1. Human Genetics, part I Liisa Kauppi (Keeney lab)
Mapping Mendelian and complex diseases
- Linkage mapping in pedigrees
- Association mapping in populations

2. Genes and Environment “Natural” mutants only
Heritability: first degree relatives of a patient at greater risk
For type I diabetes, l = 15 (6%/0.4%)
Twin studies:
Adopted (separated in infancy)
Fraternal vs. identical twins
Biological vs. non-biological siblings

3. Genes and Human Disease
Polygenic,
Reduced penetrance
Asthma
Osteoporosis
Schizophrenia
Infectious
disease
Height
Body weight
HARD
COMPLEX/MULTIFACTORIAL
DISEASE
“Common disease”
High penetrance,
Single gene
Cystic fibrosis
Pure environment
Blood type
EASY
MENDELIAN
Snakebite
Language

4. Polymorphic markers are needed for disease mapping
Microsatellites
Tandem arrays of simple repeats, for example (CA)n, n=15…27
MULTI-ALLELIC A
Single nucleotide polymorphisms (SNPs)
Abundant, perhaps 1 every 300 bp
- RFLPs G
BI-ALLELIC
Mostly non-coding

5. B allele has frequency p b allele has frequency q p + q = 1
Genotype frequencies: Hardy-Weinberg equation
B allele has frequency p b allele has frequency q p + q = 1
p (B)
q (b)
p2 (BB)
pq (Bb)
q2 (BB)
p2 (BB) + 2pq (Bb) + q2 (bb) = 1
Hardy-Weinberg equilibrium

6. How are recessive traits maintained in a population?
HWE of allele frequencies: p2 + 2pq+ q2= 1
Hypothetical example:
in Sardinia, 1 in 5 individuals have straight hair
This trait is determined by a single gene and it is recessive.
S allele = curly hair, s allele = straight hair
Frequency of s/s homozygotes is 0.2
Frequency of s allele is 0.45 (0.2)
Frequency of S allele is = 0.55
Gametes for next generation: S s
0.552=0.3
0.55 x 0.45 = 0.25
0.452=0.2
Frequencies of genotypes and alleles remain unchanged from one generation to the next.

7. HWE allows calculations of carrier frequencies for recessive traits
(with caution)
Example: Cystic fibrosis, alleles CF and cf
Incidence 1/2000 births
p2 + 2pq+ q2= 1
Frequency of cf/cf homozygotes is
Frequency of cf allele is (0.0005)
Frequency of CF allele is = 0.978
Frequency of CF/cf heterozygotes is 2 x x = 0.043

8. So what if genotypes at a locus are not in HWE?
Suggests that assumptions are not met
Example: heterozygote deficit could arise from recent admixture
p2 + 2pq+ q2= 1
Population 1
Population 2
B freq 0.9
b freq 0.1
B freq 0.1
b freq 0.9
n=1000
n=2000
B freq 0.5
b freq 0.5
HWE
expected
observed

9. Departure from HWE (heterozygote excess): the Prion protein gene and human disease
PRNP gene linked to prion diseases e.g. CJD, kuru
A common polymorphism, M129V, influences the course of these diseases: the MV heterozygous genotype is protective
Kuru acquired from ritual cannibalism was reported (1950s) in the Fore people of Papua New Guinea, where it caused up to 1% annual mortality
Departure from Hardy-Weinberg equilibrium for the M129V polymorphism is seen in Fore women over 50 (23/30 heterozygotes, P = 0.01)

10. Linkage studies - recombination in a family how often are 2 loci separated by meiotic recombination?
2 loci on same
chromosome II
Informative and
uninformative
meioses
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
III NR R
Recombination fraction  is 2/6=0.33

11. Recognizing recombinants does the disease segregate with this marker?1 I
2 5
1 6
2 1
3 4
3 1
3 2
4 1
4 2 II 6
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
III NR R
Recombination fraction  is 1/6=0.167

12. Recognizing recombinants Often samples are missing
2 1
3 4
3 1
3 2
4 1
4 2 I II
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
III NR R OR
Recombination fraction  is 1/6=0.167 or 5/6=0.833

13. Recognizing recombinants Tracing additional family members can helpII
2 1
3 4
3 1
3 2
4 1
4 2
1 5
1 6
5 6
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
III NR R
But are these identical by descent?

14. Requires a precise genetic model
Which marker is the disease locus closest to? Lod scores
Logarithm of odds (Lod) score Z
Likelihood of loci being linked
Likelihood of loci not being linked
Z = log
For the example pedigree with 1/6 recombinants:
( )5 x 0.167
(0.5)6
Z = log
= 0.632
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
Lod scores between -2 and +3 are inconclusive
Below -2  exclusion
Above +3  linkage
Requires a precise genetic model

15. Which marker is the disease locus closest to? Multi-point lod scores
chr 3p12-14
Waardenburg syndrome type 2
After Hughes et al. (1994) Nature Genet 7,

16. Multifactorial diseases (no simple Mendelian inheritance pattern)
Sib-pair analysis
2 1
3 4
3 2
4 2
3 2
3 1
4 1
Number of shared
parental alleles 2 1
1/4
1/2
probability
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.

17. Affected sib-pairs
Which loci do the affected sibs share more often than expected by chance?
3 4
2 1
3 4
2 1
Number of shared
parental alleles
Number of shared
parental alleles
Family based designs test sharing of alleles among all affected and unaffected individuals and compare the probability of transmission if linked to the affection status compared with no linkage.
3 2
3 2 2
3 2
3 2 2
3 1 1

18. Detecting linkage in pedigrees can be complicated…
One can attempt to identify recombination hotspots by analysing marker segregation in families.
-family, samples from parents and offspring
-stretches of DNA on homologous chromosomes from the father and the mother
-each of the children inherit one chromosome from the father and one from the mother
-occasionally, a child will inherit a chromosome that has undergone meiotic recombination
The problem with this approach is that -

19. … and you need lots of meioses!
-recombination events, even in hotspots, are rare
-one would have to analyse a huge number of parent-to-offspring transmissions in order to find that one chromosome that has undergone crossover
We use an alternative approach -

20. Association mapping in a population
Cases vs. controls
HLA-DR4 allele (UK)
General population
Rheumatoid arthritis
patients
36%
78%
Seek correlation between genotype and phenotype
Allele B is associated with disease D if people who have D also have B more often than predicted from B’s frequency
To test every polymorphism is too expensive

21. Linkage disequilibrium (LD) measures association between two alleles
Mutation creates new variants A G A G T
Initially, the new allele is in LD with nearby alleles
LD value = 1
Recombination reshuffles existing variation A G T X
looking at how recombination maintains genetic variability:
First, mutation creates new variation. Here are shown tracts of DNA on two homologous chromosomes, with already one base difference between them. This kind of variation is referred to as SNPs. When mutation creates another SNP site, the two alleles on the same chromosome are initially always found together. If this chromosome is successful and spreads in the population, then we speak of LD between the two SNP sites.
What recombination does is from one generation to the next, it reshuffles the combinations of alleles at different loci. If XOs occur in this interval, LD between the two loci is disrupted, and in a population not only this kind of chromosomes are found, but also these.
LD diminishes
If enough crossovers take place, the loci are in “free association”
Commonly used LD measures: D’ and r2

22. Haplotypes are sets of markers inherited as a “package”
meiotic recombination creates novel haplotypes
Markers form
haplotype
blocks in the population
instead of detecting recombinants in children, we detect recombinant molecules directly from sperm DNA
This in principles gives you an unlimited number of meiosis to study
You have millions and millions of sperm carrying the progenitor haplotypes
Among them, you have the the rare recombinants, which are fished out using allele-specific PCR methods, that I’ll describe in a moment

23. LD is a measure of allelic association in a population
2 SNP loci on the same chromosome
C/G
A/T A G C T
< 4 combinations -> LD T G
Conversely: all 4 combinations -> low or no LD
But also: population history, drift, selection…

24. Disease haplotypes shorten from one generation to the next

25. Recombination hotspots are key in shaping haplotype blocks
Perhaps at least 90% of crossovers take place at highly localized hotspots
HLA class II
Recombination
activity
Haplotype blocks
Kauppi et al. (2004) Nat Rev Genet 5,

26. How do you extract haplotypes from genotype data?
C/G
Blood DNA A T C G or ?
Other family members A T C G
Other individuals in population

27. HapMap project Examines haplotypes in four populations
Data just released:
A haplotype map of the human genome, Nature 437,
HapMap project
Examines haplotypes in four populations
DNA samples: 270 people in total
Yoruba (Nigeria): 30 parent-child trios
Whites with North and West European ancestry (USA): 30 trios
Japan: 45 unrelated individuals
China: 45 unrelated individuals
Identify “haplotype tag SNPs” to minimize genotyping effort
>3,500,000 SNPs typed in total

28. Limited within-block diversity
Example: a 8.5-kb long block on chr 2, 36 SNPs typed
In principle, could give rise to 236 different haplotypes
Only seven different haplotypes found among 120 European chromosomes

29. Recombination hotspots are widespread and account for LD structure
7q21
The International HapMap Consortium

30. Pairwise tagging A T G A G A G C T C T C G C G C A C Tags: SNP 1 SNP 32
G/C 3
T/C 4 5
A/C 6
Tags:
SNP 1
SNP 3
SNP 6
3 in total
Test for association: A T G A G A G C T C T C G C G C A C
So here’s just an illustration of tagging based on pairwise r2. 6 SNPs with 3 groups of SNPs that have high r2. This means we can pick only 3 tags without losing any information. These tags are also the tests in the association analysis.
high r2
high r2
high r2
After Carlson et al. (2004) AJHG 74:106

31. The Common-Disease Common-Variant Hypothesis
Says
disease-predisposing variants will exist at relatively high frequency (i.e. >1%) in the population.
are ancient alleles occurring on specific haplotypes.
detectable in a case-control study using tagging SNPs.
Alternative hypothesis says
disease-predisposing alleles are sporadic new mutations, perhaps around the same genes, on different haplotypes.
families with history of the same disease owe their condition to different mutations events.

32. Does same phenotype mean same genotype?
Coding SNPs, nonsynonymous or synonymous
“Regulatory” SNPs

33. Common Gene Variation in Complex Disease
Case-control studies, comparing the frequencies of common gene variants can identify susceptibility and protective alleles
Some have multiple identified genes (*)
Phenotype
IDDM*
Alzheimer dementia
Deep venous thrombosis
Colorectal cancer
NIDDM
Gene
HLA
APOE F5
APC
PPAR
Variant
DR3,4 E4
Leiden
3920A
12A

34. Other types of variation may also have a role in complex disease
common copy number polymorphisms
large scale rearrangements, deletions and insertions
microsatellite expansions, small insertion/deletions etc.