1 / 35

Autosomal linkage – 2 genes on the same chromosome

Autosomal linkage – 2 genes on the same chromosome. See page 16, 17 and page 123 in the book. Nail Patella Syndrome is caused by a dominant allele linked to the ABO blood group gene by being found on the same chromosome. (page 16).

velma
Download Presentation

Autosomal linkage – 2 genes on the same chromosome

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. Autosomal linkage – 2 genes on the same chromosome See page 16, 17 and page 123 in the book.

  2. Nail Patella Syndrome is caused by a dominant allele linked to the ABO blood group gene by being found on the same chromosome. (page 16)

  3. Alleles which are linked do not produce the expected ratios of offspring in the next generation • E.g. alleles not linked would produce the expected dihybrid ratio of _:_:_:_ in the F2 generation. • Linked alleles would produce an expected ratio of…..? (Use ABab x ABab to find out) Notice the linked AB and ab. • 1 ABAB: 2 ABab: 1 abab • However, if crossing over occurs between the linked alleles, rare recombinants can result, e.g. 70 ABAB (double dominant), 20 abab (double recessive), 4 AbAb and 6 aBaB, (the 2 rare recombinants). • The closer the genes are linked, the rarer the chance of crossing over between them. • Try SAQ 19 and 20 on page 17. A worked example is on page 123 in your book. • Hint – 2 large numbers of 4 groups of offspring and 2 very small numbers hint at crossing over of linked alleles.

  4. Epistasis– how one gene may control the expression of another. • Key definition – 2 marks – page 18 • The interaction of different gene loci, • so that one gene locus masks or suppresses the expression of another gene locus. • See next slide and page 18.

  5. Complementary Epistasis p18… • Complementary epistasis Colourless substance Yellow Orange • i.e. just Enzyme 1 = ______ • Enzyme 1 and 2 = ______ • No enzyme 1 = ______ Enzyme 1 Enzyme 2 Which offspring ratio suggests complementary epistasis? See page 129. 9:7

  6. Now complete page 19 about recessive and dominant epistasis Salvia - pure breeding pink and white produce purple F1. Interbreeding produces purple, pink and white/ 9:3:4 in the F2 generation. Is this an example of dominant or recessive epistasis? Recessive. See page 128 to explain and write down 3 genotypes for white flowers.

  7. Recessive Epistasis – recessive allele masks another gene. See page 133 to find out how eye colour is controlled by at least two genes interacting together. This example is recessive epistasis. Complete for homework.

  8. Dominant Epistasis in hens p129 2 gene loci interact for feather colour. I is dominant over i and always gives white feathers even if C for colour rather than c for no colour is present too. Write down all the genotypes which produce a) white and b) coloured hens. State the expected F2 ratio (page 129) 13:3 Dominant allele masks another gene.

  9. So, epistatic F2 expected ratios are not9 : 3 : 3 : 1 See page 20 • Recessive is suggested by 9 : 3 : 4 • Dominant is suggested by 12 : 3 : 1 or 13 : 3 • Complementary is suggested by 9 : 7 (similar) • These ratios must be learnt to help solve problems.

  10. Whiteboard Quiz! The F2 phenotypic ratio of 9:3:4 suggests? Recessive epistasis The hypostatic allele is the ________ allele locus. Masked The F2 phenotypic ratio of 45 orange flowers : 35 yellow suggests? Complementary epistasis The epistatic allele is the _______ allele locus. Controlling The F2 phenotypic ratio of 13:3 suggests? Dominant epistasis Homozygous recessive genotype at either allelelocus masking the expression of the dominant allele at the other locus is called __________ epistasis. Complementary Dominant epistasis is when……. A dominant allele locus masks a hypostatic allele locus.

  11. Inheritance of Combs in Domestic Chicken – see page 132 and q 5 + 6 Pea Rose P + rr pp + R Walnut P + R Single pp + rr

  12. Which 2 genotypes should the breeder cross to produce 25% walnut, 25% rose, 25% pea and 25% single combs? This is a 1 : 1 : 1 : 1 ratio. Therefore like the offspring of a test cross of double recessive x heterozygote parents. pprr x PpRr = single x walnut Try it! A possible explanation for crossing pea x rose combs and getting walnut combs (F1) is that CR for rose and CP for pea combs are codominant. Explain how the F2 phenotypes negate that. How many phenotypes can be present in the codominant offspring? Try crossing a walnut with another walnut parent. How many offspring?

  13. AIMS • Give back Cell Control test • Check genetic questions homework answers next Tuesday or Thursday at 11:00 • Genetics Challenge for 30 mins! • Recap last pages of booklet you did for homework – Polygenic Inheritance and the effect of the Environment on the Phenotype. • Population Genetics • Homework for tomorrow – Learn meiosis and genetics for a 30 min test question tomorrow. • Enter yourself for retake!

  14. F 213 Re-sits • These will be done through out the year, but will be entered for in June.

  15. A2 Promonitor 5 mark scheme – cell control • a) i) protein / polypeptide reject enzyme, peptide • 1 • ii) binds / forms complex with lactose; • alters structure / changes shape; • repressor molecule detaches / cannot bind; ignore where it binds 2 • iii) operator; 1 • b) i) observable characteristics of organism/expression of an allele/AW; 1 • ii) altered DNA; • altered mRNA (of repressor); • altered amino acid sequence (of repressor) / no repressor produced; • change to structure / shape (of repressor); • (repressor protein) unable to bind to / block operator; • RNA polymerase can bind to promoter; • transcription / translation continues; max 4 • 2 a) i) lysine; • valine; • tyrosine; 3 • ii) AAG; • GUU: • UAU; 3 • b) reject references to amino acids being formed but do not penalize twice • i) leucine instead of arginine / arginine replaced by leucine; 1 • ii) no change / still serine 1

  16. These should be ACG TTA GCT as on coding strand, not template! Replace codon with triplet or state mRNA codon. • 3 a) (weak) hydrogen 1 • b) i) transcription • ii) translation 2 • c) i) T G C A A T C G A • ii) U G C A A U C G A 2 • 4 a) a length of DNA; • which codes for a polypeptide; 2 • b) codon / a triplet of bases on DNA; • codes for one amino acid; • degenerate / more than one codon for many amino acids; • generally universal / used in all living organisms; • some exceptions to universal use e.g. standard stop codon codes for • tryptophan in mammalian mitochondria • non-overlapping reading / example explained; 3 • c) control the development of the body plan of an organism, including the polarity (head and tail ends) and position of organs / AW 1 • d) programmed cell death in multicellular organisms; • description of process; • example of process described; • example of imbalance described e.g. tumours or degeneration or importance of cell signaling; 2 Is this correct?

  17. Genetics Challenge! • Define autosomal linkage. • Black and spotted is dominant over brown and plain in dogs. A black (B) spotted (S) bitch had a heterozygous genotype. She was mated with a brown plain dog. • Show the expected offspring if the genes are on separate chromosomes. • Now show the expected offspring if the genes are linked on the same chromosomes. • Now show the expected offspring if the spots can only be expressed if the dog is black, i.e. if at least one B is in the genotype. Genes are not linked. • What is this an example of? (Work out parental genotypes and gametes.)

  18. Show the expected offspring if the genes are on separate chromosomes. BbSs x bbss BS Bs bS bs BS Bs bS bs x bs bs bBsS bBss bbsS bbss Phenotypes? 1 : 1 : 1 : 1 2. Now show the expected offspring if the genes are linked on the same chromosomes. BSbs x bsbs BS bs BS bs x bs bs bBSs bbss Phenotypes? How might a few rare recombinants occur?

  19. 3. Now show the expected offspring if the spots can only be expressed if the dog is black, i.e. if at least one B is in the genotype. Genes are not linked. BbSs x bbss BS Bs bS bs BS Bs bS bs x bs bs bBsS bBss bbsS bbss Phenotypes? 1 : 1 : 2 4. What is this an example of? Recessive Epistasis – the homozygous presence of a recessive allele prevents another being expressed. Remember the F2 ratio for a recessive epistasis is 9 : 3 : 4

  20. 5. A plant produces flowers that are either white, purple or dark blue. • These phenotypes could come about either through codominant alleles, or through recessive epistasis, e.g. only aa gives white, Aa or AA gives purple and B or BB gives dark blue as long as at least one A was also present. • The F2 generation produced 63 plants with dark blue flowers, 18 with purple and 31 with white. a) Which phenotypic ratio would you expect in the F2 if i) codominance and ii) negative epistasis was behind these figures? b) Calculate Chi squared using the theory which (from the data) is most likely.

  21. Negative epistasis = 9 : 3 : 4 • Total = 63 + 18 + 31 = 112 • Ratio in real numbers = 9 + 3 + 4 = 16 • 112 / 16 = 7, so 7 x 9 = 63, 7 x 3 = 21 and 7 x 4 = 28. • Observed = 63 : 18 : 31 • Expected = 63 : 21 : 28 • Chi squared = 0 + 0.428 + 0.321 = 0.749 • 3 – 1 = 2 d of f, so looking on the table, and using the critical value of 0.05 probability, any difference is due to chance and the observed is not sig diff from the expected.

  22. Define…… • Codominance. • Sex – linkage. • Autosomal – linkage + expected dihybrid F2 ratio. • Rare recombinants and what they are caused by. • Recessive epistasis + expected ratio. • Dominant epistasis + expected ratio. • Complementary epistasis + expected ratio.

  23. Collect discontinuous andcontinuous data from the class. • Count the number of peoplewith either lobedor un-lobed ears, qualitative differences. • Graph the data on the board. • Measurethe length of your middle finger in mm, quantitative differences. • Collate the numbers into groups (classes) and draw a histogram. • Label the mode and calculate the mean. • Which shows continuous and which discontinuous variation, and what is the cause of each? Check homework page 25 using page 137.

  24. Continuous inheritance is caused by the summative expression of many polygenes. Continuous variation is also caused by environmental factors.

  25. Height is another example of polygenic inheritance – show with skeleton

  26. Effect of Environment on Phenotype. What can you see?

  27. So, what has caused the “switching on” of black ear tips?

  28. How is this hare avoiding being both seen and heat loss in the biting wind?

  29. Here the effect of warming temperatures leading into Spring can be seen on these boxing hares.

  30. Explain this photo!

  31. Explain the Siamese colouring.

  32. How can the environment affect: • Height? • Bone strength? • Intelligence? • Switching on of the gene coding for lactase in E. coli? • http://www.youtube.com/watch?v=oBwtxdI1zvk

  33. How can folic acid help prevent this? Now check page 27……….. And so to population genetics!

  34. Grade Boundaries

More Related