1. Exam #1 W 2/11 at 7:30-9pm in BUR 106 Bonus posted

2. Phenotype Genotype Fig 13.5

3. The inheritance of genes on different chromo- somes is independent.

4. Y y rR Gene for seed color Gene for seed shape Approximate position of seed color and shape genes in peas Chrom. 1/7Chrom. 7/7 Fig 13.8

5. The inheritance of genes on different chromosomes is independent: independent assortment

6. Fig 13.8 meiosis I meiosis II

7. Fig 13.8 The inheritance of genes on different chromosomes is independent: independent assortment

8. Fig 13.5

9. Inheritance can be predicted by probability

10. Probability of a 4= 1/6 Probability of two 4’s in a row= 1/6x1/6=1/36

11. Probability of 3 or 4 = 1/6+1/6= 1/3

12. “and” multiply “or” add

13. Huntington’s Disease D=disease d=normal Neurological disease, symptoms begin around 40 years old.

14. Mom = ddDad = Dd d or d D or d Dd dd possible offspring 50% Huntington’s 50% Normal Mom Dad Huntington’s Disease D=disease d=normal

15. Two different people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? (Dd hh)

16. Two people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? Dd hh Probability of each outcome: Probability of Dd (Ddxdd) =.5 Probability of hh (HhxHh) =.25

17. Two people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? Dd hh Probability of each outcome: Probability of Dd (Ddxdd) =.5 Probability of hh (HhxHh) =.25 Multiply both probabilities.25 X.5 = 12.5% chance Dd hh offspring

18. Tracking two separate genes, for two separate traits, each with two alleles. Ratio of 9:3:3:1 Fig 13.5

19. Some crosses do not give the expected results Fig 13.13

20. CB 15.5 Heterozygous wild type gray w/ normal wings b + b vg + vg Homozygous wild type black w/vestigial wings b b vg vg

21. =25% 8%9%41%42% Fig 13.13

22. Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2

23. Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2 arentalecomb.

24. =25% 8%9%41%42% Fig 13.13 Why fewer recombinants than parentals?

25. These two genes are on the same chromosome Fig 13.14

26. These two genes are on the same chromosome, and close together. Fig 13.14

27. Homologous pair of chromosomes Fig 13.15

28. Fig 13.13

29. By comparing recombination frequencies, a linkage map can be constructed = ? m.u.

30. By comparing recombination frequencies, a linkage map can be constructed = 17 m.u.

31. Linkage map of Drosophila chromosome 2 Fig 13.16

32. Only 2 of the 4 chromosomes can cross-over. Fig 13.14 Recombinants

33. Linkage map of Drosophila chromosome 2 Fig 13.16

34. Yeast chromosome 3 physical distance linkage map Recombination is not completely random.

35. Fig 13.19 A single gene with 2 alleles only has a few phenotypes Traits coded for by multiple genes have a variety of phenotypes Height of males at Conn. Ag. College in 1914

36. Wheat color shows wide variation... Fig 13.20

37. ...and is coded for by three genes. Fig 13.20

38. Exam #1 W 2/11 at 7:30-9pm in BUR 106 Bonus posted