Mendelian Genetics 12/8/2018 Dihybrid Cross A breeding experiment that tracks the inheritance of two traits: 2 genes = 4 alleles Mendel’s “Law of Independent Assortment” Each pair of alleles aligns during meiosis and separates independently during gamete formation Ss Bb SB Sb sB sb Alleles independently assort during meiosis

Mendel’s next question… Will I see the same inheritance patterns and predictability in crosses involving more than 1 gene?

Parents are homozygous for both traits Dihybrid crosses f1 Traits: Seed shape & Seed color Alleles: R =round r = wrinkled Y = yellow y = green Parents are homozygous for both traits RRYY x rryy How will the alleles independently assort for each parent? To help determine how the alleles assort use the math order of operations F O I L First-Outside-Inside-Last RY RY RY RY and ry ry ry ry What would a Punnett square look like for a dihybrid cross?

Dihybrid punnett F1 ry RY RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy Mendelian Genetics 12/8/2018 Dihybrid punnett F1 Each parent can produce 4 combinations of alleles( 24 = 16) The Punnett square in a dihybrid cross will have 16 boxes Parent alleles are placed in the same way… ry Then the alleles are combined to produce the offspring; making sure to keep the alleles in the same order as the parent’s RY RrYy RrYy RrYy RrYy Genotypic ratio 16:16 RrYy Phenotypic ratio round/yellow RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy RrYy Just like with a monohybrid cross ; all genotypes are heterozygous (RrYy) and all phenotypes are dominant (Round/Yellow) for both traits

Dihybrid Cross RrYy x RrYy Mendelian Genetics 12/8/2018 Dihybrid Cross Traits: Seed shape & Seed color Alleles: R =round r = wrinkled Y = yellow y = green To help determine allele combinations use the math order of operations FOIL First-Outside-Inside-Last RrYy x RrYy RY Ry rY ry RY Ry rY ry All possible allele combinations when parents are heterozygous for both

Dihybrid punnett F2 RY Ry rY ry RY Ry rY ry RRYY RRYy RrYY RrYy Mendelian Genetics 12/8/2018 Dihybrid punnett F2 Parent alleles are placed in the same way… RY Ry rY ry Then the alleles are combined to produce the offspring; making sure to keep the alleles in the same order as the parent’s RY Ry rY ry RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Genotypic and phenotypic ratios are a bit trickier, however there is a pattern if you look closely…

Dihybrid F2 genotypic and phenotypic ratios Mendelian Genetics 12/8/2018 Dihybrid F2 genotypic and phenotypic ratios Round/Yellow: 9 RRYY 1 RRYy 2 RrYY 2 RrYy 4 Round/green: 3 RRyy 1 Rryy 2 wrinkled/Yellow:3 rrYY 1 rrYy 2 wrinkled/green: 1 rryy 1 RY Ry rY ry RRYY RRYy RrYY RrYy RRyy Rryy rrYY rrYy rryy 9:3:3:1 PHENOTYPIC RATIO

Crosses of two genes representing four alleles Dihybrid crosses

Guinea pig dihybrid crosses BBLl BbLL BbLl BBLl BBll BbLl Bbll BbLL BbLl bbLL bbLl BbLl Bbll bbLl bbll Phenotypic ratios: How many out of 16 are… 9 3 3 1 Black/Short____ Black/long____ white/Short_____ white/long_____

bL bl bL bl BL Bl bL bl bL bl bL bl How many of the offspring are: Black/Short:______ Black/long: ______ white/Short:______ white/long: ______ bL bl bL bl 6 2 BL Bl bL bl BbLL BbLl BbLL BbLl BbLl Bbll BbLl Bbll bbLL bbLl bbLL bbLl bbLl bbll bbLl bbll

bL bL bL bL Bl Bl bl bl bL bL bL bL How many of the offspring are: Black/Short:______ Black/long: ______ white/Short:______ white/long: ______ bL bL bL bL 8 Bl bl BbLl BbLl BbLl BbLl BbLl BbLl BbLl BbLl bbLl bbLl bbLl bbLl bbLl bbLl bbLl bbLl

Dominant alleles determine phenotype If you pay attention to the dominant alleles in the parent genotypes…you can determine phenotypic ratios with out having to do the squares! R = purple r = white T = tall t = dwarf RrTT x rrTt 50% or 8/16 will be purple :100% or 16/16 will be tall rrTt x RRtt 100% or 16/16 will be purple: 50% or 8/16 will be tall RrTt x rrTt 50% or 8/16 will be purple: 75% or 12/16 will be tall Rrtt x RRtt 100% or 16/16 will be purple: none Tall, 100% or 16/16 short

Let’s look at a trihybrid… Can you tell what the parent genotypes were? This is an F2 so both parents are heterozygous for the 3 traits: AaBbCc X AaBbCc

Mendelian Genetics 12/8/2018 Dihybrid Cross

copyright cmassengale

copyright cmassengale

copyright cmassengale

b. Formula: 2n (n = # of heterozygotes) Mendelian Genetics 12/8/2018 copyright cmassengale Question: How many gametes will be produced for the following allele arrangements? Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq b. Formula: 2n (n = # of heterozygotes)

Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry Mendelian Genetics 12/8/2018 Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes