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 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 plant)

4. Mendel’s conclusions 1.Genes are physical units –2 alleles for each gene –1 allele inherited from each parent

5. Genes and alleles of Pisum sativum GeneAlleles Pea color? Flower colorwhite, purple Pod shapeconstricted, inflated Pea surface? Stem heighttall, dwarf

6. 2. Principle of Dominance -One allele dominant, the other recessive -The dominant allele is expressed in the phenotype 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 ONE allele per gene –random segregation of alleles 50/50 PP pp What is the phenotype of all offspring in F1 generation? P generation

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

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

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

11. Results of Mendel’s monohybrid crosses Parental StrainsF2 progenyRatio Tall X dwarf787 tall, 277 dwarf Round seeds X wrinkled5474 round, 1850 wrinkled Yellow seeds X green6022 yellow, 2001 green Violet flowers X white705 violet, 224 white Inflated pods X constricted882 inflated, 299 constricted Green pods X yellow428 green, 152 yellow Axial flowers X terminal651 axial, 207 terminal gene = ? alleles = ?

12. In mice, black fur is dominant over white fur Cross heterozygous mouse with homozygous recessive mouse B = black b = white

13. 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  A mouse was test crossed and 7 offspring black, 2 white. What is mouse’s genotype?

14. All possible one gene crosses Parentsratio offspring

15. Pedigree Analysis Ch 11

16. 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

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

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

19. missing phenylalanine hydroxylase enzyme If plasma phe level is too high, phe is converted into a phenylpyruvate toxic to brain tissue

20. 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)

21. 1902 Archibald Garrod: One gene: one enzyme “Inborn errors of metabolism” PKU Albinism Alkaptonuria Tyrosinemia Black urine arthritis Ch 4

22. Fill in genotypes. If II,1 and II, 4 mate, what is the chance of offspring having PKU? II, 1 XII, 4p(aa) p(aa AND a girl)? How do we know this is autosomal recessive?

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

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

25. 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

26. Dominant pedigree

27. Achondroplasia -1/20,000 births Mutation in FGFR3 gene Chromosome 4Mutation in FGFR3 gene 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?

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

29. Mutation in dog FGF4 gene