210 likes | 384 Views
Chromosomal Inheritance. 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. c 2 value.
E N D
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. c2 value
S c2 = (o-e)2 e Chi-square Statistical Analysis • Determines the probability that the observed data could result from expected conditions • For each phenotype, calculate (Observed – Expected)2 / (Expected) • Add up your figures. This is the chi-square value • Degree of freedom (df) = (# of phenotypes – 1) • Find the probability in the table.
Fruit Fly Eye Color • P: Red-eyed female x White-eyed male • F1: 100 Red-eyed flies • F2: 75 Red-eyed, 25 white-eyed Write out Punnett squares for both crosses. w+ = red w = white 100 75 25
Fruit Fly Eye Color • P: Red-eyed female x White-eyed male • F1: 100 Red-eyed flies • F2: 75 Red-eyed, 25 white-eyed 50 females, 25 males 25 males 75 25
Chi-square Statistical Analysis • Our F1 cross produces 100 offspring • If we assume eye color and gender are unlinked, w+ = redw+w♀x w+w♂ w = white • Then we expect… • But we got… • Only a 0.000027% probability • Therefore, eye color and gender are linked c2 value
w+ = wild type (red) w = white eyes X-Linkage (eye color) • The gene with the white-eyed mutation is on the X-chromosome Homozygous Dominant Xw+Xw+ Hemizygous Recessive XwY
Practice Problems • 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? • Why are males more likely than females to have recessive x-linked traits?
Genetics 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: 11 one-pod, normal 12 one-pod, wrinkled 7 three-pod, normal 10three-pod, wrinkled Is this data likely if this is a case of independent assortment (unliked genes)? • What about… • 110 • 120 • 70 • 100? • What about… • 18 • 2 • 3 • 17?
A a A a A a A a B b B b B b OR B b A a A a A a b B b B b B A a A a A a A a B b B b B b B b A a A a A a B b B b b B GametesAB ab AaBb Independent Assortment Ab aB Parental AB ab Recomb.Ab aB AaBb Gene Linkage
Earlobes & Toes (Ind. Asst.) P: FFTT x fftt all FfTt (free earloes and 2nd toe longer) Testcross of F1individuals FfTtx fftt: ¼ Free, 2nd toe ¼ Free, great toe ¼ attached, 2nd toe ¼ attached, great toe
F F f f f f f T t T t t t t • Independent • Assortment • ¼ Free, 2nd • ¼ Free, Grt • ¼ Att, 2nd • ¼ Att, Grt f t Earlobes & Toes (Linked) • P: FFTT x fftt • F1: FfTt • F1 Testcross • FfTt x fftt ½ Free, 2nd Toe ½ Att, Grt Toe
F F f f f f f f f T T t t t t t t t F f T t F f T t f t Earlobes & Toes (Linked + Recombination) Recombination frequency = (# of recomb. offspring) (total offspring) 18 17 2 3 Parental Offspring FfTt (free, 2nd) fftt (attached, great) Recombinant Offspring Fftt (free, great) ffTt (attached, 2nd)
Genetic Recombination • Genes farther apart are more likely to cross over • Greater distance greater RF • Enables us to map genes on a chromosome
v v 18.5mu c OR 9.5mu 9mu 9.5mu b 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
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?