PPT-2 The Chromosomal Basis of Inheritance

15

The Chromosomal Basis

of Inheritance

▲ Figure 15.1 Where are Mendel’s hereditary factors located in the cell?

Locating Genes Along Chromosomes

T

oday, we know that genes—Mendel’s “factors”—are segments of DNA located along chromosomes. We can see the location of a particular gene by tagging chromosomes with a fluorescent dye that highlights that gene. For example, the two yellow spots in Figure 15.1 mark a specific gene on human chromosome 6. (The chromosome has duplicated, so the allele on that chromosome is present as two copies, one per sister chromatid.) However, Gregor Mendel’s “hereditary fac-tors” were purely an abstract concept when he proposed their existence in 1860. At that time, no cellular structures had been identified that could house these imaginary units, and most biologists were skeptical about Mendel’s proposed laws of inheritance.

Using improved techniques of microscopy, cytologists worked out the process of mitosis in 1875 (see the drawing at the lower left) and meiosis in the 1890s. Cytology and genetics converged as biologists began to see parallels between the behavior of Mendel’s proposed hereditary factors during sexual life cycles and the behavior of chromosomes: As shown in Figure 15.2, chromosomes and genes are both present in pairs in diploid cells, and homologous chromosomes separate and alleles segregate during the process of meiosis. Furthermore, after meiosis, fertiliza-tion restores the paired condition for both chromosomes and genes.

K e y

C o n C e p t s

15.1 Morgan showed that Mendelian

inheritance has its physical basis in the behavior of

chromosomes: Scientific inquiry

15.2 Sex-linked genes exhibit unique

patterns of inheritance

15.3 Linked genes tend to be

inherited together because they are located near each other on the same chromosome

15.4 Alterations of chromosome

number or structure cause some genetic disorders

15.5 Some inheritance patterns

are exceptions to standard Mendelian inheritance

292

C h a p t e r 15 The

Chromosomal Basis of Inheritance

293Gametes

Meiosis

The R and r alleles segregate

at anaphase I , yielding two types of daughter cells for this locus.

R y

Y

r R y

Y

r R

Two equally probable arrangements of chromosomes at metaphase I

r

Y R

y

y

R

Y

r

r

y R

Y

Y

R R

y y Y

Y r

r

r

R

y R r

Y YR

yr

Yr

yR

Anaphase I

Metaphase II

y

Y r

R y

Y R

r Each gamete

gets one long chromosome with either the R or r allele.Alleles at both loci segregate in

anaphase I , yielding four types of daughter cells, depending on the chromosome arrangement at metaphase I . Compare the arrangement of the R and r alleles relative to the Y and y alleles in anaphase I .Each gamete gets

a long and a short chromosome in one of four allele combinations.

y F 1 Generation

Y r

y 1

4

14

14

14

Y y

y

R Y r

r y r

R Y R

P Generation

Yellow-round

seeds (YYRR )Green-wrinkled seeds (yyrr )

All F 1 plants produce

yellow-round seeds (YyRr ).

Gametes

Meiosis Fertilization

Starting with two true-breeding pea

plants, we will follow two genes

through the F 1 and F 2 generations.

The two genes specify seed color (allele Y for yellow and allele y for green) and seed shape (allele R for round and allele r for wrinkled). These two genes are on different

chromosomes. (Peas have seven chromosome pairs, but only

two pairs are illustrated here.)9: 3: 3: 1

An F 1 × F 1 cross-fertilization

Fertilization

recombines the R and r alleles at random.

Fertilization results in

the 9:3:3:1

phenotypic ratio in the F 2 generation.

F 2 Generation ×

▲ Figure 15.2 The chromosomal basis of Mendel’s laws. Here we correlate the results of one of Mendel’s dihybrid crosses (see Figure 14.8) with the behavior of chromosomes during meiosis (see Figure 13.8). The arrangement of chromosomes at metaphase I of meiosis and their movement during anaphase I account, respectively, for the independent assortment and segregation of the alleles for seed color and shape. Each cell that undergoes meiosis in an F 1 plant produces two kinds of gametes. If we count the results for all cells, however, each F 1 plant produces equal numbers of all four kinds of gametes because the alternative chromosome arrangements at metaphase I are equally likely.

If you crossed an F 1 plant with a plant that was homozygous recessive for both genes (yyrr ), how

would the phenotypic ratio of the offspring compare with the 9:3:3:1 ratio seen here?

294 U n i t t h r e e Genetics

produce hundreds of offspring, and a new generation can be bred every two weeks. Morgan’s laboratory began using this convenient organism for genetic studies in 1907 and soon became known as “the fly room.”

Another advantage of the fruit fly is that it has only four pairs of chromosomes, which are easily distinguishable with a light microscope. There are three pairs of autosomes and one pair of sex chromosomes. Female fruit flies have a pair of homologous X chromosomes, and males have one X chromosome and one Y chromosome.

While Mendel could readily obtain different pea varieties from seed suppliers, Morgan was probably the first person to want different varieties of the fruit fly. He faced the te-dious task of carrying out many matings and then micro-scopically inspecting large numbers of offspring in search of naturally occurring variant individuals. After many months of this, he complained, “Two years’ work wasted. I have been breeding those flies for all that time and I’ve got nothing out of it.” Morgan persisted, however, and was finally rewarded with the discovery of a single male fly with white eyes in-stead of the usual red. The phenotype for a character most commonly observed in natural populations, such as red eyes in Drosophila , is called the wild type (Figure 15.3). Traits that are alternatives to the wild type, such as white eyes in Drosophila , are called mutant phenotypes because they are due to alleles assumed to have originated as changes, or mu-tations, in the wild-type allele.

Morgan and his students invented a notation for symbol-izing alleles in Drosophila that is still widely used for fruit flies. For a given character in flies, the gene takes its symbol from the first mutant (non–wild type) discovered. Thus, the allele for white eyes in Drosophila is symbolized by w. A su-perscript + identifies the allele for the wild-type trait: w + for the allele for red eyes, for example. Over the years, a variety of gene notation systems have been developed for different organisms. For example, human genes are usually written

Around 1902, Walter S. Sutton, Theodor Boveri, and oth-ers independently noted these parallels and began to develop the chromosome theory of inheritance . According to this theory, Mendelian genes have specific loci (positions) along chromosomes, and it is the chromosomes that undergo segregation and independent assortment. As you can see in Figure 15.2, the separation of homologs during anaphase I accounts for the segregation of the two alleles of a gene into separate gametes, and the random arrangement of chromo-some pairs at metaphase I accounts for independent assort-ment of the alleles for two or more genes located on different homolog pairs. This figure traces the same dihybrid pea cross you learned about in Figure 14.8. By carefully studying Figure 15.2, you can see how the behavior of chromosomes during meiosis in the F 1 generation and subsequent random fertiliza-tion give rise to the F 2 phenotypic ratio observed by Mendel.In correlating the behavior of chromosomes with that of genes, this chapter will extend what you learned in the past two chapters. First, we’ll describe evidence from the fruit fly that strongly supported the chromosome theory. (Although this theory made a lot of sense, it still required experimental evidence.) Next, we’ll explore the chromosomal basis for the transmission of genes from parents to offspring, including what happens when two genes are linked on the same chro-mosome. Finally, we will discuss some important exceptions

to the standard mode of inheritance.

inheritance has its physical basis in the behavior of chromosomes: Scientific inquiry

The first solid evidence associating a specific gene with a specific chromosome came early in the 20th century from the work of Thomas Hunt Morgan, an experimental em-bryologist at Columbia University. Although Morgan was initially skeptical about both Mendelian genetics and the chromosome theory, his early experiments provided con-vincing evidence that chromosomes are indeed the location of Mendel’s heritable factors.

Morgan’s Choice of Experimental Organism

Many times in the history of biology, important discover-ies have come to those insightful or lucky enough to choose an experimental organism suitable for the research prob-lem being tackled. Mendel chose the garden pea because a number of distinct varieties were available. For his work, Morgan selected a species of fruit fly, Drosophila melano-gaster , a common insect that feeds on the fungi growing on

fruit. Fruit flies are prolific breeders; a single mating will

▲ Figure 15.3 Morgan’s first mutant. Wild-type Drosophila flies have red eyes (left). Among his flies, Morgan discovered a mutant male with white eyes (right). This variation made it possible for Morgan to trace a gene for eye color to a specific chromosome (LMs).

C h a p t e r 15 The Chromosomal Basis of Inheritance 295

in all capitals, such as HD for the allele for Huntington’s disease.

Correlating Behavior of a Gene’s Alleles with Behavior of a Chromosome Pair

Morgan mated his white-eyed male fly with a red-eyed fe-male. All the F 1 offspring had red eyes, suggesting that the wild-type allele is dominant. When Morgan bred the F 1 flies to each other, he observed the classical 3:1 phenotypic ratio among the F 2 offspring. However, there was a surprising ad-ditional result: The white-eye trait showed up only in males. All the F 2 females had red eyes, while half the males had red eyes and half had white eyes. Therefore, Morgan concluded that somehow a fly’s eye color was linked to its sex. (If the eye-color gene were unrelated to sex, half of the white-eyed flies would have been male and half female.)

Recall that a female fly has two X chromosomes (XX), while a male fly has an X and a Y (XY). The correlation between the trait of white eye color and the male sex of the affected F 2 flies suggested to Morgan that the gene involved in his white-eyed mutant was located exclusively on the X chromosome, with no corresponding allele present on the Y chromosome. His reasoning can be followed in Figure 15.4. For a male, a single copy of the mutant allele would confer white eyes; since a male has only one X chromosome, there can be no wild-type allele (w +) present to mask the recessive allele. However, a female could have white eyes only if both her X chromosomes carried the recessive mutant allele (w ). This was impossible for the F 2 females in Morgan’s experiment because all the F 1 fathers had red eyes, so each F 2 female received a w + allele on the X chro-mosome inherited from her father.

Morgan’s finding of the correlation between a particular trait and an individual’s sex provided support for the chro-mosome theory of inheritance: namely, that a specific gene is carried on a specific chromosome (in this case, an eye-color gene on the X chromosome). In addition, Morgan’s work indicated that genes located on a sex chromosome exhibit unique inheritance patterns, which we will discuss in the next section. Recognizing the importance of Morgan’s early

work, many bright students were attracted to his fly room.

Experiment Thomas Hunt Morgan wanted to analyze the behavior

of two alleles of a fruit fly eye-color gene. In crosses similar to those done by Mendel with pea plants, Morgan and his colleagues mated a wild-type (red-eyed) female with a mutant white-eyed male.

Morgan then bred an F 1 red-eyed female to an F 1 red-eyed male to produce the F 2 generation.

Results The F 2 generation showed a typical Mendelian ratio of 3 red-

eyed flies : 1 white-eyed fly. However, all white-eyed flies were males; no females displayed the white-eye trait.

Conclusion All F 1 offspring had red eyes, so the mutant white-eye

trait (w ) must be recessive to the wild-type red-eye trait (w +). Since the recessive trait—white eyes—was expressed only in males in the F 2 generation, Morgan deduced that this eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome.

Source: T. H. Morgan, Sex-limited inheritance in Drosophila, Science 32:120–122

(1910).

A related Experimental Inquiry Tutorial can be assigned in MasteringBiology.

Suppose this eye-color gene were located on an auto-

2 flies in this

hypothetical cross. (Hint: Draw a Punnett square.)

1. Which one of Mendel’s laws relates to the inheritance of

alleles for a single character? Which law relates to the in-heritance of alleles for two characters in a dihybrid cross?

review the description of mei-and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel’s laws?propose a possible reason that the first

gene on a sex chromosome.

For suggested answers, see appendix a.

296 U n i t t h r e e Genetics

identified a gene on the Y chromosome required for the development of testes. They named the gene SRY , for s ex-determining r egion of Y . In the absence of SRY , the gonads develop into ovaries. The biochemical, physiological, and anatomical features that distinguish males and females are complex, and many genes are involved in their development. In fact, SRY

codes for a protein that regulates other genes.

of inheritance

As you just learned, Morgan’s discovery of a trait (white eyes) that correlated with the sex of flies was a key episode in the development of the chromosome theory of inheri-tance. Because the identity of the sex chromosomes in an individual could be inferred by observing the sex of the fly, the behavior of the two members of the pair of sex chromo-somes could be correlated with the behavior of the two al-leles of the eye-color gene. In this section, we’ll take a closer look at the role of sex chromosomes in inheritance.

The Chromosomal Basis of Sex

Although the anatomical and physiological differences be-tween women and men are numerous, the chromosomal basis for determining sex is rather simple. Humans and other mammals have two types of sex chromosomes, designated

X and Y. The Y chromosome is much smaller than the X chromosome (Figure 15.5).

A person who inherits two X chromosomes, one from each parent, usually develops as a female; a male inherits one X chromosome and one

Y chromosome (Figure 15.6a). Short segments at either end

of the Y chromosome are the only regions that are homolo-gous with regions of the X. These homologous regions allow the X and Y chromosomes in males to pair and behave like homologs during meiosis in the testes.

In mammalian testes and ovaries, the two sex chromo-somes segregate during meiosis. Each egg receives one X chromosome. In contrast, sperm fall into two categories: Half the sperm cells a male produces receive an X chro-mosome, and half receive a Y chromosome. We can trace the sex of each offspring to the events of conception: If a sperm cell bearing an X chromosome fertilizes an egg, the zygote is XX, a female; if a sperm cell containing a Y chromosome fertilizes an egg, the zygote is XY, a male (see Figure 15.6a). Thus, sex determination is a matter of chance—a fifty-fifty chance. Note that the mammalian X-Y system isn’t the only chromosomal system for deter-mining sex. Figure 15.6b–d illustrates three other systems.In humans, the anatomical signs of sex begin to emerge when the embryo is about 2 months old. Before then, the rudiments of the gonads are generic—they can develop into either testes or ovaries, depending on whether or not a Y chromosome is present. In 1990, a British research team

Y

▲ Figure 15.5 Human sex

chromosomes.

22 +The X -Y system. on whether the sperm cell contains an X chromosome or a Y.

(b)The X -0 system. In grasshoppers, cockroaches, and some

other insects, there is only one type of sex chromosome, the X. Females are XX; males have only one sex chromosome (X0). Sex of the offspring is determined by whether the sperm cell -▲ Figure 15.6 Some chromosomal systems of sex determi-nation. Numerals indicate the number of autosomes in the species pictured. In Drosophila , males are XY , but sex depends on the ratio between the number of X chromosomes and the number of autosome sets, not simply on the presence of a Y chromosome.

Researchers have sequenced the human Y chromosome and have identified 78 genes that code for about 25 proteins (some genes are duplicates). About half of these genes are expressed only in the testis, and some are required for nor-mal testicular functioning and the production of normal sperm. A gene located on either sex chromosome is called a sex-linked gene; those located on the Y chromosome are called Y-linked genes. The Y chromosome is passed along virtually intact from a father to all his sons. Because there are so few Y-linked genes, very few disorders are transferred from father to son on the Y chromosome. A rare example is that in the absence of certain Y-linked genes, an XY indi-vidual is male but does not produce normal sperm.

The human X chromosome contains approximately 1,100 genes, which are called X-linked genes. The fact that males and females inherit a different number of X chromosomes leads to a pattern of inheritance different from that pro-duced by genes located on autosomes.

Inheritance of X-Linked Genes

While most Y-linked genes help determine sex, the X chro-mosomes have genes for many characters unrelated to sex. X-linked genes in humans follow the same pattern of inheri-tance that Morgan observed for the eye-color locus he stud-ied in Drosophila (see Figure 15.4). Fathers pass X-linked alleles to all of their daughters but to none of their sons. In contrast, mothers can pass X-linked alleles to both sons and daughters, as shown in Figure 15.7 for the inheritance of a mild X-linked disorder, red-green color blindness.

If an X-linked trait is due to a recessive allele, a female will express the phenotype only if she is homozygous for that allele. Because males have only one locus, the terms homozygous and heterozygous lack meaning for describ-ing their X-linked genes; the term hemizygous is used in such cases. Any male receiving the recessive allele from his mother will express the trait. For this reason, far more males than females have X-linked recessive disorders. However, even though the chance of a female inheriting a double dose of the mutant allele is much less than the probability of a male inheriting a single dose, there are females with

X-linked disorders. For instance, color blindness is almost always inherited as an X-linked trait. A color-blind daughter may be born to a color-blind father whose mate is a car-rier (see Figure 15.7c). Because the X-linked allele for color blindness is relatively rare, though, the probability that such a man and woman will mate is low.

A number of human X-linked disorders are much more serious than color blindness, such as Duchenne muscular dystrophy, which affects about one out of 3,500 males born in the United States. The disease is characterized by a progressive weakening of the muscles and loss of coordi-nation. Affected individuals rarely live past their early 20s. Researchers have traced the disorder to the absence of a key muscle protein called dystrophin and have mapped the gene for this protein to a specific locus on the X chromosome. Hemophilia is an X-linked recessive disorder defined by the absence of one or more of the proteins required for blood clotting. When a person with hemophilia is injured, bleeding is prolonged because a firm clot is slow to form. Small cuts in the skin are usually not a problem, but bleed-ing in the muscles or joints can be painful and can lead to serious damage. In the 1800s, hemophilia was widespread

among the royal families of Europe. Queen Victoria of

▲ Figure 15.7 The transmission of X-linked recessive traits. In this diagram, red-green color blindness is used as an example. The superscript N represents the dominant allele for normal color vision carried on the X chromosome, while n represents the recessive

allele, which has a mutation for color blindness.

White boxes indicate unaffected individuals,

light orange boxes indicate carriers, and dark

orange boxes indicate color-blind individuals.

?If a color-blind woman married a man who

had normal color vision, what would be the prob-

able phenotypes of their children?

C h a p t e r15 The Chromosomal Basis of Inheritance 297

298 U n i t t h r e e Genetics

associate briefly with each other in each cell at an early stage of embryonic development. Then one of the genes, called XIST (for X -i nactive s pecific t ranscript) becomes active only on the chromosome that will become the Barr body. Multiple copies of the RNA product of this gene apparently attach to the X chromosome on which they are made, even-tually almost covering it. Interaction of this RNA with the chromosome initiates X inactivation, and the RNA products of other nearby genes help to regulate the process.

England is known to have passed the allele to several of her descendants. Subsequent intermarriage with royal family members of other nations, such as Spain and Russia, further spread this X-linked trait, and its incidence is well docu-mented in royal pedigrees. A few years ago, new genomic techniques allowed sequencing of DNA from tiny amounts isolated from the buried remains of royal family members. The genetic basis of the mutation, and how it resulted in a nonfunctional blood-clotting factor, is now understood. Today, people with hemophilia are treated as needed with intravenous injections of the protein that is missing.

X Inactivation in Female Mammals

Female mammals, including human females, inherit two X chromosomes—twice the number inherited by males— so you may wonder whether females make twice as much as males of the proteins encoded by X-linked genes. In fact, almost all of one X chromosome in each cell in female mammals becomes inactivated during early embryonic de-velopment. As a result, the cells of females and males have the same effective dose (one copy) of most X-linked genes. The inactive X in each cell of a female condenses into a compact object called a Barr body (discovered by Canadian anatomist Murray Barr), which lies along the inside of the nuclear envelope. Most of the genes of the X chromosome that forms the Barr body are not expressed. In the ovaries, however, Barr-body chromosomes are reactivated in the cells that give rise to eggs, such that following meiosis, every female gamete (egg) has an active X.

British geneticist Mary Lyon demonstrated that the selec-tion of which X chromosome will form the Barr body occurs randomly and independently in each embryonic cell pres-ent at the time of X inactivation. As a consequence, females consist of a mosaic of two types of cells: those with the ac-tive X derived from the father and those with the active X derived from the mother. After an X chromosome is inacti-vated in a particular cell, all mitotic descendants of that cell have the same inactive X. Thus, if a female is heterozygous for a sex-linked trait, about half her cells will express one allele, while the others will express the alternate allele. Figure 15.8 shows how this mosaicism results in the mot-tled coloration of a tortoiseshell cat. In humans, mosaicism can be observed in a recessive X-linked mutation that pre-vents the development of sweat glands. A woman who is heterozygous for this trait has patches of normal skin and patches of skin lacking sweat glands.

Inactivation of an X chromosome involves modifica-tion of the DNA and proteins bound to it, called histones, including attachment of methyl groups (—CH 3) to DNA nucleotides. (The regulatory role of DNA methylation is discussed in Chapter 18.) A particular region of each X

chromosome contains several genes involved in the inactiva-

tion process. The two regions, one on each X chromosome,

Early embryo:

Two cell populations in adult cat:

▲ Figure 15.8 X inactivation and the tortoiseshell cat. The tortoiseshell gene is on the X chromosome, and the tortoiseshell phe-notype requires the presence of two different alleles, one for orange fur and one for black fur. Normally, only females can have both alleles, because only they have two X chromosomes. If a female cat is hetero-zygous for the tortoiseshell gene, she is tortoiseshell. Orange patches are formed by populations of cells in which the X chromosome with the orange allele is active; black patches have cells in which the X chro-mosome with the black allele is active. (“Calico” cats also have white areas, which are determined by another gene.)

1. a white-eyed female Drosophila is mated with a red-eyed

(wild-type) male, the reciprocal cross of the one shown in Figure 15.4. What phenotypes and genotypes do you predict for the offspring? 2. neither tim nor rhoda has Duchenne muscular dystro-phy, but their firstborn son does. What is the probability that a second child will have the disease? What is the probability if the second child is a boy? a girl?Consider what you learned

if a disorder were caused by a dominant X-linked allele, how would the inheritance pattern differ from what we see for recessive X-linked disorders?

For suggested answers, see appendix a.

C h a p t e r 15 The Chromosomal Basis of Inheritance 299

The number of genes in a cell is far greater than the number of chromosomes; in fact, each chromosome (except the Y) has hundreds or thousands of genes. Genes located near each other on the same chromosome tend to be inherited together in genetic crosses; such genes are said to be genetically linked and are called linked genes . When geneticists follow linked genes in breeding experiments, the results deviate from those expected from Mendel’s law of independent assortment.

How Linkage Affects Inheritance

To see how linkage between genes affects the inheritance of two different characters, let’s examine another of Morgan’s Drosophila experiments. In this case, the characters are body color and wing size, each with two different pheno-types. Wild-type flies have gray bodies and normal-sized wings. In addition to these flies, Morgan had managed to obtain, through breeding, doubly mutant flies with black bodies and wings much smaller than normal, called vestigial wings. The mutant alleles are recessive to the wild-type al-leles, and neither gene is on a sex chromosome. In his inves-tigation of these two genes, Morgan carried out the crosses shown in Figure 15.9. The first was a P generation cross to generate F 1 dihybrid flies, and the second was a testcross.

b + b vg + vg

b b vg vg

b b vg + vg

dihybrid testcross

(gray-normal)

vestigial

vestigial normal

Experiment Morgan wanted to know whether the genes for body color and wing size are genetically linked, and if so, how this affects their

inheritance. The alleles for body color are b + (gray) and b (black), and those for wing size are vg + (normal) and vg (vestigial).Results

Conclusion Since most offspring had a parental (P generation)

phenotype, Morgan concluded that the genes for body color and wing size are genetically linked on the same chromosome. However, the production of a relatively small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between specific alleles of genes on the same chromosome.

Source: T. H. Morgan and C. J. Lynch, The linkage of two factors in Drosophila that are not sex-linked, Biological Bulletin 23:174–182 (1912).

If the parental (P generation) flies had been true-breeding

which phenotypic class(es) would be largest among the testcross offspring?

300 U n i t t h r e e Genetics

dihybrid YyRr plant.) Let’s represent the cross by the follow-

ing Punnett square:

Gametes from yellow-round type offspring

offspring

Gametes from testcross homozygous recessive parent (yyrr )

Notice in this Punnett square that one-half of the offspring are expected to inherit a phenotype that matches either of the phenotypes of the P (parental) generation originally crossed to produce the F 1 dihybrid (see Figure 15.2). These matching offspring are called parental types . But two non-parental phenotypes are also found among the offspring. Because these offspring have new combinations of seed shape and color, they are called recombinant types , or recombinants for short. When 50% of all offspring are re-combinants, as in this example, geneticists say that there is a 50% frequency of recombination. The predicted phenotypic ratios among the offspring are similar to what Mendel actu-ally found in his YyRr * yyrr crosses.

A 50% frequency of recombination in such testcrosses is observed for any two genes that are located on different chromosomes and thus cannot be linked. The physical basis of recombination between unlinked genes is the random orientation of homologous chromosomes at metaphase I of meiosis, which leads to the independent assortment of

the two unlinked genes (see Figure 13.11 and the question in the Figure 15.2 legend).

Recombination of Linked Genes: Crossing Over

Now, let’s explain the results of the Drosophila testcross in Figure 15.9. Recall that most of the offspring from the test-cross for body color and wing size had parental phenotypes. That suggested that the two genes were on the same chro-mosome, since the occurrence of parental types with a fre-quency greater than 50% indicates that the genes are linked. About 17% of offspring, however, were recombinants.

Seeing these results, Morgan proposed that some process must occasionally break the physical connection between specific alleles of genes on the same chromosome. Later experiments showed that this process, now called crossing over , accounts for the recombination of linked genes. In crossing over, which occurs while replicated homologous chromosomes are paired during prophase of meiosis I, a set of proteins orchestrates an exchange of corresponding segments of one maternal and one paternal chromatid (see

Most offspring

F 1 dihybrid female and homozygous recessive male in testcross

However, as Figure 15.9 shows, both of the combinations of traits not seen in the P generation (called nonparental phenotypes) were also produced in Morgan’s experiments, suggesting that the body-color and wing-size alleles are not always linked genetically. To understand this conclusion, we need to further explore genetic recombination , the produc-tion of offspring with combinations of traits that differ from those found in either P generation parent.*

Genetic Recombination and Linkage

Meiosis and random fertilization generate genetic variation among offspring of sexually reproducing organisms due to independent assortment of chromosomes, crossing over in meiosis I, and the possibility of any sperm fertilizing any egg (see Concept 13.4). Here we’ll examine the chromosomal basis of recombination in relation to the genetic findings of Mendel and Morgan.

Recombination of Unlinked Genes: Independent Assortment of Chromosomes

Mendel learned from crosses in which he followed two char-acters that some offspring have combinations of traits that do not match those of either parent. For example, consider a cross of a dihybrid pea plant with yellow-round seeds, het-erozygous for both seed color and seed shape (YyRr ), with a plant homozygous for both recessive alleles (with green-wrinkled seeds, yyrr ). (This acts as a testcross because the results will reveal the genotype of the gametes made in the

The resulting flies had a much higher proportion of the combinations of traits seen in the P generation flies (called parental phenotypes) than would be expected if the two genes assorted independently. Morgan thus concluded that body color and wing size are usually inherited together in specific (parental) combinations because the genes for these characters are near each other on the same chromosome:

* As you proceed, be sure to keep in mind the distinction between the terms linked genes (two or more genes on the same chromosome that tend to be in-herited together) and sex-linked gene (a single gene on a sex chromosome).

C h a p t e r 15 The

Chromosomal Basis of Inheritance 301Figure 13.9). In effect, when a single crossover occurs, end

portions of two nonsister chromatids trade places.

Figure 15.10 shows how crossing over in a dihybrid female fly resulted in recom b inant eggs and ultimately recombinant offspring in Morgan’s testcross. Most eggs had a chromosome with either the b + vg + or b vg parental genotype, but some had a recombinant chromosome (b + vg or b vg +). Fertilization of P generation (homozygous)

F 1 dihybrid testcross

Meiosis I and II

Replication

of chromosomes

Replication

of chromosomes Meiosis I

Meiosis II

Recombinant chromosomes

Homozygous recessive (black body,vestigial wings)

Wild-type F 1 dihybrid (gray body,normal wings)

Double mutant (black body,vestigial wings)

Wild type (gray body,normal wings)

Testcross offspring

b + vg +b vg

b + vg +b + vg +

b vg b vg

b + vg +b + vg +b + vg +b vg +b vg

b +

vg b vg b vg

b vg b vg b vg b vg

965Wild type (gray-normal)

Eggs

b vg

b vg

vg +

b v g +

b vg + b +vg

b +

vg b +

vg +

944Black-vestigial

b vg

b vg 206Gray-vestigial

b vg

185Black-normal

b vg

Parental-type offspring Recombinant offspring

2,300 total offspring

Recombination

frequency =391 recombinants × 100 = 17%

b +Sperm

vg

b vg b vg

b ? Figure 15.10 Chromosomal basis for recombination of linked genes. In these diagrams re-creating the testcross in Figure 15.9, we track chromosomes as well as genes. The maternal chromosomes (present in the wild-type F 1 dihybrid) are color-coded red and pink to distin-guish one homolog from the other before any meioti

c crossing over has occurred. Because crossing over be-tween the b +/b an

d vg +/vg loci occurs in some, but not all, egg-producing cells, mor

e eggs with parental-type chromosomes than with recombi-nant ones are produced in the mat-ing females. Fertilization o

f the eggs by sperm of genotype b v

g gives rise to some recombinant offspring. The recombination frequency is the per-centage of recombinant flies in the total pool of offspring.

Suppose, as in the ques-

tion at the bottom of Figure 15.9, the parental (P generation) flies were true-breeding for gray body with vestigial wings and black body with normal wings. Draw the chromosomes in each of the four possible kinds of eggs from an F 1 female, and label each chromosome as “parental” or “recombinant.”

all classes of eggs by homozygous recessive sperm (b vg ) pro-duced an offspring population in which 17% exhibited a non-parental, recombinant phenotype, reflecting combinations of alleles not seen before in either P generation parent. In the scientific skills exercise , you can use a statistical test to ana-lyze the results from an F 1 dihybrid testcross and see whether the two genes assort independently or are linked.

302 U n i t t h r e e Genetics

s c I e n t I F I c s k I l l s e x e r c I s e

Are Two Genes Linked or Unlinked? Genes that are in close prox-

imity on the same chromosome will result in the linked alleles being inherited together more often than not. But how can you tell if certain alleles are inherited together due to linkage or whether they just happen to assort together? In this exercise, you will use a simple statistical test, the chi-square (x 2) test, to analyze phenotypes of F 1 testcross progeny in order to see whether two genes are linked or unlinked.

How These Experiments Are Done If genes are unlinked and assort-ing independently, the phenotypic ratio of offspring from an F 1 testcross is expected to be 1:1:1:1 (see Figure 15.9). If the two genes are linked, however, the observed phenotypic ratio of the offspring will not match that ratio. Given that random fluctuations in the data do occur, how much must the observed numbers deviate from the expected numbers for us to conclude that the genes are not assorting independently but may instead be linked? To answer this question, scientists use a statistical test. This test, called a chi-square (x 2) test, compares an observed data set to an expected data set predicted by a hypothesis (here, that the genes are unlinked) and measures the discrepancy between the two, thus de-termining the “goodness of fit.” If the discrepancy between the observed and expected data sets is so large that it is unlikely to have occurred by random fluctuation, we say there is statistically significant evidence against the hypothesis (or, more specifically, evidence for the genes being linked). If the discrepancy is small, then our observations are well ex-plained by random variation alone. In this case, we say the observed data are consistent with our hypothesis, or that the discrepancy is statistically insignificant. Note, however, that consistency with our hypothesis is not the same as proof of our hypothesis. Also, the size of the experimental data set is important: With small data sets like this one, even if the genes are linked, discrepancies might be small by chance alone if the linkage is weak. For simplicity, we overlook the effect of sample size here.Data from the Simulated Experiment In cosmos plants, purple stem (A ) is dominant to green stem (a ), and short petals (B ) is dominant

to long petals (b ). In a simulated cross, AABB plants were crossed with aabb plants to generate F 1 dihybrids (AaBb ), which were then test-crossed (AaBb * aabb ). A total of 900 offspring plants were scored for stem color and flower petal length.Offspring from test-cross of AaBb (F 1) * aabb Purple stem/short petals (A -B -)

Green stem/short petals (aaB -)

Purple stem/long petals (A -bb )

Green stem/long petals (aabb )

Expected ratio if the genes are unlinked 1

1

1

1

Expected number of offspring (of 900)

Observed number of offspring (of 900)

220210231239

Interpret the Data

1. The results in the data table are from a simulated F 1 dihybrid test-cross. The hypothesis that the two genes are unlinked predicts the offspring phenotypic ratio will be 1:1:1:1. Using this ratio, calculate

Using the Chi-Square (x 2) Test

the expected number of each phenotype out of the 900 total off-spring, and enter the values in the data table.

2. The goodness of fit is measured by x 2. This sta-tistic measures the amounts by which the observed values

differ from their

respective predic-tions to indicate how closely the two sets of values match. The for-mula for calculating this value is

x 2=a

(o -e )2

e

where o = observed and e = expected. Calculate the x 2 value for the data using the table below. Fill out the table, carrying out the opera-tions indicated in the top row. Then add up the entries in the last column to find the x 2 value.Testcross Offspring Expected

(e )

Observed

(o )Deviation (o - e ) (o - e )2 (o - e )2/e

(A -B -) 220 (aaB -) 210 (A -bb ) 231 (aabb )

239

x 2 = Sum

3. The x 2 value means nothing on its own—it is used to find the prob-ability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability sug-gests that the observed data are not consistent with the hypothesis, and thus the hypothesis should be rejected. A standard cutoff point used by biologists is a probability of 0.05 (5%). If the probability cor-responding to the x 2 value is 0.05 or less, the differences between observed and expected values are considered statistically significant and the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically signifi-cant; the observed data are consistent with the hypothesis. To find the probability, locate your x 2 value in the x 2 Distribution Table in Appendix F. The “degrees of freedom” (df) of your data set is the number of categories (here, 4 phenotypes) minus 1, so df = 3. (a) Determine which values on the df = 3 line of the table your cal-culated x 2 value lies between. (b) The column headings for these values show the probability range for your x 2 number. Based on whether there are nonsignificant (p 7 0.05) or significant (p … 0.05) differences between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

A version of this Scientific Skills Exercise can be assigned in MasteringBiology.

▲ Cosmos plants

C h a p t e r 15 The Chromosomal Basis of Inheritance 303

New Combinations of Alleles: Variation for Natural Selection

The physical behavior of chromosomes

in offspring (see Concept 13.4). Each pair of homologous chromosomes lines up independently of other pairs during metaphase I, and crossing over prior to that, during pro-phase I, can mix and match parts of maternal and paternal homologs. Mendel’s elegant experiments show that the behavior of the abstract entities known as genes—or, more concretely, alleles of genes—also leads to variation in off-spring (see Concept 14.1). Now, putting these different ideas together, you can see that the recombinant chromosomes resulting from crossing over may bring alleles together in new combinations, and the subsequent events of meiosis distribute to gametes the recombinant chromosomes in a multitude of combinations, such as the new variants seen in Figures 15.9 and 15.10. Random fertilization then increases even further the number of variant allele combinations that can be created.

This abundance of genetic variation provides the raw ma-terial on which natural selection works. If the traits conferred by particular combinations of alleles are better suited for a given environment, organisms possessing those genotypes will be expected to thrive and leave more offspring, ensuring the continuation of their genetic complement. In the next generation, of course, the alleles will be shuffled anew. Ulti-mately, the interplay between environment and genotype will determine which genetic combinations persist over time.

Mapping the Distance Between Genes Using Recombination Data: Scientific Inquiry

The discovery of linked genes and recombination due to crossing over motivated one of Morgan’s students, Alfred H. Sturtevant, to work out a method for constructing a genetic map , an ordered list of the genetic loci along a particular chromosome.

Sturtevant hypothesized that the percentage of recom-binant offspring, the recombination frequency , calculated from experiments like the one in Figures 15.9 and 15.10, depends on the distance between genes on a chromosome. He assumed that crossing over is a random event, with the chance of crossing over approximately equal at all points along a chromosome. Based on these assumptions, Sturtevant predicted that the farther apart two genes are, the higher the probability that a crossover will occur between them and therefore the higher the recombination frequency . His reasoning was simple: The greater the distance between two genes, the more points there are between them where crossing over can occur. Using recombination data from various fruit fly crosses, Sturtevant proceeded to assign rela-tive positions to genes on the same chromosomes—that is, to map genes.

A genetic map based on recombination frequencies is called a linkage map . Figure 15.11 shows Sturtevant’s link-age map of three genes: the body-color (b ) and wing-size (vg ) genes depicted in Figure 15.10 and a third gene, called cinnabar (cn ). Cinnabar is one of many Drosophila genes affecting eye color. Cinnabar eyes, a mutant phenotype, are a brighter red than the wild-type color. The recombination frequency between cn and b is 9%; that between cn and vg , 9.5%; and that between b and vg , 17%. In other words, cross-overs between cn and b and between cn and vg are about half as frequent as crossovers between b and vg . Only a map that locates cn about midway between b and vg is consistent with these data, as you can prove to yourself by drawing alterna-tive maps. Sturtevant expressed the distances between genes in map units , defining one map unit as equivalent to a 1% recombination frequency.

In practice, the interpretation of recombination data is more complicated than this example suggests. Some genes on a chromosome are so far from each other that a crossover between them is virtually certain. The observed frequency of recombination in crosses involving two such

Application A linkage map shows the relative locations of genes

along a chromosome.

Technique A linkage map is based on the assumption that the prob-

ability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted in Figures 15.9 and 15.10. The distances between genes are expressed as map units, with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.

Results In this example, the observed recombination frequencies be-

tween three Drosophila gene pairs (b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes:

9%

Recombination frequencies Chromosome

b cn vg

The b–vg recombination frequency (17%) is slightly less than the sum of the b–cn and cn–vg frequencies (9 + 9.5 = 18.5%) because of the few times that one crossover occurs between b and cn and another crossover occurs between cn and vg . The second crossover would “cancel out” the first, reducing the observed b–vg recombination fre-quency while contributing to the frequency between each of the closer pairs of genes. The value of 18.5% (18.5 map units) is closer to the actual distance between the genes. In practice, a geneticist would add the smaller distances in constructing a map.

304 U n i t t h r e e Genetics

genes can have a maximum value of 50%, a result indistin-guishable from that for genes on different chromosomes. In this case, the physical connection between genes on the same chromosome is not reflected in the results of genetic crosses. Despite being on the same chromosome and thus being physically connected , the genes are geneti-cally unlinked ; alleles of such genes assort independently, as if they were on different chromosomes. In fact, at least two of the genes for pea characters that Mendel studied are now known to be on the same chromosome, but the dis-tance between them is so great that linkage is not observed in genetic crosses. Consequently, the two genes behaved as if they were on different chromosomes in Mendel’s ex-periments. Genes located far apart on a chromosome are mapped by adding the recombination frequencies from crosses involving closer pairs of genes lying between the two distant genes.

Using recombination data, Sturtevant and his colleagues were able to map numerous Drosophila genes in linear ar-rays. They found that the genes clustered into four groups of linked genes (linkage groups ). Light microscopy had revealed four pairs of chromosomes in Drosophila , so the linkage map provided additional evidence that genes are located on chromosomes. Each chromosome has a linear array of spe-cific genes, each gene with its own locus (Figure 15.12). Because a linkage map is based strictly on recombina-tion frequencies, it gives only an approximate picture of a chromosome. The frequency of crossing over is not actually

uniform over the length of a chromosome, as Sturtevant as-sumed, and therefore map units do not correspond to actual physical distances (in nanometers, for instance). A linkage map does portray the order of genes along a chromosome, but it does not accurately portray the precise locations of those genes. Other methods enable geneticists to construct cytogenetic maps of chromosomes, which locate genes with respect to chromosomal features, such as stained bands, that can be seen in the microscope. Technical advances over the last two decades have enormously increased the rate and affordability of DNA sequencing. Today, most researchers sequence whole genomes to map the locations of genes of a given species. The entire nucleotide sequence is the ulti-mate physical map of a chromosome, revealing the physical distances between gene loci in DNA nucleotides (see

Concept 21.1). Comparing a linkage map with such a physi-cal map or with a cytogenetic map of the same chromosome, we find that the linear order of genes is identical in all the

maps, but the spacing between genes is not.

1. When two genes are located on the same chromosome,

what is the physical basis for the production of recombi-nant offspring in a testcross between a dihybrid parent and a double-mutant (recessive) parent? 2. For each type of offspring of the testcross in Figure 15.9,

explain the relationship between its phenotype and the alleles contributed by the female parent. (it will be useful to draw out the chromosomes of each fly and follow the alleles throughout the cross.)

Genes A, B, and C are located on the same

frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes? explain.

For suggested answers, see appendix a.

structure cause some genetic disorders

As you have learned so far in this chapter, the phenotype of an organism can be affected by small-scale changes involv-ing individual genes. Random mutations are the source of all new alleles, which can lead to new phenotypic traits.

Large-scale chromosomal changes can also affect an or-ganism’s phenotype. Physical and chemical disturbances, as well as errors during meiosis, can damage chromosomes in major ways or alter their number in a cell. Large-scale chro-mosomal alterations in humans and other mammals often lead to spontaneous abortion (miscarriage) of a fetus, and individuals born with these types of genetic defects com-monly exhibit various developmental disorders. Plants may tolerate such genetic defects better than animals do.

057.5104.5

wings

16.548.567.075.5▲ Figure 15.12 A partial genetic (linkage) map of a Drosophila chromosome. This simplified map shows just seven of the genes that have been mapped on Drosophila chromosome II . (DNA sequencing has revealed over 9,000 genes on that chromosome.) The number at each gene locus indicates the number of map units between that locus and the locus for arista length (left). Notice that more than one gene can affect a given phenotypic characteristic, such as eye color.

C h a p t e r 15 The Chromosomal Basis of Inheritance 305

Abnormal Chromosome Number

Ideally, the meiotic spindle distributes chromosomes to daughter cells without error. But there is an occasional mishap, called a nondisjunction , in which the members of a pair of homologous chromosomes do not move apart properly during meiosis I or sister chromatids fail to sepa-rate during meiosis II (Figure 15.13). In nondisjunction, one gamete receives two of the same type of chromosome and another gamete receives no copy. The other chromosomes are usually distributed normally.

If either of the aberrant gametes unites with a normal one at fertilization, the zygote will also have an abnormal number of a particular chromosome, a condition known as aneuploidy . Fertilization involving a gamete that has no copy of a particular chromosome will lead to a missing chro-mosome in the zygote (so that the cell has 2n – 1 chromo-somes); the aneuploid zygote is said to be monosomic for that chromosome. If a chromosome is present in triplicate in the zygote (so that the cell has 2n + 1 chromosomes), the aneuploid cell is trisomic for that chromosome. Mitosis will subsequently transmit the anomaly to all embryonic cells. Monosomy and trisomy are estimated to occur in be-tween 10 and 25% of human conceptions, and is the main reason for pregnancy loss. If the organism survives, it usu-ally has a set of traits caused by the abnormal dose of the genes associated with the extra or missing chromosome.

Down syndrome is an example of trisomy in humans that

▲ Figure 15.13 Meiotic nondisjunction. Gametes with an abnor-mal chromosome number can arise by nondisjunction in either meiosis I or meiosis II . For simplicity, the figure does not show the spores formed by meiosis in plants. Ultimately, spores form gametes that have the defects shown. (See Figure 13.6.)

will be discussed later. Nondisjunction can also occur dur-ing mitosis. If such an error takes place early in embryonic development, then the aneuploid condition is passed along by mitosis to a large number of cells and is likely to have a substantial effect on the organism.

Some organisms have more than two complete chro-mosome sets in all somatic cells. The general term for this chromosomal alteration is polyploidy ; the specific terms triploidy (3n ) and tetraploidy (4n ) indicate three or four chromosomal sets, respectively. One way a triploid cell may arise is by the fertilization of an abnormal diploid egg produced by nondisjunction of all its chromosomes. Tetra-ploidy could result from the failure of a 2n zygote to divide after replicating its chromosomes. Subsequent normal mi-totic divisions would then produce a 4n embryo.

Polyploidy is fairly common in the plant kingdom. The spontaneous origin of polyploid individuals plays an impor-tant role in plant evolution (see Chapter 24). Many species we eat are polyploid: Bananas are triploid, wheat hexaploid (6n ), and strawberries octoploid (8n ). Polyploid animal spe-cies are much less common, but there are a few fishes and amphibians known to be polyploid. In general, polyploids are more nearly normal in appearance than aneuploids. One extra (or missing) chromosome apparently disrupts genetic balance more than does an entire extra set of chromosomes.

Alterations of Chromosome Structure

Errors in meiosis or damaging agents such as radiation can cause breakage of a chromosome, which can lead to four types of changes in chromosome structure (Figure 15.14). A deletion occurs when a chromosomal fragment is lost. The affected chromosome is then missing certain genes. The “deleted” fragment may become attached as an extra segment to a sister chromatid, producing a duplication . Alternatively, a detached fragment could attach to a nonsis-ter chromatid of a homologous chromosome. In that case, though, the “duplicated” segments might not be identical because the homologs could carry different alleles of certain genes. A chromosomal fragment may also reattach to the original chromosome but in the reverse orientation, produc-ing an inversion . A fourth possible result of chromosomal breakage is for the fragment to join a nonhomologous chro-mosome, a rearrangement called a translocation .

Deletions and duplications are especially likely to occur during meiosis. In crossing over, nonsister chromatids sometimes exchange unequal-sized segments of DNA, so that one partner gives up more genes than it receives. The products of such an unequal crossover are one chromosome with a deletion and one chromosome with a duplication.A diploid embryo that is homozygous for a large deletion (or has a single X chromosome with a large deletion, in a male) is usually missing a number of essential genes, a condi-tion typically lethal. Duplications and translocations also tend

306 U n i t t h r e e Genetics

to be harmful. In reciprocal translocations, in which segments are exchanged between nonhomologous chromosomes, and in inversions, the balance of genes is not abnormal—all genes are present in their normal doses. Nevertheless, transloca-tions and inversions can alter phenotype because a gene’s ex-pression can be influenced by its location among neighboring genes, which can have devastating effects.

Human Disorders Due to Chromosomal Alterations

Alterations of chromosome number and structure are as-sociated with a number of serious human disorders. As described earlier, nondisjunction in meiosis results in an-euploidy in gametes and any resulting zygotes. Although

the frequency of aneuploid zygotes may be quite high in humans, most of these chromosomal alterations are so di-sastrous to development that the affected embryos are spon-taneously aborted long before birth. However, some types of aneuploidy appear to upset the genetic balance less than others, where individuals with certain aneuploid conditions can survive to birth and beyond. These individuals have a set of traits—a syndrome —characteristic of the type of aneu-ploidy. Genetic disorders caused by aneuploidy can be diag-nosed before birth by fetal testing (see Figure 14.19).

Down Syndrome (Trisomy 21)

One aneuploid condition, Down syndrome , affects approxi-mately one out of every 830 children born in the United States (Figure 15.15). Down syndrome is usually the result of an extra chromosome 21, so that each body cell has a total of 47 chromosomes. Because the cells are trisomic for chromosome 21, Down syndrome is often called trisomy 21. Down syndrome includes characteristic facial features, short stature, correctable heart defects, and developmental delays. Individuals with Down syndrome have an increased chance of developing leukemia and Alzheimer’s disease but have a lower rate of high blood pressure, atherosclerosis (harden-ing of the arteries), stroke, and many types of solid tumors. Although people with Down syndrome, on average, have a life span shorter than normal, most, with proper medical treatment, live to middle age and beyond. Many live inde-pendently or at home with their families, are employed, and are valuable contributors to their communities. Almost all males and about half of females with Down syndrome are sexually underdeveloped and sterile.

The frequency of Down syndrome increases with the age of the mother. While the disorder occurs in just 0.04%

of children born to women under age 30, the risk climbs

▲ Figure 15.15 Down syndrome. The karyotype shows trisomy 21, the most common cause of Down syndrome. The child exhibits the facial features characteristic of this disorder.

to 0.92% for mothers at age 40 and is even higher for older mothers. The correlation of Down syndrome with mater-nal age has not yet been explained. Most cases result from nondisjunction during meiosis I, and some research points to an age-dependent abnormality in meiosis. Trisomies of some other chromosomes also increase in incidence with maternal age, although infants with other autosomal triso-mies rarely survive for long. Due to its low risk and its po-tential for providing useful information, prenatal screening for trisomies in the embryo is now offered to all pregnant women. In 2008, the Prenatally and Postnatally Diagnosed Conditions Awareness Act was signed into law in the United States. This law stipulates that medical practitioners give ac-curate, up-to-date information about any prenatal or post-natal diagnosis received by parents and that they connect parents with appropriate support services.

Aneuploidy of Sex Chromosomes

Aneuploid conditions involving sex chromosomes appear to upset the genetic balance less than those involving au-

tosomes. This may be because the Y chromosome carries relatively few genes. Also, extra copies of the X chromosome simply become inactivated as Barr bodies.

An extra X chromosome in a male, producing XXY, oc-curs approximately once in every 500 to 1,000 live male births. People with this disorder, called Klinefelter syndrome, have male sex organs, but the testes are abnormally small and the man is sterile. Even though the extra X is inacti-vated, some breast enlargement and other female body characteristics are common. Affected individuals may have subnormal intelligence. About 1 of every 1,000 males is born with an extra Y chromosome (XYY). These males undergo normal sexual development and do not exhibit any well-defined syndrome, but tend to be taller than average. Females with trisomy X (XXX), which occurs once in ap-proximately 1,000 live female births, are healthy and have no unusual physical features other than being slightly taller than average. Triple-X females are at risk for learning dis-abilities but are fertile. Monosomy X, which is called Turner syndrome, occurs about once in every 2,500 female births and is the only known viable monosomy in humans. Al-though these X0 individuals are phenotypically female, they are sterile because their sex organs do not mature. When provided with estrogen replacement therapy, girls with Turner syndrome do develop secondary sex characteristics. Most have normal intelligence.

Disorders Caused by Structurally Altered Chromosomes Many deletions in human chromosomes, even in a hetero-zygous state, cause severe problems. One such syndrome, known as cri du chat (“cry of the cat”), results from a spe-cific deletion in chromosome 5. A child born with this dele-tion is severely intellectually disabled, has a small head with unusual facial features, and has a cry that sounds like the mewing of a distressed cat. Such individuals usually die in infancy or early childhood.

Chromosomal translocations have been implicated in cer-tain cancers, including chronic myelogenous leukemia (CML). This disease occurs when a reciprocal translocation happens during mitosis of cells that will become white blood cells.

In these cells, the exchange of a large portion of chromo-some 22 with a small fragment from a tip of chromosome 9 produces a much shortened, easily recognized chromosome 22, called the Philadelphia chromosome(Figure 15.16). Such an exchange causes cancer by activating a gene that leads to uncontrolled cell cycle progression. (The mechanism of gene

activation will be discussed in Chapter 18.)

Normal chromosome 22

Translocated chromosome 9

Reciprocal translocation

Translocated chromosome 22

(Philadelphia chromosome)

▲ Figure 15.16 Translocation associated with chronic myelog-enous leukemia (CML). The cancerous cells in nearly all CML patients contain an abnormally short chromosome 22, the so-called Philadel-phia chromosome, and an abnormally long chromosome 9. These altered chromosomes result from the reciprocal translocation shown here, which presumably occurred in a single white blood cell precursor

undergoing mitosis and was then passed along to all descendant cells.

1. about 5% of individuals with Down syndrome have a

chromosomal translocation in which a third copy of chro-

mosome 21 is attached to chromosome 14. if this translo-

cation occurred in a parent’s gonad, how could it lead to

Down syndrome in a child?

the aBo blood type locus has been blood and a mother who has type o blood have a child

with trisomy 9 and type a blood. Using this information,

can you tell in which parent the nondisjunction oc-

curred? explain your answer. (see Figure 14.11.)

the gene that is activated on

tyrosine kinase. review the discussion of cell cycle control

in Concept 12.3, and explain how the activation of this

gene could contribute to the development of cancer.

For suggested answers, see appendix a.

C h a p t e r15 The Chromosomal Basis of Inheritance 307

308 U n i t t h r e e Genetics

came initially from crosses between normal-sized (wild-type) mice and dwarf (mutant) mice homozygous for a recessive mutation in the Ig f2 gene. The phenotypes of het-erozygous offspring (with one normal allele and one mutant) differed depending on whether the mutant allele came from the mother or the father (Figure 15.17b).

What exactly is a genomic imprint? In many cases, it seems to consist of methyl (—CH 3) groups that are added to cytosine nucleotides of one of the alleles. Such methyla-tion may silence the allele, an effect consistent with evidence that heavily methylated genes are usually inactive (see Con-cept 18.2). However, for a few genes, methylation has been shown to activate expression of the allele. This is the case for the Ig f2 gene: Methylation of certain cytosines on the paternal chromosome leads to expression of the paternal Ig f2 allele, by an indirect mechanism involving chromatin condensation.

Genomic imprinting is thought to affect only a small frac-tion of the genes in mammalian genomes, but most of the known imprinted genes are critical for embryonic develop-ment. In experiments with mice, embryos engineered to

inherit both copies of certain chromosomes from the same

exceptions to standard Mendelian inheritance

In the previous section, you learned about deviations from the usual patterns of chromosomal inheritance due to ab-normal events in meiosis and mitosis. We conclude this chapter by describing two normally occurring exceptions to Mendelian genetics, one involving genes located in the nucleus and the other involving genes located outside the nucleus. In both cases, the sex of the parent contributing an allele is a factor in the pattern of inheritance.

Genomic Imprinting

Throughout our discussions of Mendelian genetics and the chromosomal basis of inheritance, we have assumed that a given allele will have the same effect whether it was inher-ited from the mother or the father. This is probably a safe assumption most of the time. For example, when Mendel crossed purple-flowered pea plants with white-flowered pea plants, he observed the same results regardless of whether the purple-flowered parent supplied the eggs or the sperm. In recent years, however, geneticists have identified a num-ber of traits in mammals that depend on which parent passed along the alleles for those traits. Such variation in phenotype depending on whether an allele is inherited from the male or female parent is called genomic imprinting . (Note that unlike sex-linked genes, most imprinted genes are on autosomes.) Using newer DNA sequence-based

methods, over 60 imprinted genes have been identified, with hundreds more suspected.

Genomic imprinting occurs during gamete formation and results in the silencing of a particular allele of certain genes. Because these genes are imprinted differently in sperm and eggs, the offspring expresses only one allele of an imprinted gene, the one that has been inherited from either the female or the male parent. The imprints are then transmitted to all body cells during development. In each generation, the old imprints are “erased” in gamete-producing cells, and the chromosomes of the developing gametes are newly im-printed according to the sex of the individual forming the gametes. In a given species, the imprinted genes are always imprinted in the same way. For instance, a gene imprinted for maternal allele expression is always imprinted this way, generation after generation.

Consider, for example, the mouse gene for insulin-like growth factor 2 (Ig f2), one of the first imprinted genes to be identified. Although this growth factor is required for normal prenatal growth, only the paternal allele is expressed (Figure 15.17a). Evidence that the Ig f2

gene is imprinted

▲ Figure 15.17 Genomic imprinting of the mouse Igf2 gene.

parent usually die before birth, whether that parent is male or female. A few years ago, however, scientists in Japan com-bined the genetic material from two eggs in a zygote while allowing expression of the Ig f2 gene from only one of the egg nuclei. The zygote developed into an apparently healthy mouse. Normal development seems to require that embry-onic cells have exactly one active copy—not zero, not two—of certain genes. The association of improper imprinting with abnormal development and certain cancers has stimu-lated ongoing studies of how different genes are imprinted.

Inheritance of Organelle Genes

Although our focus in this chapter has been on the chro-mosomal basis of inheritance, we end with an important amendment: Not all of a eukaryotic cell’s genes are located on nuclear chromosomes, or even in the nucleus; some genes are located in organelles in the cytoplasm. Because they are outside the nucleus, these genes are sometimes called extranuclear genes or cytoplasmic genes. Mitochon-dria, as well as chloroplasts and other plastids in plants, con-tain small circular DNA molecules that carry a number of genes. These organelles reproduce themselves and transmit their genes to daughter organelles. Organelle genes are not distributed to offspring according to the same rules that di-rect the distribution of nuclear chromosomes during meio-sis, so they do not display Mendelian inheritance.

The first hint that extranuclear genes exist came from studies by the German scientist Carl Correns on the inheri-tance of yellow or white patches on the leaves of an other-wise green plant. In 1909, he observed that the coloration of the offspring was determined only by the maternal par-ent (the source of eggs) and not by the paternal parent (the source of sperm). Subsequent research showed that such coloration patterns, or variegation, are due to mutations in plastid genes that control pigmentation (Figure 15.18). In most plants, a zygote receives all its plastids from the cyto-plasm of the egg and none from the sperm, which contrib-utes little more than a haploid set of chromosomes. An egg may contain plastids with different alleles for a pigmentation gene. As the zygote develops, plastids containing wild-type or mutant pigmentation genes are distributed randomly to

▼ Figure 15.18 A painted nettle

coleus plant. The variegated

(patterned) leaves on this co-

leus plant (Solenostemon

scutellarioides) result

from mutations that

affect expression of

pigment genes lo-

cated in plastids,

which generally

are inherited

from the mater-

nal parent.daughter cells. The pattern of leaf coloration exhibited by a plant depends on the ratio of wild-type to mutant plastids in its various tissues.

Similar maternal inheritance is also the rule for mito-chondrial genes in most animals and plants, because almost all the mitochondria passed on to a zygote come from the cytoplasm of the egg. (The few mitochondria contributed by the sperm appear to be destroyed in the egg by autophagy; see Figure 6.13.) The products of most mitochondrial genes help make up the protein complexes of the electron trans-port chain and ATP synthase (see Figure 9.15). Defects in one or more of these proteins, therefore, reduce the amount of ATP the cell can make and have been shown to cause a number of rare human disorders. Because the parts of the body most susceptible to energy deprivation are the nervous system and the muscles, most mitochondrial diseases pri-marily affect these systems. For example, mitochondrial my-opathy causes weakness, intolerance of exercise, and muscle deterioration. Another mitochondrial disorder is Leber’s hereditary optic neuropathy, which can produce sudden blindness in people as young as their 20s or 30s. The four mutations found thus far to cause this disorder affect oxida-tive phosphorylation during cellular respiration, a crucial function for the cell (see Concept 9.4).

In addition to the rare diseases clearly caused by defects in mitochondrial DNA, mitochondrial mutations inherited from a person’s mother may contribute to at least some types of diabetes and heart disease, as well as to other dis-orders that commonly debilitate the elderly, such as Al-zheimer’s disease. In the course of a lifetime, new mutations gradually accumulate in our mitochondrial DNA, and some researchers think that these mutations play a role in the nor-mal aging process.

Wherever genes are located in the cell—in the nucleus or in cytoplasmic organelles—their inheritance depends on the precise replication of DNA. In the next chapter, you will

learn how this molecular reproduction occurs.

1. Gene dosage—the number of copies of a gene that are

actively being expressed—is important to proper devel-

opment. identify and describe two processes that estab-

lish the proper dosage of certain genes.

2. reciprocal crosses between two primrose varieties, a and

B, produced the following results: a female * B male S

offspring with all green (nonvariegated) leaves; B female

* a male S offspring with patterned (variegated) leaves.

explain these results.

Mitochondrial genes are critical to the

caused by mutations in these genes are generally not

lethal. Why not?

For suggested answers, see appendix a.

C h a p t e r15 The Chromosomal Basis of Inheritance 309

SuMMAry OF KEy COnCEpTS

C O N C E P T

15.1

Morgan showed that Mendelian inheritance has its physical basis in the behavior of chromosomes: (pp. 294–295)

? Morgan’s work with an eye color gene in chromosome theory of inheritance located on chromosomes and that the behavior of chromosomes during meiosis accounts for Mendel’s laws.

? What characteristic of the sex chromosomes allowed Morgan to

correlate their behavior with that of the alleles of the eye-color gene?

C O N C E P T

15.2

Chapter Review

15

C h a p t e r 15 The Chromosomal Basis of Inheritance 311

white-oval plants, and 1,000 F 2 progeny are obtained. How many F 2 plants of each of the four phenotypes do you expect? 9. You design Drosophila crosses to provide recombination data

for gene a , which is located on the chromosome shown in Fig-ure 15.12. Gene a has recombination frequencies of 14% with the vestigial-wing locus and 26% with the brown-eye locus. Approximately where is a located along the chromosome?LEvEL 3: SynTHESiS/EvALuATiOn

10. Banana plants, which are triploid, are seedless and therefore

sterile. Propose a possible explanation.11. EvOLuTiOn COnnECTiOn

Crossing over is thought to be evolutionarily advantageous because it continually shuffles genetic alleles into novel combinations. Until recently, it was thought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosome with which to pair up prior to crossing over. However, when the Y chromosome was sequenced, eight large regions were found to be internally homologous to each other, and quite a few of the 78 genes rep-resent duplicates. (Y chromosome researcher David Page has called it a “hall of mirrors.”) What might be a benefit of these regions?

12. SCiEnTiFiC inQuiry

D r a w I t Assume you are mapping genes A, B, C, and D in Drosophila . You know that these genes are linked on the same chromosome, and you determine the recombination frequen-cies between each pair of genes to be as follows: A–B , 8%; A–C , 28%; A–D, 25%; B–C, 20%; B–D, 33%.

(a) Describe how you determined the recombination frequen-cies for each pair of genes.

(b) D raw a chromosome map based on your data.13. WriTE ABOuT A THEME: inFOrMATiOn

The continuity of life is based on heritable information in the form of DNA. In a short essay (100–150 words), relate the structure and behavior of chromosomes to inheritance in both asexually and sexually reproducing species.TEST yOur unDErSTAnDinG

LEvEL 1: KnOWLEDGE/COMprEHEnSiOn

1. A man with hemophilia (a recessive, sex-linked condition) has

a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will have hemophilia? That a son will have hemo-philia? If the couple has four sons, what is the probability that all four will be born with hemophilia?

2. Pseudohypertrophic muscular dystrophy is an inherited disor-der that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to apparently normal parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.

3. A wild-type fruit fly (heterozygous for gray body color and

normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild-type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

4. A planet is inhabited by creatures that reproduce with the

same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not “intelligent,” Earth scientists are able to do some con-trolled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae-upturned snout, 47; antennae-downturned snout, 2; no antennae-downturned snout, 48; no antennae-upturned snout, 3. Calculate the re-combination frequencies for both experiments.LEvEL 2: AppLiCATiOn/AnALySiS

5. Using the information from problem 4, scientists do a further

testcross using a heterozygote for height and nose morphol-ogy. The offspring are tall-upturned snout, 40; dwarf-upturned snout, 9; dwarf-downturned snout, 42; tall-downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

6. A wild-type fruit fly (heterozygous for gray body color and

red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body-color, wing-size, and eye-color genes on the chromosome?

7. Assume that genes A and B are on the same chromosome and

are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phe-notypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

8. Two genes of a flower, one controlling blue (B ) versus white

(b ) petals and the other controlling round (R ) versus oval (r ) stamens, are linked and are 10 map units apart. You cross a homozygous blue-oval plant with a homozygous white-round plant. The resulting F 1 progeny are crossed with homozygous

14. SynTHESiZE yOur KnOWLEDGE

Butterflies have an X-Y sex determina-tion system that is different from that of flies or humans. Female butterflies may be either XY or XO, while but-terflies with two or more X chromo-somes are males. This photograph shows a tiger swal-lowtail gynandro-morph , which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual-looking butterfly.

For selected answers, see Appendix A.

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