1. 1 Mendelian Genetics

2. 2 Gregor Mendel The Father of Genetics

3. 3 Gene A discrete unit of hereditary information consisting of a specific nucleotide sequence in DNA (or RNA in some viruses)

4. 4 Alleles Alleles are alternate forms of a gene. Examples: tall and short for plant height or purple or white for flower color. Every trait has at least two alleles- one from each parent. The location of an allele on a chromosome is known as its locus (loci = plural form).

5. 5 Genotype The letters that represent (symbolize) the trait being investigated. The genetic make- up of an organism. Examples: Bb, BB, bb

6. 6 Phenotype The actual representation of the genes. The Physical appearance or traits in an organism resulting from its genetic makeup (what you see). Examples: tall, purple flower or white flower, blond hair, freckles, etc.

7. 7 Dominant The allele that is fully expressed in an organism (observed). Represented by capital letters. Tall = T

8. 8 Recessive The allele that is masked by the dominant allele. Represented by lower case letters. Short = t

9. 9 Homozygous When both alleles (letters) are the same. BB = Homozygous Dominant bb = Homozygous recessive

10. 10 Heterozygous When the alleles (letters) are different. One upper case letter and one that is lower case. Bb = Heterozygous

11. 11 Example Problem Round = R wrinkled = r If a plant has round seeds, do we know what its genotype is? It could be RR or Rr If a plant has wrinkled seeds, do we know what its genotype is? Yes, it is rr.

12. 12 Punnett Square A method for finding predicted outcomes and probabilities for offspring from any cross. A chart for predicting the traits of offspring.

13. Some more terms: P-generation is the parental generation. The p-generation produce the F1 generation. The F1 generation crossed with itself produces the F2 generation. 13

14. Example Problem 14

15. In foxes, red coat color is determined by the dominant gene R; silver-black coat is determined by the recessive gene r. A homozygous (pure) red male is crossed with a silver-black female. (The P generation). 15 1.What is the genotype of the female?

16. What are the genotype percentages of their offspring? First…make a Punnett square for showing your work 16

17. 17 Example Problem r r RR Rr

18. 18 Assignment Section 32-3

19. 19 Assignment 33-4 & 33-5

20. 20 #1 b b Bb Bb bbBb bb B = Brown b = blue 1 point

21. 21 #1 1/2 or 50% chance of blue-eyed. 1/2 or 50% chance of Brown-eyed. 1 point

22. 22 #2 t t Tt Tt ttTt tt T = Tall t = short 1 point

23. 23 #2 What fraction of offspring would be tall? 1/2 or 50% would be Tall. 1 point

24. 24 #3 w W Ww Ww WW ww W = White w = black 1 point

25. 25 #3a What fraction of the offspring will be white? 3/4 or 75% will be white. 1 point

26. 26 #3b What fraction of the offspring will be black? 1/4 or 25% will be black. 1 point

27. 27 #3c What fraction of each genotype will you get? 1/4 or 25% will be WW 1/2 or 50% will be Ww 1/4 or 25% will be ww. 1 point

28. 28 #3d What fraction of each phenotype will you get? 3/4 or 75% will be White. 1/4 or 25% will be black. 1 point

29. 29 #4 W W RR RW RR = Red WW = White RW = Roan 1 point

30. 30 #4 Give the fraction of each color of offspring? 1/1 or 100% will be Roan - RW. 1 point

31. 31 #5 W W RW RW WWRW WW RR = Red WW = White RW = Roan 1 point

32. 32 #5 Give the fraction of each color of offspring? 1/2 or 50% will be Roan - RW. 1/2 or 50% will be White - WW. 1 point

33. 33 #1a g G Gg Gg GG gg G = Green g = red 1 point

34. 34 #1a Give the fraction of each genotype of offspring? 1/4 or 25% will be GG. 1/2 or 50% will be Gg. 1/4 or 25% will be gg. 1 point

35. 35 #1b g g Gg Gg ggGg gg G = Green g = red 1 point

36. 36 #1b Give the fraction of each genotype of offspring? 1/2 or 50% will be Gg. 1/2 or 50% will be gg. 1 point

37. 37 #2a What is the fraction of each phenotype? 3/4 or 75% will be Green. 1/4 or 25% will be red. 1 point

38. 38 #2b What is the fraction of each phenotype. 1/2 or 50% will be Green 1/2 or 50% will be red. 1 point

39. 39 # 3 The mother had to be heterozygous or Bb since the couple had a blue eyed child. b B bb bb Bb bb 1 point

40. 40 # 4 There would be a 50% chance that the 2nd child from the couple would have a brown eyes. 1 point

41. 41 # 5 Spotted = S white = s Mother = ss since she is white. Father’s Genotype would be Ss. Father’s Phenotype would be Spotted The couple had two spotted and two white kittens. 2 points

42. 42 # 6 Man has blue eyes - bb. G-Ma has blue eyes - bb. Woman has brown eyes - Bb. 50% of children would be Bb. (Brown) 50% of children would be bb. (Blue) 5 points

43. 43 Please put a score on top of their paper. Put the number correct out of 28

44. 44 Punnett Squares Why are punnett squares useful? We can use a punnett square to predict the probable genotypes and phenotypes for offspring from a genetic cross. Genotype = What is inside the genes, the make-up. Phenotype = The outward expression of the genes.