2. W Three hourly exams plus final exam (450 pts), â You will have 1.5 hours to complete each exam, â You will be allowed one (1) 11” x 8.5” crib sheet, both sides, for each exam, â Exams - 150 points each, Final Exam cumulative. W Quizzes will be given every Wednesday (total 100 pts), â will cover the basics of the assigned reading (including that day's assignment), â quizzes 12.5 points each, ~15 minutes, No Make-up Quizzes, absolutely no exceptions, â No Make-up Quizzes, absolutely no exceptions, â can drop two (2) lowest quiz scores (except 1&2).  Total course points - 550 Grades …may seem really hard. …should be relatively easy.

3. Know This… 1 1/2 1 1/41/21/4 1/2 1 Assignment: Correlate this with the observed phenotype.

4. Mendel’s Results, F2 Dihybrid P generation cross: YYRR x yyrr F1 generation cross: YyRr x YyRr Y_ R_ = 315 yyR_= 108 Y_rr= 101 yyrr= 32 = 9 = 3 = 1

5. Forked-Line Method 1/4 YY 1/2 Yy 1/4 yy 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 x 1/4 = 1/16 YYRR 1/4 x 1/2 = 1/8 YYRr 1/4 x 1/4 = 1/16 YYrr 1/2 x 1/4 = 1/8 YyRR 1/2 x 1/2 = 1/4 YyRr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyrr

6. Genotypes Y--R-- 1/4 YY 1/2 Yy 1/4 RR 1/2 Rr 1/4 RR 1/2 Rr yellow/round 1/4 x 1/4 = 1/16 YYRR 1/16 + 1/4 x 1/2 = 1/8 YYRr 2/16 + 1/2 x 1/4 = 1/8 YyRR 2/16 + 1/2 x 1/2 = 1/4 YyRr 4/16 = 9/16

7. Genotypes Y--rr 1/4 YY 1/2 Yy 1/4 yy 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 x 1/4 = 1/16 YYRR 1/4 x 1/2 = 1/8 YYRr 1/4 x 1/4 = 1/16 YYrr 1/2 x 1/4 = 1/8 YyRR 1/2 x 1/2 = 1/4 YyRr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyrr

8. Genotypes Y--rr 1/4 YY 1/2 Yy 1/4 rr yellow/wrinkled 1/4 x 1/4 = 1/16 YYrr 1/16 + 1/2 x 1/4 = 1/8 Yyrr 2/16 = 3/16

9. Genotypes yyR-- 1/4 YY 1/2 Yy 1/4 yy 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 x 1/4 = 1/16 YYRR 1/4 x 1/2 = 1/8 YYRr 1/4 x 1/4 = 1/16 YYrr 1/2 x 1/4 = 1/8 YyRR 1/2 x 1/2 = 1/4 YyRr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyrr

10. Genotypes yyR-- 1/4 yy 1/2 Rr 1/4 RR green/round 1/4 x 1/2 = 1/8 yyRr 2/16 + 1/4 x 1/4 = 1/16 yyRR 1/16 = 3/16

11. Genotypes yyrr 1/4 YY 1/2 Yy 1/4 yy 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 RR 1/2 Rr 1/4 rr 1/4 x 1/4 = 1/16 YYRR 1/4 x 1/2 = 1/8 YYRr 1/4 x 1/4 = 1/16 YYrr 1/2 x 1/4 = 1/8 YyRR 1/2 x 1/2 = 1/4 YyRr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyrr

12. Genotypes yyrr 1/4 yy 1/4 rr green/wrinkled 1/4 x 1/4 = 1/16 yyrr 1/16

13. F2 via Forked Line Y--R--yellow/round9/16 Y--rryellow/wrinkled3/16 yyR--green/round3/16 yyrrgreen/wrinkled1/16

14. Why use Forked-Line Method? Based on a classic dihybrid cross (YyRr x YyRr), what is the probability that an organism in the F2 generation will have round seeds and breed true for green cotyledons?

15. OK? YR Yr yR yr YYrrYyrr yyRRyyRr YYRRYYRr YyRRYyRr YyRRYyRr YyrryyRryyrr 3/16 p = 0.1875

16. Better 1/4 yy 1/4 RR 1/2 Rr 1/4 rr 1/4 x 1/4 = 1/16 yyRR 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyrr 3/16 p = 0.1875

17. Best (?) 1/4 yy 3/4 R_1/4 x 3/4 = 3/16 yyR_ Sum Law: 1/4 RR + 1/2 Rr

18. Forked-Line Method (phenotypes) 3/4 yellow 1/4 green 3/4 round 1/4 wrinkled 3/4 round 1/4 wrinkled 9/16 yellow round 3/16 yellow wrinkled 3/16 green round 1/16 green wrinkled

19. Example P Rr YY x rrYy Probability Rr YY in offspring; 1/2 Rr 1/2 rr 1/2 YY 1/2 Yy 1/4 RrYY

20. Example P Rr Yy x RRYy Probability of Rr Yy in offspring; 1/2 Yy 1/2 Rr 1/2 RR 1/4 RrYy 1/4 YY 1/4 yy

21. Using Probability Lecture 3 Example YY x Yy Ss x Ss YYSs x YySs Independent Assortment YY or Yy SS Ss ss gametes.5.25 probability (p) Y_ = 1(p) S_ =.75Product Rule: (p) Y_S_ =.75 (p) ss =.25Product Rule: (p) Y_ss =.25 Random Segregation

22. Humans ? Is it possible to ascertain the mode of inheritance of genes in organisms where designed crosses and the production of large numbers of offspring are not practical? Pedegree: an orderly diagram of a families relevant genetic features.

23. Albinism is a recessive trait in humans. Assignment: figure out this pedigree.

24. From Previous Page

25. Symbols

26. More Symbols

27. And more… 2

28. Where Do you Start? Aa Aa Aa or AA Aa Recessive Trait? or Dominant Trait? aa

29. What More Can You Say? Aa Aa Aa or AA Aa Recessive Trait aa

30. Predictions What if you were a genetic counselor? What are the odds that this individual carries the trait?

31. Predictions What if you were a genetic counselor? What are the odds that this individual carries the trait?

32. Conditional Probability ? 1/4 AA Aa 1/4 Aa1/4 aa1/4 Aa 1 : 1 : 1 (p)Aa = 2/3 =.66 Monohybrid Cross

33. Conditional Probability …is the probability of an event occurring given that another event also occurs... P(event) without the condition p(condition)

34. Conditional Probability Example: With a 6-sided die, what is the probability of rolling a 2, given that an even number is rolled on the die: p(2 roll | even #) = p(2 roll) p(even#) p(2 roll | even #) = 1/6 1/2 = 1/3

35. probability without the condition probability of the condition ? Aa p(probability of A_) p(probability of being a heterozygote) = 1/2 = 3/4 = 2/3 p( heterozygous | A_ ) = 1/2 3/4

36. Conditional Probability Use the formula, Or use a Punnett Square, Or... Aa A a AAAa aa 1/4 AA1/4 Aa1/4 aa1/4 Aa 1 : 1 : 1 p(event) without the condition p(condition) p(A|B) =

37. Kidney Disease If 1 and 2 had an offspring, what is the probability that their first kid would show the phenotype?

38. A Simplification Unless otherwise specified (or the pedigree suggests otherwise), the traits that we will track will be rare, We will assume a p = 0 that a non-familial mate carries the trait.

39. Kidney Disease non-familial mates: from outside of the family, if k is the recessive trait, then these individuals are KK.

40. Kidney Disease p(P1) heterozygous 1/2 = 1/8 =.125 x p(P2) heterozygous 1x x p(FF) homozygous recessive 1/4x If 1 and 2 had an offspring, what is the probability that their first kid would show the phenotype?

41. Kidney Disease If 1 and 2 had an offspring, what is the probability that the first kid would be a boy, and show the phenotype?

42. Kidney Disease p(P1) heterozygous 1/2 = 1/16 x p(P2) heterozygous 1x x p(FF) homozygous recessive 1/4x x 1/2 p (boy) And, what is the probability that this boy is a carrier? If 1 and 2 had an offspring, what is the probability that the first kid would be a boy, and show the phenotype?

43. Practice #1 Round (R) and Yellow (Y) are dominant.

44. Practice #2

45. Practice #3

46. Questions Don’t rely on the answers in the back of the book to solve your problems… Don’t just solve them, but understand the principles needed to solve them.

50. a b c d e f

53. Assignments Read from Chapter 3, 3.6 (pp. 100-105), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 - 4.14, 4.16, 4.19 - 4.20.