GENETIC DATA ANALYSIS SUM RULE The combined probability

GENETIC DATA ANALYSIS

SUM RULE- The combined probability of two events that are mutually exclusive is the sum (add!) 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

GENETIC EXAMPLE: MONOHYBRID CROSS What is the probability that the F 2 offspring has the dominant phenotype (is either GG or Gg)? P: F 1: GG x gg = Gg Gg x Gg = F 2: 1/4 GG: 1/2 Gg: 1/4 gg 3 out of 4 dominant

1. MONOHYBRID COMPLETE DOMINANCE RATIO P: Homozygous Cross BB x BB = 4 BB : 0 Bb : 0 bb bb x bb= 0 BB : 0 Bb : 4 bb F 2: Heterozygous Cross Bb x Bb= 1 BB : 3 Bb: 1 bb 3 dominant : 1 recessive 3: 1 RATIO

2. DIHYBRID COMPLETE DOMINANCE RATIO Heterozygous= 9 : 3 : 1 Ratio � 9 dominant in both loci B � 3 dominant in first locus, recessive in second locus � 3 dominant in second locus, recessive in first locus � 1 recessive in both loci 9+3+3+1= 16 (possible offspring genotypes) 9 Smooth, Yellow (SY) 3 wrinkle, Yellow (s. Y) 3 Smooth, green (Sy) 1 wrinkle, green (sy)

3. TRIHYBRID, TETRAHYBRID, ETC Doing a punnet square would be excessive, so use the ratio! Monohybrid 3: 1 3/4 B 1/4 b Dihybrid 9: 3: 3: 1 3/4 × 3/4 = 9/16 BD 3/4 × 1/4 = 3/16 Bd 1/4 × 3/4 = 3/16 b. D 1/4 × 1/4 = 1/16 bd Trihybrid 27: 9: 9: 9: 3: 3: 3: 1 3/4 × 3/4 = 27/64 BDR 3/4 × 1/4 = 9/64 BDr 3/4 × 1/4 × 3/4 = 9/64 Bd. R 1/4 × 3/4 = 9/64 b. DR 3/4 × 1/4 = 3/64 Bdr 1/4 × 3/4 × 1/4 = 3/64 b. Dr 1/4 × 3/4 = 3/64 bd. R 1/4 × 1/4 = 1/64 bdr

PRODUCT RULE-The probability of several independent events is the product (multiply!) of the individual probabilities. Clue: look for “and” Q: You roll two dice. What’s the probability of getting a ‘ 2’ on the first one and a ‘ 5’ on the second one. A: 1/6 x 1/6 = 1/36

GENETIC EXAMPLE: P: AA bb CC DD ee ff x aa BB cc dd EE FF F 1: Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee Ff Q: What proportion of F 2 progeny will be AA bb Cc DD ee Ff ? A: 1/4 * 1/2 * 1/4 * 1/2 = 1/1024

CHI SQUARED ANALYSIS

CHI SQUARED ANALYSIS Determines if data is “good” FORMULA: X 2 = Σ (O – E)2 E O: observed values E: expected values

CHI SQUARED NULL HYPOTHESES Null Hypothesis: there is no significant difference between the expected and observed result 1. � If accepted, you are claiming that mating is random and so is segregation and independent assortment � P Value is greater than 0. 05 Reject Null Hypothesis: 2. • you are seeing that the deviation between observed and expected is very far apart…something non-random must be occurring…. • P Value is less than 0. 05

DEGREES OF FREEDOM Must determine the degrees of freedom Degrees of freedom: # of phenotypic possibilities in your cross minus 1

LET’S LOOK AT A FRUIT FLY CROSS x P: Black body, eyeless F 1: all wild

F 1 x F 1 5610 1896 1881 622

RESULTS ANALYSIS 1. Determine the cross that you were using Dihybrid Complete Dominance 2. What is the expected outcome of this cross? 9/16 wild type (yellow body, eyes) � 3/16 normal body eyeless (yellow body, no eyes) � 3/16 black body wild eyes (black body, no eyes) � 1/16 black body eyeless (black body, eyes) �

3. CONDUCT THE ANALYSIS: CREATE A PERFECT, “EXPECTED” 9: 3: 3: 1 RATIO A. Divide total individuals by 16 10009/16 = 626 B. Multiply by perfect ratio# 626*9= 5, 634 for 2 domani loci (wild type) Phenotype Observed Expected Wild 5, 610 5, 634 (=626*9) Eyeless 1, 881 1, 878 (=626*3) Black Body 1, 896 1, 878 (=626*3) Eyeless, Black Body 622 626 (=626*1) TOTAL 10, 009

4. Apply the Chi Squared formula to each phenotype: (5610 - 5630)2/ 5630 = 0. 07 (1881 - 1877)2/ 1877 = 0. 01 (1896 - 1877 )2/ 1877 = 0. 20 (622 - 626) 2/ 626 = 0. 02 2=. 30 5. Determine Degrees of Freedom = # of phenotypes – 1 4 -1= 3 6. Find the value closest to chi squared on Chi Squared Distribution Table, ex. 0. 35

CHI-SQUARE DISTRIBUTION TABLE Reject Null Hypothesis Accept Null Hypothesis Probability (p) Degrees of Freedom 0. 95 0. 90 0. 80 0. 70 0. 50 0. 30 0. 20 0. 10 0. 05 0. 01 0. 001 1 0. 004 0. 02 0. 06 0. 15 0. 46 1. 07 1. 64 2. 71 3. 84 6. 64 10. 83 2 0. 10 0. 21 0. 45 0. 71 1. 39 2. 41 3. 22 4. 60 5. 99 9. 21 13. 82 3 0. 35 0. 58 1. 01 1. 42 2. 37 3. 66 4. 64 6. 25 7. 82 11. 34 16. 27 4 0. 71 1. 06 1. 65 2. 20 3. 36 4. 88 5. 99 7. 78 9. 49 13. 38 18. 47 5 1. 14 1. 61 2. 34 3. 00 4. 35 6. 06 7. 29 9. 24 11. 07 15. 09 20. 52 6 1. 63 2. 20 3. 07 3. 83 5. 35 7. 23 8. 56 10. 64 12. 59 16. 81 22. 46 7 2. 17 2. 83 3. 82 4. 67 6. 35 8. 38 9. 80 12. 02 14. 07 18. 48 24. 32 8 2. 73 3. 49 4. 59 5. 53 7. 34 9. 52 11. 03 13. 36 15. 51 20. 09 26. 12 9 3. 32 4. 17 5. 38 6. 39 8. 34 10. 66 12. 24 14. 68 16. 92 21. 67 27. 88 10 3. 94 4. 86 6. 18 7. 27 9. 34 11. 78 13. 44 15. 99 18. 31 23. 21 29. 59

6. CONCLUDE CHI SQUARED ANALYSIS With degrees of freedom, my chi square value is , which gives me a p value of %, I therefore my null hypothesis “With 3 degrees of freedom, my chi squared value is 0. 35, which gives a P value of 95%, therefore I accept my null hypothesis. “ This means that 95% of the time when our observed data is this close to our expected data, this deviation is due to random chance. • P value is greater than 0. 05, therefore accept our null hypothesis •

EXCEPTIONS TO MENDELIAN RATIOS

EXCEPTIONS TO MENDELIAN RATIO 1. 2. 3. 4. 5. Lethal Alleles Epistasis Unusual sex linkage Sex influenced inheritance Genetic Anticipation

1. LETHAL ALLELES: MANX CATS

LETHAL ALLELES F 1: Mm F 1: 1 MM x Mm 2 Mm 1 mm F 2: 1 Lethal: 2 Manx: 1 Normal F 2 phenotypic ratio is 2: 1 instead of 3: 1

OTHER LETHAL MUTATIONS Achondroplasia (humans)- short limbed dwarfism Yellow body color (domestic mice) Curly wings (Drosophila flies)

2. EPISTASIS Genetic interaction between two (or more) loci One gene modifies the phenotypic effects of another gene Normal dihybrid ratio is altered from 9: 3: 3: 1 to 9: 3: 4 � 9 Wild � 3 Black *novel/new genotype � 4 Albino

Agouti: wild type

P: BB CC agouti x Bb Cc F 1: agouti bb cc albino Simple dominant phenotype? F 2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc albino agouti albino black F 2 Phen. ratio: 9 agouti : 3 black : 4 albino novel phenotype

Epistasis Normal dihybrid ratio is altered from 9: 3: 3: 1 to 9: 3: 4 C and B gene have an epistatic interaction

PRACTICE PROBLEM 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 F 1 offspring are black. F 1 offspring are crossed (Bb Ee x Bb Ee). Puppies appear in the ratio: 9/16 black; 3/16 chocolate; 4/16=1/4 yellow. What genotypes correspond to these three phenotypes? 9/16 B- E- 3/16 B- ee 3/16 bb E_ 1/16 eebb bb E- and bb ee

3. UNUSUAL SEX LINKAGE

4. ENVIRONMENT DEPENDENT ALLELES PHENOTYPES ARE NOT ALWAYS A DIRECT REFLECTION OF GENOTYPES 1. 2. Temperature-sensitive alleles- Siamese color pattern Nutritional effects: phenylketonuria

SIAMESE OR “ HIMALAYAN” • Different allele of the C locus that causes lighter color is Temperature sensitive. • Cooler the body area, the darker the color

5. GENETIC ANTICIPATION Causes later generations in a family to be more severely affected by a disease Increases the # of triplet repeats in the fragile area of the X chromosome through the generations Ex. Huntington disease Fragile-X syndrome Kennedy disease Myotonic muscular dystrophy

*Huntingtons Disease is a Dominant Disorder