Probability of inheriting a specific allele

1. Probability of inheriting a specific allele Remember: 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.

2. Probabilities for second filial generation (Crossing first filial generation) Example 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 = 0.25 P(WW) = P(W) ♂ × P(W) ♀ = 0.5×0.5 = 0.25 P(RW) = P(R) ♂ × P(W) ♀ + P(W) ♂ × P(R) ♀ = 0.5×0.5 + 0.5×0.5 = 0.25 + 0.25 = 0.5 P(Red) = P(RR) = 0.25 P(White) = P(WW) = 0.25 P(Pink) = P(RW) = 0.5

3. 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 = 0.5 But the phenotypic probabilities are different: P(Blue) = P(bb) = 0.25 P(Brown) = P(BB) + P(Bb) = 0.25 + 0.5 = 0.75

4. Crosses Involving Two Genes A a B b Possible gametes A B A b a B a b A cross that involves two independent traits is termed dihybrid cross. Example: Suppose Gene 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 chromosomes Here is a doubly heterozygous individual:

5. Probabilities in gametogenesis A a B b Possible gametes A B A b a B a b Note: Each gamete is equally probable What is the probability of each gamete? P(AB) = P(A) × P(B) = 0.5 × 0.5 = 0.25 P(Ab) = P(A) × P(b) = 0.5 × 0.5 = 0.25 P(aB) = P(a) × P(B) = 0.5 × 0.5 = 0.25 P(ab) = P(a) × P(b) = 0.5 × 0.5 = 0.25

6. Crossing two double heterozygotes AaBb × AaBb AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb What are the possible genotypes of the offspring? Answer: AABB , AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabb What are the probabilities of an offspring being each genotype?

7. Probabilities of genotypes in dihybrid cross What are the probabilities of each genotype? P(AABB) = P(AB) ♂ × P(AB) ♀= 0.25 × 0.25 = 0.0625 P(AABb) = P(AB) ♂ × P(Ab) ♀ +P(Ab) ♂ × P(AB) ♀= 0.25×0.25 + 0.25×0.25 = 0.125 P(AAbb)= P(Ab) ♂ × P(Ab) ♀= 0.25 × 0.25 = 0.0625 P(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)= 0.125 P(Aabb)= 0.125 P(aaBB)= 0.0625 P(aaBb)= 0.125 P(aabb)= 0.0625

8. Probabilities of phenotypes in dihybrid cross If you are familiar with it, note that these probabilities give you the classic 9 : 3 : 3 : 1 ratio What are the probabilities of each phenotype? P(“AB”) = P(AABB) + P(AABb) + P(AaBB) + P(AaBb) = 0.0625 + 0.125 + 0.125 + 0.25 = 0.5625 P(“Ab”) = P(AAbb) + P(Aabb) = 0.0625 + 0.125 = 0.1875 P(“aB”) = P(aaBB) + P(aaBb) = 0.0625 + 0.125 = 0.1875 P(“ab”) = P(aabb) = 0.0625

9. Exercise Cross AaBb father with an aaBb mother: What are the probabilities of each genotype and phenotype? Father’s gametes: P(AB) = 0.25; P(Ab) = 0.25; P(aB) = 0.25; P(ab) = 0.25 Mother’s gametes: P(aB) = 0.5; P(ab) = 0.5 Genotypic probabilities: P(AaBB) = P(AB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125 = 1/8 P(AaBb) = P(AB) ♂ × P(ab) ♀ + P(Ab) ♂ × P(aB) ♀= 0.25 × 0.5 + 0.25 × 0.5 = 0.25 = 1/4 P(Aabb) = P(Ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125= 1/8 P(aaBB) = P(aB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125= 1/8 P(aaBb) = P(aB) ♂ × P(ab) ♀ + P(ab) ♂ × P(aB) ♀= 0.25 × 0.5 + 0.25 × 0.5= 0.25 = 1/4 P(aabb) = P(ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125= 1/8 Phenotypic probabilities: P(“AB”) = P(AaBB) + P(AaBb) = 0.125 + 0.25 = 0.375 = 3/8 P(“Ab”) = P(Aabb) = 0.125 = 1/8 P(“aB”) = P(aaBB) + P(aaBb) = 0.125 + 0.25 = 0.375 = 3/8 P(“ab”) = P(aabb) = 0.125 = 1/8

10. As a Punnet Square Mother’s gametes aB ab Father’s gametes AB AaBB (“AB”) AaBb (“AB”) Ab AaBb (“AB”) Aabb (“Ab”) aB aaBB (“aB”) aaBb (“aB”) ab aaBb (“aB”) aabb (“ab”)

11. Binomial Distribution Consider crossing heterozygous black guinea pigs (Bb) among themselves. We could ask the following questions: 1. What is the probability of the first four offspring being alternately white and black? 2. What is the probability among the four offspring of producing three black and one white in any order?

12. Binomial FormulaBernoulli Trial: Random experiments are called Bernoulli trials if • the same experiment is repeated several times • there are only two possible outcomes (success and failure) on each trial • the repeated trials are independent • the probability of each outcome remains the same for each trial

13. Bernoulli trials can always be represented by a tree diagram. Let the outcome 'success' be denoted by S and the outcome 'failure', by F. If P(S) = p, and P(F) = q, then p + q = 1.The tree diagram for the experiment repeated twice is: p S p S F q p S q F q F The binomial probability formula: the probability of getting k successes in n trials is given by the formula:

14. Multiple offspring We now consider the case of parents producing 100 offspring? The production of each offspring is completely independent from that of the others. The questions we ask here are 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.)