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 Key: R = red flower W = white flower Cross: RR x WW 3 WW R R Genotypic Ratio: Phenotypic Ratio: Answer on next slide…

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

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

INCOMPLETE DOMINANCE 6 Practice: Cross a pink snapdragon with a red snapdragon. Give the expected phenotypic and genotypic ratios. Key: R = red W = white RW = pink Cross: _RW x RR__ Genotypic Ratio: __2RR: 2RW: 0WW__ Phenotypic Ratio: __2 red: 2 pink: 0 white_ R W RR RRRR RW RW

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 Key: B = black feathers W = white feathers Cross: BB x WW 8 WW B B Genotypic Ratio: Phenotypic Ratio: Answer on next slide…

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

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

CODOMINANCE 11 Practice: Cross two black and white feathered chickens. Give the expected phenotypic and genotypic ratios of the offspring. Key: B = black W = white BW = black and white Cross: _BW x BW__ Genotypic Ratio: __1BB: 2BW: 1WW_ Phenotypic Ratio: _1 black: 2 black and white: 1 white_ B W BW BBBW BW WW

 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:  Alleles: R round r wrinkled Y yellow y green 14 RrYy x RrYy RY Ry rY ry All possible gamete combinations

DIHYBRID CROSS 15RYRyrYry RYRy rY ry Gametes are placed in the punnett square

DIHYBRID CROSS 16RYRyrYry RYRy rY ry Put R’s back together and Y’s back together RRYY Continue filling in the Punnett Square….

DIHYBRID CROSS 17 RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy RYRyrYryRY Ry rY ry Genotypic Ratio: 1 RRYY: 2 RRYy: 2 RrYY: 1 RRyy: 4 RrYy: 2 Rryy: 2 rrYy: 1 rrYY 1 rryy

DIHYBRID CROSS 18 RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Phenotypic Ratio: Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 RYRyrYryRY Ry rY ry

 If a pea plant with genotype RRYy (round, yellow peas) is crossed with a pea plant with genotype rrYy (wrinkled, yellow peas), what would the results be?  Key:  Cross:  Gametes:  Genotype:  Phenotype: 19

OTHER PATTERNS OF INHERITANCE 20

 “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

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