1. Chromosomal Basis of Inheritance Chapter 15

2. Slide 2 of 36 Mendel & Chromosomes  Today we know that Mendel’s “hereditary factors” are located on chromosomes  So we can link Mendelian genetics to modern genetics through the genes that lie on the chromosome  Chromosome Theory of Inheritance  Mendelian genes have specific loci (positions) on chromosomes  Chromosomes undergo Segregation & Independent Assortment

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4. Slide 4 of 36 Converting to Morgan speak  Phenotype  Dominant = wild type  Recessive = mutant  Wild-type – normal or typical  W+  Mutant type – not normal  W  For example, vg+ = wild type for body size (full body) vg = mutant (vestigial body, smaller size)

5. Slide 5 of 36 T.H. Morgan studied flies  He studied Drosophila Melanogaster – Fruit fly  They _____ like flies  Take 2 weeks to breed  Hundreds of offspring per brood  Only 4 pairs of chromosomes  3 autosomes and 1pair of sex chromosomes

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7. Slide 7 of 36 Morgan’s Results  He crossed wild type (red eyes) with a mutant (white eyes), but did not get Mendelian results, or did he?  Gender or sex differences  Called Sex-linked gene or trait  ONLY F 2 MALES = white eyes !!

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9. Slide 9 of 36 Sex-linked Inheritance

10. Slide 10 of 36 Now…  Cross the other 3 combinations  What are the other combinations?  Are there any patterns?

11. Slide 11 of 36 What happens to produce?  All males are mutant, but all females are wild-type?  All females are wild type, but only 50% of males are?  50% are wild-type, 50% are mutant?

12. Slide 12 of 36 Think about it…  IF the mother is homozygous dominant, then sons are   IF the mother is heterozygous, then sons are   IF the mother is homozygous recessive, then sons are

13. Slide 13 of 36 Think about it… (Page 2)  IF the father is wild-type, then daughters are   IF the father is mutant, then daughters are   So mothers determine ______ & wild type fathers produce ______ daughters

14. Slide 14 of 36 Known Sex-Linked Disorders  Duchenne muscular dystrophy  Progressively weakening of muscles and loss of coordination  Hemophilia  Blood that is unable to clot normally  Due to absence of proteins required for proper clotting

15. Slide 15 of 36 Sex-linked Disorders?  Which gender do you think is afflicted at a higher rate?  Why?

16. Slide 16 of 36 Hmm?  If XX is female & XY is male, but the Y chromosome contains virtually no genetic material, do females have more genetic information than males?

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18. Slide 18 of 36 Females = X Inactivation  Although females receive 2 copies of alleles, one chromosome becomes inactivated during embryonic development  Due to XX  Chromosome inactivation is Random  Inactivation is due to methylation  So BOTH females and males are operating on only 1 sex chromosome  Barr Body – inactivated chromosome condenses  Lies on the inside of the nuclear envelope

19. Slide 19 of 36 More fly stuff 2 Characters: body color & wing size Body Color b+ = Grey (wild type) b = black (mutant) Wing Size vg+ = normal wings vg = vestigial wings (Reduced wing size)

20. Slide 20 of 36 Vocabulary  Linked Genes – located on same chromosome  Tend to be inherited together  Genetic Recombination – Offspring with new combination of genes inherited from parents  Parental Phenotype - at least one of the parental phenotypes  Recombinants – NOT either of the parental phenotypes

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22. Slide 22 of 36 What should have happened?  What should have been the ratio if the characters were inherited via a Mendelian pattern?  How do the recombinants form?

23. Slide 23 of 36 Results  2,300 offspring  Far higher proportion of parental phenotypes than expected from independent assortment  Genes are inherited together  There were also recombinants or non-parental phenotypes as well  Conclusion = Partial linkage & Genetic recombination (recombinants or recombinant types)

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25. Slide 25 of 36 Recombination frequency  Calculate by (Total Recombinants / Offspring * 100)  If the genes are located on different chromosomes, then the recombination frequency should be 50%  In the flies, the recombinant frequency was less than 50%; it was about 17%  Evidence of that the 2 genes lied on the same chromosome  So some linkage but incomplete  More recombinants = less linkage

26. Slide 26 of 36 Linkage Maps  Crossing Over explains why some linked genes get separated during meiosis  Crossing Over occurs in Meiosis I  Farther apart 2 genes = Higher P(Crossing Over)  Linkage Map – genetic map based on the percentage of cross-over events  Map unit – 1% recombination frequency  Used ONLY for relative distances on the chromosome

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28. Slide 28 of 36 Explain  We know that Mendel’s seed color and flower color were on the SAME chromosome, but they did not behave as linked genes. Explain.

29. Slide 29 of 36 Chromosomal Abnormalities  Nondisjunction – mishap where pairs of homologs do not move apart properly during meiosis  Could happen in Meiosis I or when Sister chromatids fail to separate correctly in Meiosis II  One gamete receives 2 of the same type of chromosome, while another receives no copy

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31. Slide 31 of 36 Abnormal NUMBERS of chromosomes  Aneuploidy – abnormal number of chromosomes  Nondisjunction could result in a cell with 2n+1  Here this cell would be considered aneuploid, and considered trisomic for that individual chromosome  -somy = different number of an INDIVIDUAL chromosomes  Trisomy – 3 copies of a chromosome  2n + 1  Monosomy – only 1 copy of a chromosome  2n – 1

32. Slide 32 of 36 Abnormal Number of Chromosome SETS  Alteration of an ENTIRE CHROMOSOMAL SET  Called polyploidy  Triploid = 3n  Tetraploid = 4n  Polyploidy plants are fairly common  animals are less common  Polyploids are more normal than aneuploids  Hence, 1 chromosome extra or fewer is more disruptive, than an entire set of chromosomes extra or fewer

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34. Slide 34 of 36 Down Syndrome  1 of 700  Trisomy 21 (each cell has 47 chromosomes, not 46)  Risk increases with maternal age

35. Slide 35 of 36 Klinefelter’s Syndrome  Male have extra X chromosome  Possess male sex organs, but are sterile  IF Female, 3 chromosomes (XXX) = healthy & normal

36. Slide 36 of 36 Turner Syndrome  Female with only 1 X  Only viable monosomy in humans