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Assignments Read from Chapter 3, 3.6 (pp. 100-106), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9,

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Presentation on theme: "Assignments Read from Chapter 3, 3.6 (pp. 100-106), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9,"— Presentation transcript:

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3 Assignments Read from Chapter 3, 3.6 (pp ), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions , 4.6, 4.7, 4.9, , a,b,c,d. Exam Week from Friday… –One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.

4 Sex Determination Systems Different mechanisms of sex selection exist: » XX / XO (O = null), ZW / ZZ (female ZW, Male ZZ), haplo / diplo (males are haploid), XX / XY (most mammals).

5 Sex Chromosomes most mammals…... ‘X’ and ‘Y’ chromosomes that determine the sex of an individual in many organisms, Females: XX Males: XY

6 Differential Region Differential Region Paring Region XY: male XX: female aaA hemizygous: condition where gene is present in only one dose (one allele).

7 X Linkage …the pattern of inheritance resulting from genes located on the X chromosome. X-Linked Genes… …refers specifically to genes on the X- chromosome, with no homologs on the Y chromosome.

8 Blue Female Pink Male x P Gametes or Blue is dominant.

9 Gametes or F1 Blue FemaleBlue Male

10 F1 Blue FemaleBlue Male x Gametes or

11 Gametes or F2 Blue FemaleBlue MaleBlue FemalePink Male

12 F2 Blue FemaleBlue MaleBlue FemalePink Male 3 : 1 Blue to Pink 1 : 1 Female to Male

13 Pink Female Blue Male x P Gametes or

14 F1 Blue FemalePink Male Gametes or

15 F2 Pink FemalePink MaleBlue FemaleBlue Male Gametes or

16 F2 Pink FemalePink MaleBlue FemaleBlue Male : 1 Female to Male 1 : 1 Pink to Blue

17 Sex Linkage to Ponder Female is homozygous recessive X-linked gene, –what percentage of male offspring will express? –what percentage of female offspring will express if, mate is hemizygous for the recessive allele? mate is hemizygous for the dominant allele? Repeat at home with female heterozygous X- linked gene!

18 Sex-Linked vs. Autosomal autosomal chromosome: non-sex linked chromosome, autosomal gene: a gene on an autosomal chromosome, autosomes segregate identically in reciprocal crosses.

19 X-Linked Recessive Traits Characteristics Many more males than females show the phenotype, –female must have both parents carrying the allele, –male only needs a mother with the allele, Very few (or none) of the offspring of affected males show the disorder, –all of his daughters are carriers, roughly half of the sons born to these daughters are carriers.

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21 X-Linked Dominant Affected males married to unaffected females pass the phenotype to their daughters, but not to their sons, Heterozygous females married to unaffected males pass the phenotype to half their sons and daughters, Homozygous dominant females pass the phenotype on to all their sons and daughters.

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23 Autosomal Dominant Phenotypes appear in every generation, Affected males and females pass the phenotype to equal proportions of their sons and daughters.

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26 Recessive?---> Yes! Pedigree for Very Rare Trait ? = kid with trait Autosomal?X-Linked?---> Yes! 1/2 1/2 x 1/2 x ?1/2 = 1/8 ? x 1/2 = 1/16 (p)boy

27 X-Linked Dominant examples (OMIM) HYPOPHOSPHATEMIA: “Vitamin-D resistant Rickett’s”, LISSENCEPHALY: “smooth brain”, FRAGILE SITE MENTAL RETARDATION : mild retardation, RETT Syndrome: neurological disorder, More on OMIM…

28 Genetics: … in the News

29 Linkage Genes linked on the same chromosome may segregate together.

30 A b 2n = 4 Independent Assortment a A A B B b a B a b

31 Meiosis No Cross Over Parent Cell Daughter Cells Have Parental Chromosomes Aa Bb A B a b a b A B 2n = 1

32 Meiosis With Cross Over Parent Cell Daughter Cells Have Recombinant Chromosomes Aa Bb A B A b a b a B 2n = 1

33 Dihybrid Cross yellow/round green/wrinkled GGWW x ggww GW gw GgWw phenotype genotype gametes genotype P F1

34 Gamate Formation in F1 Dihybrids P: GGWW x ggww, Independent Assortment G g W w GWGwgWgw alleles gametes probability.25 F1 Genotype: GgWw

35 How do you test for assortment of alleles? GWGwgWgw.25 F1: GgWw Test Cross: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.

36 Test Cross GgWw x ggww GW (.25) Gw (.25) gW (.25) gw (.25) G gww (.25) GgWw (.25) ggWw (.25) ggww (.25) gw (1) x x x x

37 Test Cross GgWw x ggww GW (.25) Gw (.25) gW (.25) gw (.25) gw (1) Ggww (.25) GgWw (.25) ggWw (.25) ggww (.25) P P F1 parental types GgWw and gwgw R R recombinant types Ggww and ggWw x x x x

38 Recombination Frequency …or Linkage Ratio: the percentage of recombinant types, –if 50%, then the genes are not linked, –if less than 50%, then linkage is observed.

39 Linkage Genes closely located on the same chromosome do not recombine, –unless crossing over occurs, The recombination frequency gives an estimate of the distance between the genes.

40 Recombination Frequencies Genes that are adjacent have a recombination frequency near 0%, Genes that are very far apart on a chromosome have a recombination frequency of 50%, The relative distance between linked genes influences the amount of recombination observed.

41 AB In this example, there is a 2/10 chance of recombination. ab AC In this example, there is a 4/10 chance of recombination. ac homologs

42 Linkage Ratio P GGWW x ggww Testcross F1: GgWw x ggww # recombinant # total progeny GWGwgWgw ???? x 100 = Linkage Ratio Units: % = mu (map units) - or - % = cm (centimorgan) determine

43 Fly Crosses (simple 3-point mapping) (white eyes, minature, yellow body) In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu, In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu, When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu. Study Figs 4.2, 4.3, and 4.5

44 Simple Mapping white eyes x miniature = 36.9 mu, white eyes x yellow body = 0.5 mu, miniature x yellow body = 38 mu, my 38 mu 36.9 mu w 0.5 mu

45 Do We have to Learn More Mapping Techniques? Yes, –three point mapping, Why, –Certainty of Gene Order, –Double crossovers, –To answer Cyril Napp’s questions, –and, for example: over 4000 known human diseases have a genetic component, knowing the protein produced at specific loci facilitates the treatment and testing.

46 cis “coupling”

47 trans “repulsion”

48 Classical Mapping Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. target What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2?

49 Gene Order It is often difficult to assign the order of genes based on two-point crosses due to uncertainty derived from sampling error. A x B = 37.8 mu, A x C = 0.5 mu, B x C = 37.6 mu,

50 Double Crossovers More than one crossover event can occur in a single tetrad between non-sister chromatids, –if recombination occurs between genes A and B 30% of the time (p = 0.3), then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or nearly one map unit. If there is a double cross over, does recombination occur? –how does it affect our estimation of distance between genes?

51 Classical Mapping model organisms Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. target What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2?

52 Classical mapping in humans requires pedigrees…

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54 Three Point Testcross Triple Heterozygous (AaBbCc ) x Triple Homozygous Recessive (aabbcc)

55 Three Point Mapping Requirements The genotype of the organism producing the gametes must be heterozygous at all three loci, You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring, You must look at enough offspring so that all crossover classes are represented.

56 w g d

57 Representing linked genes w g d x w g d P Testcross = WwGgDd = wwggdd

58 Phenotypic Classes tri-hybrid cross?

59 W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II

60 W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d I Total = 500 Region I: x 100 = 20.8 mu

61 W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d II Total = 500 Region II: x 100 = 10.0 mu 20.8 mu

62 W G D w g d W-gg-D wwG-dd 4 2 Recombinants, double crossover Total = mu20.8 mu 0.1 x = /500 = NO GOOD! Coefficient of Coincidence = Observed Expected Interference = 1 - Coefficient of Coincidence

63 Interference …the effect a crossing over event has on a second crossing over event in an adjacent region of the chromatid, –(positive) interference: decreases the probability of a second crossing over, most common in eukaryotes, –negative interference: increases the probability of a second crossing over.

64 Gene Order in Three Point Crosses Find - either - double cross-over phenotype…based on the recombination frequencies, Two parental alleles, and one cross over allele will be present, The cross over allele fits in the middle... –

65 # Which one is the “odd” one? A C B a c b III A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- aabbcc

66 A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # aabbcc1786 Region I A C B a c b I x 100 = 28.4 mu

67 A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # aabbcc1786 Region II A C B a c b 28.4 mu x 100 = 18.5 mu II 18.5 mu

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69 Genetics In the News

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71 Fig Molecular Markers (RFLP) EcoRI cleavage sites

72 Fig Fig. 4.20a Molecular Mapping Markers

73 Fig. 4.20b

74 p Fluorescent dyes are often used to label DNA so that the positions of DAN fragments in a gel can be identified.

75 Assignments Read from Chapter 3, 3.6 (pp ), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions , 4.6, 4.7, 4.9, , a,b,c,d. Exam Friday, Oct. 16 th, –One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.

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