oeb50 readings probs x-linkage

the spindle breaks down, and, depending on the species, either a nuclear envelope briefly forms around each group of chromosomes or the chromosomes enter the second meiotic division after only a limited uncoiling.

The Second Meiotic Division: EquationThe second meiotic division (meiosis II) is

sometimes called the equational division because the chromosome number remains the same in each cell before and after the second division. In some species, the chromosomes pass directly from telophase I to prophase II with-out loss of condensation; in others, there is a brief pause between the two meiotic divisions and the chromosomes may “decondense” (uncoil) somewhat. Chromosome replication never takes place between the two divisions; the chromo-somes present at the beginning of the second division are identical to those present at the end of the first.

After a short prophase (prophase II) and the formation of second-division spindles, the cen-tromeres of the chromosomes in each nucleus become aligned on the central plane of the spin-dle at metaphase II (Figure 4.10C). In ana-phase II the protein holding the sister

centromeres to-gether breaks down. As a result, the sister centromeres ap-pear to split longitu-dinally, and the chromatids of each chromosome move to opposite poles of the spindle (Fig-ure 4.10D). Once the centromeres split at anaphase II, each chromatid is consid-ered to be a separate chromosome.

Telophase II (Figure 4.10E) is marked by a transi-tion to the inter-phase condition of the chromosomes in the four haploid nuclei, accompanied by division of the cytoplasm. Thus, the second meiotic divi-sion superficially

resembles a mitotic division. However, there is an important difference:

The chromatids of a chromosome are usu-ally not genetically identical along their entire length because of crossing over associated with the formation of chiasmata during pro-phase of the first division.

4.4 Sex-Chromosome InheritanceThe first rigorous experimental proof that genes are parts of chromosomes was obtained in experiments concerned with the pattern of transmission of the sex chromosomes, the chromosomes responsible for determination of the separate sexes in some plants and in nearly all animals. We will examine these results in this section.

Chromosomal Determination of SexThe sex chromosomes are an exception to

the rule that all chromosomes of diploid organ-isms are present in pairs of morphologically similar homologs. As early as 1891, microscopic analysis had shown that one of the chromo-somes in males of some insect species does not have a homolog. This unpaired chromosome was called the X chromosome, and it was pres-ent in all somatic cells of the males but in only half the sperm cells. The biological significance of these observations became clear when females of the same species were shown to have two X chromosomes.

In other species in which the females have two X chromosomes, the male has one X chro-mosome along with a morphologically unmatched chromosome. The unmatched chromosome is referred to as the Y chromo-some, and it pairs with the X chromosome during meiosis in males, usually along only part of its length because of a limited region of homology. The difference in chromosomal constitution between males and females is a chromosomal mechanism for determining sex at the time of fertilization. Whereas every egg cell contains an X chromosome, half of the sperm cells contain an X chromosome and the rest contain a Y chromosome. Fertilization of an X-bearing egg by an X-bearing sperm results in an XX zygote, which normally devel-ops into a female; and fertilization by a Y-bearing sperm results in an XY zygote, which normally develops into a male (FIGURE 4.12). The result is a criss-cross pattern of inheritance of the X chromosome in which a male receives

4.4 Sex-Chromosome Inheritance 131

Metaphase II

Anaphase II

Telophase II

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132 CHAPTER 4 Chromosomes and Sex-Chromosome Inheritance

X

X Y

Female (!) Male (")

SpermEggs

X X

X YX X

X X Y

"(XY)!(XX)

Sex of the progeny is determined by the sex chromosome contained in the sperm.

The result ofmeiosis is onetype of eggand two typesof sperm.

Fertilization

Meiosis

Progeny

FIGURE 4.12 The chromosomal basis of sex determination in mammals, many insects, and other animals.

his X chromosome from his mother and trans-mits it only to his daughters.

The XX–XY type of chromosomal sex deter-mination is found in mammals, including human beings, in many insects, and in other animals, as well as in some flowering plants. The female is called the homogametic sex because only one type of gamete (X-bearing) is produced, and the male is called the heteroga-metic sex because two different types of gam-etes (X-bearing and Y-bearing) are produced. When the union of gametes in fertilization is random, a sex ratio at fertilization of 1 : 1 is expected because males produce equal numbers of X-bearing and Y-bearing sperm.

The X and Y chromosomes together consti-tute the sex chromosomes; this term distin-guishes them from other pairs of chromosomes, which are called autosomes. Although the sex chromosomes control the developmental switch that determines the earliest stages of female or male development, the developmental process itself requires many genes scattered throughout the genome, including genes on the autosomes. The X chromosome also contains many genes with functions unrelated to sexual differentia-tion, as we will see in the next section. In most organisms, including human beings, the Y chro-mosome carries few genes other than those related to male determination.

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X-Linked InheritanceThe compelling evidence that genes are

located in chromosomes came from the study of a Drosophila gene for white eyes, which proved to be present in the X chromosome. Recall that in Mendel’s crosses, it did not matter which trait was present in the male parent and which in the female parent. Reciprocal crosses gave the same result. One of the earliest excep-tions to this rule was found by Thomas Hunt Morgan in 1910 in an early study of a mutant fruit fly that had white eyes. The wildtype eye color is a brick-red combination of red and brown pigments (FIGURE 4.13). Although white eyes can result from certain combinations of autosomal genes that eliminate the pigments individually, the white-eye mutation that Mor-gan studied results in a metabolic block that knocks out both pigments simultaneously. (The gene codes for a transmembrane protein that is necessary to transport the eye pigment precur-sors into the pigment cells.)

Morgan’s study started with a single male with white eyes that appeared in a wildtype laboratory population that had been maintained for many generations. In a mating of this male with wildtype females, all of the F1 progeny of both sexes had red eyes, which showed that the allele for white eyes is recessive. In the F2 prog-eny from the mating of F1 males with females, Morgan observed 2459 red-eyed females, 1011 red-eyed males, and 782 white-eyed males. It was clear that the white-eyed phenotype was


(A)

FIGURE 4.13 Drawings (A) of a male and a female fruit fly, Drosophila melanogaster. The photographs (B) show the eyes of a wildtype red-eyed male and a mutant white-eyed male. [Illustrations © Carolina Biological Supply Company. Used with permis-sion. Photographs courtesy of E. R. Lozovsky.]

Male (") Female (!) Red eyes White eyes

(B)

MORGAN’S GROUP partying in the fly room at Columbia University in 1919, to celebrate A. H. Sturtevant’s return from military service in World War I. T. H. Morgan is on the far right in the back row, with H. J. Muller beside him. Sturtevant is in the foreground, leaning back in his chair, and C. B. Bridges is in shirtsleeves sitting next to an apelike creature dubbed “Pithecanthropus,” a dummy made up for the occasion. [Courtesy of the Archives, California Institute of Technology.]

somehow connected with sex, because all of the white-eyed flies were males.

On the other hand, white eyes were not restricted to males. For example, when red-eyed F1 females from the cross of wildtype ! ! white " were backcrossed with their white-eyed fathers, the progeny consisted of both red-eyed and white-eyed females and red-eyed and white-eyed males in approximately equal numbers.

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A key observation came from the mating of white-eyed females with wildtype males. All the female progeny had wildtype eyes, but all the male progeny had white eyes. This is the recip-rocal of the original cross of wildtype ! ! white ", which had given only wildtype females and wildtype males, so the reciprocal crosses gave different results.

Morgan realized that reciprocal crosses would yield different results if the allele for white eyes were present in the X chromosome. A gene on the X chromosome is said to be X-linked. The normal chromosome comple-ment of Drosophila melanogaster is shown in FIG-URE 4.14. Females have an XX chromosome complement, whereas males are XY. Morgan’s hypothesis was that an X chromosome contains either a wildtype w" allele or a mutant w allele and that the Y chromosome does not contain a counterpart of the white gene. Using the white allele present in an X chromosome to represent the entire X chromosome, we can write the geno-type of a white-eyed male as wY and that of a

wildtype male as w"Y. Because the w allele is recessive, white-eyed females are of genotype ww and wildtype females are either heterozygous w"w or homozygous w"w". The implications of this model for reciprocal crosses are shown in FIG-URE 4.15. The mating wildtype ! ! white " is Cross A, and that of white ! ! wildtype " is Cross B.

The X-linked mode of inheritance does account for the different phenotypic ratios observed in the F1 and F2 progeny from the crosses. The characteristics of X-linked inheri-tance can be summarized as follows: 1. Reciprocal crosses resulting in differ-

ent phenotypic ratios in the sexes often indicate X-linked inheritance. In the case of white eyes in Drosophila, the cross of a red-eyed female with a white-eyed male yields all red-eyed progeny (Figure 4.15, Cross A), whereas the cross of a white-eyed female with a red-eyed male yields red-eyed female progeny and white-eyed male progeny (Figure 4.15, Cross B).

2. Heterozygous females transmit each X-linked allele to approximately half of their daughters and half of their sons; this is illustrated in the F2 generation of Cross B in Figure 4.15.

3. Males that inherit an X-linked recessive allele exhibit the recessive trait because the Y chromosome does not contain a wildtype counterpart of the gene. Affected males transmit the recessive allele to all of their daughters but none of their sons; this principle is illustrated in the F1 generation of Cross A in Fig-ure 4.15. Any male that is not affected must carry the wildtype allele in his X chromosome.

The essence of X-linked inheritance is captured in the Punnett square in FIGURE 4.16: A male trans-mits his X chromosome only to his daughters, whereas a female transmits an X chromosome to offspring of both sexes.

Pedigree Characteristics of Human X-Linked InheritanceAn example of a human trait with an

X-linked pattern of inheritance is hemophilia A, a severe disorder of blood clotting determined by a recessive allele. Affected persons lack a blood-clotting protein called factor VIII that is needed for normal clotting, and they suffer excessive, often life-threatening bleeding after injury. A famous pedigree of hemophilia starts with Queen

X

X

Male

Female

X

Y

FIGURE 4.14 The diploid chromosome complements of a male and a female Drosophila melanogaster. The centromere of the X chromosome is nearly terminal, but that of the Y chromosome divides the chromosome into two unequal arms. The large autosomes (chromosomes 2 and 3, shown in blue and green) are not easily distinguishable in these types of cells. The tiny autosome (chromosome 4, shown in yellow) appears as a dot.

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Victoria of England (FIGURE 4.17). One of her sons, Leopold, was hemophilic, and two of her daugh-ters were heterozygous carriers of the gene. Two of Victoria’s granddaughters were also carriers, and by marriage they introduced the gene into the royal families of Russia and Spain. The heir to the Russian throne of the Romanovs, Tsarev-ich Alexis, was afflicted with the condition. He inherited the gene from his mother, the Tsarina Alexandra, one of Victoria’s granddaughters. The Tsar, the Tsarina, Alexis, and his four sisters were all executed by the Russian Bolsheviks in the 1918 revolution. Ironically, the present royal family of England is descended from a normal son of Victoria and is free of the disease.

X-linked inheritance in human pedigrees shows several characteristics that distinguish it from other modes of genetic transmission. 1. For any rare trait due to an X-linked

recessive allele, the affected individuals

Eggs

! "

! "

X

X

X

Y

XX XY

XX XY

SpermXY(")

XX(!)

#

The female (!) to male (") ratio is 1:1.

FIGURE 4.16 In chromosomal sex determination, each son gets his X chromosome from his mother and his Y chromosome from his father.

Parents:

F1 :

F2 :

Red-eyedfemale (XX)

Cross A

White-eyedmale (XY)

#

Red-eyed females and males

1/4 Red-eyed females

1/4 Red-eyedfemales

1/4 Red-eyedmales

1/4 White-eyedmales

w+

w+

w+

w+

w

w+

w

w+

w+

w

w+

w

White-eyedfemale (XX)

Cross B

Red-eyedmale (XY)

#

Red-eyed femalesand white-eyed males

1/4 Red-eyed females

1/4 White-eyedfemales

1/4 Red-eyedmales

1/4 White-eyedmales

w+

w+

w

w

w

w+

w

w+w

w

w

w

Mating F1 progeny Mating F1 progeny

FIGURE 4.15 Chromosomal interpretation of the results obtained in F1 and F2 progenies in crosses of Drosophila. Cross A is a mating of a wildtype (red-eyed) female with a white-eyed male. Cross B is the reciprocal mating of a white-eyed female with a red-eyed male. In the X chromosome, the wildtype w+ allele is shown in red, the mutant w allele in white. The Y chromosome does not carry either allele of the w gene.

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are exclusively, or almost exclusively, male. There is an excess of males because females carrying the rare X-linked recessive are almost exclusively het-erozygous and so do not express the mutant phenotype.

2. Affected males who reproduce have normal sons. This follows from the fact that a male transmits his X chromosome only to his daughters.

Queen Victoria of England(1819–1901)

Prince Albert(1819–1861)

Beatrice(1857–1944)

Arthur(1850–1942)

Helena(1846–1923)

Alice(1843–1878)

PrincessVictoria

(1840–1901)

King George V(1865–1936)

King George VI(1895–1952)

Carrier female Hemophilic male Died at birth, sex unrecorded

Queen Elizabeth II

(1926–)

Edward VIIKing of England

(1841–1910)

AlfredDuke of Edinburgh

(1844–1900)

LeopoldDuke of Albany

(1853–1884)Louise

(1848–1939)

Alexandra,Tsarina of Russia(1872–1918)

Marie(1899–1918)

Tatiana(1897–1918)

Anastasia(1901–1918)

Olga(1895–1918)

Alexis(1904–1918)

Nicholas II Tsar of Russia(1868–1918)

FIGURE 4.17 Genetic transmission of hemophilia A among the descendants of Queen Victoria of England, including her granddaughter, Tsarina Alexandra of Russia, and Alexandra’s five children. The photograph is that of Tsar Nicholas II, Tsarina Alexandra, and their children. Tsarevich Alexis was afflicted with hemophilia. [Source: Culver Pictures. Photo courtesy of the Trustees of the Boston Public Library, Print Department.]

3. A woman whose father was affected has normal sons and affected sons in the ratio 1 : 1. This is true because any daughter of an affected male must be heterozygous for the recessive allele.

Heterogametic FemalesIn some organisms, the homogametic and

heterogametic sexes are reversed; that is, the males are XX and the females are XY. This type of sex determination is found in birds, in some reptiles and fish, and in moths and butterflies. The reversal of XX and XY in the sexes results in an opposite pattern of nonreciprocal inheritance of X-linked genes. To distinguish sex determina-tion in these organisms from the usual XX–XY mechanism, the sex chromosome constitution in the homogametic sex is designated ZZ and that in the heterogametic sex as WZ. Hence in organ-isms with heterogametic females, the chromo-somal constitution of the females is designated WZ and that of the male ZZ.

A specific example of Z-linked inheritance in chickens is shown in FIGURE 4.18. A few breeds of chickens have feathers with alternating

Barred !(WZ)

Nonbarred "(ZZ)

Nonbarred !!(WZ)

and

#

Barred ""(ZZ)

(A)

Nonbarred !(WZ)

Barred "(ZZ)

and

#

Barred !!(WZ)

Barred ""(ZZ)

(B)

FIGURE 4.18 Barred feathers in chickens, a classic example showing that chromosomal sex determination in birds is the reverse of that in mammals. In birds, females are the heterogametic sex. The Z chromosome carrying the dominant barred mutation is indicated in red.

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ccccccccccoooooooooonnnnnnnnnnnnnnnnnnnneeeeeeeeeeccccccccccttttttttttttttttiiiiiiiiiiiiiiiiioooooooooonnnnnnnnnn

The White-Eyed MaleMorgan’s genetic analysis of the white-

eye mutation marks the beginning of Drosoph i l agenetics. It is in the nature of sci-ence that as knowledge in-creases, the terms used to describe

things change also. This paper affords an example, because the term sex-limited inheritance is used today to mean some-thing completely different from Morgan’s usage. What Morgan was referring to is now called X-linked inheritance or sex-linked inheritance. To avoid confusion, we have taken the liberty of substituting the modern equivalent wherever appropriate. Morgan was also unaware that Drosoph-ila males had a Y chromosome. He thought that females were XX and males X, as in grasshoppers (see the Carothers paper). We have also supplied the missing Y chromo-some. On the other hand, Morgan’s gene symbols have been retained as in the origi-nal. He uses R for the wildtype allele for red eyes and W for the recessive allele for white eyes. This is a curious departure from the convention, already introduced by Mendel, that dominant and recessive alleles should be represented by the same symbol. Today we use w for the recessive allele and w" for the dominant allele.

In a pedigree culture of Drosophilathat had been running for nearly a

year through a considerable number of generations, a male appeared with white eyes. The normal flies have bril-liant red eyes. The white-eyed male, bred to his red-eyed sisters, produced 1,237 red-eyed offspring. . . . The F1

hybrids, inbred, produced

2,459 red-eyed females 1,011 red-eyed males 782 white-eyed males

No white-eyed females appeared.The new character showed itself to be sex-linked in the sense that it was transmitted only to the grandsons. But that the character is not incompatible with femaleness is shown by the following experiment. The white-eyed male (mutant) was later crossed with some of his daughters (F1), and produced

129 red-eyed females 132 red-eyed males 88 white-eyed females 86 white-eyed males

The results show that the new character, white eyes, can be carried over to the females by a suitable cross, and is in consequence in this sense not limited to one sex. It will be noted that the four classes of individuals occur in approximately

equal numbers (25 percent). . . . The results just described can be accounted for by the following hypothesis. Assume that all of the spermatozoa of the white-eyed male carry the “factor” for white eyes “W”; that half of the spermatozoa carry a sex factor “X,” [and] the other half lack it, i.e., the male is het-erozygous for sex. [The male is actu-ally XY.] Thus, the symbol for the male is “WXY”, and for his two kinds of spermatozoa WX—Y. Assume that all of the eggs of the red-eyed female carry the red-eyed “factor” R; and

that all of the eggs (after meiosis) carry one X each, the symbol for the red-eyed female

will be therefore RRXX and that for her eggs will be RX. . . . The hypoth-esis just utilized to explain these results first obtained can be tested in several ways. [There follow four types of crosses, each yielding the expected result.] . . . In order to obtain these results it is necessary to assume that, when the two classes of spermatozoa are formed in the RXY male, R and X go together. . . . The fact is that this R and X are com-bined and have never existed apart.

Source: T. H. Morgan, Science 32 (1910): 120–122.

Thomas Hunt Morgan 1910Columbia University, New York, New YorkSex-Limited Inheritance in Drosophila

No white-eyedfemales appeared.

transverse bands of light and dark color, resulting in a phenotype known as barred. In other breeds the feathers are uniformly colored and non-barred. Reciprocal crosses between true-breeding barred and true-breeding nonbarred breeds give the results shown, which indicate that the gene that determines barring must be dominant and must be located in the Z chromosome.

Nondisjunction as Proof of the Chromosome Theory of HeredityThe parallel between the inheritance of the

Drosophila white mutation and the genetic trans-mission of the X chromosome supported the chromosome theory of heredity that genes are

parts of chromosomes. Other experiments with Drosophila provided the definitive proof.

One of Morgan’s students, Calvin Bridges, discovered rare exceptions to the expected pat-tern of inheritance in crosses with several X-linked genes. For example, when white-eyed Drosophila females were mated with red-eyed males, most of the progeny consisted of the expected red-eyed females and white-eyed males. However, about 1 in every 2000 F1 flies was an exception, either a white-eyed female or a red-eyed male. Bridges showed that these rare exceptional offspring resulted from occasional failure of the two X chromosomes in the mother to separate from each other during meiosis—a

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phenomenon called nondisjunction. The con-sequence of nondisjunction of the X chromo-some is the formation of some eggs with two X chromosomes and others with none. Four classes of zygotes are expected from the fertiliza-tion of these abnormal eggs (FIGURE 4.19). Ani-mals with no X chromosome are not detected because embryos that lack an X are not viable; likewise, most progeny with three X chromo-somes die early in development. Microscopic examination of the chromosomes of the excep-tional progeny from the cross white ! ! wild-type " showed that the exceptional white-eyed females had two X chromosomes plus a Y chro-mosome, and the exceptional red-eyed males had a single X but were lacking a Y. The latter, with a sex-chromosome constitution denoted XO, were sterile males.

These and related experiments demon-strated conclusively the validity of the chromo-some theory of heredity.

Chromosome theory of heredity: Genes are physically located within chromosomes.

Bridges’s evidence for the chromosome theory was that exceptional behavior on the part of chromosomes is precisely paralleled by excep-tional inheritance of their genes. This proof of the chromosome theory ranks among the most important and elegant experiments in genetics.

Sex Determination in DrosophilaIn the XX–XY mechanism of sex determina-

tion, the Y chromosome is associated with the male. In some organisms, including humans, this association occurs because the presence of the

FIGURE 4.19 Results of nondisjunction of the X chromosomes in the first meiotic division in a female Drosophila.

Normal sperm

Nondisjunctional eggs

Red-eyed male

Red-eyed female(usually dies)

XXXWhite-eyed female

XXY

Red-eyed male (sterile)X

DiesY

w

w+w

w

ww

w

w

w

White-eyed female

#

w+

w+

w+

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Y chromosome triggers events in embryonic development that result in the male sexual char-acteristics (Chapter 8). Drosophila is unusual among organisms with an XX–XY type of sex determination because the Y chromosome, although associated with maleness, is not male determining. This is demonstrated by the finding, shown in FIGURE 4.20, that XXY embryos in Dro-sophila develop into morphologically normal, fertile females, whereas XO embryos develop into morphologically normal, but sterile, males. (The O is written in the formula XO to emphasize that a sex chromosome is missing.) The sterility of XO males shows that the Y chromosome, though not necessary for male development, is essential for male fertility; in fact, the Drosophila Y chromosome contains six genes required for the formation of normal sperm.

The genetic determination of sex in Drosoph-ila depends on an X-linked gene known as Sex-lethal (Sxl) because some mutant alleles are lethal when present in males. The Sxl gene is active in normal females and inactive in normal males. The product of Sxl is the Sex-lethal protein SXL, which is an RNA-binding protein that binds with the RNA transcripts of several genes and causes female-specific RNA processing. In the absence of SXL, these transcripts undergo male-specific processing. Hence the targets of SXL are genes whose protein products are normally made only in females. One of the targets of SXL is the Sxl gene itself, so a small burst of SXL activity early in female development leads to self-sus-taining production of SXL because of feedback in the processing of more Sxl transcripts.

What makes for the early burst of SXL is the amount of product produced by a small number of X-linked genes known as X-linked signal

elements or XSE (Figure 4.20). If the amount of XSE is high, the early burst occurs and the organ-ism becomes female; whereas if the amount of XSE is low, the early burst does not occur and the organism becomes male (Figure 4.20). For almost 100 years geneticists believed that sex was deter-mined by the ratio of the number of X chromo-somes (X) relative to the number of sets of autosomes (A). The correlation is perfect. In nor-mal females (XXAA) the ratio equals 1, in normal males (XAA) it equals 1/2, and in rare individuals of composition XXAAA the ratio is 2/3 and the organism develops as an intersex. Furthermore, rare XA haploids initially develop as female, although they eventually die, which again sup-ports the idea that an embryo with a ratio of X : A of 1/2 develops as male, an embryo with an X : A ratio of 1 develops as a female, and an embryo with an X : A ratio between 1/2 and 1 develops as an intersex.

While the correlation with X : A is perfect, however, recent data show that the hypothesis is wrong. Sex in Drosophila is based on Sxl reacting to the amount of XSE, which in normal flies is determined by the number of X chromosomes. The autosomes do play a role but not in sex determination. Their role is in determining when in early development the commitment to sexual differentiation takes place. In normal XXAA or XAA embryos with two sets of autosomes, com-mitment takes place after 15 cycles of division. In XA haploid embryos with one set of autosomes, commitment takes place one cell cycle later when the amount of XSE is sufficient to induce Sxl. In XXAAA embryos, some cells commit in cycle 13 and have too little XSE to induce Sxe, whereas other cells commit in cycle 14 and have enough XSE to induce Sxl. The result is an


Amount ofprotein fromX-linked signalelements (XSE)

High(equivalent to two X chromosomes)

Low(equivalent to oneX chromosome)

Intermediate

Sxl ON

Sxl ON

Sxl OFF

Sxl OFF

Some cells

SXL proteinproduced

SXL proteinproduced insome cells,not in others

SXL proteinnot produced

Other cells

Female-specifictranscripts

Normalfemale

Intersex

Normalmale

Female-specifictranscripts in some cells, male-specifictranscripts in other cells

Male-specifictranscripts

FIGURE 4.20 Early steps in the genetic control of sex determination in Drosophila through the activity of the Sex-lethal gene Sxl.

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embryo that is a mosaic of male and female cells, and the phenotypic result is an intersex.

Sex determination in Drosophila is an excel-lent example of why correlation does not neces-sarily mean causation, and of the principle that ideas in science, even those that are long entrenched and widely believed, sometimes turn out to be wrong.

4.5 Probability in the Prediction of Progeny Distributions

Genetic transmission includes a large compo-nent of chance. A particular gamete from an Aa organism might or might not include the A allele, depending on chance. A particular gam-ete from an Aa Bb organism might or might not include both the A and B alleles, depending on the chance orientation of the chromosomes on the metaphase I plate. Genetic ratios result not only from the chance assortment of genes into gametes, but also from the chance combination of gametes into zygotes. Although exact predic-tions are not possible for any particular event, it is possible to determine the probability that a particular event will be realized, as we saw in Chapter 3. In this section, we consider some additional probability methods used in inter-preting genetic data.

Using the Binomial Distribution in GeneticsThe addition rule of probability deals with

outcomes of a genetic cross that are mutually exclusive. Outcomes are “mutually exclusive” if they are incompatible in the sense that they cannot occur at the same time. For example, the possible sex distributions among three children consist of four mutually exclusive outcomes: zero, one, two, or three girls. These outcomes have probabilities 1!8, 3!8, 3!8, and 1!8, respec-tively. The addition rule states that the overall probability of any combination of mutually exclusive events is equal to the sum of the prob-abilities of the events taken separately. Forexample, the probability that a sibship of size three contains at least one girl includes theoutcomes one, two, and three girls, so theoverall probability of at least one girl equals 3!8 " 3!8 " 1!8 # 7!8.

The multiplication rule of probability deals with outcomes of a genetic cross that are inde-pendent. Any two possible outcomes are inde-pendent if the knowledge that one outcome is actually realized provides no information about whether the other is realized also. For example, in a sequence of births, the sex of any particular child is not affected by the sex of any sibling

born earlier and has no influence whatsoever on the distribution of sexes of any siblings born later. Each successive birth is independent of all the others. When possible outcomes are inde-pendent, the multiplication rule states that the probability of any combination of outcomes being realized equals the product of the proba-bilities of all the individual outcomes taken separately. For example, the probability that a sibship of three children will consist of three girls equals 1!2 ! 1!2 ! 1!2, because the probability of each birth resulting in a girl is 1!2, and the successive births are independent.

Probability calculations in genetics fre-quently use the addition and multiplication rules together. For example, the probability that all three children in a family will be of the same sex uses both the addition and the multiplica-tion rules. The probability that all three will be girls is (1!2)(1!2)(1!2) # 1!8, and the probabil-ity that all three will be boys is also 1!8. Because these outcomes are mutually exclusive (a sib-ship of size three cannot include three boys and three girls), the probability of either three girls or three boys is the sum of the two probabilities, or 1!8 " 1!8 # 1!4. The other possible out-comes for sibships of size three are that two of the children will be girls and the other a boy, and that two will be boys and the other a girl. For each of these outcomes, three different orders of birth are possible—for example, GGB, GBG, and BGG—each having a probability of 1!2 ! 1!2 ! 1!2 # 1!8. The probability of two girls and a boy, disregarding birth order, is the sum of the probabilities for the three possi-ble orders, or 3!8; likewise, the probability of two boys and a girl is also 3!8. Therefore, the distribution of probabilities for the sex ratio in families with three children is

GGG GGB GBB BBB GBG BGB BGG BBG

(1/2)3 " 3(1/2)2(1/2) " 3(1/2)(1/2)2 " (1/2)3 #

1/8 " 3/8 " 3/8 " 1/8 #1

The sex ratio information in this display can be obtained more directly by expanding the bino-mial expression (p " q)n, in which p is the prob-ability of the birth of a girl (1!2), q is the probability of the birth of a boy (1!2), and n is the number of children. In the present example,

(p " q)3 # 1p3 " 3p2q " 3pq2 " 1q3

in which the red numerals are the possible number of birth orders for each sex distribu-tion. Similarly, the binomial distribution of

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probabilities for the sex ratios in families of five children is

( p " q)5 # 1p5 " 5p4q " 10p3q2 "

10p2q3 " 5pq4 " 1q5

Each term tells us the probability of a particular combination. For example, the third term is the probability of three girls (p3) and two boys (q2) in a family that has five children:

10(1!2)3(1!2)2 # 10!32 # 5!16

There are n " 1 terms in a binomial expan-sion to the power n. The exponents of p decrease by one from n in the first term to 0 in the last term, and the exponents of q increase by one from 0 in the first term to n in the last term. The coefficients generated by successive values of ncan be arranged in a regular triangle known as Pascal’s triangle, shown for n # 0 to 10 in FIGURE 4.21. The horizontal rows of the triangle are symmetrical; each row begins and ends with a 1, and each other entry can be obtained as the sum of the two numbers on either side of it in the row above.

To generalize just a bit, if the probability of event A is p, that of event B is q, and the two events are independent and mutually exclusive, then the probability that A will be realized four times and B two times (in a specified order) is, by the multiplication rule, p4q2. Usually we are interested in a combination of events regardless of their order, such as “four A and two B.” In this case, we multiply the probability that the combi-nation 4A : 2B will be realized in any specified order by the number of possible orders. The number of different combinations of six events, four of type A and two of type B, is given by the coefficient of p4q2 in the expansion of (p A " q B)6. This coefficient can be found in the row for n # 6 in Pascal’s triangle, as the fifth entry from the left. (It is the fifth because the successive entries are the coefficients of p0q6, p1q5, p2q4, p3q3, p4q2, p5q1, and p6q0.) Hence, the overall probability of four realizations of A and two real-izations of B in six trials is given by 15p4q2.

The general rule for repeated trials of events with constant probabilities is as follows:

If the probability of event A is p and the prob-ability of the alternative event B is q, the probability that, in n trials, event A is realized s times and event B is realized t times is

n!$s!t! psqt (1)

in which s " t # n and p " q # 1. The sym-bol n! is read as “n factorial,” and it stands for

the product of all positive integers from 1 through n (that is, n! # 1 ! 2 ! 3 ! · · · ! n). Values n factorial from n # 0 through n # 15 are given in TABLE 4.2. The value 0! # 1 is defined arbitrarily to generalize its use in math-ematical formulas. The magnitude of n! increases very rapidly; 15! is more than a trillion.

Equation (1) applies even when either s or t equals 0 because 0! # 1. (Remember also that any number raised to the zero power equals 1.) Any individual term in the expansion of the binomial (p " q)n is given by Equation (1) for the appropriate values of s and t. In Pascal’s tri-angle, successive entries in the nth row are the values of n!!(s! t!) for s # 0, 1, 2, . . . , n.

Let us consider a specific application of Equation (1), in which we calculate the proba-bility that a mating between two heterozygous parents will yield exactly the expected 3 : 1 ratio of the dominant and recessive traits among sib-ships of a particular size. The probability p of a child showing the dominant trait is 3!4, and the probability q of a child showing the recessive

Coefficientsn

6 20 11

0 1

1 1

10

1

33

10

2 2 11

44

1515

3 11

55

4 6 11

66

5

10

7

8

9

1 1

35 7 11 3521 217

56 28 8 11 7028 568

84 84 36 9 11 12636 1269

120 210 120 45 10 11 21045 25210

FIGURE 4.21 Pascal’s triangle. The numbers in the nth row are the coefficients of the terms in the expansion of the polynomial (p " q)n.

4.5 Probability in the Prediction of Progeny Distributions 141

FactorialsTable 4.2

n n! n n!0 1 8 40,3201 1 9 362,8802 2 10 3,628,8003 6 11 39,916,8004 24 12 479,001,6005 120 13 6,227,020,8006 720 14 87,178,291,2007 5040 15 1,307,674,368,000

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Analysis and Applications 149

whereas st!/st! and st!/st genotypes have wildtype brick-red eyes. Flies homozygous for bw have brown eyes, whereas bw!/ bw! and bw!/ b genotypes have wildtype brick-red eyes. The double homozygous genotype st/st; bw/bw results in white eyes. In a test of independent assortment of these two genes, a geneticist crosses st!/st; bw!/bw females with st!/st; bw!/bw males. Among 240 progeny there are 150 flies with wildtype eyes, 36 with scarlet eyes, 46 with brown eyes, and 8 with white eyes.(a) Under the null hypothesis that these genes undergo

independent assortment, what are the expected numbers in each phenotypic class?

(b) What is the value of the chi-square in a test of good-ness of fit between the observed values and the expected values based on the null hypothesis of inde-pendent assortment?

(c) How many degrees of freedom does this chi-square value have?

(d) What is the P-value for the chi-square value in this goodness-of-fit test? Does this P-value support the null hypothesis of independent assortment or should the hypothesis be rejected?

(e) Would a greater chi-square value increase or decrease the P-value?

Answer(a) Because this is a dihybrid cross st!/st; bw!/bw " st!/

st; bw!/bw, the ratio of 9 : 3 : 3 : 1 of the offspring phenotypes should be expected if the genes assort independently. We can calculate the expected num-ber of flies in each class of the progeny:

Phenotype Progeny Expected numberwildtype st!/–; bw!/– (9/16) " 240 # 135scarlet st/st; bw!/– (3/16) " 240 # 45brown st!/–; bw/bw (3/16) " 240 # 45white st/st; bw/bw (1/16) " 240 # 15

(b) The chi-square is calculated as

$2 #(observed - expected)2

expected!

where the sum is over all classes of progeny. In this case,

$2 #(150 %135)2

135!

(36 % 45)2

45!

(46 % 45)2

45!

(8 %15)2

15

# 6.76

(c) The number of degrees of freedom equals the number of classes of data minus 1. In this case there are four classes of data; thus there are three degrees of freedom.

(d) The P-value for a chi-square value of 6.76 with 3 degrees of freedom equals 0.08. This P-value is greater than 0.05. Therefore, we should not reject the null hypothesis of independent assortment.

(e) A greater chi-square value would decrease the P-value.

The mutation for bar-shaped eyes in Drosophila melanogaster shows the following features of genetic transmission: (a) The mating of bar-eyed males with wildtype females

produces wildtype sons and bar-eyed daughters. (b) The bar-eyed females from the mating in part a,

when mated with wildtype males, yield a 1 : 1 ratio of bar : wildtype sons and a 1 : 1 ratio of bar : wildtype daughters.

What mode of inheritance do these characteristics suggest?

Answer Because the sexes are affected unequally in the prog-eny of the mating in a, some association with the sex chro-mosomes is suggested. The progeny from mating a provide the important clue. Because a male receives his X-chromo-some from his mother, then the observation that all males are wildtype suggests that the gene for bar eyes is on the X-chromosome. The fact that all daughters are affected is also consistent with X-linkage, provided that the bar muta-tion is dominant. Mating b confirms the hypothesis of a dominant X-linked gene, because the females from a mat-ing a would have the genotype bar/! and so would produce the observed progeny. The data are therefore consistent with the bar mutation being an X-linked dominant.

4.1 The diagrams below illustrate a pair of homologous chromosomes in prophase I of meiosis. Which dia-gram corresponds to each stage: leptotene, zygotene, pachytene, diplotene, and diakinesis?(A) (B)

(D) (E)

(C)

4.2 The diagrams below illustrate anaphase in an organ-ism that has two pairs of homologous chromosomes. Identify the stages as mitosis, meiosis I, or meiosis II.

(A) (B) (C)

4.3 What is the probability that four offspring from the mating Aa " Aa consist of exactly 3 A% and 1 aa?

4.4 A woman who is heterozygous for both a phenylke-tonuria mutation and an X-linked hemophilia muta-tion has a child with a phenotypically normal man who is also heterozygous for a phenylketonuria

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mutation. What is the probability that the child will be affected by both diseases? (Assume that they are equally likely to have a boy or a girl.)

4.5 In the accompanying pedigree the purple symbols represent individuals affected with X-linked Du-chenne muscular dystrophy. What is the probability that the woman III-3 is a heterozygous carrier?

I

1 2 3 4

1 2 3 4 5

II

III

1 2

4.6 A chromosomally normal woman and a chromosom-ally normal man have a son whose sex-chromosome constitution is XYY. In which parent, and in which meiotic division, did the nondisjunction take place?

4.7 Drosophila virilis is a diploid organism with 6 pairs of chromosomes (12 chromosomes altogether). How many chromatids and chromosomes are present in the following stages of cell division:(a) Metaphase of mitosis?(b) Metaphase I of meiosis?(c) Metaphase II of meiosis?

4.8 In a trihybrid cross with genes that undergo indepen-dent assortment: (a) What is the expected proportion of triply hetero-

zygous offspring in the F2 generation?(b) What is the expected proportion of triply homo-

zygous offspring in the F2 generation?4.9 The diploid gekkonid lizard Gonatodes taniae from

Venezuela has a somatic chromosome number of 16. If the centromeres of the 8 homologous pairs are des-ignated as Aa, Bb, Cc, Dd, Ee, Ff, Gg, and Hh:(a) How many different combinations of centro-

meres could be produced during meiosis?(b) What is the probability that a gamete will contain

only those centromeres designated by capital letters?

4.10 Among sibships consisting of six children, and assum-ing a sex ratio of 1 : 1:(a) What is the proportion with no girls?(b) What is the proportion with exactly one girl?(c) What is the proportion with exactly two girls?(d) What is the proportion with exactly three girls?(e) What is the proportion with three or more boys?

4.11 A litter of cats includes eight kittens. What is the probability that it contains an even number each of males and females? Assume an equal likelihood of male and female kittens, and for purposes of this problem regard 0 as an even number. Does the answer surprise you? Why?

4.12 A horticulturalist crossed a true-breeding onion plant with red bulbs to a true-breeding onion plant with white bulbs. All of the F1 plants had white

bulbs. When seeds resulting from self-fertilization of the F1 plants were grown, onion bulbs were recovered in the ratio of 12 white bulbs : 3 red bulbs : 1 yellow bulb. Propose a hypothesis to explain these observations.

4.13 Among 160 progeny in the F2 generation of a dihy-brid cross, a geneticist observes four distinct pheno-types in the ratio 91: 21 : 37 : 11. She believes this result may be consistent with a ratio of 9 : 3 : 3 : 1. To test this hypothesis, she calculates the chi-square value. Does the test support her hypothesis, or should she reject it?

4.14 What is the value of the chi-square that tests good-ness of fit between the observed numbers 40 : 60 as compared with the expected numbers 50 : 50?

4.15 What is the probability that a sibship of seven chil-dren includes at least one boy and at least one girl? Assume that a sex ratio is 1 : 1.

4.16 How many genotypes are possible for an autosomal gene with six alleles? How many genotypes are pos-sible with an X-linked gene with six alleles?

4.17 The growth habit of the Virginia groundnut Arachis hypogaea can be “runner“ (spreading) or “bunch“ (compact). Two pure-breeding strains of groundnuts with the bunch growth habit are crossed. The F1 plants have the runner growth habit. If the plants are allowed to self-fertilize, the F2 progeny ratio is 9 run-ner: 7 bunch. What genetic hypothesis can account for these observations?

4.18 In some human pedigrees, the blue/brown eye color variation segregates like a single-gene difference with the brown allele dominant to the blue allele. Two brown-eyed individuals each of whom had a blue-eyed parent mate. Assuming that the trait segregates like a single-gene difference in this pedigree:(a) What is the probability that the first child will

have brown eyes?(b) If a brown-eyed child results, what is the prob-

ability that the child will be heterozygous?(c) What is the probability that this couple will have

three children with blue eyes? That none of the three children will have blue eyes?

4.19 In the pedigree below, the male I-2 is affected with red-green color blindness owing to an X-linked mutation. What is the probability that male IV-1 is color blind? (Assume that the only possible source of the color blind-ness mutation in the pedigree is that from male I-2.)

I

II

III

IV

1 2

1 2 3 4 5

1 2 3

1

4

?

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4.20 The male II-1 in the pedigree shown here is affected with a trait due to a rare X-linked recessive allele. Use the principles of conditional probability to calculate the probability that the female III-1 is a heterozygous carrier.

I

II

III

IV

1 2 3

1 2

1 2 3

1 2

4.21 The pedigree shown below is the same as that in Problem 4.20 but with additional information based on molecular analysis. The male II-1 is again affected with a trait due to a rare X-linked recessive allele. The bands in the gel are restriction fragments that identify the wildtype allele (the shorter fragment S corre-sponding to the band near the bottom of the gel) or the mutant allele (the longer fragment L correspond-ing to the band near the top of the gel). From the information given in the pedigree and the gels, cal-culate the probability that the female III-1 is a het-erozygous carrier.

I

1 2 3

1 2

1 2 3

II

III

IV

1 2

LS S

S

S

L

S

4.22 In the pedigree shown here, the males I-2 and III-2 are affected with a trait due to a common X-linked recessive allele. What is the probability that individ-ual III-1 is heterozygous?

I

II

III

IV

1 2

1 2

1 2

1 2 3

4.23 Yellow body color in Drosophila is determined by the recessive allele y of an X-linked gene, and the wild-type gray body color is determined by the y! allele. What genotype and phenotype ratios would be expected from the following crosses?

(a) yellow male " wildtype female(b) yellow female " wildtype male(c) daughter from mating in part a " wildtype male(d) daughter from mating in part a " yellow male

4.24 The accompanying pedigree and gel diagram showthe molecular phenotypes for an RFLP with two alleles. What mode of inheritance does the pedigree suggest? Based on this hypothesis, and using A1 to represent the RFLP allele associated with the 3-kb band and A2 to represent that associated with the 8-kb band, deduce the genotype of each individual in the pedigree.

12 kb9 kb

6 kb

3 kb

1 kb

III-1 III-2 III-3

II-6II-5II-4II-3II-2II-1

I-1 I-2

III-4

4.25 Tall, yellow-flowered crocus is mated with short, white-flowered crocus. Both varieties are true breed-ing. All the F1 plants are backcrossed with the short, white-flowered variety. This backcross yielded 800 progeny in the proportions 234 tall yellow, 203 tall white, 175 short yellow, and 188 short white plants. Does the observed result fit the genetic hypothesis of 1 : 1 : 1 : 1 segregation as assessed by a $2 test?

4.26 Attached-X chromosomes in Drosophila are formed from two X chromosomes attached to a common cen-tromere. Females of genotype C(1)RM/Y, in which C(1)RM denotes the attached-X chromosomes, pro-duce C(1)RM-bearing and Y-bearing gametes in equal proportions. What progeny is expected to result from the mating between a male carrying the X-linked allele, y, for yellow body and an attached-X female with wildtype body? How does this result dif-fer from the typical pattern of X-linked inheritance? (Note: Drosophila zygotes containing three X chromo-somes or no X chromosomes do not survive.)

4.27 A rare dominant autosomal mutation W results in wooly, curled hair in some European pedigrees. A woman with wooly hair with blood group O marries a man with straight hair (ww) with blood group AB. The genes are in different chromosomes.(a) What are the chances that the mating will pro-

duce a wooly-haired group B child?(b) What are the chances that the mating will pro-

duce a straight-haired group B child?(c) If three straight-haired children with blood group

A children are born to these parents, what are

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appearance of the daughter chromosomes as they separate from each other in anaphase (FIGURE 8.5). A chromosome with its centromere about in the middle is a metacentric chromosome; the arms are of approximately equal length, and each daughter chromo-some forms a V shape at ana-phase. When the centromere is somewhat off-center, the chro-mosome is a submetacen-tric chromo some, and each daughter chromosome forms a J shape at anaphase. A chro-mosome with the centromere very close to one end appears I-shaped at anaphase because the arms are grossly unequal in length; such a chromosome is acrocentric.

Chromosomes with a sin-gle centromere are usually the only ones that are reliably transmitted from parental cells to daughter cells or from parental organisms to their progeny. When a cell divides, spindle fibers attach to the centromere of each chromosome and pull the sister chromatids to opposite poles. A chromosome that lacks a centromere is an acentric chromosome. Acentric chromosomes are genetically unstable because they cannot be maneuvered properly during cell division and are lost. Occasionally, a chromosome arises that has two centromeres and is said to be dicentric. A dicentric chromosome is also genetically unstable because it is not transmitted in a pre-dictable fashion. The dicentric chromosome is frequently lost from a cell when the two centro-meres proceed to opposite poles in the course of cell division; in this case, the chromosome is stretched and forms a bridge between the daugh-ter cells. This bridge may not be included in either daughter nucleus, or it may break, with the result that each daughter nucleus receives a broken chromosome. We will consider one mechanism by which dicentric and acentric chromosomes are formed when we discuss inversions.

Although most dicentric chromosomes are genetically unstable, if the two centromeres are close enough together, they can frequently behave as a single unit and be transmitted nor-mally. This principle was important in the

evolution of human chromosome 2. Among higher primates, chimpanzees and human beings have 23 pairs of chromosomes that are similar in morph ology and G-banding pattern, but chimpanzees have no obvious homolog of human chromosome 2, a large metacentric chromosome. Instead, chimpanzees have two medium-sized acrocentric chromosomes not found in the human genome. The cause of this situation is shown in FIGURE 8.6. The G-banding patterns indicate that human chromosome 2 was formed by fusion of the telomeres between the short arms of two acrocentric chromosomes that, in chimpanzees, remain acrocentrics. The fusion created a dicentric chromosome with its two centromeres close together. The new chro-mosome must have been sufficiently stable to be retained in the human lineage. The present-day human chromosome 2 has only a single func-tional centromere. The chromosome fusion reduced the chromosome number in the human lineage from 48, which is characteristic of the great apes (chimpanzee, gorilla, and orangutan), to the number 46.

Dosage Compensation of X-Linked GenesFor all organisms with XX–XY sex determi-

nation, there is a problem of the dosage of genes on the X and Y chromosome, because females

8.1 The Human Karyotype 261

Spindle fibers

Metacentric(V-shaped)

Submetacentric(J-shaped)

Acrocentric(I-shaped)

FIGURE 8.5 Three possible shapes of monocentric chromosomes in anaphase as determined by the position of the centromere (shown in dark blue).

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262 CHAPTER 8 Human Karyotypes and Chromosome Behavior

females, so that both X chromosomes together have the same transcriptional activity as that of the single X chromo-some in males.

The mechanism of dosage compensation in mammals is seemingly simple. In the early cleavage divisions of the embryo, at roughly the 64–128 cell stage, one and only one X chromo-some in each cell, chosen at random, remains genetically active, and all the other X chromo-somes that may be present in the cell undergo a process of X inactivation. Any X chromosome that is inactivated in a particular somatic cell remains inactive in all the descendants of that cell (FIGURE 8.7); hence the inactive state of anX chromosome is inherited from parental cell to daughter cell.

The process of X inactivation takes place in all embryos with two or more X chromosomes, including normal XX females. The inactivation process is one of chromosome condensation initiated at a site called XIC (for X-inactivation center) near the centromere on the long arm between Xq11.2 and Xq21.1. Cytologically, this site marks a “bend” in the chromosome in mitotic metaphase, which is thought to be a visible manifestation of the X inactivation. The finding that the Xq11.2–Xq21.1 region is pres-ent in all structurally abnormal X chromosomes is taken as one line of evidence for its essential

have two copies of this chromosome whereas males have only one. (There is less of a problem with Y-linked genes, because the Y chromosome is largely heterochromatic and carries relatively few genes.) In most organisms, a mechanism of dosage compensation has evolved in which the unequal dosage in the sexes is corrected either by increasing the activity of genes in the X chromosome in males or by reducing the activity of genes in the X chromosome in females. Opposite extremes of dosage compen-sation are illustrated by the fruit fly Drosophila melanogaster and the nematode worm Cae-norhabditis elegans:

• In males of D. melanogaster, a complex including at least one male-specific pro-tein is recruited to the X chromosome, which alters its chromatin structure, most likely by acetylation of a specific lysine in histone H4. This results in an increase in the transcriptional activity of the single X chromosome in males to a level equal to the sum of that of both X chromosomes in females.

• In females of C. elegans, a protein com-plex is recruited to the X chromosomes, which decreases the level of transcrip-tion of each X chromosome by half. This results in a decrease in the level of tran-scription of each X chromosome in

G

GG

A

AC

T

C

C

TT

G

GG

ATT

GG

A

CC

G

GG

AA

C

T

C

C

TT

G

GG

ATT

GG

A

CC

Telomere fusionjoins the DNA strands at thepositions shown;the sequencesbetween the brackets are lost.

A chromosomewith twofunctionalcentromerescannot persistfor very long.

No cytological indication of theinactivated centromere remains.

The alphoidsequencesoriginallyassociated withthe centromereof the long arm are mutated andinactive.Telomere sequences in the

short arm of each of two acrocentric chromosomes in an ancestral species; note the normal 3’ overhangs due to telomerase activity. (Not drawn to scale.)

Likely banded appearance of the ancestral acrocentric chromosomes.

Telomeres Fusion oftelomeresof short arms

Centromereinactivation

Dicentricchromosome

Modern humanchromosome 2

Shortarm

Longarm

FIGURE 8.6 Human ancestors had 24 pairs of chromosomes rather than 23. In the evolution of the human genome, two acrocentric chromo-somes fused to create human chromosome 2.

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function in X inactivation, because structurally abnormal X chromosomes lacking this region would not be inactivated and therefore cause lethality and not be observed. Another line of evidence is that the XIC includes a transcribed region designated Xist (for X-inactivation–specific transcript). Transcription of Xist is the earliest event observed in X inactivation, and Xist is the only gene known to be transcribed only from the inactive X chromosome. Remarkably, the spliced transcript of Xist does not contain an open reading frame encoding a protein. It appears to function as a structural RNA, and as transcription of Xist continues, the spliced tran-script progressively coats the inactive X chro-mosome, spreading outward from the XIC. Thereafter, other molecular changes take place along the inactive X chromosome that are typi-cally associated with gene silencing, including heavy cytosine methylation in 5! regulatory regions, deacetylation and methylation of his-tones, and aggregation of heterochromatin-specific DNA-binding proteins. In mouse embryos in which the Xist homolog has been disrupted, X inactivation does not take place, which demonstrates that Xist is essential for inactivation.

X-chromosome inactivation has two conse-quences. First, it results in dosage compensa-tion. It equalizes the number of active copies of X-linked genes in females and males. Although a female has two X chromosomes and a male has only one, because of inactivation of one X chromosome in each of the somatic cells of a female, both sexes have the same number of active X chromosomes. The mechanism of dos-age compensation by means of X inactivation was originally proposed by Mary Lyon and is called the single-active-X principle (or Lyon hypothesis).

The second consequence of X-chromosome inactivation is that a normal female is a sort of mosaic for X-linked genes. That is, each somatic cell expresses the genes in only one X chromo-some, but the X chromosome that is active genetically differs from one cell to the next. This mosaicism can be observed directly in females that are heterozygous for X-linked alleles that determine different forms of an enzyme, A and B: When cells from a heterozygous female are individually cultured in the laboratory, half of the clones are found to produce only the A form of the enzyme and the other half to produce only the B form. Mosaicism can also be observed

Zygote

Early embryonic cleavage divisions

Both Xchromosomesactive

Inactive X

Somatic cells in female

Active X

Somewhat later in cell development, random X inactivation in each cell lineage

Once a particular X chromosome is inactivated, it remains the inactive X in all descendant cells.

FIGURE 8.7 Schematic diagram of somatic cells of a normal female, showing that the female is a mosaic for X-linked genes. The two X chromosomes are shown in red and blue. An active X is depicted as a straight chromosome, an inactive X as a tangle. Each cell has just one active X, but the particular X that remains active is a matter of chance. In human beings, the inactivation includes all but a few genes in the tip of the short arm.


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ccccccccoooooooonnnnnnnnnnnnnnnneeeeeeeecccccccctttttttttttttiiiiiiiiiiiiiioooooooonnnnnnnn

How do organisms solve the problem that females possess two X chromosomes

whereas males have only one? Unless there were some type of correction (called dosage compensation), the unequal

number would mean that for all the genes in the X chromosome, cells in females would have twice as much gene product as cells in males. It would be dif-ficult for the developing organism to cope with such a large difference in dosage for so many genes. The problem of dosage compensation has been solved by different organisms in different ways. The hypoth-esis put forward in this paper is that in the mouse (and, by inference, in other mammals), the mechanism is very simple: One of the X chromosomes, chosen at ran-dom in each cell lineage early in develop-ment, becomes inactivated and remains inactivated in all descendant cells in the lineage. In certain cells, the inactive X chromosome becomes visible in interphase as a deeply staining “sex-chromatin body.” We now know that there are a few genes in the short arm of the X chromo-some that are not inactivated. There is also good evidence that the inactivation of the rest of the X chromosome takes place sequentially from an “X- inactivation cen-ter.” We also know that in marsupial

mammals, such as the kangaroo, it is always the paternal X chromosome that is inactivated.

It has been suggested that the so-called sex- chro matin body is com-posed of one hetero pyk notic [that is, deeply staining during interphase] X chromosome. . . . The present com-munication suggests that evidence of mouse genetics indicates: (1) that the heteropyknotic X chro-mosome can be either paternal or maternal in origin, in different cells of the same animal; (2) that it is genetically inactivated. The evi-dence has two main parts. First, the normal phenotype of XO fe-males in the mouse shows that only one active X chromo-some is necessary for normal develop-ment, including sexual development. The second piece of evidence concerns the mosaic phenotype of female mice heterozygous for some sex-linked [X-linked] mutants. All sex-linked mutants so far known affecting coat colour cause a “mottled” or “dappled” phenotype, with patches of normal and mutant colour. . . . It is here sug-gested that this mosaic phenotype is due to the inactivation of one or other X chromosome early in embryonic development. If this is true, pigment

cells descended from the cells in which the chromosome carrying the mutant gene was inactivated will give rise to a normal-coloured patch and those in which the chromosome carrying the normal gene was in activated will give rise to a mutant-coloured patch. . . . Thus this hypothesis predicts that for all sex-linked genes of the mouse in which the phenotype is due to local-

ized gene action the het-erozygote will have a mosaic appearance. . . . The genetic evidence does not indicate at what early stage of embryonic develop-ment the inactivation of the one X chromosome occurs. . . . The sex-chromatin body is

thought to be formed from one X chromosome in the rat and in the opossum. If this should prove to be the case in all mammals, then all female mammals heterozygous for sex-linked mutant genes would be expected to show the same phenomena as those in the mouse. The coat of the tortoise-shell cat, being a mosaic of the black and yellow colours of the two homozygous genotypes, fulfills this expectation.

Source: M. F. Lyon, Nature 190 (1961): 372.

Mary F. Lyon 1961Medical Research Council, Harwell, EnglandGene Action in the X Chromosome of the Mouse (Mus musculus L.)

The coat of the tortoiseshell cat, being

a mosaic of the black and yellow colours

of the two homozygous genotypes, fulfills this

expectation.

Lyonization of an X Chromosome


directly in women who are heterozygous for an X-linked recessive mutation that results in the absence of sweat glands; these women exhibit large patches of skin in which sweat glands are present (these patches are derived from embry-onic cells in which the normal X chromosome remained active and the mutant X was inacti-vated) and other large patches of skin in which sweat glands are absent (these patches are derived from embryonic cells in which the nor-mal X chromosome was inactivated and the mutant X remained active).

Although all mammals use X-chromosome inactivation for dosage compensation in females,

the choice of which chromosome to inactivate is not always random. In marsupial mammals, which include the kangaroo, the koala, and the wombat, the X chromosome that is inactivated is always the one contributed by the father. The result is that female marsupials are not genetic mosaics of paternal and maternal X-linked genes. A female marsupial expresses the X-linked genes that she inherited from her mother.

The Calico Cat In nonmarsupial mammals, the result of

random X inactivation in females can some-times be observed in the external phenotype.

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One example is the “calico” pattern of coat coloration in female cats. Two alleles affecting coat color are pres-ent in the X chromosome in cats. One allele results in an orange coat color (some-times referred to as yellow), the other in a black coat color. Because he has only one X chromosome, a nor-mal male has either the orange or the black allele. A female can be heterozygous for orange and black, and in this case the coat color is “calico”—a mosaic of orange and black patches. The orange and black patches result from X-chromosome inactivation. In cell lineages in which the X chromosome bearing the orange allele is inactivated, the X chromosome with the black allele is active and so the fur is black. In cell lineages in which the X chromosome with the black allele is inacti-vated, the orange allele in the active X chromo-some results in orange fur (FIGURE 8.8).

The white patches have a completely dif-ferent explanation. The white patches are due to an autosomal gene S for white spotting, which prevents pigment formation in the cell lineages in which it is expressed. Why the S gene is expressed in some cell lineages and not others is not known. Homozygous SS cats have more white than heterozygous Ss cats. Ginger, the female cat in Figure 8.8, is homo-zygous SS.

Pseudoautosomal InheritanceThe silencing of genes in the inactive X

chromosome evolved gradually as the X and Y chromosomes progressively diverged from their ancestral chromosomes and the Y chromosome began to lose the function of most of its genes. The gene inactivation in the inactive X chromo-some therefore affects individual genes and blocks of genes, and some genes in the inactive X are not silenced. Some of the genes that escape X inactivation have functional homologs in the Y chromosome, whereas others do not.

Two continuous regions that escape X inacti-vation are found at the tips of the long and short arms. These are regions in which the Y chromo-some does retain functional homologous genes. These regions of homology enable the X and Y chromosomes to synapse in spermatogenesis, and

a crossover takes place that holds the chromo-somes together to ensure their proper segrega-tion during anaphase I. The regions of shared X–Y homology define the pseudoautosomal regions: PARp is a 2.7-Mb region at the terminus of the short arms and PARq is a 0.3-Mb region at the terminus of the long arms. Because cross-ing over regularly takes place at least in PARp, the rate of recombination per nucleotide pair is at least twentyfold greater PARp in than in the autosomes.

The pedigree patterns of inheritance of genes in the pseudoautosomal regions are indis-tinguishable from patterns characteristic of autosomal inheritance. The reason is that a mutant allele in a pseudoautosomal region is neither completely X-linked nor completely Y-linked, but can move back and forth between the X and Y chromosomes because of recombi-nation in the pseudoautosomal region. A gene that shows an autosome-like pattern of inheri-tance, but that is known from molecular studies to reside in a pseudoautosomal region, is said to show pseudoautosomal inheritance.

Active Genes in the “Inactive” X ChromosomeIn addition to genes in the pseudoautoso-

mal regions PARp and PARq, some other genes in the “inactive” X chromosome also show tran-scriptional activity. In particular, a study of the level of transcription of 624 genes along the human X chromosome has shown that about 15 percent of the X-linked genes escape inacti-vation to some degree. The “inactive X” is there-fore not completely silenced. The location of the partially active genes along the chromosome is

FIGURE 8.8 This cat’s name is Ginger. She is a female that is heterozygous for the orange and black coat color alleles and shows the classic “calico” pattern of patches of orange and black fur.


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shown in FIGURE 8.9. The transcribed genes occur in large blocks that tend to be located in the distal portions of the arms, especially the short arm, a pattern suggesting that escape from X inactivation may be correlated with distance from Xist.

In any event, most of the genes that escape inactivation have levels of transcription that range from 15–50 percent of those observed for their homologs in the active X chromosome; hence the level of activity is robust. The number of genes that escape complete X inactivation, and their levels of transcription, could readily account for some of the differences in expression of traits between males and females, for phenotypic vari-ation among females heterozygous for such X-linked mutations as hemophilia A, and for phenotypic variation among individuals with abnormal numbers of X chromosomes.

Gene Content and Evolution of the Y ChromosomeComparative cytogenetic and molecular

studies suggest that the X and Y chromo-somes began their existence as a pair of

ordinary autosomes in the common ancestor of modern mammals and birds. They started to diverge in DNA sequence and gene con-tent at about the same time that the evolu-tionary lineage of mammals diverged from that of birds, some 300–350 million years ago (MY). One must assume that prior to this time, recombination took place at normal levels throughout the entire proto-X and proto-Y chromosomes and that their gene contents were identical.

In the human genome as it exists today, the Y chromosome includes genes that encode about 26 distinct proteins, many of which are important for male fertility. The Y chromosome is also populated with many repeat sequences, some extremely long, which can undergo gene conversion and serve as sites for homologous recombination. One of the key genes in the Y chromosome is the master sex-determining gene SRY at position Yp11.3. This band is in the short arm of the Y, near but not included in the pseudoautosomal region. The gene SRY codes for a protein transcription factor, the testis-determining factor (TDF), which directs embryonic development toward the male sex by inducing the undifferentiated embryonic genital ridge, the precursor of the gonad, to develop as a testis. A transcription factor is a protein that stimulates transcription of its tar-get genes.

Once SRY had evolved as a sex-determin-ing mechanism, the Y chromosome began to diverge in DNA sequence from the X chromo-some, and the region of possible X–Y recombi-nation became progressively restricted to the telomeric regions. In regions with no X–Y recombination, there is a steady selection pres-sure for genes in the Y chromosome to undergo mutational degeneration into nonfunctional states. This results from the forced heterozy-gosity of the Y chromosome, which allows multiple deleterious mutations to accumulate through time because there is no opportunity for recombination to regenerate Y chromo-somes that are free of deleterious mutations. Hence any Y-linked gene whose function is nonessential will tend to degenerate gradually because of the accumulation of mutations, and at the same time there will be selection pres-sure for dosage compensation of the homolo-gous gene in the X chromosome. Eventually, only the dosage-compensated X-linked gene will remain functional.

Apparently blocks of genes were removed from the region of X–Y recombination in large chunks. Molecular evidence for this conclusion

Xp

Xcen

Xq

Pseudoautosomalregion

Active genes

Inactivatedgenes

Gene activity may differ from one inactive X chromosome to another.

About 15% of X-linked genes escape inactivation to some degree.

The preponderance of genes that escape X inactivation are in the short arm concentrated near the pseudoauto-somal region.

About 10% of X-linked genes show such variable patterns of inactivation.

FIGURE 8.9 Distribution along the human X chromosome of the approximately 15 percent of genes that escape complete transcrip-tional silencing in the inactive X chromosome. [Data from L. Carrel and H. F. Willard, Nature 434 (2005): 400–404.]

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is summarized in FIGURE 8.10. Shown at the left are the locations of some protein-coding sequences in the short arm of the mod-ern X chromosome from band Xp11 to the telomere. All of these genes have homologous sequences in the modern Y chromosome, as shown at the right. The amount of sequence divergence between the X and Y homologs shows a remarkable pattern. In the synonymous positions of the codons, where a nucleotide sub-stitution can occur without changing the encoded amino acid, the sequence divergence (measured, roughly speaking, as the proportion of synonymous sites that differ between the X and Y homologs) for the genes GYG2–AMELX is 0.07–0.11, for TB4X–UTX it is 0.23–0.36, for SMCX it is 0.52, and for other genes outside the region shown it is "0.94. Because the evolu-tionary rate of nucleotide substitutions at syn-onymous sites in mammalian genes is known, we can say that these levels of divergence cor-respond to divergence times of 30–50 million years (MY) for GYG2–AMELX, 80–130 MY for TB4X–UTX, 130–170 MY for SMCX, and

300–350 MY for other genes outside the region shown. The simplest explanation is that these were the times when successive blocks of genes were removed from the region of X–Y recombi-nation by chromosome rearrangements, namely, inversions, in which a region some-where in the interior of the Y chromosome becomes reversed in orientation. One way in which such an inversion can take place is by means of homologous recombination between two repeated DNA sequences that are present in opposite orientations at different locations in the Y chromosome, a mechanism that is examined in greater detail in Section 8.4. Whether the inversions actually took place via homologous recombination or were generated by some other process of breakage and reunion is not known. However, the inversions happened, the evolu-tionary reconstruction in Figure 8.10 shows that the 130–170 MY inversion breakpoint was adja-cent to UTY, the 80–130 MY inversion break-point adjacent to AMELY, and the 30–50 MY year breakpoint adjacent to the present-day pseudoautosomal region on the short arm. As

PARp

PARp PARp PARp PARp PARp

PARq

SRYZFY

ZFY

PRKY

PRKY

AMELY

AMELY ARSEP

ARSEP

ARSDP

ARSDP

GYG2P

GYG2P

DFFRY

DFFRY

DBY

DBY

CASKP

CASKP

UTY

UTY

TB4Y

TB4Y

KALP

KALP

STSP

STSPPRKY

AMELY

ARSEPARSDPGYG2P

KALPSTSP

SMCY

SMCY

EIF1AY

EIF1AYZFY

DFFRYDBY

CASKPUTY

TB4Y

PRKY

AMELY

ARSEPARSDPGYG2P

KALPSTSP

EIF1AY

SMCX

ZFX

DFFRXDBXCASKUTX

TB4X

PRKX

AMELX

ARSEARSDGYG2

KAL1STS

EIF1AX

Modern Ychromosome

Modern Xchromosome

Inversion(30–50 MY)



BREAK

BREAK

BREAK

Ancient Ychromosome(300–350 MY)

Shading indicates region of X–Y recombination.

Xp11

FIGURE 8.10 Progressive shortening of the mammalian X!Y pseudoautosomal region through time due to inversions in the Y chromosome, inferred from DNA sequence data. The arrows denote the distal (nearest the telemere) breakpoint of each successive inversion interrupting the pseudoauto-somal region. [Data from B. T. Lahn and D. C. Page, Science 286 (1999): 964–967.]


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each of these inversions was fixed in the evolv-ing Y chromosome, it removed the correspond-ing block of genes from the region of X–Y recombination, so that the rate of sequence divergence between the X and Y homologs accelerated. Other rearrangements in the Y chromosome, not able to be traced from these data, led to some additional scrambling of the gene order in the Y chromosome.

8.2 Chromosome Abnormalities in Human Pregnancies

Approximately 15 percent of all recognized pregnancies in human beings terminate in spontaneous abortion, and in about half of all spontaneous abortions, the fetus has a major chromosome abnormality. TABLE 8.2 summarizes the average rates of chromosome abnormality found per 100,000 recognized pregnancies in several studies. The term trisomy refers to an otherwise diploid organism that has an extra

copy of an individual chromosome. Many of the spontaneously aborted fetuses have trisomy of one of the autosomes. Triploids, which have three sets of chromosomes (total count 69), and tetraploids, which have four sets of chromo-somes (total count 92), are also common in spontaneous abortions. Triploids and tetraploids are examples of euploid conditions, because they have the same relative gene dosage as found in the diploid. In contrast, relative gene dosage is upset in a trisomic, because three cop-ies of the genes located in the trisomic chromo-some are present, whereas two copies of the genes in the other chromosomes are present. Such unbalanced chromosome complements are said to be aneuploid. Although it is not apparent in the data in Table 8.2, in most organ-isms, euploid abnormalities generally have less severe phenotypic effects than an euploid abnor-malities. In Table 8.2 the term balanced transloca-tion refers to a euploid condition in which nonhomologous chromosomes have an inter-change of part, but all of the parts are present; the term unbalanced translocation refers to an aneuploid condition in which some part of the genome is missing. The much greater survivor-ship of the balanced translocation indicates that a euploid chromosomal abnormality is generally less harmful than an aneuploid chromosome abnormality.

When an otherwise diploid organism has a missing copy of an individual chromosome, the condition is known as monosomy. In most organisms, chromosome loss (resulting in monosomy) is a more frequent event than chromosome gain (resulting in trisomy). How-ever, monosomies are conspicuously absent in the data on spontaneous abortions in Table 8.2. Their absence is undoubtedly due to another feature of monosomy:

A missing copy of a chromosome (monosomy) usually results in more harmful effects than an extra copy of the same chromosome (trisomy).

In human fertilizations, monosomic zygotes are probably created in even greater numbers than trisomic zygotes, but monosomy is not found among aborted fetuses in Table 8.2 because the abortions take place so early in development that the pregnancy goes unrecognized by the mother. Data relevant to very early abortions come from medical records of women attempting to become pregnant who while trying to con-ceive undergo a pregnancy hormone test every day. The hormone assayed is human chorionic gonadotropin, a glycoprotein first produced by

Chromosome abnormalities per 100,000 recognized human pregnancies

Table 8.2

15,000 spontaneous abortions 85,000 live births

Trisomy1

A: 2 3

B: 4 5 C: 6–12 13 D: 14 15 16 E: 17 18 F: 19–20

G: 21 22

0 159 53 95 0 561 128 275 318 1229 10 223 52 350 424

0 0 0 0 0 0 17 0 0 0 0 13 0 113 0

Sex chromosomes XYY XXY XO XXX

4 4 1350 21

46 44 8 44

Translocations Balanced Unbalanced

14 225

164 52

Polyploid Triploid Tetraploid

1275 450

0 0

Other (mosaics, etc.) 280 49Total 7500 550

!"#"$

!#$

!"#"$

!"#"$

!#$

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8.2 Chromosome Abnormalities in Human Pregnancies 271

abnormal gamete in Down syndrome is usually the egg. Furthermore, the risk of non-disjunction of chromosome 21 increases dramatically with the age of the mother, result-ing in a risk of Down syn-drome that reaches 6 percent in mothers age 45 and older (FIGURE 8.11). For this reason, many physicians recommend that older women who are pregnant have cells from the fetus tested in order to detect Down syndrome prenatally. This can be done 15–16 weeks after fertilization by amnio-centesis, in which cells of the developing fetus are obtained by insertion of a fine needle through the wall of the uterus and into the sac of fluid (the amnion) that contains the fetus, or it can be done at 10–11 weeks after fertilization by a procedure called chorionic villus sampling (CVS), which uses cells from a zygote-derived embryonic membrane (the chorion) associated with the placenta. Early diagnosis is desirable, but CVS has about a threefold higher risk of induc-ing a miscarriage than does amniocentesis.

About 3 percent of all cases of Down syn-drome are due not to simple nondisjunction, but to an abnormality in chromosome structure. In these cases the risk of recurrence of the syn-drome in subsequent children is very high—up to 20 percent of births. The high risk is caused by a chromosomal translocation in one of the par-ents, in which chromosome 21 has been broken and become attached to another chromosome. This situation is considered in Section 8.5.

Trisomic SegregationIn a trisomic organism, the segregation of

chromosomes in meiosis is upset because the trisomic chromosome has two pairing partners instead of one. The behavior of the chromo-somes in meiosis depends on the manner in which the homologous chromosome arms pair and on the chiasmata formed between them. In some cells, the three chromosomes form a trivalent in which distinct parts of one

Inci

denc

e of

Dow

n S

yndr

ome

(per

cent

of l

ive

birt

hs)

Maternal age, years

0.2

10.0

3.0

2.0

1.0

0.3

0.1

0.03

15 20 25 30 40 45 5035

FIGURE 8.11 Frequency of Down syndrome (number of cases per 100 live births) related to age of mother. The graph is based on 438 Down syndrome births (among 330,859 total births) in Sweden in the period 1968 to 1970. [Data from E. B. Hook and A. Lindsjö, Am. J. Hum. Genet. 30 (1978): 19–27.]

chromosome are paired with homologous parts of each of the others (FIGURE 8.12, part A). In metaphase, the trivalent is usually oriented with two centromeres pointing toward one pole and the other centromere pointing toward the other pole. The result is that at the end of both meiotic divisions, one pair of gametes contains two copies of the trisomic chromosome, and the other pair of gametes contains only a single copy. Alternatively, the trisomic chromosome can form one normal bivalent and one univalent, or unpaired chromosome, as shown in Figure 8.12B. In anaphase I, the bivalentdisjoins normally and the univalent usually pro-ceeds randomly to one pole or the other. Again, the end result is the formation of two products of meiosis that contain two copies of the trisomic chromosome and two products of meiosis that contain one copy. To state the matter in another way, a trisomic organism with three copies of a chromosome (say, CCC) will produce gametes among which half contain two copies (CC) and

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half contain one (C). When mated with a chro-mosomally normal individual, a trisomic is therefore expected to produce trisomic or nor-mal progeny in a ratio of 1 : 1. This theoretical expectation is borne out in experimental organ-isms by matings of trisomics and in human beings by the finding that, in the few known cases when a person with Down syndrome reproduces, the frequency of Down syndrome in the offspring is approximately 50 percent.

Sex-Chromosome Abnormalities Sex-chromosome trisomies, as a group, are

even more frequent among newborns than is tri-somy 21 (Table 8.2). There are two reasons why extra sex chromosomes have phenotypic effects that are relatively mild compared with those of autosomal trisomies. First, the single-active-X principle results in the silencing of most X-linked

genes in all but one X chromosome in each somatic cell. Second, the Y chromosome contains relatively few functional genes.

The four most common types of sex-chromosome abnormalities are described below. The karyotypes are given in the con-ventional fashion, with the total number of chromosomes listed first, followed by the sex chromosomes that are present. For example, in the designation 47,XXX the number 47 refers to the total number of chromosomes, and XXX indicates that the person has three X chromosomes.

• 47,XXX This condition is often called simply trisomy-X. People with the karyotype 47,XXX are female. Most such women are mentally and phenotypic-ally normal and give no physical or behavioral indication that an extra

(B)(A)

Two centromeresgo to one pole; onecentromere goes toother pole.

Chromosomes form one bivalent and one univalent

Chromosomesform a trivalent

Result: Two gametes with two copies of chromosome,two gametes with one copy of chromosome.

Spindle fibers

Bivalent

UnivalentTrivalent

Metaphase I

Anaphase I

Final products of meiosis Final products of meiosis

FIGURE 8.12 Meiotic pairing in a trisomic. (A) Forma tion of a trivalent. (B) Formation of a univalent and a bivalent. Both types of pairing result in one pair of gametes containing two copies of the trisomic chromo some and the other pair of gametes containing one copy of the trisomic chromo some.

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X chromosome is present. Mild mental impairment may be somewhat more fre-quent than among 46,XX females.

• 47,XYY This condition is often called the double-Y karyotype. These people are male and tend to be taller than average, but they are otherwise phenotypically normal. At one time it was thought that 47,XYY males developed severe person-ality disorders and were at a high risk of committing crimes of violence, a belief based on the finding of 47,XYY among violent criminals. Further study indi-cated that the frequency of 47,XYY males in the general population was much higher than had been supposed. Some 47,XYY males have slightly impaired mental function. Although their rate of criminality is higher than that of normal males, the crimes are mainly nonviolent crimes such as petty theft. The majority of 47,XYY males are phenotypically and psychologically nor-mal, have mental capabilities in the normal range, and have no criminal convictions.

• 47,XXY This condition is called Kline-felter syndrome. Affected persons are male. They tend to be tall, do not undergo normal sexual maturation, are sterile, and in some cases have enlarge-ment of the breasts. Mild mental impair-ment is common.

• 45,X Monosomy of the X chromosome in females is called Turner syndrome. Affected persons are phenotypically female, but short in stature and without sexual maturation. Mental abilities are typically within the normal range. Note in Table 8.2 that more than 99 percent of 45,X fetuses undergo spontaneous abor-tion. This indicates that the condition usually has profound deleterious effects on development. Table 8.2 also shows that monosomy of the X chromosome is much more frequent than trisomy, which is almost certainly true of autosomes as well, in spite of the absence of monoso-mics in recognized pregnancies.

Environmental Effects on NondisjunctionBecause a large fraction of aneuploid zygotes

result in miscarriage or in live births with congeni-tal defects or mental retardation, the identification of environmental hazards that may increase the incidence of meiotic errors is of great importance. Environmental risk factors that have been

suggested include radiation, smoking, alcohol consumption, oral contraceptives, fertility drugs, environmental pollutants, pesticides, and so forth. When significant effects have been found, they are usually small and not always reproducible, in part due to confounding effects of other factors such as maternal age. In view of the maternal-age effect, the female sex hormone estrogen and mol-ecules resembling estrogen have long been under suspicion. In view of this background, it was no great surprise to learn that modest concentrations of a common estrogen mimic known as bisphenol A [technical name 2,2-(4,4-dihydroxy-diphenol)propane] caused about an eightfold increase in the incidence of aneuploidy in mice.

Bisphenol A is the basic subunit of polycar-bonate plastic products widely used as a can liner in the food and beverage industry. In its polymerized form it may be completely harm-less, but the monomers can leach out of plastic products that are damaged. The effect of bisphe-nol A was discovered by chance in studies of aneuploidy in the mouse. In the course of microscopic studies, the researchers noted a sudden increase in the incidence of meiotic cells in which one or more chromosomes failed to align properly on the metaphase II plate. At the same time the incidence of aneuploid oocytes increased from 1.4 percent to 11.6 percent.

The investigators succeeded in tracing the cause to the inadvertent use of a harsh alkaline detergent to wash the polycarbonate cages and water bottles. They suspected that the detergent was causing bisphenol A monomers to leach out of the plastic, and follow-up experiments with bisphenol A monomers demonstrated its effect on aneuploidy directly. Although the studies were carried out in female mice, the similarity of the meiotic process in female humans is a matter of concern. More generally the study raises a warning flag about environmental agents that may increase the rate of aneuploidy, particularly agents that have estrogen-like activity.

8.3 Chromosomal Deletions and Duplications

We now turn to the genetic implications of aberrations in chromosome structure. In the his-tory of genetics, chromosome abnormalities were originally discovered through their genetic effects, which, though confusing at first, even-tually came to be understood and were later confirmed directly by microscopic observations. There are several principal types of structural aberrations—deletions, duplications, inversions,

8.3 Chromosomal Deletions and Duplications 273

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8.1 What is the biological function of the obligatory crossover that occurs between the X and Y chromo-some during meiosis in the pseudoautosomal region?

8.2 What is the genetic consequence of the obligatory crossover that occurs between the X and Y chromo-some during meiosis in the pseudoautosomal region?

8.3 In certain types of white blood cells, the inactive X chromosome in females can be observed microscopi-cally as a densely staining heterochromatic body in the nucleus of the cells during interphase. This densely staining blob is called a Barr body. In cells with more than one inactivated X chromosome, the number of Barr bodies equals the number of X chro-mosomes that are inactivated. How many Barr bodies would be present in white blood cells of an individual of karyotype 49, XXXYY?

8.4 How many Barr bodies would be present in each of the following human conditions?(a) Klinefelter syndrome(b) Turner syndrome(c) Down syndrome(d) 47, XYY(e) 47, XXX

8.5 Why is it not completely correct to refer to the X chromosome that forms a Barr body as “inactive” in a normal human female?

8.6 A spontaneously aborted human fetus is found to have 45 chromosomes. What is the most probable karyotype? Had the fetus survived, what genetic dis-order would it have had?

8.7 A spontaneously aborted human fetus was found to have the karyotype 92, XXYY. If the zygote resulted

crossing over within the inversion loop. In the simplified version, only the inverted region is shown paired, the nor-mal regions unpaired. In the case of a paracentric inver-sion, this means that the centromeres are on opposite sides. (In real cells, of course, the inverted region forms a loop.) In this diagram it is easy to see that a crossover at the position of the arrow results in a dicentric chromo-some and an acentric fragment (in addition to the normal and inverted chromosomes that do not participate in the crossover). Deduce the consequence of a two-strand dou-ble crossover when the first one takes place as shown and the second one takes place between the genes C and D.

Answer The result is one normal chromosome and one inverted chromosome (these derive from chromatids not involved in either crossover), a normal chromosome car-rying the alleles A B c D E, and an inverted chromosome carrying the alleles a b C d e. Note that the two-strand dou-ble crossover has transferred an allele between the normal and the inverted chromosomes.

Indicate what types of gametes would be expected, and in what proportions, from an organism that is:(a) Trisomic for a single chromosome.(b) Monosomic for a single chromosome.

ANSWER

(a) In a trisomic organism, either two of the chromosomes will form a pair and the other remain unpaired, or all three chromosomes will come together, each chromo-some pairing along part of its length with both of the others. These situations are shown in the illustration. The anaphase I segregation determines the chromo-somes included in the gametes. In either type of pair-ing, two gametes receive two copies of the chromosome and two gametes receive one copy. Therefore, trisomic segregation leads to disomic or monosomic gametes in an expected ratio of 1 : 1.

(b) In a monosomic organism, the monosomic chromo-some must remain unpaired. At anaphase I, it goes to one pole or the other. The end result is two gametes containing no copies of the chromosome and two gametes containing one copy. Therefore, monosomic segregation leads to nullisomic (missing the chromo-some) gametes or monosomic gametes in an expected ratio of 1 : 1.

aa

AA

BB

CC

DD E

E

bb

ee

cc

dd

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from the fusion of normal X-bearing and Y-bearing gametes, what might have happened to the chromo-somes in the zygote to result in this karyotype?

8.8 Color blindness in human beings is an X-linked trait. A man who is color blind has a 45, X (Turner syn-drome) daughter who is also color blind. Did the nondisjunction that led to the 45, X child occur in the mother or the father? Explain the evidence support-ing your answer.

8.9 The long arm of the normal human Y chromosome contains a large region that is duplicated. In some mutant Y chromosomes the region between the duplications is missing, and only one copy of the duplication is present. How can this observation be explained? Does the observation tell you anything about how the duplications are oriented in the chromosome?

8.10 A chromosome has the gene sequence A B C D E F G. What is the sequence following a C-through-E inver-sion? Following a C-through-E deletion? Two chro-mosomes with the sequences A B C D E F G and M N O P Q R S T U V undergo a reciprocal translocation after breaks in E F and S T. What are the possible products? Which products are genetically stable?

8.11 A recessive mutation in the human genome results in a condition called anhidrotic ectodermal dysplasia, which is associated with an absence of sweat glands. The condition can be detected by studies of the elec-trical conductivity of the skin, because skin without sweat glands has a lower electrical conductivity (higher resistance) than normal skin. In kinships in which the recessive allele is segregating, affected males are found to show low conductance uniformly across their skin surface, as do affected females. How-ever, heterozygous females show a mosaic pattern with normal conductance in some patches of skin and low conductance in others. The pattern of tissue lacking sweat glands is different for each mosaic female examined. How could this pattern of gene expression be explained?

8.12 A phenotypically normal woman has a child with Down syndrome. The woman is found to have 45 chromosomes. What kind of chromosome abnormal-ity can account for these observations? How many chromosomes does the affected child have? How does this differ from the usual chromosome number and karyotype of a child with Down syndrome?

8.13 The vast majority of progeny from either 47, XXX females or 47, XYY males are karyotypically normal 46, XX or 46, XY. Is this finding the one expected?

8.14 Genes a, b, c, d, e, and f are closely linked in a chromo-some, but their order is unknown. Three deletions in the region are found to uncover recessive alleles of the genes as follows: Deletion 1 uncovers a, b, and d. Deletion 2 uncovers a, d, c, and e. Deletion 3 uncovers e and f. What is the order of the genes? In this prob-lem, you will see that there is enough information to order most, but not all, of the genes. Suggest what experiments you might carry out to complete the ordering.

8.15 Recessive genes a, b, c, d, e, and f are closely linked in a chromosome, but their order is unknown. Four deletions in the region are examined. One deletion uncovers a, c, and f; another uncovers b and c; the third uncovers d and f; and the fourth uncovers d, e, and f. What is the order of the genes?

8.16 Recessive genes a, b, c, d, e, and f are closely linked in a chromosome, but their order is unknown. Three deletions in the region are examined. One deletion uncovers a, d, and e; another uncovers c, d, and f; and the third uncovers b and c. What is the order of the genes?

8.17 Four strains of Drosophila melanogaster are isolated from different localities. The banding patterns of a particular region of salivary chromosome 2 have the following configurations (each letter denotes a band).(a) a b f e d c g h i j (c) a b f e h g i d c j(b) a b c d e f g h i j (d) a b f e h g c d i j

Assuming that part c represents the ancestral sequence, deduce the evolutionary ancestry of the other chromosomes.

8.18 Two species of Australian grasshoppers coexist side by side. In meiosis, each has 5 pairs of chromosomes. When the species are crossed, the chromosomes in the hybrid form 3 pairs of chromosomes and one group of 4. What alteration in chromosome structure could account for these results?

8.19 Why are translocation heterozygotes semisterile? Why are translocation homozygotes fully fertile? If a translocation homozygote is crossed with an indi-vidual that has normal chromosomes, what fraction of the F1 progeny is expected to be semisterile?

8.20 Asiatic wild cotton (Gossypium arboreum) has 13 pairs of chromosomes, and an American wild species (G. thurberi) also has 13 pairs of chromosomes. Inter-specific crosses between the species are sterile because of highly irregular chromosome pairing in meiosis. The American cultivated cotton (G. hirsutum) has 26 pairs of chromosomes and is fully fertile. Crosses between G. arboreum and G. hirsutum and crosses between G. thurberi and G. hirsutum produce plants that, in meiosis, exhibit 13 pairs of chromosomes (bivalents) and 13 chromosomes with no pairing partner (univalents). What do these cytological data tell us about the genetic origin of present-day Ameri-can cultivated cotton?

8.21 From how many possible pathways of hybridization and endoreduplication could the diploid species with genomes AA, BB, and CC yield an allohexaploid with the genome AABBCC? Each of A, B, and C stands for one monoploid set of chromosomes. (One such path-way, for example, is (AA ! BB) ! CC, which means that A and B underwent hybridization and polyploidi-zation to yield the allotetraploid AABB, which then underwent hybridization and polyploidization with CC to yield the allohexaploid AABBCC.)

8.22 What types of hybridization and endoreduplication could account for a polyploid species with the genome composition AABBBBCC, where the diploid ancestors have genome compositions AA, BB, and CC,

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oeb50 readings probs x-linkage

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