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MENDEL’S EXPERIMENTS. CONTRASTING TRAITS OF PEA PLANTS.

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Presentation on theme: "MENDEL’S EXPERIMENTS. CONTRASTING TRAITS OF PEA PLANTS."— Presentation transcript:

1 MENDEL’S EXPERIMENTS

2 CONTRASTING TRAITS OF PEA PLANTS

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4 Mendel’s Principle of Dominance: When an individual is hybrid for a certain trait, only one of the two genes expresses itself. The gene that expresses itself is the dominant gene, and the other whose expression is masked is the recessive gene.

5 Mendel’s Principle of Segregation: The two genes that determine a particular trait segregate (separate) when gametes form. Half of the gametes recieve one of the genes from each pair, and the other half recieve the other gene of the pair.

6 Mendel’s Principle of Independent Assortment: During gamete formation, gene pairs that control different traits separate (segregate) independently of one another. (This is true for genes located on different chromosomes. Linked genes do not obey this principle.)

7 MONOHYBRID CROSS Example: A cross between a pure round seeded and wrinkled seeded pea plants Parents: RR X rr Gametes: Rr 1/1 F1 Generation: 1/1 Rr Phenotype:round seeded Genotype:heterozygous round seeded F1 Cross: Rr X Rr F1 Gametes: ½ R, ½ r X ½ R, ½ r fertilization F2 Generation: ¼ RR; ¼ Rr ; ¼ Rr ; ¼ rr Phenotypes :Types = 2 n = 2 1 = 2 round and wrinkled(Ratio: ¾ round, ¼ wrinkled) Genotypes :Types = 3 n = 3 1 = 3 RR ( homozygous round) Rr (heterozygous round) rr ( homozygous wrinkled) 1:2:1 3:1

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11 A DIHYBRID CROSS Example: A cross between a pure round yellow seeded and wrinkled green seeded pea plants Parents: RRYY X rryy Gametes: RY ry1/1 F1 Generation: 1/1 RrYy Phenotype:round yellow seeded Genotype:heterozygous round yellow seeded F1 Cross: Rr Yy X Rr Yy F1 Gametes: 1/4 RY, 1/4 Ry 1/4 RY, 1/4 Ry X 1/4 rY, 1/ 4 ry fertilization

12 GAMETES RYRyrYry RYRRYYRRYyRrYYRrYy RyRR YyRRyyRrYyRryy rYRrYYRrYyrrYYrrYy ryRrYyRryyrrYyrryy

13 Phenotypes : Types = 2 n = 2 2 = 4 types 1.Round yellow (R-Y) 9/16 2.Round green (R-y) 3/16 3.Wrinkled yellow (r-Y) 3/16 4.Wrinkled green (r-y) 1/16 9: 3: 3: 1 PHENOTYPE RATIO

14 Genotypes : Types = 3 n = 3 2 = 9 possible genotypes R-Y R-y r-Y r-y 1/16 RRYY 1/16 RRyy 1/16 rrYY 1/16 rryy 2/16RRYy 2/16 Rryy 2/16rrYy 2/16 RrYY 4/16 RrYy 1:2.:2:4:1:2:1:2:1 RATIO

15 EXCEPTIONS TO MENDEL’S PATTERNS OF INHERITANCE INCOMPLETE DOMINANCE, CODOMINANCE, MULTIPLE ALLELES

16 Example: A cross between red and white flowered snapdragon. P: RR (red) X WW (white) F 1 : RW(pink) F 1 Cross: RW X RW F 2 : RR RW RW WW ¼ RED 2/4 PINK ¼ WHITE 1:2:1 RATIO INCOMPLETE DOMINANCE

17 Incomplete Dominance in Snapdragon

18 RRRWWW

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20 EXAMPLE : Four o’clock flower (akşam sefası) EXAMPLE : Feather Color Andulisian Chicken P: BB(black) X WW (white) F1: BW(blue) F1 Cross:BW X BW F2: BB (black); BW(blue);BW(blue); WW(white)

21 CODOMINANCE In codominance both alleles for a trait are dominant, and organisms produced from these crosses have both characteristics of the trait.

22 Example: A cross between M blood-grouped and N blood-grouped couple. P: MM X NN F1: MN F1 Cross: MN X MN F2: 1/4 MM ; MN ; MN ; 1/4 NN *2/4 *(Both M and N proteins are produced, they are not dominant over each other) CODOMINANCE

23 GENOTYPE PHENOTYPEANTIGENSANTIBODIES MM M M NN N N MN MN M, N NO ANTIBODIES ARE PRODUCED

24 Codominance in horse coat color Example: A cross between reddish and white horses results in roan horses which have both red and white hairs in their coat.

25 P: C R C R (red) X C W C W (white) F1: C R C W ( roan coat) F1 Cross: C R C W X C R C W F2:C R C R ; C R C W ; C R C W ; C W C W red roan coat white 1 : 2 : 1 Codominance in cattle coat color Phenotype ratio = Genotype ratio = 1: 2 : 1

26 MULTIPLE ALLELES PhenotypeGenotypeAntigensAntibodies AAA or A0AAnti-b BBB or B0BAnti-a AB A, BNone 000None Anti-a, Anti-b Rh +RR, RrRh- Rh -rr-Anti-Rh

27 RECIPIENT A B AB 0 DONOR A,0 B,0 A,B,O 0  0 is the universal donor, AB is the universal receipent.  Rh (+) can receive both Rh(+) and Rh (-) blood but Rh (-) receives only Rh(-).

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30 To calculate the types of genotypes in a population for a trait having multiple alleles, use the formula n (n+1) n: number of alleles 2


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