Lecture 2 Human Genetics

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Transcript Lecture 2 Human Genetics

Sir Archibald E Garrod 1857-1936
1902 – alcaptonuria – black urine - (Madness of King George)
Alcaptonuria
Rare disease
Blackening of urine on exposure of air
Alcapton or homogentisic acid (2, 5-dihydrophenylacetic acid)
Alcapton is black (normally metabolized to colorless)
But…only shows up in families with consanguinity
“Inborn Errors of Metabolism”
Recessive
Mendelian
So how are we going to use genetics
to study humans?
Phenotypes are too complicated (and ambiguous)
The breakthrough came with the realization
that DNA markers would be useful
Two common types of DNA
variants
There are no dominance and recessiveness issues
All of the markers are co-dominant
They are always present in every generation
PCR is relatively cheap
The expense is in finding the polymorphic markers
Dad heterozygous
at genes A, B and C
Mom heterozygous
at genes A, B and C
(and different from Dad)
Offspring get EITHER
A1 or A2 from Dad
MENDEL’S FIRST LAW
Offspring getting A1 from
Dad can get EITHER B1 or
B2 from Dad
MENDEL’S SECOND LAW
Offspring get only B1 and C1
Or B2 and C2 from Dad
LINKAGE
Recall: Mapping in Drosophila
Ly Sb br
+ + br
+ + +
+ + br
Mapping in humans just like any other organism
Distance in cM = # recombinant gametes/Total
Two Problems….
The Major Problem: Low Numbers
The not quite as major problem: The “Phase”
Dealing with low numbers…..
Sir Ronald Fisher, 1890 - 1962
The solution comes from the “Likelihood Ratio”
Consider two genes, and at least 1 heterozygous parent
Analyze the genotypes of the family members
Consider some recombination frequency 
L = Likelihood that the genotypes arose from linked genes given 
Likelihood that the genotypes arose due to segregation of unlinked genes
Dd 12
dd 22
Dad is “Non-informative”
Mom is “Informative”
Dd 12
If linkage, was Mom:
dd 12
D
Dd 22
Dd 22
d
dd 12
D
d
1
2
OR
2
1
4 Parental
1 Recombinant
1 parental
4 Recombinant
Calculate the likelihood given both phases and then average
D
d
D
d
1
2
OR
2
1
4 Parental
1 Recombinant
1 parental
4 Recombinant
L = [(0.9)4 (0.1) + (0.1)4 (0.9)]/ 2
(0.5)5
Z (the lod score..log of odds) = log10 L = 0.021
What if there were no crossovers?
L =
[(0.9)5 + (0.1)5 ]/ 2
(0.5)5
Z = log10 L = 0.97
So crossover has a penalty and lowers the lod score (0.021<0.97)
A lod score of 0.97 means the genotypes are 10 times more likely to arise
by segregation of linked genes (=0.1) than by segregation of unlinked genes
What if we calculate for  = 0.2
D
d
D
d
1
2
OR
2
1
1 parental
4 Recombinant
4 Parental
1 Recombinant
L = [(0.8)4 (0.2) + (0.2)4 (0.8)]/ 2
(0.5)5
Z = log10 L = -0.549 (note lod scores can be negative)
For no crossovers at  of 0.2
Z = log10 L = 0.72
Therefore…..
The lod score varies with 
The lod score varies with data (no crossovers vs 1 crossover)
What should we expect? What should be our confidence?
Z = 3
This is when it is 1000 times more likely that genotypes arose from
linked genes than from unlinked genes
Calculate the lod score for any two markers at every  and plot the results
Mapping DNA markers
• The principle is the same except that the markers are co-dominant
• We saw before that markers are “informative” in meiosis when doubly heterozygous
• The utility of any marker will depend on the degree of heterozygosity in the population
• For best microsatellites the heterozygosity ranges from 10-30%
• Need sufficient number to cover the human genome
• The human genome is ~3000 cM in genetic distance
• For = 0.01 (1 cM resolution) you need 3000 evenly spaced informative markers
• First step is to identify as many microsattelite markers as possible
• You have no idea, a priori, where thy map relative to each other. So Map them!
• Where do we get the DNA for doing the mapping?
CEPH Centre d'Etude du Polymorphisme Humain
• Maintain the DNA database of ~10,000 informative microsattelite markers
• DNA from each individual of 50 large 3-generation families
• Generated immortalized lymphocytes from each individual (cell lines)
• Represents a renewable resource of DNA
• Computed 25,924,717 lod scores between all pairs of marker loci on each pair
of segregating chromosomes
From these data, and the principles of lod score mapping we generate
the human genetic map
Note that the genetic map
is different for men and women
Recombination frequency is higher in meiosis
in women