1. Basic Mendelian Crosses
Purebreeding Parents 1 1

2. Basic Mendelian Crosses
Testcross 2 2

3. Basic Mendelian Crosses
F1 x F1 Cross 3 3

4. Forked Line (Branch) Diagrams
Gametes possible from AaBbccDd individual 4 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 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 Aa
1/2 bb 1/1 Cc p (aa bb Cc)?
1/4 aa
1/2 bb 1/1 Cc 6 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) = p (d) = 1/2
p (a) and (b) and (c) and (d) = 1/2 * 1/2 * 1 * 1/2 = 1/8 7 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 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 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
Ex. (3)3 = 27 10 10

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

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 : = 4 possible (AB, Ab, aB, ab)
Ex. AaBbCc : = 8 possible (ABC, ABc, Abc, etc.)
Human chromosomes : = > 8 x 10 6 12

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 13

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

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

16. Binomial Expansion: Generating Comprehensive Ratios
Example for AaBbCcDdEe x AaBbCcDdEe cross
(p+q)5 = p p4q p3q p2q pq4 + q5
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 16

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

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

19. Binomial Expansion: Determining Phenotypic Ratios
If order is specified, do not use the coefficient (only one way posible)
If no order given, use coefficient to determine # of different ways.
Probability: p(A-B-C-) = 27/(43) = 27/64
p (A-B-cc) =
p(two dominant and one recessive) = 19 19

20. Binomial Expansion: Determining Phenotypic Ratios20 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 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 22

23. Sample Problem 23 23

24. Answer to Sample Problem (Part 1)24 24

25. Answer to Sample Problem (Part 2)25 25

26. Pedigree Analysis Symbols Used Note sibship line,
Consanguineous marriage, second cousins or closer. 26 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 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
Note: Two unaffected parents can have affected child; Affected individuals often have no affected children; Consanguineous marriages frequently result in expression of rare recessives. For autosomal traits, males and females are affected equally. 28 28

29. Things to Look for with Autosomal Recessive Traits
Note: Two unaffected parents can have affected child; Affected individuals often have no affected children; Consanguineous marriages frequently result in expression of rare recessives. For autosomal traits, males and females are affected equally. 29 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
Hypercholesterolemia: due to insufficient cholesterol receptors so cholesterol is not pulled from blood stream.
Note: autosomal, so again seen equally in males and females
dominant alleles appear in every generation when traced backward from a person expressing the phenotype.
Most affected individuals will be heterozygous if the trait is rare. An affected parent will have 50% chance of affected child. 30 30

31. Things to Look for with Autosomal Dominant Traits
Note: Affected individuals can have unaffected children; For autosomal traits, males and females are affected equally. 31 31

32. Are these autosomal traits dominant or recessive?32 32

33. Are these autosomal traits dominant or recessive?33 33

34. Are these autosomal traits dominant or recessive?34 34

35. Are these autosomal traits dominant or recessive?35 35

36. Are these autosomal traits dominant or recessive?36 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 37

38. What to look for with X- linked Traits
Dominant Recessive Hemizygous
XAXA or XAXa XaXa XAY XaY
Recessive Dominant
Affected daughter Affected father
has affected father passes it on to
and carrier mother all daughters 38 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 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 40

41. Sample X- linked Recessive Pedigree41 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 42

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

44. Y- linked (Holandric) Pedigrees
Traits are only seen in males and passed on from father to son.
X = normal, Y = abnormal 44 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
II3 = aa; I1 and 2 = Aa
II1? = AA or Aa (p = 1/2 each) 45 45

46. Sample Pedigree Problem46 46

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

48. Sample Pedigree Problem
Autosomal trait (rare recessive)
P II 1 = Aa = 1
P III 1 = Aa = 1/2 AA x Aa - Aa
P IV 1 = Aa = 1/4 1/2 x 1/2
P V1 = aa? pIV1 = Aa = 1/4 x p IV 2 = Aa = 1/4 x p aa from Aa x Aa = 1/4 = 1/64 48 48

49. Sample Pedigree Problem from LabII
III
Probability of carrier?
16 17
If 16 married 17, what is the probability of an affected child? 49 49