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Probability of inheriting a specific allele

Probability of inheriting a specific allele. Remember: The specific allele which ends up being inherited from a parent to an offspring is random.

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Probability of inheriting a specific allele

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  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. A a B b Possible gametes A B A b a B a b Crosses Involving Two Genes 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. A a B b Possible gametes A B A b a B a b Probabilities in gametogenesis 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 Note: Each gamete is equally probable

  6. Crossing two double heterozygotes 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 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 If you are familiar with it, note that these probabilities give you the classic 9 : 3 : 3 : 1 ratio

  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

  11. Multiple offspring 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?

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