1 / 42

Exam #3 W 12/5 at 7-8:30pm in ETC 2.108 for the 9am class and ECJ 1.202 for the noon class

Exam #3 W 12/5 at 7-8:30pm in ETC 2.108 for the 9am class and ECJ 1.202 for the noon class Review T 12/4 at 5pm in WRW 102 Homework #4 due 12/3 (if needed) Bonus #2 is posted. Grades: A = 88.5+ B = 77.5 C = 65.5 Exam avg x 0.88 + homework + bonus = grade.

teril
Download Presentation

Exam #3 W 12/5 at 7-8:30pm in ETC 2.108 for the 9am class and ECJ 1.202 for the noon class

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Exam #3 W 12/5 at 7-8:30pmin ETC 2.108 for the 9am class and ECJ 1.202 for the noon class • Review T 12/4 at 5pm in WRW 102 • Homework #4 due 12/3 (if needed) • Bonus #2 is posted

  2. Grades: A = 88.5+ B = 77.5 C = 65.5 Exam avg x 0.88 + homework + bonus = grade

  3. Tracking two separate genes, for two separate traits, each with two alleles. Ratio of 9:3:3:1 Fig 3.4

  4. Box 2.2 Crossing-over Meiosis: In humans, crossing-over and independent assortment lead to over 1 trillion possible unique gametes. (1,000,000,000,000) Meiosis I (Ind. Assort.) Meiosis II 4 Haploid cells, each unique

  5. Crossing over produces new allelic combinations Fig 4.3

  6. Homologous pair of chromosomes

  7. Longer regions have more crossovers and thus higher recombinant frequencies Fig 4.10

  8. Some crosses do not give the expected results

  9. =25% 42% 41% 9% 8%

  10. These two genes are on the same chromosome

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

  12. Another test

  13. 42 recombinants out of 381 offspring = 42/381 11% recombination

  14. The probability of crossing over can be used to determine the spatial relationship of different genes Fig 4.9

  15. Double recombinants arise from two crossovers Recombinant Fig 4.11

  16. Double recombinants can show gene order Fig4.12

  17. What is the relationship between these 3 genes? What order and how far apart? similar to pg 141

  18. What is the relationship between these 3 genes?What order and how far apart? similar to pg 141

  19. Double crossover similar to pg 141

  20. Which order produces the double crossover?

  21. Which order produces the double crossover?

  22. We have the order.What is the distance? similar to pg 141

  23. Recombinants between st and ss: (50+52+5+3)/755=14.6% similar to pg 141

  24. Recombinants between ss and e: (43+41+5+3)/755=12.2% similar to pg 141

  25. Put it all together… 26.8 m.u. st ss e 14.6 m.u. 12.2 m.u.

  26. Linkage map of Drosophila chromosome 2

  27. Recombination is not completely random. physical distance linkagemap Yeast chromosome 3 Fig 4.13 and 20

  28. Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple. Figs 1.15-17

  29. The relationship between genes and traits is often complex Complexities include: • Complex relationships between alleles • Multiple genes controlling one trait • One gene controlling multiple traits • Environmental effects

  30. The colors of peppers are determined by the interaction of several genes

  31. Complexity of inheritance leads to genetic diversity. Fig 3.14+.16

  32. Eye color: One trait controlled by multiple genes

  33. Seven alleles and their interactions in leaf patterning of clover Fig 6.7

  34. The relationship between genes and traits is often complex Complexities include: • Complex relationships between alleles • Multiple genes controlling one trait • One gene controlling multiple traits • Environmental effects

  35. 1 gene controlling many traits

  36. S=sickle-cell H=normal Sickle-Cell Anemia Mom = HS Dad = HS Dad H or S possible offspring 75% Normal 25% Sickle-cell HS HH H or S Mom HS SS

  37. Coincidence of malaria and sickle-cell anemia

  38. S=sickle-cell H=normal Sickle-Cell Anemia Mom = HS Dad = HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible Dad H or S HS HH H or S Mom HS SS

  39. The relationship between genes and traits is often complex Complexities include: • Complex relationships between alleles • Multiple genes controlling one trait • One gene controlling multiple traits • Environmental effects

  40. Next we will look at what DNA can tell us about the origins of Homo sapiens.

  41. Exam #3 W 12/5 at 7-8:30pmin ETC 2.108 for the 9am class and ECJ 1.202 for the noon class • Review T 12/4 at 5pm in WRW 102 • Homework #4 due 12/3 (if needed) • Bonus #2 is posted

More Related