1. Dihybrid Crosses

2. 1.Principle of Dominance – dominant trait hides or masks the recessive trait 2.Law of Segregation – when gametes form, the two alleles responsible for the individual’s traits separate from each other; the alleles then recombine in fertilization 3.Law of Independent Assortment - Alleles for different traits are distributed to sex cells independently of one another. Recall Mendel’s Laws

3. Mendel’s Discoveries: Law of Independent Assortment Alleles for different genes segregate independently during meiosis – the formation of gametes In other words: If a gamete gets T or t, this doesn’t affect whether it gets B or b. Any combo is possible: Parent: TtBb Possible gametes:TBTbtBtb (each gamete has a trait for eye color & height) This creates genetic diversity between gametes, and therefore a greater diversity of offspring.

4. Dihybrid Cross 4 The Law of Independent Assortment can be applied using a Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Di = two A different letter will represent each trait. These follow Mendel’s laws, so for each trait there will be a dominant (capital) and a recessive (lowercase) allele. Mendel’s “Law of Independent Assortment” ▫a. Each pair of alleles segregates independently during gamete formation ▫b. Formula: 2 n (n = number of heterozygous pairs)

5. Determining the number of gametes 2 n = G n= number of heterozygous pairs or hybrid pairs G= the number of different gametes the parent can produce. The value of G will inform how many “slots” each parent must have within the Punnett square. RRYY RRYy RrYy 2 0 = 1 2 1 = 2 2 2 = 4 1 combo. possible 2 combos possible 4 combos possible You will work this equation for each parent.

6. Determining the alleles for each gamete To determine the alleles, you will use the FOIL method. F – first O – outer I – inner L – last Example: BbDd BD Bd bD bd I

7. Question: How many gametes will be produced for the following allele arrangements? What are the possible gametes? 7 Remember: 2 n (n = # of heterozygotes) & FOIL 1.RrYy 2. Aabb 3.mmtt

8. Answer: 8 1.RrYy: 2 n = 2 2 = 4 gamete 2. Aabb : 2 n = 2 1 = 2 gametes 3. mmtt: 2 n = 2 0 = 1 gametes RY Ry rY ry Ab ab mt

9. So what will the steps look like? Step 1: Trait(s) Step 2: Alleles Step 3: Cross **Step 4: Gametes (2 n ) for each parent Step 5: Punnett Step 6 Genotype Step 7 Phenotype

10. Cross two heterozygous round, heterozygous yellow pea plant 10 Traits: Seed shape & Seed color Alleles: R = round, r = wrinkled, Y = yellow, y = green RrYy x RrYy RY Ry rY ry You will cross All possible gamete combinations.

11. Dihybrid Cross 11 RYRyrYry RY Ry rY ry

12. 12 Phenotype: Round/Yellow: Round/green: wrinkled/Yellow: wrinkled/green: Dihybrid Cross Genotype: RRYY – RRYy – RrYY – RrYy – RRyy - Rryy - rrYY - rrYy - rryy - l ll llll l ll l ll l 1:1:2:4:1:1:2:1:2:1 9 3 1 9:3:3:1

13. Cross a homozygous brown eyed, dimpled cheeked female with a blue eyed male without dimples. (dimples are dominant) Step 1: Trait(s) – eye color, dimples Step 2: Alleles B = brown eyes, b = blue eyes D = dimples, d = no dimples Step 3: Cross BBDD x bbdd **Step 4: Gametes (2 n ) for each parent/FOIL BBDD – 2 0 = 1 bbdd – 2 0 = 1 Step 5: Punnett Step 6 Genotype 1 BbDd – 100% 1:0 Step 7 Phenotype 1 brown eyed with dimples - 100% 1:0 BD bd BD bd BbDd

14. Cross a heterozygous brown eyed, homozygous no dimpled man with a heterozygous brown eyed, heterozygous dimpled woman. Write this on your paper! Step 1: Trait(s) – _________ _________ Step 2: Alleles Step 3: Cross **Step 4: Gametes (2 n ) for each parent/FOIL

15. Step 5 Punnett Square x BDBdbDbd Bd bd Step 6 Genotype/Ratio BBDd– BBdd – BbDd – Bbdd – bbDd – bbdd – 1:1:2:2:1:1 Step 7 Phenotype/Ratio Brown eyed, dim- Brown eyed, no dim- Blue eyed, dim- Blue eyed, no dim- 3:3:1:1 BBDdBBddBbDdBbdd BbDdBbddbbDdbbdd I II I III I

16. Some hints on dihybrid crosses The number of traits is the key indicator. If there are two traits, it’s a dihybrid! Sometimes the crosses can be simple, and other times complex. Multiply the number of possible gametes and this will tell the number of possible offspring. In this class you will have no more than 16 possible offspring. (4 x 4 = 16 - The Punnett square will be 4 x 4!)

17. The End

19. Step 5 Punnett Square x BDBdbDbd BD Bd bD bd Step 6 Genotype/Ratio 1 BBDD 2 BBDd 2 BbDD 4 BbDd 1 BBdd 2 bbDD 2 bbDd 1 bbdd 1:2:2:4:1:2:1:2:1 BBDDBBDdBbDDBbDd BBDdBBddBbDdBBDD BbDDBbDdbbDDbbDd BbDdBbddbbDdbbdd

20. Step 5 Punnett Square Step 7 Phenotype/Ratio 9 brown eyes w/dimples 3 brown eyes w/o dimples 3 blue eyes w/dimples 1 blue eyes w/o dimples 9:3:3:1