14 mendel text

Mendel and the Gene Idea ,[object Object],[object Object],[object Object]

[object Object],[object Object]

[object Object],[object Object],Figure 14.1

Mendel’s Experimental, Quantitative Approach ,[object Object],[object Object],[object Object]

[object Object],Figure 14.2 1 5 4 3 2 Removed stamens from purple flower Transferred sperm- bearing pollen from stamens of white flower to egg- bearing carpel of purple flower Parental generation (P) Pollinated carpel matured into pod Carpel (female) Stamens (male) Planted seeds from pod Examined offspring: all purple flowers First generation offspring (F 1 ) APPLICATION By crossing (mating) two true-breeding varieties of an organism, scientists can study patterns of inheritance. In this example, Mendel crossed pea plants that varied in flower color. TECHNIQUE TECHNIQUE When pollen from a white flower fertilizes eggs of a purple flower, the first-generation hybrids all have purple flowers. The result is the same for the reciprocal cross, the transfer of pollen from purple flowers to white flowers. TECHNIQUE RESULTS

[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object]

The Law of Segregation ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],Figure 14.3 P Generation (true-breeding parents) Purple flowers White flowers  F 1 Generation (hybrids) All plants had purple flowers F 2 Generation EXPERIMENT True-breeding purple-flowered pea plants and white-flowered pea plants were crossed (symbolized by ). The resulting F 1 hybrids were allowed to self-pollinate or were cross- pollinated with other F 1 hybrids. Flower color was then observed in the F 2 generation. RESULTS Both purple-flowered plants and white- flowered plants appeared in the F 2 generation. In Mendel’s experiment, 705 plants had purple flowers, and 224 had white flowers, a ratio of about 3 purple : 1 white.

[object Object],[object Object],Table 14.1

Mendel’s Model ,[object Object],[object Object],[object Object]

[object Object],[object Object],Figure 14.4 Allele for purple flowers Locus for flower-color gene Homologous pair of chromosomes Allele for white flowers

[object Object],Figure 14.5 P Generation F 1 Generation F 2 Generation P p P p P p P p Pp PP pp Pp Appearance: Genetic makeup: Purple flowers PP White flowers pp Purple flowers Pp Appearance: Genetic makeup: Gametes: Gametes: F 1 sperm F 1 eggs 1 / 2 1 / 2  Each true-breeding plant of the parental generation has identical alleles, PP or pp . Gametes (circles) each contain only one allele for the flower-color gene. In this case, every gamete produced by one parent has the same allele. Union of the parental gametes produces F 1 hybrids having a Pp combination. Because the purple- flower allele is dominant, all these hybrids have purple flowers. When the hybrid plants produce gametes, the two alleles segregate, half the gametes receiving the P allele and the other half the p allele. 3 : 1 Random combination of the gametes results in the 3:1 ratio that Mendel observed in the F 2 generation. This box, a Punnett square, shows all possible combinations of alleles in offspring that result from an F 1  F 1 ( Pp  Pp ) cross. Each square represents an equally probable product of fertilization. For example, the bottom left box shows the genetic combination resulting from a p egg fertilized by a P sperm.

Useful Genetic Vocabulary ,[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],Figure 14.6 3 1 1 2 1 Phenotype Purple Purple Purple White Genotype PP (homozygous) Pp (heterozygous) Pp (heterozygous) pp (homozygous) Ratio 3:1 Ratio 1:2:1

The Testcross ,[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],Figure 14.7  Dominant phenotype, unknown genotype: PP or Pp ? Recessive phenotype, known genotype: pp If PP , then all offspring purple: If Pp , then 1 ⁄ 2 offspring purple and 1 ⁄ 2 offspring white: p p P P Pp Pp Pp Pp pp pp Pp Pp P p p p APPLICATION An organism that exhibits a dominant trait, such as purple flowers in pea plants, can be either homozygous for the dominant allele or heterozygous. To determine the organism’s genotype, geneticists can perform a testcross. TECHNIQUE In a testcross, the individual with the unknown genotype is crossed with a homozygous individual expressing the recessive trait (white flowers in this example). By observing the phenotypes of the offspring resulting from this cross, we can deduce the genotype of the purple-flowered parent. RESULTS

The Law of Independent Assortment ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],Figure 14.8 YYRR P Generation Gametes YR yr  yyrr YyRr Hypothesis of dependent assortment Hypothesis of independent assortment F 2 Generation (predicted offspring) 1 ⁄ 2 YR YR yr 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 yr YYRR YyRr yyrr YyRr 3 ⁄ 4 1 ⁄ 4 Sperm Eggs Phenotypic ratio 3:1 YR 1 ⁄ 4 Yr 1 ⁄ 4 yR 1 ⁄ 4 yr 1 ⁄ 4 9 ⁄ 16 3 ⁄ 16 3 ⁄ 16 1 ⁄ 16 YYRR YYRr YyRR YyRr Yyrr YyRr YYrr YYrr YyRR YyRr yyRR yyRr yyrr yyRr Yyrr YyRr Phenotypic ratio 9:3:3:1 315 108 101 32 Phenotypic ratio approximately 9:3:3:1 F 1 Generation Eggs YR Yr yR yr 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 Sperm RESULTS CONCLUSION The results support the hypothesis of independent assortment. The alleles for seed color and seed shape sort into gametes independently of each other. EXPERIMENT Two true-breeding pea plants— one with yellow-round seeds and the other with green-wrinkled seeds—were crossed, producing dihybrid F 1 plants. Self-pollination of the F 1 dihybrids, which are heterozygous for both characters, produced the F 2 generation. The two hypotheses predict different phenotypic ratios. Note that yellow color ( Y ) and round shape ( R ) are dominant.

[object Object],[object Object], Rr Segregation of alleles into eggs Rr Segregation of alleles into sperm R r r R R R R 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 1 ⁄ 2 r r R r r Sperm  Eggs Figure 14.9

Solving Complex Genetics Problems with the Rules of Probability ,[object Object],[object Object]

The Spectrum of Dominance ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],Figure 14.10 P Generation F 1 Generation F 2 Generation Red C R C R Gametes C R C W  White C W C W Pink C R C W Sperm C R C R C R C w C R C R Gametes 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 1 ⁄ 2 Eggs 1 ⁄ 2 C R C R C R C W C W C W C R C W

Multiple Alleles ,[object Object],[object Object],Table 14.2

Pleiotropy ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],Figure 14.11 BC bC Bc bc 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 BC bC Bc bc 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 1 ⁄ 4 BBCc BbCc BBcc Bbcc Bbcc bbcc bbCc BbCc BbCC bbCC BbCc bbCc BBCC BbCC BBCc BbCc 9 ⁄ 16 3 ⁄ 16 4 ⁄ 16 BbCc BbCc  Sperm Eggs

[object Object],[object Object],Figure 14.12  AaBbCc AaBbCc aabbcc Aabbcc AaBbcc AaBbCc AABbCc AABBCc AABBCC 20 ⁄ 64 15 ⁄ 64 6 ⁄ 64 1 ⁄ 64 Fraction of progeny

Nature and Nurture: The Environmental Impact on Phenotype ,[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object]

Skin tones – Nat. Geo. ‘ Yet with the effects of human migrations and cultural habits, people in one place can show tremendous variation in skin tone – like students from the Washington International Primary School.” ‘Unmasking Skin,’ Joel L. Swerdlow, National Geographic , Nov. 2002 p46-47.

Q 664 abcnews www.abcnews.com , Science page, "We're all the same," 9/10/98 What the facts show is that there are differences among us, but they stem from culture, not race. Q 664

Acts 17:26 Acts 17:26 And hath made of one blood all nations of men for to dwell on all the face of the earth, and hath determined the times before appointed, and the bounds of their habitation;

Pedigree Analysis ,[object Object],[object Object]

[object Object],[object Object],Figure 14.14 A, B ,[object Object],[object Object],Ww ww ww Ww ww Ww Ww ww ww Ww WW or Ww ww First generation (grandparents) Second generation (parents plus aunts and uncles) Third generation (two sisters) Ff Ff ff Ff ff Ff Ff ff Ff FF or Ff ff FF or Ff Widow’s peak No Widow’s peak Attached earlobe Free earlobe (a) Dominant trait (widow’s peak) (b) Recessive trait (attached earlobe)

Recessively Inherited Disorders ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Cystic Fibrosis ,[object Object],[object Object],[object Object]

Sickle-Cell Disease ,[object Object],[object Object],[object Object],[object Object],[object Object]

Mating of Close Relatives ,[object Object],[object Object],[object Object]

Dominantly Inherited Disorders ,[object Object],[object Object]

[object Object],[object Object],[object Object],Figure 14.16

Multifactorial Disorders ,[object Object],[object Object],[object Object],[object Object]

Genetic Testing and Counseling ,[object Object],[object Object]

Counseling Based on Mendelian Genetics and Probability Rules ,[object Object],[object Object]

Tests for Identifying Carriers ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],Figure 14.17 A, B (a) Amniocentesis Amniotic fluid withdrawn Fetus Placenta Uterus Cervix Centrifugation A sample of amniotic fluid can be taken starting at the 14th to 16th week of pregnancy. (b) Chorionic villus sampling (CVS) Fluid Fetal cells Biochemical tests can be Performed immediately on the amniotic fluid or later on the cultured cells. Fetal cells must be cultured for several weeks to obtain sufficient numbers for karyotyping. Several weeks Biochemical tests Several hours Fetal cells Placenta Chorionic viIIi A sample of chorionic villus tissue can be taken as early as the 8th to 10th week of pregnancy. Suction tube Inserted through cervix Fetus Karyotyping and biochemical tests can be performed on the fetal cells immediately, providing results within a day or so. Karyotyping

Newborn Screening ,[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],Figure 15.1

[object Object],Figure 15.2 Yellow-round seeds ( YYRR ) Green-wrinkled seeds ( yyrr ) Meiosis Fertilization Gametes All F 1 plants produce yellow-round seeds ( YyRr ) P Generation F 1 Generation Meiosis Two equally probable arrangements of chromosomes at metaphase I LAW OF SEGREGATION LAW OF INDEPENDENT ASSORTMENT Anaphase I Metaphase II Fertilization among the F 1 plants 9 : 3 : 3 : 1 1 4 1 4 1 4 1 4 YR yr yr yR Gametes Y R R Y y r r y R Y y r R y Y r R y Y r R Y r y r R Y y R Y r y R Y Y R R Y r y r y R y r Y r Y r Y r Y R y R y R y r Y F 2 Generation Starting with two true-breeding pea plants, we follow two genes through the F 1 and F 2 generations. The two genes specify seed color (allele Y for yellow and allele y for green) and seed shape (allele R for round and allele r for wrinkled). These two genes are on different chromosomes. (Peas have seven chromosome pairs, but only two pairs are illustrated here.) The R and r alleles segregate at anaphase I, yielding two types of daughter cells for this locus. 1 Each gamete gets one long chromosome with either the R or r allele. 2 Fertilization 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 1 Each gamete gets a long and a short chromosome in one of four allele combinations. 2 Fertilization results in the 9:3:3:1 phenotypic ratio in the F 2 generation. 3

Morgan’s Experimental Evidence: Scientific Inquiry ,[object Object],[object Object]

Morgan’s Choice of Experimental Organism ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],Figure 15.3

Correlating Behavior of a Gene’s Alleles with Behavior of a Chromosome Pair ,[object Object],[object Object],[object Object]

[object Object],[object Object],Figure 15.4 The F 2 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 F 1 red-eyed female to an F 1 red-eyed male to produce the F 2 generation. RESULTS P Generation F 1 Generation X F 2 Generation Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F 1 offspring all had red eyes. EXPERIMENT

CONCLUSION Since all F 1 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 F 2 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 F 1 Generation F 2 Generation Ova (eggs) Ova (eggs) Sperm Sperm X X X X Y W W + W + W W + W + W + W + W + W + W + W + W W + W W W

How Linkage Affects Inheritance: Scientific Inquiry ,[object Object],[object Object]

[object Object],[object Object],Double mutant (black body, vestigial wings) Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) P Generation (homozygous) b + b + vg + vg + x b b vg vg F 1 dihybrid (wild type) (gray body, normal wings) b + b vg + vg b b vg vg TESTCROSS x b + vg + b vg b + vg b vg + b vg b + b vg + vg b b vg vg b + b vg vg b b vg + vg 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal Sperm Parental-type offspring Recombinant (nonparental-type) offspring RESULTS EXPERIMENT Morgan first mated true-breeding wild-type flies with black, vestigial-winged flies to produce heterozygous F 1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F 1 dihybrid females with black, vestigial-winged males, producing 2,300 F 2 offspring, which he “scored” (classified according to phenotype). CONCLUSION If these two genes were on different chromosomes, the alleles from the F 1 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. Figure 15.5 Double mutant (black body, vestigial wings) Double mutant (black body, vestigial wings)

[object Object],[object Object],[object Object],Parents in testcross b + vg + b vg b + vg + b vg b vg b vg b vg b vg Most offspring X or

Recombination of Unlinked Genes: Independent Assortment of Chromosomes ,[object Object],[object Object],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 yr

Recombination of Linked Genes: Crossing Over ,[object Object],[object Object]

[object Object],[object Object],Figure 15.6 Testcross parents Gray body, normal wings (F 1 dihybrid) b + vg + b vg Replication of chromosomes b + vg b + vg + b vg vg 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 combinations.  Recombinant chromosome b + vg + b vg b + vg b vg + b vg Sperm b vg b vg Replication of chromosomes vg vg b b b vg b vg Meiosis I and II: Even if crossing over occurs, no new allele combinations are produced. Ova Gametes Testcross offspring Sperm b + vg + b vg b + vg b vg + 965 Wild type (gray-normal) b + vg + b vg b vg b vg b vg b vg + b + vg + b vg + 944 Black- vestigial 206 Gray- vestigial 185 Black- normal Recombination frequency = 391 recombinants 2,300 total offspring  100 = 17% Parental-type offspring Recombinant offspring Ova b vg Black body, vestigial wings (double mutant) b

Linkage Mapping: Using Recombination Data: Scientific Inquiry ,[object Object],[object Object],[object Object]

[object Object],[object Object],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. Figure 15.7

[object Object],[object Object],Figure 15.8 Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Long aristae (appendages on head) Gray body Red eyes Normal wings Red eyes Wild-type phenotypes II Y I X IV III 0 48.5 57.5 67.0 104.5

[object Object],[object Object],Figure 15.9a (a) The X-Y system 44 + XY 44 + XX Parents 22 + X 22 + Y 22 + XY Sperm Ova 44 + XX 44 + XY Zygotes (offspring)

[object Object],[object Object],Figure 15.9b–d 22 + XX 22 + X 76 + ZZ 76 + ZW 16 (Haploid) 16 (Diploid) (b) The X–0 system (c) The Z–W system (d) The haplo-diploid system

Inheritance of Sex-Linked Genes ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],Figure 15.10a–c X A X A X a Y X a Y X A X a X A Y X A Y X A Y a X A X A Ova Sperm X A X a X A Y Ova X A X a X A X A X A Y X a Y X a Y A X A Y Sperm X A X a X a Y   Ova X a Y X A X a X A Y X a Y X a Y a X A X a 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) Sperm

X inactivation in Female Mammals ,[object Object],[object Object]

[object Object],[object Object],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 Active X Black fur Inactive X Figure 15.11

Abnormal Chromosome Number ,[object Object],[object Object],[object Object],Figure 15.12a, b Meiosis I Nondisjunction Meiosis II Nondisjunction Gametes n + 1 n + 1 n  1 n – 1 n + 1 n – 1 n n Number of chromosomes Nondisjunction of homologous chromosomes in meiosis I Nondisjunction of sister chromatids in meiosis II (a) (b)

Alterations of Chromosome Structure ,[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],Figure 15.14a–d A B C D E F G H Deletion A B C E G H F A B C D E F G H Duplication A B C B D E C F G H A A M N O P Q R B C D E F G H B C D E F G H Inversion Reciprocal translocation A B P Q R M N O C D E F G H A D C B E F H G (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 ,[object Object],[object Object]

Down Syndrome ,[object Object],[object Object],Figure 15.15

Aneuploidy of Sex Chromosomes ,[object Object],[object Object]

Disorders Caused by Structurally Altered Chromosomes ,[object Object],[object Object]

[object Object],[object Object],Figure 15.16 Normal chromosome 9 Reciprocal translocation Translocated chromosome 9 Philadelphia chromosome Normal chromosome 22 Translocated chromosome 22

Genomic Imprinting ,[object Object],[object Object]

[object Object],[object Object],Figure 15.17a, b (a) A wild-type mouse is homozygous for the normal igf2 allele. Normal Igf2 allele (expressed) Normal Igf2 allele with imprint (not expressed) Paternal chromosome Maternal chromosome Wild-type mouse (normal size) Normal Igf2 allele Paternal Maternal Mutant lgf2 allele Mutant lgf2 allele Paternal Maternal Dwarf mouse Normal Igf2 allele with imprint 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 ,[object Object],[object Object]

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