1. GENETICS & EVOLUTION: CHROMOSOMAL INHERITANCE & MUTATION Chapter 7.2

2. Overview  Chromosomal Inheritance  Sex-linked Genes  Gene linkage and analysis  Mutations  Gene Mutations  Chromosomal Abberations

3. Overview: Locating Genes Along Chromosomes  Mendel’s “hereditary factors” were genes, though this wasn’t known at the time  Today we can show that genes are located on chromosomes  The location of a particular gene can be seen by tagging isolated chromosomes with a fluorescent dye that highlights the gene

4. Fig. 15-1

5. Mendelian inheritance has its physical basis in the behavior of chromosomes  Mitosis and meiosis were first described in the late 1800s  The chromosome theory of inheritance states:  Mendelian genes have specific loci (positions) on chromosomes  Chromosomes undergo segregation and independent assortment  The behavior of chromosomes during meiosis was said to account for Mendel’s laws of segregation and independent assortment

6. Fig. 15-2a P Generation Gametes Meiosis Fertilization Yellow-round seeds (YYRR) Green-wrinkled seeds ( yyrr) All F 1 plants produce yellow-round seeds (YyRr) y y y r r r Y Y YR R R 

7. Fig. 15-2b 0.5 mm Meiosis Metaphase I Anaphase I Metaphase II Gametes LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. 1 4 yr 1 4 Yr 1 4 YR 3 3 F 1 Generation 1 4 yR R R R R R R R R R R R R Y Y Y Y Y Y Y Y Y Y YY y rr r r r r r r r r r r y y y y y y y y y y y All F 1 plants produce yellow-round seeds (YyRr) 1 22 1

8. Fig. 15-2c F 2 Generation An F 1  F 1 cross-fertilization 9 : 3 : 1 3 3

9. Fig. 15-2 P Generation Yellow-round seeds (YYRR) Y F 1 Generation Y R R R Y  r r r y y y Meiosis Fertilization Gametes Green-wrinkled seeds ( yyrr) All F 1 plants produce yellow-round seeds ( YyRr ) R R Y Y r r y y Meiosis R R Y Y r r y y Metaphase I Y Y RR r r y y Anaphase I r r y Y Metaphase II R Y R y y y y R R Y Y r r r r y Y Y R R yR Yr yr YR 1/41/4 1/41/4 1/41/4 1/41/4 F 2 Generation Gametes An F 1  F 1 cross-fertilization 9 : 3 : 1 LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. 1 2 3 3 2 1

10. The Chromosomal Basis of Sex  In humans and other mammals, there are two varieties of sex chromosomes: a larger X chromosome and a smaller Y chromosome  Only the ends of the Y chromosome have regions that are homologous with the X chromosome  The SRY gene on the Y chromosome codes for the development of testes

11. Fig. 15-5 X Y

12.  Females are XX, and males are XY  Each ovum contains an X chromosome, while a sperm may contain either an X or a Y chromosome  Other animals have different methods of sex determination

13. Fig. 15-6 44 + XY Parents 44 + XX 22 + X 22 + X 22 + Y or + 44 + XX or Sperm Egg 44 + XY Zygotes (offspring) (a) The X-Y system 22 + XX 22 + X (b) The X-0 system 76 + ZW 76 + ZZ (c) The Z-W system 32 (Diploid) 16 (Haploid) (d) The haplo-diploid system

14. 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  In humans, sex-linked usually refers to a gene on the larger X chromosomeSex-linked genes follow specific patterns of inheritance  For a recessive sex-linked trait to be expressed  A female needs two copies of the allele  A male needs only one copy of the allele  Sex-linked recessive disorders are much more common in males than in females

15. Fig. 15-7 (a)(b) (c) XNXNXNXN XnYXnY XNXnXNXn   XNYXNYXNXnXNXn  XnYXnY YXnXn Sperm Y XNXN Y XnXn XNXnXNXn Eggs XNXN XNXN XNXnXNXn XNYXNY XNYXNY XNXN XnXn XNXNXNXN XnXNXnXN XNYXNY XnYXnY XNXN XnXn XNXnXNXn XnXnXnXn XNYXNY XnYXnY

16.  Some disorders caused by recessive alleles on the X chromosome in humans:  Color blindness  Duchenne muscular dystrophy  Hemophilia

17. X Inactivation in Female Mammals  In mammalian females, one of the two X chromosomes in each cell is randomly inactivated during embryonic development  The inactive X condenses into a Barr body  If a female is heterozygous for a particular gene located on the X chromosome, she will be a mosaic for that character

18. Fig. 15-8 X chromosomes Early embryo: Allele for orange fur Allele for black fur Cell division and X chromosome inactivation Two cell populations in adult cat: Active X Inactive X Black furOrange fur

19.  Each chromosome has hundreds or thousands of genes  Genes located on the same chromosome that tend to be inherited together are called linked genes  Morgan did other experiments with fruit flies to see how linkage affects inheritance of two characters  Morgan crossed flies that differed in traits of body color and wing size Linked genes tend to be inherited together

20. How Linkage Affects Inheritance  Morgan found that body color and wing size are usually inherited together in specific combinations (parental phenotypes)  He noted that these genes do not assort independently, and reasoned that they were on the same chromosome  However, nonparental phenotypes were also produced  Understanding this result involves exploring genetic recombination, the production of offspring with combinations of traits differing from either parent

21. Fig. 15-UN1 b + vg + Parents in testcross Most offspring b + vg + b vg  or

22. Fig. 15-9-1 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg b + b + vg + vg +

23. Fig. 15-9-2 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg

24. Fig. 15-9-3 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg Testcross offspring Eggs b + vg + b vg b + vg b vg + Black- normal Gray- vestigial Black- vestigial Wild type (gray-normal) b vg Sperm b + b vg + vg b b vg vg b + b vg vg b b vg + vg

25. Fig. 15-9-4 EXPERIMENT P Generation (homozygous) RESULTS Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg Testcross offspring Eggs b + vg + b vg b + vg b vg + Black- normal Gray- vestigial Black- vestigial Wild type (gray-normal) b vg Sperm b + b vg + vg b b vg vg b + b vg vg b b vg + vg PREDICTED RATIOS If genes are located on different chromosomes: If genes are located on the same chromosome and parental alleles are always inherited together: 1 1 1 1 1 1 0 0 965 944206 185 : : : : : : : : :

26. Genetic Recombination and Linkage  The genetic findings of Mendel and Morgan relate to the chromosomal basis of recombination

27. Recombination of Unlinked Genes: Independent Assortment of Chromosomes  Mendel observed that combinations of traits in some offspring differ from either parent  Offspring with a phenotype matching one of the parental phenotypes are called parental types  Offspring with nonparental phenotypes (new combinations of traits) are called recombinant types, or recombinants  A 50% frequency of recombination is observed for any two genes on different chromosomes

28. Fig. 15-UN2 YyRr Gametes from green- wrinkled homozygous recessive parent ( yyrr ) Gametes from yellow-round heterozygous parent (YyRr) Parental- type offspring Recombinant offspring yr yyrrYyrr yyRr YRyr Yr yR

29. Recombination of Linked Genes: Crossing Over  Morgan discovered that genes can be linked, but the linkage was incomplete, as evident from recombinant phenotypes  Morgan proposed that some process must sometimes break the physical connection between genes on the same chromosome  That mechanism was the crossing over of homologous chromosomes

30. Fig. 15-10 Testcross parents Replication of chromo- somes Gray body, normal wings (F 1 dihybrid) Black body, vestigial wings (double mutant) Replication of chromo- somes b + vg + b vg b + vg + b + vg b vg + b vg Recombinant chromosomes Meiosis I and II Meiosis I Meiosis II b vg + b + vg b vg b + vg + Eggs Testcross offspring 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal b + vg + b vg b + vg b vg b vg + Sperm b vg Parental-type offspringRecombinant offspring Recombination frequency = 391 recombinants 2,300 total offspring  100 = 17%

31. Fig. 15-10a Testcross parents Replication of chromo- somes Gray body, normal wings (F 1 dihybrid) Black body, vestigial wings (double mutant) Replication of chromo- somes b + vg + b vg b + vg + b + vg b vg + b vg Recombinant chromosomes Meiosis I and II Meiosis I Meiosis II Eggs Sperm b + vg + b vg b + vg b vg b vg +

32. Fig. 15-10b Testcross offspring 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal b + vg + b vg b + vg b vg b + vg + Sperm b vg Parental-type offspring Recombinant offspring Recombination frequency = 391 recombinants 2,300 total offspring  100 = 17% b vg b + vg b vg + Eggs Recombinant chromosomes

33. Mapping the Distance Between Genes Using Recombination Data  Alfred Sturtevant, one of Morgan’s students, constructed a genetic map, an ordered list of the genetic loci along a particular chromosome  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”

34.  A linkage map is a genetic map of a chromosome based on recombination frequencies  Distances between genes can be expressed as map units; one map unit, or centimorgan, represents a 1% recombination frequency  Map units indicate relative distance and order, not precise locations of genes

35. Fig. 15-11 RESULTS Recombination frequencies Chromosome 9%9.5% 17% bcnvg

36.  Genes that are far apart on the same chromosome can have a recombination frequency near 50%  Such genes are physically linked, but genetically unlinked, and behave as if found on different chromosomes  Sturtevant used recombination frequencies to make linkage maps of fruit fly genes  Using methods like chromosomal banding, geneticists can develop cytogenetic maps of chromosomes  Cytogenetic maps indicate the positions of genes with respect to chromosomal features

37. Fig. 15-12 Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Red eyes Normal wings Red eyes Gray body Long aristae (appendages on head) Wild-type phenotypes 048.557.567.0104.5

38. Abnormal Chromosome Number  Large-scale chromosomal alterations often lead to spontaneous abortions (miscarriages) or cause a variety of developmental disorders  In nondisjunction, pairs of homologous chromosomes do not separate normally during meiosis  As a result, one gamete receives two of the same type of chromosome, and another gamete receives no copy

39. Fig. 15-13-1 Meiosis I (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II Nondisjunction

40. Fig. 15-13-2 Meiosis I Nondisjunction (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II Meiosis II Nondisjunction

41. Fig. 15-13-3 Meiosis I Nondisjunction (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II Meiosis II Nondisjunction Gametes Number of chromosomes n + 1 n – 1 nn

42.  Aneuploidy results from the fertilization of gametes in which nondisjunction occurred  Offspring with this condition have an abnormal number of a particular chromosome  A monosomic zygote has only one copy of a particular chromosome  A trisomic zygote has three copies of a particular chromosome

43.  Polyploidy is a condition in which an organism has more than two complete sets of chromosomes  Triploidy (3n) is three sets of chromosomes  Tetraploidy (4n) is four sets of chromosomes  Polyploidy is common in plants, but not animals  Polyploids are more normal in appearance than aneuploids

44. Fig. 15-14

45. Mutagens  Spontaneous mutations can occur during DNA replication, recombination, or repair  Mutagens are physical or chemical agents that can cause mutations  Mutations are changes in the genetic material of a cell or virus  Gene mutation  Chromosomal abberation

46. Causes of Mutations  Spontaneous mutation  DNA can undergo a chemical change  Movement of transposons from one chromosomal location to another  Replication Errors 1 in 1,000,000,000 replications DNA polymerase Proofreads new strands Generally corrects errors  Induced mutation:  Mutagens such as radiation, organic chemicals  Many mutagens are also carcinogens (cancer causing)  Environmental Mutagens Ultraviolet Radiation Tobacco Smoke

47. Effect of Mutations on Protein Activity  Point Mutations  Involve change in a single DNA nucleotide  Changes one codon to a different codon  Affects on protein vary: Nonfunctional Reduced functionality Unaffected  Frameshift Mutations  One or two nucleotides are either inserted or deleted from DNA  Protein always rendered nonfunctional Normal :THE CAT ATE THE RAT After deletion:THE ATA TET HER AT After insertion:THE CCA TAT ETH ERA T

48. Point mutations can affect protein structure and function  Point mutations are chemical changes in just one base pair of a gene  The change of a single nucleotide in a DNA template strand can lead to the production of an abnormal protein

49. Fig. 17-22 Wild-type hemoglobin DNA mRNA Mutant hemoglobin DNA mRNA 3 3 3 3 3 3 5 5 5 5 5 5 CCTT T T G G A A A A AA A GG U Normal hemoglobinSickle-cell hemoglobin Glu Val

50. Types of Point Mutations  Point mutations within a gene can be divided into two general categories  Base-pair substitutions  Base-pair insertions or deletions

51. Fig. 17-23 Wild-type 3 DNA template strand 5 5 5 3 3 Stop Carboxyl end Amino end Protein mRNA 3 3 3 5 5 5 A instead of G U instead of C Silent (no effect on amino acid sequence) Stop T instead of C 3 3 3 5 5 5 A instead of G Stop Missense A instead of T U instead of A 3 3 3 5 5 5 Stop Nonsense No frameshift, but one amino acid missing (3 base-pair deletion) Frameshift causing extensive missense (1 base-pair deletion) Frameshift causing immediate nonsense (1 base-pair insertion) 5 5 5 3 3 3 Stop missing 3 3 3 5 5 5 Stop 5 5 5 3 3 3 Extra U Extra A (a) Base-pair substitution(b) Base-pair insertion or deletion

52. Fig. 17-23a Wild type 3 DNA template strand 3 3 5 5 5 mRNA Protein Amino end Stop Carboxyl end A instead of G 3 3 3 U instead of C 5 5 5 Stop Silent (no effect on amino acid sequence)

53. Fig. 17-23b Wild type DNA template strand 3 5 mRNA Protein 5 Amino end Stop Carboxyl end 5 3 3 T instead of C A instead of G 3 3 3 5 5 5 Stop Missense

54. Fig. 17-23c Wild type DNA template strand 3 5 mRNA Protein 5 Amino end Stop Carboxyl end 5 3 3 A instead of T U instead of A 3 3 3 5 5 5 Stop Nonsense

55. Fig. 17-23d Wild type DNA template strand 3 5 mRNA Protein 5 Amino end Stop Carboxyl end 5 3 3 Extra A Extra U 3 3 3 5 5 5 Stop Frameshift causing immediate nonsense (1 base-pair insertion)

56. Fig. 17-23e Wild type DNA template strand 3 5 mRNA Protein 5 Amino end Stop Carboxyl end 5 3 3 missing 3 3 3 5 5 5 Frameshift causing extensive missense (1 base-pair deletion)

57. Fig. 17-23f Wild type DNA template strand 3 5 mRNA Protein 5 Amino end Stop Carboxyl end 5 3 3 missing 3 3 3 5 5 5 No frameshift, but one amino acid missing (3 base-pair deletion) Stop

58. Substitutions  A base-pair substitution replaces one nucleotide and its partner with another pair of nucleotides  Silent mutations have no effect on the amino acid produced by a codon because of redundancy in the genetic code  Missense mutations still code for an amino acid, but not necessarily the right amino acid  Nonsense mutations change an amino acid codon into a stop codon, nearly always leading to a nonfunctional protein

59. Insertions and Deletions  Insertions and deletions are additions or losses of nucleotide pairs in a gene  These mutations have a disastrous effect on the resulting protein more often than substitutions do  Insertion or deletion of nucleotides may alter the reading frame, producing a frameshift mutation

60. Alterations of Chromosome Structure  Breakage of a chromosome can lead to four types of changes in chromosome structure:  Deletion removes a chromosomal segment  Duplication repeats a segment  Inversion reverses a segment within a chromosome  Translocation moves a segment from one chromosome to another

61. Fig. 15-15 Deletion A B C D E F G HA B C E F G H (a) (b) (c) (d) Duplication Inversion Reciprocal translocation A B C D E F G H A B C B C D E F G H A D C B E F G H M N O C D E F G H M N O P Q RA B P Q R

62. Human Disorders Due to Chromosomal Alterations  Alterations of chromosome number and structure are associated with some serious disorders  Some types of aneuploidy appear to upset the genetic balance less than others, resulting in individuals surviving to birth and beyond  These surviving individuals have a set of symptoms, or syndrome, characteristic of the type of aneuploidy

63. Down Syndrome (Trisomy 21)  Down syndrome is an aneuploid condition that results from three copies of chromosome 21  It affects about one out of every 700 children born in the United States  The frequency of Down syndrome increases with the age of the mother, a correlation that has not been explained

64. Fig. 15-16

65. 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  Monosomy X, called Turner syndrome, produces X0 females, who are sterile; it is the only known viable monosomy in humans

66. Changes in Sex Chromosome a. Turner syndromeb. Klinefelter syndrome Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. a: Courtesy UNC Medical Illustration and Photography; b: Courtesy Stefan D. Schwarz, http://klinefeltersyndrome.org

67. Disorders Caused by Structurally Altered Chromosomes  The syndrome cri du chat (“cry of the cat”), results from a specific deletion in chromosome 5  A child born with this syndrome is mentally retarded and has a catlike cry; individuals usually die in infancy or early childhood  Certain cancers, including chronic myelogenous leukemia (CML), are caused by translocations of chromosomes

68. Fig. 15-17 Normal chromosome 9 Normal chromosome 22 Reciprocal translocation Translocated chromosome 9 Translocated chromosome 22 (Philadelphia chromosome)

69. 69 Carcinogenesis  Development of cancer involves a series of mutations  Proto-oncogenes – Stimulate cell cycle  Tumor suppressor genes – inhibit cell cycle  Mutation in oncogene and tumor suppressor gene: Stimulates cell cycle uncontrollably Leads to tumor formation

70. 70 Cell Signaling Pathway  Cell signaling pathway that stimulates a mutated tumor suppressor gene Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. receptor inhibiting growth factor cytoplasm plasma membrane signal transducers transcription factor nucleus protein that is unable to inhibit the cell cycle or promote apoptosis mutated tumor suppressor gene

71. 71 Cell Signaling Pathway Cell signaling pathway that stimulates a proto- oncogene Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. receptor stimulating growth factor cytoplasm plasma membrane signal transducers transcription factor nucleus protein that overstimulates the cell cycle oncogene

72. Some inheritance patterns are exceptions to the standard chromosome theory  There are two normal exceptions to Mendelian genetics  One exception involves genes located in the nucleus, and the other exception involves genes located outside the nucleus

73. Genomic Imprinting  For a few mammalian traits, the phenotype depends on which parent passed along the alleles for those traits  Such variation in phenotype is called genomic imprinting  Genomic imprinting involves the silencing of certain genes that are “stamped” with an imprint during gamete production

74. Fig. 15-18a Normal Igf2 allele is expressed Paternal chromosome Maternal chromosome (a) Homozygote Wild-type mouse (normal size) Normal Igf2 allele is not expressed

75. Fig. 15-18b Mutant Igf2 allele inherited from mother Mutant Igf2 allele inherited from father Normal size mouse (wild type) Dwarf mouse (mutant) Normal Igf2 allele is expressed Mutant Igf2 allele is expressed Mutant Igf2 allele is not expressed Normal Igf2 allele is not expressed (b) Heterozygotes

76. Fig. 15-18 Normal Igf2 allele is expressed Paternal chromosome Maternal chromosome Normal Igf2 allele is not expressed Mutant Igf2 allele inherited from mother (a) Homozygote Wild-type mouse (normal size) Mutant Igf2 allele inherited from father Normal size mouse (wild type) Dwarf mouse (mutant) Normal Igf2 allele is expressed Mutant Igf2 allele is expressed Mutant Igf2 allele is not expressed Normal Igf2 allele is not expressed (b) Heterozygotes

77.  It appears that imprinting is the result of the methylation (addition of –CH 3 ) of DNA  Genomic imprinting is thought to affect only a small fraction of mammalian genes  Most imprinted genes are critical for embryonic development

78. Fig. 15-UN3

79. Inheritance of Organelle Genes  Extranuclear genes (or cytoplasmic genes) are genes found in organelles in the cytoplasm  Mitochondria, chloroplasts, and other plant plastids carry small circular DNA molecules  Extranuclear genes are inherited maternally because the zygote’s cytoplasm comes from the egg  The first evidence of extranuclear genes came from studies on the inheritance of yellow or white patches on leaves of an otherwise green plant

80.  Some defects in mitochondrial genes prevent cells from making enough ATP and result in diseases that affect the muscular and nervous systems  For example, mitochondrial myopathy and Leber’s hereditary optic neuropathy

81. You should now be able to: 1. Explain the chromosomal theory of inheritance and its discovery 2. Explain why sex-linked diseases are more common in human males than females 3. Distinguish between sex-linked genes and linked genes 4. Explain how meiosis accounts for recombinant phenotypes 5. Explain how linkage maps are constructed

82. 6. Explain how nondisjunction can lead to aneuploidy 7. Define trisomy, triploidy, and polyploidy 8. Define mutation 9. Distinguish between different gene mutations 10. Distinguish among deletions, duplications, inversions, and translocations 11. Explain genomic imprinting 12. Explain why extranuclear genes are not inherited in a Mendelian fashion

83. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Morgan’s Experimental Evidence Imagine that Morgan had used a grasshopper (2N = 24 and an XX, XO sex determination system). Predict where the first mutant would have been discovered. a. on the O chromosome of a male b. on the X chromosome of a male c. on the X chromosome of a female d. on the Y chromosome of a male

84. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. The Chromosomal Basis of Sex Think about bees and ants, groups in which males are haploid. Which of the following are accurate statements about bee and ant males when they are compared to species in which males are XY and diploid for the autosomes? a. Bee males have half the DNA of bee females whereas human males have nearly the same amount of DNA that human females have. b. Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males. c. Human and Drosophila males have sons but bee males do not. d. Inheritance in bees is like inheritance of sex-linked characteristics in humans. e. none of the above

85. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. The Chromosomal Basis of Sex In some Drosophila species 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.

86. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. The Chromosomal Basis of Sex Imagine that a deleterious recessive allele occurs on the W chromosome of a chicken (2N = 78). Where would it be most likely to appear first in a genetics experiment? a. in a male because there is no possibility of the presence of a normal, dominant allele b. in a male because it is haploid c. in a female because there is no possibility of the presence of a normal, dominant allele d. in a female because all alleles on the W chromosomes are dominant to those on the Z chromosome e. none of the above

87. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Inheritance of Sex-Linked Genes 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? a. The phenotype of O-Y males is orange because the functional allele O converts eumelanin into phaeomelanin. b. The phenotype of o-Y males is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin. c. The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin. d. 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 non-functional allele o does not convert eumelanin into phaeomelanin. e. The phenotype of O-Y males is orange because the non-functional 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.

88. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Inheritance of Sex-Linked Genes 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? a. The phenotype of O-Y females is orange because the functional allele O converts eumelanin into phaeomelanin. b. The phenotype of o-Y females is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin. c. The phenotype of OO and Oo females is orange because the functional allele O converts eumelanin into phaeomelanin. d. The phenotype of Oo females 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 non-functional allele o does not convert eumelanin into phaeomelanin. e. The phenotype of O-Y females is orange because the non-functional 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.

89. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. X Inactivation in Female Mammals Imagine two species of mammals 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? a. Both species will have similar sized patches of orange and black/brown fur. b. Species A will have smaller patches of orange or black/brown fur than will species B. c. The females of both species will show the dominant fur color, orange.

90. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Mapping the Distance between Genes 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. a. The recombination rate between locus A and locus C is either 2% or 68%. b. The recombination rate between locus A and locus C is probably 2%. c. The recombination rate between locus A and locus C is either 2% or 50%. d. The recombination rate between locus A and locus C is either 2% or 39%. e. The recombination rate between locus A and locus C cannot be predicted.

91. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Triploid species are usually sterile (unable to reproduce) whereas tetraploids are often fertile. Which of the following are likely good explanations of these facts? a. In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. b. In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. c. In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. d. In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.

92. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Chromosomal rearrangements can occur after chromosomes break. Which of the following statements are most accurate with respect to alterations in chromosome structure? a. Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates. b. Translocations and inversions are not deleterious because no genes are lost in the organism. c. Chromosomal rearrangements are more likely to occur during mitosis than during meiosis. d. An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than is one that is homozygous for a duplication of that same gene because loss of a function can be lethal.

93. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings. Imagine that you could create medical policy for a country. In this country it is known that the frequency of Down syndrome babies increases with increasing age of the mother and that the severity of characteristics varies enormously and unpredictably among affected individuals. Furthermore, financial resources are severely limited, both for testing of pregnant women and for supplemental training of Down syndrome children. The graph on the next slide shows the incidence of Down syndrome as a function of maternal age. Which of the following policies would you implement? a. No testing of pregnant women should be conducted and all the health care money should be used for training of Down syndrome children. b. The health care system should provide testing only for women over 30. c. The health care system should provide testing only for women over 40. d. The health care system should require termination of all Down syndrome fetuses. e. The health care system should provide training for the 30% most seriously affected children only.

94. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

95. The lawyer for a defendant in a paternity suit asked for DNA testing of a baby girl. Which of the following set of results would demonstrate that the purported father was not actually the genetic father of the child? a. The mitochondrial DNA of the child and “father” did not match. b. DNA sequencing of chromosome #5 of the child and “father” did not match. c. The mitochondrial DNA of the child and “mother” did not match. d. DNA sequencing of chromosome #5 of the child and “mother” did not match. e. The mitochondrial DNA of the child and “father” matched but the mitochondrial DNA of the child and “mother” did not.