1 / 45

Unit 3: Genetics

Unit 3: Genetics. The Cell Cycle + DNA structure/function Mitosis and Meiosis Mendelian Genetics (aka - fun with Punnett squares) DNA replication. Yesterday’s Exit Ticket. g. G. G. G. Gg. gg. g. Gg. Gg. g. Gg. gg. g. Gg. Gg. g.

locke
Download Presentation

Unit 3: Genetics

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. Unit 3: Genetics • The Cell Cycle + DNA structure/function • Mitosis and Meiosis • Mendelian Genetics (aka - fun with Punnett squares) • DNA replication

  2. Yesterday’s Exit Ticket g G G G Gg gg g Gg Gg g Gg gg g Gg Gg g • Create and complete two testcross Punnet squares: (assume G=green and g=yellow) gg x Gggg x GG ½ green all green ½ yellow • Why homozygous recessive for testcross? • Clear and easy determination of unknown’s genotype: 1:1 = heterozygote; all dominant = homozygote

  3. Today’s Agenda: • Mendel and multiple characters • Exceptions to Mendel • Sex-linked traits • Gene linkage

  4. How To Punnet Squares https://www.youtube.com/watch?v=Y1PCwxUDTl8

  5. 2. Probability and genetic outcomes Fig. 14-8 What about multiple characters? Are they inherited together or separately? For the purposes of example, consider the following two characters: 1. Seed color: • Possible phenotypes = Yellow OR green • Yellow is dominant to green 2. Seed shape: • Possible phenotypes = Round OR wrinkled • Round is dominant to wrinkled F1 Generation YyRr Hypothesis of dependent assortment Hypothesis of independent assortment Predictions Predicted offspring of F2 generation Sperm Sperm or 1/4 1/4 1/4 yr 1/4 YR yr YR yR 1/2 1/2 Yr 1/4 YR YYRr YYRR YyRR YyRr 1/2 YR YyRr YYRR 1/4 Yr Eggs YYRr YYrr Yyrr YyRr Eggs 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 Yyrr yyRr YyRr yyrr 3/16 1/16 9/16 3/16 Phenotypic ratio 9:3:3:1 RESULTS

  6. 2. Probability and genetic outcomes Fig. 14-8 What about multiple characters? Are they inherited together or separately? YYRR yyrr EXPERIMENT P Generation Gametes yr YR  F1 Generation YyRr Hypothesis of dependent assortment Hypothesis of independent assortment Predictions Important Vocab Note: A MONOHYBRID cross deals with one gene e.g. Aa x Aa A DIHYBRID cross deals with two genes e.g. AaBb x AaBb Predicted offspring of F2 generation Sperm Sperm or 1/4 1/4 1/4 yr 1/4 YR yr YR yR 1/2 1/2 Yr 1/4 YR YYRr YYRR YyRR YyRr 1/2 YR YyRr YYRR 1/4 Yr Eggs YYRr YYrr Yyrr YyRr Eggs 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 Yyrr yyRr YyRr yyrr 3/16 1/16 9/16 3/16 Phenotypic ratio 9:3:3:1 RESULTS

  7. Fig. 14-8 YYRR yyrr EXPERIMENT P Generation Gametes yr YR  F1 Generation Suppose that two F1 individuals are crossed. Consider two mutually exclusive hypotheses about inheritance: 1. Strict dependent assortment = inherited allele combinations are ALWAYS preserved in the gametes an individual produces 2. Independent assortment = all possible combinations of inherited alleles of different genes are equally likely in an individual’s gametes

  8. 2. Probability and genetic outcomes Fig. 14-8 YYRR yyrr EXPERIMENT P Generation Gametes yr YR  F1 Generation YyRr Hypothesis of independent assortment Hypothesis of dependent assortment Sperm or 1/4 1/4 1/4 yr 1/4 YR yR Yr Sperm YR yr 1/2 1/2 1/4 YR YYRr YYRR YyRR YyRr Predicted offspring of F2 generation 1/2 YR YyRr YYRR 1/4 Yr Eggs YYRr YYrr Yyrr YyRr Eggs 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 Yyrr yyRr YyRr yyrr 3/16 1/16 9/16 3/16 Phenotypic ratio 9:3:3:1 RESULTS 315 108 101 32 Phenotypic ratio approximately 9:3:3:1

  9. 2. Probability and genetic outcomes Fig. 14-8 YYRR yyrr EXPERIMENT P Generation Gametes yr YR  F1 Generation YyRr Hypothesis of independent assortment Hypothesis of dependent assortment Sperm or 1/4 1/4 1/4 yr 1/4 YR yR Yr Sperm YR yr 1/2 1/2 1/4 YR YYRr YYRR YyRR YyRr Predicted offspring of F2 generation 1/2 YR YyRr YYRR 1/4 Yr Eggs YYRr YYrr Yyrr YyRr Eggs 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 Yyrr yyRr YyRr yyrr 3/16 1/16 9/16 3/16 Phenotypic ratio 9:3:3:1 RESULTS 315 108 101 32 Phenotypic ratio approximately 9:3:3:1

  10. Fig. 15-2b F1 Generation: 2 possiblearrangements ofchromosomes All F1 plants produce yellow-round seeds (YyRr) 0.5 mm LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. R R y y r r LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. Y Y Meiosis r r R R Metaphase I Y y Y y 1 1 r R r R Anaphase I Y Y y y Metaphase II r R R r 2 2 y Y y Y y Y Y y Y y Y y Gametes r R r R r r R R yr yR 4 YR 4 4 Yr 4 1 1 1 1 3 3

  11. 2. Probability and genetic outcomes • Mendel’s “law” of independent assortment = alleles for each character segregate independently during gamete formation • Given what YOU know about the relationship between genes and chromosomes (which Mendel did NOT), when would this “law” be violated?

  12. 2. Probability and genetic outcomes Y R r y Y R r y

  13. Today’s Agenda: • Mendel and multiple characters • Exceptions to Mendel • Sex-linked traits • Gene linkage

  14. If only it were all so simple… The view provided by (my simplified presentation of) Mendel’s pea experiments: • one gene  one character(e.g., flower color gene  color of flower) • one allele  one phenotype(e.g., Pallele  purple flower) • two alleles of each gene, one completely dominant, the other recessive(e.g., P dominant to p)

  15. 3. Extending the Mendelian model • Patterns of inheritance different from those discussed so far can be caused in many ways. Just to name a few: • Lack of complete dominance by one allele • A gene has more than two alleles • A gene produces multiple phenotypes • Multiple genes affect a single phenotype • Environmental circumstances affect the phenotype To learn more about all of these, take EBIO 2070!

  16. For now, the simplest exceptions: Genes on sex chromosomes Gene linkage X Y

  17. REMINDER: A complete single set of human chromosomes includes: • 22 autosomes (non-sex chromosomes) • 1 sex chromosome (diploid cells have 44 autosomes and 2 sex chromosomes)

  18. Humans and many other species have chromosomal sex determination In the human system, females have two “X” chromosomes, males have one “X” and one “Y” X Y Fig. 15-5

  19. Other forms of chromosomal sex determination in the animal kingdom… 76 + ZW 76 + ZZ Fig. 15-6c The Z-W system

  20. What consequences might sex chromosomes have for patterns of inheritance and gene expression? 22 pairs of chromosomes + 22 pairs of chromosomes+ X Y X X

  21. Who determines the sex of our offspring? Diploid Parent Cell Gametes XX XY X X X Y Dad determines a child’s sex!

  22. Patterns of inheritance in mammals (and other XY systems) Diploid Parent Cell Gametes XAXa XAY XA Xa XA Y

  23. Patterns of inheritance in mammals (and other XY systems) Dad is who determines a child’s sex Diploid Parent Cell Gametes XX XYA X X X YA

  24. Patterns of gene expressionin mammals (and other XY systems) Male-pattern baldness SRY gene: Testes formation

  25. Santhi’s Story http://www.ibnlive.com/videos/28851/how-are-athletes-gender-tested.html Santhi Soundarajan won the silver medal in the 800-meter race at the 2006 Asian Games in Doha, Qatar. Following her silver medal performance, she was stripped of her medal. Santhi has female genitalia but her genotype is XY. Speakequal.com

  26. Patterns of gene expressionin mammals (and other XY systems)

  27. Genes on chromosomes b) Sex-linked traits • Breeding fruit flies (Drosophila melanogaster) • Rapid breeders • Males = XY; Females = XX • For Drosphila: recessive alleles = “mutant” (b) dominant alleles = “wild type” (b+) News.wisc.edu

  28. Fig. 15-3

  29. P Generation (true breeding) F1 Generation All offspring had red eyes One of Morgan’s experiments (think back to Mendel’s peas): • Character: eye color • Phenotypes: red or white  Is the allele for white eyes dominant or recessive?

  30. Then, cross the F1 offspring with each other, and what does the F2 generation look like?  3:1 ratio of red : white 2:1:1 ratio of red female : red male : white male

  31. The best explanation for the pattern of inheritance seen in the F2 generation is: • The eye color gene is on an autosome • The eye color gene is sex-linked, on the X chromosome • The eye color gene is sex-linked, on the Y chromosome • There is not enough information to discriminate between hypotheses (a) through (c)

  32. Genes on chromosomes Fig. 15-4c b) Original discoveries CONCLUSION R r P X X  Generation Y X r r Sperm Eggs All females XRXr All males XRY F1 R R R Generation r R Sperm Eggs R R R F2 Generation r r r r R

  33. Genes on chromosomes Fig. 15-4c b) Original discoveries CONCLUSION R r P X X  Generation Y X r r Sperm Eggs All females XRXr All males XRY F1 R R R Generation r R Sperm Eggs R R Females all red: ½ XRXr ½ XRXR Males half red (XRY) and half white (XrY) R F2 Generation r r r r R

  34. For now, the simplest exceptions: Genes on sex chromosomes Gene linkage X Y

  35. Gene Linkage and Fruit Flies https://www.youtube.com/watch?v=-_UcDhzjOio

  36. Fig. 15-2b All F1 plants produce yellow-round seeds (YyRr) LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. R y r Y Meiosis r R Metaphase I Y y 1 r R Anaphase I Y y Metaphase II R r 2 y Y Y y Y y r r R R yR 4 4 Yr 1 1 3

  37. Fig. 14-8  In crosses involving two characters, sometimes you get outcomes that were intermediate between these two hypotheses. F1 Generation YyRr Hypothesis of independent assortment Hypothesis of dependent assortment Sperm or 1/4 1/4 1/4 yr 1/4 YR yR Yr Sperm YR yr 1/2 1/2 1/4 YR YYRr YYRR YyRR YyRr Predicted offspring of F2 generation 1/2 YR YyRr YYRR 1/4 Yr Eggs YYRr YYrr Yyrr YyRr Eggs 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 Yyrr yyRr YyRr yyrr 3/16 1/16 9/16 3/16 Phenotypic ratio 9:3:3:1

  38. Example • Morgan crossed flies to study the characters of body color and wing size  Genes for both are located on autosomes

  39. Fig. 15-9-1 P Generation (homozygous) EXPERIMENT Double mutant (black body, vestigial wings) Wild type (gray body, normal wings)  b b vg vg b+ b+ vg+ vg+ F1 generation ?

  40. Fig. 14-8 F1 dihybrid (wild type phenotype) x F1 dihybrid (wild type phenotype) b+ b vg+ vg b+ b vg+ vg Hypothesis of dependent assortment Hypothesis of independent assortment b+b?v+v? bbvv b+b?v+v? b+b?vv bbv+v? bbvv 3 : 1 9 : 3 : 3 : 1 Observed (approx.): 8 : 2 : 2 : 4

  41. Why would some genes be inherited neither completely together nor completely independently? Gene linkage • Each chromosome has hundreds or thousands of genes • Genes located on the same chromosome that tend to be inherited together are called linked genes • Occasional crossing over leads to occasional, but not common, recombinant chromosomes

  42. Sources of genetic variation crossing over Recombination of Linked Genes: Crossing Over Prophase I of meiosis Nonsister chromatids held together during synapsis Pair of homologs Chiasma Centromere Anaphase I  Crossing over during Prophase I of meiosis is the mechanism for recombining alleles

  43. Gene linkage Fig. 15-UN1 b vg b+vg+  Parents in testcross b vg b vg b vg b+vg+ Most offspring or b vg b vg

  44. 3. Gene linkage Fig. 15-10b Recombinant chromosomes b+vg+ b+vg bvg+ bvg Eggs 944 Black- vestigial 965 Wild type (gray-normal) 185 Black- normal 206 Gray- vestigial Testcross offspring bvg b+vg+ bvg bvg+ b+vg bvg bvg bvg bvg Sperm Recombinant offspring Parental-type offspring 5 : 5 : 1 : 1

  45. Today’s Exit Ticket An x-linked recessive allele b produces red-green color blindness in humans. A normal-sighted woman whose father was color-blind marries a color-blind man. • What genotypes are possible for the mother of the colorblind man? • What are the chances that the first child from this marriage will be a color-blind boy? • Of the girls produced by these parents, what proportion can be expected to be color-blind? • Of all the children (sex unspecified) of these parents, what proportion can be expected to have normal color vision?

More Related