1.  Tuesday March 24 By the end of class today, I should be able to: 1. Summarize cell division (m itosis & meiosis) 2. Define & apply key genetics terms 3. Predict the outcome of specific genetic crosses.

2. Announcements  Remember: put your phones away!

3. Mitosis & Meiosis Comprehension Check If a parent cell undergoes mitosis, how many daughter cells will result? Two If that parent cell began with ten chromosomes, how many chromosomes will each daughter cells have? Ten If a parent cell undergoes meiosis I & II, how many daughter cells will result? Four If that parent cell began with 10 chromosomes, how many chromosomes will each daughter cell have? Five

4. How Are Features Passed Along? 4

5. 5 Gregor Mendel (1822-1884) Monk given credit for discovering the Laws governing Inheritance of Traits “Father of Genetics”

6. 6 Mendel’s Pea Plant Experiments

7. 7 Mendel’s Experimental Methods  Mendel hand-pollinated flowers using a paintbrush  He could snip the stamens to prevent self- pollination  He traced traits through the several generations

8. 8 How Mendel Began Mendel produced pure strains by allowing the plants to self- pollinate for several generations

9. 9

10. 10 Mendel’s Experimental Results

11. 11 Generation “Gap”  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)  From breeding individuals from the P 1 generation  F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation)  From breeding individuals from the F 1 generation

12. 12 Following the Generations P (parental) Cross 2 Pure Plants TT x tt F1 Results in all Hybrids Tt F2 Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt

13. 13 Genetics Terms  Genes- DNAinstructions for producing a trait  Genes- DNA instructions for producing a trait  Alleles alternative forms of the same gene that codefor the same trait(i.e. flower color)  Alleles - alternative forms of the same gene that code for the same trait (i.e. flower color)

14. 14 Genetics Terms  Homozygous condition, both alleles are the same (e.g. PP; pp)  H eterozygous condition, the alleles differ (e.g. Pp)  For a given Mendelian trait, there is a dominant allele (expressed when present) and a recessive allele (only expressed when dominant absent)  In a heterozygous condition, a Mendelian trait would consist of a dominant allele (D) and a recessive allele (d)

16. 16 Genetics Terms  Genotype: The kind of alleles (genes) an organism has  Phenotype: What the trait looks like Genotype: Pp Phenotype: Purple

17. 17 Human Mendelian Traits

18. 18 Human Mendelian Traits Recessive phenotype: Albinism Dominant phenotype: Achondroplasia

19. 19 Punnett Square Used to help solve genetics problems

20. 20 How to Make Punnett Square

21. 21 Types of Genetic Crosses  Monohybrid cross - cross involving a single trait (e.g. flower color)  Dihybrid cross - cross involving two traits (e.g. flower color & plant height)  Polyhybrid cross - cross involving three or more traits (e.g. flower color, plant height, & pod shape…)

22. 22  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Round seeds x Wrinkled seeds RR x rr P1 Monohybrid Cross R R rr Rr Genotype:Rr Genotype: Rr PhenotypeRound Phenotype: Round Genotypic Ratio:All heterozygous Genotypic Ratio: All heterozygous Phenotypic Ratio: All round

23. 23  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Round seeds x Round seeds Rr x Rr F1 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

24. 24 …And Now the Test Cross  Mendel then crossed a pure & a heterozygous from his F 2 generation  This is known as an F 2 or test cross  There are two possible testcrosses: Homozygous dominant x Heterozygous Homozygous recessive x Heterozygous

25. 25  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Round seeds x Round seeds  RR x Rr F2 Monohybrid Cross (1st) R R rR RR RrRR Rr Genotype:RR, Rr Genotype: RR, Rr PhenotypeRound Phenotype: Round Genotypic Ratio:1:1 Genotypic Ratio: 1:1 Phenotypic Ratio: Round

26. 26  Trait: Seed Shape  Alleles: R – Roundr – Wrinkled  Cross: Wrinkled seeds x Round seeds rr x Rr F2 Monohybrid Test Cross r r rR Rr rrRr rr Genotype:Rr, rr Genotype: Rr, rr PhenotypeRound & Wrinkled Phenotype: Round & Wrinkled G. Ratio:1:1 G. Ratio: 1:1 P.Ratio: 1:1

27.  27 Mendel’s Laws (Genetics Notes Part II)

28. 28 Law of Dominance If a dominant trait is present, it will be expressed and the recessive will not! e.g. RR x rr yields all Rr (R = round seeds)

29. 29 Law of Segregation “  A “Tt” parent makes an egg with “T” AND a different egg with “t”

30. 30 Applying the Law of Segregation

31. 31  Mendel: Alleles for different traits (like Tt and Pp) are distributed to sex cells (& offspring) independently of one another.  Dominant Tall plants can have Dominant Red flowers OR recessive white flowers.  This law applies when using dihybrid crosses. Law of Independent Assortment

32. 32 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 round green: 3 round yellow: 3 wrinkled green: 1 wrinkled yellow

33. 33 Incomplete Dominance RrRrRrRr R Rr All Rr = pink (heterozygous pink) produces the F 1 generation r Alleles combine as unique phenotype Alleles combine as unique phenotype Example: the protein for red and protein for white are produced in snapdragons and make an intermediate color of pink.

34. 34 Codominance  Two alleles are expressed (multiple alleles) in heterozygous individuals. All alleles equally “strong”  Example: Coat color in horses  Red x white

35. 35 Genetic Practice Problems

36. 36  tall (TT) x dwarf (tt) pea plants T T tt

37. 37 Solution: T T tt Tt All Tt = tall (heterozygous tall) produces the F 1 generation tall (TT) vs. dwarf (tt) pea plants

38. 38 Breed the F 1 generation  tall (Tt) vs. tall (Tt) pea plants T t Tt

39. 39 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

40. 40 Dihybrid Cross RYRyrYry RYRy rY ry R= round seed r= wrinkled seed Y= yellow seed y= green seed

41. 41 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

42. 42 Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1