Transmission (Classical, Mendelian) Genetics Ch 11

1. Transmission (Classical, Mendelian) Genetics Ch 11 • Gregor Mendel • Experiments in Plant Hybridization, 1865 • Controlled experiments, mathematical analysis

2. Pisum sativum, the garden pea • What makes this a good model organism? • easy • 100s offspring per cross • short generation time • can self fertilize or cross • Paint pollen (sperm) from one plant onto the female parts of another (emasculated plant)

4. Mendel’s conclusions • Genes physical units • 2 alleles/gene • 1 allele inherited from each parent

5. Genes and alleles of Pisum sativum Gene Alleles • Pea color ? • Flower color white, purple • Pod shape constricted, inflated • Pea surface ? • Stem height tall, dwarf

6. 2. Principle of Dominance • The dominant allele is expressed in the phenotype • Recessive is not • Gene for flower color • P allele = purple • p allele = white • GENOTYPES PHENOTYPE • Homozygous dominant = • Heterozygous = • Homozygous recessive =

7. 3. Random segregation of alleles into gametes • gamete receives 1allele/gene • random segregation of alleles 50/50 P generation PP pp What is the phenotype of all offspring in F1 generation?

8. Note that the P generation is true breeding P P p p Genotype Phenotype

9. How did Mendel do it? Monohybrid cross YY yy Which allele is dominant? Genotype of the f1 generation?

10. Cross 2 f1 plants (or let one self-fertilize) What is the ratio of phenotypes?

11. Results of Mendel’s monohybrid crosses Parental Strains F2 progeny Ratio Tall X dwarf 787 tall, 277 dwarf Round seeds X wrinkled 5474 round, 1850 wrinkled Yellow seeds X green 6022 yellow, 2001 green Violet flowers X white 705 violet, 224 white Inflated pods X constricted 882 inflated, 299 constricted Green pods X yellow 428 green, 152 yellow Axial flowers X terminal 651 axial, 207 terminal gene = ? alleles = ?

12. Test cross (one gene) • In cats, white is the dominant fur color. fur, what are its 2 possible genotypes? How can a test cross be used to determine genotype? \

13. All possible one gene crosses Parents ratio offspring

14. Autosomal recessive inheritance (bb) • unaffected parents can have affected offspring • May “skip” generation • Two affected parents cannot have unaffected child • Not sex related

15. Autosomal recessive traits • Sickle cell disease • Albinism • Cystic fibrosis • O blood type

16. Phenylketonuria (Ch.4) • PKU (1/12,000) Mutation in gene encoding phenylalanine hydroxylase enzyme needed for phe metabolism Chromosome 12 12q24.1

17. phenylalanine hydroxylase block toxic to brain tissue

18. All US babies tested at birth (Guthrie test) These contain phe

19. Why are PKU babies normal when born? Pleiotropy = multiple effects from one gene no tyrosine (little melanin) slow growth retardation blue eyes low adrenaline

20. Ch 4 1902 Archibald Garrod: One gene: one enzyme “Inborn errors of metabolism” PKU Albinism Alkaptonuria Tyrosinemia Black urine arthritis

21. 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) • p(aa AND a girl)?

22. If III-3 and II-1 matep (normal child) Product rule: p (affected boy)?

23. All people have harmful recessive alleles, small chance That 2 people with same rare alleles will mate Consanguinous marriage? Bedoin intermarriage

24. Autosomal dominant disordersAa and AA =affectedaa =unaffected • Tend to show up in every generation • 2 affected parents can have unaffected child • 2 unaffected parents cannot have an affected child

25. Dominant pedigree

26. Achondroplasia -1/20,000 births • Mutation in FGFR3 gene Chromosome 4 • Affects cartilage growth needed for bone lengthening • Affected individuals Aa  why not AA? • Most cases spontaneous (associated with increasing paternal age) what is the genotype of parents in this case?

27. P(III, 3 and III, 5 have a child of normal height) P ( II, 3 and III, 7 have a boy with achondroplasia)

28. Dihybrid cross – 2 genes Ch. 11 Mendel’s Law of Independent assortment- each allele for a trait is inherited independently of other alleles Seeds: G = yellow allele g = green allele gene? W = round allele w = wrinkled allele gene?

29. Parents = GGWW X ggwwphenotype?gametes? F1 genotype ? F1 phenotype ? F1 Gametes?

32. Note that each gene gives the 3:1 ratio of a monohybrid cross Yellow/green ratio = Round/ wrinkled =

33. Forked line method for phenotypes GgWw X GgWw

34. 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:

35. Probability Product rule- the 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)

36. A female lilac tabby X A male black siamese genotype: ww bb CC dd Aa X ww Bb CsCs Dd Aa p( black tabby cat) p (brown solid color cat) W = white B = black C = solid (not siamese) D = not dilute (full color) A = striped (tabby)

37. Trihybrid cross AaBbCc X AaBbCc p(A-B-cc) AabbCcDD X AaBbCcDd p(triply recessive)