1IB 201: Review QuestionA phenotype ratio of 9:3:3:1 in theoffspring of a mating between twoindividuals that are heterozygousfor two traits occurs when:A. the genes reside on the samechromosomeB. each gene contains twomutationsC. the gene pairs assortindependently during meiosisD. only recessive traits are scoredE. none of the aboveCourse OutlineGenetic Data Analysis:Probability & StatisticsDeviations from Mendelism: Epistasis; Unusual Modes of InheritanceChromosomal Inheritance:Chromosomal Abnormalities; Sex DeterminationMapping:Gene and Genome MappingTraits Affected by Genes & EnvironmentQuantitative TraitsGenes in PopulationsGenetic Mechanisms of Evolution; Population Genetics of Diseaseand Disease resistanceGenetic Data Analysis IISome simple rules of probabilitySum RuleThe combined probability of twoevents that are mutually exclusive isthe sum of the individualprobabilities. Clue: look for “or”Q: What’s the probability of rolling a ‘five’or a ‘six’ on one six-sided die?A: 1/6 + 1/6 = 1/32Genetic Example: Monohybrid CrossP: GG x ggF1: Gg Gg x Gg ==>F2: 1/4 GG: 1/2 Gg: 1/4 ggWhat is the probability that the F2offspring has the dominant phenotype (iseither GG or Gg)?1/4 GG + 1/2 Gg = 3/4 G-Genetic Example 2: Dihybrid CrossP: GG ww x gg WWF1: Gg Ww Gg Ww x Gg Ww ==>F2: 9/16 G-W- 3/16 G-ww 3/16 ggW- 1/16 ggwwWhat is the probability that an F2 offspring will have thedominant phenotype (G-ww or ggW-) for only one of thetwo traits?3/16 G-ww + 3/16 ggW- = 6/16=3/8Product RuleThe probability of severalindependent events is the productof the individual probabilities.Two events are independent if theoccurrence of the first event has noeffect on the probability of the secondevent. Clue: look for “and”.Q: You roll two dice. What’s theprobability of getting a ‘two’ on thefirst one and a ‘five’ on the secondone.A: 1/6 * 1/6 = 1/36Genetic example of product rule P: AA bb CC DD ee ff x aa BB cc dd EE FFF1: Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee FfQ: What proportion of F2 progeny will be AA bb Cc DD ee Ff ?A: 1/4 * 1/4 * 1/2 * 1/4 * 1/4 * 1/2 = 1/10243Deviations from MendelismLethal AllelesEpistasisUnusual sex linkageSex influenced inheritanceGenetic AnticipationManx CatsLethal allelesF1: Mm x MmF1: 1 MM 2 Mm 1 mmF2: 1 Lethal: 2 Manx: 1 NormalF2 phenotypic ratio: 2:1 instead of 3:1Other lethal mutationsAchondroplasia (humans)Yellow body color (domestic mice)Curly wings (Drosophila)4EpistasisGenetic interaction between two (or more)loci.One gene modifies the phenotypic effectsof another gene.Agouti:wild typeF2 Phen. ratio: 9 agouti : 3 black : 4 albino agouti F1: agoutiP: agouti albinoF2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb ccSimple dominant phenotype?novel phenotype BB CC x bb ccblackalbinoalbinoBb CcEpistasisNormaldihybrid ratiois altered from9:3:3:1 to9:3:4C and Bgene havean epistaticinteraction5Epistasis Locus 1 Locus 2BB Bb bb CC Cc ccagouti agouti black no effect no effect albinoBiochemical modelCC or Cc: tyrosinase is produced (involved inproduction of melanin)BB or Bb: controls distribution of the pigmentC enzyme present?B enzyme present?Colorless precursor ->Yes: CC or Cc some melaninproducedYes: BB or Bb agoutiNo: cc no melaninproducedNo: bb blackYellow BrownXF2 generationF1 generationParentalgenerationRed9/16Yellow3/16Brown3/16Green1/16RedSelf-fertilizationFigure 10.18bCrosses between pure lines produce novel colors. R-Y- rrY- R-yy rryy R-Y- rrYY RRyyCodominance?Genotype Color Explanation of colorR-Y-rrY-R-yyRedYellowBrownRed pigment + no chlorophyllYellow pigment + no chlorophyllRed pigment + chlorophyllrryy Green Yellow pigment + chlorophyllGene 1 Gene 2R = Redr = Yellow(-) = R or rY = Absence of green (no chlorophyll)y = Presence of green (+ chlorophyll)(-) = Y or yFigure 10.18cModel to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color. yellow ----------------> green -----------------> brown pepper yy? chlorophyll R? red pigmentyellow -------------------> yellow ------------------> orange Y? No chlorophyll R? red pigment6Practice ProblemIn Labrador retrievers, coat color is controlled by twoloci each with two alleles B,b and E,e respectively.When pure breeding Black labs with genotype BB EEare crossed with pure breeding yellow labs ofgenotype bb ee the resulting F1 offspring are black.F1 offspring are crossed (Bb Ee x Bb Ee). Puppiesappear in the ratio:9/16 black; 9/16 black; 3/16 chocolate3/16 chocolate;; 4/16=1/4 yellow.What genotypes correspond to these threephenotypes?9/16 B- E- 3/16 B- ee 3/16 bb E- 1/16 eebbB- E- B- ee bb E- and bb eeOther kinds of epistasis1/16 aabb9/16 A-B-3/16 A-bb 3/16 aaB-Hint: usually given numbers, not fractions 27 agouti; 12 albino; 9 black 28 agouti; 11 albino; 4 blackPractice ProblemIn the summer squash (Cucurbita pepo) fruit shape isdetermined by two genes. Two different true-breedingspherical types were crossed. The F1's were all disk,and the F2's segregated 35 disk, 25 spherical and 4long. Explain these results.What’s the first step? Notice novel phenotype: disk, long. What’s the next step?Notice there are three F2 phenotypes. What kind ofinheritance will give three F2 phenotypes?Genetic Model?Incomplete, codominanceEpistasisExpected F2 ratio?1:2:1Variation on 9:3:3:1Practice Problem, cont.In the summer squash (Cucurbita pepo) spherical fruit isrecessive to disk, True-breeding spherical types from differentgeographic regions were crossed. The F1's were disk, and theF2's segregated 35 disk, 25 spherical and 4 long. Explainthese results.Are the phenotypic ratios closer 1:2:1 or to avariant of 9:3:3:1 ?If phenotypic ratios closer to a variant of 9:3:3:1,then what variant is it?Total # of individuals = 35 + 25 + 4 = 6464/16 = 49*4 = 366*4 = 241*4 = 4Phenotypic ratio close to 9:6:17Practice Problem, cont.In the summer squash (Cucurbita pepo) spherical fruit isrecessive to disk, True-breeding spherical types from differentgeographic regions were crossed. The F1's were disk, and theF2's segregated 35 disk, 25 spherical and 4 long. Explainthese results.If phenotypic ratios are close to 9:6:1, then what arethe genotypes associated with each phenotype?35 disk 25 spherical 4