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INHERITANCE OF TWO OR MORE INDEPENDENT GENES

INHERITANCE OF TWO OR MORE INDEPENDENT GENES. vg + (1/2). e + vg + (1/4). e + (1/2). vg (1/2). e + vg (1/4). e vg + (1/4). vg + (1/2). e (1/2). e vg (1/4). vg (1/2).

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INHERITANCE OF TWO OR MORE INDEPENDENT GENES

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  1. INHERITANCE OF TWO OR MORE INDEPENDENT GENES

  2. vg+(1/2) e+ vg+(1/4) e+(1/2) vg(1/2) e+ vg(1/4) e vg+(1/4) vg+(1/2) e (1/2) e vg(1/4) vg(1/2) In Drosophilamelanogaster, body colourisdetermined by the e gene: the recessive allele isresponsible for the blackcolour of the body, the dominant allele e+ isresponsible for the grey body. Vestigialwings are determined by the recessive allele vg, normalwings are determined by the dominant allele vg+. Thesetwogenes are independent. Ifdihybridflies for thesetwogenes are crossed and resultingprogenyiscomposed by 368 individuals, howmanyindividuals are present in everyphenotypicclass? e+e+ e e vg+vg+ vgvg ParentalGenotypes: e+e vg+vg X e+e vg+vg Gametes

  3. Phenotypes for gene e Phenotypes for gene vgPhenotipicalclasses (crosse+e X e+e) (crossvg+vg X vg+vg) vg+ (3/4) e+ vg+ (¾ X ¾= 9/16) e+ (3/4) e+ vg (¾ X ¼ = 3/16) vg (1/4) e vg + (¾ X ¼ = 3/16) vg+ (3/4) e (1/4) e vg (¼ X ¼=1/16) vg (1/4) 9/16 di 368 = 207 individualse+ vg+brownbody and normal wings 3/16 di 368 = 69 individualse+vgbrown body and vestigialwings 3/16 di 368 = 69 individualse vg+ black body and normalwings 1/16 di 368 = 23 individualse vgblack body and vestigialwings

  4. In guineapigs, the Rgene determinatesfurrough or smooth, while the Bgene controlsfurcolor. Determine the genotype of individualsused for the crosses and verifyyourhypothesis with χ2 test. n° of individuals in the progeny • a)The parentshavedifferentphenotypes and the progenyisall Black and Rough: • Black and Rough are dominantcharacters (B, black; R, rough) • The parents are homozygous: BB RR X bb rr • We do notperfomedstatistic test becausewe do nothavedifferencesbetweenexpectedprogeny and observedprogeny.

  5. In guineapigs, the Rgene determinatesfurrough or smooth, while the Bgene controlsfurcolor. Determine the genotype of individualsused for the crosses and verifyyourhypothesis with χ2 test. n° of individuals in the progeny b) Weknowthat Black and Rough are dominant traits. The crossedindividualshave the samephenotypesbut in the progenywehave the recessive characters, thusbothparents are double heterozygous: BbRr X Bb Rr In the progenyweexcpect: 9/16 BR (Black rough): 3/16 Br (blacksmooth): 3/16 bR (brownrough): 1/16 br(brownsmooth)

  6. c2 c2 = 0,49 FreedomDegree= 4 – 1 = 3

  7. Rejected Accepted P >90%, weaccept the hypotesis

  8. In guineapigs, the Rgene determinatesfurrough or smooth, while the Bgene controlsfurcolor. Determine the genotype of individualsused for the crosses and verifyyourhypothesis with χ2 test. n° of individuals in the progeny c) The parentshavedifferentphenotypes. The secondparentishomozygous recessivefor bothgenes(bb rr). The first parenthas the dominantphenotypes for bothcharactersB - R -. In the progenyweobserveclasses of individuals with recessive phenotypesthus the first indiviudual id double heterozygous: BbRr. In the progenyweexpect 4 phenotypicclasses with ratio:

  9. c2 c2 = 0,38 FreedomDegree = 4 – 1 = 3 P =>90%, weaccept the hypothesis

  10. In guineapigs, the Rgene determinatesfurrough or smooth, while the Bgene controlsfurcolor. Determine the genotype of individualsused for the crosses and verifyyourhypothesis with χ2 test. n° of individuals in the progeny d) The first individualishomozygous recessive for color (bb) and he has the dominantphenotype for Rough gene (R -). The secondindividualisdominant for bothcharacters (B - R -). In the progenyweobserve the classes with recessive phenotypes for brownanssmoothhair, thus the originalindividualshavea recessive allele to donate to the progeny:b b R r X B b R r

  11. With branchdiagramdetermineexpectedphenotypicclasses: Gene B Gene R phenotypicClasses bb X BbRr x Rr R (3/4) B R (1/2 x 3/4 = 3/8) blackrough B (1/2) r (1/4) B r (1/2 x 1/4 = 1/8) blacksmooth R (3/4) b R (1/2 x 3/4 = 3/8) brownrough b (1/2) r (1/4) b r (1/2 x 1/4 = 1/8) brownsmooth

  12. c2 c2 = 0,87 Degrees of freedom= 4 – 1 = 3 P =80-90%, weaccept the hypothesis

  13. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height Numero di individuidella progenie • Color: the crossedindividualshave the samephenotypesbut in the progenywehavealso recessive phenotypeindividuals: bothparents are heterozygous (Ww) and Red (W) isdominant over white(w). • Height: wecannotestablishwichis the dominant allele since the characterdoesnot segregate. All the progenyistall. Bothparentscould be recessive (dd) or D D x D - o or D - x D D

  14. c2 c2 = 0,44 Degrees of freedom= 2 – 1 = 1 P =50-70%, we can accept the hypothesis

  15. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height n° of individuals in the progeny b) The crossingplantshavetwodifferentphenotypes. In the progenyweobserveplantsred and white: weknowthatredisdominant over whitethus the genotypes areWw X w w Height: sincewehaveonlytallplantsweestabishthattallisdominant over short. In the progenythereisn’tsegregation of characterthen the tallplantishomozygous. The genotypes are: D D X d d

  16. WwDD X ww dd Weexpect: ½ WD (red and tall) and½ wD(white and tall). Verifichiamo con il test del c2 se possiamo accettare l’ipotesi. c2 = 0,12 Degrees of freedom= 2 – 1 = 1 P = 70-80%, weaccept the hypothesis

  17. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height n° of individuals in the progeny c) The crossedplantshavedifferentphenotypes. In the progenyweobservedsegregation of the character: weknowthatredisdominant over white and weassessthat the cross isbetween a heterozygous and a homozygous recessive:W w X w w Height: Tallisdominant over short thusalso for height the cross isbetween a heterozygous and a homozygousrecessive D d X d d

  18. Ww D d X w w d d Weexpect 4 phenotypicclasses with ratio 1 : 1 : 1 : 1. test c2 c2 = 0,31 Degrees of freedom= 4 – 1 = 3 P =>90%, we can accept the hypothesis

  19. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height n° of individuals in the progeny d) In the progenywehave recessive phenotypicclasses. We can suppose that the cross isbetween: ww D d X W w D d

  20. Whatare phenotypicclassesexpected? Gene W Gene D Phenotypicclasses ww X WwDd x Dd D (3/4) W D (1/2 x 3/4 = 3/8) RedTall W (1/2) d(1/4) W d (1/2 x 1/4 = 1/8) Red short D (3/4) w D (1/2 x 3/4 = 3/8) White Tall w (1/2) d (1/4) w d (1/2 x 1/4 = 1/8) White short Weexpect 4 phenotypicclasses:3/8 : 1/8 : 3/8 : 1/8

  21. c2 = 0,31 Degrees of freedom= 4 – 1 = 3 P =70-80%, Hypothesisaccepted

  22. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant height n° of individuals in the progeny e)In the progenyweobserve recessive phenotypes. The individuals must be double heterozygous: Ww D d X W w D d From a dihybrid cross weexpect 4 phenotipicclasses: 9/16 WD (Redtall) : 3/16 Wd(Red short) : 3/16 wD(whitetall) : 1/16 wd(white short).

  23. c2 = 0,34 Degrees of freedom= 4 – 1 = 3 P >90%, hypothesisaccepted

  24. Crossing a pure line of melons whit white and spherical fruits with an other pure line of melons with yellow and flat fruits the f1 progeny is melons with white and spherical fruits. Crossing two plants of F1 we have: Determine the genotypes of individuals of P1, P2 e F1, make the hypothesis of traits segregation and verify the results with X2 P1 P2 RR GG rrgg X RrGg F1 • 148 Plants with white and spherical fruits • 52 Plants with yellow and spherical fruits F2 • 49 Plants with white and flat fruits • 23 Plants with yellow and flat fruits From first cross weknowthatsphericalisdominant over flat and whiteisdominant over yellow.WE decide to nameR the gene that control the shapemelon and G the gene that control the color melon: the first parentsishomozygousdominant and the secondparentishomozygous recessive. The progenywill be heterozygous for bothgenes.

  25. In F2 wehave 4 phenotipicclasses. Thisis a dihybrid cross and weexpectfrequencies9 : 3 : 3 : 1. test c2. c2 = 2,38 Degrees fo freedom= 4 – 1 = 3 P=30-50%, hypothesisaccepted

  26. Using the branch diagram, determine the gametes produced by individuals with the following genotype (A, B and C genes are independent): a) Aa Bb CC Gene A Aa A (1/2) a (1/2) Gene B Bb B (1/2) b (1/2) B (1/2) b (1/2) Gene C CC C (1) C (1) C (1) C (1) Gametes ABC (½ X ½ X 1= 1/4) AbC (½ X ½ X 1= 1/4) aBC (½ X ½ X 1= 1/4) abC (½ X ½ X 1= 1/4)

  27. b) Aa bb Cc Dd Gene B bb b (1) b (1) Gene A Aa A (1/2) a (1/2) Gene C Cc C (1/2) c (1/2) C(1/2) c(1/2) Gametes AbCD (½ X 1 X ½ X ½ = 1/8) AbCd (½ X 1 X ½ X ½ = 1/8) AbcD (½ X 1 X ½ X ½ = 1/8) Abcd (½ X 1 X ½ X ½ = 1/8) abCD (½ X 1 X ½ X ½ = 1/8) abCd (½ X 1 X ½ X ½ = 1/8) abcD (½ X 1 X ½ X ½ = 1/8) abcd (½ X 1 X ½ X ½ = 1/8) Gene D Dd D (1/2) d (1/2) D (1/2) d (1/2) D (1/2) d (1/2) D(1/2) d(1/2)

  28. c) Aa Bb Cc Gene A Aa A (1/2) a (1/2) Gametes ABC (½ X ½ X ½ = 1/8) ABc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8) aBC (½ X ½ X ½ = 1/8) aBc (½ X ½ X ½ = 1/8) abC (½ X ½ X ½ = 1/8) abc (½ X ½ X ½ = 1/8) Gene B Bb B (1/2) b (1/2) B (1/2) b (1/2) Gene C Cc C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) C(1/2) c(1/2)

  29. For the following cross, determine the phenotypic classes expected in the progeny and their frequencies (A, B and C genes are independent) AaBbCc x aabbcc Phenotypes for Gene C Cc X cc C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) C (1/2) c (1/2) Phenotypes for Gene A Aa X aa A (1/2) a (1/2) Phenotypes for Gene B Bb X bb B (1/2) b (1/2) B (1/2) b (1/2) FinalPhenotypes ABC (½ X ½ X ½ = 1/8) ABc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8) aBC (½ X ½ X ½ = 1/8) aBc (½ X ½ X ½ = 1/8) AbC (½ X ½ X ½ = 1/8) Abc (½ X ½ X ½ = 1/8)

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