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The Chromosomal Basis of Inheritance
Chapter 15 The Chromosomal Basis of Inheritance
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Overview: Locating Genes Along Chromosomes
Mendel’s “hereditary factors” were genes 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 © 2011 Pearson Education, Inc.
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Figure 15.1 Figure 15.1 Where are Mendel’s hereditary factors located in the cell?
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Concept 15.1: 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 can account for Mendel’s laws of segregation and independent assortment © 2011 Pearson Education, Inc.
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Figure 15.2 The chromosomal basis of Mendel’s laws.
P Generation Yellow-round seeds (YYRR) Green-wrinkled seeds (yyrr) Y r y Y R R r y Meiosis Fertilization R Y y r Gametes All F1 plants produce yellow-round seeds (YyRr). F1 Generation R R y y r r Y Y Meiosis 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. R r r R Metaphase I Y y Y y 1 1 R r r R Anaphase I Y y Y y Figure 15.2 The chromosomal basis of Mendel’s laws. R r r R 2 2 Y y Metaphase II Y y Y y Y y Y Y y y Gametes R R r r r r R R 1/4 YR 1/4 yr 1/4 Yr 1/4 yR F2 Generation An F1 F1 cross-fertilization 3 Fertilization recombines the R and r alleles at random. 3 Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation. 9 : 3 : 3 : 1
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P Generation Yellow-round seeds (YYRR) Green-wrinkled seeds (yyrr)
Figure 15.2a P Generation Yellow-round seeds (YYRR) Green-wrinkled seeds (yyrr) Y y r R Y R r y Meiosis Fertilization R Y y r Gametes Figure 15.2 The chromosomal basis of Mendel’s laws.
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F1 Generation All F1 plants produce yellow-round seeds (YyRr).
Figure 15.2b All F1 plants produce yellow-round seeds (YyRr). F1 Generation R R y y r r Y Y 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. Meiosis R r r R Metaphase I Y y Y y 1 1 R r r R Anaphase I Y y Y y Figure 15.2 The chromosomal basis of Mendel’s laws. R r r R Metaphase II 2 2 Y y Y y Y y Y Y y Y y y Gametes R R r r r r R R 1/4 YR 1/4 yr 1/4 Yr 1/4 yR
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LAW OF INDEPENDENT ASSORTMENT
Figure 15.2c LAW OF SEGREGATION LAW OF INDEPENDENT ASSORTMENT F2 Generation An F1 F1 cross-fertilization 3 Fertilization recombines the R and r alleles at random. 3 Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation. 9 : 3 : 3 : 1 Figure 15.2 The chromosomal basis of Mendel’s laws.
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Morgan’s Experimental Evidence: Scientific Inquiry
The first solid evidence associating a specific gene with a specific chromosome came from Thomas Hunt Morgan, an embryologist Morgan’s experiments with fruit flies provided convincing evidence that chromosomes are the location of Mendel’s heritable factors © 2011 Pearson Education, Inc.
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Morgan’s Choice of Experimental Organism
Several characteristics make fruit flies a convenient organism for genetic studies They produce many offspring A generation can be bred every two weeks They have only four pairs of chromosomes © 2011 Pearson Education, Inc.
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Traits alternative to the wild type are called mutant phenotypes
Morgan noted wild type, or normal, phenotypes that were common in the fly populations Traits alternative to the wild type are called mutant phenotypes © 2011 Pearson Education, Inc.
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Figure 15.3 Figure 15.3 Morgan’s first mutant.
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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-eyed mutant allele must be located on the X chromosome Morgan’s finding supported the chromosome theory of inheritance © 2011 Pearson Education, Inc.
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All offspring had red eyes.
Figure 15.4 EXPERIMENT P Generation F1 Generation All offspring had red eyes. RESULTS F2 Generation CONCLUSION P Generation w w X X X Y w w Sperm Eggs F1 Generation w w w Figure 15.4 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? w w Sperm Eggs w w F2 Generation w w w w w w
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P Generation F1 Generation F2 Generation
Figure 15.4a EXPERIMENT P Generation F1 Generation All offspring had red eyes. RESULTS Figure 15.4 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? F2 Generation
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P Generation F1 Generation F2 Generation
Figure 15.4b CONCLUSION P Generation w w X X X Y w w Sperm Eggs F1 Generation w w w w w Sperm Figure 15.4 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? Eggs w w F2 Generation w w w w w w
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Concept 15.2: Sex-linked genes exhibit unique patterns of inheritance
In humans and some other animals, there is a chromosomal basis of sex determination © 2011 Pearson Education, Inc.
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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 corresponding regions of the X chromosome The SRY gene on the Y chromosome codes for a protein that directs the development of male anatomical features © 2011 Pearson Education, Inc.
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Figure 15.5 X Y Figure 15.5 Human sex chromosomes.
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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 © 2011 Pearson Education, Inc.
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(d) The haplo-diploid system
Figure 15.6 44 XY Parents 44 XX 22 X or 22 Y 22 X Sperm Egg 44 XX or 44 XY Zygotes (offspring) (a) The X-Y system 22 XX 22 X (b) The X-0 system 76 ZW 76 ZZ Figure 15.6 Some chromosomal systems of sex determination. (c) The Z-W system 32 (Diploid) 16 (Haploid) (d) The haplo-diploid system
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Genes on the X chromosome are called X-linked genes
A gene that is located on either sex chromosome is called a sex-linked gene Genes on the Y chromosome are called Y-linked genes; there are few of these Genes on the X chromosome are called X-linked genes © 2011 Pearson Education, Inc.
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Inheritance of X-Linked Genes
X chromosomes have genes for many characters unrelated to sex, whereas the Y chromosome mainly encodes genes related to sex determination © 2011 Pearson Education, Inc.
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X-linked genes follow specific patterns of inheritance
For a recessive X-linked trait to be expressed A female needs two copies of the allele (homozygous) A male needs only one copy of the allele (hemizygous) X-linked recessive disorders are much more common in males than in females © 2011 Pearson Education, Inc.
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XNXN XnY XNXn XNY XNXn XnY Sperm Xn Y Sperm XN Y Sperm Xn Y Eggs XN
Figure 15.7 XNXN XnY XNXn XNY XNXn XnY Sperm Xn Y Sperm XN Y Sperm Xn Y Eggs XN XNXn XNY Eggs XN XNXN XNY Eggs XN XNXn XNY XN XNXn XNY Xn XNXn XnY Xn XnXn XnY Figure 15.7 The transmission of X-linked recessive traits. (a) (b) (c)
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Some disorders caused by recessive alleles on the X chromosome in humans
Color blindness (mostly X-linked) Duchenne muscular dystrophy Hemophilia © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Cell division and X chromosome inactivation
Figure 15.8 X chromosomes Allele for orange fur Early embryo: Allele for black fur Cell division and X chromosome inactivation Two cell populations in adult cat: Active X Inactive X Active X Black fur Orange fur Figure 15.8 X inactivation and the tortoiseshell cat.
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Concept 15.3: 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 (except the Y chromosome) Genes located on the same chromosome that tend to be inherited together are called linked genes © 2011 Pearson Education, Inc.
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How Linkage Affects Inheritance
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 © 2011 Pearson Education, Inc.
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P Generation (homozygous) Double mutant (black body, vestigial wings)
Figure EXPERIMENT P Generation (homozygous) Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) b b vg vg b b vg vg Figure 15.9 Inquiry: How does linkage between two genes affect inheritance of characters?
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P Generation (homozygous) Double mutant (black body, vestigial wings)
Figure EXPERIMENT P Generation (homozygous) Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) b b vg vg b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS b b vg vg b b vg vg Figure 15.9 Inquiry: How does linkage between two genes affect inheritance of characters?
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Wild type (gray-normal)
Figure EXPERIMENT P Generation (homozygous) Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) b b vg vg b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS b b vg vg b b vg vg Testcross offspring Eggs b vg b vg b vg b vg Wild type (gray-normal) Black- vestigial Gray- vestigial Black- normal b vg Figure 15.9 Inquiry: How does linkage between two genes affect inheritance of characters? Sperm b b vg vg b b vg vg b b vg vg b b vg vg
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Wild type (gray-normal)
Figure EXPERIMENT P Generation (homozygous) Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) b b vg vg b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS b b vg vg b b vg vg Testcross offspring Eggs b vg b vg b vg b vg Wild type (gray-normal) Black- vestigial Gray- vestigial Black- normal b vg Figure 15.9 Inquiry: How does linkage between two genes affect inheritance of characters? 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: 1 : 1 : 1 : 1 If genes are located on the same chromosome and parental alleles are always inherited together: 1 : 1 : : RESULTS 965 : 944 : 206 : 185
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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 © 2011 Pearson Education, Inc.
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F1 dihybrid female and homozygous recessive male in testcross
Figure 15.UN01 b+ vg+ b vg F1 dihybrid female and homozygous recessive male in testcross b vg b vg b+ vg+ b vg Figure 15.UN01 In-text figure, p. 294 Most offspring or b vg b vg
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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 © 2011 Pearson Education, Inc.
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Genetic Recombination and Linkage
The genetic findings of Mendel and Morgan relate to the chromosomal basis of recombination © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Gametes from yellow-round dihybrid parent (YyRr)
Figure 15.UN02 Gametes from yellow-round dihybrid parent (YyRr) YR yr Yr yR Gametes from green- wrinkled homozygous recessive parent (yyrr) yr YyRr yyrr Yyrr yyRr Figure 15.UN02 In-text figure, p. 294 Parental- type offspring Recombinant offspring
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Recombination of Linked Genes: Crossing Over
Morgan discovered that genes can be linked, but the linkage was incomplete, because some recombinant phenotypes were observed He proposed that some process must occasionally break the physical connection between genes on the same chromosome That mechanism was the crossing over of homologous chromosomes © 2011 Pearson Education, Inc.
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Animation: Crossing Over Right-click slide / select”Play”
© 2011 Pearson Education, Inc. 42
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Figure 15.10 Chromosomal basis for recombination of linked genes.
Testcross parents Gray body, normal wings (F1 dihybrid) Black body, vestigial wings (double mutant) b vg b vg b vg b vg Replication of chromosomes Replication of chromosomes b vg b vg b vg b vg b vg b vg b vg b vg Meiosis I b vg Meiosis I and II b vg b vg b vg Meiosis II Recombinant chromosomes bvg b vg b vg b vg Figure Chromosomal basis for recombination of linked genes. 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 b vg b vg b vg b vg Sperm Parental-type offspring Recombinant offspring Recombination frequency 391 recombinants 2,300 total offspring 17%
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Gray body, normal wings (F1 dihybrid)
Figure 15.10a Testcross parents Gray body, normal wings (F1 dihybrid) Black body, vestigial wings (double mutant) b vg b vg b vg b vg Replication of chromosomes Replication of chromosomes b vg b vg b vg b vg b vg b vg b vg b vg Meiosis I b vg Meiosis I and II b vg Figure Chromosomal basis for recombination of linked genes. b vg b vg Meiosis II Recombinant chromosomes bvg b vg b vg b vg b vg Eggs Sperm
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965 Wild type (gray-normal) 391 recombinants 2,300 total offspring
Figure 15.10b Recombinant chromosomes bvg 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 b vg b vg b vg b vg Figure Chromosomal basis for recombination of linked genes. Sperm Parental-type offspring Recombinant offspring Recombination frequency 391 recombinants 2,300 total offspring 17%
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New Combinations of Alleles: Variation for Normal Selection
Recombinant chromosomes bring alleles together in new combinations in gametes Random fertilization increases even further the number of variant combinations that can be produced This abundance of genetic variation is the raw material upon which natural selection works © 2011 Pearson Education, Inc.
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Mapping the Distance Between Genes Using Recombination Data: Scientific Inquiry
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 © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Recombination frequencies
Figure 15.11 RESULTS Recombination frequencies 9% 9.5% Chromosome 17% Figure Research Method: Constructing a Linkage Map b cn vg
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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 © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Long aristae (appendages on head) Gray body Red eyes Normal wings
Figure 15.12 Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes 48.5 57.5 67.0 104.5 Figure A partial genetic (linkage) map of a Drosophila chromosome. Long aristae (appendages on head) Gray body Red eyes Normal wings Red eyes Wild-type phenotypes
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Concept 15.4: Alterations of chromosome number or structure cause some genetic disorders
Large-scale chromosomal alterations in humans and other mammals often lead to spontaneous abortions (miscarriages) or cause a variety of developmental disorders Plants tolerate such genetic changes better than animals do © 2011 Pearson Education, Inc.
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Abnormal Chromosome Number
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 © 2011 Pearson Education, Inc.
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Meiosis I Nondisjunction Figure 15.13-1
Figure Meiotic nondisjunction.
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Meiosis I Nondisjunction Meiosis II Non- disjunction Figure 15.13-2
Figure Meiotic nondisjunction.
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Nondisjunction of homo- logous chromosomes in meiosis I (a)
Figure Meiosis I Nondisjunction Meiosis II Non- disjunction Gametes Figure Meiotic nondisjunction. n 1 n 1 n 1 n 1 n 1 n 1 n n Number of chromosomes Nondisjunction of homo- logous chromosomes in meiosis I (a) Nondisjunction of sister chromatids in meiosis II (b)
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Aneuploidy results from the fertilization of gametes in which nondisjunction occurred
Offspring with this condition have an abnormal number of a particular chromosome © 2011 Pearson Education, Inc.
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A monosomic zygote has only one copy of a particular chromosome
A trisomic zygote has three copies of a particular chromosome © 2011 Pearson Education, Inc.
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Polyploidy is common in plants, but not animals
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 © 2011 Pearson Education, Inc.
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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 orientation of a segment within a chromosome Translocation moves a segment from one chromosome to another © 2011 Pearson Education, Inc.
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A deletion removes a chromosomal segment.
Figure 15.14 (a) Deletion A B C D E F G H A deletion removes a chromosomal segment. A B C E F G H (b) Duplication A B C D E F G H A duplication repeats a segment. A B C D E F G H (c) Inversion A B C D E F G H An inversion reverses a segment within a chromosome. Figure Alterations of chromosome structure. A D C B E F G H (d) Translocation A B C D E F G H M N O P Q R A translocation moves a segment from one chromosome to a nonhomologous chromosome. M N O C H F E D G A B P Q R
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A deletion removes a chromosomal segment.
Figure 15.14a (a) Deletion A C D E F G H B A deletion removes a chromosomal segment. A B C E F G H (b) Duplication A B C D E F G H Figure Alterations of chromosome structure. A duplication repeats a segment. A B C D E F G H
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An inversion reverses a segment within a chromosome.
Figure 15.14b (c) Inversion A B C D E F G H An inversion reverses a segment within a chromosome. A D C B E F G H (d) Translocation A B C D E F G H M N O P Q R Figure Alterations of chromosome structure. A translocation moves a segment from one chromosome to a nonhomologous chromosome. G M N O C H F E D A B P Q R
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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 © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Figure 15.15 Figure Down syndrome.
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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 © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Translocated chromosome 22 (Philadelphia chromosome)
Figure 15.16 Normal chromosome 9 Normal chromosome 22 Reciprocal translocation Figure Translocation associated with chronic myelogenous leukemia (CML). Translocated chromosome 9 Translocated chromosome 22 (Philadelphia chromosome)
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Concept 15.5: Some inheritance patterns are exceptions to standard Mendelian inheritance
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 In both cases, the sex of the parent contributing an allele is a factor in the pattern of inheritance © 2011 Pearson Education, Inc.
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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 © 2011 Pearson Education, Inc.
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Normal Igf2 allele is expressed.
Figure 15.17 Normal Igf2 allele is expressed. Paternal chromosome Maternal chromosome Normal-sized mouse (wild type) Normal Igf2 allele is not expressed. (a) Homozygote Mutant Igf2 allele inherited from mother Mutant Igf2 allele inherited from father Normal-sized mouse (wild type) Dwarf mouse (mutant) Figure Genomic imprinting of the mouse Igf2 gene. Normal Igf2 allele is expressed. Mutant Igf2 allele is expressed. Mutant Igf2 allele is not expressed. Normal Igf2 allele is not expressed. (b) Heterozygotes
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Normal Igf2 allele is expressed.
Figure 15.17a Normal Igf2 allele is expressed. Paternal chromosome Maternal chromosome Normal-sized mouse (wild type) Normal Igf2 allele is not expressed. Figure Genomic imprinting of the mouse Igf2 gene. (a) Homozygote
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Mutant Igf2 allele inherited from mother
Figure 15.17b Mutant Igf2 allele inherited from mother Mutant Igf2 allele inherited from father Normal-sized mouse (wild type) Dwarf mouse (mutant) Normal Igf2 allele is expressed. Mutant Igf2 allele is expressed. Figure Genomic imprinting of the mouse Igf2 gene. Mutant Igf2 allele is not expressed. Normal Igf2 allele is not expressed. (b) Heterozygotes
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Most imprinted genes are critical for embryonic development
It appears that imprinting is the result of the methylation (addition of —CH3) of cysteine nucleotides Genomic imprinting is thought to affect only a small fraction of mammalian genes Most imprinted genes are critical for embryonic development © 2011 Pearson Education, Inc.
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Inheritance of Organelle Genes
Extranuclear genes (or cytoplasmic genes) are 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 © 2011 Pearson Education, Inc.
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Figure 15.18 Figure Variegated leaves from English holly (Ilex aquifolium).
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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 © 2011 Pearson Education, Inc.
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Sperm Egg C c B b D A d E a P generation gametes e F f
Figure 15.UN03 Sperm Egg C c B b D A d E a P generation gametes e F f The alleles of unlinked genes are either on separate chromosomes (such as d and e) or so far apart on the same chromosome (c and f) that they assort independently. This F1 cell has 2n 6 chromo- somes and is heterozygous for all six genes shown (AaBbCcDdEeFf). Red maternal; blue paternal. D e C B d Figure 15.UN03 In-text figure, p. 294 A Genes on the same chromo- some whose alleles are so close together that they do not assort independently (such as a, b, and c) are said to be genetically linked. Each chromosome has hundreds or thousands of genes. Four (A, B, C, F) are shown on this one. E F f a c b
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Figure 15.UN04 Figure 15.UN04 In-text figure, p. 294
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Figure 15.UN05 Figure 15.UN05 In-text figure, p. 294
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Figure 15.UN06 Figure 15.UN06 In-text figure, p. 294
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The Chromosomal Basis of Inheritance
Chapter 15 The Chromosomal Basis of Inheritance Questions prepared by Janet Lanza University of Arkansas at Little Rock Louise Paquin McDaniel College
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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 genes and chromosomes. It allowed scientists to see the DNA present within chromosomes. It led to the discovery of mitochondria. It showed genes functioning to direct the formation of enzymes. Answer: b
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c) vg+vgse+se × vgvgsese d) +2 × +3
Morgan and his colleagues worked out a set of symbols to represent fly genotypes. Which of the following are representative? a) AaBb × AaBb b) 46, XY or 46, XX c) vg+vgse+se × vgvgsese d) +2 × +3 Answer: c
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Morgan’s Experimental Evidence Imagine that Morgan had chosen a different organism for his genetics experiments. What kind of species would have made a better choice than fruit flies? Answer: This question is designed to make students think about the process of genetic experimentation. Choosing a plant that could be self-pollinated would allow the experimenter to self-pollinate the plant and expose recessive alleles more quickly. Choosing a species with more genetic diversity could make recessive alleles be detected more quickly. A shorter generation time would be beneficial in that genetic tests could be conducted more quickly. Choosing a species with many small chromosomes would have made light microscopy difficult.
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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. 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.
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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? Bee males have half the DNA of bee females, whereas human males have nearly the same amount of DNA that human females have. 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: 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.
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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. 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.
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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? in a male because there is no possibility of the presence of a normal, dominant allele in a male because it is haploid in a female because there is no possibility of the presence of a normal, dominant allele in a female because all alleles on the W chromosomes are dominant to those on the Z chromosome none of the above Answer: E This should be a very easy question if students have studied Figure Although it is easy, it should reinforce the idea that there are different methods of sex determination. The correct answer is e. Answer a is incorrect because males do not have a W chromosome. Answer b is incorrect because males are diploid. Answer c is incorrect because females could have a dominant allele on the Z chromosome. Answer d is incorrect because the question stem stated that the allele in question is recessive.
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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? 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 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. 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.
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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? 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.
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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. 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).
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Triploid species are usually sterile (unable to reproduce), whereas tetraploids are often fertile. Which of the following are likely good explanations of these facts? 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 (Chapter 13) 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.
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Chromosomal rearrangements can occur after chromosomes break
Chromosomal rearrangements can occur after chromosomes break. Which of the following statements are 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 is 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.
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Imagine that you could create medical policy for a country
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. What kind of policy regarding fetal testing would you implement? Answer: This question should stimulate a spirited discussion. It will be important to emphasize that financial resources are limited. We often act as if such funds are, or should be, unlimited. Some things to consider: medical testing is likely to be much cheaper than training, fewer than 4% of fetuses are affected in the highest risk group, and Down syndrome incidence increases somewhat sharply at certain maternal ages. The website contains a lot of information.
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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? The mitochondrial DNA of the child and “father” did not match. DNA sequencing of chromosome 5 of the child and “father” did not match. The mitochondrial DNA of the child and mother did not match. DNA sequencing of chromosome 5 of the child and mother did not match. The mitochondrial DNA of the child and “father” matched but the mitochondrial DNA of the child and mother did not. Answer: b The purpose of this question is to help students understand that mitochondria are inherited from the mother. Mitochondrial DNA between a mother and child should match but there is no expectation of a match between father and child. Autosomal DNA should match both parents. Therefore, only answer b would demonstrate that the man was not the father of the child in question. A follow-up question could be which answer supports the idea that the man was the father. The fact that none of the answers support this idea should help students realize this is not an either-or situation.
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