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Transmission (Classical, Mendelian) Genetics Ch 11 • Gregor Mendel – Experiments in Plant Hybridization, 1865 • Simple, controlled, data collection, mathematical analysis Pisum sativum, the garden pea • • • • • What makes this a good model organism? easy to grow hundreds of offspring per cross short generation time can self fertilize or cross – Paint pollen (sperm) from one plant onto the female parts of another (emasculated) Mendel’s conclusions 1. Unit factors in pairs - Genes are physical units – 2 alleles for each gene – 1 allele inherited from each parent Genes and alleles of Pisum sativum Gene • Pea color • Flower color • Pod shape • Pea surface • Stem height Alleles ? white, purple constricted, inflated ? tall, dwarf 2. Principle of Dominance - One allele is dominant the other is recessive - The dominant allele is expressed in the phenotype Gene for flower color P allele = purple p allele = white GENOTYPES Homozygous dominant = Heterozygous = Homozygous recessive = PHENOTYPE 3. Random segregation of alleles into gametes – gamete receives ONE allele per gene – random segregation of alleles 50/50 P generation PP pp What is the phenotype of all offspring in F1 generation? Note that the P generation is true breeding P p Genotype Phenotype p P How did Mendel do it? The Monohybrid cross YY yy Which allele is dominant? What is the genotype of the f1 generation? Cross 2 f1 plants (or let one self-fertilize) What is the ratio of phenotypes? Results of Mendel’s monohybrid crosses Parental Strains Tall X dwarf Round seeds X wrinkled Yellow seeds X green Violet flowers X white Inflated pods X constricted Green pods X yellow Axial flowers X terminal gene = ? alleles = ? F2 progeny Ratio 787 tall, 277 dwarf 5474 round, 1850 wrinkled 6022 yellow, 2001 green 705 violet, 224 white 882 inflated, 299 constricted 428 green, 152 yellow 651 axial, 207 terminal Test cross (one gene) • A mouse has black fur, what are its 2 possible genotypes? Test cross mouse to homozygous recessive mouse If black mouse is BB If black mouse is Bb Mouse was test crossed and 7 offspring were black 2 were white. What is mouse’s genotype? Autosomal recessive inheritance (bb) • unaffected parents can have affected offspring • May “skip” a generation • Two affected parents cannot have an unaffected child • Not sex related Examples of autosomal recessive traits • • • • Sickle cell disease Albinism Cystic fibrosis O blood type Phenylketonuria (Ch. 4 pg 73) • PKU (1/12,000) Mutation in gene encoding phenylalanine hydroxylase enzyme needed for phe metabolism missing phenylalanine hydroxylase enzyme If plasma phe level is too high, phe is converted into a phenylpyruvate toxic to brain tissue Why are these babies normal when born? Pleiotropic effects no tyrosine (little melanin) slow growth retardation blue eyes low adrenaline No nutrasweet low phe diet ($5K/yr) Page 68 1902 Archibald Garrod: One gene: one enzyme “Inborn errors of metabolism” PKU Albinism Alkaptonuria Tyrosinemia Black urine arthritis One gene/one enzyme • Garrod’s work on alkaptonuria “Inborn Errors of Metabolism” 1902 Autosomal recessive metabolic disease Fill in genotypes. If II,1 and II, 4 mate, what is the chance of offspring having PKU? How do we know this is autosomal recessive? II, 1 X II, 4 •p(aa AND a girl)? p(aa) If III-3 and II-1 mate p (normal child) p (affected boy)? All people have harmful recessive alleles, small chance That 2 people with same rare alleles will mate Consanguinous marriage increases the chance Bedoin intermarriage Autosomal dominant disorders Aa and AA =affected aa =unaffected •Tend to show up in every generation •2 affected parents can have unaffected child •2 unaffected parents cannot have an affected child Dominant pedigree Achondroplasia -1/20,000 births • Mutation in one allele of FGFR3 gene Chromosome 4 • Affects cartilage growth needed for bone lengthening • Most affected individuals are Aa why? • Most cases are spontaneous (Parents are aa X aa) P(III, 3 and III, 5 have a child of normal height) P ( II, 3 and III, 7 have a boy with achondroplasia) Fruit fly nomenclature pg 317 box 12.1 Red eyes is wildtype phenotype, brown is mutant bw+ = wildtype allele bw = brown allele genotype phenotype red brown • Try it: • Wingless is recessive mutant (wg allele) • Genotype of wildtype, heterozygote, mutant? Autosomal Sex-linked Genetics Home Reference page (National Library of Medicine) • Collagen • Blood clotting factor • Red blood cell enzyme • Dystrophin muscle protein • Color vision gene Sex-linked genes Ch 12 pg 314 – 317, 326 - 328 • Human Female = XX – two alleles for each X-linked gene – normal application of recessive and dominance X HX H X HX h X hXh • Human Male XY XHY XhY X-linked genes • Hemophilia (recessive) 1/5000 males – Mutation in gene for clotting factor Xq28 •Mate III 13 with III 1 Probability of a hemophiliac son? •Mate IV 2 with homozygous normal female p (hemophilia)? *Criss cross inheritance of X linked traits Dihybrid cross – 2 genes Mendel’s Law of Independent assortment each allele for a trait is inherited independently of other alleles Seeds: G = yellow allele g = green allele W = round allele w = wrinkled allele gene? gene? Parents = GGWW X phenotype? gametes? F1 genotype ? F1 phenotype ? F1 Gametes? ggww Note that each gene gives the 3:1 ratio of a monohybrid cross Yellow/green ratio = Round/ wrinkled = Forked line method for phenotypes GgWw X GgWw Test cross A pea is round and yellow. What is its genotype? G-W- X ggww Note the cross of the “unknown” to a homozygous recessive If all yellow and round: If all yellow and some wrinkled: If all round and some green: If 1:1:1:1: Probability Product rulethe probability that two outcomes occur simultaneously is product of their individual probabilities assumes independent assortment of genes GgWw X GgWw • What is the probability of a yellow AND wrinkled? p(G-ww) Trihybrid cross AaBbCc X AaBbCc p(A-B-cc) AabbCcDD X AaBbCcDd p(triply recessive) Modified Mendelian Ratios 1. INCOMPLETE DOMINANCE R = red flower (snapdragon) R’ = white flower * allele symbols do not connote dominance * phenotypic ratio = genotypic ratio = ? P CrCr X Cw Cw F1 F2 Incomplete dominance 2. Codominance Each allele encodes separate gene product distinct in phenotype of heterozygote L gene for human blood cell surface protein LM = M antigen LN = N antigen • A man with the M bloodtype has a child with a woman of the MN bloodtype • Expected ratio of offspring? 3. Multiple alleles (more than 2 alleles for gene in population) • Example: Blood Groups Karl Landsteiner 1900s Chromosome 9 I gene ABO blood system = polymorphic I gene Blood type A B AB O genotype IAIA or IAi IBIB or IBi ? ii What is the mechanism of inheritance of A, B, AB, O? Autosomal or sex chromosome? ABO dominance and codominance 4. Dominance series – C series/ rabbits c+ = full color cch = chinchilla (hypomorphic) ch = himalayan (hypomorphic) c = albino (apomorphic allele = nonfunctional) Chinchill a Himalayan Albino Genotype cch cch cch ch ch c c+ cch c+ = full color cch = chinchilla (hypomorphic) ch = himalayan (hypomorphic) c = albino (apomorphic allele = nonfunctional) phenotype? 5. Lethal alleles MM = normal spine MM’ = manx cat (no tail) M’M’ = lethal Cross two manx, what is ratio of phenotypes in offspring? How do breeders obtain manx cats? 6. Epistasis- gene product interactions. Table 13.4 page 355 (look at 4 phenotypic classes and fewer than 4) • A product of one gene influences, or masks, the expression of another gene(s) • Modification of dihybrid cross ratio AaBb X AaBb 9:3:3:1 Epistasis in Cats • W = white w = not white • B = black b = brown Mate 2 heterozygous cats What is the expected ratio? Epistasis in labrador retrievers • B and E color genes (labs) B black E color b brown e no color (yellow) ee is epistatic Cross two double heterozygotes Phenotypes of parents? Phenotypes of offspring? ratio? Polydactyly, dominant 7. Penetrance • % individuals that exhibit phenotype corresponding to genotype Pp 5,5 pp 6, 5 6, 6 8. Expressivity (ex. Piebald spotting) – the extent to which a trait is exhibited osteogenesis imperfecta pg. 359 Penetrance AND expressivity • • • • NF-1 = Neurofibromatosis1 Autosomal dominant trait N50 – 80% penetrance Expressivity – Pigmented skin to tumors on nerve CT coverings (neurofibromas) on skin, eyes, organs, face – Speech, blood pressure, spine curvature, headaches 9. Quantitative (multifactorial) traits • Vary continuously – Weight, height, IQ Gene expression also affected by: • • • • Sex (baldness) Temperature (melanin in Siamese cats) Chemicals (PKU) Diet (height, cancer) + many other factors!