﻿ Answers to Problems - Thompson & Thompson Genetics in Medicine, 8th Edition

Thompson & Thompson Genetics in Medicine, 8th Edition

Answers to Problems

Chapter 2 Introduction to the Human Genome

1. (a) A or a.

(b) i. At meiosis I. ii. At meiosis II.

2. Meiotic nondisjunction.

3. (image)23 × (image)23; you would be female.

4. (a) 23; 46.

(b) 23; 23.

(c) At fertilization; at S phase of the next cell cycle.

5. Chromosome 1, ≈9 genes/Mb; chromosome 13, ≈3-4 genes/Mb; chromosome 18, ≈4 genes/Mb; chromosome 19, ≈19 genes/Mb; chromosome 21, ≈5 genes/Mb; chromosome 22, ≈10 genes/Mb. Because of the higher density of genes, one would expect that a chromosome abnormality of chromosome 19 would have a greater impact on phenotype than an abnormality of chromosome 18. Similarly, chromosome 22 defects are expected to be more deleterious than those of chromosome 21.

Chapter 3 The Human Genome: Gene Structure and Function

1. There are several possible sequences because of the degeneracy of the genetic code. One possible sequence of the double-stranded DNA is


RNA polymerase “reads” the bottom (3′ to 5′) strand. The sequence of the resulting mRNA would be 5′ AAA AGA CAU CAU UAU CUA 3′.

The mutants represent the following kinds of mutations:

Mutant 1: single-nucleotide substitution in fifth codon; for example, UAU → UGU.

Mutant 2: frameshift mutation, deletion in first nucleotide of third codon.

Mutant 3: frameshift mutation, insertion of G between first and second codons.

Mutant 4: in-frame deletion of three codons (nine nucleotides), beginning at the third base.

2. The sequence of the haploid human genome consists of nearly 3 billion nucleotides, organized into 24 types of human chromosome. Chromosomes contain chromatin, consisting of nucleosomes. Chromosomes contain G bands that contain several thousand kilobase pairs of DNA (or several million base pairs) and hundreds of genes, each containing (usually) both introns and exons. The exons are a series of codons, each of which is three base pairs in length. Each gene contains a promoter at its 5′ end that directs transcription of the gene under appropriate conditions.

3. A mutation in a promoter could interfere with or eliminate transcription of the gene. Mutation of the initiator codon would prevent normal translation. Mutations at splice sites can interfere with the normal process of RNA splicing, leading to abnormal mRNAs. A 1-bp deletion in the coding sequence would lead to a frameshift mutation, thus changing the frame in which the genetic code is read; this would alter the encoded amino acids and change the sequence of the protein. (See examples in Chapter 11.) A mutation in the stop codon would allow translation to continue beyond its normal stopping point, thus adding new, incorrect amino acids to the end of the encoded protein.

4. Mutations in introns can influence RNA splicing, thus leading to an abnormally spliced mRNA (see Chapter 11). Alu or L1 sequences can be involved in abnormal recombination events between different copies of the repeat, thus deleting or rearranging genes. L1 repeats can also actively transpose around the genome, potentially inserting into a functional gene and disrupting its normal function. Locus control regions influence the proper expression of genes in time and space; deletion of a locus control region can thus disrupt normal expression of a gene (see Chapter 11). Pseudogenes are, generally, nonfunctional copies of genes; thus, in most instances, mutations in a pseudogene would not be expected to contribute to disease, although there are rare exceptions.

5. RNA splicing generates a mature RNA from the primary RNA transcript by combining segments of exonic RNA and eliminating RNA from introns. RNA splicing is a critical step in normal gene expression in all tissues of the body and operates at the level of RNA. Thus the genomic DNA is unchanged. In contrast, in somatic rearrangement, segments of genomic DNA are rearranged to eliminate certain sequences and generate mature genes during lymphocyte precursor cell development as part of the normal process of generating immunoglobulin and T-cell receptor diversity. Somatic rearrangement is a highly specialized process, specific only to these genes and specific cell types.

6. Variation in epigenetic modifications might lead to overexpression or underexpression of a gene or genes. DNA methylation might lead to epigenetic silencing of a gene. miRNAs can be involved in regulating the expression of other genes, and mutations in such an miRNA might be expected to alter gene expression patterns. The product of lncRNA genes are RNAs that can be involved in epigenetic regulation or other regulatory pathways; deletion or improper expression of such an miRNA might therefore lead to abnormalities of developmental pathways.

7. Genomic imprinting involves epigenetic silencing of an allele (or alleles at a number of closely located genes) based solely on parental origin due to epigenetic marks inherited through the germline. X inactivation involves epigenetic silencing of alleles along much of an entire chromosome based not on parental origin, but rather on a random choice of one or the other X chromosome at the time of initiation of the process in early embryonic development.

Chapter 4 Human Genetic Diversity: Mutation and Polymorphism

1. (a) CNV.

(b) Indel.

(c) A mutation in a splice site.

(d) An inversion.

(e) A SNP (or in/del) in a noncoding region or intron, or a SNP that leads to a synonymous substitution.

2. Assuming 20 years is one generation, 41 mutations/9 million alleles/2 generations = ≈2.3 × 10−6 mutations/generation at the aniridia locus. The estimate is based on assumptions that ascertained cases result from new mutation, that the disease is fully penetrant, that all new mutants are liveborn (and ascertained), and that there is only a single locus at which mutations can lead to aniridia. If there are multiple loci, the estimated rate is too high. If some mutations are not ascertained (because of lack of penetrance or death in utero), the estimated rate might be too low.

3. A microsatellite polymorphism, because microsatellite polymorphisms typically have more alleles, which provides greater capacity to distinguish genomes. A single SNP or indel would only have two alleles.

4. Based on information in this chapter, each cell division leads to less than one new single-bp mutation per genome. Rounding this up a bit to be one mutation per cell division, then one would expect at most 100 single-bp differences between cells at the end points of the two lineages mentioned. The rate of CNV de novo changes is much higher, and thus one would expect many, many such differences between the two lineages. Improved technologies now allow sequencing of single-cell genomes (i.e., rather than sequencing DNA from a collection of millions of cells). Thus it will now be possible to determine the answer to this question experimentally, rather than just theoretically.

5. Different types of mutations are sensitive to maternal or paternal age. Both single-bp mutations and CNVs show an increase in frequency with increasing age of the father. In contrast, meiotic nondisjunction for many chromosomes (including chromosome 21) shows an increase with increasing age of the mother. The rate of mutation (per bp) varies greatly in different locations around the genome; hot spots of mutation show much higher rates, although the basis for this is poorly understood. Intrachromosomal homologous recombination can lead to copy number variation in gene families or to deletion/duplications for regions flanked by homologous sequences (e.g., segmental duplications). Overall, the rate of mutation can be influenced also by genetic variation, both at a population level and in specific parental genomes. In any individual genome, this may influence where one falls in the ranges observed in typical genomes, as summarized in the Box on page 55.

Chapter 5 Principles of Clinical Cytogenetics and Genome Analysis

1. (a) Forty-six chromosomes, male; one of the chromosome 18s has a shorter long arm than is normal.

(b) To determine whether the abnormality is de novo or inherited from a balanced carrier parent.

(c) Forty-six chromosomes, male, only one normal 7 and one normal 18, plus a reciprocal translocation between chromosomes 7 and 18. This is a balanced karyotype. For meiotic pairing and segregation, see text, particularly Figure 5-12.

(d) The del(18q) chromosome is the der(18) translocation chromosome, 18pter → 18q12::7q35 → 7qter. The boy's karyotype is unbalanced; he is monosomic for the distal long arm of 18 and trisomic for the distal long arm of 7. Given the number of genes on chromosomes 7 and 18 (see Fig. 2-7), one would predict that the boy is monosomic for approximately 100 genes on chromosome 18 and trisomic for approximately 100 genes on chromosome 7.

2. (a) Approximately 95%.

(b) No increased risk, but prenatal diagnosis may be offered.

3. Postzygotic nondisjunction, in an early mitotic division. Although the clinical course cannot be predicted with complete accuracy, it is likely that she will be somewhat less severely affected than would a nonmosaic trisomy 21 child.

4. (a) Abnormal phenotype, unless the marker is exceptionally small and restricted only to the centromeric sequences themselves. Gametes may be normal or abnormal; prenatal diagnosis indicated.

(b) Abnormal phenotype (trisomy 13; see Chapter 6); will not reproduce.

(c) Abnormal phenotype in proband and approximately 50% of offspring.

(d) Normal phenotype, but risk for unbalanced offspring (see text).

(e) Normal phenotype, but risk for unbalanced offspring, depending on the size of the inverted segment (see text).

5. (a) Not indicated.

(b) Fetal karyotyping indicated; at risk for trisomy 21, in particular.

(c) Karyotype indicated for the child to determine whether it is trisomy 21 or translocation Down syndrome. If it is translocation, parental karyotypes are indicated.

(d) Not indicated, unless other clinical findings might suggest a contiguous gene syndrome (see Chapter 6).

(e) Karyotype indicated for the boys to rule out deletion or other chromosomal abnormality. If clinical findings indicate possibility of fragile X syndrome, a specific DNA diagnostic test would be indicated.

6. (a) Paracentric inversion of the X chromosome, between bands Xq21 and Xq26, determined by karyotyping.

(b) Terminal deletion of 1p in a female, determined by karyotyping.

(c) Female with deletion within band q11.2 of chromosome 15, determined by in situ hybridization with probes for the SNRPN gene and D15S10 locus.

(d) Female with interstitial deletion of chromosome 15, between bands q11 and q13, determined by karyotyping. In situ hybridization analysis confirmed deletion of sequences within 15q11.2, with use of probes for the SNRPN gene and D15S10 locus.

(e) Female with deletion of sequences in band 1q36.3, determined by array CGH with the three BAC probes indicated.

(f) Male with an extra marker chromosome, determined by karyotyping. Marker was identified as an r(8) chromosome by in situ hybridization with a probe for D8Z1 at the centromere.

(g) Female with Down syndrome, with a 13q;21q Robertsonian translocation in addition to two normal chromosome 21s, determined by karyotyping.

(h) Presumably normal male carrier of a 13q;21q Robertsonian translocation, in addition to a single normal chromosome 21 (and a single normal chromosome 13), as determined by karyotyping.

7. (a) For Figure 5-6C: 46,XY,dup(X)(q28). The increased ratio of sequences in Xq28 indicates a duplication.

(b) For Figure 5-9C: 47,XX,+21. The individual is a female, because the intensity of sequences on the X are equivalent to all of the autosomes (other than 21). (The scattering of very low-intensity signals from the Y sequences is simply background noise.)

Chapter 6 The Chromosomal and Genomic Basis of Disease: Disorders of the Autosomes and the Sex Chromosomes

1. Theoretically, X and XX gametes in equal proportions; expected XX, XY, XXX, and XXY offspring (25% each). In actuality, XXX women have virtually all chromosomally normal offspring, XX and XY, implying that XX gametes are at a significant disadvantage or are lost.

2. It is possible that the relevant region of chromosome 9 has very few genes and that the inversion does not interfere with gene structure or function. Carriers are not genetically imbalanced. Their potential risk might be to the offspring, as seen with other pericentric inversions. However, the flanking regions of 9p and 9q are so large (i.e., most of those chromosome arms) that duplication or deletion resulting from meiotic crossing over may be incompatible with life. Alternatively, centromeric regions of chromosomes are relatively poor in recombination; thus there may be very few crossovers in this region, and the inv(9) could pass to the next generation unchanged.

3. No. XYY can result only from meiosis II nondisjunction in the male, whereas XXY can result from nondisjunction at meiosis I in the male or at either division in the female.

4. Translocation of Y chromosome material containing the sex-determining region (and the SRY gene) to the X chromosome or to an autosome.

5. The small r(X) may contain genes that normally would undergo X inactivation but fail to do this in this abnormal chromosome lacking an X inactivation center. Such genes would show biallelic expression and be expressed at higher levels than seen in typical males (one X) or typical females (one active X, one inactive X). This abnormal gene expression may underlie the intellectual disability.

In the second family, the larger r(X) contains the X inactivation center. Thus one predicts that X inactivation would proceed normally and the r(X) would be the inactive X in all cells (due to secondary cell selection; see Fig. 6-13B). The phenotype is somewhat uncertain, however, because this individual may be deficient for genes that would typically escape X inactivation and would be expressed biallelically; some features of Turner syndrome may therefore be present.

6. 46,XX; autosomal recessive; prenatal diagnosis possible; need for clinical attention in neonatal period to determine sex and to forestall salt-losing crises.

7. (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).

8. Question for discussion. See text for possible explanations.

9. (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 age-related 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

1. (b) Autosomal recessive; 1 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. DNA-based carrier testing will answer the question definitively.

2. (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).

3. 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.

4. (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.

5. All are possible except (c), which is unlikely if the parents are completely unaffected.

6. (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%.

7. 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 ancestor is known as the coefficient of inbreeding (F). In the accompanying figure, suppose individual I-1 is heterozygous for alleles 1 and 2, whereas individual I-2 is a 3,4 heterozygote. The chance II-2 inherits the 2 allele is image, and the chance III-2 inherits it from II-2 is image, so the chance III-2 inherits allele 2 from I-1 is image × image = image. Similarly, the chance IV-1 is a carrier of allele 2 inherited from I-1 is image × image × image =image, and the chance III-2 and IV-1 are both heterozygotes for 2 inherited from I-1 is image × image =image. The probability their child could be a 22 homozygote inherited from I-1 as a common ancestor is therefore image × image = image. Repeat this calculation for allele 1 in I-1, and for either allele 3 or 4 in II-1, which means the probability the child could be homozygous 1,12,23,3, or 4,4 is equal to 4 × image = image 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.


Figure for Chapter 7, question 7.

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 (image)n−1. So, in this example, the loop A-D-H-K-L-I-E-A contains seven unique nodes, n = 7. Add all the coefficients for each loop together to find F. For the pedigree, then:




8. AD is most likely. Vertical transmission, including male to male, from generation to generation, males and females affected.

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.

9. 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 × 1 × 10−6, and therefore the odds both parents are carriers versus only the mother is 0.25/(0.5 × 10−6) = 5 × 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

1. (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 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.

2. Male to male transmission can disprove X-linkage; other criteria of multifactorial inheritance can be examined, as in the text.

3. 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

1. 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 1 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 1 or allele 2 or … allele 5. Therefore 71.5% of individuals would be heterozygous at this locus.

2. 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 = image × 2.66% = 1.33% and the chance a sensitized Rh−/− mother will have a Rh+/− child when her mate is Rh+/− = image × 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.

3. (a) Assume there are 100 individuals in the population, carrying 200 alleles at a particular locus. The frequency of A is (2 × image) + (image) = 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.

4. (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 = 2 (1 − q)q = 0.04, or q2 − q + 0.02 = 0 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%.

5. Only (d) is in equilibrium. Possible explanations include selection for or against particular genotypes, nonrandom mating, and recent migration.

6. (a) Abby has a image chance of being a carrier. Andrew has approximately a image chance of being a carrier. Therefore their risk for having an affected child is image × image × image, or image.

(b) image × image × image = image.

(c) image × image × image = imageimage × image × image = image.

7. (a) Facioscapulohumeral muscular dystrophy: q = image, 2pq = image. Friedreich ataxia: q = image, 2pq = image. 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 1 in 25,000, then, assuming half of the population is male, the frequency in males must be 1 in 12,500, so q = image, 2pq = image.

(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.

8. Mutant allele frequencies are approximately image and image. Two possible explanations for the difference in 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

1. The HD and MNSs loci map far enough apart on chromosome 4 to be unlinked, even though they are syntenic.

2. 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 5 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.

3. 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 do not know what the phase was in the father simply by inspection). We do not know the phase in individuals IV-3 or IV-4 because we do 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.

4. (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.

(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.

5. 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.

6. 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.

7. In 10-7A, D = 0, so D′ = 0.

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.

8. 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

1. 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 populations, including Melanesians. Parents with this genotype cannot transmit a −−/−− genotype to their offspring.

2. 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 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 7.

3. 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 normal-sized 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.

4. These two mutations affect different globin chains. The expected offspring are image normal, image Hb M Saskatoon heterozygotes with methemoglobinemia, image Hb M Boston heterozygotes with methemoglobinemia, and image 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.

5. image × image × image = image.

6. image.

7. 8, 1, 2, 7, 10, 4, 9, 5, 6, and 3.

8. 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.

9. 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

1. 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.

2. 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).

3. 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.

4. 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 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.

5. 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 image × 5% to 15% = 2.5% to 7.5%.

6. 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.

7. 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).

8. 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).

9. 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, 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.

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 disease-causing 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 do 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 RNA-mediated 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

1. 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 DNA-binding 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.

2. 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 do 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.

3. 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.

4. Unresponsive patients probably have alleles that do 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.

5. (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 (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.

6. 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 do 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.

7. 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

1. Before determination, an embryo can lose one or more cells, and the remaining cells can undergo 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.

2. a–3, b–2, c–4, d–1.

3. a–4, b–3, c–5, d–2, e–1.

4. 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.

5. 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

1. 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.

2. 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.

3. 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).

4. 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.

5. 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 ATMBARD1BRIP1CDH1CHEK2PALB2PTEN, 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 disease-causing 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.

6. 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.

Chapter 16 Risk Assessment and Genetic Counseling

1. (a) Prior risk, image; posterior risk (two normal brothers), image.

(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.

2. (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. 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 1.

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 do 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 image.


Figure for Chapter 16, question 2.

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%.

3. (image)13 for 13 successive male births.

(image)13 × 2 for 13 consecutive births of the same sex. (The 2 arises because this is the chance of 13 consecutive male births or 13 consecutive female births, before any children are born.)

image. The probability of a boy is image 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).

4. (a) Use the first equation, I = µ + image. To solve for H and substitute it for H in the second equation, H = 2µ + image + 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 image × 0.9 = 0.45.

(b) Substitute f = 0 into the equations and you get I = 3µ and H = 4µ.

(c) 0.147.


Figure for Chapter 16, question 4.

5. (a) The prior risk that either Ira or Margie is a cystic fibrosis carrier is image; therefore the probability that both are carriers is image × image = image.

(b) Their risk for having an affected child in any pregnancy is image × image = image.

(c) Bayesian analysis is carried out.

Thus the chance that Ira's and Margie's next child will be affected is image × image = image.


Both Carriers

Not Both Carriers





 (3 normal children)




image × (image)3 = 0.19

image = 0.56


0.19/(0.19 + 0.56) = image

0.56/0.75 = ≈image

6. The child's prior probability of carrying a mutant myotonic dystrophy gene is image. If it is assumed that he has a image chance of being asymptomatic, even if he carries the mutant gene, then his chance of carrying it and showing no symptoms is image. 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).

7. (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.

8. 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.

9. The seven scenarios are shown in the table.

Conditional Probability Calculation


*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


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


Step One of Dummy Consultand Method


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 image the probability that her mother, II-1, is a carrier = image × imageimage; the prior probability that she is not a carrier is 1 − (image) = image (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

III-2 Not a Carrier

Prior probability



Conditional (2 unaffected sons)



Joint probability



Posterior probability



Thus the posterior probability that III-2 is a carrier using the dummy consultand method, given she has two unaffected sons, is image, 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 image (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 image (50%). (We thank Susan Hodge from Columbia University for pointing out this problem with the dummy consultand method.)

Chapter 17 Prenatal Diagnosis and Screening

1. c, e, f, i and j, d, h, g, b, i (and, in part, j), and a.

2. 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.

3. No, not necessarily; the problem could be maternal cell contamination.

4. 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.

5. (a) Approximately 15% (see Table 5-2).

(b) At least 50% are chromosomally abnormal.

(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.

6. (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.

7. 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.

8. 600,000 women, 1000 pregnancies affected.

Assume everyone is willing to participate in the sequential screening. Of 1000 true positives, first-trimester 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 second-trimester 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 do 970 invasive tests in affected pregnancies while also doing 29,950 + 28,452 = 58,402 invasive tests in unaffected pregnancies.

Thus you will do 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 do 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


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 do not carry a particular marker whereas a PPV is the actual (or absolute) risk for someone who carries the marker.


Deep Venous Thrombosis in the Legs, Oral Contraceptive Use, and Factor V Leiden


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%.

3. 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.

4. 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?

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?


Carbamazepine-Induced TEN or SJS


SJS, Stevens-Johnson syndrome; TEN, toxic epidermal necrolysis.

Sensitivity = 44/44 = 100%.

Specificity = 98/101 = 97%.

Positive predictive value = 44/47 = 94%.

6. 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

1. 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 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 do you do about testing the asymptomatic older sib?

2. 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.

3. 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.

4. 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.