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

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

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

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

GARDEN PEA POLLENATION

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

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

MENDELIAN ANALYSIS: 1 GENE Fcrosses are reciprocal

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

MENDELIAN ANALYSIS: 1 GENE

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

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

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

MENDELIAN ANALYSIS: 1 GENE

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

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

MENDELIAN ANALYSIS: TEST CROSS

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

MENDELIAN ANALYSIS: 2 GENES

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

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

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

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 !

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 

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

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 

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

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

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

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

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

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...

STATISTICS: CHI-SQUARE ANALYSIS F go to table 2-2 on page 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%)

STATISTICS: CHI-SQUARE ANALYSIS

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)

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