Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Remember: The specific allele which ends up being inherited from a parent to an offspring is random. Since a parent has two alleles, and each allele is equally likely to be inherited by an offspring, the probability that an offspring will get a specific allele = 0.5.This is equally true for both parents.Probability of inheriting a specific alleleExample 1: Suppose you breed a pair of plants with pink flowers. What are the possible genotypes and phenotypes of the offspring? RR (Red), RW (Pink), and WW (White)Question: What is the probability of getting these genotypes? Phenotypes? Answer:P(RR) = P(R) ♂ × P(R) ♀ = 0.5×0.5 = 0.25P(WW) = P(W) ♂ × P(W) ♀ = 0.5×0.5 = 0.25P(RW) = P(R) ♂ × P(W) ♀ + P(W) ♂ × P(R) ♀ = 0.5×0.5 + 0.5×0.5 = 0.25 + 0.25 = 0.5P(Red) = P(RR) = 0.25 P(White) = P(WW) = 0.25 P(Pink) = P(RW) = 0.5Probabilities for second filial generation(Crossing first filial generation)Example 2: What if we did the same thing with eye color? The genotype probabilities are exactly the same!P(BB) = P(B) ♂ × P(B) ♀ = 0.5×0.5 = 0.25 P(bb) = P(b) ♂ × P(b) ♀ = 0.5×0.5 = 0.25 P(Bb) = P(B) ♂ × P(b) ♀ + P(b) ♂ × P(B) ♀ = 0.5×0.5 + 0.5×0.5 = 0.25 + 0.25 = 0.5But the phenotypic probabilities are different:P(Blue) = P(bb) = 0.25P(Brown) = P(BB) + P(Bb) = 0.25 + 0.5 = 0.75A cross that involves two independent traits is termed dihybrid cross.Example: SupposeGene 1 has alleles A and a (A is dominant over a)Gene 2 has alleles B and b (B is dominant over b)The genes are on different chromosomesHere is a doubly heterozygous individual:Crosses Involving Two GenesA a B bPossible gametesA B A b a B a bWhat is the probability of each gamete? Probabilities in gametogenesisA a B bPossible gametesA B A b a B a bP(AB) = P(A) × P(B) = 0.5 × 0.5 = 0.25P(Ab) = P(A) × P(b) = 0.5 × 0.5 = 0.25P(aB) = P(a) × P(B) = 0.5 × 0.5 = 0.25P(ab) = P(a) × P(b) = 0.5 × 0.5 = 0.25Note: Each gameteis equally probableCrossing two double heterozygotesAaBb × AaBbWhat are the possible genotypes of the offspring?Answer: AABB , AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabbWhat are the probabilities of an offspring being each genotype?AB Ab aB abAB AABB AABb AaBB AaBbAb AABb AAbb AaBb AabbaB AaBB AaBb aaBB aaBbab AaBb Aabb aaBb aabbWhat are the probabilities of each genotype?Probabilities of genotypes in dihybrid crossP(AABB) = P(AB) ♂ × P(AB) ♀ = 0.25 × 0.25 = 0.0625P(AABb) = P(AB) ♂ × P(Ab) ♀ + P(Ab) ♂ × P(AB) ♀ = 0.25×0.25 + 0.25×0.25 = 0.125P(AAbb) = P(Ab) ♂ × P(Ab) ♀ = 0.25 × 0.25 = 0.0625P(AaBb) = P(AB) ♂ × P(ab) ♀ + P(ab) ♂ × P(AB) ♀ + P(Ab) ♂ × P(aB) ♀ + P(aB) ♂ × P(Ab) ♀ = 0.25 (add them all up)P(AaBB) = 0.125P(Aabb) = 0.125P(aaBB) = 0.0625P(aaBb) = 0.125P(aabb) = 0.0625What are the probabilities of each phenotype?Probabilities of phenotypes in dihybrid crossP(“AB”) = P(AABB) + P(AABb) + P(AaBB) + P(AaBb) = 0.0625 + 0.125 + 0.125 + 0.25 = 0.5625P(“Ab”) = P(AAbb) + P(Aabb) = 0.0625 + 0.125 = 0.1875P(“aB”) = P(aaBB) + P(aaBb) = 0.0625 + 0.125 = 0.1875P(“ab”) = P(aabb) = 0.0625If you are familiar with it, note that these probabilities give you the classic 9 : 3 : 3 : 1ratioCross AaBb father with an aaBb mother:What are the probabilities of each genotype and phenotype?Exercise Genotypic probabilities:P(AaBB) = P(AB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125 = 1/8P(AaBb) = P(AB) ♂ × P(ab) ♀ + P(Ab) ♂ × P(aB) ♀ = 0.25 × 0.5 + 0.25 × 0.5 = 0.25 = 1/4P(Aabb) = P(Ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125 = 1/8P(aaBB) = P(aB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125 = 1/8P(aaBb) = P(aB) ♂ × P(ab) ♀ + P(ab) ♂ × P(aB) ♀ = 0.25 × 0.5 + 0.25 × 0.5 = 0.25 = 1/4P(aabb) = P(ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125 = 1/8Father’s gametes: P(AB) = 0.25; P(Ab) = 0.25; P(aB) = 0.25; P(ab) = 0.25Mother’s gametes: P(aB) = 0.5; P(ab) = 0.5Phenotypic probabilities:P(“AB”) = P(AaBB) + P(AaBb) = 0.125 + 0.25 = 0.375 = 3/8P(“Ab”) = P(Aabb) = 0.125 = 1/8P(“aB”) = P(aaBB) + P(aaBb) = 0.125 + 0.25 = 0.375 = 3/8P(“ab”) = P(aabb) = 0.125 = 1/8As a Punnet SquareMother’s gametesaB abFather’s gametesAB AaBB (“AB”)AaBb (“AB”)Ab AaBb (“AB”)Aabb (“Ab”)aB aaBB (“aB”)aaBb (“aB”)ab aaBb (“aB”)aabb (“ab”)What if we take our parents are breed them so they produce 100 offspring?Each offspring is completely independent from the others. The previous values still describe the probability of any individual offspring having a specific genotype or phenotype.But how many of the 100 offspring should have each genotype? Each phenotype? What is the probability that heterozygous parents will have 100 brown-eyed offspring? 99 brown-eyed offspring and 1 blue-eyed offspring? 98 brown-eyed offspring and 2 blue-eyed offspring? (etc.) How do we calculate this?Multiple