16 16 13 probability what is the probability of

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Unformatted text preview: ances it is a heart] - (1/13x1/4) [Chances it is not both] = Important Special Case of the Sum Rule: If A and B are mutual exclusive events, the probability one of them occur, P(A or B) is P[A + B] = P(A) + P(B) if you roll a dice, what is the probability of getting a 5 or 6? (1/6) + (1/6) = 1/3 Probability • What is the probability of drawing an Ace or a King from this deck of cards? 1/13 (getting an Ace) + 1/13 (getting a King) = 2/13 IV. Principle of Independent Assortment in a Dihybrid Cross Monohybrid = one gene= heterozygote = Aa Dihybrid = two genes = heterozygous for both genes = AaBb Dihybrid Cross: T genes wo Why is it a pure breeding line? Homozygous X P RRyy round green rrYY wrinkled yellow F1 RrYy round yellow F2 315 round yellow 108 round green 101 wrinkled yellow 32 wrinkled green The thing for his success is the fact he looked for the ratio which is 9:3:3:1. Dihybrid Cross → 9:3:3:1 Phenotypic Ratio RrYy X RrYy 315 round yellow 9 108 round green 3 101 wrinkled yellow 3 32 wrinkled green 1 556 seeds 16 R-YR-yy rrYrryy homozygous recessive • Mendel added up the numbers of individuals in certain F2 phenotypic classes to determine if the monohybrid 3:1 F2 ratios were still present. round : wrinkled 315 + 108 : 101 + 32 yellow : green 315 + 101 : 108 + 32 423 : 133 416 : 140 3:1 3:1 Explanation of the 9:3:3:1 • The 9:3:3:1 ratio of a dihybrid cross can be predicted because we can consider each trait separately. -Monohybrid phenotypes: 3/4 round and 1/4 wrinkled 3/4 yellow and 1/4 green -Dihybrid phenotypes (product rule): Round and yellow: they are assorted independently R_Y_ = 3/4 X 3/4 = 9/16 Wrinkled and yellow: rrY_ = 1/4 X 3/4 = 3/16 Round and green: R_yy = 3/4 X 1/4 =3/16 Wrinkled and green: rryy = 1/4 X 1/4 = 1/16 Mendel’s Principle of Independent Assortment •  Two hypotheses for gene assortment in a dihybrid cross: (b) Hypothesis: Independent Assortment (a) Hypothesis: Dependent assortment P Generation RRYY rryy RRYY Gametes RY ry (self) heterozygous for both Eggs 1/ 2 1/ 2 ry RY 1/ 2 RY 1/ 2 RrYy genes 1/ 4 Eggs Sperm 1/ 4 ry F2 Generation 1/ 4 1/ 4 ry RRYY RrYY RRYy RrYy Sperm 1/ 4 RrYY RrYy (self) RY rrYY rY 1/ 4 RRYy Rryy rrYy Rryy rryy Actual results support hypothesis Ry 1/ 4 RrYy RrYy RRyy rrYy Actual results contradict this hypothesis 1/ 4 RY rY Ry dihybrid cross: mating of parental varieties differing in 2 characteristics ry Gametes RY RrYy F1 Generation rryy ry 9/ 16 3/ 16 3/ 16 1/ 16 Yellow round Green round Yellow wrinkled Green wrinkled Punnett square illustrating the genotypes underlying a 9 : 3 : 3 : 1 ratio if you only look at the green color, how many round and wrinkled do you have? (the ratio). 3:1, because they are independently assorted. same with yellow, 3:1 Dihybrid T Cross est If the Independent Assortment Hypothesis is right, how many types of gamete F1 plant RrYy can produce? In what ratio? The dihybrid should make four types of gametes in equal numbers RY rY Ry ry Test cross Rr Yy round yellow F1 rr Yy wrinkled yellow X Rr Yy round yellow rr yy wrinkled green you see 1:1:1:1 because they are independently assorted and make the four types of gametes which can be equally distributed 31 27 Rr yy round green 26 rr yy wrinkled green 26 Result → 1 : 1 : 1 : 1 ratio if the unknown parent is RrYy Summary One Trait (one gene) 3:1 1:1...
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