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Biology 2250 Principles of Genetics

Biology 2250 Principles of Genetics. Announcements Lab 3 Information: B2250 (Innes) webpage download and print before lab. Virtual fly: log in and practice http://biologylab.awlonline.com/. Weekly Online Quizzes.

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Biology 2250 Principles of Genetics

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  1. Biology 2250Principles of Genetics Announcements Lab 3 Information: B2250 (Innes) webpage download and print before lab. Virtual fly: log in and practice http://biologylab.awlonline.com/

  2. Weekly Online Quizzes Marks Oct. 14 - Oct. 22 Example Quiz 2** for logging in Oct. 21- Oct. 24 Quiz 1 2 Oct. 28 Quiz 2 2 Nov. 4 Quiz 3 2 Nov. 10 Quiz 4 2

  3. B2250Readings and Problems Ch. 4 p. 100 – 112 Prob: 10, 11, 12, 18, 19 Ch. 5 p. 118 – 129 Prob: 1 – 3, 5, 6, 7, 8, 9 Ch. 6 p. 148 – 165 Prob: 1, 2, 3, 10

  4. Mendelian Genetics    Topics: -Transmission of DNA during cell division Mitosis and Meiosis - Segregation - Sex linkage (problem: how to get a white-eyed female) - Inheritance and probability - Independent Assortment - Mendelian genetics in humans - Linkage - Gene mapping - Tetrad Analysis (mapping in fungi) - Extensions to Mendelian Genetics - Gene mutation - Chromosome mutation - Quantitative and population genetics

  5. Sex-linked Inheritance Correlation between inheritance of genes and sex

  6. Drosophila melanogaster(T. H. Morgan) White eye (mutant) Red eye (wild) X

  7. Drosophila Red Eye White eye

  8. Cross A red female Xwhite male F1 all red F2 red : white 3 : 1 white all male red 2 : 1 female : male No white-eyed females

  9. How to obtain a white-eyed female? Cross B white female Xred male F1 females males F2 females males 1 : 1 : 1 : 1

  10. Xw+Xw+ Xw Y Cross A w+w+X wY F1 ww+ w+Y ww+ XwY w Y w ww wY w+ ww+ w+Y

  11. Mendelian Inheritance Determining mode of inheritance: - single gene or more complicated - recessive or dominant - sex linked or autosomal Approach: cross parents observed progeny compare with expected

  12. Principle of SegregationImplications Answer questions on inheritance: - mode of inheritance (dominant, recessive sex-linked) - paternity - hybridization

  13. Mendel’s First Law ½ A gametes Aa Meiosis: diploid nucleus divides produces haploid nuclei Equal segregation of two members of a gene pair ½ a gametes P(a) = ½ P(A) = ½

  14. Rules of Probability Product rule (AND): probability of 2 independent events occurring simultaneously 2. Sum Rule (OR): probability of either one of two mutually exclusive events

  15. Probability **Coin toss: P (T) = P(H) = 1/2 P(T, T, T) = P(T) and P(T) and P(T) = (½)3 = ½ x ½ x ½ 2. One die: P (6) or P (5) = 1/6 + 1/6 **http://shazam.econ.ubc.ca/flip/

  16. equal Segregation P(A) = ½ P(a) = ½ F2 sperm Self F1Aa X Aa 1/2 A 1/2 a 1/2 A 1/4 AA 1/4 Aa P(AA) = ½ * ½ 1/2 a 1/4 Aa 1/4 aa eggs Prob. (AA or Aa) = 1/4 + 2/4 = ¾ Prob. (aa) = ¼

  17. Two Characters Monohybrid Cross parents differ for a single character (single gene ); seed shape Dihybrid Cross parents differ for two characteristics (two genes)

  18. Dihybrid Two Characters: 1. Seed colour yellowgreen Yy 2. Seed shape Round wrinkled Rr 4 phenotypes

  19. Dihybrid P Gametes RRyy X rrYY Ry rY RrYy DIHYBRID F1

  20. F1 Dihybrid ----->F2 F1 RrYy RrYy X RrYy F2 9 315 round, yellow 3 108 round, green 3 101 wrinkled, yellow 1 32 wrinkled, green Total 556

  21. Individual Characters 1. Seed shape round : wrinkled 423 : 133 3 : 1 (¾ : ¼) 2. Seed colour yellow : green 416 : 140 3 : 1

  22. Conclusion * 3 : 1 monohybrid ratio for each character * 9 : 3 : 3 : 1 phenotypic ratio a random combination of 2 independent 3:1 ratios

  23. Two Independent Genes F2 seed shape 3/4 1/4 colour round wrinkled yellow 3/4 9/16 3/16 green 1/4 3/16 1/16 F2 Phenotypes

  24. Applying Probability to Genetics Dihybrid: RrYy Hypothesis: mechanism for putting R or r into a gamete is independent of the mechanism for putting Y or y into a gamete

  25. Gametes from Dihybrid Dihybrid: RrYy (F1) Principle of segregation during gamete formation: Yy -------> P(Y) = P(y) = 1/2 Rr ------->P(R) = P(r) = 1/2

  26. Gametes from dihybrid 4 gamete types RrYy : probability Y and R 1/2 * 1/2 = ¼ YR Y and r 1/2 * 1/2 = ¼ Yr y and R 1/2 * 1/2 = ¼ yR y and r 1/2 * 1/2 = ¼ yr

  27. F1 gametes produce F2 F1 gametes YyRr X YyRr ¼ YR Yr yR yr ¼ YR1/16 RRYY Yr yR yr gametes F2

  28. Sperm F2 Fig. 6-7 F2 4 Gametes 9 Genotypes 4 Phenotypes Eggs

  29. Mendel’s Second Law Independent assortment: during gamete formation, the segregation of one gene pair is independent of other gene pairs.

  30. (Genes) Meiosis I A Correlation of genes and Chromosomes during Meiosis I a b B A A OR a a b B

  31. Producing the F2 F1 YyRr X YyRr 1. F1 Gametes produce F2 2. Genotypes 3. Phenotypes F2

  32. Independent Assortment Male gametes Two gene systems: 1. Gametes from dihybrid 4 x 4 = 16 YyRr: ¼ YR Yr yR yr ¼ YR1/16YYRR Yr yR yr Female gametes F2

  33. Independent Assortment YyRr X YyRr 2. F2 Genotypes 3 x 3 = 9 ¼ RR ½ Rr ¼ rr ¼ YY1/16 YYRR ½ Yy ¼ yy F2

  34. Independent Assortment YyRr X YyRr 3. F2 Phenotypes 2 x 2 = 4 ¾ R- ¼ rr ¾ Y-9/16 R-Y- ¼ yy

  35. F1 YyRr x YyRr YY RR YY Rr Y-R- Yy RR Yy Rr YY rr Y-rr Yy rr yy RR yyR- yy Rr yyrr yy rr 9 Genotypes 4 phenotypes

  36. Independent Assortment F1 AaBb X AaBb F2 9 A-B- 3 A-bb 3 aaB- 1 aabb 4 phenotypes

  37. Independent Assortment Test Cross AaBb X aabb gametes ab 1/4 AB AaBb 1/4 Ab Aabb 1/4 aB aaBb 1/4 ab aabb 4 phenotypes 4 genotypes

  38. Independent Assortment Inferred F1 gamete types AB Fig 6-6 ab Ab aB Interchromosomal Recombination

  39. Independent Assortment Any number of independent genes: Genes Phenotypes Genotypes 1 2 3 2 4 9 3 8 27 n 2n 3n

  40. Mendelian Genetics in Humans Determining mode of inheritance Problems: 1. long generation time 2. can not control mating Alternative: * information from matings that have already occurred “Pedigree”

  41. Human Pedigrees Pedigree analysis: trace inheritance of disease or condition provide clues for mode of inheritance (dominant vs. recessive) (autosomal vs. sex linked) however, some pedigrees ambiguous

  42. Human Pedigrees 1. Ambiguous: 2. Unambiguous: Normal female Normal male Affected female

  43. Clues (non sex-linked) Recessive: 1. individual expressing trait has two normal parents 2. two affected parents can not have an unaffected child.

  44. Rare Recessive Rare = AA A- (AA or Aa) Cousins (inbreeding)

  45. Clues Dominant: 1. every affected person has at least one affected parent 2. each generation will have affected individuals

  46. Dominant Not AA All genotypes known

  47. Examples Recessive: - phenylketonuria (PKU) - hemophilia (sex linked) - cystic fibrosis - albinism Dominant: - huntingtons chorea - brachydactyly (short fingers) - polydactyly (extra fingers) - achondroblasia (dwarf)

  48. http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=OMIM

  49. Brachydactyly Bb bb Bb short fingers bb normal

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