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1 Mendel’s Laws. 2 Law of Dominance In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation.

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Presentation on theme: "1 Mendel’s Laws. 2 Law of Dominance In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation."— Presentation transcript:

1 1 Mendel’s Laws

2 2 Law of Dominance In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. All the offspring will be heterozygous and express only the dominant trait. RR x rr yields all Rr (round seeds)

3 3 Law of Dominance

4 4 Law of Segregation During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other. Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring.

5 5 Applying the Law of Segregation

6 6 P

7 7 Please pick up a copy of dihybrid practice on the front desk. Get out a sheet of notebook paper.

8 8 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.

9 9 Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Mendel’s “Law of Independent Assortment” a. Each pair of alleles segregates independently during gamete formation

10 10

11 11 Dihybrid Cross Traits: Seed shape & Seed color Alleles: Alleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry All possible gamete combinations

12 12 Dihybrid Cross RYRyrYry RYRy rY ry

13 13 Dihybrid Cross RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RYRyrYryRY Ry rY ry

14 14 Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1

15 15 Test Cross A mating between an individual of unknown genotype and a homozygous recessive individual. Example: bbC__ x bbcc BB = brown eyes Bb = brown eyes bb = blue eyes CC = curly hair Cc = curly hair cc = straight hair bCb___bc

16 16 Test Cross Possible results: bCb___bcbbCc CbCb___bc bbccor c

17 17 Summary of Mendel’s laws LAW PARENT CROSS OFFSPRING DOMINANCE TT x tt tall x short 100% Tt tall SEGREGATION Tt x Tt tall x tall 75% tall 25% short INDEPENDENT ASSORTMENT RrGg x RrGg round & green x round & green 9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods

18 18 Please get out the llama dihybrid cross from yesterday and a piece of notebook paper.

19 19 Incomplete Dominance and Codominance

20 20 Incomplete Dominance F1 hybrids in betweenphenotypes F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example:snapdragons (flower) Example: snapdragons (flower) red (RR) x white (WW) RR = red flower WW = white flower R R WW

21 21 Incomplete Dominance RWRWRWRW R RW All RW = pink (heterozygous pink) produces the F 1 generation W

22 22 Incomplete Dominance

23 23 Codominance Two alleles are simultaneously expressed (multiple alleles) in heterozygous individuals. Example: blood type 1.type A= I A I A or I A i 2.type B= I B I B or I B i 3.type AB= I A I B 4.type O= ii

24 24 Codominance Problem Example:homozygous male Type B (I B I B ) x heterozygous female Type A (I A i) IAIBIAIB IBiIBi IAIBIAIB IBiIBi 1/2 = I A I B 1/2 = I B i IBIB IAIA i IBIB

25 25 Another Codominance Problem Example:Example: male Type O (ii) x female type AB (I A I B ) IAiIAiIBiIBi IAiIAiIBiIBi 1/2 = I A i 1/2 = I B i i IAIA IBIB i

26 26 Codominance Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents? boy - type O (ii) X girl - type AB (I A I B )

27 27 Codominance Answer: IAIBIAIB ii Parents: genotypes genotypes = I A i and I B i phenotypes phenotypes = A and B IBIB IAIA i i


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