Biology Campbell 12 ed-115-414-229-248.pdf

15 The Chromosomal Basis of Inheritance KEY CONCEPTS 15.1 Mendelian inheritance has its physical basis in the behavior of chromosomes p. 295 15.2 Sex-linked genes exhibit unique patterns of inheritance p. 298 15.3 Linked genes tend to be inherited together because they are located near each...

15 The Chromosomal Basis of Inheritance KEY CONCEPTS 15.1 Mendelian inheritance has its physical basis in the behavior of chromosomes p. 295 15.2 Sex-linked genes exhibit unique patterns of inheritance p. 298 15.3 Linked genes tend to be inherited together because they are located near each other on the same chromosome p. 301 15.4 Alterations of chromosome number or structure cause some genetic disorders p. 306 15.5 Some inheritance patterns are exceptions to standard Mendelian inheritance p. 310 Study Tip Draw chromosomes: As you work on the figure legend questions in this chapter, draw the chromosomes for the crosses described in the questions. Below is a starting sketch for the What If? question that follows Figure 15.3. (Here you are including sex chromosomes to keep track of the offspring.) w+ w X Y X X w X Sperm w Y F1 Generation X Go to Mastering Biology For Students (in eText and Study Area) • Get Ready for Chapter 15 • Animation: The Chromosomal Basis of Independent Assortment • Animation: Linked Genes and Crossing Over For Instructors to Assign (in Item Library) • Tutorial: Pedigrees and Sex-Linkage • Tutorial: Recombination and Linkage Mapping 294 What is the relationship between genes and chromosomes? Chromosome Genes are located on chromosomes. In diploid cells, chromosomes and genes are present in pairs. w P Generation w+ Eggs w+ Figure 15.1 The four yellow dots mark the locations of a specific gene, tagged with a fluorescent yellow dye, on a pair of homologous chromosomes. The chromosomes have duplicated, so each chromosome has two sister chromatids, each with a copy of the gene. This provides a visual demonstration that genes—Mendel’s “factors”—are segments of DNA located along chromosomes. Maternal cell Chromosomes duplicate before cell division. Each duplicated chromosome has two copies of each allele, one on each sister chromatid. Alleles (alternative versions of a gene) Pair of homologous chromosomes Paternal cell Sister chromatids of one duplicated chromosome During meiosis I, homologous chromosomes separate and alleles segregate. In meiosis II, sister chromatids separate. Egg Meiosis I and II Sperm Genes are passed on as discrete units. Each chromosome Fertilization has one version of a gene (one allele). Maternal chromosome Offspring inherit one allele from each parent. Pair of homologous chromosomes (one from each parent) Paternal chromosome Homologous chromosomes each have one allele of a given gene at the same locus. The chromosomes in this example have different alleles. CONCEPT 15.1 Mendelian inheritance has its physical basis in the behavior of chromosomes When Gregor Mendel proposed the existence of “hereditary factors” in 1860, no cellular structures had been identified that could house these imaginary units, and most biologists were skeptical. When the processes of mitosis and meiosis were worked out later that century, biologists saw parallels between the behavior of Mendel’s proposed factors (genes) during sexual life cycles and the behavior of chromosomes: As shown in Figure 15.1, chromosomes and genes are both present in pairs in diploid cells, and homologous chromosomes separate and alleles segregate during the process of meiosis. After meiosis, fertilization restores the paired condition for both chromosomes and genes. Around 1902, Walter S. Sutton, Theodor Boveri, and others independently noted these parallels and began to develop the chromosome theory of inheritance. According to this theory, Mendelian genes have specific loci (sites) along chromosomes, and it is the chromosomes that undergo segregation and independent assortment. The first solid evidence associating a specific gene with a specific chromosome came early in the 1900s from the work of Thomas Hunt Morgan, an experimental embryologist at Columbia University. Although Morgan was initially doubtful about both Mendelian genetics and the chromosome theory, his early experiments provided convincing 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 discoveries have come to those insightful or lucky enough to choose an experimental organism suitable for the research problem 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 melanogaster, a common insect that feeds on the fungi growing on fruit. Fruit flies are prolific breeders; a single mating will 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 tedious task of carrying out many matings and then microscopically . Figure 15.2 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. Wild type (red eyes) Mutant (white eyes) 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 instead 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 15.2). 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 mutations, in the wild-type allele. Morgan and his students invented a notation for symbolizing 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 superscript + 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 in all capitals, such as HTT for the gene involved in Huntington’s disease. (More than one letter may be used for an allele, as in HTT.) Correlating Behavior of a Gene’s Alleles with Behavior of a Chromosome Pair: Scientific Inquiry Morgan mated his white-eyed male fly with a red-eyed female. All the F1 offspring had red eyes, suggesting that the wild-type allele is dominant. When Morgan bred the F1 flies to each other, he observed the classical 3:1 phenotypic ratio among the F2 offspring. However, there was a surprising additional result: The white-eye trait showed up only in males. All the F2 females had red eyes, while half the males had red eyes and half had white eyes. Therefore, Morgan concluded CHAPTER 15 The Chromosomal Basis of Inheritance 295 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 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 F2 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 15.3. 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 F2 females in Morgan’s experiment because all the F1 fathers had red eyes, so each F2 female received a w+ allele on the X chromosome inherited from her father. Morgan’s finding of the correlation between a particular trait and an individual’s sex provided support for the chromosome theory of inheritance: namely, that a specific gene is carried on a specific chromosome. In this case, an eye color gene is carried on the X chromosome. Figure 15.4 illustrates the relationship between the chromosome theory of inheritance and Mendel’s laws. 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 chromosome pairs at metaphase I accounts for independent assortment 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 15.4, you can see how the behavior of chromosomes during meiosis in the F1 generation and subsequent random fertilization give rise to the F2 phenotypic ratio observed by Mendel. In addition to showing that a specific gene is carried on a specific chromosome, 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. CONCEPT CHECK 15.1 1. WHAT IF? Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found in a male. 2. Which one of Mendel’s laws describes the inheritance of alleles for a single character? Which law relates to the inheritance of alleles for two characters in a dihybrid cross? 3. MAKE CONNECTIONS Review the description of meiosis (see Figure 13.8) and Mendel’s laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel’s laws? For suggested answers, see Appendix A. 296 UNIT THREE Genetics ▼ Figure 15.3 Inquiry In a cross between a wild-type female fruit fly and a mutant white-eyed male, what color eyes will the F1 and F2 offspring have? 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. P Generation 3 F1 Generation All offspring had red eyes. Morgan then bred an F1 red-eyed female to an F1 red-eyed male to produce the F2 generation. Results The F2 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. F2 Generation Conclusion All F1 offspring had red eyes, so the mutant white-eye trait (w) must be recessive to the wild-type redeye trait (w+). Since the recessive trait—white eyes—was expressed only in males in the F2 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. P Generation X X w1 3 w1 X Y w Sperm Eggs F1 Generation w1 w1 w1 w w1 Eggs F2 Generation w w1 Sperm w1 w1 w1 w w w w1 Data from T. H. Morgan, Sex-limited inheritance in Drosophila, Science 32: 120–122 (1910). Instructors: A related Experimental Inquiry Tutorial can be assigned in Mastering Biology. WHAT IF? Suppose this eye color gene were located on an autosome rather than a sex chromosome. Predict the phenotypes (including sex) of the F2 flies in this hypothetical cross. (Hint: Draw Punnett squares, including chromosomes, for the F1 and F2 generations; see the Study Tip.) . Figure 15.4 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 for, respectively, the independent assortment and segregation of the alleles for seed color and shape. Each cell that undergoes meiosis in an F1 plant produces two kinds of gametes. If we count the results for all cells, however, each F1 plant produces equal numbers of all four kinds of gametes because the alternative chromosome arrangements at metaphase I are equally likely. P Generation Starting with two true-breeding pea Y plants, we will follow two genes through the F1 and F2 generations. R R Y 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 chromosome pairs. (Peas have seven chromosome pairs, but only R Y two pairs are illustrated here.) Gametes Yellow round seeds (YYRR) Mastering Biology Animation: The Chromosomal Basis of Independent Assortment MP3 Tutor: The Chromosomal Basis of Inheritance Green wrinkled seeds (yyrr) r 3 y r y Meiosis Fertilization y r All F1 plants produce yellow round seeds (YyRr). F1 Generation R R y r y r Y Y Meiosis LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. As an example, follow the fate of the long chromosome pair (carrying R and r). Read the numbered explanations below. R r Y y Two equally probable arrangements of chromosomes at metaphase I r R Y y LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. As an example, follow both the long and short chromosome pair along both paths. Read the numbered explanations below. 1 The R and r alleles segregate 1 Alleles of each gene segregate in at anaphase I, yielding two types of daughter cells for this gene. R r Y y R Y y 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. Anaphase I r R r R Y y Metaphase II 2 Each gamete gets one long chromosome with either the R or r allele. Gametes r y Y y Y Y R R 1 4 2 Each gamete gets YR r r r 1 4 F2 Generation Y Y y r yr 1 4 Yr y y R R 1 4 yR An F1 3 F1 cross-fertilization 3 Fertilization recombines the R and r alleles at random. a long and a short chromosome in one of four allele combinations. 3 Fertilization results in 9 :3 :3 :1 the 9:3:3:1 phenotypic ratio in the F2 generation. ? If you crossed an F1 plant above 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? CHAPTER 15 The Chromosomal Basis of Inheritance 297 CONCEPT 15.2 Sex-linked genes exhibit unique patterns of inheritance Morgan’s discovery of a trait (white eyes) that correlated with the sex of flies provided important support for the chromosome theory of inheritance. 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 chromosomes could be correlated with the behavior of the two alleles 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 sex has traditionally been described as binary— male or female—we are coming to understand that this classification may be too simplistic. Here, we use the term sex to refer to classification into a group with a shared set of anatomical and physiological traits. In this sense, sex in many species is determined largely by inheritance of sex chromosomes. (The term gender, previously used as a synonym of sex, is now more often used to refer to an individual’s own experience of identifying as male, female, or otherwise.) 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 anatomy we associ. Figure 15.5 Human sex ate with the “female” sex, chromosomes (duplicated). while “male” properties are associated with the inheriX tance of one X chromosome and one Y chromosome (Figure 15.6a). Short segY ments at either end of the Y chromosome are the only regions that are homologous with regions on 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 chromosomes 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 chromosome, 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, in general, 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 determining sex. Figure 15.6b–d illustrates three other systems. 298 UNIT THREE Genetics . Figure 15.6 Some chromosomal systems of sex determination. Numerals indicate the number of autosomes in the species pictured. In Drosophila, males are XY, but sex depends on the ratio of the number of X chromosomes to the number of autosome sets, not simply on the presence of a Y chromosome. In some species (not shown here), sex is determined not by chromosomes but by environmental factors such as temperature. 44 + XY Parents 44 + XX + 22 + X 22 + 22 + or Y X Sperm 44 + XX Egg or 44 + XY Zygotes (offspring) (a) The X-Y system. In mammals, the sex of an offspring depends on whether the sperm cell contains an X chromosome or a Y. 22 + XX 22 + X (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 contains an X chromosome or no sex chromosome. 76 + ZW 76 + ZZ (c) The Z-W system. In birds, some fishes, and some insects, the sex chromosomes present in the egg (not the sperm) differ, and thus determine the sex of offspring. The sex chromosomes are designated Z and W. Females are ZW and males are ZZ. 32 (Diploid) 16 (Haploid) (d) The haplo-diploid system. There are no sex chromosomes in most species of bees and ants. Females develop from fertilized eggs and are thus diploid. Males develop from unfertilized eggs and are haploid; they have no fathers. 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, and depending on what genes are active. A gene on the Y chromosome—called SRY, for sex-determining region of Y—is required for the development of testes. In the absence of SRY, the gonads develop into ovaries, even in an XY embryo. very few Y-linked genes, many of which help determine sex, the X chromosomes have numerous genes for characters unrelated to sex. X-linked genes in humans follow the same pattern of inheritance that Morgan observed for the eye color locus he studied in Drosophila (see Figure 15.3). 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 describing 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 colorblind father whose mate is a carrier (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 5,000 males born in the United States. The disease is characterized by a progressive weakening of the muscles and loss of coordination. Affected individuals’ life expectancy is into the mid-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. Since the gene is known, gene therapy is being explored. A gene located on either sex chromosome is called a sex-linked gene. The human X chromosome contains approximately 1,100 genes, which are called X-linked genes, while genes located on the Y chromosome are called Y-linked genes. On the human Y chromosome, researchers 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 normal testicular functioning and the production of normal sperm. 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. The development of female gonads in humans requires a gene called WNT4 (on chromosome 1, an autosome), which encodes a protein that promotes ovary development. An embryo that is XY but has extra copies of the WNT4 gene can develop rudimentary female gonads. Overall, sex is determined by the interactions of a network of gene products like these. The biochemical, physiological, and anatomical features associated with “males” and “females” are turning out to be more complicated than previously thought, with many genes involved in their development. Because of the complexity of this process, many variations exist: Some individuals vary in the number of sex chromosomes in their cells (see Concept 15.4), and others are born with intermediate sexual (intersex) characteristics, or with anatomical features that do not match an individual’s sense of their own gender (transgender individuals). Sex determination is an active area of research that will likely yield a more sophisticated understanding in years to come. Inheritance of 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 produced by genes located on autosomes. While there are . 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 causing color blindness. White boxes indicate unaffected individuals, light orange boxes indicate carriers, and dark orange boxes indicate color-blind individuals. XNXN 3 Xn X nY Y XNXn XN Sperm 3 ? If a color-blind woman married a man who had normal color vision, what would be the probable phenotypes of their children? Mastering Biology Animation: X-Linked Genes in “MendAliens” XNY Y XNXn Xn Sperm XnY Y Eggs XN XNXn XNY Eggs XN XNXn XNY Eggs XN XNXn XNY XN XNXn XNY Xn XNXn XnY Xn XnXn XnY (a) A father with color-blindness will transmit the mutant allele to all daughters but to no sons. When the mother is a dominant homozygote, the daughters will have the normal phenotype but will be carriers of the mutation. (b) If a carrier mates with a male who has normal color vision, there is a 50% chance that each daughter will be a carrier like her mother and a 50% chance that each son will have the disorder. CHAPTER 15 Sperm (c) If a carrier mates with a color-blind male, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who have normal color vision will be carriers, whereas males who have normal color vision will be free of the recessive allele. The Chromosomal Basis of Inheritance 299 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 bleeding 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 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 documented 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 many of the proteins encoded by X-linked genes as males. In fact, almost all of one X chromosome in each cell in female mammals becomes inactivated during early embryonic development. As a result, the cells of females and males have the same effective dose (one active 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, resulting in every female gamete (egg) having an active X after meiosis. British geneticist Mary Lyon demonstrated that the selection of which X chromosome will form the Barr body occurs randomly and independently in each embryonic cell present at the time of X inactivation. As a consequence, females consist of a mosaic of two types of cells: those with the active X derived from the father and those with the active X derived from the mother. After an X chromosome is inactivated 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 of her cells will express one allele, while the others will express the alternate allele. Figure 15.8 shows how this mosaicism results in the patchy coloration of a tortoiseshell cat. In humans, mosaicism can be observed in a recessive X-linked mutation that prevents 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 modification of the DNA and proteins bound to it called histones, including attachment of methyl groups ( ¬ CH3) to DNA nucleotides. 300 UNIT THREE Genetics . Figure 15.8 X inactivation and the tortoiseshell cat. The tortoiseshell gene is on the X chromosome, and the tortoiseshell phenotype 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 heterozygous 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 chromosome with the black allele is active. (“Calico” cats also have white areas, which are determined by another gene.) X chromosomes Early embryo: Two cell populations in adult cat: Allele for orange fur Allele for black fur A few cell divisions and X chromosome inactivation Active X Inactive X Active X Black fur Orange fur Mastering Biology HHMI Animation: X Inactivation (The regulatory role of DNA methylation is discussed in Concept 18.2.) A particular region of each X chromosome contains several genes involved in the inactivation process. The two regions, one on each X chromosome, associate briefly with each other in each cell at an early stage of embryonic development. Then one of the genes, called XIST (for X-inactive specific transcript), becomes active only on the chromosome that will become the Barr body (the inactive X). Multiple copies of the RNA product of this gene apparently attach to the X chromosome on which they are made, eventually almost covering it. Interaction of this RNA with the chromosome initiates X inactivation, and the RNA products of nearby genes help to regulate the process. CONCEPT CHECK 15.2 1. A white-eyed female Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of the one shown in Figure 15.3. What phenotypes and genotypes do you predict for the offspring from this cross? 2. Neither Tim nor Shonda has Duchenne muscular dystrophy, 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? 3. MAKE CONNECTIONS Consider what you learned about dominant and recessive alleles in Concept 14.1. 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. CONCEPT How Linkage Affects Inheritance 15.3 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 phenotypes. 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 alleles, and neither gene is on a sex chromosome. In his investigation of these two genes, Morgan carried out the crosses shown in Figure 15.9. The first was a P generation cross to generate F1 dihybrid flies, and the second was essentially a testcross. Linked genes tend to be inherited together because they are located near each other on the same chromosome 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. ▼ Figure 15.9 Inquiry How does linkage between two genes affect inheritance of characters? 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). Morgan mated true-breeding P (parental) P Generation generation flies—wild-type flies with black, (homozygous) Double mutant vestigial-winged flies—to produce hetero3 (black body, 1 1 Wild type zygous F1 dihybrids (b b vg vg), all of vestigial wings) (gray body, which are wild-type in appearance. normal wings) b b vg vg b1 b1 vg1 vg1 He then mated wild-type F1 dihybrid females with homozygous recessive males. This testcross will reveal the genotype of the eggs made by the dihybrid female. F1 dihybrid testcross Wild-type F1 dihybrid (gray body, normal wings) Homozygous recessive (black body, vestigial wings) 3 b1 b vg1 vg The male’s sperm contributes only recessive alleles, so the phenotype of the offspring reflects the genotype of the female’s eggs. b b vg vg Testcross offspring Note: Although only females (with pointed abdomens) are shown, half the offspring in each class would be males (with rounded abdomens). Eggs b1 vg1 b vg b1 vg b vg1 Wild type (gray normal) Black vestigial Gray vestigial Black normal b1 b vg1 vg b b vg vg b1 b vg vg b b vg1 vg b vg Sperm Predicted ratios of testcross offspring Results if genes are located on different chromosomes: 1 : 1 : 1 : 1 if genes are located on the same chromosome and parental alleles are always inherited together: 1 : 1 : 0 : 0 965 : 944 : 206 : 185 Data from Morgan’s experiment: 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. Data from 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). WHAT IF? If the parental (P generation) flies had been true-breeding for gray body with vestigial wings and black body with normal wings, which phenotypic class(es) would be largest among the testcross offspring? CHAPTER 15 The Chromosomal Basis of Inheritance 301 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 are linked; they are near each other on the same chromosome: F1 dihybrid female and homozygous recessive male in testcross b vg b1 vg1 3 b vg b vg b1 vg1 b vg Most offspring or b vg b vg 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 inherited together) and sex-linked gene (a single gene on a sex chromosome). 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 production 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 of alleles 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 characters 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, heterozygous for both seed color and seed shape (YyRr), with a plant homozygous for both recessive alleles (with green wrinkled seeds, yyrr). (This cross acts as a testcross because the results will reveal the genotype not only of the dihybrid YyRr plant, 302 UNIT THREE Genetics which we know, but of the gametes made in that plant.) Let’s represent the cross by the following Punnett square: Gametes from yellow round dihybrid parent (YyRr) Gametes from green wrinkled homozygous recessive parent (yyrr) YR yr Yr yR YyRr yyrr Yyrr yyRr yr Parentaltype offspring Recombinant offspring 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 F1 dihybrid (see Figure 15.2). These matching offspring are called parental types (short for phenotypes). But two nonparental 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 recombinants, 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 actually 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 15.4 legend). Recombination of Linked Genes: Crossing Over Now, let’s explain the results of the Drosophila testcross in Figure 15.9. Recall that most (83%) of the offspring from the testcross for body color and wing size had parental phenotypes. That suggested that the two genes were on the same chromosome, since the occurrence of parental types with a frequency 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 breaks the DNA molecules of one maternal and one paternal chromatid and rejoins each to the other c Figure 15.10 Chromosomal P generation Wild type basis for recombination of linked (homozygous) (gray body, genes. In these diagrams re-creating normal wings) the testcross in Figure 15.9, we track b1 vg1 chromosomes as well as genes. The maternal chromosomes (those present in the wild-type F1 dihybrid) b1 vg1 are color-coded red and pink to distinguish one homolog from the other before any meiotic crossing over has occurred. Because crossing over between the b+/b and vg+/vg F1 dihybrid testcross Wild-type F1 dihybrid loci occurs in some, but not most, (gray body, egg-producing cells, more eggs normal wings) with parental-type chromosomes b1 vg1 than with recombinant ones are produced in the mating females. Fertilization of the eggs by sperm b vg of genotype b vg gives rise to Replication some recombinant offspring. The of chromosomes recombination frequency is the b1vg1 percentage of recombinant flies in the total pool of offspring. b1 vg1 DRAW IT Suppose, as in the question b vg at the bottom of Figure 15.9, the parental (P generation) flies were b vg true-breeding for gray body with Meiosis I vestigial wings and black body with (including normal wings. Draw the chromosomes crossing over) b1 vg1 in each of the four possible kinds of eggs from an F1 female, and label each chromosome as “parental” or b1 vg “recombinant.” b vg1 Mastering Biology Animation: Linked Genes and Crossing Over Double mutant (black body, vestigial wings) b vg b vg Homozygous recessive (black body, vestigial wings) b vg b vg Replication of chromosomes b vg b vg b vg b vg Meiosis I and II b vg Meiosis II Eggs Testcross offspring b1vg1 965 Wild type (gray normal) Recombinant chromosomes b vg b1 vg 185 Black normal 206 Gray vestigial 944 Black vestigial b vg1 b1 vg1 b vg b1 vg b vg1 b vg b vg b vg b vg Parental-type offspring b vg Sperm Recombinant offspring Recombination = 391 recombinants 3 100 = 17% frequency 2,300 total offspring (see Figure 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 recombinant 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 eggs had a recombinant chromosome (b+ vg or b vg+). Fertilization of all classes of eggs by homozygous recessive sperm (b vg) produced an offspring population in which 17% exhibited a nonparental, 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 analyze the results from an F1 dihybrid testcross and see whether the two genes assort independently or are linked. CHAPTER 15 The Chromosomal Basis of Inheritance 303 Scientific Skills Exercise Using the Chi-Square (x2) Test Are Two Genes Linked or Unlinked? Genes that are in close proximity 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 randomly? In this exercise, you’ll use a simple statistical test, the chi-square (x2) test, to analyze phenotypes of F1 testcross progeny in order to see whether two genes are linked or unlinked. How These Experiments Are Done If genes are unlinked and assorting independently, the phenotypic ratio of offspring from an F1 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 (x2) test, compares an observed data set with an expected data set predicted by a hypothesis (here, that the genes are unlinked) and measures the discrepancy between the two, thus determining 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 explained by random variation alone. In this case, we say that 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 F1 dihybrids (AaBb), which were then testcrossed (AaBb * aabb). A total of 900 offspring plants were scored for stem color and flower petal length. Offspring from testcross of AaBb (F1) 3 aabb Purple stem/short petals (A-B-)* Green stem/short petals (aaB-) Purple stem/long petals (A-bb) Green stem/long petals (aabb) 1 1 1 1 220 210 231 239 Expected ratio if the genes are unlinked Expected number of offspring (of 900) Observed number of offspring (of 900) *If the phenotype is dominant, a dash is used for the second allele; it could be either the dominant or recessive allele. 304 UNIT THREE Genetics Cosmos plants INTERPRET THE DATA 1. The results in the data table are from a simulated F1 dihybrid testcross. The hypothesis that the two genes are unlinked predicts that the offspring phenotypic ratio will be 1:1:1:1. Using this ratio, calculate the expected number of each phenotype out of the 900 total offspring, and enter the values in that data table. 2. The goodness of fit is measured by x2. This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is x2 5 S (o 2 e)2 e where g = sum of, o = observed and e = expected. Calculate the x2 value for the data using the table below. Fill out that table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the x2 value. Testcross Expected Observed Deviation Offspring (e) (o) (o 2 e) (o 2 e)2 (o 2 e)2/e A2B2 220 aaB2 210 A2bb 231 aabb 239 x2 = Sum 3. The x2 value means nothing on its own—it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests 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 corresponding to the x2 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 significant; the observed data are consistent with the hypothesis that the genes are unlinked. To find the probability, locate your x2 value in the x2 distribution table in Appendix D. 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 calculated x2 value lies between. (b) The column headings for these values show the probability range for your x2 number. Based on whether there are nonsignificant (p . 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? Instructors: A version of this Scientific Skills Exercise can be assigned in Mastering Biology. New Combinations of Alleles: Variation for Natural Selection EVOLUTION The physical behavior of chromosomes during meiosis contributes to the generation of variation 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 prophase 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 offspring (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 material 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. Ultimately, the interplay between environment and phenotype (and thus 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 recombinant 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 relative 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 linkage 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, crossovers 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 alternative 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 ▼ Figure 15.11 Research Method Constructing a Linkage Map Application A linkage map shows the relative locations of genes along a chromosome. Technique A linkage map is based on the assumption that the probability 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 between three Drosophila gene pairs (b and cn, 9%; cn and vg, 9.5%; and b and vg, 17%) best fit a linear order in which cn is positioned about halfway between the other two genes: Recombination frequencies 9% Chromosome 9.5% 17% 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 frequency 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. CHAPTER 15 The Chromosomal Basis of Inheritance 305 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 genes can have a maximum value of 50%, a result indistinguishable 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 genetically 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 distance 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 experiments. 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 arrays. 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 specific genes, each gene with its own locus (Figure 15.12). Because a linkage map is based strictly on recombination 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 assumed, 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 recent decades have enormously increased the rate and affordability of DNA sequencing. Therefore, today, most researchers sequence whole genomes to map the locations of genes of a given species. The entire nucleotide sequence is the ultimate physical map of a chromosome, revealing the physical distances in DNA nucleotides between gene loci (see Concept 21.1). Comparing a linkage map with such a physical 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. CONCEPT CHECK 15.3 1. When two genes are located on the same chromosome, what is the physical basis for the production of recombinant offspring in a testcross between a dihybrid parent and a double-mutant (recessive) parent? 2. VISUAL SKILLS 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.) 3. WHAT IF? Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and that between A and C is 12%. Can you determine the linear order of these genes? Explain. For suggested answers, see Appendix A. CONCEPT . Figure 15.12 A partial genetic (linkage) map of a Drosophila chromosome. This partial map shows just seven mapped genes on Drosophila chromosome II. (DNA I sequencing has revealed over 9,000 genes on that chromosome.) The number at each X IV Y II III gene locus is the number of map units from the arista length gene (far left). Mutant phenotypes Short aristae 0 Maroon eyes 16.5 Long Red aristae eyes (appendages on head) 306 Black body UNIT THREE Vestigial wings Down- Brown curved eyes wings 57.5 67.0 75.5 104.5 Cinnabar eyes 48.5 Gray body Red eyes Normal wings Wild-type phenotypes Genetics Normal wings Red eyes 15.4 Alterations of chromosome number or 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 involving 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 organism’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 chromosomal alterations in humans and other mammals often lead to spontaneous abortion (miscarriage) of a fetus, and individuals born with these types of genetic defects commonly exhibit various developmental disorders. Plants appear to tolerate such genetic defects better than animals do. Mastering Biology Interview with Terry Orr-Weaver: Why are humans so bad at meiosis? 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 separate 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 particul