2. 1 Gregor Johann Mendel  Austrian monk  Studied the inheritance of traits in pea plants  Developed the laws of inheritance  Mendel's work was not recognized until the turn of the 20th century

3. 2 Gregor Johann Mendel  Between 1856 and 1863, Mendel cultivated and tested some 28,000 pea plants  He found that the plants' offspring retained traits of the parents  Called the “ Father of Genetics"

4. 3 Site of Gregor Mendel ’ s experimental garden in the Czech Republic

5. 4 Important Terminology  Genotype - gene combination for a trait (e.g. RR, Rr, rr)  Phenotype - the physical feature resulting from a genotype (e.g. red, white)

6. 5 More Terminology  Dominant allele-  Dominant allele- An allele that always expresses its phenotypic effect if present  Recessive allele- The allele that is masked by the dominant

7. 6 Genotype & Phenotype in Flowers Genotype of alleles: R = red flower r = yellow flower All genes occur in pairs, so 2 alleles affect a characteristic Possible combinations are: GenotypesRR Rrrr PhenotypesRED RED YELLOW

8. 7 Genotypes  Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure  Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure  Heterozygous genotype - gene combination of one dominant & one recessive allele (e.g. Rr); also called hybrid

9. 8 Homozygous Dominant – Home = same Dominant = Uppercase – Example Homozygous Dominant for Green Color GG – Homozygous Recessive – Home = same Recessive = Lowercase – Example Homozygous Recessive for albinism aa – Heterozygous – Hetero = different – Example Heterozygous for Tongue Rolling Rr

10. 9 Odds and Ends Dominant will mask the recessive in all normal situations – Example Heterozygous for Tongue Rolling – Dominant mask the recessive – Genotype = Genetic make upRr – Phenotype = Physical make up Tongue Roller

11. 10 Practice Write the following: 1. Heterozygous for Widows Peak 2. Homozygous Dominant for Tongue Rolling 3. Homozygous Recessive for Widows Peak

12. 11 Mendel’s Laws copyright cmassengale

13. 12 Law of Segregation During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other. 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. copyright cmassengale

14. 13 Applying the Law of Segregation copyright cmassengale

15. 14 Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another. Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law can be illustrated using dihybrid crosses. This law can be illustrated using dihybrid crosses. copyright cmassengale

16. 15 Generation “ Gap ” Parental P 1 Generation = the parental generation in a breeding experiment. Parental P 1 Generation = the parental generation in a breeding experiment. F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation) F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation) – From breeding individuals from the P 1 generation F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) – From breeding individuals from the F 1 generation

17. 16 Following the Generations Cross 2 Pure Plants TT x tt Results in all Hybrids Tt Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt

18. 17 Breed the P 1 generation tall (TT) x dwarf (tt) pea plants tall (TT) x dwarf (tt) pea plants T T tt

19. 18 Solution: T T tt Tt All Tt = tall (heterozygous tall) produces the F 1 generation tall (TT) vs. dwarf (tt) pea plants

20. 19 Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants tall (Tt) vs. tall (Tt) pea plants T t Tt

21. 20 Solution: TT Tt tt T t Tt produces the F 2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype 3:1 phenotype tall (Tt) x tall (Tt) pea plants

22. 21 Monohybrid Crosses

23. 22 Results of Monohybrid Crosses Inheritable factors or genes are responsible for all heritable characteristics Inheritable factors or genes are responsible for all heritable characteristics Phenotype is based on Genotype Phenotype is based on Genotype Each trait is based on two genes, one from the mother and the other from the father Each trait is based on two genes, one from the mother and the other from the father True-breeding individuals are homozygous ( both alleles) are the same True-breeding individuals are homozygous ( both alleles) are the same copyright cmassengale

24. 23 Trait: Seed Shape Trait: Seed Shape Alleles: R – Roundr – Wrinkled Alleles: R – Roundr – Wrinkled Cross: Round seeds x Wrinkled seeds Cross: Round seeds x Wrinkled seeds RR x rr RR x rr P 1 Monohybrid Cross R R rr Rr Genotype:Rr Genotype: Rr PhenotypeRound Phenotype: Round Genotypic Probability:100% Genotypic Probability: 100% Phenotypic Probability: 100%

25. 24 P 1 Monohybrid Cross Review  Homozygous dominant x Homozygous recessive  Offspring all Heterozygous (hybrids)  Offspring called F 1 generation  Genotypic & Phenotypic ratio is ALL ALIKE

26. 25 Trait: Seed Shape Trait: Seed Shape Alleles: R – Roundr – Wrinkled Alleles: R – Roundr – Wrinkled Cross: Round seeds x Round seeds Cross: Round seeds x Round seeds Rr x Rr Rr x Rr F 1 Monohybrid Cross R r rR RR rrRr Genotype:RR, Rr, rr Genotype: RR, Rr, rr PhenotypeRound & wrinkled Phenotype: Round & wrinkled G.Ratio:1:2:1 G.Ratio: 1:2:1 P.Ratio: 3:1

27. 26 F 1 Monohybrid Cross Review  Heterozygous x heterozygous  Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr  Offspring called F 2 generation  Genotypic ratio is 1:2:1  Phenotypic Ratio is 3:1

28. 27 Monohybrid Problems A heterozygous man with Widows peak is crossed with a female without widows peak. Cross the individuals Give Genotype, Phenotype and Probability for each

29. 28 Practice Problems 1.In pigs, white color is dominant & black color is recessive. Show the results of the following crosses. Provide genotype & phenotype. a. A homozygous male white pig is mated with a homozygous black pig. b. A male heterozygous white pig is crossed with a female heterozygous white pig. c. A male heterozygous white pig is mated with a homozygous black pig. 2. The gene for brown eye color is dominant to the allele for blue eyes. A man with blue eyes marries a brown eyed woman whose mother had blue eyes. What are the genotypes & phenotypes of their children?

30. 29 Monohybrid Quiz 1. In leghorn chickens, colored feathers are due to a dominant allele, white feathers are recessive. Cross a homozygous white chicken with a heterozygous chicken. Provide the ratio of possible genotypes & phenotypes. 2. The hornless trait in cattle is dominant & the horned trait is recessive. A hornless bull is mated to 3 cows. Cow A is horned & gives birth to a hornless calf. Cow B is horned & gives birth to a horned calf. Cow C is hornless & gives birth to a horned calf. What are the genotypes of the bull, the 3 cows, & the 3 calves? This does not require a Punnett square only the genotypes of your final answer.