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The Chromosomal Basis of Inheritance

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Presentation on theme: "The Chromosomal Basis of Inheritance"— Presentation transcript:

1 The Chromosomal Basis of Inheritance
Chp. 15

2 Genes are located on… CHROMOSOMES!

3

4 Human Genome Project

5 Chromosomal Basis of Mendel’s Laws…
Page 275

6 Drosophila melanogaster
Thomas Hunt MORGAN – first to locate a specific gene on a specific chromosome FRUIT FLY Drosophila melanogaster

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8 FEMALE MALE

9 WILD TYPE (red eyes) MUTANT (white eyes)

10 Drosophila allele symbols
Gene symbol comes from mutant Ex: white eyes  w Wild type (normal phenotype) is dsignated with a “+” Ex: normal (red) eyes  w+ If mutant is recessive, use lower case… If mutant is dominant to wild type, use upper case…

11 White eyed male crossed with a wild-type female…
All F1 had red (wild-type) eyes F2 had 3 wild type : 1 white BUT… ONLY MALES had WHITE eyes Thus, eye color “linked” to sex

12 Gene for white eye color located on the “X” chromosome*
Symbols: Xw+ = wild type Xw = white eye *Called a Sex-Linked Gene

13 PRACTICE: Punnett Squares with Sex Linked Genes
P Generation = wild-type female & white eyed male Xw+ Xw+ x Xw Y F1 = ? Xw+ Xw Y Xw+ Xw Xw+ Xw Xw+ Y Xw+ Y

14 PRACTICE: Punnett Squares with Sex Linked Genes
P Generation = wild-type female & white eyed male Xw+ Xw+ x Xw Y F1 = Xw+ Xw and Xw+ Y (all wild type) F2 = Xw+ Xw Xw+ Xw+ Xw+ Xw Y Xw+ Y Xw Y

15 Linked Genes Linked Genes = genes on same chromosome
Tend to be inherited together black bodies and vestigial wings Wild type

16 F1 = b+ b vg+ vg b+ vg+ b+ vg+ b vg b vg Black body & vestigial wing
Wild type b+ vg+ b+ vg+ b vg b vg b+ b+ vg+ vg+ b b vg vg Gametes: b+ vg b vg b+ vg+ b vg F1 = b+ b vg+ vg

17 If on different chromosomes (independent assortment), then
Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg x b b vg vg b+ b b b vg+ vg vg vg Gametes: b+vg+; b+vg; b vg+; b vg b vg

18 If on different chromosomes (independent assortment), then
Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg x b b vg 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 Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial : : :

19 F1 = b+ b vg+ vg b+ vg+ b+ vg+ b vg b vg Black body & vestigial wing
Wild type b+ vg+ b+ vg+ b vg b vg b+ b+ vg+ vg+ b b vg vg Gametes: b+ vg b vg b+ vg+ b vg F1 = b+ b vg+ vg

20 If on same chromosome with NO CROSSOVER, then:
Test cross of F1 If on same chromosome with NO CROSSOVER, then: b+ b vg+ vg x b b vg vg b+ vg+ b vg b vg b vg Gametes: b+ vg+ or b vg b vg

21 If on same chromosome with NO CROSSOVER, then:
Test cross of F1 If on same chromosome with NO CROSSOVER, then: b+ b vg+ vg x b b vg 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 Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial

22 If on same chromosome with CROSSOVER, then:
Test cross of F1 If on same chromosome with CROSSOVER, then: b+ b vg+ vg x b b vg vg b vg b+ vg b+ vg+ b vg+ b vg b vg b vg Gametes: b+ vg+ or b vg b vg b+ vg or b vg+

23 If on same chromosome with CROSSOVER, then:
Test cross of F1 If on same chromosome with CROSSOVER, then: b+ b vg+ vg x b b vg vg Recombinants b+ vg+ b+ vg b vg+ b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg Parental Types Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial RATIO ???

24 Parental Types = 1909 flies Recombinants = 391 flies % Recombinants 391 recomb. = .17 or 2300 total %

25 b vg 17 map units

26 Linkage Map: uses recombination frequencies to map relative location of genes on chromosomes
1 map unit = 1 % recombination freq. ex: b-vg = 17% b-cn = 9% cn-vg = 9.5%

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28 Other chromosomal maps:
Cytogenic map – actually pinpoints genes on physical location of chromosome (bands) DNA sequencing/physical map – gives order of nucleotides for a gene and intergenic sequences in # of b.p. (base pairs)

29 PRACTICE In tomatoes, round fruit shape (O) is dominant to elongated (o), and smooth skin (S) is dominant to fuzzy skin (s). Test crosses of F1 individuals heterozygous for these pairs of alleles gave the following results: 12 elongated-smooth 123 round-smooth 133 elongated-fuzzy 12 round-fuzzy Are these genes linked? Calculate the % recombination and the map distance between the two genes.

30 PRACTICE parental recombinants
In tomatoes, round fruit shape (O) is dominant to elongated (o), and smooth skin (S) is dominant to fuzzy skin (s). Test crosses of F1 individuals heterozygous for these pairs of alleles gave the following results: 12 elongated-smooth 123 round-smooth 133 elongated-fuzzy 12 round-fuzzy Calculate the % recombination and the map distance between the two genes. 24 / 280 = .086  8.6% 8.6 map units parental recombinants

31 PRACTICE The cross-over percentages between linked genes are given below: A – B = 40% C – D = 10% B – D = 10% B – C = 20% A – C = 20% What is the sequence of genes on the chromosome? (draw a map and label distance between genes) 20 10 10 A C D B

32 PRACTICE 3. Recombination frequency is given below for several gene pairs. Create a linkage map for these genes, and show the map unit distances between loci (genes). j, k = 12% k, l = 6% j, m = 9% l, m = 15% 9 6 6 m j l k

33 Sex Chromosomes and sex-linked genes:
XX = female XY = male Father’s gamete determines sex of child Presence of a Y chromosome (SRY genes) allows development of testes/male characteristics

34 Inheritance of sex-linked genes
Sex-linked gene = gene carried on sex chromosome (usually X) Females (XX) only express recessive sex-linked phenotypes if homozygous recessive for the trait Males (XY) will express what ever allele is present on the X chromosome = hemizygous

35 PRACTICE What are the possible phenotypes of the offspring from a woman who is a carrier for a recessive sex-linked allele and a man who is affected by the recessive disorder? 1 normal female: 1 affected female: 1 normal male: 1 affected male

36 PRACTICE Two normal color-sighted individuals produce the following family (see pedigree). Fill in the probably genotypes of the numbered individuals. Solid symbols represent color blindness. 1 2 5 3 4 6 7

37 PRACTICE XAY XAXa 1 2 5 3 4 6 7 XAXA XAXa XAY XaY XAXa
Two normal color-sighted individuals produce the following family (see pedigree). Fill in the probably genotypes of the numbered individuals. Solid symbols represent color blindness. XAY XAXa 1 2 5 3 4 6 7 XAXA XAXa XAY XaY XAXa

38 Sex-linked Disorders in Humans
Duchenne Muscular Dystrophy

39 Sex-linked Disorders in Humans
Duchenne Muscular Dystrophy Hemophilia

40 Sex-linked Disorders in Humans
Duchenne Muscular Dystrophy Hemophilia Fragile X

41 Sex-linked Disorders in Humans
Duchenne Muscular Dystrophy Hemophilia Fragile X (Baldness & red-green color-blindness)

42 X Inactivation: females have two X chromosomes, but only need one active X
One X condenses in each cell during embryonic development  Barr body Females are a “mosaic” if heterozygous for a sex-linked trait ex: Calico cats

43 Chromosomal Alterations
Aneuploidy – 1 more/less chromosome Due to NONDISJUNCTION: separation of homologous chromosomes (Anaphase I) or sister chromatids (Anaphase II) fails

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45 Chromosomal Alterations
Aneuploidy – 1 more/less chromosome Due to NONDISJUNCTION: separation of homologous chromosomes (Anaphase I) or sister chromatids (Anaphase II) fails Trisomy = 1 extra chromosome (2n + 1) Monosomy = 1 less chromosome (2n – 1) HUMANS – cannot have more than 47 or less than 45 chromosomes & NEED AT LEAST ONE “X” to survive

46 Aneuploid Disorders Down Syndrome: Trisomy 21
Klinefelter Syndrome: XXY Trisomy X: XXX Turner Syndrome: Monosomy X (X0) Only viable monosomy in humans!

47 Polyploidy Polyploidy = more than two complete sets of chromosomes (nondisjunction) TRIPLOIDY = 3n Humans: n = haploid = 1 set = 23 chromosomes 2n = diploid = 2 sets = 46 chromosomes 3n = triploid = 3 sets = 69 chromosomes COMMON IN PLANT KINGDOM

48 Activity: Polyploid Plants

49 Alterations of Chromosome Structure

50 Prader Willi & Angelman Syndrome

51

52 Cri du chat Deletion on chromosome #5

53 CML Translocation (22 & 9)  “Philadelphia Chromosome”

54 on the first chromosome?
PRACTICE: Two non-homologous chromosomes have genes in the following order: A-B-C-D-E-F-G-H-I-J & M-N-O-P-Q-R-S-T What chromosome alterations have occurred if daughter cells have a gene sequence of A-B-C-O-P-Q-G-J-I-H on the first chromosome?

55 on the first chromosome?
PRACTICE: Two non-homologous chromosomes have genes in the following order: A-B-C-D-E-F-G-H-I-J & M-N-O-P-Q-R-S-T deletion inversion translocation What chromosome alterations have occurred if daughter cells have a gene sequence of A-B-C-O-P-Q-G-J-I-H on the first chromosome?

56 Genomic Imprinting When it matters which parent you inherited the allele from… Occurs during formation of gametes Methyl groups (-CH3) added to DNA and “silence” alleles When offspring produce own gametes, parental imprinting is erased & alleles re-imprinted according to sex of offspring Ex: insulin–like growth factor 2

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58 Genomic Imprinting

59 “Extranuclear Genes” Mitochondria (mtDNA), chloroplasts, etc..
inherited from mother through the egg


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