Genes Are located on chromosomes Can be visualized using certain techniques

The chromosome theory of inheritance states that Mendelian inheritance has its physical basis in the behavior of chromosomes The behavior of chromosomes during meiosis was said to account for Mendel’s laws of segregation and independent assortment The chromosome theory of inheritance states that Mendelian genes have specific loci on chromosomes that undergo segregation and independent assortment

Fertilization among the F1 plants The chromosomal basis of Mendel’s laws Yellow-round seeds (YYRR) Green-wrinkled seeds (yyrr) Meiosis Fertilization Gametes All F1 plants produce yellow-round seeds (YyRr) P Generation F1 Generation Two equally probable arrangements of chromosomes at metaphase I LAW OF SEGREGATION LAW OF INDEPENDENT ASSORTMENT Anaphase I Metaphase II Fertilization among the F1 plants 9 : 3 : 1 1 4 YR yr yR Y R y r F2 Generation Starting with two true-breeding pea plants, we follow two genes through the F1 and F2 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.) The R and r alleles segregate at anaphase I, yielding two types of daughter cells for this locus. Each gamete gets one long chromosome with either the R or r allele. 2 recombines the R and r alleles at random. 3 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 in the cells on the left and right Each gamete gets a long and a short chromosome in one of four allele combinations. Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation.

Morgan worked with fruit flies Morgan’s Experiment Thomas Hunt Morgan Provided convincing evidence that chromosomes are the location of Mendel’s heritable factors Morgan worked with fruit flies Because they breed at a high rate A new generation can be bred every two weeks They have only four pairs of chromosomes

Morgan first observed and noted Wild type (normal), phenotypes that were common in the fly populations Traits alternative to the wild type Are called mutant phenotypes

Correlating Behavior of a Gene’s Alleles with Behavior of a Chromosome Pair In one experiment Morgan mated male flies with white eyes (mutant) with female flies with red eyes (wild type) The F1 generation all had red eyes The F2 generation showed the 3:1 red:white eye ratio, but only males had white eyes

Morgan determined That the white-eye mutant allele must be located on the X chromosome The F2 generation showed a typical Mendelian 3:1 ratio of red eyes to white eyes. However, no females displayed the white-eye trait; they all had red eyes. Half the males had white eyes, and half had red eyes. Morgan then bred an F1 red-eyed female to an F1 red-eyed male to produce the F2 generation. RESULTS P Generation F1 X F2 Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes. EXPERIMENT

CONCLUSION Since all F1 offspring had red eyes, 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 F2 generation, Morgan hypothesized that the eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome, as diagrammed here. P Generation F1 F2 Ova (eggs) Sperm X Y W W+

Each chromosome has hundreds or thousands of genes Morgan’s discovery that transmission of the X chromosome in fruit flies correlates with inheritance of the eye-color trait Was the first solid evidence indicating that a specific gene is associated with a specific chromosome Linked genes tend to be inherited together because they are located near each other on the same chromosome Each chromosome has hundreds or thousands of genes

Recombinant (nonparental-type) Morgan crossed flies That differed in traits of two different characters Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) P Generation (homozygous) b+ b+ vg+ vg+ x b b vg vg F1 dihybrid (wild type) b+ b vg+ vg TESTCROSS b+vg+ b vg b+ vg b vg+ b+ b vg vg b b vg+ vg 965 (gray-normal) 944 Black- vestigial 206 Gray- 185 normal Sperm Parental-type offspring Recombinant (nonparental-type) RESULTS EXPERIMENT Morgan first mated true-breeding wild-type flies with black, vestigial-winged flies to produce heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to phenotype). CONCLUSION If these two genes were on different chromosomes, the alleles from the F1 dihybrid would sort into gametes independently, and we would expect to see equal numbers of the four types of offspring. If these two genes were on the same chromosome, we would expect each allele combination, B+ vg+ and b vg, to stay together as gametes formed. In this case, only offspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morgan concluded that the genes for body color and wing size are located on the same chromosome. However, the production of a small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between genes on the same chromosome.

Morgan determined that: Genes that are close together on the same chromosome are linked and do not assort independently Unlinked genes are either on separate chromosomes of are far apart on the same chromosome and assort independently Parents in testcross b+ vg+ b vg b+ vg+ b vg b vg Most offspring X or

Recombination of Unlinked Genes: Independent Assortment of Chromosomes When Mendel followed the inheritance of two characters He observed that some offspring have combinations of traits that do not match either parent in the P generation Gametes from green- wrinkled homozygous recessive parent (yyrr) Gametes from yellow-round heterozygous parent (YyRr) Parental- type offspring Recombinant offspring YyRr yyrr Yyrr yyRr YR yr Yr yR

Recombinant offspring Are those that show new combinations of the parental traits When 50% of all offspring are recombinants Geneticists say that there is a 50% frequency of recombination

Recombination of Linked Genes: Crossing Over Morgan discovered that genes can be linked But due to the appearance of recombinant phenotypes, the linkage appeared incomplete Morgan proposed that Some process must occasionally break the physical connection between genes on the same chromosome Crossing over of homologous chromosomes was the mechanism

Linked genes Exhibit recombination frequencies less than 50%  = Testcross parents Gray body, normal wings (F1 dihybrid) b+ vg+ b vg Replication of chromosomes Meiosis I: Crossing over between b and vg loci produces new allele combinations. Meiosis II: Segregation of chromatids produces recombinant gametes with the new allele  Recombinant chromosome b+vg+ b   vg b+ vg b vg+ b vg Sperm Meiosis I and II: Even if crossing over occurs, no new allele combinations are produced. Ova Gametes offspring b+  vg+ b   vg b+   vg b   vg+ 965 Wild type (gray-normal) b  vg b  vg+ 944 Black- vestigial 206 Gray- 185 normal Recombination frequency = 391 recombinants 2,300 total offspring 100 = 17% Parental-type offspring Recombinant offspring Black body, vestigial wings (double mutant)

Linkage Mapping: Using Recombination Data: Scientific Inquiry A genetic map Is an ordered list of the genetic loci along a particular chromosome Can be developed using recombination frequencies The farther apart genes are on a chromosome The more likely they are to be separated during crossing over

A linkage map Is the actual map of a chromosome based on recombination frequencies Recombination frequencies 9% 9.5% 17% b cn vg Chromosome The b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover would “cancel out” the first and thus reduce the observed b–vg recombination frequency. In this example, the observed recombination frequencies between 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: RESULTS A linkage map shows the relative locations of genes along a chromosome. APPLICATION 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 Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.

Many fruit fly genes Were mapped initially using recombination frequencies Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown Long aristae (appendages on head) Gray Red Normal Wild-type phenotypes II Y I X IV III 48.5 57.5 67.0 104.5

The Chromosomal Basis of Sex An organism’s sex Is an inherited phenotypic character determined by the presence or absence of certain chromosomes In humans and other mammals There are two varieties of sex chromosomes, X and Y

(d) The haplo-diploid system Different systems of sex determination Are found in other organisms 22 + XX X 76 + ZZ ZW 16 (Haploid) (Diploid) (b) The X–0 system (c) The Z–W system (d) The haplo-diploid system

Inheritance of Sex-Linked Genes The sex chromosomes Have genes for many characters unrelated to sex A gene located on either sex chromosome Is called a sex-linked gene Some recessive alleles found on the X chromosome in humans cause certain types of disorders Color blindness, Duchenne muscular dystrophy and hemophilia

Sex-linked genes Follow specific patterns of inheritance XAXA XaY Xa Y Ova Sperm XaXA  XaXa A father with the disorder 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. If a carrier mates with a male of normal phenotype, 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. If a carrier mates with a male who has the disorder, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who do not have the disorder will be carriers, where as males without the disorder will be completely free of the recessive allele. (a) (b) (c)

X inactivation in Female Mammals In mammalian females One of the two X chromosomes in each cell is randomly inactivated during embryonic development

If a female is heterozygous for a particular gene located on the X chromosome She will be a mosaic for that character Two cell populations in adult cat: Active X Orange fur Inactive X Early embryo: X chromosomes Allele for black fur Cell division and X chromosome inactivation Black Figure 15.11 Allele for orange fur

Large-scale chromosomal alterations Alterations of chromosome number or structure cause some genetic disorders Large-scale chromosomal alterations Often lead to spontaneous abortions or cause a variety of developmental disorders Aneuploidy Results from the fertilization of gametes in which nondisjunction occurred Is a condition in which offspring have an abnormal number of a particular chromosome

Abnormal Chromosome Number When nondisjunction occurs Pairs of homologous chromosomes do not separate normally during meiosis Gametes contain two copies or no copies of a particular chromosome Meiosis I Nondisjunction Meiosis II Gametes n + 1 n  1 n – 1 n –1 n Number of chromosomes Nondisjunction of homologous chromosomes in meiosis I Nondisjunction of sister chromatids in meiosis II (a) (b)

If a zygote is monosomic If a zygote is trisomic It has three copies of a particular chromosome If a zygote is monosomic It has only one copy of a particular chromosome

Polyploidy Is a condition in which there are more than two complete sets of chromosomes in an organism

Alterations of Chromosome Structure B C D E F G H Deletion Duplication M N O P Q R Inversion Reciprocal translocation (a) A deletion removes a chromosomal segment. (b) A duplication repeats a segment. (c) An inversion reverses a segment within a chromosome. (d) A translocation moves a segment from one chromosome to another, nonhomologous one. In a reciprocal   translocation, the most common type, nonhomologous chromosomes exchange fragments. Nonreciprocal translocations also occur, in which a chromosome transfers a fragment without receiving a fragment in return.

Human Disorders Due to Chromosomal Alterations Alterations of chromosome number and structure Are associated with a number of serious human disorders

Down Syndrome Down syndrome Is usually the result of an extra chromosome 21, trisomy 21

Aneuploidy of Sex Chromosomes Nondisjunction of sex chromosomes Produces a variety of aneuploid conditions Klinefelter syndrome Is the result of an extra chromosome in a male, producing XXY individuals Turner syndrome Is the result of monosomy X, producing an X0 karyotype

Two normal exceptions to Mendelian genetics include Some inheritance patterns are exceptions to the standard chromosome theory Two normal exceptions to Mendelian genetics include Genes located in the nucleus Genes located outside the nucleus

Genomic Imprinting In mammals The phenotypic effects of certain genes depend on which allele is inherited from the mother and which is inherited from the father

(a) A wild-type mouse is homozygous for the normal igf2 allele. Genomic imprinting Involves the silencing of certain genes that are “stamped” with an imprint during gamete production (a) A wild-type mouse is homozygous for the normal igf2 allele. Normal Igf2 allele (expressed) with imprint (not expressed) Paternal chromosome Maternal Wild-type mouse (normal size) Mutant lgf2 allele Dwarf mouse Normal size mouse (b) When a normal Igf2 allele is inherited from the father, heterozygous mice grow to normal size. But when a mutant allele is inherited from the father, heterozygous mice have the dwarf phenotype.

Inheritance of Organelle Genes The inheritance of traits controlled by genes present in the chloroplasts or mitochondria Some diseases affecting the muscular and nervous systems Are caused by defects in mitochondrial genes that prevent cells from making enough ATP Depends solely on the maternal parent because the zygote’s cytoplasm comes from the egg

Animations and Videos Independent Assortment and Gamete Diversity Alleles That Do Not Assort Independently Meiosis - Nondisjunction – 1 Meiosis - Nondisjunction – 2 Karyotype Animation Bozeman - Chi-squared Test Speciation by Ploidy Changes in Chromosome Structure

Animations and Videos Chapter Quiz Questions – 1 Chapter Quiz Questions - 2

Why did the improvement of microscopy techniques in the late 1800s set the stage for the emergence of modern genetics? It revealed new and unanticipated features of Mendel’s pea plant varieties. It allowed the study of meiosis and mitosis, revealing parallels between behaviors of the Mendelian concept of the gene and the movement/pairing of chromosomes. It allowed scientists to see the nucleotide sequence of DNA. It led to the discovery of mitochondria. It showed genes functioning to direct the formation of enzymes. Answer: B 39

Why did the improvement of microscopy techniques in the late 1800s set the stage for the emergence of modern genetics? It revealed new and unanticipated features of Mendel’s pea plant varieties. It allowed the study of meiosis and mitosis, revealing parallels between behaviors of the Mendelian concept of the gene and the movement/pairing of chromosomes. It allowed scientists to see the nucleotide sequence of DNA. It led to the discovery of mitochondria. It showed genes functioning to direct the formation of enzymes. 40

Morgan and his colleagues worked out a set of symbols to represent fly genotypes. Which of the following is representative? AaBb  AaBb 46 or 46w w or w on X 2  3 Answer: C 41

Morgan and his colleagues worked out a set of symbols to represent fly genotypes. Which of the following is representative? AaBb  AaBb 46 or 46w w or w on X 2  3 42

on the O chromosome of a male on the X chromosome of a male Imagine that Morgan had used a grasshopper (2n  24, and sex is determined as follows: male has X, and female has XX) to study sex linkage. Predict where the first mutant would have been discovered. on the O chromosome of a male on the X chromosome of a male on the X chromosome of a female on the Y chromosome of a male Answer: B This question is designed to show students that there are a variety of sex determination mechanisms and to help students understand the significance of Morgan’s work. Answer A is incorrect because there is no O chromosome—the X chromosome has no pairing partner. Answer C is incorrect because a mutant is likely recessive and is likely to be masked by a dominant on the other X chromosome in a female. Answer D is incorrect because grasshopper males do not have a Y chromosome. Answer B is correct because a mutant on the male’s single X chromosome would not be masked by a normal allele on a second X chromosome. 43

on the O chromosome of a male on the X chromosome of a male Imagine that Morgan had used a grasshopper (2n  24, and sex is determined as follows: male has X, and female has XX) to study sex linkage. Predict where the first mutant would have been discovered. on the O chromosome of a male on the X chromosome of a male on the X chromosome of a female on the Y chromosome of a male 44

Think about bees, which have no X and Y sex chromosomes Think about bees, which have no X and Y sex chromosomes. Males are haploid, whereas fertilization results in females, as diploid cells become females. Which of the following are accurate statements about bee males when they are compared to species in which males are XY and diploid for the autosomes? Bee males have half the DNA of bee females, whereas human males have nearly the same amount of DNA as human females. Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males. Human and Drosophila males have sons, but bee males do not. Inheritance in bees is like inheritance of sex-linked characteristics in humans. none of the above Answer: A, B, or C The point of this question is that there are a variety of sex determination mechanisms among different species. Answers A, B, and C are correct. Answer A is correct because bee males are haploid, but the only difference in the amount of DNA in human males and females is that the Y chromosome is slightly smaller than the X chromosome. Answer B is correct because deleterious alleles from the whole genome will affect male bees, but deleterious recessives on the autosomes in Drosophila males can be masked by a dominant allele. Answer C is correct because, in humans and Drosophila, half of a male’s sperm carry a Y chromosome and cause the production of male offspring, but in bees and ants males are haploid and any egg the sperm fertilizes develops into a female. Answer D is incorrect because male bees have no sons. 45

Think about bees, which have no X and Y sex chromosomes Think about bees, which have no X and Y sex chromosomes. Males are haploid, whereas fertilization results in females, as diploid cells become females. Which of the following are accurate statements about bee males when they are compared to species in which males are XY and diploid for the autosomes? Bee males have half the DNA of bee females, whereas human males have nearly the same amount of DNA as human females. Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males. Human and Drosophila males have sons, but bee males do not. Inheritance in bees is like inheritance of sex-linked characteristics in humans. none of the above 46

This mutation occurs in all offspring of a male with the mutation. Determination of sex in Drosophila is similar to that in humans. In some species of Drosophila, there are genes on the Y chromosome that do not occur on the X chromosome. Imagine that a mutation of one gene on the Y chromosome reduces the size by half of individuals with the mutation. Which of the following statements is accurate with regard to this situation? This mutation occurs in all offspring of a male with the mutation. This mutation occurs in all male but no female offspring of a male with the mutation. This mutation occurs in all offspring of a female with the mutation. This mutation occurs in all male but no female offspring of a female with the mutation. This mutation occurs in all offspring of both males and females with the mutation. Answer: B The point of this question is to help students understand sex linkage by looking at the effect of genes on the Y chromosome. Answer A is incorrect because half of the sperm of the mutant male will contain an X chromosome and produce normal size daughters. Answer C is incorrect because females do not have a Y chromosome and cannot carry the mutation. Answer D is incorrect because, again, females do not have a Y chromosome and thus cannot carry the mutation. Answer E is incorrect because females cannot carry the mutation (no Y chromosome) and only the half of the eggs that are fertilized by a Y-bearing sperm will receive the mutation (i.e., only half of the male offspring). Answer B is correct because all the sons of the mutant male receive his Y chromosome. 47

This mutation occurs in all offspring of a male with the mutation. Determination of sex in Drosophila is similar to that in humans. In some species of Drosophila, there are genes on the Y chromosome that do not occur on the X chromosome. Imagine that a mutation of one gene on the Y chromosome reduces the size by half of individuals with the mutation. Which of the following statements is accurate with regard to this situation? This mutation occurs in all offspring of a male with the mutation. This mutation occurs in all male but no female offspring of a male with the mutation. This mutation occurs in all offspring of a female with the mutation. This mutation occurs in all male but no female offspring of a female with the mutation. This mutation occurs in all offspring of both males and females with the mutation. 48

In cats, a sex-linked gene affects coat color In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate? The phenotype of o-Y males is black/brown because the nonfunctional allele o does not convert eumelanin into phaeomelanin. The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of Oo males is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the nonfunctional allele o does not convert eumelanin into phaeomelanin. The phenotype of O-Y males is orange because the nonfunctional allele O does not convert eumelanin into phaeomelanin, while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin. Answer: A This focuses on the color of males and the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment). Male genotypes will be either O-Y or o-Y, with phenotypes of either orange or black/brown, respectively. In O-Y males, the eumelanin is converted to phaeomelanin, and in o-Y males, the eumelanin is unchanged. To answer this question, a student must know that males have only one copy of the gene and must understand that a functional allele produces an enzyme that catalyzes the chemical reaction. 49

In cats, a sex-linked gene affects coat color In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate? The phenotype of o-Y males is black/brown because the nonfunctional allele o does not convert eumelanin into phaeomelanin. The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of Oo males is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the nonfunctional allele o does not convert eumelanin into phaeomelanin. The phenotype of O-Y males is orange because the nonfunctional allele O does not convert eumelanin into phaeomelanin, while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin. 50

Imagine two species of cats that differ in the timing of Barr body formation during development. Both species have genes that determine coat color, O for the dominant orange fur and o for the recessive black/brown fur, on the X chromosome. In species A, the Barr body forms during week 1 of a 6-month pregnancy, whereas in species B, the Barr body forms during week 3 of a 5-month pregnancy. What would you predict about the coloration of heterozygous females (Oo) in the two species? Both species will have similar sized patches of orange and black/brown fur. Species A will have smaller patches of orange or black/brown fur than will species B. The females of both species will show the dominant fur color, orange. Answer: B This question relates genetics, embryonic development, and Barr body formation. The sizes of the patches should be related to the timing of Barr body formation in development. Early formation of the Barr body would mean that larger clusters of cells are affected by which X chromosome is inactivated compared to later Barr body formation. 51

Imagine two species of cats that differ in the timing of Barr body formation during development. Both species have genes that determine coat color, O for the dominant orange fur and o for the recessive black/brown fur, on the X chromosome. In species A, the Barr body forms during week 1 of a 6-month pregnancy, whereas in species B, the Barr body forms during week 3 of a 5-month pregnancy. What would you predict about the coloration of heterozygous females (Oo) in the two species? Both species will have similar sized patches of orange and black/brown fur. Species A will have smaller patches of orange or black/brown fur than will species B. The females of both species will show the dominant fur color, orange. 52

Imagine a species with three loci thought to be on the same chromosome Imagine a species with three loci thought to be on the same chromosome. The recombination rate between locus A and locus B is 35%, and the recombination rate between locus B and locus C is 33%. Predict the recombination rate between A and C. The recombination rate between locus A and locus C is either 2% or 68%. The recombination rate between locus A and locus C is probably 2%. The recombination rate between locus A and locus C is either 2% or 50%. The recombination rate between locus A and locus C is either 2% or 39%. The recombination rate between locus A and locus C cannot be predicted. Answer: C The recombination rate between loci A and B is 35%. Locus C can be either between A and B or on the opposite side of B from A. If locus C is in between locus A and locus B (ACB), the distance between locus A and C would be 2%. Locus C cannot be on the other side of A from B (CAB) because the recombination rate would have to be higher than 35%. Thus far, answers A, B, C, and D could be correct. If locus C is on the other side of locus B from locus A (ABC), adding the two recombination rates gives a prediction of 68%, but the maximum recombination rate between two loci is 50%, the same frequency of recombinants that are on different chromosomes. Using this information, answer A cannot be correct because a recombination rate cannot be greater than 50%. Answer B could be right, but it is not the only possible placement and so answer B is incomplete. Answer C is the best answer (2% if C is between A and B, and 50% if C is on the other side of B than A). 53

Imagine a species with three loci thought to be on the same chromosome Imagine a species with three loci thought to be on the same chromosome. The recombination rate between locus A and locus B is 35%, and the recombination rate between locus B and locus C is 33%. Predict the recombination rate between A and C. The recombination rate between locus A and locus C is either 2% or 68%. The recombination rate between locus A and locus C is probably 2%. The recombination rate between locus A and locus C is either 2% or 50%. The recombination rate between locus A and locus C is either 2% or 39%. The recombination rate between locus A and locus C cannot be predicted. 54

Triploid species are usually sterile (unable to reproduce), whereas tetraploids are often fertile. Which of the following is likely a good explanation of these facts? (Hint: Synapsis.) In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. Answer: B The point of this question is to make students think about mitosis and meiosis in relation to polyploids. To answer this question, students should draw chromosomes of a triploid and a tetraploid as they go through mitosis and meiosis. Answers A and C are incorrect because chromosomes do not synapse during mitosis. Answer D is incorrect because tetraploids do have partners at synapsis but triploids do not. Answer B is correct—one-third of the chromosomes do not have a partner. 55

Triploid species are usually sterile (unable to reproduce), whereas tetraploids are often fertile. Which of the following is likely a good explanation of these facts? (Hint: Synapsis.) In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. 56

Chromosomal rearrangements can occur after chromosomes break Chromosomal rearrangements can occur after chromosomes break. Which of the following statements is most accurate with respect to alterations in chromosome structure? Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates. Translocations and inversions are not deleterious because no genes are lost in the organism. Chromosomal rearrangements are more likely to occur during mitosis than during meiosis. An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than one that is homozygous for a duplication of that same gene because loss of a function can be lethal. Answer: D Chromosomal rearrangements are important in evolution. For example, duplications provide raw material on which natural selection can act (e.g., the globin genes are thought to have arisen via gene duplication). This question will make students think about the consequences of chromosomal rearrangements. Answer A is not correct and should be nixed by students because there is no information in the chapter that would lead to this conclusion. Answers B and C directly contradict material in the text and are therefore incorrect. 57

Chromosomal rearrangements can occur after chromosomes break Chromosomal rearrangements can occur after chromosomes break. Which of the following statements is most accurate with respect to alterations in chromosome structure? Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates. Translocations and inversions are not deleterious because no genes are lost in the organism. Chromosomal rearrangements are more likely to occur during mitosis than during meiosis. An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than one that is homozygous for a duplication of that same gene because loss of a function can be lethal. 58

Which of the following statements about crossing over is false? It accounts for the recombination of linked genes. It occurs while replicated homologs are paired during prophase I of meiosis. Portions of sister chromatids change places. It breaks the physical connection between specific alleles on the same chromosome. Recombinants may result. Answer: C

Which of the following statements about crossing over is false? It accounts for the recombination of linked genes. It occurs while replicated homologs are paired during prophase I of meiosis. Portions of sister chromatids change places. It breaks the physical connection between specific alleles on the same chromosome. Recombinants may result.

Which of the following is a type of chromosomal alteration that differ from all of the others? aneuploidy polyploidy triploidy tetraploidy octaploidy Answer: A

Which of the following is a type of chromosomal alteration that differs from all of the others? aneuploidy polyploidy triploidy tetraploidy octaploidy Answer: A