(BQ) Part 2 book Thompson & Thompson genetics in medicine presents the following contents: Developmental genetics and birth defects, cancer genetics and genomics, risk assessment and genetic counseling, prenatal diagnosis and screening, application of genomics to medicine and personalized health care, ethical and social issues in genetics and genomics.
C H A P T E R 12 The Molecular, Biochemical, and Cellular Basis of Genetic Disease In this chapter, we extend our examination of the molecular and biochemical basis of genetic disease beyond the hemoglobinopathies to include other diseases and the abnormalities in gene and protein function that cause them In Chapter 11, we presented an outline of the general mechanisms by which mutations cause disease (see Fig 11-1) and reviewed the steps at which mutations can disrupt the synthesis or function of a protein (see Table 11-2) Those outlines provide a framework for understanding the pathogenesis of all genetic disease However, mutations in other classes of proteins often disrupt cell and organ function by processes that differ from those illustrated by the hemoglobinopathies, and we explore them in this chapter To illustrate these other types of disease mechanisms, we examine here well-known disorders such as phenylketonuria, cystic fibrosis, familial hypercholesterolemia, Duchenne muscular dystrophy, and Alzheimer disease In some instances, less common disorders are included because they best demonstrate a specific principle The importance of selecting representative disorders becomes apparent when one considers that to date, mutations in almost 3000 genes have been associated with a clinical phenotype In the coming decade, one anticipates that many more of the approximately 20,000 to 25,000 coding genes in the human genome will be shown to be associated with both monogenic and genetically complex diseases DISEASES DUE TO MUTATIONS IN DIFFERENT CLASSES OF PROTEINS Proteins carry out an astounding number of different functions, some of which are presented in Figure 12-1 Mutations in virtually every functional class of protein can lead to genetic disorders In this chapter, we describe important genetic diseases that affect representative proteins selected from the groups shown in Figure 12-1; many other of the proteins listed, as well as the diseases associated with them, are described in the Cases section Housekeeping Proteins and Specialty Proteins in Genetic Disease Proteins can be separated into two general classes on the basis of their pattern of expression: housekeeping proteins, which are present in virtually every cell and have fundamental roles in the maintenance of cell structure and function; and tissue-specific specialty proteins, which are produced in only one or a limited number of cell types and have unique functions that contribute to the individuality of the cells in which they are expressed Most cell types in humans express 10,000 to 15,000 protein-coding genes Knowledge of the tissues in which a protein is expressed, particularly at high levels, is often useful in understanding the pathogenesis of a disease Two broad generalizations can be made about the relationship between the site of a protein’s expression and the site of disease • First (and somewhat intuitively), mutation in a tissuespecific protein most often produces a disease restricted to that tissue However, there may be secondary effects on other tissues, and in some cases mutations in tissue-specific proteins may cause abnormalities primarily in organs that not express the protein at all; ironically, the tissue expressing the mutant protein may be left entirely unaffected by the pathological process This situation is exemplified by phenylketonuria, discussed in depth in the next section Phenylketonuria is due to the absence of phenylalanine hydroxylase (PAH) activity in the liver, but it is the brain (which expresses very little of this enzyme), and not the liver, that is damaged by the high blood levels of phenylalanine resulting from the lack of hepatic PAH Consequently, one cannot necessarily infer that disease in an organ results from mutation in a gene expressed principally or only in that organ, or in that organ at all • Second, although housekeeping proteins are expressed in most or all tissues, the clinical effects of mutations in housekeeping proteins are frequently limited to one or just a few tissues, for at least two reasons In 215 216 THOMPSON & THOMPSON GENETICS IN MEDICINE ORGANELLES NUCLEUS Mitochondria Oxidative phosphorylation • ND1 protein of electron transport chain - Leber hereditary optic neuropathy Translation of mitochondrial proteins • tRNAleu - MELAS • 12S RNA - sensorineural deafness Peroxisomes Peroxisome biogenesis • 12 proteins - Zellweger syndrome Lysosomes Lysosomal enzymes • Hexosaminidase A - Tay-Sachs disease • α-L-iduronidase deficiency - Hurler syndrome Developmental transcription factors • Pax6 -aniridia Genome integrity • BRCA1, BRCA2 - breast cancer • DNA mismatch repair proteins - hereditary nonpolyposis colon cancer RNA translation regulation • FMRP (RNA binding to suppress translation) - fragile X syndrome Chromatin-associated proteins • MeCP2 (transcriptional repression) - Rett syndrome Tumor suppressors • Rb protein - retinoblastoma Oncogenes • BCR-Abl oncogene - chronic myelogenous leukemia EXTRACELLULAR PROTEINS Transport • β-globin - sickle cell disease - b-thalassemia Morphogens • Sonic hedgehog - holoprosencephaly Protease inhibition • α1-Antitrypsin - emphysema, liver disease Hemostasis • Factor VIII - hemophilia A Hormones • Insulin - rare forms of type diabetes mellitus Extracellular matrix • Collagen type - osteogenesis imperfecta Inflammation, infection response • Complement factor H - age-related macular degeneration CELL SURFACE CYTOPLASM Metabolic enzymes • Phenylalanine hydroxylase - PKU • Adenosine deaminase - severe combined immunodeficiency Cytoskeleton • Dystrophin - Duchenne muscular dystrophy Hormone receptors • Androgen receptor - androgen insensitivity Growth factor receptors • FGFR3 receptor - achondroplasia Metabolic receptors • LDL receptor - hypercholesterolemia Ion transport • CFTR - cystic fibrosis Antigen presentation • HLA locus DQβ1 - type diabetes mellitus Figure 12-1 Examples of the classes of proteins associated with diseases with a strong genetic component (most are monogenic), and the part of the cell in which those proteins normally function CFTR, Cystic fibrosis transmembrane regulator; FMRP, fragile X mental retardation protein; HLA, human leukocyte antigen; LDL, low-density lipoprotein; MELAS, mitochondrial encephalomyopathy with lactic acidosis and strokelike episodes; PKU, phenylketonuria most such instances, a single or a few tissue(s) may be affected because the housekeeping protein in question is normally expressed abundantly there and serves a specialty function in that tissue This situation is illustrated by Tay-Sachs disease, as discussed later; the mutant enzyme in this disorder is hexos aminidase A, which is expressed in virtually all cells, but its absence leads to a fatal neurodegeneration, leaving non-neuronal cell types unscathed In other instances, another protein with overlapping biological activity may also be expressed in the unaffected tissue, thereby lessening the impact of the loss of function of the mutant gene, a situation known as genetic redundancy Unexpectedly, even mutations in genes that one might consider as essential to every cell, such as actin, can result in viable offspring DISEASES INVOLVING ENZYMES Enzymes are the catalysts that mediate the efficient conversion of a substrate to a product The diversity of substrates on which enzymes act is huge Accordingly, the human genome contains more than 5000 genes that encode enzymes, and there are hundreds of human diseases—the so-called enzymopathies—that involve enzyme defects We first discuss one of the best-known groups of inborn errors of metabolism, the hyperphenylalaninemias CHAPTER 12 — The Molecular, Biochemical, and Cellular Basis of Genetic Disease 217 GTP GTP-cyclohydrolase DHNP Protein (diet, endogenous) 6-PT synthase Sepiapterin reductase Phenylalanine phe Tyrosine 4αOHBH4 DHPR qBH2 BH4 tyr CO2 + H2O phe hydroxylase Phenylalanine hydroxylase BH4 6-PT tyr BH4 L-dopa tyr hydroxylase PCD trp BH4 5-OH trp trp hydroxylase dopamine NE E serotonin Figure 12-2 The biochemical pathways affected in the hyperphenylalaninemias BH4, tetrahydro- biopterin; 4αOHBH4, 4α-hydroxytetrahydrobiopterin; qBH2, quinonoid dihydrobiopterin, the oxidized product of the hydroxylation reactions, which is reduced to BH4 by dihydropteridine reductase (DHPR); PCD, pterin 4α-carbinolamine dehydratase; phe, phenylalanine; tyr, tyrosine; trp, tryptophan; GTP, guanosine triphosphate; DHNP, dihydroneopterin triphosphate; 6-PT, 6-pyruvoyltetrahydropterin; L-dopa, L-dihydroxyphenylalanine; NE, norepinephrine; E, epinephrine; 5-OH trp, 5-hydroxytryptophan TABLE 12-1 Locus Heterogeneity in the Hyperphenylalaninemias Biochemical Defect Incidence/106 Births Enzyme Affected Treatment Mutations in the Gene Encoding Phenylalanine Hydroxylase Classic PKU Variant PKU Non-PKU hyperphenylalaninemia 5-350 (depending on the population) Less than classic PKU PAH Low-phenylalanine diet* PAH 15-75 PAH Low-phenylalanine diet (less restrictive than that required to treat PKU* None, or a much less restrictive low-phenylalanine diet* Mutations in Genes Encoding Enzymes of Tetrahydrobiopterin Metabolism Impaired BH4 recycling a 9% R243Q 18% E6nt–96a>g 14% Figure 12-4 The nature and identity of PAH mutations in popu- lations of European and Asian descent (the latter from China, Korea, and Japan) The one-letter amino acid code is used (see Table 3-1) See Sources & Acknowledgments normal, and the hyperphenylalaninemia results from a defect in one of the steps in the biosynthesis or regeneration of BH4, the cofactor for PAH (see Table 12-1 and Fig 12-2) The association of a single biochemical phenotype, such as hyperphenylalaninemia, with mutations in different genes, is an example of locus heterogeneity (see Table 11-1) The proteins encoded by genes that manifest locus heterogeneity generally act at different steps in a single biochemical pathway, another principle of genetic disease illustrated by the genes associated with hyperphenylalaninemia (see Fig 12-2) BH4deficient patients were first recognized because they developed profound neurological problems in early life, despite the successful administration of a lowphenylalanine diet This poor outcome is due in part to the requirement for the BH4 cofactor of two other enzymes, tyrosine hydroxylase and tryptophan hydroxylase These hydroxylases are critical for the synthesis of the monoamine neurotransmitters dopamine, norepinephrine, epinephrine, and serotonin (see Fig 12-2) 220 THOMPSON & THOMPSON GENETICS IN MEDICINE The locus heterogeneity of hyperphenylalaninemia is of great significance because the treatment of patients with a defect in BH4 metabolism differs markedly from subjects with mutations in PAH, in two ways First, because the PAH enzyme of individuals with BH4 defects is itself normal, its activity can be restored by large doses of oral BH4, leading to a reduction in their plasma phenylalanine levels This practice highlights the principle of product replacement in the treatment of some genetic disorders (see Chapter 13) Consequently, phenylalanine restriction can be significantly relaxed in the diet of patients with defects in BH4 metabolism, and some patients actually tolerate a normal (i.e., a phenylalanine-unrestricted) diet Second, one must also try to normalize the neurotransmitters in the brains of these patients by administering the products of tyrosine hydroxylase and tryptophan hydroxylase, L-dopa and 5-hydroxytryptophan, respectively (see Fig 12-2 and Table 12-1) Remarkably, mutations in sepiapterin reductase, an enzyme in the BH4 synthesis pathway, not cause hyperphenylalaninemia In this case, only doparesponsive dystonia is seen, due to impaired synthesis of dopamine and serotonin (see Fig 12-2) It is thought that alternative pathways exist for the final step in BH4 synthesis, bypassing the sepiapterin reductase deficiency in peripheral tissues, an example of genetic redundancy For these reasons, all hyperphenylalaninemic infants must be screened to determine whether their hyperphenylalaninemia is the result of an abnormality in PAH or in BH4 metabolism The hyperphenylalaninemias thus illustrate the critical importance of obtaining a specific molecular diagnosis in all patients with a genetic disease phenotype—the underlying genetic defect may not be what one first suspects, and the treatment can vary accordingly Tetrahydrobiopterin Responsiveness in PAH Mutations. Many hyperphenylalaninemia patients with mutations in the PAH gene (rather than in BH4 metabolism) will also respond to large oral doses of BH4 cofactor, with a substantial decrease in plasma phenylalanine BH4 supplementation is therefore an important adjunct therapy for PKU patients of this type, allowing them a less restricted dietary intake of phenylalanine The patients most likely to respond are those with significant residual PAH activity (i.e., patients with variant PKU and non-PKU hyperphenylalaninemia), but even a minority of patients with classic PKU are also responsive The presence of residual PAH activity does not, however, necessarily guarantee an effect of BH4 administration on plasma phenylalanine levels Rather, the degree of BH4 responsiveness will depend on the specific properties of each mutant PAH protein, reflecting the allelic heterogeneity underlying PAH mutations The provision of increased amounts of a cofactor is a general strategy that has been employed for the treatment of many inborn errors of enzyme metabolism, as discussed further in Chapter 13 In the general case, a cofactor comes into contact with the protein component of an enzyme (termed an apoenzyme) to form the active holoenzyme, which consists of both the cofactor and the otherwise inactive apoenzyme Illustrating this strategy, BH4 supplementation has been shown to exert its beneficial effect through one or more mechanisms, all of which result from the increased amount of the cofactor that is brought into contact with the mutant PAH apoenzyme These mechanisms include stabilization of the mutant enzyme, protection of the enzyme from degradation by the cell, and increase in the cofactor supply for a mutant enzyme that has a low affinity for BH4 Newborn Screening. PKU is the prototype of genetic diseases for which mass newborn screening is justified (see Chapter 18) because it is relatively common in some populations (up to approximately in 2900 live births), mass screening is feasible, failure to treat has severe consequences (profound developmental delay), and treatment is effective if begun early in life To allow time for the postnatal increase in blood phenylalanine levels to occur, the test is performed after 24 hours of age Blood from a heel prick is assayed in a central laboratory for blood phenylalanine levels and measurement of the phenylalanine-to-tyrosine ratio Positive test results must be confirmed quickly because delays in treatment beyond weeks postnatally have profound effects on intellectual outcome Maternal Phenylketonuria. Originally, the low-phenyl- alanine diet was discontinued in mid-childhood for most patients with PKU Subsequently, however, it was discovered that almost all offspring of women with PKU not receiving treatment are clinically abnormal; most are severely delayed developmentally, and many have microcephaly, growth impairment, and malformations, particularly of the heart As predicted by principles of mendelian inheritance, all of these children are heterozygotes Thus their neurodevelopmental delay is not due to their own genetic constitution but to the highly teratogenic effect of elevated levels of phenylalanine in the maternal circulation Accordingly, it is imperative that women with PKU who are planning pregnancies commence a low-phenylalanine diet before conceiving Lysosomal Storage Diseases: A Unique Class of Enzymopathies Lysosomes are membrane-bound organelles containing an array of hydrolytic enzymes involved in the degradation of a variety of biological macromolecules Mutations in these hydrolases are unique because they lead to the accumulation of their substrates inside the CHAPTER 12 — The Molecular, Biochemical, and Cellular Basis of Genetic Disease lysosome, where the substrates remain trapped because their large size prevents their egress from the organelle Their accumulation and sometimes toxicity interferes with normal cell function, eventually causing cell death Moreover, the substrate accumulation underlies one uniform clinical feature of these diseases—their unrelenting progression In most of these conditions, substrate storage increases the mass of the affected tissues and organs When the brain is affected, the picture is one of neurodegeneration The clinical phenotypes are very distinct and often make the diagnosis of a storage disease straightforward More than 50 lysosomal hydrolase or lysosomal membrane transport deficiencies, almost all inherited as autosomal recessive conditions, have been described Historically, these diseases were untreatable However, bone marrow transplantation and enzyme replacement therapy have dramatically improved the prognosis of these conditions (see Chapter 13) Tay-Sachs Disease Tay-Sachs disease (Case 43) is one of a group of heterogeneous lysosomal storage diseases, the GM2 gangliosidoses, that result from the inability to degrade a sphingolipid, GM2 ganglioside (Fig 12-5) The biochem ical lesion is a marked deficiency of hexosaminidase A (hex A) Although the enzyme is ubiquitous, the disease has its clinical impact almost solely on the brain, the predominant site of GM2 ganglioside synthesis Catalytically active hex A is the product of a three-gene system (see Fig 12-5) These genes encode the α and β subunits of the enzyme (the HEXA and HEXB genes, respectively) and an activator protein that must associate with the substrate and the enzyme before the enzyme can cleave the terminal N-acetyl-β-galactosamine residue from the ganglioside The clinical manifestations of defects in the three genes are indistinguishable, but they can be differentiated by enzymatic analysis Mutations in the HEXA gene affect the α subunit and disrupt hex A activity to cause Tay-Sachs disease (or less severe variants of hex A deficiency) Defects in the HEXB gene or in the gene encoding the activator protein impair the activity of both hex A and hex B (see Fig 12-5) to produce Sandhoff disease or activator protein deficiency (which is very rare), respectively The clinical course of Tay-Sachs disease is tragic Affected infants appear normal until approximately to months of age but then gradually undergo progressive neurological deterioration until death at to years The effects of neuronal death can be seen directly in the form of the so-called cherry-red spot in the The GM2 gangliosidoses Disease Tay-Sachs disease and later-onset variants Affected gene Polypeptide Sandhoff disease and later-onset variants Activator deficiency α (chr 15) β (chr 5) activator (chr 5) α subunit β subunit activator Hex A: αβ Isozyme: subunits Hex B: ββ activator Active enzyme complex 221 αβ GM2 ganglioside N-acetylgalactosamine - galactose - glucose - ceramide Cleavage site NANA Figure 12-5 The three-gene system required for hexosaminidase A activity and the diseases that result from defects in each of the genes The function of the activator protein is to bind the ganglioside substrate and present it to the enzyme Hex A, Hexosaminidase A; hex B, hexosaminidase B; NANA, N-acetyl neuraminic acid See Sources & Acknowledgments 222 THOMPSON & THOMPSON GENETICS IN MEDICINE – Arg – Ile – Ser – Try – Gly – Pro – Asp – Normal HEXA allele Tay-Sachs allele CGT ATA TCC TAT CGT ATA TCT ATC GCC CCT CTA TGC GAC CCC TGA C – Arg – Ile – Ser – Ile – Leu – Cys – Pro – Stop Altered reading frame Figure 12-6 Four-base insertion (TATC) in the hexosaminidase A (hex A) gene in Tay-Sachs disease, leading to a frameshift mutation This mutation is the major cause of Tay-Sachs disease in Ashkenazi Jews No detectable hex A protein is made, accounting for the complete enzyme deficiency observed in these infantile-onset patients retina (Case 43) In contrast, HEXA alleles associated with some residual activity lead to later-onset forms of neurological disease, with manifestations including lower motor neuron dysfunction and ataxia due to spinocerebellar degeneration In contrast to the infantile disease, vision and intelligence usually remain normal, although psychosis develops in one third of these patients Finally, pseudodeficiency alleles (discussed next) not cause disease at all Hex A Pseudodeficiency Alleles and Their Clinical Significance. An unexpected consequence of screening for Tay-Sachs carriers in the Ashkenazi Jewish population was the discovery of a unique class of hex A alleles, the so-called pseudodeficiency alleles Although the two pseudodeficiency alleles are clinically benign, individuals identified as pseudodeficient in screening tests are genetic compounds with a pseudodeficiency allele on one chromosome and a common Tay-Sachs mutation on the other chromosome These individuals have a low level of hex A activity (approximately 20% of controls) that is adequate to prevent GM2 ganglioside accumulation in the brain The importance of hex A pseudodeficiency alleles is twofold First, they complicate prenatal diagnosis because a pseudodeficient fetus could be incorrectly diagnosed as affected More generally, the recognition of the hex A pseudodeficiency alleles indicates that screening programs for other genetic diseases must recognize that comparable alleles may exist at other loci and may confound the correct characterization of individuals in screening or diagnostic tests Population Genetics. In many single-gene diseases, some alleles are found at higher frequency in some populations than in others (see Chapter 9) This situation is illustrated by Tay-Sachs disease, in which three alleles account for 99% of the mutations found in Ashkenazi Jewish patients, the most common of which (Fig 12-6) accounts for 80% of cases Approximately in 27 Ashkenazi Jews is a carrier of a Tay-Sachs allele, and the incidence of affected infants is 100 times higher than in other populations A founder effect or heterozygote advantage is the most likely explanation for this high frequency (see Chapter 9) Because most Ashkenazi Jewish carriers will have one of the three common alleles, a practical benefit of the molecular characterization of the disease in this population is the degree to which carrier screening has been simplified Altered Protein Function due to Abnormal Post-translational Modification A Loss of Glycosylation: I-Cell Disease Some proteins have information contained in their primary amino acid sequence that directs them to their subcellular residence, whereas others are localized on the basis of post-translational modifications This latter mechanism is true of the acid hydrolases found in lysosomes, but this form of cellular trafficking was unrecognized until the discovery of I-cell disease, a severe autosomal recessive lysosomal storage disease The disorder has a range of phenotypic effects involving facial features, skeletal changes, growth retardation, and intellectual disability and survival of less than 10 years (Fig 12-7) The cytoplasm of cultured skin fibroblasts from I-cell patients contains numerous abnormal lysosomes, or inclusions, (hence the term inclusion cells or I cells) In I-cell disease, the cellular levels of many lysosomal acid hydrolases are greatly diminished, and instead they are found in excess in body fluids This unusual situation arises because the hydrolases in these patients have not been properly modified post-translationally A typical hydrolase is a glycoprotein, the sugar moiety containing mannose residues, some of which are phosphorylated The mannose-6-phosphate residues are essential for recognition of the hydrolases by receptors on the cell and lysosomal membrane surface In I-cell disease, there is a defect in the enzyme that transfers a phosphate group to the mannose residues The fact that many enzymes are affected is consistent with the diversity of clinical abnormalities seen in these patients CHAPTER 12 — The Molecular, Biochemical, and Cellular Basis of Genetic Disease 223 may be amenable to chemical therapies that reduce the excessive glycosylation Loss of Protein Function due to Impaired Binding or Metabolism of Cofactors Figure 12-7 I-cell disease facies and habitus in an 18-month-old girl See Sources & Acknowledgments Gains of Glycosylation: Mutations That Create New (Abnormal) Glycosylation Sites In contrast to the failure of protein glycosylation exemplified by I-cell disease, it has been shown that an unexpectedly high proportion (approximately 1.5%) of the missense mutations that cause human disease may be associated with abnormal gains of N-glycosylation due to mutations creating new consensus N-glycosylation sites in the mutant proteins That such novel sites can actually lead to inappropriate glycosylation of the mutant protein, with pathogenic consequences, is highlighted by the rare autosomal recessive disorder, mendelian susceptibility to mycobacterial disease (MSMD) MSMD patients have defects in any one of a number of genes that regulate the defense against some infections Consequently, they are susceptible to disseminated infections upon exposure to moderately virulent mycobacterial species, such as the bacillus Calmette-Guérin (BCG) used throughout the world as a vaccine against tuberculosis, or to nontuberculous environmental bacteria that not normally cause illness Some MSMD patients carry missense mutations in the gene for interferon-γ receptor (IFNGR2) that generate novel N-glycosylation sites in the mutant IFNGR2 protein These novel sites lead to the synthesis of an abnormally large, overly glycosylated receptor The mutant receptors reach the cell surface but fail to respond to interferon-γ Mutations leading to gains of glycosylation have also been found to lead to a loss of protein function in several other monogenic disorders The discovery that removal of the abnormal polysaccharides restores function to the mutant IFNGR2 proteins in MSMD offers hope that disorders of this type Some proteins acquire biological activity only after they associate with cofactors, such as BH4 in the case of PAH, as discussed earlier Mutations that interfere with cofactor synthesis, binding, transport, or removal from a protein (when ligand binding is covalent) are also known For many of these mutant proteins, an increase in the intracellular concentration of the cofactor is frequently capable of restoring some residual activity to the mutant enzyme, for example by increasing the stability of the mutant protein Consequently, enzyme defects of this type are among the most responsive of genetic disorders to specific biochemical therapy because the cofactor or its precursor is often a water-soluble vitamin that can be administered safely in large amounts (see Chapter 13) Impaired Cofactor Binding: Homocystinuria due to Cystathionine Synthase Deficiency Homocystinuria due to cystathionine synthase deficiency (Fig 12-8) was one of the first aminoacidopathies to be recognized The clinical phenotype of this autosomal recessive condition is often dramatic The most common features include dislocation of the lens, intellectual disability, osteoporosis, long bones, and thromboembolism of both veins and arteries, a phenotype that can be confused with Marfan syndrome, a disorder of connective tissue (Case 30) The accumulation of homocysteine is believed to be central to most, if not all, of the pathology Homocystinuria was one of the first genetic diseases shown to be vitamin responsive; pyridoxal phosphate is the cofactor of the enzyme, and the administration of large amounts of pyridoxine, the vitamin precursor of the cofactor, often ameliorates the biochemical abnormality and the clinical disease (see Chapter 13) In many patients, the affinity of the mutant enzyme for pyridoxal phosphate is reduced, indicating that altered conformation of the protein impairs cofactor binding Not all cases of homocystinuria result from mutations in cystathionine synthase Mutations in five dif ferent enzymes of cobalamin (vitamin B12) or folate metabolism can also lead to increased levels of homocysteine in body fluids These mutations impair the provision of the vitamin B12 cofactor, methylcobalamin (methyl-B12), or of methyl-H4-folate (see Fig 12-8) and thus represent another example (like the defects in BH4 synthesis that lead to hyperphenylalaninemia) of genetic diseases due to defects in the biogenesis of enzyme cofactors The clinical manifestation of these disorders is variable but includes megaloblastic anemia, developmental delay, and failure to thrive These conditions, all 224 THOMPSON & THOMPSON GENETICS IN MEDICINE Cystathionine synthase Methionine Homocysteine Cystathionine Methionine synthase Vitamin B6 Methyl-B12 H4-folate Cysteine Pyridoxal phosphate Methyl-H4-folate Figure 12-8 Genetic defects in pathways that impinge on cystathionine synthase, or in that enzyme itself, and cause homocystinuria Classic homocystinuria is due to defective cystathionine synthase Several different defects in the intracellular metabolism of cobalamins (not shown) lead to a decrease in the synthesis of methylcobalamin (methyl-B12) and thus in the function of methionine synthase Defects in methylene-H4-folate reductase (not shown) decrease the abundance of methylH4-folate, which also impairs the function of methionine synthase Some patients with cystathionine synthase abnormalities respond to large doses of vitamin B6, increasing the synthesis of pyridoxal phosphate, thereby increasing cystathionine synthase activity and treating the disease (see Chapter 13) of which are autosomal recessive, are often partially or completely treatable with high doses of vitamin B12 Mutations of an Enzyme Inhibitor: α1-Antitrypsin Deficiency α1-Antitrypsin (α1AT) deficiency is an important autosomal recessive condition associated with a substantial risk for chronic obstructive lung disease (emphysema) (Fig 12-9) and cirrhosis of the liver The α1AT protein belongs to a major family of protease inhibitors, the serine protease inhibitors or serpins; SERPINA1 is the formal gene name Notwithstanding the specificity suggested by its name, α1AT actually inhibits a wide spectrum of proteases, particularly elastase released from neutrophils in the lower respiratory tract In white populations, α1AT deficiency affects approximately in 6700 persons, and approximately 4% are carriers A dozen or so α1AT alleles are associated with an increased risk for lung or liver disease, but only the Z allele (Glu342Lys) is relatively common The reason for the relatively high frequency of the Z allele in white populations is unknown, but analysis of DNA haplotypes suggests a single origin with subsequent spread throughout northern Europe Given the increased risk for emphysema, α1AT deficiency is an important public health problem, affecting an estimated 60,000 persons in the United States alone The α1AT gene is expressed principally in the liver, which normally secretes α1AT into plasma Approximately 17% of Z/Z homozygotes present with neonatal jaundice, and approximately 20% of this group subsequently develop cirrhosis The liver disease associated with the Z allele is thought to result from a novel property of the mutant protein—its tendency to aggregate, trapping it within the rough endoplasmic reticulum (ER) of hepatocytes The molecular basis of the Z protein aggregation is a consequence of structural changes in the protein that predispose to the formation of long beadlike necklaces of mutant α1AT polymers Figure 12-9 The effect of smoking on the survival of patients with α1-antitrypsin deficiency The curves show the cumulative probability of survival to specified ages of smokers, with or without α1-antitrypsin deficiency See Sources & Acknowledgments Cumulative probability of survival 1.0 All females (mostly M/M) 0.8 0.6 Z/Z nonsmokers All males (mostly M/M) 0.4 Z/Z smokers 0.2 20 30 40 50 60 70 Age (years) 80 90 100 518 ANSWERS TO PROBLEMS (a) None; the short arms of all acrocentric chromosomes are believed to be identical and contain multiple copies of rRNA genes (b) None if the deletion involves only heterochromatin (Yq12) A more proximal deletion might delete genes important in spermatogenesis (see Fig 6-9) (c) Cri du chat syndrome, severity depending on the amount of DNA deleted (see Fig 6-6) (d) Some features of Turner syndrome, but with normal stature; the Xq− chromosome is preferentially inactivated in all cells (provided that the X inactivation center is not deleted), thus reducing the potential severity of such a deletion Different parts of the genome contain different density of genes Thus deletion of the same amount of DNA on different chromosomes might delete a vastly different number of genes, thus leading to different phenotypic expectations (see Fig 2-7) Question for discussion See text for possible explanations (a) A 1% risk is often quoted, but the risk is probably not greater than the population age-related risk (b) Age-related risk is greater than 1% (c) No increased risk if the niece with Down syndrome has trisomy 21, but if the niece carries a Robertsonian translocation, the consultand may be a carrier and at high risk (d) 10% to 15%; see text (e) Only a few percent; see text The woman’s agerelated risk may be relevant 10 46,XX,rob(21;21)(q10;q10) or 46,XX,der(21;21) (q10;q10) (There is no need to add +21 to the karyotype because the 46 designates that she must have a normal 21 in addition to the translocation.) 11 Crossing over leads to either balanced gametes or nonviable gametes (see Fig 5-13) Thus liveborn offspring are genomically balanced CHAPTER 7 Patterns of Single-Gene Inheritance (b) Autosomal recessive; in 4, assuming same paternity as her first child (c) Calvin and Cathy are obligate heterozygotes Given that Calvin and Cathy are first cousins, it is also very likely that they inherited their mutant allele through Betty and Barbara from the same grandparent Thus Betty and Barbara are very likely to be carriers, but it is not obligatory It is theoretically possible that Cathy inherited her CF allele from Bob and that Calvin inherited his from his father, Barbara’s husband DNAbased carrier testing will answer the question definitively (a) Heterozygous at each of two loci; for example, A/a B/b (b) George and Grace are likely carriers for an autosomal recessive form of deafness; Horace is either a homozygote or compound heterozygote at this same deafness locus Gilbert and Gisele are both homozygotes or compound heterozygotes for deafness due to mutations at a deafness locus as well The fact that all of Horace and Hedy’s children are deaf suggests that the deafness locus in Gilbert and Gisele’s family and the locus in George and Grace’s family are the same locus Isaac and Ingrid, however, although deaf, are deaf due to each being homozygous or compound heterozygote at two different deafness loci, so all of their children are double heterozygotes (as shown in part a of this question) Variable expressivity—d; uniparental disomy—i; consanguinity—j; inbreeding—c; X-linked dominant inheritance—g; new mutation—e; allelic heterogeneity—h; locus heterogeneity—a; homozygosity for an autosomal dominant trait—b; pleiotropy—f (b) They are homozygous (c) 100% for a son of Elise; virtually zero for a daughter unless Elise’s partner has hemophilia A (d) Enid is an obligatory carrier (heterozygote for hemophilia A mutation) because she has an affected father but is herself not affected, so the chance a son will be affected is 50% The probability a daughter will be a carrier is 50%, but the daughter’s chance of being affected is virtually zero unless Enid’s partner is himself affected with hemophilia A, which would give a 50% chance of being affected, or if there is some very unusual circumstance of highly skewed X inactivation or if her daughter has Turner syndrome with a single maternal X carrying the mutant hemophilia A gene All are possible except (c), which is unlikely if the parents are completely unaffected (a) New mutation or germline mosaicism in one of the parents (b) Mutation rate at the NF1 locus if truly a new mutation; if one of the parents is a germline mosaic, the risk in the next pregnancy is a function of the fraction of gametes carrying the mutation, which is unknown (c) Mutation rate at the NF1 locus if truly a new mutation; if the father is a germline mosaic for NF1, the risk in the next pregnancy is a function of the fraction of sperm carrying the mutation, which is unknown (d) 50% The consultand and her partner are first cousins once removed The probability that a child of this mating will be homozygous at any given locus for an allele inherited through each parent from a common ANSWERS TO PROBLEMS ancestor is known as the coefficient of inbreeding (F) In the accompanying figure, suppose individual I-1 is heterozygous for alleles and 2, whereas individual I-2 is a 3,4 heterozygote The chance II-2 inherits the allele is , and the chance III-2 inherits it from II-2 is , so the chance III-2 inherits allele from I-1 is × = Similarly, the chance IV-1 is a carrier of 2 allele inherited from I-1 is × × = , and the chance III-2 and IV-1 are both heterozygotes for inherited from I-1 is × = 32 The probability their child could be a 22 homozygote inherited from I-1 as a common ancestor is therefore × 32 = 1128 Repeat this calculation for allele in I-1, and for either allele or in II-1, which means the probability the child could be homozygous 1,1, 2,2, 3,3, or 4,4 is equal to × 1128 = 32 because there are four ways the child could be a homozygote for an allele inherited from either of the two common ancestors This is the coefficient of inbreeding A simple way to calculate F in simple pedigrees like this is the path method, in which one determines all the paths by which an allele from a common ancestor can be transmitted to the individual whose coefficient of inbreeding one is seeking to calculate Form all the paths connecting all the pertinent individuals in this pedigree (see Figure) Each path that gives a closed loop is a consanguineous path There are two closed loops: A-D-H-K-L-I-E-A and B-D-H-K-L-I-E-B To calculate F, count all the “nodes” (the dots representing each of the individuals) in each of the closed loops, counting each node only once Call that n The coefficient of inbreeding due to that loop is then given by ( )n−1 So, in this example, the loop A-D-H-K-L-I-E-A contains seven unique nodes, n = Add all the coefficients for each loop together to find F For the pedigree, then: (1 2)n−1 = (1 2)6 = (1 64)n−1 = (1 64)6 = 64 for loop A-D-H-K-L-I-E-A 64 for loop B-D-H-K-L-I-E-B and thus F = 32 AD is most likely Vertical transmission, including male to male, from generation to generation, males and females affected A C D AR and XR are possible but unlikely AR would require that both of the spouses of the two affected individuals in generations I and II be carriers, which is unlikely unless the pedigree comes from a genetic isolate (so-called pseudodominant inheritance of a recessive disorder due to high frequency of carriers in the population) XR would require that the same two women be carriers and, in addition, that there be something unusual in the X inactivation pattern for the female in generation III to be affected while neither of the females in generation II (who are both obligatory carriers) is affected Mitochondrial and XD inheritance are incompatible There is male to male transmission, which eliminates both of these modes of inheritance In addition, there are female offspring of affected males who are not affected The probability a child of two carriers of cystic fibrosis will be affected is 0.25, based on autosomal recessive inheritance The probability a child will be affected by cystic fibrosis whose mother is a carrier, but whose father is not, is the chance of inheriting the mother’s mutant allele and a new mutation in the sperm, which is 0.5 × × 10−6, and therefore the odds both parents are carriers versus only the mother is 0.25/(0.5 × 10−6) = × 107 The odds in favor of both parents being carriers are overwhelming Indeed, the probability of misattributed paternity, with the biological father being a carrier, dwarfs the probability of a new mutation CHAPTER 8 Complex Inheritance of Common Multifactorial Disorders (a) Autosomal dominant with reduced penetrance If it were truly multifactorial, the risk for more distantly related relatives would drop by more than 50% with each increase in the degree of relatedness (b) In dominant disease, a study of multiple families with the condition would reveal the expected 50% ratio of affected to unaffected in the children of an affected individual (after correcting A B E I H D D F B A B E H G H I E I J L K L 519 K Figure for Chapter 7, question L K 520 ANSWERS TO PROBLEMS for bias of ascertainment of the families) In multifactorial inheritance, there would be fewer than the expected 50% affected in the children of an affected individual Male to male transmission can disprove X-linkage; other criteria of multifactorial inheritance can be examined, as in the text For autosomal recessive but not for multifactorial inheritance, all of the affected individuals in a family tend to be in the same sibship, with unaffected parents, whereas diseases with multifactorial inheritance can present as affected parents with affected children It is generally rare for a parent of children with an autosomal recessive disorder to be himself or herself affected because it would require a homozygote or compound heterozygote–affected parent to mate with a carrier of a mutant allele at that same locus There can be an increased incidence of such rare matings, however, when there is assortative mating or if the couple is consanguineous or comes from an inbred population CHAPTER 9 Genetic Variation in Populations One way of determining this is to reverse the question and ask instead what proportion of individuals would be homozygous Then the proportion that is heterozygous is minus the proportion that is homozygous The frequency of homozygotes for the first allele would be 0.40 × 0.40 = 0.16, 0.30 × 0.30 = 0.09 for the second allele, 0.15 × 0.15 = 0.0225 for allele three, etc Adding these up for the five alleles (0.16 + 0.09 + 0.0225 + 0.01 + 0.0025 = 0.285) means 28.5% of individuals would be homozygous for allele or allele or … allele Therefore 71.5% of individuals would be heterozygous at this locus q = 0.26, p = ≈0.74, p2 = ≈0.55, 2pq = ≈0.38, q2 = ≈0.07 Frequency of Rh−/− genotype in mother = 0.07 Frequency of Rh+/+ in father = 0.55 Frequency of Rh+/− in father = 0.38 First pregnancy: Probability of Rh−/− mother × Rh+/+ father mating = 0.07 × 0.55 = 3.8% Probability of Rh−/− mother × Rh+/− father mating = 0.07 × 0.38 = 2.66% Second pregnancy: All second pregnancies of Rh−/− mother × Rh+/+ father will be sensitized by the first pregnancy = 3.8%, and all are at risk for Rh incompatibility in the next pregnancy Only half of the first pregnancies of an Rh−/− mother by Rh+/− father will have sensitized the mother (Rh+/−), so the risk for a sensitized Rh−/− mother with an Rh+/− partner in the second pregnancy = × 2.66% = 1.33% and the chance a sensitized Rh−/− mother will have a Rh+/− child when her mate is Rh+/− = × 1.33% = 0.66% Total at risk for incompatibility 3.8% + 0.66% ≈ 4.5% in the population at the time of a second pregnancy in the absence of any prophylaxis (a) Assume there are 100 individuals in the population, carrying 200 alleles at a particular locus The frequency of A is (2 × 81 200 ) + (18 200) = 0.9 and frequency of a = 0.1 (b) The genotype frequencies will be the same as in this generation, assuming Hardy-Weinberg equilibrium (c) Frequency of A/a by A/a matings = 0.18 × 0.18 = ≈0.0324 (a) When q is small, p = ≈1, and so 2pq = ≈2q Therefore, if 2pq = 0.04, then the allele frequency of β-thalassemia q = ≈0.02 (You can also calculate q exactly by letting 2pq = (1 − q)q = 0.04, or q2 − q + 0.02 = and solve the quadratic equation.) (b) Assuming only heterozygotes for β-thalassemia are likely to reproduce (a reasonable assumption because the fitness in β-thalassemia is quite low), then 0.04 × 0.04 = 0.0016 = 0.16% of matings will be between heterozygotes (c) Incidence of affected fetuses or newborns = 0.04% assuming no increased fetal losses in β-thalassemia, which is a reasonable assumption because the disorder has a postnatal onset (d) Incidence of carriers among the offspring of couples both found to be heterozygous is 50% Only (d) is in equilibrium Possible explanations include selection for or against particular genotypes, nonrandom mating, and recent migration (a) Abby has a chance of being a carrier Andrew has approximately a 1150 chance of being a carrier Therefore their risk for having an affected child is × 1150 × , or 900 (b) × × = 24 (c) × 22 × = 1132; × × = 24 (a) Facioscapulohumeral muscular dystrophy: q = 1 50,000 , 2pq = 25,000 Friedreich ataxia: q = 158, 2pq = 79 Duchenne muscular dystrophy is X-linked recessive and occurs mostly in males, so we will ignore any of the rare females affected If it occurs in the population at a frequency of in 25,000, then, assuming half of the population is male, the frequency in males must be in 12,500, so q = 112,500, 2pq = 6,250 (b) The autosomal dominant and X-linked disorders would increase rapidly, within one generation, to reach a new balance The autosomal recessive disorders would increase also, but only very slowly, because the majority of the mutant alleles are not subject to selection Mutant allele frequencies are approximately 26 and 316 Two possible explanations for the difference in ANSWERS TO PROBLEMS allele frequency could be (1) founder effect (or more generally genetic drift) in the early Quebec population when it was very small and inbred, resulting in an increased mutant allele frequency, or (2) environmental conditions of unknown type that provided a heterozygote advantage in Quebec that raised the allele frequency through increased fitness of heterozygote carriers CHAPTER 10 Identifying the Genetic Basis for Human Disease The HD and MNSs loci map far enough apart on chromosome to be unlinked, even though they are syntenic The LOD scores indicate that this polymorphism in the α-globin gene locus is closely linked to the polycystic kidney disease gene The peak LOD score, 25.85, occurs at cM The odds in favor of linkage at this distance compared with no linkage at all are 1025.85 : 1 (i.e., almost 1026 : 1) The data in the second study indicate that there is no linkage between the disease gene and the polymorphism in this family Thus there is genetic heterogeneity in this disorder, and linkage information can therefore be used for diagnosis only if there is previous evidence that the disease in that particular family is linked to the polymorphism Every parent who passed on the cataract was also informative at the γ-crystallin locus, that is, was heterozygous for the polymorphic alleles at this locus The phase is known by inspecting the pedigree in individuals IV-7 and IV-8 because these two individuals received both the cataract allele and the A allele at the γ-crystallin locus from their father (but note, we not know what the phase was in the father simply by inspection) We not know the phase in individuals IV-3 or IV-4 because we not know if they inherited the cataract mutation along with the A or the B allele at the γ-crystallin locus from their mother Phase is also known in individuals V-1, V-2, V-6, and V-7 The cataract seems to cosegregate with the “A” allele There are no crossovers A complete LOD score analysis should be performed In addition, one might examine the γ-crystallin gene itself for mutations in affected persons because it would be a reasonable candidate for being the gene in which mutations could cause cataracts (a) The phase in the mother is probably B-WAS (where WAS is the disease-causing allele), according to the genotype of the affected boy This phase can be determined with only 95% certainty because there is a 5% chance that a crossover occurred in the meiosis leading to the affected boy On the basis of this information, there is a (0.95 × 0.95) + (0.05 × 0.05) = 0.905 chance that the fetus (who is male) will be unaffected 521 (b) This surprising result (assuming paternity is as stated) indicates that the mother has inherited the A allele (and the WAS allele) from her mother and her phase is therefore A-WAS, not B-WAS Thus there must have been a crossover in the meiosis leading to the affected boy To confirm this, one should examine polymorphisms on either side of this one on the X chromosome to make sure that the segregation patterns are consistent with a crossover On the basis of this new information, there is now a 95% chance that the fetus in the current pregnancy is affected No, because you would not know if II-2 inherited the mutant allele D along with the A from her father or the A from her mother Phase becomes unknown again, as in Figure 10-10A Yes, phase is known in the mother of the two affected boys because she must have received the mutant factor VIII allele (h) and the M allele at the polymorphic locus on the X she received from her father In 10-7A, D = 0, so D′ = In 10-7B, D= −0.05, and because D < 0, F = smaller of freq(A)freq(S) versus freq (a)freq(s), so F = (0.5) (0.1) = 0.05 versus (0.5)(0.9) = 0.45, so F = 0.05 and D′ = −0.05/0.05 = −1, reflecting the complete linkage disequilibrium In 10-7C, D = −0.04, F = 0.05 again, and D′ = −0.8, reflecting a high degree but not perfect LD Odds ratio for the variant and disease = (a/b)/(c/d) = ad/bc Relative risk = [a/(a + b)]/[c/(c + d)] = a(c + d)/c(a + b) With three times as many controls, the odds ratio = a(3b)/c(3d) = 3ad/3bc = ad/bc, which is unchanged from the previous odds ratio Relative risk = [a/(a + 3b)]/[c/(c + 3d)] = a(c + 3d)/c(a + 3b) which is not the same as the previously calculated relative risk CHAPTER 11 The Molecular Basis of Genetic Disease The pedigree should contain the following information: Hydrops fetalis is due to a total absence of α chains The parents each must have the genotype αα/−− The α− genotype is common in some pop ulations, including Melanesians Parents with this genotype cannot transmit a −−/−− genotype to their offspring Except in isolated populations, patients with β-thalassemia will often be genetic compounds because there are usually many alleles present in a population in which β-thalassemia is common In isolated populations, the chance that a patient is a true homozygote of a single allele is greater than it 522 ANSWERS TO PROBLEMS would be in a population in which thalassemia is rare In the latter group, more “private mutations” might be expected (ones found solely or almost solely in a single pedigree) A patient is more likely to have identical alleles if he or she belongs to a geographical isolate with a high frequency of a single allele or a few alleles, or if his or her parents are consanguineous See text in Chapter Three bands on the RNA blot could indicate, among other possibilities, that (a) one allele is producing two mRNAs, one normal in size and the other abnormal, and the other allele is producing one mRNA of abnormal size; (b) both alleles are making a normalsized transcript and an abnormal transcript, but the aberrant ones are of different sizes; or (c) one allele is producing three mRNAs of different sizes, and the other allele is making no transcripts Scenario (c) is highly improbable, if possible at all Two mRNAs from a single allele could result from a splicing defect that allows the normal mRNA to be made, but at reduced efficiency, while leading to the synthesis of another transcript of abnormal size, which results from either the incorporation of intron sequences in the mRNA or the loss of exon sequences from the mRNA In this case, the other abnormal band comes from the other allele A larger band from the other allele could result from a splicing defect or an insertion, whereas a smaller band could be due to a splicing defect or a deletion Hb E is caused by an allele from which both a normal and a shortened transcript are made (see Fig 11-10); the normal mRNA makes up 40% of the total β-globin mRNA, producing only a mild anemia These two mutations affect different globin chains The expected offspring are normal, Hb M Saskatoon heterozygotes with methemoglobinemia, Hb M Boston heterozygotes with methemoglobin4 emia, and double heterozygotes with four hemoglobin types: normal, both types of Hb M, and a type with abnormalities in both chains In the double heterozygotes, the clinical consequences are unknown— probably more severe methemoglobinemia × × = 8, 1, 2, 7, 10, 4, 9, 5, 6, and Exceptions to this rule can arise, for example, from splice site mutations that lead to the mis-splicing of an exon The exon may be excluded from the mRNA, generating an in-frame deletion of the protein sequence or causing a change in the reading frame, leading to the inclusion of different amino acids in the protein sequence Approximately two thirds of the couples to whom such infants were born did not know about thalassemia or the prevention programs Approximately 20% refused abortion, and false paternity was identified in 13% of cases CHAPTER 12 The Molecular, Biochemical, and Cellular Basis of Genetic Disease Three types of mutations could explain a mutant protein that is 50 kD larger than the normal polypeptide: A mutation in the normal stop codon that allows translation to continue A splice mutation that results in the inclusion of intron sequences in the coding region The intron sequences would have to be free of stop codons for sufficient length to allow the extra 50 kD of translation An insertion, with an open reading frame, into the coding sequence For any of these, approximately 500 extra residues would be added to the protein if the average molecular weight of an amino acid is approximately 100 Five hundred amino acids would be encoded by 1500 nucleotides Autosomal dominant PCSK9 gain-of-function mutations that cause familial hypercholesterolemia are genocopies of autosomal dominant loss-of-function mutations in the LDL receptor gene (LDLR), because a genocopy is a genotype that determines a phenotype closely similar to that determined by a different genotype (for comparison, see Glossary for the definition of phenocopy) A nucleotide substitution that changes one amino acid residue to another should be termed a putative pathogenic mutation, and possibly a polymorphism, unless (1) it has been demonstrated, through a functional assay of the protein, that the change impairs the function to a degree consistent with the phenotype of the patient, or (2) instead of or in addition to a functional assay, it can be demonstrated that the nucleotide change is found only on mutant chromosomes, which can be identified by haplotype analysis in the population of patients and their parents and not on normal chromosomes in this population The fact that the nucleotide change is only rarely observed in the normal population and found with significantly higher frequency in a mutant population is strong supportive evidence but not proof that the substitution is a pathogenic mutation If Johnny has CF, the chances are approximately 0.85 × 0.85, or 70%, that he has a previously described mutation that could be readily identified by DNA analysis His parents are from northern Europe; therefore the probability that he is homozygous for the ΔF508 mutation is 0.7 × 0.7, or 50%, because approximately 70% of CF carriers in northern Europe have this mutation If he does not have the ΔF508 mutation, he could certainly still have CF, because approximately 30% of the alleles (in the northern European population, at least) are not ANSWERS TO PROBLEMS ΔF508 Steps to DNA diagnosis for CF include the following: (1) look directly for the ΔF508 mutation; if it is not present, (2) look for other mutations common in the specific population; (3) then look directly for other mutations based on probabilities suggested by the haplotype data; (4) if all efforts to identify a mutation fail (or if time does not allow), perform linkage analysis with polymorphic DNA markers closely linked to CF James may have a new mutation on the X chromosome because Joe inherited the same X chromosome from his mother, and the deletion was present in neither Joe nor his mother If this is the case, there is no risk for recurrence Alternatively, the mother may be a mosaic, and the mosaicism includes her germline In this case, there is a definite risk that the mutant X could be inherited by another son or passed to a carrier daughter Approximately 5% to 15% of cases of this type appear to be due to maternal germline mosaicism Thus the risk is half of this figure for her male offspring because the chance that a son will inherit the mutant X is × 5% to 15% = 2.5% to 7.5% For DMD, as a classic X-linked recessive disease that is lethal in males, one third of cases are predicted to be new mutations The large size of the gene is likely to account for the high mutation rate at this locus (i.e., it is a large target for mutation) The ethnic origin of the patient will have no effect on either of these phenomena A DMD female like T.N might have the disease because she carries a DMD gene mutation on the X chromosome inherited from her mother T.N could show clinical symptoms if her paternal X (carrying a normal allele at this locus) was subject to nonrandom inactivation in most or all cells An alternative explanation would be that she has Turner syndrome and that her only X chromosome (inherited from her mother) carries a DMD gene mutation A third explanation would be that she has a balanced X;autosome translocation that disrupts the DMD gene on the translocated X chromosome Although her normal X chromosome carries a normal allele at the DMD locus, balanced X;autosome translocations show nonrandom inactivation of the structurally normal X due to secondary cell selection (see Chapter 6) The limited number of amino acids that have been observed to substitute for glycine in collagen mutants reflects the nature of the genetic code Single- nucleotide substitutions at the three positions of the glycine codons allow only a limited number of missense mutations (see Table 3-1) Two bands of G6PD on electrophoresis of a red cell lysate indicate that the woman has a different G6PD allele on each X chromosome and that each allele is being expressed in her red cell population However, 523 no single cell expresses both alleles because of X inactivation Males have only a single X chromosome and thus express only one G6PD allele A female with two bands could have two normal alleles with different electrophoretic mobility, one normal allele and one mutant allele with different electrophoretic mobility, or two mutant alleles with different electrophoretic mobility Because the two common deficiency alleles (A− and B−) migrate to the same position as the common normal-activity alleles (A and B), the woman is unlikely to have a common deficiency allele at both loci Apart from that, one cannot say much about the possible pathological significance of the two bands without measuring the enzymatic activity If one of the alleles has low activity, she would be at risk for hemolysis to the extent that the high-activity allele is inactivated as a result of X inactivation 10 The box in Chapter 12 entitled “Mutant Enzymes and Disease: General Concepts” lists the possible causes of loss of multiple enzyme activities: they may share a cofactor whose synthesis or transport is defective; they may share a subunit encoded by the mutant gene; they may be processed by a common enzyme whose activity is critical to their becoming active; or they may normally be located in the same organelle, and a defect in biological processes of the organelle can affect all four enzymes For example, they may not be imported normally into the organelle and may be degraded in the cytoplasm Almost all enzymopathies are recessive (see text), and most genes are autosomal 11 Haploinsufficiency Thus, in some situations, the contributions of both alleles are required to provide a sufficient amount of protein to prevent disease An example of haploinsufficiency is provided by heterozygous carriers of LDL receptor deficiency 12 This situation is well illustrated by diseases due to mutations in mtDNA or in the nuclear genome that impair the function of the oxidative phosphorylation complex Nearly all cells have mitochondria, and therefore oxidative phosphorylation occurs in nearly all cells, yet the phenotypes associated with defects in oxidative phosphorylation damage only a subset of organs, particularly the neuromuscular system with its high energy requirements 13 One example is phenylketonuria, in which intellectual disability is the only significant pathological effect of deficiency of phenylalanine hydroxylase, which is found not in the brain but solely in the liver and kidneys, organs that are unaffected by this biochemical defect Hypercholesterolemia due to deficiency of the LDL receptor is another example Although the LDL receptor is found in many cell types, its hepatic deficiency is primarily responsible for the increase in LDL cholesterol levels in blood 524 ANSWERS TO PROBLEMS 14 There are two defining characteristics of these alleles: the hex A activity that they encode is sufficiently reduced to allow their detection in screening assays (when the other allele is a common Tay-Sachs mutation with virtually no activity), and their hex A activity is nevertheless adequate to prevent the accumulation of the natural substrate (GM2 ganglioside) There are probably only a few substitutions in the hex A protein that would reduce activity to only a modest degree (i.e., without crippling the protein more substantially) Thus the region of residues 247 to 249 appears to be relatively tolerant of substitutions, at least of Trp for Arg Substitutions that more dramatically alter the charge or bulk of the residues at these positions may well be diseasecausing alleles 15 A gain-of-function mutation leads to an abnormal increase in the activities performed by the wild-type protein Consequently, the overall integrity of the protein and each of its functional domains must remain intact despite the gain-of-function mutation In addition, of course, the mutation must confer the gain of function Consequently, the mutation must nothing to disturb the normal properties of the protein and must enhance at least one of them, if not more, to confer the gain of function Mutations other than missense mutations (e.g., deletions, insertions, premature stops) are almost uniformly highly disruptive to protein structure 16 The presence of three common alleles for Tay-Sachs disease in the Ashkenazi population seems likely to be due either to a heterozygote advantage or to genetic drift (one form of which is the founder effect, as explained in Chapter 9) The high frequency of these alleles might also be due to gene flow, although the population of origin of the three common mutations is not apparent, making this explanation seem less likely (in contrast, say, to the evidence indicating that the most common PKU alleles in many populations around the world are of Celtic origin) 17 As with any genetically complex disorder, the other sources of genetic variation in Alzheimer disease (AD) may include the following: (1) additional AD loci, with lower effect sizes, that have not yet been identified; (2) synergistic effects between known AD genes (or between known genes and environmental risks) that may have a bigger effect than each of the genes or environments individually; (3) genes that harbor multiple very rare mutations of large effect, but which are undetectable by genome-wide association studies methods because each mutation occurs on a different SNP background 18 The two forms of myotonic dystrophy are characterized by an expansion of a CUG trinucleotide in the RNA, which is thought to lead to an RNAmediated pathogenesis According to this model, the greatly enhanced number of CUG repeats binds an abnormally large fraction of RNA-binding proteins, including, for example, regulators of splicing, thereby depleting the cell of these critical proteins One approach to therapy might be to prevent this binding This might be achieved by introducing, perhaps by gene transfer (see Chapter 13), a viral vector that expresses a GAC trinucleotide repeat, which would bind to the CUG repeat sequences in the RNA and block the binding of the RNA-binding proteins to the expanded CUG repeats Expression of too large a number of GAC repeat–containing molecules might itself have undesirable side effects, however, including binding to CUG codons that encode leucine, blocking their translation CHAPTER 13 The Treatment of Genetic Disease Unresponsive patients may have mutations that drastically impair the synthesis of a functional gene product Responsive patients may have mutations in the regulatory region of the gene The effects of these mutations may be counteracted by the administration of IFN-γ These mutations could be in the DNAbinding site that responds to the interferon stimulus or in some other regulatory element that participates in the response to IFN-γ Alternatively, responsive patients may produce a defective cytochrome b polypeptide that retains a small degree of residual function The production of more of this mutant protein, in response to IFN-γ, increases the oxidase activity slightly but significantly An enzyme that is normally intracellular can function extracellularly if the substrate is in equilibrium between the intracellular and extracellular fluids and if the product is either nonessential inside the cell or in a similar equilibrium state Thus enzymes with substrates and products that not fit these criteria would not be suitable for this strategy This approach may not work for phenylalanine hydroxylase because of its need for tetrahydrobiopterin However, if tetrahydrobiopterin could diffuse freely across the polyethylene glycol layer around the enzyme, the administration of tetrahydrobiopterin orally may suffice This strategy would not work for storage diseases because the substrate of the enzyme is trapped inside the lysosome In Lesch-Nyhan syndrome, the most important pathological process is in the brain, and the enzyme in the extracellular fluid would not be able to cross the blood-brain barrier Tay-Sachs disease could not be treated in this way because of the nondiffusibility of the substrate from the lysosome Rhonda’s mutations prevent the production of any LDL receptor Thus the combination of a bile acid– binding resin and a drug (e.g., lovastatin) to inhibit cholesterol synthesis would have no effect on increasing the synthesis of LDL receptors The boy must have one or two mutant alleles that produce a receptor with some residual function, and the increased expression of these mutant receptors on the surface of the hepatocyte reduces the plasma LDL-bound cholesterol Unresponsive patients probably have alleles that not make any protein, that decrease its cellular abundance in some other way (e.g., make an unstable protein), or that disrupt the conformation of the protein so extensively that its pyridoxal-phosphate binding site has no affinity for the cofactor, even at high concentrations The answer to the second part of this question is less straightforward The answer given here is based on the generalization that most patients with a rare autosomal recessive disease are likely to have two different alleles, which assumes that there is no mutational hot spot in the gene and that the patients are not descended from a “founder” and are not members of an ethnic group in which the disease has a high frequency In this context, Tom is likely to have two alleles that are responsive; first cousins with the same recessive disease are likely to share only one allele, so that Allan is likely to have one responsive allele that he shares with Tom and another allele that is either unresponsive or that responds more poorly to the cofactor than Tom’s other allele (a) You need both a promoter that will allow the synthesis of sufficient levels of the mRNA in the target tissue of choice and the phenylalanine hydroxylase cDNA In reality, you also need a vector to deliver the “gene” into the cell, but this aspect of the problem has not been dealt with much in the text (b) A phenylalanine hydroxylase “gene” will probably be effective in any tissue that has a good blood supply for the delivery of phenylalanine and an adequate source of the cofactor of the enzyme, tetrahydrobiopterin The promoter would have to be capable of driving transcription in the target tissue chosen for the treatment (c) Any mutation that severely reduces the abundance of the protein in the cell but has no effect on transcription This group includes those mutations that impair translation or that render the protein highly unstable The thalassemias include examples of all these types (d) Liver cells are capable of making tetrahydrobiopterin, whereas other cells may not be The target cell for the gene transfer should thus be capable of making this cofactor; otherwise, the enzyme will not function unless the cofactor is administered in large amounts (e) Human phenylalanine hydroxylase probably exists as a homodimer or homotrimer In patients whose alleles produce a mutant polypeptide ANSWERS TO PROBLEMS 525 (versus none at all), these alleles may manifest a dominant negative effect on the product of the transferred gene This effect could be overcome by making a gene construct that produces more of the normal phenylalanine hydroxylase protein (thus diluting out the effect of the mutant polypeptide) or by transferring the gene into a cell type that does not normally express phenylalanine hydroxylase and that would therefore not be subject to the dominant negative effect One must consider the kinds of mutations that decrease the abundance of a protein but that are associated with residual function One class of such mutations are those that decrease the abundance of the mRNA but not alter the protein sequence (i.e., each protein molecule produced has normal activity, but there are fewer molecules) Mutations of this type might include enhancer or promoter mutations, splice mutations, or others that destabilize the mRNA In this case, one could consider strategies to increase expression from the normal allele and perhaps also the mutant allele, as is done with hereditary angioedema, in which danazol administration increases the expression of the product from both the wild-type and mutant alleles A second class of such mutations are those within the coding sequence that destabilize the protein but still allow some residual function Here, a strategy to increase the stability or the function of the mutant protein should be considered For example, if the affected protein has a cofactor, one could administer increased amounts of the cofactor, provided such an approach would not have unacceptable side effects The drug can facilitate the skipping of a premature stop codon, allowing the translational apparatus to misincorporate an amino acid that has a codon comparable to that of the mutant termination codon This treatment might allow the synthesis of a protein of normal size in both patients In the responsive patient, the nonsense codon is located in a functionally “neutral” part of the protein, and the amino acid that is substituted in place of the nonsense codon allowed normal folding, processing, and function of the “corrected” protein In the nonresponsive patient, however, the nonsense mutation is located in codon 117, which in wild-type CFTR is an Arg residue (see Fig 12-15) This Arg residue contributes to the Cl− channel of CFTR In this unresponsive patient, the drug did not lead to incorporation of Arg at this position, and the Cl− channel had defective conduction as a result CHAPTER 14 Developmental Genetics and Birth Defects Before determination, an embryo can lose one or more cells, and the remaining cells can undergo 526 ANSWERS TO PROBLEMS specification and ultimately develop into a complete embryo Once cells are determined, however, mosaic development takes place—an embryonic tissue will follow its developmental program regardless of what happens elsewhere in the embryo Regulative development means that an embryonic cell can be removed by blastomere biopsy for the purpose of preimplantation diagnosis without harming the rest of the embryo a–3, b–2, c–4, d–1 a–4, b–3, c–5, d–2, e–1 Mature T or B cells that have somatically rearranged their T-cell receptor or immunoglobulin loci would not be appropriate This change is not epigenetic; it is a permanent alteration of the DNA sequence itself Animals derived from a single nucleus from a mature T or B cell are incapable of mounting an appropriately broad immune response Consider issues of regulation versus simple capacity to carry out a biochemical reaction Also, consider dominant negative effects of transcription factors, taking into account the frequent binary nature of such factors (DNA-binding and activation domains) CHAPTER 15 Cancer Genetics and Genomics Approximately 15% of unilateral retinoblastoma is actually the heritable form but affecting only one eye You need family history; careful examination of both parents’ retinas, looking for signs of a scar that could have been a spontaneously regressed retinoblastoma; cytogenetic analysis if the tumor is associated with other malformations and, very importantly, seek to find a mutation in the child in DNA from peripheral blood to see if it is a germline mutation If the child carries a germline mutation, it is a heritable retinoblastoma, and the child is at risk for tumor in the other eye, in the pineal gland, and for sarcomas later in life, particularly associated with radiation therapy With the mutation in hand, the parents can be tested to see if one or the other is a nonpenetrant carrier, and prenatal diagnosis could be offered for future pregnancies Even if no mutation is found in a parent, given that a parent could be a germline mosaic with some increase in recurrence risk over simply the rate of new mutation, prenatal diagnosis can be offered using the mutation found in the affected child If prenatal diagnosis is not used or if the fetus carries a mutation and the parents choose to allow the pregnancy to go to term, the newborn would need examination under anesthesia as soon as possible after birth and then at frequent intervals after that, with institution of therapy as soon as a tumor is found If the child is not clearly heterozygous for a pathogenic mutation in his or her blood, it is still possible there is somatic mosaicism and the child is still at increased risk for tumor in the other eye or, more generally, for sarcomas later in life Sequencing of the tumor itself may show a mutation that could have been easily missed but could be specifically looked for at low levels using next-generation sequencing of the peripheral blood DNA Colorectal cancer seems to require a number of sequential mutations in several genes, a process that may take longer than one (in hereditary) or two (in sporadic) mutations in the retinoblastoma gene Age dependence may also reflect the number, timing, and rate of cell divisions in colon cells and in retinoblasts A cell line with i(17q) is monosomic for 17p and trisomic for 17q Thus formation of the isochromosome leads to loss of heterozygosity for genes on 17p This may be particularly important because one or more tumor suppressor genes (such as TP53) are present on 17p; a “second hit” on the other copy of TP53 would lead to complete loss of the p53 protein function In addition, a number of proto-oncogenes map to 17q It is possible that increasing their dosage may confer a growth advantage on cells containing the i(17q) The chief concern is the need to reduce radiation exposure to the lowest possible level because of the risk for cancer in children with this genetic defect Although most (>95%) breast cancer appears to follow multifactorial inheritance, there are two known genes (BRCA1 and BRCA2) in which mutations confer a substantial increased lifetime risk for breast cancer (fivefold to sevenfold) inherited in an autosomal dominant manner Certain mutations in certain other genes, such as ATM, BARD1, BRIP1, CDH1, CHEK2, PALB2, PTEN, and TP53, among others, increase the lifetime breast cancer risk significantly over the 12% background risk in the population as well, but generally not to the extent seen with mutations in BRCA1 or BRCA2 In the absence of a gene mutation in a hereditary breast cancer gene, the empirical risk figures are consistent with an overall multifactorial model with admixture of dominant forms of the disease with somewhat reduced lifetime penetrance; thus there is an approximately twofold increased risk for breast cancer in any woman with a first-degree female relative with breast cancer Direct mutation detection could be performed if desired by the probands in Wanda’s and Wilma’s families, and if a mutation were found in BRCA1, BRCA2, or one of the other genes that cause substantial increased risk for breast cancer, a direct test for cancer risk could be offered to their relatives More recently, one leading breast cancer researcher suggested that population-wide screening for diseasecausing mutations in BRCA1 or BRCA2 should be initiated independent of family history, either restricted to high-risk ethnic groups or, more widely, to the entire population ANSWERS TO PROBLEMS It is likely that many activated oncogenes, if inherited in the germline, would disrupt normal development and be incompatible with survival There are a few exceptions, such as activating RET mutations in MEN2 and activating MET mutations in hereditary papillary kidney cancer These activated oncogenes must have tissue-specific oncogenic effects without affecting development Although it is not known why such specific types of cancers occur in individuals who inherit germline mutations in these oncogenes, one plausible theory is that other genes expressed in most of the tissues of the body counteract the effect of these activating mutations, thereby allowing normal development and suppressing oncogenic effects in most of the tissues in heterozygotes Write down an abbreviated pedigree (see illustration), and calculate all the possible prior probabilities There are four scenarios: In A, Nathan is a new mutation with probability µ In B, Molly is the new mutation—but because Lucy is not a carrier, Molly can only carry a new mutation and did not inherit the mutation; her prior probability is 2µ (not 4µ) because the new mutation could have occurred on either her paternal or her maternal X chromosome In C, Lucy is a carrier As shown earlier in this chapter in the Box describing the calculation for the probability that any female is a carrier of an X-linked lethal disorder, Lucy’s prior probability is 4µ Molly inherits the mutant gene, but Martha does not, so the probability her two sons would be unaffected is essentially In D, Lucy is a carrier, as is Molly, but so is Martha, and yet she did not pass the mutant gene to her two sons (We not consider all the other combinations of carrier states; because they are so unlikely, they can be ignored For example, the possibility that Lucy is a mutation carrier, but Molly does not inherit a mutation from Lucy, and then Nathan is another new mutation is vanishingly small because the joint probability of such an event would require two new mutations and would contain µ2 terms in the joint probability that are too small to contribute to the posterior probability.) The conditional probabilities can then be calculated from these various joint probabilities For Molly, she is a carrier in situations B, C, and D, so her probability of being a carrier is 13 21 CHAPTER 16 Risk Assessment and Genetic Counseling (a) Prior risk, ; posterior risk (two normal brothers), 110 (b) Zero, unless the autosomal dominant form can show nonpenetrance, in which case there is a very small probability that Rosemary, Dorothy, and Elsie would all be nonpenetrant carriers Without knowing the penetrance, we cannot calculate the exact risk that Elsie is heterozygous (a) Restrict your attention and conditional probability calculations to those women for whom we have conditional probability information that could alter their carrier risk These individuals are the maternal grandmother (Lucy, see pedigree), who has an affected grandson and two unaffected grandsons, her daughter Molly, who has an affected son, and Martha, who has two unaffected sons Maud does not contribute any additional information because she has no sons A B C D 4µ Lucy Lucy µ Nathan Prior Nathan new mutation = µ Joint = µ 1/2 Molly Martha 4µ Lucy 2µ Molly Molly Martha 1/2 Lucy 1/2 Martha 1/2 Nathan 527 Nathan 1/2 Molly 1/2 1/2 Martha 1/2 1/2 Nathan Prior Molly new mutation = 2µ Prior Lucy a carrier = 4µ Prior Lucy a carrier = 4µ Conditional Nathan inherits = 1/2 Conditional Molly inherits = 1/2 Nathan inherits = 1/2 Martha does not inherit = 1/2 Conditional Molly inherits = 1/2 Nathan inherits = 1/2 Martha inherits = 1/2 Two boys unaffected = 1/2 x 1/2 Joint = 2µ/2 = µ Joint = 4µ/8 = µ/2 Joint = 4µ/32 = µ/8 Figure for Chapter 16, question 528 ANSWERS TO PROBLEMS Similarly, Molly’s mother, Lucy, 5/21; Norma and Nancy, 13/42; Olive and Odette, 13/84; Martha, 1/21; Nora and Nellie, 1/42; Maud, 5/42; Naomi, 5/84 (b) To have a 2% risk for having an affected son, a woman must have an 8% chance of being a carrier; thus Martha, Nora, and Nellie would not be obvious candidates for prenatal diagnosis by DNA analysis because their carrier risk is less than 8% ( )13 for 13 successive male births ( )13 × for 13 consecutive births of the same sex (The arises because this is the chance of 13 consecutive male births or 13 consecutive female births, before any children are born.) 1 The probability of a boy is for each pregnancy, regardless of how many previous boys were born (assuming there is straightforward chromosome segregation, no abnormality in sexual development that would alter the underlying 50% to 50% segregation of the X and Y chromosomes during spermatogenesis, and no sex-specific lethal gene carried by a parent) (a) Use the first equation, I = µ + H To solve for H and substitute it for H in the second equation, H = 2µ + H + If Solve for I, I = 3µ/(1 − f) Substituting 0.7 for f gives: The incidence of affected males I = 10µ The incidence of carrier females H = 18µ Chance next son will be affected is × 0.9 = 0.45 (b) Substitute f = into the equations and you get I = 3µ and H = 4µ (c) 0.147 Prior 18µ Conditional 1/2 Joint Posterior 1–18µ = ~1 µ 9µ 0.9 µ 0.1 Figure for Chapter 16, question (a) The prior risk that either Ira or Margie is a cystic fibrosis carrier is ; therefore the probability that both are carriers is × = (b) Their risk for having an affected child in any pregnancy is × = (c) Bayesian analysis is carried out Thus the chance that Ira’s and Margie’s next child will be affected is × = 116 Not Both Carriers Both Carriers Prior Conditional (3 normal children) Joint Posterior ( 4) × ( 4)3 = 0.19 0.19/(0.19 + 0.56) = 9 = 0.56 0.56/0.75 = ≈ The child’s prior probability of carrying a mutant myotonic dystrophy gene is If it is assumed that he has a chance of being asymptomatic, even if he carries the mutant gene, then his chance of carrying it and showing no symptoms is Testing is a complex issue Many would think that testing an asymptomatic child for an incurable illness with adult onset is improper because the child should be allowed to make that decision for himself or herself (see Chapter 19) (a) Yes; autosomal recessive, autosomal dominant (new mutation), X-linked recessive, and multifactorial inheritance and a chromosome disorder all need to be considered, as would nongenetic factors such as prenatal teratogen exposure and intrauterine infection A careful physical examination and laboratory testing are required for a proper assessment of risks for this couple (b) This increases suspicion that the disorder is autosomal recessive, but the possibility of consanguinity does not prove autosomal recessive inheritance, and all other causes must still be investigated thoroughly (c) This fact certainly supports the likelihood that the problem has a genetic explanation The pedigree pattern would be consistent with autosomal recessive inheritance only if the sister’s husband were carrying the same defect (possible if he is from the same village, for example) An X-linked recessive pattern (particularly if the affected children are all boys) or a chromosome defect (e.g., the mothers of the affected children having balanced translocations with unbalanced karyotypes in the affected children) ought to be considered The mother and her son should receive a genetic evaluation appropriate to the clinical findings, such as karyotyping and fragile X analysis The woman needs genetic counseling She has a 50% risk for passing the mutant NF1 gene to her offspring The fact that she carries a new mutation only reduces the recurrence risk elsewhere in the family The seven scenarios are shown in the table ANSWERS TO PROBLEMS Conditional Probability Calculation Female Carrier Status Joint Probabilities* Situation I-1 II-1 II-3 III-2 A B1 No No No Yes (new mutation) No No No No µ B2 No Yes (new mutation) No Yes {2à ì } × [1] × [ × ( )2] = µ/8 Yes No C1 Yes C2 Yes Yes Yes No No {2à ì } × [1] × [ ] = µ/2 {4µ × × } × [ 12 ] ì [ 12 ] = à/4 {4à ì × } × [ × ( )2] × [ ] = µ/16 C3 Yes Yes No Yes {4µ × × } × [ 12 ] × [ 12 ì ( )2] = à/16 C4 Yes Yes Yes Yes {4à ì ì } × [ × ( )2] × [ × ( )2] = µ/64 *The joint probabilities for the core individuals in the pedigree (I-1, II-1, and III-1) are enclosed in curly brackets { }, and the probabilities for individuals II-3 and III-2 are shown in square brackets [ ] See Figure 19-7 The scenarios in which III-2 is a carrier are B2, C3, and C4 Her posterior probability of being a carrier is therefore µ + µ 16 + µ 64 µ + µ + µ + µ + µ 16 + µ 16 + µ 64 10 Make II-1 the dummy consultand Proceeding as if III-2 and her two unaffected children were not present, the risk that II-1 is a carrier is covered in situations B, C1, and C2 in the accompanying table, giving a posterior probability of µ+µ 2+µ = 13 21 µ+µ+µ 2+µ = × 13 21 = 13 42 ; the prior probability that she is not a carrier is − (13 42 ) = 29 42 (see Table for step two) Then we use another round of conditional probability to see what effect the two unaffected sons of III-2 have, to determine the posterior risk that III-2 is a carrier Step Two of Dummy Consultand Method III-2 Carrier Prior probability Conditional (2 unaffected sons) Joint probability Posterior probability 13 29 42 ( 2) 13 13 III-2 Not a Carrier 168 129 42 29 13 42 116 Thus the posterior probability that III-2 is a carrier using the dummy consultand method, given she has two unaffected sons, is 13 129 , the same as when we used the approach in Table 16-3 So far, so good Some consider the dummy consultand method to be faster than the comprehensive approach of drawing out all the scenarios, but it is also easy to misapply, resulting in miscalculation For example, the dummy consultand method, as outlined here, gives the correct result only for the consultand III-2 herself and not necessarily for other females in the pedigree For example, the 13 21 (62%) carrier risk for individual II-1, calculated in the first step of the two-step dummy consultand method, which ignores the information for individual III-2, is actually incorrect The correct result for II-1 is the posterior probability of all the situations except A in the conditional probability calculation table, which equals 65 129 (50%) (We thank Susan Hodge from Columbia University for pointing out this problem with the dummy consultand method.) CHAPTER 17 Prenatal Diagnosis and Screening Step One of Dummy Consultand Method Female Carrier Status Situation I-1 II-1 II-3 Joint Probabilities A B No No No No {1 × × µ} = µ C1 Yes No Yes (new mutation) Yes No {4à ì ì = µ/2 C2 Yes Yes Yes {4µ × × } × [ × ( )2] = à/8 {1 ì 2à ì 529 }=à } ì [ 12 ] One can then use this calculation as a starting point to determine that the prior probability that III-2 is a carrier, ignoring her two unaffected sons, is the probability that her mother, II-1, is a carrier c, e, f, i and j, d, h, g, b, i (and, in part, j), and a The child can have only Down syndrome or monosomy 21, which is almost always lethal Thus they should receive counseling and consider other alternatives for having children No, not necessarily; the problem could be maternal cell contamination The level of maternal serum alpha-fetoprotein (MSAFP) is typically elevated when the fetus has an open neural tube defect The levels of MSAFP and unconjugated estriol are usually reduced and the human chorionic gonadotropin level is usually elevated when the fetus has Down syndrome (a) Approximately 15% (see Table 5-2) (b) At least 50% are chromosomally abnormal 530 ANSWERS TO PROBLEMS (c) Prenatal diagnosis or karyotyping of the parents is usually not indicated after a single miscarriage; most practitioners would offer parental chromosome analysis and prenatal diagnosis after three spontaneous unexplained miscarriages (although some practitioners suggest offering testing after only two), provided there are no other indications (a) Yes Given that her CPK levels indicate she is a carrier of DMD and she had an affected brother, she must have inherited the mutation from her mother because her brother could not have received the mutation from her father The phase can be determined from analysis of her father, who has to have transmitted a normal X chromosome to his daughter, the consultand (b) Yes A male fetus that receives her father’s allele linked to the DMD locus will be unaffected If a male fetus receives her mother’s allele linked to DMD, it will be affected This, of course, assumes no recombination between the microsatellite marker and the mutation in the DMD gene in the transmitted chromosome (c) First, the consultand should have DMD gene testing The most common mutations in DMD are deletions (and less commonly duplications), although point mutations are also possible (see Chapter 12) The advent of powerful new sequencing technologies and new methods for determining deletions and duplications, such as the multiplex ligation-dependent polymerase assay (MLPA) or copy number measurements by comparative genome hybridization, has made carrier detection for DMD much more sensitive than in the past, when it was limited by the very large size of the gene and difficulties determining a partial gene deletion in a female with two copies of the gene Question for discussion Consider issues of sensitivity and specificity of each of the different forms of testing, the psychosocial issues of prenatal diagnosis and termination at different stages of pregnancy, and risk for complications of the two invasive methods 600,000 women, 1000 pregnancies affected Assume everyone is willing to participate in the sequential screening Of 1000 true positives, firsttrimester screening will identify 840 high-risk “positives” (84%) who undergo CVS; 160 are low risk, and they get a second-trimester screen Of these 160, 130 (81%) are positive and undergo amniocentesis and are found to have an affected fetus; 30 affected pregnancies are missed Of the 599,000 unaffected false positives in the first-trimester screen, 29,950 positives need CVS The remaining 569,050 are low risk and get a secondtrimester screen You get 28,452 positives in the second-trimester screen who undergo amniocentesis; the remaining 540,598 unaffected pregnancies are reassured In summary, with sequential screening, you will detect 970 of the 1000 (97%) and will miss 30 (3%) You will 970 invasive tests in affected pregnancies while also doing 29,950 + 28,452 = 58,402 invasive tests in unaffected pregnancies Thus you will 62 invasive tests to detect each affected pregnancy This compares to the situation if you just offered invasive testing to everyone Depending on the uptake, you will miss some fraction of affected fetuses If the uptake were 97% (very, very unlikely for an invasive test), you would end up doing 582,000 invasive tests to find 970 affected pregnancies You would miss the same 30 affected pregnancies as with the sequential testing but would have to a 10 times greater number of invasive tests to achieve the same detection rate CHAPTER 18 Application of Genomics to Medicine and Personalized Health Care Idiopathic Cerebral Vein Thrombosis and Factor V Leiden iCVT Genotype Homozygous FVL Heterozygous FVL Wild-type Total Affected Unaffected Total 15 18 624 48,748 950,610 999,982 625 48,750 950,625 1,000,000 FVL, Factor V Leiden; iCVT, idiopathic cerebral vein thrombosis You would expect 625 FVL homozygotes and 48,750 heterozygotes Relative risk for iCVT in FVL homozygotes = (1/625)/ (15/950,625) = ≈101 Relative risk for iCVT in FVL heterozygotes = (2/48,750)/(15/950,625) = ≈3 Sensitivity of testing positive for either one or two FVL alleles = 3/18 = 17% Positive predictive value for homozygotes = 1/625 = 0.16% Positive predictive value for heterozygotes = 2/48,748 = 0.004% Although the relative risks are elevated with FVL, particularly when the individual is homozygous for the allele, the disorder itself is very rare and thus the PPV is low This example highlights the concept that a relative risk is always a comparison to people who not carry a particular marker whereas a PPV is the actual (or absolute) risk for someone who carries the marker ANSWERS TO PROBLEMS Distinguish between SS homozygotes and AS heterozygotes What harm might accrue from the identification of AS individuals by newborn screening? What does identification of a newborn with SS or AS tell you about the genotypes of the parents and genetic risks for future offspring to the parents? Deep Venous Thrombosis in the Legs, Oral Contraceptive Use, and Factor V Leiden DVT Genotype Homozygous FVL Heterozygous FVL Wild-type Total Affected Unaffected Total 58 39 100 59 4,825 95,025 99,000 62 4,875 95,063 100,000 DVT, Deep venous thrombosis; FVL, factor V Leiden You would expect ≈62 FVL homozygotes and 4875 heterozygotes Relative risk for DVT in FVL homozygotes taking oral contraceptives (OCs) = ≈118 Relative risk for DVT in FVL heterozygotes taking OCs = ≈30 Sensitivity of testing positive for either one or two FVL alleles = 62% Positive predictive value for homozygotes = 3/62 = ≈5% Positive predictive value for heterozygotes = 58/4875 = 1.2% Note that DVT is more common than the example of idiopathic cerebral vein thrombosis given in question 1, whereas the relative risks for homozygotes are of similar magnitude (101 vs 118); thus the PPV of testing homozygotes is accordingly much higher but is still only 5% You should first explain to the parents that the test is a routine one performed on all newborns and that the results, as in many screening tests, are often falsely positive The parents should also be told that the test result may be a true positive, and if it is, a more accurate and definitive test needs to be done before we will know what the child’s condition really is and what treatment will be required The child should be brought in as soon as possible for examination and the appropriate samples obtained to confirm the elevated phenylalanine level to determine if the child has classic or variant PKU or hyperphenylalaninemia, and to test for abnormalities in tetrabiopterin metabolism Once a diagnosis is made, dietary phenylalanine restriction is instituted to bring blood phenylalanine levels down below the range considered toxic (>300 µmol/L) The child must then be observed for dietary adjustments to be made to keep the blood phenylalanine levels under control Questions to consider in formulating your response are as follows: Consider the benefits of preventing disease by knowing a newborn’s genotype at the β-globin locus Can knowing the genotype help prevent pneumococcal sepsis or other complications of sickle cell anemia? 531 Carbamazepine-Induced TEN or SJS HLA-B*1502 allele present + − Total TEN or SJS Affected Unaffected Total 44 44 98 101 47 98 145 SJS, Stevens-Johnson syndrome; TEN, toxic epidermal necrolysis Sensitivity = 44/44 = 100% Specificity = 98/101 = 97% Positive predictive value = 44/47 = 94% Terfenadine blocks the HERG cardiac-specific potassium channel encoded by KCNH2 Various alleles in the coding portion of KCNH2 are associated with prolongation of the QT interval on electrocardiography, which is associated with sudden death Terfenadine is metabolized by the cytochrome P450 enzyme CYP3A4, which has numerous alleles associated with reduced metabolism Itraconazole is an antifungal that blocks CYP3A4 cytochrome and increases serum levels of drugs metabolized by this cytochrome Grapefruit juice contains a series of naturally occurring compounds, furanocoumarins, that are thought to interfere with CYP3A4 metabolism of numerous drugs, including terfenadine Caffeine is unlikely to be involved in that caffeine has very little effect on CYP3A4, which has only a minor role in caffeine metabolism Most caffeine is metabolized by CYP1A2 CHAPTER 19 Ethical and Social Issues in Genetics and Genomics The first consideration is testing the boy for an incurable disease Because the boy has symptoms and the family is seeking a diagnosis, this is not the same situation as if an asymptomatic child is being considered for predictive testing for an adult-onset disorder, such as classic myotonic dystrophy However, because Huntington disease in a child is overwhelmingly the result of an expansion of an enlarged triplet repeat in one of the parents, usually the father, finding a markedly enlarged expansion in the child will automatically raise the possibility that one of the parents, probably the father, is a carrier of a repeat that is 532 ANSWERS TO PROBLEMS enlarged enough to cause adult-onset Huntington disease in him Thus, by testing the child, one might inadvertently discover something about a parent’s risk Testing should therefore be carried out with informed consent from the parents Other issues: If one of the parents carries the HD gene, what you about testing the asymptomatic older sib? To justify screening, one must show that the good that comes from screening, the beneficence of the testing, outweighs the harm Consider the issue of autonomy because implicit in the act of informing families that their child has a chromosomal abnormality is the fact that the child cannot decide whether she or he wants such testing later in life How predictive is the test? Are we making a diagnosis of a possible chronic disability that may or may not develop or, if it does, may vary in severity and for which there is little the parents can do? One might ask whether there are effective interventions for the abnormalities in learning and behavior that occur in some individuals with sex chromosome anomalies In fact, there is evidence that informing the parents and providing educational and psychological intervention before major problems arise proves beneficial There is also, however, the concern about “the self-fulfilling prophecy,” that telling parents there might be a problem increases the risk that there will be a problem by altering parental attitudes toward the child There is a large amount of literature on this subject that is worth investigating and reading See, for instance: Bender BG, Harmon RJ, Linden MG, Robinson A: Psychosocial adaptation of 39 adolescents with sex chromosome abnormalities Pediatrics 96(pt 1): 302-308, 1995 Puck MH: Some considerations bearing on the doctrine of self-fulfilling prophecy in sex chromosome aneuploidy Am J Med Genet 9:129-137, 1981 Robinson A, Bender BG, Borelli JB, et al: Sex chromosomal aneuploidy: prospective and longitudinal studies Birth Defects Orig Artic Ser 22:23-71, 1986 You must consider the extent to which withholding information constitutes “a serious threat to another person’s health or safety.” In these different disorders, consider how serious the threat is and whether there is any effective intervention if the relative were informed of his or her risk Give your rationale for picking the disorders you choose Consider such factors as how great a threat to is health the disorder, whether the disorder is likely to remain undiscovered and a potential cause of serious illness if not found before symptoms develop by sequencing, how predictive is finding a gene mutation for the disease, and how effective, how invasive, and how risky are any interventions An initial (and somewhat controversial) list of 56 such disorders as proposed by a committee of the American College of Medical Genetics and Genomics can be found in: Green RC, Berg JS, Grody WW, et al: ACMG recommendations for reporting of incidental findings in clinical exome and genome sequencing Genet Med 15:565-574, 2013 A framework for considering potentially pathogenic sequence variants detected by whole-exome or whole-genome sequencing can be found in: Richards S, Aziz N, Bale S, et al: Standards and guidelines for the interpretation of sequence variants: a joint consensus recommendation of the American College of Medical Genetics and Genomics and the Association for Molecular Pathology Genet Med doi:10.1038/gim.2015.30, 2015 ... monoamine neurotransmitters dopamine, norepinephrine, epinephrine, and serotonin (see Fig 12- 2) 22 0 THOMPSON & THOMPSON GENETICS IN MEDICINE The locus heterogeneity of hyperphenylalaninemia... conditions, all 22 4 THOMPSON & THOMPSON GENETICS IN MEDICINE Cystathionine synthase Methionine Homocysteine Cystathionine Methionine synthase Vitamin B6 Methyl-B 12 H4-folate Cysteine Pyridoxal... deleting the Phe codon CF, Cystic fibrosis; MSD, membrane-spanning domain; NBD, nucleotide-binding domain; R-domain, regulatory domain See Sources & Acknowledgments 23 2 THOMPSON & THOMPSON GENETICS