1 / 49

Basic Mendelian Crosses

Basic Mendelian Crosses. Purebreeding Parents. 1. Basic Mendelian Crosses. Testcross. 2. Basic Mendelian Crosses. F1 x F1 Cross. 3. Forked Line (Branch) Diagrams. Gametes possible from AaBbccDd individual. 4. Forked-Line (Branch) Diagrams. Phenotypes possible from cross:

axel
Download Presentation

Basic Mendelian Crosses

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Basic Mendelian Crosses Purebreeding Parents 1

  2. Basic Mendelian Crosses Testcross 2

  3. Basic Mendelian Crosses F1 x F1 Cross 3

  4. Forked Line (Branch) Diagrams Gametes possible from AaBbccDd individual 4

  5. Forked-Line (Branch) Diagrams Phenotypes possible from cross: AaBbCcDd x AaBbCcDd p(A- B- C- D-) = 81/256 P(A- bb C- dd) = 9/256 5

  6. Forked-Line (Branch) Diagrams Genotypes possible from cross Aa Bb CC x Aa bb cc 1/2 Bb 1/1 Cc 1/4 AA 1/2 bb 1/1 Cc p (AA bb Cc)? 1/2 Bb 1/1 Cc 1/2 Aa 1/2 bb 1/1 Cc p (aa bb Cc)? 1/2 Bb 1/1 Cc 1/4 aa 1/2 bb 1/1 Cc 6

  7. Mathematical Method Answers specific question, not all inclusive Individual with genotype Aa Bb cc Dd What is the probability of gamete with abcd ? p (a) = 1/2 p (b) = 1/2 p (c) = 1 p (d) = 1/2 p (a) and (b) and (c) and (d) = 1/2 * 1/2 * 1 * 1/2 = 1/8 7

  8. Mathematical Method Cross between Aa Bb Cc x Aa Bb Cc What is the probability of offspring with genotype: Aa BB cc? p (Aa) = 1/2 p (BB) = 1/4 p (cc) = 1/4 p (Aa) and (BB) and (cc) = 1/2 * 1/4 * 1/4 = 1/32 AA BB Cc? 8

  9. Mathematical Method Cross between Aa Bb Cc Dd EE x aa Bb CC dd Ee p (aa bb CC dd Ee)? p (Aa Bb Cc Dd Ee)? 9

  10. Mathematical Method When ratios for all genes are consistent: Ex. Aa Bb Cc x Aa Bb Cc # Phenotypes = (2)n 2 = # phenotypes for each gene n = # monohybrid crosses (genes) Ex. (2)3 = 8 # Genotypes = (3)n 3 = # genotypes for each gene n = # monohybrid crosses (genes) Ex. (3)3 = 27 10

  11. Calculating Number of Genotypes Possible # genotypes possible = 3 n 3 = Genotypes possible for each gene - AA, Aa, aa n = # heterozygous gene pairs

  12. Calculating Number of Gametes Possible # gametes possible = 2 n 2 = diploid with 2 copies of each gene/chromosome n = # heterozygous gene pairs Ex. AaBb : 2 2 = 4 possible (AB, Ab, aB, ab) Ex. AaBbCc : 2 3 = 8 possible (ABC, ABc, Abc, etc.) Human chromosomes : 2 23= > 8 x 10 6

  13. Binomial Expansion: Uses Predict comprehensive phenotypic ratios ( 9:3:3:1, etc.) Determine probability of particular categories Only applicable if: Ratio for every trait (gene) is the same (ex. AaBb x AaBb - 3:1) Not applicable if: Ratios vary (ex. AaBb x Aabb - 3:1 for A-:aa; 1:1 for B-:bb) 13

  14. Binomial Expansion: Generating Comprehensive Ratios Probabilities: p = dominant phenotype, q = recessive phenotype N = # genes X = # of dominant (N-X) = # of recessives 14

  15. Binomial Expansion: Generating Comprehensive Ratios Example based on phenotypes from AaBb x AaBb cross 2 A-B-: A-bb: aaB-: aabb 15

  16. Binomial Expansion: Generating Comprehensive Ratios Example for AaBbCcDdEe x AaBbCcDdEe cross (p+q)5 = p5 + 5 p4q + 10 p3q2 + 10 p2q3 + 5 pq4 + q5 5! = 5 * 4 * 3 * 2 * 1 = 4! 1! (4*3*2*1) * 1 5! = 5* 4 * 3 * 2 * 1 = 3! 2! (3 *2*1) * (2*1) 16

  17. Binomial Expansion: Ways to Determine Coefficients Pascal’s Pyramid p q p6 + p5q + p4q2 + p3q3 + p2q4 + pq5 + q6 17

  18. Binomial Expansion: Ways to Determine Coefficients Direct method (shortcut): Example: (p+q)4 = 18

  19. Binomial Expansion: Determining Phenotypic Ratios Probability: p(A-B-C-) = 27/(43) = 27/64 p (A-B-cc) = p(two dominant and one recessive) = 19

  20. Binomial Expansion: Determining Phenotypic Ratios 20

  21. Binomial Expansion: Determining Probability p = probability of one event (ex. girl) q = probability of alternative event (ex. boy) Probability that in N trials, you will get X girls and (N-X) boys = N! (pX) (q)(N-X) Note: Only use coefficient X! (N-X)! when order is not considered Example: probability of 2 girls and 4 boys in a family of 6? 6! (1/2)2 (1/2) 4 = 2! 4! 21

  22. Examples of Applying Binomial Two brown-eyed parents mate and have a blue-eyed child. What are the parents genotypes? Which allele for color is dominant? If two individuals with the same genotype had four children, what is the probability of them having all blue eyes? What is the probability of them having two brown and two blue? What is the probability of the first two being brown and the rest blue? In how many different orders could this occur? What is the probability of their fifth child having blue eyes? 22

  23. Sample Problem 23

  24. Answer to Sample Problem (Part 1) 24

  25. Answer to Sample Problem (Part 2) 25

  26. Pedigree Analysis Symbols Used 26

  27. Pedigree Analysis Sample Pedigree For rare conditions: assume those outside family are homozygous or hemizygous normal; Genetic counselor: p (carrier in population) ex. CF carrier = 0.05 Arrow indicates propositus 27

  28. Pedigree for Autosomal Recessive Trait Usually loss-of-function Albinism: absence of pigment in skin, eyes, hair Cystic fibrosis: thick mucus that blocks lungs, glands Sickle cell anemia: abnormal hemoglobin, blockage Xeroderma pigmentosum: No nucleotide excision repair 28

  29. Things to Look for with Autosomal Recessive Traits 29

  30. Pedigree for Autosomal Dominant Trait Insufficient product, interference with normal, or gain-of-function Achondroplasia: dwarfism, defect in long bone growth Brachydactyly: shortened fingers Hypercholesterolemia: high cholesterol, heart disease Huntington disease: nervous system degeneration 30

  31. Things to Look for with Autosomal Dominant Traits 31

  32. Are these autosomal traits dominant or recessive? 32

  33. Are these autosomal traits dominant or recessive? 33

  34. Are these autosomal traits dominant or recessive? 34

  35. Are these autosomal traits dominant or recessive? 35

  36. Are these autosomal traits dominant or recessive? 36

  37. Pedigrees for X- linked Traits Recessive Color blindness: insensitivity to red or green light Hemophilia: defective clotting, A or B type Muscular dystrophy: Duschenne type, muscle wasting Dominant Hypophosphatemia: rickets (bowlegged) 37

  38. What to look for with X- linked Traits DominantRecessiveHemizygous XAXA or XAXa XaXa XAYXaY Recessive Dominant Affected daughter Affected father has affected father passes it on to and carrier mother all daughters 38

  39. What to look for with X- linked Recessive Traits Mother can carry and pass on the recessive allele More males express these traits Mother to son inheritance Affected daughters must have affected fathers Criss-cross pattern of inheritance 39

  40. Sample X- linked Recessive Pedigree X = normal, X = abnormal Males: 1/2 normal, 1/2 affected Females: 1/2 normal, 1/2 carriers 40

  41. Sample X- linked Recessive Pedigree 41

  42. What to look for with X- linked Dominant Traits Affected mother will pass it on to 50% both daughters and sons Affected father will only pass it on to all daughters 42

  43. Sample X- linked Dominant Pedigree X = normal, X = abnormal Overall: 1/2 affected, 1/2 normal 43

  44. Y- linked (Holandric) Pedigrees Traits are only seen in males and passed on from father to son. X = normal, Y = abnormal 44

  45. Sample Pedigree Problem Adherent Earlobes (recessive autosomal) Genotypes: II 3 = I 1 and 2 = p II 1 is Aa = p III 2 is Aa = p III 2 and III 4 would have aa child = 4 45

  46. Sample Pedigree Problem 46

  47. Sample Pedigree Problem Huntington Disease (same family, later date) 47

  48. Sample Pedigree Problem Autosomal trait (rare recessive) 48

  49. Sample Pedigree Problem from Lab I II III Probability of carrier? 1-4 8 9 15 16 17 If 16 married 17, what is the probability of an affected child? 49

More Related