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Biology 2250 Principles of Genetics Announcements Lab 3 Information: B2250 (Innes) webpage Lab 3 Information: B2250 (Innes) webpage download and print.

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Presentation on theme: "Biology 2250 Principles of Genetics Announcements Lab 3 Information: B2250 (Innes) webpage Lab 3 Information: B2250 (Innes) webpage download and print."— Presentation transcript:

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

2 B2250 Readings 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

3 Weekly Online Quizzes Marks Marks Oct Oct. 25 Example Quiz 2** for logging in Oct. 21- Oct. 25 Quiz 1 2 Oct. 28 Quiz 2 2 Nov. 4 Quiz 3 2 Nov. 10 Quiz 4 2

4 Weekly Online Quizzes Results Example quiz: Example quiz: Quiz 1: Answers:

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

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

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

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

9 Mendel’s Second Law Independent assortment: during gamete formation, the segregation of one gene pair is independent of other gene pairs. during gamete formation, the segregation of one gene pair is independent of other gene pairs. Genes independent because they are on different chromosomes Genes independent because they are on different chromosomes

10 Independent Assortment F 1 AaBb X AaBb F 1 AaBb X AaBb F 2 9 A-B- F 2 9 A-B- 3 A-bb 3 A-bb 3 aaB- 3 aaB- 1 aabb 1 aabb 4 phenotypes AABBAaBbAaBBAABb Genotypes Aabb, AAbb aaBb, aaBB

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

12 Fig 6-6 Independent Assortment Interchromosomal Recombination AB AB ab ab Ab Ab aB aB Inferred F 1 gamete types

13 A a B b A a b B Meiosis I OR (Genes) Correlation of genes and Chromosomes during meiosis A a 4 gamete types

14 Linkage of Genes - Many more genes than chromosomes - Many more genes than chromosomes - Some genes must be linked on the same chromosome; therefore not independent - Some genes must be linked on the same chromosome; therefore not independent

15 Complete Linkage P A B a b A B a b F 1 A B a b a b F 1 gametes A B a b a b X dihybrid ABAB AaBb parental

16 Recombinant Gametes ? Crossing over: - exchange between homologous chromosomes - exchange between homologous chromosomes

17 Crossing over in meiosis I Meiosis I - homologous chromosomes pair - homologous chromosomes pair - reciprocal exchange between non-sister - reciprocal exchange between non-sister chromatids chromatids Ch 4 meiosis animation:

18 Crossing over in meiosis I (animation)

19 Gamete Types F 1 A B a b AaBb a b AaBb gametes A B AB Parental a b ab Parental a b ab Parental A b Ab Recomb. A b Ab Recomb. a B aB Recomb. a B aB Recomb. X

20 Two Ways to produce dihybrid A B a b X 1 Note: Chromatids omitted AABB x aabb  AaBb A b a B X 2 AAbb x aaBB  AaBb

21 1. Ways to produce dihybrid A B a b A B AaBb A B AaBb a b (dihybrid ) a b (dihybrid )Gametes: AB P AB P ab P ab P Ab R Ab R aB R aB R X P Cis Note: Chromatids omitted

22 2. Ways to produce dihybrid A b a B A b a B AaBb A b trans AaBb A b trans (dihybrid ) a B (dihybrid ) a B Gametes: Gametes: P Ab P Ab P aB P aB R AB R AB R ab R ab X P

23 Two ways to produce dihybrid A B a b A b a B cis A B AaBb A b trans a b (dihybrid ) a B a b (dihybrid ) a BGametes: AB P Ab AB P Ab ab P aB ab P aB Ab R AB Ab R AB aB R ab aB R ab XX P

24 Fig 6-6 Independent Assortment Linkage Fig 6-11 InterchromosomalIntrachromosomal

25 Example Test Cross AaBb X aabb ab Exp. Obs. ab Exp. Obs. AB AaBb R AB AaBb R Ab Aabb P Ab Aabb P aB aaBb P aB aaBb P ab aabb R ab aabb R How to distinguish: Parental  high freq. Recombinant  low freq.

26 Example (cont.) Gametes: AB R Gametes: AB R Ab P Ab P aB P aB P ab R ab R Therefore dihybrid: Therefore dihybrid: A b (trans) A b (trans) a B a B

27 Linkage Maps Genes close together on same chromosome: - smaller chance of crossovers - smaller chance of crossovers between them between them - fewer recombinants - fewer recombinantsTherefore: percentage recombination can be percentage recombination can be used to generate a linkage map used to generate a linkage map

28 Linkage maps A B large # of recomb. a b C D small number of recombinants c d Alfred Sturtevant (1913)

29 Linkage maps example Testcross progeny: P AaBb 2146 P AaBb 2146 R Aabb 43 R Aabb 43 R aaBb 22 R aaBb 22 P aabb 2302 P aabb 2302 Total map units Total map units = 1.4 % RF A 1.4 mu B

30 Additivity of map distances separate maps A B A C separate maps A B A C combine maps C A B or Locus or Locus A C B (pl. loci) A C B (pl. loci)

31 Linkage Deviations from independent assortment Dihybrid  gametes 2 parent (noncrossover) common 2 parent (noncrossover) common 2 recombinant (crossover) rare 2 recombinant (crossover) rare % recombinants a function of distance between genes genes % RF = map distance

32 Linkage maps Tomato Drosophila Linkage group = chromosome

33 Summary Mendelian Genetics: Mendelian Genetics: Monohybrid cross (segregation): Monohybrid cross (segregation): - ratios (3:1, 1:2:1, 1:1) - ratios (3:1, 1:2:1, 1:1) - dominance, recessive - dominance, recessive - autosomal, sex-linked - autosomal, sex-linked - probability - probability - pedigrees - pedigrees Dihybrid Cross (Indep. Assort.): - ratios (9:3:3:1, 1:1:1:1) - ratios (9:3:3:1, 1:1:1:1) - linkage (deviation from I.A.) - linkage (deviation from I.A.) - recombination - recombination - linkage maps - linkage maps

34 Gametes Number of Genes Number of Different Number of Genes Number of Different Gametes Gametes monohybrid 1 (Aa) 2 dihybrid 2 (AaBb) 4 trihybrid 3 (AaBbCc) ?

35 Three Point Test Cross AaBbCc X aabbcc AaBbCc X aabbcc ABC ABC ABc abc ABc abc AbC AbC Abc Abc aBC aBC aBc aBc abC abC abc abc 8 gamete types Trihybrid

36 Three Point Test Cross C ABC C ABC B c ABc c ABcA C AbC C AbC b c Abc c Abca Trihybrid Gametes

37 Three Point Test Cross AaBbCc 3 genes: AaBbCc 3 genes: Possibilities: Possibilities: 1. All unlinked 1. All unlinked 2. Two linked; one unlinked 2. Two linked; one unlinked 3. Three linked 3. Three linked Trihybrid 1 2 3

38 Three genes 1. Eye colour 2.Wing 3. Wing Wild (+) mutant v cv ct

39 Three Point Test Cross Three recessive mutants of Drosophila: +, v vermilion eyes +, cv crossveinless +, cv crossveinless +, ct cut wing +, ct cut wing P +/+ cv/cv ct/ct X v/v +/+ +/+ P +/+ cv/cv ct/ct X v/v +/+ +/+

40 Three Point Test Cross P +/+ cv/cv ct/ct x v/v +/+ +/+ P +/+ cv/cv ct/ct x v/v +/+ +/+ Gametes + cv ct v + + F 1 trihybrid v/+ cv/+ ct/+

41 Three Point Test Cross F 1 v/+ cv/+ ct/+ x v/v cv/cv ct/ct v cv ct v cv ct 8 gamete types one gamete type 8 gamete types one gamete type

42 8 gamete types F 1 v/+ cv/+ ct/+ v Parental + cv ct 592 Parental v cv ct 40 v cv ct 89 Recombinant v + ct 3 + cv (most frequent) Parental = non crossover

43 8 gamete types F 1 v/+ cv/+ ct/+ v cv ct 592 v cv ct 40 v cv ct v + ct 3 + cv Recombinant Recombinant Parental Parental Examine two genes at a time

44 8 gamete types F 1 v/+ cv/+ ct/+ v cv ct 592 v cv ct 40 v cv ct v + ct 3 + cv Recombinant Recombinant Parental Parental

45 8 gamete types F 1 v/+ cv/+ ct/+ v cv ct 592 v cv ct 40 v cv ct v + ct 3 + cv Recombinant Recombinant Parental Parental

46 Calculate Recombination Fraction 1. v - cv R v cv R R / 1448 = 18.5 % 268 / 1448 = 18.5 % 2. v - ct R R v ct R v ct /1448 = 13.2 % 191/1448 = 13.2 % 3. ct - cv R ct R + cv R + cv /1448 = 6.4 % 93/1448 = 6.4 %

47 Three point test cross Observations: all 3 RF < 50 % 3 genes on same chromosome all 3 RF < 50 % 3 genes on same chromosome v-----cv largest distance ct in middle v-----cv largest distance ct in middle map v ct cv = cv ct v map v ct cv = cv ct v = 19.6 > 18.5 !! Why ? = 19.6 > 18.5 !! Why ?

48 Three Point Test Cross P +/+ ct/ct cv/cv x v/v +/+ +/+ P +/+ ct/ct cv/cv x v/v +/+ +/+ gametes + ct cv v + + gametes + ct cv v + + F 1 trihybrid v ct cv + ct cv Correct gene order

49 Three Point Test Cross Double crossover class rarest: v---cv v---cv P v + + v + P + ct cv + cv R v ct + v + R + + cv + cv XX XX

50 Three Point test cross 1. Double crossovers not counted in v--cv RF 1. Double crossovers not counted in v--cv RF 2. Double crossovers generate P types (with 2. Double crossovers generate P types (with respect to v--cv) respect to v--cv) 3. Double crossovers not detected as 3. Double crossovers not detected as recombinants recombinantsConsequence: underestimate of v----cv map distance underestimate of v----cv map distance Greater distance of genes  greater error

51 Double recombinant class: (3 + 5) x 2 = 16 (3 + 5) x 2 = = = /1448 = /1448 = 19.6 NOTE: double crossovers detected because of middle gene (ct) because of middle gene (ct)


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