MENDEL’S EXPERIMENTS. CONTRASTING TRAITS OF PEA PLANTS.

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MENDEL’S EXPERIMENTS

Transcript of MENDEL’S EXPERIMENTS. CONTRASTING TRAITS OF PEA PLANTS.

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MENDEL’S EXPERIMENTS

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CONTRASTING TRAITS

OF PEA PLANTS

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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.

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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.

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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.)

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MONOHYBRID CROSSExample: A cross between a pure round seeded and wrinkled seeded pea plants

Parents: RR X rr

Gametes: R r1/1 1/1

F1 Generation: 1/1 RrPhenotype: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 = 2n = 21 = 2 round and wrinkled(Ratio: ¾ round, ¼ wrinkled)

Genotypes :Types = 3n = 31 = 3 RR ( homozygous round)

Rr (heterozygous round)

rr ( homozygous wrinkled)

1:2:1

3:1

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A DIHYBRID CROSSExample: A cross between a pure round yellow seeded and wrinkled green seeded pea plants

Parents: RRYY X rryy

Gametes: RY ry1/1 1/1

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 1/4 rY , 1/ 4 ry

fertilization

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GAMETES

RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RR Yy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

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Phenotypes :

Types = 2n = 22 = 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

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Genotypes :

Types = 3n = 32 = 9 possible genotypes

R-Y R-y r-Y r-y

1/16 RRYY 1/16 RRyy 1/16 rrYY 1/16 rryy

2/16 RRYy 2/16 Rryy 2/16rrYy

2/16 RrYY

4/16 RrYy

1:2.:2:4:1:2:1:2:1

RATIO

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EXCEPTIONS TO MENDEL’S PATTERNS OF INHERITANCE

INCOMPLETE DOMINANCE,

CODOMINANCE,

MULTIPLE ALLELES

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Example: A cross between red and white flowered snapdragon.

P: RR (red) X WW (white)

F1: RW(pink)

F1 Cross: RW X RW

F2 : RR RW RW WW

¼ RED 2/4 PINK ¼ WHITE

1:2:1 RATIO

INCOMPLETE DOMINANCE

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Incomplete Dominance in Snapdragon

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RR RW WW

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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)

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

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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

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GENOTYPE PHENOTYPE ANTIGENS ANTIBODIES

MM M M -----

NN N N -----

MN MN M, N -----

NO ANTIBODIES ARE PRODUCED

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

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P: CR CR (red) X CW CW (white)

F1: CR CW ( roan coat)

F1 Cross: C R C W X C R C W

F2:C R CR ; CR CW; CR CW ; C WC W

red roan coat white 1 : 2 : 1

Codominance in cattle coat color

Phenotype ratio = Genotype ratio = 1: 2 : 1

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MULTIPLE ALLELES

Phenotype Genotype Antigens Antibodies

A AA or A0 A Anti-b

B BB or B0 B Anti-a

AB AB A, B None

0 00 None Anti-a, Anti-b

Rh + RR, Rr Rh -

Rh - rr - Anti-Rh

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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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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