1. Adding this Class Both sections of this class are currently full.Both sections of this class are currently full. A lottery will be held for any spaces that become open during the first week of class.A lottery will be held for any spaces that become open during the first week of class. To participate in the lottery, put your name, phone number or email address and preferred lab section on an index card and submit it to the instructor.To participate in the lottery, put your name, phone number or email address and preferred lab section on an index card and submit it to the instructor. You will be contacted if space is available. You do not need to be present to win!You will be contacted if space is available. You do not need to be present to win! Lottery will be held on Thursday of this week.Lottery will be held on Thursday of this week. Name Phone or Email Lab Section: AM or PM

2. Patterns of Inheritance What patterns can be observed when traits are passed to the next generation?

3. Use of the Garden Pea for Genetics Experiments

4. Principles of Heredity F2: 5474 round: 1850 wrinkled (3/4 round to 1/4 wrinkled) Mendel’s Experiment with Peas Round seed x Wrinkled seed F1: All round seed coats F1 round plants x F1 round plants

5. Principles of Heredity Mendel needed to explain 1.Why 1.Why one trait seemed to disappear in the first generation. 2. Why the same trait reappeared in the second generation in one-fourth of the offspring.

6. Principles of Heredity Mendel proposed: 1.Each 1.Each trait is governed by two factors – now called genes. 2. Genes are found in alternative forms called alleles. 3. Some alleles are dominant and mask alleles that are recessive.

7. Principles of Heredity HomozygousDominantHomozygousRecessive Heterozygous Mendel’s Experiment with Peas Round seed x Wrinkled seed Round seed x Wrinkled seed RR rr RR rr F1: All round seed coats Rr Rr

8. RRR R Homozygous parents can only pass one form of an allele to their offspring.

9. RrRr Heterozygous parents can pass either of two forms of an allele to their offspring.

10. Principles of Heredity Additional Genetic Terms Genotype: alleles carried by an individual eg. RR, Rr, rr Phenotype: physical characteristic or appearance of an individual eg. Round, wrinkled eg. Round, wrinkled

11. Mendel’s Principle of Genetic Segregation In the formation of gametes, the members of a pair of alleles separate separate (or segregate) cleanly from each other so that only one member is included in each gamete. Each gamete has an equal probability of containing either member of the allele pair.

12. Genetic Segregation Parentals: RR x rr R R r r RR r r r r F1 x F1: Rr x Rr R r R r ½ R ½ r ½ R ½ r ½ R ½ r ¼ RR ¼ Rr ¼ rr RrRr Rr Rr

13. Genetic Segregation Genotypic Ratio: ¼ RR + ½ Rr + ¼ rr Phenotypic Ratio: ¾ Round + ¼ Wrinkled

14. Seven Traits used by Mendel in Genetic Studies

15. What Is a Gene? Agene is a segment of DNA that directs the synthesis of a specific protein. DNADNA is transcribed into RNA which is translated into protein.

16. Molecular Basis for Dominant and Recessive Alleles Dominant Allele Codes for a functional protein Recessive Allele Codes for a non- functional protein or prevents any protein product from forming

17. Principles of Heredity Mendel’s Experiment with Peas Round Yellow x Wrinkled Green F1: All round yellow seed coats F2: 315 round, yellow 9/16 108 round, green 3/16 101 wrinkled, yellow 3/16 32 wrinkled, green 1/16 F2: 315 round, yellow 9/16 108 round, green 3/16 101 wrinkled, yellow 3/16 32 wrinkled, green 1/16 F1 plants x F1 plants

18. Principles of Heredity Mendel needed to explain 1.Why 1.Why non-parental combinations appeared in the F2 offspring. 2. Why the ratio of phenotypes in the F2 generation was 9:3:3:1.

19. Mendel’s Principle of Independent Assortment When gametes are formed, the alleles of one gene segregate independently of the alleles of another gene gene producing equal proportions of all possible gamete types.

20. Genetic Segregation + Independent Assortment Parentals: Parentals: RRYY x rryy RRYY x rryy RY RY RY RY ry ry ry ry RY RY RY RY ry ry ry ry ry RY RrYy F1: 100% RrYy, round, yellow

21. F1 x F1: RrYy x RrYy RY Ry rY ry RY Ry rY ry ¼ RY ¼ Ry ¼ rY ¼ ry ¼ RY ¼ Ry ¼ rY ¼ ry 1/16 RRYY1/16 RRYy1/16 RrYY1/16 RrYy 1/16 RRYy1/16 RRyy1/16 RrYy1/16 Rryy 1/16 RrYY1/16 RrYy1/16 rrYY1/16 rrYy 1/16 RrYy1/16 Rryy1/16 rrYy1/16 rryy

22. F2 Genotypes and Phenotypes PhenotypesGenotypesRoundYellow 1/16 RRYY + 2/16 RRYy + 2/16 RrYY + 4/16 RrYy Total = 9/16 R_Y_ RoundGreen 1/16 RRyy+ 2/16 Rryy Total = 3/16 R_yy Wrinkled Yellow 1/16 rrYY+ 2/16 rrYy Total = 3/16 rrY_ Wrinkled Green 1/16 rryy

23. Meiotic Segregation explains Independent Assortment

24. Solving Genetics Problems 1.Convert 1.Convert parental phenotypes to genotypes 2.Use 2.Use Punnett Square to determine genotypes of offspring 3.Convert 3.Convert offspring genotypes to phenotypes

25. Using Probability in Genetic Analysis 1. Probability (P) of an event (E) occurring: P(E) = Number of ways that event E can occur Total number of possible outcomes Eg. P(Rr) from cross Rr x Rr 2 ways to get Rr genotype 4 possible outcomes P(Rr) = 2/4 = 1/2

26. Using Probability in Genetic Analysis 2. Addition Rule of Probability – used in an “either/or” situation Eg. P(Rr or RR) from cross Rr x Rr 2 ways to get Rr genotype 1 way to get RR genotype 4 possible outcomes P(Rr or RR) = 2/4 + 1/4 = 3/4 P(E 1 P(E 1 or E 2 ) E 2 ) = P(E 1 ) P(E 1 ) + P(E 2 )

27. Using Probability in Genetic Analysis 3. Multiplication Rule of Probability – used in an “and” situation Eg. P(wrinkled, yellow) from cross RrYy x RrYy P(rr and Y_) = 1/4 x 3/4 = 3/16 P(E 1 P(E 1 and E 2 ) E 2 ) = P(E 1 ) P(E 1 ) X P(E 2 )

28. Using Probability in Genetic Analysis 4. Conditional Probability: Calculating Calculating the probability that each individual has a particular genotype Eg. Jack and Jill do not have PKU. Each has a sibling with the disease. What is the probability that Jack and Jill will have a child with PKU?

29. Pp P p P pPPPp Pp pp X Using Probability in Genetic Analysis 4. Conditional Probability Jack is P_, Jill is P_ Jack is P_, Jill is P_ Parents of Jack or Jill: Pp x Pp Parents of Jack or Jill: Pp x Pp P(Pp) = 2 ways to get Pp 3 possible genotypes P(Jack is Pp) =2/3 P (Jill is Pp) = 2/3

30. Using Probability in Genetic Analysis 4. Conditional Probability P(child with PKU)= P(child without PKU)= 1-1/9 = 8/9 P(Jack is Pp) x P(Jill is Pp) x P(child is pp) = 2/3 x 2/3 x 1/4 = 1/9

31. JackJill P_ child Probability 1/3 PP 11/9 2/3 Pp 12/9 1/3 PP 12/9 2/3 Pp 3/43/9 Total=8/9 To calculate probability of child without PKU, look at all possibilities for Jack and Jill.

32. Using Probability in Genetic Analysis 5. Ordered Events: use Multiplication Rule 5. Ordered Events: use Multiplication Rule For Jack and Jill, what is the probability that the first child will have PKU, the second child will not have PKU and the third child will have PKU? P(pp) x P(P_) x P(pp) = 1/9 x 8/9 x 1/9 = 8/729

33. Using Probability in Genetic Analysis 6. Binomial Rule of Probability – used for unordered events a = probability of event X (occurrence of one event) b = probability of event Y = 1-a (occurrence of alternate event) n = total s = number of times event X occurs t = number of times event Y occurs (s + t = n) P = n! (a s b t ) s! t!

34. Using Probability in Genetic Analysis 6. Binomial Rule of Probability ! = factorial= number multiplied by each lower number until reaching 1 5! = 5 x 4 x 3 x 2 x 1 1! =1 3! = 3 x 2 x 1 = 3 x 2! 0! = 1 2! = 2 x 1

35. Using Probability in Genetic Analysis 6. Binomial Rule of Probability Out of 3 children born to Jack and Jill, what is the probability that 2 will have PKU? 3! (1/9) 2 (8/9) 1 = 3 x 2! (1/81) (8/9)= 24 3! (1/9) 2 (8/9) 1 = 3 x 2! (1/81) (8/9)= 24 2! 1! 2! 1! 729 n=3, a=1/9, s=2, b=8/9, t=1

36. Using Probability in Genetic Analysis The same result can be obtained using the multiplicative rule if all possible birth orders for families of three are considered: 1 st child 2 nd child 3 rd child Probability PKU=1/9 No= 8/9 PKU=1/98/729 No=8/9PKU=1/9PKU=1/98/729 PKU=1/9PKU=1/9No=8/98/729 8/729 + 8/729 + 8/729 = 24/729

37. Chi-Square Goodness of Fit Test To evaluate how well data fits an expected genetic ratio

38. Chi-square Test for Goodness of Fit for 9:3:3:1 Ratio Phenotype Observed Number ExpectedNumber (Fraction x Total) O-E (O-E) 2 E E Round, yellow 315 Round, green 108 Wrinkled, yellow 101 Wrinkled, green 32 Total556 9/16 x 556 = 313 3/16 x 556 = 104 1/16 x 556 = 35 2 4 -3 -3 4 16 9 9.0128.0128.154.154.087.087.257.257 X 2 =.511 df=degrees of freedom= number of phenotypes – 1 = 4-1=3 p value from table on page 1-17: p>.5 from table in Pierce:.975 > p >.9 Data supports hypothesis for any p>0.05

39. X2X2 p

40. Sex Determination Sex Chromosomes : homologous chromosomes that differ in size and genetic composition between males and females Sex Chromosomes : homologous chromosomes that differ in size and genetic composition between males and females HumanChromosomeXY Size and gene number Larger 2142 genes Smaller 326 genes GeneticComposition Multiple genes unrelated to gender Single gene influences gender: TDF = testis determining factor

41. Sex Determination Autosome: any chromosome that is not a sex chromosome Autosome: any chromosome that is not a sex chromosome Humans have 22 pairs of autosomes and 1 pair of sex chromosomes Humans have 22 pairs of autosomes and 1 pair of sex chromosomes

42. Human Karyotype showing homologous chromosome pairs

43. Sex Determination Female Male XX x XY XX x XY ½ X ½ Y ½ X ½ X ¼ XX ¼ XY Phenotypic Ratio of Offspring ½ Female + ½ Male ½ Female + ½ Male

44. X-linked Genes Hemophilic Male Non-hemophilic Female Female (father is hemophilic) (father is hemophilic) ½ XhXhXhXh ½ Y ½ X H X H ½ XhXhXhXh ¼ X H X h ¼ X h X h ¼ X H Y ¼ X h Y Phenotypic Ratio of Offspring ¼ hemophilic males + ¼ non-hemophilic males ¼ hemophilic females + ¼ non-hemophilic females X h Y X h Y x XHXhXHXhXHXhXHXh

45. Terms Specific to Sex-linked Genes Females Homogametic: only X found in gametes Can be homozygous or heterozygous for traits on X- chromosome Males Heterogametic: either X or Y in gametes Hemizygous for traits on the X-chromosome

46. Patterns of Inheritance Autosomal DominantAutosomal Dominant Autosomal RecessiveAutosomal Recessive X-linked DominantX-linked Dominant X-linked RecessiveX-linked Recessive Y-Linked InheritanceY-Linked Inheritance

47. Pedigree for Huntington’s Disease, an Autosomal Dominant Trait

48. Pedigree for Albinism, an Autosomal Recessive Trait

50. Pedigree for Colorblindness, an X-linked Recessive Trait