity increases the risks for birth defects in general and confers a high risk of sharing predisposing genes. Having a third-degree relative with a multifactorial disorder does not increase recurrence risk above the population risk and there is only a slight risk with a second-degree relative with mild disease.

Which one of the following would be expected to increase recurrence risk the most for a multifactorial disorder? (A) The disorder occurs in a third-degree relative. (B) There is one affected second cousin. (C) There is a first cousin with mild disease. (D) The parents are first cousins.

*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.* 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.* 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.* 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.* 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 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 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 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 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.* 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 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.* 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 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 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.*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.* 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 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 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 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 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 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.* 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 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.* 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 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 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 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 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 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 A.* In a deletion, a portion of the chromosome is lost, leaving only one copy of that area on the homologous chromosome. Since there is only one copy left on the normal homolog, that chromosome is monosomic for the deleted area.

A chromosome deletion results in which of the following? (A) a chromosome monosomic for the deleted area (B) a chromosome disomic for the deleted area (C) a chromosome trisomic for the deleted area (D) a chromosome tetrasomic for the deleted area

*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 A.* The light Giemsa negative G-bands are GC-rich and contain more genes than the AT-rich G positive G bands and the equivalent Giemsa negative R-bands. C bands are heterochromatic and do not contain coding sequences.

In a balanced reciprocal translocation in which two chromosomes exchange pieces, a breakpoint in which one of the following would be most likely to cause gene disruption and thus an abnormal phenotype? (A) Giemsa negative G-band (B) Giemsa positive G-band (C) Giemsa negative R-band (D) C-band

*The answer is D.* The male patient described above has a history of excessive bleeding and hemarthroses, suggesting a diagnosis of hemophilia A or B. Both of these diseases are X-linked recessive coagulation factor deficiencies. The probability that his sister will give birth to an affected child can be calculated by multiplying the following probabilities: - The probability (p1) that the sister is a carrier = 0.5. The patients father does not carry the mutation on his X chromosome, because he would be affected by the disease if he did. That means the mother carries the mutation on one of her two X chromosomes. This gives the daughter e 50% chance of having inherited the mutated x chromosome end thus being a carrier. - The probability (p2) that the offspring of a female carrier will inherit the X chromosome with the hemophilia gene = 0.5. Assuming the daughter is a carrier, the probability of passing on the hemophilia-carrying version of her two x chromosomes is 50%, since only one of her two x chromosomes is passed to her offspring. - The probability (p3) that his sister will have a male child = 0.5. If the sister's child is female then the child could be a carrier of the disease, but would not be affected by it if a male child inherits the mutated X chromosome. he will have the disease The probability that the sister will have an affected son is thus the probability that all three of the above events take place (i.e., the product of their individual probabilities): p1 x p2 x p3 = 1/2 x 1/2 x 1/2 = 1/8

A 14-year-old boy experiences severe prolonged bleeding following a tooth extraction. He also has a history of multiple episodes of painful joint swelling following minor trauma. His parents have no bleeding problems. Evaluation reveals that the patient has an inherited disorder and that one of his parents is a genetic carrier. Now his older sister, who does not have this condition, it pregnant (sex of the child unknown) and asks about the risk that her child will be affected. The best probable estimate that her child will have the disease is: A. Near 0 B. 1/2 C. 1/4 D. 1.8 E. 1/16 F. 1/32

*The answer is G.* This patient most likely has congenital coarctetion of the aorta, which typically affects the region of the aorta just distal to the left subctavian artery. Although traditionally classified into preductal and postductal types, the signs and symptoms of aortic coarctation depend more on the age at presentation, which varies according to the severity of the stenosis. Severe coarctation usually presents in infancy with differential cyanosis affecting the lower extremities as long as the ductus arteriosus remains patent. On ductal closure, these neonates can develop signs of heart failure and shock. More moderate stenosis often presents in childhood or adolescence with symptoms of lower extremity claudication (eg, pain and cramping with exercise), blood pressure discrepancy between the upper and lower extremities, and delayed or diminished femoral pulses. Continuous murmurs and pulsatile intercostal collaterals can also develop secondary to restricted circulation. Congenital aortic coarctation occurs in up to 10% of patients with Tumor syndrome (45.XO).

A 12-year-old girl is brought to the physician by her parents, who are concerned about her loss of interest. in playing sports at school. During a recent competition, she walked off the field in the middle of the game, complaining about the pain in her logs. The patient has no other medical conditions and takes no medications. Her vaccination schedule is up-to-date. Physical examination shows pulsatile vessels within the intercostal spaces and diminished femoral pulses relative to brachial pulses. This patient's symptoms are most likely associated with which of the following conditions? A DiGeorge syndrome B. Down syndrome C. Friedreich ataxia D. Kartagener syndrome E. Marfan syndrome F. Tuberous sclerosis G. Turner syndrome

*The answer is C.* The McKusick number for cystic fibrosis begins with 2, indicating an autosomal recessive disorder. The genotype of the affected woman with cystic fibrosis is therefore best represented as the two lowercase letters cc. Her parents are obligate carriers for the disorder (genotypes Cc), and one of her grandparents must also be a carrier (barring new mutations). Her first cousin then has a 1/4 chance of being a carrier, since one of their common grandparents is a carrier, one of his parents has a 1/2 chance of being a carrier, and he has a 1/2 chance of inheriting the c allele from his parent. The affected woman can only transmit c alleles to her fetus, while her cousin has 1/2 chance of transmitting his c allele if it is present. Thus, the probability that the first child will have cystic fibrosis is 1/4 (cousin is carrier) × 1/2 (cousin transmits c allele) = 1/8 (fetus has cc genotype).

A woman with cystic fibrosis marries her first cousin. What is the risk that their first child will have cystic fibrosis? a. 1/2 b. 1/4 c. 1/8 d. 1/16 e. 1/32

*The answer is E.* The common forms of color blindness are X-linked recessive, as indicated by the initial 3 of the McKusick number. The couple's daughters will be obligate carriers—that is, carriers implied by the pedigree. Using a lowercase c to represent the recessive color blindness allele, the woman is XCXC, while her husband is XcY. The Punnett square indicates that all daughters will be carriers (XcXC), while sons will be normal (XCY). Note again that loci on the X chromosome cannot be transmitted from father to son, since the son receives the father's Y chromosome.

A woman with no history of color blindness marries a color-blind man. What are the risks for this couple of having a son or daughter who is color-blind? a. 100% b. 75% c. 50% d. 25% e. Virtually 0

*The answer is C.* Genomic imprinting refers to the phenomenon in which an offspring's genes are expressed in a parent-specific manner. This occurs via an epigenetic process that alters the phenotype of an organism independent of the genetic code. Specifically, this epigenetic process involves DNA methylation, which can silence gene expression without altering the genetic sequence. DNA methylation is carried out by DNA methyltransferases that transfer methyl groups from methyl group donors (such as S-adenosyl-methionine) to cytosine residues in the DNA molecule.

You are working for an institute involved in studying the molecular processes associated with genetic transmission. Your experiments focus on studying a specific nuclear enzyme that transfers a methyl group from S-adenosyl-methionine to a cytosine residue in a DNA molecule. This enzyme is implicated in which of the following processes? A. Aneuploidy B. Anticipation C. Imprinting D. Meiotic nondisjunction E. Pleiotropy

*The answer is A.* FISH analysis of the child's chromosomes showed that the Prader-Willi/Angelman (PWA) locus on chromosome 15 was not present. Any unbalanced rearrangement of chromosome 15 that would have been inherited to cause Prader-Willi in the child would have to come from the father, since it is the paternally inherited deletion of chromosome 15 that causes most cases of Prader-Willi syndrome. The father could not be

You see a 4-year-old boy in clinic whom you believe has Prader Willi syndrome. You request cytogenetic studies and the child is found to have an unbalanced 14;15 translocation. Fluorescent in situ hybridization (FISH) confirms that the Prader-Willi/Angelman area on chromosome 15 is deleted. You request cytogenetic studies of the parents and one of them is found to have a balanced translocation. Which of the following are the most likely cytogenetic findings? (A) The father has a balanced 14;15 translocation. B) The father's Prader-Willi/Angelman locus is found by FISH to be deleted. (C) The mother has a balanced 14;15 translocation. (D) The mother's Prader-Willi/Angelman locus is found by FISH to be deleted.

*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 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 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 D.* The gene will not be eliminated from the population because it is "protected" in heterozygotes who are not affected and can pass the gene on to succeeding generations.

An autosomal recessive gene is lethal in homozygotes. In a population at Hardy-Weinberg equilibrium, how many generations will it take to eliminate the gene from the population (A) 1 (B) 25 (C) 100 (D) It will not be eliminated from the population.

*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.* The threshold of liability is reached when the environmental and genetic factors that contribute to a trait reach a level where the trait occurs.

Which of the following best describes the threshold of liability? (A) maximal risk for a trait with a bell-shaped population distribution (B) a high discordance rate in dizygotic twin pairs (C) a low concordance rate in monozygotic twin pairs (D) minimal level of defect-causing genes and environmental factors for trait occurrence

*The answer is D.* Although certain Robertsonian translocations involving chromosome 21 and other translocations involving chromosome 21 can result in Down syndrome, the most common cytogenetic finding in Down syndrome is three copies of chromosome 21, or trisomy for chromosome 21. Trisomy 21 is caused by nondisjunction of chromosome 21 during meiosis. A nondisjunction of chromosome 21 in mitosis is rare and would lead to mosaicism for trisomy 21.

Which of the following is the most common cause of Down syndrome? (A) Robertsonian translocations (B) 21;21 balanced reciprocal translocation (C) nondisjunction in mitosis (D) nondisjunction in meiosis

*The answer is E.* Aneuploidy involves extra or missing chromosomes that do not arise as increments of the haploid chromosome number n. Polyploidy involves multiples of n, such as triploidy (3n = 69,XXX) or tetraploidy (4n = 92,XXXX). Diploidy (46,XX) and haploidy (23,X) are normal karyotypes in gametes and somatic cells, respectively. A 90,XX karyotype represents tetraploidy with two missing X chromosomes, which has been seen in one patient who had features that resembled those of Turner's syndrome.

Which of the following karyotypes is an example of aneuploidy? a. 46,XX b. 23,X c. 69,XXX d. 92,XXXX e. 90,XX

*The answer is A.* A 47,XYY karyotype is usually only detected incidentally to some other indication for study since there is not an abnormal phenotype associated with it. Males with this karyotype are just as likely to be viable as those with a normal 46,XY karyotype. A 47,XX or XY, 18 karyotype can result in a liveborn, but the majority of fetuses with this karyotype spontaneously abort. Triploids (69 chromosomes) are rarely liveborn and even then do not usually survive beyond a couple of hours. A large percentage of first trimester spontaneous abortions have a 47,XX or XY, 16 karyotype.

Which of the following karyotypes is most likely to result in a viable (capable of being born alive) outcome? (A) 47,XYY (B) 47,16 (C) 69,XXX (D) 47,XY,18

*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