Course Outline IB 201 Review Question A phenotype ratio of 9 3 3 1 in the offspring of a mating between two individuals that are heterozygous for two traits occurs when A the genes reside on the same chromosome B each gene contains two mutations C the gene pairs assort independently during meiosis D only recessive traits are scored E none of the above Genetic Data Analysis Probability Statistics Deviations from Mendelism Epistasis Unusual Modes of Inheritance Chromosomal Inheritance Chromosomal Abnormalities Sex Determination Mapping Gene and Genome Mapping Traits Affected by Genes Environment Quantitative Traits Genes in Populations Genetic Mechanisms of Evolution Population Genetics of Disease and Disease resistance Sum Rule Genetic Data Analysis II Some simple rules of probability The combined probability of two events that are mutually exclusive is the sum of the individual probabilities 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 3 1 Genetic Example Monohybrid Cross P F1 GG x gg Gg Gg x Gg F2 1 4 GG 1 2 Gg 1 4 gg Genetic Example 2 Dihybrid Cross P GG ww x gg WW F1 Gg Ww Gg Ww x Gg Ww F2 9 16 G W 3 16 G ww 3 16 ggW 1 16 ggww What is the probability that the F2 offspring has the dominant phenotype is either GG or Gg What is the probability that an F2 offspring will have the dominant phenotype G ww or ggW for only one of the two traits 1 4 GG 1 2 Gg 3 4 G 3 16 G ww 3 16 ggW 6 16 3 8 Product Rule The probability of several independent events is the product of the individual probabilities Two events are independent if the occurrence of the first event has no effect on the probability of the second event Clue look for and Q You roll two dice What s the probability of getting a two on the first one and a five on the second one A 1 6 1 6 1 36 Genetic example of product rule P AA bb CC DD ee ff x aa BB cc dd EE FF F1 Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee Ff Q 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 1024 2 Deviations from Mendelism Manx Cats Lethal Alleles Epistasis Unusual sex linkage Sex influenced inheritance Genetic Anticipation Lethal alleles F1 Mm x Other lethal mutations Achondroplasia humans Mm F1 1 MM 2 Mm F2 1 Lethal 2 Manx 1 mm 1 Normal Yellow body color domestic mice Curly wings Drosophila F2 phenotypic ratio 2 1 instead of 3 1 3 Epistasis Agouti wild type Genetic interaction between two or more loci One gene modifies the phenotypic effects of another gene P F1 BB CC agouti x Bb Cc agouti bb cc albino Simple dominant phenotype Epistasis Normal dihybrid ratio is altered from 9 3 3 1 to 9 3 4 F2 9 16 B C 3 16 bb C 3 16 B cc 1 16 bb cc albino agouti albino black F2 Phen ratio 9 agouti 3 black 4 albino C and B gene have an epistatic interaction novel phenotype 4 Biochemical model Epistasis C enzyme present Colorless precursor Locus 1 BB Bb agouti agouti bb black Locus 2 CC Cc cc no effect no effect albino B enzyme present Yes CC or Cc some melanin produced Yes BB or Bb agouti No cc no melanin produced No bb black CC or Cc tyrosinase is produced involved in production of melanin BB or Bb controls distribution of the pigment Figure 10 18b Figure 10 18c Crosses between pure lines produce novel colors Model to explain 9 3 3 1 pattern observed above Two genes interact to produce pepper color Parental generation rrYY X Yellow yellow Genotype RRyy R Y yy Brown brown pepper Colorgreen Explanation of color chlorophyll Red rrY R pigment Red red pigment no chlorophyll Yellow pigment no chlorophyll yellow yellow Yellow orange F1 generation Codominance R Y R yy Brown rryy Y No chlorophyll Green Red pigment chlorophyll R red pigment Yellow pigment chlorophyll Red Self fertilization F2 generation R YRed 9 16 rrYYellow 3 16 R yy rryy Brown 3 16 Green 1 16 Gene 1 Gene 2 R Red Y Absence of green no chlorophyll r Yellow y Presence of green chlorophyll R or r Y or y 5 Practice Problem Other kinds of epistasis 9 16 A B 3 16 A bb 3 16 aaB 1 16 aabb In Labrador retrievers coat color is controlled by two loci each with two alleles B b and E e respectively When pure breeding Black labs with genotype BB EE are crossed with pure breeding yellow labs of genotype bb ee the resulting F1 offspring are black F1 offspring are crossed Bb Ee x Bb Ee Puppies appear in the ratio 9 16 black 3 16 chocolate chocolate 9 16 B EB E 3 16 B ee B ee 4 16 1 4 yellow 3 16 bb E 1 16 eebb bb E and bb ee What genotypes correspond to these three phenotypes Practice Problem In the summer squash Cucurbita pepo fruit shape is determined by two genes Two different true breeding spherical types were crossed The F1 s were all disk and the F2 s segregated 35 disk 25 spherical and 4 long 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 of inheritance will give three F2 phenotypes Genetic Model Incomplete codominance Epistasis Expected F2 ratio 1 2 1 Variation on 9 3 3 1 Hint usually given numbers not fractions 27 agouti 12 albino 9 black 28 agouti 11 albino 4 black Practice Problem cont In the summer squash Cucurbita pepo spherical fruit is recessive to disk True breeding spherical types from different geographic regions were crossed The F1 s were disk and the F2 s segregated 35 disk 25 spherical and 4 long Explain these results Are the phenotypic ratios closer 1 2 1 or to a variant 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 64 64 16 4 9 4 36 6 4 24 1 4 4 Phenotypic ratio close to 9 6 1 6 Practice Problem cont In the summer squash Cucurbita pepo spherical fruit is recessive to disk True breeding spherical types from different geographic regions were crossed The F1 s were disk and the F2 s segregated 35 disk 25 spherical and 4 long Explain these results If phenotypic ratios are close to 9 6 1 then what are the genotypes associated with each phenotype 35 disk 9 16 A B 25 spherical 3 16 A bb 3 16 aa B Sex Linkage mammals flies XAXa Diploid Adults Gametes Xa XA XA Y XA …