Dihybrid Crosses Recall Mendels Laws 1 Principle of

Dihybrid Crosses

Recall Mendel’s Laws 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.

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: Tt. Bb Possible gametes: TB Tb t. B tb (each gamete has a trait for eye color & height) �This creates genetic diversity between gametes, and therefore a greater diversity of offspring.

Dihybrid Cross 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” 4 ▫ a. Each pair of alleles segregates independently during gamete formation ▫ b. Formula: 2 n (n = number of heterozygous pairs)

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 20 = 1 RRYy 21 = 2 1 combo. possible 2 combos possible Rr. Yy 22 = 4 4 combos possible You will work this equation for each parent.

Determining the alleles for each gamete • To determine the alleles, you will use the FOIL method. • • F – first BD O – outer Bd I – inner b. D L – last bd Example: I Bb. Dd

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

8 Answer: 1. Rr. Yy: 2 n= 22 = 4 gamete RY Ry r. Y ry 2. Aabb : 2 n = 21 = 2 gametes Ab ab 3. mmtt: 2 n = 20 = 1 gametes mt

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 �Traits: Seed shape & Seed color �Alleles: R = round, r = wrinkled, Y = yellow, y = green Rr. Yy x Rr. Yy RY Ry r. Y ry You will cross All possible gamete combinations.

11 Dihybrid Cross RY RY Ry r. Y ry

12 Dihybrid Cross Genotype: RRYY – RRYy – Rr. YY – Rr. Yy – RRyy Rryy rr. YY rr. Yy rryy - l l ll llll l 1: 1: 2: 4: 1: 1: 2: 1 Phenotype: Round/Yellow: Round/green: wrinkled/Yellow: wrinkled/green: 9 3 3 1 9: 3: 3: 1

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 – 20 = 1 BD bd bbdd – 20 = 1 Step 5: Punnett bd BD Bb. Dd • Step 6 Genotype 1 Bb. Dd – 100% 1: 0 • Step 7 Phenotype 1 brown eyed with dimples - 100% 1: 0

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

• Step 5 Punnett Square x Bd bd Step 6 Genotype/Ratio BD Bd b. D bd BBDd BBdd Bb. Dd Bbdd bb. Dd bbdd BBDd– BBdd – Bb. Dd – Bbdd – bb. Dd – bbdd – 1: 1: 2: 2: 1: 1 I I II II I I Step 7 Phenotype/Ratio Brown eyed, dim. Brown eyed, no dim. Blue eyed, no dim 3: 3: 1: 1 III I I

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!)

The End

• Step 5 Punnett Square x BD Bd b. D bd BD Bd Step 6 Genotype/Ratio b. D bd BBDD BBDd Bb. DD Bb. Dd BBdd Bb. Dd BBDD Bb. Dd bb. DD bb. Dd 1 BBDD 2 BBDd 2 Bb. DD 4 Bb. Dd 1 BBdd 2 bb. DD 2 bb. Dd 1 bbdd Bb. Dd bbdd 1: 2: 2: 4: 1: 2: 1 Bbdd

• 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