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

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

3. Non linked genes R R r r Y Y y y R R r r y y Y Y RYryRyrY Meiosis 1 Gametes

4. R = allele for round r = allele for wrinkled Y = allele for yellow y = allele for green Original cross RRYYXrryy gametesAll RYAll ry F1F1 All RrYy Second cross RrYy x gametesRYRyrYry F 1 Self-fertilised RYRyrYry RYRyrYry RY Ry rY ry rRyY RrYYRrYyRRYYRRYy rRYY RRyYRRyyRryy rRyy rRYy RryY rryy rrYyrrYY rryY F2 (phenotypic ratio) 9 round yellow 3 round green 3 wrinkled yellow 1 wrinkled green 9:3:3:1 ratio

5. 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 RRYY rryy RRyy rrYY Original parents Recombinants

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 RSRsrS rs RS Rs rS rs RRSS RRSs RrSS RrSsrrss RRSsRrSS RrSs RRss RrSsRrss RrSsrrSS rrSs RrssrrSs

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 HTHthT ht HT Ht hT ht HHTT HHTt HhTT HhTthhtt HHTtHhTT HhTt HHtt HhTtHhtt HhTthhTT hhTt HhtthhTt

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 PNPnpN pn PN Pn pN pn PPNN PPNn PpNN PpNnppnn PPNn PpNN PpNn PPnnPpNn Ppnn PpNn ppNN ppNn PpnnppNn

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 X rS Rs RrSs

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 X ht HT HhTt

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 X pn PN PpNn Pn Ppnn pN ppNn pn ppnn

13. Linked genes 1 RY ry HOMOLOGOUS PAIR Genotype = RrYy DNA REPLICATION RY RY ry ry

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

15. Linked genes 3 Ht hT HOMOLOGOUS PAIR Genotype = HhTt DNA REPLICATION Ht Ht hT hT

16. Crossing over

18. 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

19. 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 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 RS rs RS RRSS RrSs rs RrSs rrss

20. 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) X ht HT HhTt ht hhtt

21. Linked genes 5 Meiosis 1 Ht Ht hT hT Gametes Ht hT Large numbers of parental gametes Crossing over X X HT ht Small numbers of recombinant gametes

22. 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 recombinant phenotypes purple normal (33) and cut twisted (37) by crossing over X pn Pn Ppnn pN ppNn

23. Gametes and Ratios Linked genesNon linked genes RrYy (R+Y) RY, ryRY,Ry,rY,ry PpTt (P+T) PT, ptPT,Pt,pT,pt HhSs (H +s) Hs, hSHS,Hs,hS.hs RrYy X RrYy 3 : 1 9:3:3:1 RrYy X rryy 1:1 1:1:1:1 HhSs X hhss 1:1 1:1:1:1 PpTt X PpTt 3:1 9:3:3:1

24. 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 R Y x X R X r 1:1:1:1 ratio

25. Genetics Strategy Dihybrid Two characteristics, four phenotypes Colour and Size A = red, a = green S = short, s = long Non linkage AaBb x AaBb 4 phenotypes 9:3:3:1 AaBb x aabb 4 phenotypes 1:1:1:1 Linkage AaBb x AaBb 2 phenotypes 3:1 AaBb x aabb 2 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)