1. Balanced Translocation detected by FISH

2. Red-
Chrom. 5
probe
Green-
Chrom. 8
probe

4. 2D Protein Gels

6. MS-peptide size signature: match to all predicted proteins

7. X 2 3
Parent X
Gamete X
Child

8. 2 3 X Parent X Gamete X Child A1 B1 C1 D1 E1 D2 E2 A2 B2 C2 A2 B1 C1

9. 2 3 X Parent X Gamete X Child A1 B1 C1 D1 E1 D2 E2 A2 B2 C2 A2 B1 C1NR
Gamete R A2 B1 C1 E1 X
Child

10. co-inherited (linked)
Positional Cloning by Recombination Mapping
Follow the mutation
2. Follow which DNAs are
co-inherited (linked)

11. To determine disease gene presence or absence (genotype)
Positional Cloning by Recombination Mapping
Follow the mutation
To determine disease gene
presence or absence (genotype)
from phenotype you must
first establish
Dominant / recessive
Aurosomal / sex-linked

12. SINGLE GENE DEFECTS Modes of Inheritance Unaffected Male
To deduce who (likely) has one or two copies of mutant gene
Unaffected Male
Affected Female

13. +/+
D/+
+/+
D/+
AUTOSOMAL DOMINANT

14. a/+
a/+
+/Y
x/+
x/+
+/Y
a/a
x/Y
RECESSIVE
RECESSIVE
X-LINKED
AUTOSOMAL

16. co-inherited (linked) Use DNA sequences that differ
Positional Cloning by Recombination Mapping
2. Follow which DNAs are
co-inherited (linked)
Use DNA sequences that differ
among individuals within a family-
Polymorphisms.

17. VNTR / STRP DETECTION

19. A1 B1 C1 X 2 3
Parent A2 B2 C2 A2 B1 C1 X
Gamete A2 B1 C1 X
Child

20. Meiosis
Meiosis
Meiosis
A2 B2
A1 B1

21. Recombination Mapping
Measures distance between 2 sites on a chromosome according to frequency of recombination
Distance between 2 DNA markers or
Distance between a “disease gene”
and a DNA marker

22. No fixed proportional
Conversion between
Genetic distance (cM)
and
Physical distance (kb, Mb)

23. FAMILY A A1 D A2 + NR R

24. FAMILY B A1 + A1 D A2 + A2 D NR NR NR NR NR R R R R R R NR

25. INFORMATIVE MEIOSIS Ideally:-
unambiguous inheritance of mutation and markers
(requires heterozygosity for each in parent)
knowledge of which alleles linked in parent (phase)

26. Assign numbers to results of linkage analysis
to deal with non-ideal meioses
to sum data from many meioses in a family
to sum data from several families

27. 
If unlinked:-
If linked and RF = 
1/2
Likelihood of R
1 - 
Likelihood of NR
1/2
Family A has 1 recombinant and 5 Non-Recombinants 
Likelihood, given linkage of
Or given unlinked:- 5
= (1- ) 
L ( )   6
L (1/2) = (1/2)
Z = Lod = log { L ( ) / L (1/2)}   Z

28. Z = 3
Lod q

29. FAMILY B A1 + A1 D A2 + A2 D NR NR NR NR NR R R R R R R NR

30.  Family B:- Disease gene may be linked to A1 or A2
Consider equally likely
50% chance Family B has 1 R and 5 NR
50% chance Family B has 5 R and 1 NR
L ( ) 
L (1/2) = (1/2) 6
Z = Lod = log { L ( ) / L (1/2)} Z
= (1- ) 5
1/2 { }
(1- )
1/2 { }

31. Phase known  Z
Phase unknown  Z

32. For family “A” with meioses 1, 2, 3, 4 …..
Z = Z1 + Z2 + Z3 + Z4 +…..
For multiple families, “A”, “B”, “C”, “D”…..
Z = Z(A) + Z(B) + Z(C) + Z(D) + ….
Assumption:
same gene responsible for disease in all families
Problem:
locus heterogeneity

33. Z = 3
Lod q

35. LINKAGE DISEQUILIBRIUM
Many
generations

36. PCR test DNA segments
Strachan p126

38. Testing for specific mutations

39. ARMS 3’ mis-match of primer
Strachan p128

40. Strachan p128 cont.

41. TaqMan
Strachan p129

42. OLA
Glick p217

43. Glick p217 cont.