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AP Biology Chapter 15 The Chromosome Theory of Inheritance.

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1 AP Biology Chapter 15 The Chromosome Theory of Inheritance

2  Mendel’s “factors” are now known to be genes—segments of chromosomes.  Where the chromosomes go during Meiosis determines which traits end up in each of the gametes.

3  First described by Walter S. Sutton in 1902:  Genes have specific loci (positions) along chromosomes  It is the chromosomes that undergo segregation and independent assortment

4 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  The Chromosomal Basis of Mendel’s Laws:

5 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)

6 F 2 Generation An F 1  F 1 cross-fertilization 9 : 3 : Fertilization recombines the R and r alleles at random. On a different chromosome, fertilization recombines theY and y alleles also. YyRr X YyRr produces 9:3:3:1 ratio. So, Mendel’s ratios can be explained by examining the behavior of the chromosomes during meiosis.

7  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.

8  Thomas Hunt Morgan chose Drosophila melanogaster, a common insect that feeds on the fungi growing on fruit. Why?  They are prolific breeders (a single mating can produce hundreds of offspring)  They can be bred every two weeks  It has only 4 pairs of chromosomes  There are many types of easily identified mutants which differ from the normal (wild) type.

9  When Thomas Hunt Morgan studied fruit flies, he was looking for naturally occurring variants. After years of study, he finally found one male fruit fly with white eyes instead of the usual red.  The allele for the mutant trait is written as a lower case letter (ex: white eyes is w). The wild-type fly (normal phenotype) is shown with the same letter with a superscript + : w +

10 EXPERIMENT P Generation F1F1 All offspring had red eyes  In one experiment, Morgan mated a white-eyed male with a red-eyed female. All of the F 1 offspring had red eyes. What does this tell us about the red eye trait? Correlating behavior of an allele with behavior of the chromosome

11 Fig. 15-4b RESULTS Generation F2F2 When Morgan bred the F 1 flies to each other, he observed the classical 3:1 phenotypic ratio in the F 2 offspring. However, surprisingly, the white-eyed trait showed up only in the males! Somehow, the fly’s eye color is related to its sex. F 2 Results:

12 Eggs F1F1 CONCLUSION Generation P X X w Sperm X Y Eggs Sperm Generation F2F2  w w w w w w w w w w w w w w w

13  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 X and Y chromosomes → The SRY gene on the Y chromosome codes for the development of the testes.

14 (a) The X-Y system 44 + XY 44 + XX Parents 44 + XY 44 + XX 22 + X 22 + X 22 + Y or Sperm Egg + Zygotes (offspring) In mammals, the sex of an offspring depends on whether the sperm cell contains an X chromosome or a Y.

15 Fig. 15-6b (b) The X-0 system 22 + XX 22 + X In grasshoppers, cockroaches, and some other insects, there is only one type of sex chromosome, the X. Females are XX, males have only one sex chromosome (XO). Sex of the offspring is determined by whether the sperm cell contains an X chromosome or no sex chromosome.

16 Fig. 15-6c (c) The Z-W system 76 + ZW 76 + ZZ In birds, some fishes, and some insects, the sex chromosomes present in the egg (not the sperm) determine the sex of offspring. The sex chromosomes are designated Z and W. Females are ZW and males are ZZ.

17 Fig. 15-6d (d) The haplo-diploid system 32 (Diploid) 16 (Haploid) There are no sex chromosomes in most species of bees and ants. Females develop from fertilized eggs and are thus diploid. Males develop from unfertilized eggs and are haploid; they have no fathers.

18  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 chromosome

19  Sex-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 Inheritance Patterns of Sex Chromosomes.

20 Fig (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 The Transmission of Sex-linked recessive traits. A color-blind father will transmit the mutant allele to all daughters but not to sons. Daughters are carriers. If a carrier mates with a male who has normal color vision, there is a 50% chance the son will be color-blind. If a carrier mates with a color-blind male, there is a 50% chance their child will be color- blind.

21  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.  The Barr body lies along the inside of the nuclear envelope. Most of the genes in the Barr body are not expressed. Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

22 If a female is heterozygous for a particular gene located on the X chromosome, she will be a mosaic for that character

23 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 The tortoiseshell gene is on the X chromosome in cats. The tortoiseshell color requires the presence of two alleles: one orange and one black. These are located on the X chromosome. If a female (XX) is heterozygous, the orange and black patches are present in populations of cells with that activated gene.

24  Genes located near each other on the same chromosome tend to be inherited together.  These are called linked genes.  These are not to be confused with sex-linked traits (traits that come from the sex chromosomes)

25  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 Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings Wild type—normal wings/color Ebony body Vestigial (short) wings

26  In the fruit fly Drosophila melanogaster, flies reared in the laboratory occasionally exhibit mutations in their genes.  Two such mutations, affecting body color and wing structure are linked. Morgan did experiments studying these two traits. Grey body and normal wings are dominant traits. Black body and vestigial wings are recessive

27 Fig. 15-UN1 b + vg + Parents in testcross Most offspring b + vg + b vg  or Heterozygous for grey body, normal wings X Homozygous for black body, vestigial wings Grey body, normal wings Black body, vestigical wings

28 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg b + b + vg + vg +

29 Fig 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

30 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

31 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: : : : : : : : : :

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

33  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 Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings Genetic Recombination Original phenotypes Recombinant phenotypes

34  nt/chp10/ html nt/chp10/ html

35  When traits appear that are different from either one of the parents, it is due to independent assortment when genes are not on the same chromosome.  Parental types: resemble the parents  Recombinants: contain new combinations of genes Parental types Recombinants If genes are located on different chromosomes, there will be a 50% recombination rate.

36 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 Recombinants

37  Recombinations in traits that are located on the same chromosome (linked genes) are due to crossing over.

38

39  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

40

41  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 Genetic Linkage Maps

42 RESULTS Recombination frequencies Chromosome 9%9.5% 17% bcnvg A linkage map from Drosophila: Genes that are far apart on a chromosome can have recombination frequencies close to 50%. (These behave almost the same as if they were on difference chromosomes)

43  Determine which traits are like the original parents (parental traits). Determine which traits are new combinations of the genes (these are the recombinants)  Figure out the total number of recombinant offspring and divide by the total number of offspring X 100 = Recombination % What is the recombination frequencies of the b and vg genes?

44  A wild-type fruit fly heterozygous for gray body color and normal wings, b + b vg + vg, is mated with a black fly with vestigial wings bbvgvg. The offspring have the following phenotypic distribution:  Wild-type (gray, normal wings): 778  Black-vestigial: 785  Black-normal wings: 158  Gray-vestigial: 162  What is the recombination frequency between these genes for body color and wing size?

45  Number recombinants = 320  Total offspring = 1883  Recombinant frequency = 320/1883 X 100 =  17%

46  Determine the sequence of genes along a chromosome based on the following recombination frequencies:  A—B, 8 map units  A—C, 28 map units  A—D, 25 map units  B—C, 20 map units  B—D, 33 map units

47  D A B C

48  Large-scale chromosomal alterations often lead to spontaneous abortions (miscarriages) or cause a variety of developmental disorders Children with Down’s Syndrome/ Trisomy 21

49  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  Offspring with this condition is called aneuploidy

50 Fig Meiosis I (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II Nondisjunction

51 Fig Meiosis I Nondisjunction (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II Meiosis II Nondisjunction

52 Fig 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

53  A monosomic zygote has only one copy of a particular chromosome  A trisomic zygote has three copies of a particular chromosome Monosomic, Trisomic Zygotes Turner’s Syndrome Karyotype: XO (monosomic zygote) Down’s Syndrome Karyotype: Trisomy 21

54  Polyploidy is a condition in which an organism has more than two complete sets of chromosomes  Polyploidy is common in plants, but not animals. In plants, it may result in hybrids that are more vigorous. Polyploidy can result when a 2N zygote fails to divide after replicating its chromosomes. This will produce a 4N embryo. (Note the hybrid vigor in the middle plant) Polyploidy

55  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 Cri du chat results from a deletion of a portion of chromosome 5.

56 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

57 Normal chromosome 9 Normal chromosome 22 Reciprocal translocation Translocated chromosome 9 Translocated chromosome 22 (Philadelphia chromosome) Translocation The cancerous cells in nearly all CML (chromic myelogenous leukemia) patients contain an abnormally short chromosome 22, the so-called Philadelphia chromosome, and an abnormally long chromosome 9. These altered chromosomes result from translocation.


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