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Dihybrid Cross

Dihybrid Cross. A cross between two true-breeding parents that possess different forms (alleles) of two genes. true-breeding plant with wrinkled green seeds. true-breeding plant with round yellow seeds. X. Y. y. Y. Y. y. y. r. r. r. R. R. R. Non linked genes. DNA REPLICATION.

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Dihybrid Cross

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  1. Dihybrid Cross A cross between two true-breeding parents that possess different forms (alleles) of two genes true-breeding plant with wrinkled green seeds true-breeding plant with round yellow seeds X

  2. Y y Y Y y y r r r R R R Non linked genes DNA REPLICATION HOMOLOGOUS PAIRS Genotype = RrYy

  3. Y Y y Y Y y y y ry rY Ry RY r r r r R R R R Non linked genes Meiosis 1 Gametes

  4. R = allele for round r = allele for wrinkled Y = allele for yellow y = allele for green RRYY X rryy Original cross gametes All RY All ry F1 All RrYy Second cross RrYy x RrYy F1 Self-fertilised gametes RY Ry rY ry RY Ry rY ry F2 (phenotypic ratio) 9 round yellow 3 round green 3 wrinkled yellow 1 wrinkled green 9:3:3:1 ratio RY Ry rY ry RRYY RRYy RrYY RrYy RY RRyY RRyy RryY Rryy Ry rrYY rrYy rY rRYY rRYy rryY rryy ry rRyY rRyy

  5. Recombination • In a dihybrid cross two of the F2 phenotypes resemble the original parents • Two display new combinations • The process by which new combinations of parental characteristics arise is called recombination • The individuals possessing them are called recombinants Original parents rryy RRYY Recombinants RRyy rrYY

  6. Mendel’s second law • The principle of independent assortment During gamete formation, the alleles of a gene segregate into different gametes independently of the segregation of the two alleles of another gene

  7. Dihybrid cross 2 Let R = Red let r = white Let S = straight let s = curly RrSs x RrSs Possible gametes RS,Rs,rS,rs x RS,Rs,rS,rs 9/16 red straight 3/16 red curly 3/16 white straight 1/16 white curly 800 offspring red straight = 450 red curly = 150 white straight = 150 white curly = 50 X rs RS Rs rS RS RRSS RRSs RrSS RrSs RRSs RRss Rs RrSs Rrss RrSs rrSS RrSS rrSs rS rs RrSs Rrss rrSs rrss

  8. Dihybrid cross 3 Let H= hairless, h = hairy Let T = tall, t = dwarf HhTt x HhTt Possible gametes HT,Ht,hT,ht x HT,Ht,hT,ht 9/16 hairless tall 3/16 hairless dwarf 3/16 hairy tall 1/16 hairy dwarf 1280 offspring Hairless tall = 720 Hairless dwarf = 240 Hairy tall = 240 Hairy dwarf = 80 X ht HT Ht hT HT HHTT HHTt HhTT HhTt HHTt HHtt Ht HhTt Hhtt HhTt hhTT HhTT hhTt hT ht HhTt Hhtt hhTt hhtt

  9. Dihybrid cross 4 Let P = purple, p = cut Let N = normal, n = twisted PpNn x PpNn Possible gametes PN,Pn,pN,pn x PN,Pn,pN,pn 6400 offspring Homozygous (4/16) = 1600 Purple (12/16) = 4800 Cut (12/16) = 1600 Twisted(4/16) = 1600 X pn PN Pn pN PN PpNN PPNN PPNn PpNn Pn PPNn Ppnn PPnn PpNn PpNn ppNn PpNN pN ppNN pn PpNn Ppnn ppNn ppnn

  10. Dihybrid Cross 4 • Let R = Red let r = white • Let S = straight let s = curly • rrSS x RRss • Gametes • rS x Rs • F1 • 100% RrSs • 100% Red straight

  11. Dihybrid Test cross 1 • Let H= hairless, h = hairy • Let T = tall, t = dwarf HHTT x hhtt • Gametes • HT x ht • F1 100% HhTt • F1 100% hairless tall

  12. Dihybrid test cross 2 • Let P = purple, p = cut • Let N = normal, n = twisted • PpNn x ppnn • PN,Pn,pN,pn x pn • 1000 offspring • in 1:1:1:1ratio • Purple normal (¼) = 250 • Purple twisted (¼) = 250 • Cut normal (¼) = 250 • Cut twisted (¼) = 250

  13. R r R R r r Y y Y Y y y Linked genes 1 HOMOLOGOUS PAIR DNA REPLICATION Genotype = RrYy

  14. R R r r y Y Y y RY ry Ry rY Linked genes 2 Gametes Meiosis 1 Large numbers of parental gametes X Small numbers of recombinant gametes Crossing over X

  15. H h H H h h t T t t T T Linked genes 3 HOMOLOGOUS PAIR DNA REPLICATION Genotype = HhTt

  16. Crossing over

  17. Frequency of crossing over Chiasmata can occur at anypoint along a chromosome More crossing over (recombination) occurs between two distantly located genes than two that are close together Only cross over 1 would break the link between A and B or a and b Any one of crossovers 2,3,4 and 5 would break the link between genes B/b and C/c

  18. Linked genes 4 • Let R = Red let r = white • Let S = straight let s = curly • The genes are linkedR and S , r and s • RrSs x RrSs • Possible gametes • RS , rs x RS , rs • 6400 offspring • ¾ Red straight = 4800 • ¼ white curly = 1600 • Small numbers of red curly and white straight by crossing over

  19. Linked genes test cross • Let H= hairless, h = hairy • Let T = tall, t = dwarf • Genes T and H and t and h are linked on the same chromosome HhTt x hhtt • Possible gametes • HT , ht x ht • 1000 offspring • in 1:1 ratio • 50% hairless tall = 500 ( 460) • 50% hairy dwarf = 500 ( 445) • Small number of recombinants by crossing over • Hairless dwarf ( 45) • Hairy dwarf (50)

  20. H H h h T t t T Ht hT HT ht Linked genes 5 Gametes Meiosis 1 Large numbers of parental gametes X Small numbers of recombinant gametes Crossing over X

  21. Linked genes 6 • Let P = purple, p = cut • Let N = normal, n = twisted • P and n and p and N are on the same chromosome • PpNn x ppnn • Possible gametes • Pn , pN x pn • 1280 offspring • 50% purple twisted = 610 • 50% cut normal = 600 • Small numbers of recombinantphenotypespurple normal (33) and cut twisted (37) by crossing over

  22. Gametes and Ratios

  23. Genetics Strategy • Monohybrid • One characteristic (two alleles, two phenotypes) • e.g. Eye colour-Red eyes and White eyes • Dominance – Rr x Rr 3 red : 1 white Rr x rr 1:1 ratio • Co dominance – two alleles, three phenotypes RR (red) RW (pink) and WW (white) • Multiple alleles – more than two alleles e.g. ABO blood groups • Sex linkage – show X and Y chromosome XRY x XRXr1:1:1:1 ratio

  24. Genetics Strategy • Dihybrid • Two characteristics, four phenotypes • Colour and Size • A = red, a = green S = short, s = long • Non linkage • AaBb x AaBb4 phenotypes 9:3:3:1 • AaBb x aabb4 phenotypes 1:1:1:1 • Linkage • AaBb x AaBb2 phenotypes 3:1 • AaBb x aabb2 phenotypes 1:1 Red short Red long Green short Green long Red short (A+B) Green long (a+b) 0R Red long (A+b) Green short (a+B)

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