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

Exam #1 W 2/11 at 7:30-9pm in BUR 106 Bonus posted. Fig 13.5. Phenotype. Genotype. The inheritance of genes on different chromo-somes is independent. Fig 13.5. Approximate position of seed color and shape genes in peas. Gene for seed color. y. Y. Fig 13.8. r. R. Gene for seed shape.

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

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

  2. Fig 13.5 Phenotype Genotype

  3. The inheritance of genes on different chromo-somes is independent. Fig 13.5

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

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

  6. Fig 13.8 meiosis I meiosis II

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

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

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

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

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

  20. CB 15.5 Heterozygous wild typegray w/ normal wings b+ b vg+ vg Homozygous wild typeblack w/vestigial wings bb vg vg

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

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

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

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

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

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

  27. Fig 13.15 Homologous pair of chromosomes

  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. Fig 13.16 Linkage map of Drosophila chromosome 2

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

  33. Fig 13.16 Linkage map of Drosophila chromosome 2

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

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

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

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

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

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