2. Chi-square Statistical Analysis You and a sibling flip a coin to see who has to take out the trash. Your sibling grows skeptical of the legitimacy of your coin because “tails” always seems to win. You decide to test it out by flipping the coin 20 times. Poss. Outcomes Tails Heads Obs. 13 7 Exp. 10 (o-e) 2 /e.9 1.8  2 value

3. Chi-square Statistical Analysis Determines the probability that observed data could result from expected conditions 1.For each phenotype, calculate (Observed – Expected) 2 / (Expected) 2.Add up your figures. This is the chi-square value 3.Degree of freedom (df) = (# of phenotypes – 1) 4.Find the probability in the table.      (o-e) 2 e 

4. Fruit Fly Eye Color P:Red-eyed female x White-eyed male P:Red-eyed female x White-eyed male F 1 :100 Red-eyed flies F 1 :100 Red-eyed flies F 2 : 75 Red-eyed, 25 white-eyed F 2 : 75 Red-eyed, 25 white-eyed Write out Punnett squares for both crosses. 7525 100 w + = red w = white

5. Fruit Fly Eye Color P:Red-eyed female x White-eyed male P:Red-eyed female x White-eyed male F 1 :100 Red-eyed flies F 1 :100 Red-eyed flies F 2 : 75 Red-eyed, 25 white-eyed F 2 : 75 Red-eyed, 25 white-eyed 7525 25 males 50 females, 25 males

6. Chi-square Statistical Analysis Our F 1 cross produces 100 offspring If we assume eye color and gender are unlinked, w + = red w + w ♀ x w + w ♂ w = white Then we expect… But we got… ExpectedPhenotype 37.5 Red ♀ 12.5 White ♀ 37.5 Red ♂ 12.5 White ♂ Observed 50 0 25 Calc. 4.2 12.5 4.2 12.5 33  2 value Only a 0.000027% probability Therefore, eye color and gender are linked

7. X-Linkage The gene with the white-eyed mutation is on the X-chromosome Homozygous Dominant X w+ X w+ Hemizygous Recessive X w Y

8. Practice Problems 1. Colorblindness is due to a recessive x-linked allele. What are the chances of a normal male and a carrier female having a colorblind son as their first child? 2. Why are males more likely than females to have recessive x-linked traits?

9. Practice Problems In sesame plants, the one-pod condition, (A) is dominant to the 3- pod condition (a), and normal leaf (B) is dominant to wrinkled leaf (b). An AaBb plant is testcrossed to produce the following offspring: 11one-pod, normal 12one-pod, wrinkled 7three-pod, normal 10three-pod, wrinkled Is this data likely if this is a case of independent assortment? What about… 110 120 70 100? What about… 18 2 3 17?

10. A B a b A B a b A B a b A B a b A B a b A B a b A B a b Aa Bb a B A b A B a b A B a b A B a b Aa Bb Gametes AB ab ParentalABab Recomb. Ab aB AaBb AaBb Aa Bb ORAbaB

11. Earlobes & Toes (Independent Assortment) P:FFTT x fftt all FfTt (free earloes and 2 nd toe longer) Testcross of F 1 individuals FfTt x fftt: ¼ Free, 2 nd toe ¼ Free, great toe ¼ attached, 2 nd toe ¼ attached, great toe

12. Earlobes & Toes (Linked) P:FFTT x fftt F 1 :FfTt F 1 Testcross FfTt x fftt F T f t f t f t F T f t f t f t ½ Att, Grt Toe ½ Free, 2 nd Toe Independent Assortment ¼ Free, 2 nd ¼ Free, Grt ¼ Att, 2 nd ¼ Att, Grt

13. Earlobes & Toes (Linked + Recombination) F T f t f t f t F T f t f t f t T f F t F t T f f t f t Parental Offspring FfTt (free, 2 nd) fftt (attached, great) Recombinant Offspring Fftt (free, great) ffTt (attached, 2 nd ) Recombination frequency = (# of recomb. offspring) (total offspring) 181723

14. Genetic Recombination Genes farther apart are more likely to cross over Greater distance  greater RF Enables us to map genes on a chromosome

16. Genetic Recombination b / c= 9% (9 map units or centimorgans) v / c= 9.5% (9.5 mu) b / v= 17% (17 mu) According to our map, b/v = 18.5 mu (not 17) Double crossover leads to lower than expected recombinant frequencies b c 9mu v v OR 9.5mu 18.5mu

17. Review Construct a linkage map given the following information: a-b  7.2 % b-c  3.5 % a-c  3.9 % b-d  9.6 % c-d  6.3 % How many recombinant organisms would you expect to find out of 1000 offspring from a cross between AaDd and aadd organisms?