1 / 89

Chromosome # 7

Figure 14-8 The Cause of Cystic Fibrosis. Section 14-1. Chromosome # 7. CFTR gene. The most common allele that causes cystic fibrosis is missing 3 DNA bases. As a result, the amino acid phenylalanine is missing from the CFTR protein.

alagan
Download Presentation

Chromosome # 7

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. Figure 14-8 The Cause of Cystic Fibrosis Section 14-1 Chromosome # 7 CFTR gene The most common allele that causes cystic fibrosis is missing 3 DNA bases. As a result, the amino acid phenylalanine is missing from the CFTR protein. Normal CFTR is a chloride ion channel in cell membranes. Abnormal CFTR cannot be transported to the cell membrane. The cells in the person’s airways are unable to transport chloride ions. As a result, the airways become clogged with a thick mucus.

  2. Lipid (gangliosides) builds up in brain, Mental retardation. Blindness.

  3. Achondroplasia

  4. 1 pair of chromosomes results in two sex cells R r XR Xr

  5. 1 pair of chromosomes results in two sex cells XR Xr XR Xr

  6. 2 pairs of chromosomes results in 4 sex cells XR Xr XT Xt XR Xr XT Xt XR Xr Xt XT

  7. 3 pairs of chromosomes results in 8 sex cells XR Xr XT Xt The number of possible sex cell =2N XS Xs RTS rts rtS RTs rTs RtS Rts rTS

  8. Non Linked Genes

  9. Linked Genes

  10. Figure 15.3 Morgan’s first mutant

  11. Figure 15.4 In a cross between a wild-type female fruit fly and a mutant white-eyed male, what color eyes will the F1 and F2 offspringhave? EXPERIMENT Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes. P Generation X F1 Generation Morgan then bred an F1 red-eyed female to an F1 red-eyed male to Produce the F2 generation. RESULTS The F2 generation showed a typical Mendelian 3:1 ratio of red eyes to white eyes. However, no females displayed the white-eye trait; they all had red eyes. Half the males had white eyes, and half had red eyes. F2 Generation

  12. CONCLUSION Since all F1 offspring had red eyes, the mutant white-eye trait (w) must be recessive to the wild-type red-eye trait (w+). Since the recessive trait—white eyes—was expressed only in males in the F2 generation, Morgan hypothesized that the eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome, as diagrammed here. W+ W X X P Generation X X Y W+ W Ova (eggs) Sperm W+ W+ F1 Generation W+ W W+ Ova (eggs) Sperm F2 Generation W+ W+ W+ W+ W W W W+

  13. b+ vg+ bvg X Parents in testcross b vg b vg b+ vg+ bvg Most offspring or b vg b vg

  14. Problems 4, 11, 13, 14 on page 277

  15. + Problems 4, 5 on page 291 There is a 17% cross over rate between b and vg

  16. Do Lab Simulation at this site http://www.midpac.edu/~biology/Intro%20Biology/PH%20Biology%20Lab%20Simulations/inheritance/intro.html Do Problem 5 in the Genetics X Files at the end – print out answers to turn in!

  17. Given the following crossing over information : cn - b=9%, b-vg=19%, cn-vg=10%, map out the chromosome Cn b vg This map does not work out, explain why draw one that does

  18. I IV X Y III II Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes 0 48.5 57.5 67.0 104.5 Long aristae (appendages on head) Red eyes Gray body Normal wings Red eyes Wild-type phenotypes Figure 15.8 A partial genetic (linkage) map of a Drosophila chromosome

  19. Problems 8, 9, on page 292

  20. 1. Describe how nondisjunction in meiosis I and in meiosis II differ. Meiosis I: homologous chromosomes fail to separate; meiosis II: sister chromatids fail to separate.

  21. 2. How is it possible for nondisjunction to occur and for some normal gametes to be produced? If nondisjunction occurs in meiosis II, two of the four resulting gametes will be normal haploid.

  22. Nondisjunction is when homologous pairs of chromosomes fail to segregate It can result in TRISOMY where instead of having a normal pair of chromosomes, a person ends up with three chromosomes. Trisomy 21 = Down’s Syndrome XXX=super female XYY=super male (Jacob Syndrome)

More Related