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The Chromosomal Basis of Inheritance Chp. 15 Genes are located on… CHROMOSOMES! Human Genome Project Chromosomal Basis of Mendel’s Laws… Page 275 Thomas Hunt MORGAN – first to locate a specific gene on a specific chromosome Drosophila melanogaster FEMALE MALE WILD TYPE (red eyes) MUTANT (white eyes) 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… 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 Gene for white eye color located on the “X” chromosome* Symbols: Xw+ = wild type Xw = white eye *Called a Sex-Linked Gene PRACTICE: Punnett Squares with Sex Linked Genes • P Generation = wild-type female & white eyed male Xw+ Xw+ x Xw Y • F1 = ? Xw+ Xw+ Xw Xw+ Xw Xw+ Xw Y Xw+ Y Xw+ Y PRACTICE: Punnett Squares with Sex Linked Genes • P Generation = wild-type female & white eyed male Xw+ Xw+ x Xw Y • F1 = • F2 = Xw+ Xw and Xw+ Y (all wild type) Xw+ Y Xw+ Xw Xw+ Xw+ Xw+ Xw Xw+ Y Xw Y Linked Genes • Linked Genes = genes on same chromosome – Tend to be inherited together Wild type black bodies and vestigial wings Wild type Black body & vestigial wing b+ b+ b b vg+ vg+ vg vg b+ b+ vg+ vg+ Gametes: b+ vg+ b b vg vg b vg b+ b vg+ vg F1 = b+ b vg+ vg Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg b+ x b b b vg vg b vg+ b vg Gametes: b+vg+; b+vg; b vg+; b vg vg b vg vg Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg x b b vg vg b+ vg+ b vg Body: Wing: b+ vg b vg+ b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg Normal Normal 1 Normal Vestigial : 1 Black Normal : 1 Black Vestigial : 1 Wild type Black body & vestigial wing b+ b+ b b vg+ vg+ vg vg b+ b+ vg+ vg+ Gametes: b+ vg+ b b vg vg b vg b+ b vg+ vg F1 = b+ b vg+ vg Test cross of F1 If on same chromosome with NO CROSSOVER, then: b+ b vg+ vg x b b vg vg b+ b b b vg+ vg vg vg Gametes: b+ vg+ or b vg b vg 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 Body: Wing: b+ vg b vg+ b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg Normal Normal Normal Vestigial Black Normal Black Vestigial Test cross of F1 If on same chromosome with CROSSOVER, then: b+ b vg+ vg x b b vg vg b+ b b b vg vg+ vg+ vg vg vg Gametes: b+ vg+ or b vg b+ vg or b vg+ b vg 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 vg b+ b vg+ vg b+ b Parental vg vg b b vg+ vg b b vg vg Types Body: Wing: Normal Normal Normal Vestigial Black Normal RATIO ??? Black Vestigial Parental Types 965 + 944 = 1909 flies Recombinants 206 + 185 = 391 flies % Recombinants 391 recomb. = .17 or 2300 total 17% b vg 17 map units 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% • 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) PRACTICE 1. 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. PRACTICE 1. 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 parental recombinants 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 PRACTICE 2. 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 A 10 10 C D B 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 m 6 6 j l k 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 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 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 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 3 4 5 6 7 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. XAY 1 3 XAXA 5 XAY 2 XAXa 4 XAXa 6 7 XaY XAXa Sex-linked Disorders in Humans • Duchenne Muscular Dystrophy Sex-linked Disorders in Humans • Duchenne Muscular Dystrophy • Hemophilia Sex-linked Disorders in Humans • Duchenne Muscular Dystrophy • Hemophilia • Fragile X Sex-linked Disorders in Humans • • • • Duchenne Muscular Dystrophy Hemophilia Fragile X (Baldness & red-green color-blindness) 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 sexlinked trait ex: Calico cats Chromosomal Alterations • Aneuploidy – 1 more/less chromosome – Due to NONDISJUNCTION: separation of homologous chromosomes (Anaphase I) or sister chromatids (Anaphase II) fails 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 Aneuploid Disorders • Down Syndrome: Trisomy 21 • Klinefelter Syndrome: XXY • Trisomy X: XXX • Turner Syndrome: Monosomy X (X0) – Only viable monosomy in humans! 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 Activity: Polyploid Plants Alterations of Chromosome Structure Prader Willi & Angelman Syndrome Cri du chat • Deletion on chromosome #5 CML • Translocation (22 & 9) “Philadelphia 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? 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? 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 reimprinted according to sex of offspring – Ex: insulin–like growth factor 2 Genomic Imprinting “Extranuclear Genes” • Mitochondria (mtDNA), chloroplasts, etc.. inherited from mother through the egg