1. Simple Inheritance: with distinct heterozygotes Dominant and recessive alleles are represented with capital and small letters. Ex. legends T_, tall tt, dwrrf Lz_, Lozenge lz/lz, disc If neither allele is dominant, it’s best to use another designation and all genotypes must be defined in the legend Ex: B’B’ or B 1 B 1 Black B’B, or B 1 B 2 Gray BB, or B 2 B 2 white

2. Incomplete Dominance Co-dominance R’R’ red RR white R’R pink R’R’ red RR white R’R roan

3. PROBABILITIES OF INDEPENDENT EVENTS P (A, B, C) = P (A) P(B) P(C) 0 ≤ P ≤ 1

4. Legends for use in probability examples: A _ = Pigmented T _= Taster aa = albino t t= non - taster D’D’= lethal D’D= dwarf D D = normal

5. What is the probability that an Aa, Tt, D’ D, individual will produce an A, t, D gamete? Answer: 1/2 X 1/2 X 1/2 = 1/8 since the three traits are independent

6. What fraction of the children produced by matings between Aa, Tt, D’D parents will have the same phenotype as their parents?

7. Answer P (pigmented, A_)= 3/4 P (taster, T_)= 3/4 P (dwarf, D’D)= 2/3 P (A_,T_,D’D) = 3/4 X 3/4 X 2/3 =3/8

8. Conditional Probabilities take advantage of extra information!

9. What is the probability that I-1 and II - 1in the pedigree below are heterozygous for the single gene trait in this family? I II

10. For I-1: Since two normal parents have an affected child, the trait must be recessive and each parent must be heterozygous.

11. I II Aa A_ aa Two of 3 NORMAL progeny in this case can be expected to be heterozygous ! 1/4 AA: 2/4Aa: 1/4aa II-1is normal: she must be AA or Aa

12. Answer The trait must be recessive; but since daughter II - 1 has the dominant phenotype, the probability she is heterozygous is 2/3.

13. Question What fraction of the progeny from marriages between Cc, Tt females and Cc, Tt males OR will be tasters OR pigmented?

14. Tasters pigmented tasters Pigmented 1 1

15. Answer The “ OR ” tells us to add; - the probability of a taster, T_, is 3/4 the same for pigmented, C_ - since pigmented tasters, C_ T_, are counted twice we must subtract them once 3/4 + 3/4 - (3/4) (3/4) = 24 _ 9 15 16 16 16 =

16. Probabilities of simultaneous events P (A or B) = P(A) + P(B) - P(AB)

17. FAMILY OUTCOMES; BINOMIAL DISTRIBUTION P(x, y) = n! (p x q y ) x!y!

18. Remember that: 6! = 6 X 5 X 4 X 3 X 2 X 1 0! = 1 by definition (N) 0 = 1

19. If two tasters, each of whom had a non-taster parent, marry and have 4 children, what is the probability 3 will be tasters and one will be a non-taster? Typical binomial expansion question:

20. Answer The cross is Tt X Tt, so the odds any child is a taster is 3/4. The probability for a specific order with three tasters and one nontaster is ( 3/4 ) 3 ( 1/4 ) 1 The number of possible orders is 4 ! 3 ! 1 ! 4 ! 3 ! 1 ! ( 3/4 ) 3 ( 1/4 ) 1 = 27/64 Together this becomes:

21. Binomial “at least one” example: P (at least 1) = 1-P(0)

22. Albinism If two parents who are both heterozygous for albinism have 5 children, what is the probability at least 1 child will be albino?

23. Answer Probability at least 1 albino child: = 1 - P (no albinos) = 1 - (P) all normal = 1 - ( 3/4 ) 5 = 1 - ( 243 /1024 ) = 0.763