1 / 36

Lectures posted online after lecture.

Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture. CB 14.3. In many instances there is a unique pattern of inheritance. Traits disappear and reappear in new ratios. CB 14.6. Genotype. Phenotype. Human blood types. CB tbl 14.2.

afleming
Télécharger la présentation

Lectures posted online after lecture.

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. Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture.

  2. CB 14.3 In many instances there is a unique pattern of inheritance. Traits disappear and reappear in new ratios.

  3. CB 14.6 Genotype Phenotype

  4. Human blood types CB tbl 14.2 AA or AO BB or BO AB OO

  5. Looking at the past: If Frank has B blood type, his Dad has A blood type, And his Mom has B blood type… Should Frank be worried?

  6. Mom=B blood BB or BO Dad=A blood AA or AO possible genotypes all A / 50% A and 50% O all B / 50% B and 50% O Gametes Frank can be BO = B blood …no worries

  7. We can also predict the future Fig 2.12

  8. Inheritance of blood types Mom = AB Dad = AB

  9. Inheritance of blood types Mom = AB Dad = AB A or B Gametes: A or B

  10. Inheritance of blood types Mom = AB Dad = AB A or B A or B Gametes: Dad A or B Chance of each phenotype for each offspring 25% AA 50% AB 25% BB AA A or B AB Mom AB BB

  11. CB14.7 Testcross:determining dominant/ recessive and zygosity

  12. Sickle-cell anemia is caused by a point mutation CB5.21

  13. Sickled and normal red blood cells

  14. Sickle-Cell Anemia: A dominant or recessive allele? S=sickle-cell H=normal Mom = HS Dad = HS Dad H or S possible offspring 75% Normal 25% Sickle-cell HS HH H or S Mom HS SS

  15. Coincidence of malaria and sickle-cell anemia CB 23.13

  16. Sickle-Cell Anemia: A dominant or recessive allele? S=sickle-cell H=normal 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

  17. Variation in Peas

  18. CB 14.8 Phenotype Genotype

  19. The inheritance of genes on different chromo-somes is independent. CB 14.8

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

  21. CB 15.2 The inheritance of genes on different chromosomes is independent:independent assortment

  22. CB 15.2 meiosis I meiosis II

  23. The inheritance of genes on different chromosomes is independent:independent assortment CB 15.2

  24. CB 14.8

  25. CB 14.9 Inheritance can be predicted by probability

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

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

  28. “and” multiply “or” add

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

  30. 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

  31. 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)

  32. 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

  33. 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

  34. CB 14.11 Many traits are coded for by more than one gene.

  35. Eye color: One trait controlled by multiple genes

  36. Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture.

More Related