1. MENDELIAN GENETICS 1

2. GREGOR JOHANN MENDEL  Austrian monk  Studied the inheritance of traits in pea plants  Developed the laws of inheritance  Mendel's work was not recognized until the turn of the 20th century  Between 1856 and 1863, Mendel cultivated and tested some 28,000 pea plants  He found that the plants' offspring retained traits of the parents  Called the “Father of Genetics" 2

3. SITE OF MENDEL’S EXPERIMENTAL GARDEN IN THE CZECH REPUBLIC NOTE THE RATIO OF FLOWER COLORS… 3

4. GENETIC TERMINOLOGY  Trait - any characteristic that can be passed from parent to offspring  Heredity - passing of traits from parent to offspring  Genetics - study of heredity 4

5. MENDEL’S PEA PLANT EXPERIMENTS 5

6. MENDEL’S EXPERIMENTAL METHODS  Mendel hand-pollinated flowers using a paintbrush  He could snip the stamens to prevent self-pollination  Covered each flower with a cloth bag  He traced 8 traits through the several generations  Ultimately he produced pure strains for each  6

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9. Once the pure strains were made; Mendel then began systematically crossing plants, this time however, he let the plants self-pollinate. Why? He observed the offspring produced for each trait, in each generation, for any patterns that appeared. Mendel’s questions:  If traits are known, can the inheritance of the traits in offspring be predicted?  What is the ‘mechanism’ for inheritance? 9

10. GENERATION “GAP”  Parental P 1 Generation = the parental generation in a breeding experiment, the pure strain  F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation)  F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) 10

11. 11 WHY no white? What happened to the white trait? White trait reappeared in this generation, so there must be some ‘factor’ in the purple flower that masked it

12. MENDEL’S CONCLUSIONS  Dominant – trait that was expressed if present; represented by a capital letter (R)  Recessive – trait that is expresses only if the dominant is not present; represented by a lowercase letter (r)  Factors: sequence of DNA that codes for a trait, today we know those are the GENES  Alleles - two forms of a gene (dominant & recessive) Ex: Trait=Flower Color Alleles=Purple/White Alleles=Purple/White Regardless of the trait, results were the same and could be predicted in future generations…Mendel developed his first law of genetics; the Law of Dominance 12

13. LAW OF DOMINANCE 13 In a cross of parents(P) that are pure for contrasting traits, only one form of the trait will appear in the next generation(F 1 ). All the offspring will be heterozygous and express only the dominant trait. RR x rr yields all Rr (round seeds)

14. EIGHT PEA PLANT TRAITS GENES ALLELES GENES ALLELES  Seed shape --- Round (R) or Wrinkled (r)  Seed Color ---- Yellow (Y) or Green ( y )  Pod Shape --- Smooth (S) or wrinkled ( s )  Pod Color --- Green (G) or Yellow (g)  Seed Coat Color ---Gray (G) or White (g)  Flower position---Axial (A) or Terminal (a)  Plant Height --- Tall (T) or Short (t)  Flower Color --- Purple (P) or white ( p ) 14

15. 15 RRrr Rr RR Rr x Rr Rr rrRr

16. PREDICTABLE GENERATIONS 16 Cross 2 Pure Plants TT x tt Results in all Hybrids Tt Cross 2 Hybrids get 3 Tall & 1 Short TT,Tt,Tt,tt

17. 17 Gg GG gg P generation F1F1 GG Gg gg F2F2 All Gg, yellow ¾ yellow ¼ GG, 2/4 Gg ¼ green ¼ gg

18. GENOTYPES  Genotype - allele combination for a trait  Homozygous – allele combination involving two of the same alleles; can be either  Homozygous dominant: RR  Homozygous recessive: rr  Heterozygous – allele combination of one dominant & one recessive allele  Heterozygous – allele combination of one dominant & one recessive allele  Heterozygous: Rr 18

19.  Phenotype - the physical features resulting from the genotype; am I… Phenotype also refers to your chemical features as well; enzymes and functional proteins are also determined by genes and allele combinations 19 PURPLE or, am I WHITE

20. GENOTYPE & PHENOTYPE IN SEED COLOR 20 Genotype of alleles: R = yellow seed r = green seed In the F 2 The possible combinations are: GenotypesRR Rrrr PhenotypesYELLOW YELLOW GREEN Phenotypes YELLOW YELLOW GREEN BUNNY RABBIT

21. PUNNETT SQUARES  Diagram that is used to predict an outcome of a particular cross or breeding experiment  Illustrate Mendel’s second law Law of Segregation 21

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23. MONOHYBRID CROSSES CROSSES INVOLVING 1 GENE: 2 ALLELES 23

24. P 1 MONOHYBRID CROSS  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Round seeds x Wrinkled seeds  RR x rr 24 R R rr Rr Genotype:Rr Ratio: 1:1 or 100% Genotype: Rr Ratio: 1:1 or 100% PhenotypeRound Ratio: 1:1 or 100% Phenotype: Round Ratio: 1:1 or 100%

25. F 1 MONOHYBRID CROSS  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Round seeds x Round seeds Rr x Rr 25 R r rR RR rrRr Genotype:RR, Rr, rr Genotype: RR, Rr, rr Ratio:1:2:1 Ratio: 1:2:1 Phenotypes Phenotypes: Round & wrinkled Ratio: 3:1

26. P 1 MONOHYBRID CROSS REVIEW  Homozygous dominant x Homozygous recessive  Offspring all Heterozygous (hybrids)  Offspring called F 1 generation  Genotypic & Phenotypic ratio is ALL ALIKE 26

27. F 1 MONOHYBRID CROSS REVIEW  Heterozygous x heterozygous  Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr  Offspring called F 2 generation  Genotypic ratio is 1:2:1  Phenotypic Ratio is 3:1 27

28. RESULTS OF MONOHYBRID CROSSES  Inheritable factors or genes are responsible for all heritable characteristics  Phenotype is based on Genotype  Each trait is based on two genes, one from the mother and the other from the father  True-breeding individuals are homozygous ( both alleles) are the same 28

29.  Alleles for different traits are distributed to sex cells (& offspring) independently of one another.  This law can be illustrated using dihybrid crosses. 29

30. DIHYBRID CROSS  A breeding experiment that tracks the inheritance of two traits.  Mendel’s “Law of Independent Assortment”  a. Each pair of alleles segregates independently during gamete formation  b. Formula: 2 n (n = # of heterozygotes) 30

31. QUESTION: HOW MANY GAMETES WILL BE PRODUCED FOR THE FOLLOWING ALLELE ARRANGEMENTS?  Remember: 2 n (n = # of heterozygotes)  1.RrYy  2.AaBbCCDd  3.MmNnOoPPQQRrssTtQq copyright cmassengale 31

32. ANSWER: 32 1. RrYy: 2 n = 2 2 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2 n = 2 3 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2 n = 2 6 = 64 gametes

33. DIHYBRID CROSS  Traits: Seed shape & Seed color  Alleles:  Alleles: R round Y yellow r wrinkled y green  To help determine allele combinations use the math order of operations FOIL First-Outside-Inside-Last 33 RrYy x RrYy RY Ry rY ry All possible allele combinations

34. DIHYBRID CROSS 34RYRyrYry RYRy rY ry Rryy RRYy RrYY RrYy RRYy Rryy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy

35. DIHYBRID CROSS 35 RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Round/Yellow: 9 RRYY 1 RRYy 2 RrYY 2 RrYy 4 Round/green: 3 RRyy 1 Rryy 2 wrinkled/Yellow: 3 rrYY 1 rrYy 2 wrinkled/green: 1 Rryy 1 9:3:3:1 phenotypic ratio RYRyrYryRY Ry rY ry

36. DIHYBRID CROSS 36 Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1

37. GENETIC PRACTICE PROBLEMS copyright cmassengale 37

38. BREED THE P 1 GENERATION  tall (TT) x dwarf (tt) pea plants copyright cmassengale 38 T T tt

39. SOLUTION: copyright cmassengale 39 T T tt Tt All Tt = tall (heterozygous tall) produces the F 1 generation tall (TT) vs. dwarf (tt) pea plants

40. BREED THE F 1 GENERATION  tall (Tt) vs. tall (Tt) pea plants copyright cmassengale 40 T t Tt

41. SOLUTION: copyright cmassengale 41 TT Tt tt T t Tt produces the F 2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype 3:1 phenotype tall (Tt) x tall (Tt) pea plants