1. Download Lab. Handouts Download Lab. #3 and Print out 3 files:
Quiz #2 Marks posted on Webpage

2. Mendelian Genetics Topics: -Transmission of DNA during cell division 
Mitosis and Meiosis
- Segregation
- Sex linkage (problem: how to get a white-eyed female)
- Inheritance and probability
- Mendelian genetics in humans
- Independent Assortment
- Linkage
- Gene mapping
- 3 point test cross    
- Tetrad Analysis (mapping in fungi)
- Extensions to Mendelian Genetics
- Gene mutation
- Chromosome mutation
- Quantitative and population genetics  

3. Linkage Chapter 6 - recombination - linkage maps
Ch. 6 p. 148 – Prob: 1-5, 7, 8, 10, 11, 14

4. Linkage of Genes - Many more genes than chromosomes
- Some genes must be linked on the same chromosome; therefore not independent

5. Independent Assortment
Linkage
Fig 6-6
Fig 6-11
Interchromosomal
Intrachromosomal

6. Two ways to produce dihybrid
A B a b A b a B
cis A B AaBb A b trans
a b (dihybrid ) a B
Gametes:
AB P Ab
ab P aB
Ab R AB
aB R ab P X X

7. Example Test Cross AaBb X aabb ab Exp. Obs. AB AaBb 25 10 R
Ab Aabb P
aB aaBb P
ab aabb R
How to distinguish:
Parental high freq.
Recombinant
low freq.

8. Gametes: AB R Ab P aB P ab R Therefore dihybrid: A b (trans) a B
Example (cont.)
Gametes: AB R
Ab P
aB P
ab R
Therefore dihybrid:
A b (trans)
a B

9. Linkage Maps Genes close together on same chromosome:
- smaller chance of crossovers
between them
- fewer recombinants
Therefore:
percentage recombination can be
used to generate a linkage map

10. A B large # of recomb. a b C D small number of recombinants c d
Linkage maps
A B large # of recomb.
a b
C D small number of recombinants
c d

11. Linkage maps example Testcross progeny: P AaBb 2146 R Aabb 43
Total map units 65
= 1.4 % RF
4513
A mu B

12. Additivity of map distances
separate maps A B A C
combine maps C A B
or Locus
A C B (pl. loci)

13. Linkage Deviations from independent assortment Dihybrid gametes
2 parent (noncrossover) common
2 recombinant (crossover) rare
% recombinants a function of distance between
genes
% RF = map distance

14. Gametes Number of Genes Number of Different Gametes
monohybrid 1 (Aa)
dihybrid (AaBb)
trihybrid (AaBbCc)

15. AaBbCc X aabbcc ABC ABc abc AbC Abc aBC aBc abC abc
Three Point Test Cross
Trihybrid
AaBbCc X aabbcc
ABC
ABc abc
AbC
Abc
aBC
aBc
abC
abc
8 gamete types 11

16. C ABC B c ABc A C AbC b c Abc a
Three Point Test Cross
Trihybrid Gametes
C ABC B
c ABc A
C AbC b
c Abc a 11

17. Three Point Test Cross Trihybrid AaBbCc 3 genes: Possibilities:
1. All unlinked
2. Two linked; one unlinked
3. Three linked
- order ? A---B---C
B---C---A
B---A---C
Trihybrid 11

18. Three Point Test Cross +, cv crossveinless +, ct cut wing
Three recessive mutants of
Drosophila: , v vermilion eyes
+, cv crossveinless
+, ct cut wing
P +/+ cv/cv ct/ct X v/v +/+ +/+ 11

19. Three Point Test Cross P +/+ cv/cv ct/ct x v/v +/+ +/+
Gametes cv ct v + +
F1 trihybrid v/+ cv/+ ct/+ 11

20. Three Point Test Cross F1 v/+ cv/+ ct/+ x v/v cv/cv ct/ct v cv ct
8 gamete types one gamete type 11

21. 8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45
v ct
+ cv
1448
Parental (most frequent)
Recombinant 11

22. 8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45
v ct
+ cv
1448
Parental
Recombinant
268
Recombinant
Parental
268
1448
= 18.5 % 11

23. 8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45
v ct
+ cv
1448
Parental
Parental
Recombinant
191
Recombinant
191
1448
= 13.2 % 11

24. 8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45
v ct
+ cv
1448
Parental
Recombinant 93
Parental
Recombinant 93
1448
= 6.4 % 11

25. Calculate Recombination Fraction
v cv R v cv R
268 / = %
v ct R
R v ct
191/ = %
3. ct - cv R ct
R + cv
93/ = % 11

26. Three point test cross Observations:
all 3 RF < 50 % genes on same chromosome
v-----cv largest distance ct in middle
map v ct cv = cv ct v
= > !! Why ? 12

27. Three Point Test Cross P +/+ ct/ct cv/cv x v/v +/+ +/+
gametes ct cv v
F1 trihybrid v
ct cv 11

28. Three Point Test Cross Double crossover class rarest: v---cv
P v v
P ct cv cv
R v ct v
R cv cv X X X X 3 13

29. Three Point test cross 1. Double crossovers not counted in v--cv RF
2. Double crossovers generate P types (with
respect to v--cv)
3. Double crossovers not detected as
recombinants
Consequence:
underestimate of v----cv map distance
Greater distance of genes  greater error 14

30. Double recombinant class: (3 + 5) x 2 = 16 268 + 16 = 284
284/1448 = 19.6
NOTE: double crossovers detected
because of middle gene (ct) 15

31. Mapping Function Genes close together on chromosome
-RF good estimate of map distance
Genes far apart on chromosome
- RF underestimates true map distance due
to undetected multiple crossovers 17 27

32. Mapping Function m = avg. # crossovers per meiosis
(linear with true map distance)
if m = 1 (1 cross over for every meiosis)
then 50 % recombinants produced
Therefore:
map units (mu) = m x 50 17 27

33. Mapping Function Mapping function: - relates RF to true map distance
(better estimate for genes separated by
large distances)
m = -ln (1 - 2RF) mu = m x 50
Mapping function 17 27

34. Mapping Function
m = -ln(1 - 2RF)

35. Mapping Function example
1. RF = 18.5 % m = mu = 23.1
2. RF = 6.4 % m = mu = 6.8
Summary:
- short distances: use RF
- long distances: use mapping function 17 27

36. Linkage Other Points: 1. No crossing over in male Drosophila
male: AaBb A B  gametes AB, ab
a b
use female dihybrid: AaBb x aabb
O O 16 26

37. Linkage 2. Linkage of genes on the X chromosome: AaBb x --Y O O
Male progeny:
AB Y
Ab Y male progeny direct
aB Y measure of female meiotic
ab Y products 17 27

38. Fungal Genetics Fungi: important organisms in the ecosystem
- decomposers
- pathogens
important for humans
- food
(Biology 4040 – Mycology) 19 29

39. Fun Facts About Fungi

40. Fungi

42. Neurospora crassa (bread mold)
Morphological mutants
Biochemical mutants (one gene, one enzyme)

43. Linkage Map
Neurospora crassa Linkage group I

44. Fungus Life Cycle vegetative stage haploid +, - mating types
brief diploid stage  meiosis n n +
spores +
meiosis - 2n n - n

45. Gamete Pool Gametes: Products of many meioses all pooled together A B
a b AB AB ab ab AB ab
P A B ab ab AB ab Ab AB Gamete
P a b AB aB ab ab AB AB pool
R a B ab AB AB ab
R A b 18 28

46. Tetrad Analysis Some Fungi and algae: 4 products of a single
meiosis can be recovered
Advantages:
1. haploid organism - no dominance
2. examine a single meiosis - test cross not needed
3. small, easy to culture
4. Tetrad Analysis - map gene to centromere 19 29

47. Ascus with ascospores 30

48. Tetrad Analysis Types of Tetrads:
1. Unordered - 4 products mixed together
2. Ordered (linear) - 4 products lined up, each
haploid nucleus can be traced
back through meiosis
3. Octads - mitotic division after meiosis
8 products (2 x 4) 20 31

50. Linear Tetrad Analysis
Life Cycle:
+ = a+ a a a a + a + +
Meiosis +
Diploid
Haploid +
Mating: a x  a /+
n n n
4 haploid
products

51. Linear Tetrad Analysis
mitosis a a a a + a + + + + +
4 haploid
products
8 haploid
spores
(Octad)

52. Linear Tetrad Analysis
Two types of asci:
1. no crossover----> first division segregation (MI)
2. crossover between
gene and centromere-----> second division
segregation (MII) 21 33

53. Mapping gene to centromere
First Division a a a a a a + + + +
No Crossover + + 22 34

54. First division segregation
meiosis A a

55. Mapping gene to centromere
Second division a a a + a + + a + a + +
crossover 22 34

56. Second division segregation
** recombinant

57. 1st and 2nd Division segregation
First Division
Second division a a a a a a a + a a a + + + + + a + +
No Crossover a + + + +
Crossover

58. Mapping gene to centromere
a a a
a a a
a a a
a a a
Total = 100
MI = 86
MII = 14 I II 23 35

59. Mapping gene to centromere
MI = MII = 14
14/100 = 14 % of meioses showed a crossover
½ of the crossover products
recombinant
RF = ½ x 14 % = 7 % a
7 m.u. 24 36

60. Unordered Tetrad Analysis
1. still products of a single meiosis
2. can not map gene to centromere
3. linear tetrads can be analyzed as unordered
4. map distance between linked genes a b a b a+ b+
meiosis X a+ b+
n n n 25 37

61. Three kinds of unordered tetrads:b
a b
a b+
a b a+ b+
a b+
a b
a b+
meiosis
a+b+
a+b
a+b
a+b+
a+b
a+b+
1. Parental Ditype
2. Nonparental Ditype PD
NPD T
3. Tetratype 26 38

62. PD
a b
a+ b+ T
NPD
a b
a b+
a+ b+
a+ b
a b+
a+ b

63. Unlinked genes PD = NPD a b + + X a b + + meiosis a + + b a/+, b/+ PD25 37

64. Unordered Tetrads Unlinked Genes: PD = NPD
NPD----> all products recombinant
T > ½ products recombinant
PD-----> all parental type
PD = 58
RF = ½ T + NPD T = 40
T + NPD + PD NPD = 2
RF = m.u. 27 39

65. Tetrad Analysis Types of Tetrads:
1. Ordered (linear): map gene to centromere
2. Unordered: map genes 20 31

66. Linkage: Summary Recombination: generates variation
(inter and intrachromosomal)
Genetic maps:
- genes linked on the same chromosome
- location of new genes relative to genes
already mapped 28 40

67. Linkage: Summary Hunting for genes (Human Diseases)
- genetic markers: DNA variation
- co-inheritance with diseases using pedigree
information
- recombinants used to estimate linkage
- MUN Medical Genetics 28 40

68. Linkage vs. Association
Association - test if a disease and a marker allele show correlated occurrence in a population
Linkage – test if disease and a marker allele show correlated transmission within a pedigree
4,4
1,1
1,4
2,4
3,4
1,2
Linkage
3,3
1,3
3,4
1,3
3,3
2,3
1,4
2,2
2,4
4,4
Association

69. Genetic linkage in humans
Nail- patella syndrome
Rare disease inherited as a dominant

70. Genetic linkage in humans
ABO Blood group marker:
Alleles: A, B, O
Genotypes: OO, AO, BO, AB
Two Genes: N-P syndrome; Blood Group
Evidence for linkage

71. Genetic linkage in humans