INCOMPLETE DOMINANCE AND CODOMINANCE 1 INCOMPLETE DOMINANCE Neither

INCOMPLETE DOMINANCE AND CODOMINANCE 1

INCOMPLETE DOMINANCE Neither allele has “complete” dominance over the other; heterozygous phenotype is a blend of the 2 homozygous phenotypes � Ex: snapdragons R = red W = white RW = pink � 2

INCOMPLETE DOMINANCE Cross a red (RR) snapdragon with a White (WW) snapdragon. Give the genotypic and phenotypic ratios. Key: R = red flower W = white flower Cross: RR x WW Genotypic Ratio: W W R Phenotypic Ratio: R Answer on next slide… 3

INCOMPLETE DOMINANCE Key: R = red W = white Cross: RR x WW Genotypic Ratio: W W 0 RR: 4 RW: 0 WW R RW RW Phenotypic Ratio: 0 Red: 4 Pink: 0 White 4

INCOMPLETE DOMINANCE Practice: Cross a pink snapdragon with a red snapdragon. Give the expected phenotypic and genotypic ratios. Key: _________ Cross: _________ Genotypic Ratio: ____________ Phenotypic Ratio: ____________ 5

INCOMPLETE DOMINANCE Practice: Cross a pink snapdragon with a red snapdragon. Give the expected phenotypic and genotypic ratios. R Key: R = red W = white RW = pink Cross: _RW x RR__ R R RR RR W RW RW Genotypic Ratio: __2 RR: 2 RW: 0 WW__ Phenotypic Ratio: __2 red: 2 pink: 0 white_ 6

CODOMINANCE Codominance – Both alleles share dominance and are always expressed if present. � Ex: In chickens � B = black feathers W = white feathers BW = black AND white feathers 7

CODOMINANCE Cross a black (B) chicken with a white (W) chicken. Give the genotypic and phenotypic ratios. Key: B = black feathers W = white feathers Cross: BB x WW W W Genotypic Ratio: B Phenotypic Ratio: B Answer on next slide… 8

CODOMINANCE Key: B = black W = white Cross: BB x WW W B B W BW BW Genotypic Ratio: 0 BB: 4 BW: 0 WW Phenotypic Ratio: 0 black: 4 black and white: 0 white 9

CODOMINANCE Practice: Cross two black and white feathered chickens. Give the expected phenotypic and genotypic ratios of the offspring. Key: _________ Cross: _________ Genotypic Ratio: ____________ Phenotypic Ratio: ____________ 10

CODOMINANCE Practice: Cross two black and white feathered chickens. Give the expected phenotypic and genotypic ratios of the offspring. B Key: B = black W = white BW = black and white Cross: _BW x BW__ W B BB BW W W Genotypic Ratio: __1 BB: 2 BW: 1 WW_ Phenotypic Ratio: _1 black: 2 black and white: 1 white_ 11

LAW OF INDEPENDENT ASSORTMENT Alleles for different traits are distributed to sex cells (& offspring) independently of one another. � This law can be illustrated using dihybrid crosses. � 12

DIHYBRID CROSS Tracks the inheritance of two traits. Ex. Seed shape and seed color � Each pair of alleles segregates independently during gamete formation � 13

DIHYBRID CROSS Traits: Seed shape & Seed color � Alleles: R round r wrinkled Y yellow y green � � Rr. Yy RY Ry r. Y ry x Rr. Yy RY Ry r. Y ry All possible gamete combinations 14

DIHYBRID CROSS Gametes are placed in the punnett square RY Ry r. Y ry 15

DIHYBRID CROSS Put R’s back together and Y’s back together RY RY Ry r. Y ry RRYY Ry r. Y ry Continue filling in the Punnett Square…. 16

DIHYBRID CROSS RY Ry r. Y ry RY RRYy Rr. YY Rr. Yy Ry RRYy RRyy Rr. Yy Rryy r. Y Rr. Yy rr. YY rr. Yy Rryy rr. Yy rryy ry Genotypic Ratio: 1 RRYY: 2 RRYy: 2 Rr. YY: 1 RRyy: 4 Rr. Yy: 2 Rryy: 2 rr. Yy: 1 rr. YY 1 rryy 17

DIHYBRID CROSS RY Ry r. Y ry RY RRYy Rr. YY Rr. Yy Ry RRYy RRyy Rr. Yy Rryy r. Y Rr. Yy rr. YY rr. Yy ry Rr. Yy Rryy rr. Yy rryy Phenotypic Ratio: Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 18

DIHYBRID PRACTICE � If a pea plant with genotype RRYy (round, yellow peas) is crossed with a pea plant with genotype rr. Yy (wrinkled, yellow peas), what would the results be? Key: � Cross: � Gametes: � Genotype: � Phenotype: � 19

OTHER PATTERNS OF INHERITANCE 20

POLYGENIC TRAITS � � “Many genes” ; Many traits are controlled by more than one gene; have a variety of choices for expression. Ex: _hair color, eye color, skin tone___ 21

MULTIPLE ALLELES Genes that have more than 2 alleles � Ex: blood group. � There are 3 possible alleles for this gene. � 22