Dihybrid Cross A cross between two truebreeding parents

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

Non linked genes HOMOLOGOUS PAIRS Y R r DNA REPLICATION R Y y r Genotype = Rr. Yy r y y

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

Original cross RRYY gametes All RY F 1 X R = allele for round rryy r = allele for wrinkled All ry Y = allele for yellow y = allele for green All Rr. Yy Second cross gametes RY Rr. Yy x Ry r. Y ry Rr. Yy RY F 1 Self-fertilised Ry r. Y ry F 2 RY Ry r. Y ry RY RRYy Rr. YY Rr. Yy Ry RRy. Y RRyy Rry. Y Rryy r. Y r. RYy rr. YY rr. Yy ry r. Ry. Y r. Ryy rry. Y rryy (phenotypic ratio) 9 round yellow 3 round green 3 wrinkled yellow 1 wrinkled green 9: 3: 3: 1 ratio

Recombination • In a dihybrid cross two of the F 2 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 rr. YY

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

• Let R = Red let r = white • Let S = straight let s = curly Rr. Ss x Rr. Ss • Possible gametes • RS, Rs, r. S, rs x RS, Rs, r. S, 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 Dihybrid cross 2 X RS Rs r. S rs RRSS RRSs Rr. SS Rr. Ss Rs RRSs RRss r. S Rr. Ss rs Rr. Ss RS Rrss Rr. Ss Rrss rr. SS rr. Ss rrss

• Let H = hairless, h = hairy • Let T = tall, t = dwarf Hh. Tt x Hh. Tt • Possible gametes • HT, Ht, h. T, ht x HT, Ht, h. T, 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 Dihybrid cross 3 X HT Ht Ht HHTT HHTt Hh. TT Hh. Tt HHtt Hh. Tt Hhtt h. T Hh. TT ht Hh. Tt Hhtt h. T ht HT hh. Tt hhtt

• Let P = purple, p = cut • Let N = normal, n = twisted Pp. Nn x Pp. Nn • Possible gametes • PN, Pn, p. N, pn x PN, Pn, p. N, pn • 6400 offspring • Homozygous (4/16) = 1600 • Purple (12/16) = 4800 • Cut (12/16) = 1600 • Twisted(4/16) = 1600 Dihybrid cross 4 X PN Pn PN p. N pn PPNn Pp. NN Pp. Nn PPnn Pp. Nn Ppnn p. N Pp. Nn pp. Nn pn Pp. Nn Ppnn PPNN Pn PPNn pp. NN pp. Nn ppnn

Dihybrid Cross 4 • • Let R = Red let r = white Let S = straight let s = curly rr. SS x RRss Gametes r. S x Rs F 1 100% Rr. Ss 100% Red straight X Rs r. S Rr. Ss

Dihybrid Test cross 1 • Let H = hairless, h = hairy • Let T = tall, t = dwarf HHTT x hhtt • Gametes • HT x ht • F 1 100% Hh. Tt • F 1 100% hairless tall X ht HT Hh. Tt

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

Linked genes 1 HOMOLOGOUS PAIR R Y r y Genotype = Rr. Yy DNA REPLICATION R Y r y

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

Linked genes 3 HOMOLOGOUS PAIR H t h T Genotype = Hh. Tt DNA REPLICATION H t h T

Crossing over

Frequency of crossing over Chiasmata can occur at any point 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

Linked genes 4 • Let R = Red let r = white • Let S = straight let s = curly • The genes are linked R and S , r and s • Rr. Ss x Rr. Ss • 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 RS rs RS RRSS Rr. Ss rs Rr. Ss rrss

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 Hh. Tt 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) X ht HT Hh. Tt ht hhtt

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

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 • Pp. Nn x ppnn • Possible gametes • Pn , p. N x pn • 1280 offspring • 50% purple twisted = 610 • 50% cut normal = 600 • Small numbers of recombinant phenotypes purple normal (33) and cut twisted (37) by crossing over X pn Pn Ppnn p. N pp. Nn

Gametes and Ratios Linked genes Non linked genes Rr. Yy (R+Y) RY, ry RY, Ry, r. Y, ry Pp. Tt (P+T) PT, pt PT, Pt, p. T, pt Hh. Ss (H +s) Hs, h. S HS, Hs, h. S. hs Rr. Yy X Rr. Yy 3: 1 9: 3: 3: 1 Rr. Yy X rryy 1: 1: 1: 1 3: 1 9: 3: 3: 1 Hh. Ss X hhss Pp. Tt X Pp. Tt

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 X RY x X RX r 1: 1: 1: 1 ratio

Genetics Strategy • • • Dihybrid Two characteristics, four phenotypes Colour and Size A = red, a = green S = short, s = long Red short Non linkage Aa. Bb x Aa. Bb 4 phenotypes 9: 3: 3: 1 Red long Green short Aa. Bb x aabb 4 phenotypes 1: 1: 1: 1 Green long Linkage Aa. Bb x Aa. Bb 2 phenotypes 3: 1 Red short (A+B) Green long (a+b) Aa. Bb x aabb 2 phenotypes 1: 1 0 R Red long (A+b) Green short (a+B)