1. LECTURE 02: PATTERNS OF INHERITANCE I Fannouncements Fkey concepts FMendelian analysis: E1 gene E2 genes E n genes Fstatistics Echi-square

2. CHAPTER 2: KEY CONCEPTS ( most of which I assume you already know) Fexistence of genes inferred by observing standard progeny ratios derived from controlled matings Fdiscrete phenotypes can have single gene basis Fin a diploid cell, each gene is represented twice, one allele on each chromosome pair

3. CHAPTER 2: MORE KEY CONCEPTS ( most of which I assume you already know) Finheritance patterns based on behavior of chromosomes during meiosis Falleles of a gene segregate into different gametes Fgene pairs on different chromosomes assort independently Fgenes on sex chromosomes show unique inheritance patterns

4. MENDELIAN ANALYSIS Fwhat organism to study? FMendel used garden pea ( Pisum sativum ) Favailable cheap from local merchants Feasy to grow, small space, fast generation time Flots of phenotypic variation Fcould be manipulated for controlled pollenation: self and outcross

5. GARDEN PEA POLLENATION

6. MENDELIAN ANALYSIS Fwhat characteristics to study study? FMendel observed 7 different characters Fone at a time

7. MENDELIAN ANALYSIS: 1 GENE F e.g. : purple vs white flowers Fmust start with true-breeding or pure-breeding lines Fhow do you know? selfing  1 phenotype Fnomenclature: Fdistinguish genotype & phenotype Findividuals are homozygotes or heterozygotes Fgenerations designated P, F 1, F 2... Fdetermine genotypic & phenotypic ratios

8. MENDELIAN ANALYSIS: 1 GENE Fcrosses are reciprocal

9. MENDELIAN ANALYSIS: 1 GENE PF1F2PF1F2 purple  x white   purple  x purple   ¾ purple + ¼ white white  x purple   purple  x purple   ¾ purple + ¼ white Fnot blending because we see a return of white

10. MENDELIAN ANALYSIS: 1 GENE

11. 1.particulate hereditary determinants (genes) 2.alleles  different phenotypes; dominance 3.alleles of a gene segregate into different gametes 4.gametes receive 1 allele with equal probability 5.union of gametes at fertilization is random

12. MENDELIAN ANALYSIS: 1 GENE random union alleles segregate 1 allele, = probability particulate genes alleles  phenotypes

13. MENDELIAN ANALYSIS: 1 GENE P gametes F 1 gametes F 2 ratio purple white P/P  x p/p  P  p all purple P/p  x P/p  ½ P + ½ p  ½ P + ½ p ¾ purple ¼ white ¼ P/P + ½ P/p + ¼ p/p 1 : 2 : 1

14. MENDELIAN ANALYSIS: 1 GENE

15. MENDELIAN ANALYSIS: CROSSES 1 gene: e.g., F 1 yellow heterozygotes Y/y x Y/y  monohybrid cross 1 gene: e.g., yellow unknown x green homozygote Y/ ? x y/y  test cross

16. MENDELIAN ANALYSIS: CROSSES 1 gene: e.g., yellow heterozygote x green homozygote Y/y x y/y  ½ Y/y & ½ y/y or yellow homozygote x green homozygote Y/Y x y/y  all Y/y 1 gene: e.g., F 1 yellow heterozygotes Y/y x Y/y  monohybrid cross

17. MENDELIAN ANALYSIS: TEST CROSS

18. MENDELIAN ANALYSIS: CROSSES 2 genes: e.g., F 1 round yellow heterozygotes R/r Y/y x R/r Y/y  dihybrid cross 1 gene: e.g., F 1 yellow heterozygotes Y/y x Y/y  monohybrid cross 1 gene: e.g., yellow unknown x green homozygote Y/ ? x y/y  test cross

19. MENDELIAN ANALYSIS: 2 GENES

22. 1.particulate hereditary determinants (genes) 2.alleles  different phenotypes; dominance 3.alleles of a gene segregate into different gametes 4.gametes receive 1 allele with equal probability 5.union of gametes at fertilization is random 6.alleles of different genes* assort independently into different gametes * gene pairs on different chromosomes

23. MENDELIAN ANALYSIS: 2 LAWS 1.particulate hereditary determinants (genes) 2.alleles  different phenotypes; dominance 3.alleles of a gene segregate into different gametes 4.gametes receive 1 allele with equal probability 5.union of gametes at fertilization is random 6.alleles of different genes* assort independently into different gametes * = gene pairs on different chromosomes

24. MENDELIAN ANALYSIS: 1  n GENES F3 methods of working out expected outcomes of controlled breeding experiments: 1. Punnet square 2. tree method (long) – genotypes 3. tree method (short) – phenotypes

25. MENDELIAN ANALYSIS: PUNNET SQUARE Fall pairings of  and  gametes F= probabilities of all pairings Fsome pairings occur >1 F  different P s for different genotypes & phenotypes F1 gene, 2 x 2 = 4 cells F2 genes, 4 x 4 = 16 cells F3 genes, 8 x 8 = 64 cells... Ftoo much work !

26. MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a 1/4 A/A 1/2 A/a 1/4 a/a GENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a 

27. MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a 1/4 A/A 1/2 A/a 1/4 a/a GENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a  PHENOTYPE 3/4 A 1/4 a

28. MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a ; B, b... 1/4 B/B 1/4 A/A 1/2 B/b 1/4 b/b 1/4 B/B 1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B 1/4 a/a 1/2 B/b 1/4 b/b GENOTYPE 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b 

29. MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a ; B, b... 1/4 B/B 1/4 A/A 1/2 B/b 1/4 b/b 1/4 B/B 1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B 1/4 a/a 1/2 B/b 1/4 b/b GENOTYPE 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b  PHENOTYPE 9/16 AB 3/16 Ab 3/16 aB 1/16 ab

30. MENDELIAN ANALYSIS: SHORT TREE 2 genes, alleles A, a ; B, b... ¾ B ¾ A ¼ b ¾ B ¼ a ¼ b much easier... can extend to >2 genes  PHENOTYPE 9/16 AB 3/16 Ab 1/16 ab

31. formulae for n genes, dominance, n hybrid crosses … MENDELIAN ANALYSIS: 1  n GENES 1. # possible different types of gametes = 2 n 2. # possible different genotypes of progeny = 3 n 3. frequency of least common genotype = (1/4) n 4. # possible different phenotypes of progeny = 2 n

32. STUFF TO THINK ABOUT FMendel’s choice of characters was critical Fdid he chose 7 characters that appeared to reside on different chromosomes by chance ? Fwhat would happen if some were linked? Fall exhibited dominant/recessive relationships

33. GENETIC RATIOS AND RULES Fsum rule: the probability of either of two mutually exclusive events occurring is the sum of the probabilities of the individual events... OR A/a  x A/a  ½ A + ½ a  ½ A + ½ a P( A/a ) = ¼ + ¼ = ½ Fproduct rule: the probability of independent events occurring together is the product of the probabilities of the individual events... AND A/a  x A/a  ½ A + ½ a  ½ A + ½ a P( a/a ) = ½ x ½ = ¼

34. STATISTICS: CHI-SQUARE ANALYSIS F observe 4 phenotypes in roughly 9:3:3:1 ratio F generate a hypothesis F what does it mean ? Ftest hypothesis...

35. STATISTICS: CHI-SQUARE ANALYSIS F go to table 2-2 on page 41... F 4 groups... 3 degrees of freedom (df = 3) F chose the probability that you are willing to accept for mistakenly rejecting a hypothesis that is in fact true... (  = 0.05 or 5%)

36. STATISTICS: CHI-SQUARE ANALYSIS

37. F express this as: 0.95 > P (  2 c = 0.511) > 0.90 F the data do not deviate significantly from a 9:3:3:1 ratio F we do not reject our H 0 (alternatively we could reject ) F “seed color and seed shape fit an unlinked, 2 gene classic Mendelian model with complete dominance” F the probability of deviation by chance from this model lies between 90 and 95% ( i.e., the biological explanation is supported by data)

38. READING & PRACTICE PROBLEMS F chapter 2... not done yet but, F summary F key terms F solved problems 1, 2 F questions 1 - 4, 9, 12, 13