OVERVIEW

1
OVERVIEW

• Wrap up phase discussion- computing LOD scores
  with unknown phase
• How do we determine the model of inheritance?
  Autosomal dominant, autosomal recessive, or
  sex-linked?
• How are good markers chosen?

2
Phase UnknownComputing the LOD score
No Disease
Colorblind
1
2
Colorblind Hemophilia
HC/Y
HC/hc
Hc/hC
OR
1
2
3
4
5
6
HC
hc/Y
Hc/Y
HC/Y
hc/Y
hc/Y
HC/Y
What is the genetic distance between these genes?
Could this computation be done without the
grandparents?
3
Y statistics

• In the previous example, the recombination
  fraction r 1/6 or 5/6 depending on the phase.
• Note that it is always possible to bin the
  progeny into two groups, but that the difficulty
  lies in labeling the recombinant group
• Early geneticists thought that such pedigrees did
  not yield information on linkage.
• However, assuming k of n recombinants, Bernstein
  (1931) pointed out that the product y k(n-k) is
  the same in either phase.
• Note that y is largest for r ½ and zero for r0
• Bernstein published mean values of y for
  different parameters N and r. These enabled
  researchers to estimate recombination fractions
  for phase-unknown data.

4
Modified LOD score

• The likelihood ratio function can also be
  modified to deal with unknown phase.
• Recall that the LOD score is a ratio of two
  likelihoods L(q) and L(q½) with
• If phase is unknown, k cannot be computed. But
  since the two phases are equally likely

5
How do we determine which model of inheritance to
use?

• FOR AUTOSOMAL DOMINANCE
• Since diseases are typically rare, the frequency
  of allele D is assumed low.
• Most affected individuals are therefore expected
  to have genotype Dd rather than DD.
• Matings between an affected and unaffected will
  be type Dd x dd.
• In this case the probability of affected
  offspring will be 50 (see next table).
• This prediction can be used as a test of
  autosomal dominance.

6
Six possible mating types autosomal dominance
7
Test 1 the binomial test

• The binomial distribution is the discrete
  probability distribution of the number of
  successes k in a sequence of n independent yes/no
  experiments, each of which yields success with
  probability p.
• We want the associated p-value of observing k
  when p 1/2

8
Example Opalescent dentine(Neel and Schull
1954)
Examined 112 offspring of an affected parent 52
were similarly affected as the parent The other
60 were normal Are these observations consistent
with autosomal dominance?
9
Test 2 Maximum likelihood

• The hypothesis of p1/2 can also be tested by
  considering the likelihood of the data L(p) as a
  function of parameter p.
• Given k of n offspring are affected
• Let L1 be the maximum value of this function for
  0p1, and let L0 be the value at p1/2.
• The likelihood ratio statistic l 2(ln L1 ln
  L0)What are the lower and upper bounds of l?
• If the hypothesis that p1/2 is true, l can be
  shown to be distributed as c21.

10
Maximizing the likelihood

• Computing L1 requires finding the value of p
  which maximizes L.
• Solution Set the derivative 0 and solve for p.

Ignoring then choose k constant
These terms equal zero only at the extremes of p
11
Plots of L vs. p for two different k
k 50 n 100
k 5 n 100
12
Maximizing the likelihood (continued)

• The maximum likelihood estimate of p k/n. This
  is used to compute L1
• The p-value is given by the probability that a
  chi-square with 1 DOF will exceed 2(lnL1 lnL0)

13
Back to dentin example

• Example for the opalescent dentin data
• Thus the likelihood ratio statistic l 0.5719.
  This gives a p-value of 0.4495 (see next slide).
• This value is very close to that calculated with
  the binomial CDF.

14
Chi-square with k degrees of freedom
15
Autosomal recessive disorders

• The entire preceding discussion was focused on
  testing for autosomal dominance. The autosomal
  recessive model is harder to test because of
  so-called ascertainment bias, which occurs as
  follows
• For a dominant disorder, we are examining Dd x
  dd, which means that families can be selected
  based on the phenotypes of the parents.
• For a recessive disorder, the mating one would
  like to examine is Dd x Dd, with an expected
  segregation ratio of ¼.
• Since these parents are phenotypically normal, we
  must select families based on the presence of
  diseased children.
• However, this ascertainment procedure will miss
  families with the Dd x Dd mating type that by
  chance had no affected children.

16
Autosomal recessive disorders

• The need to account for the incomplete selection
  of a mating type in segregation analysis was
  pointed out by Fisher in a classic 1934 paper
  entitled The effect of methods of ascertainment
  upon the estimation of frequencies.
• It is a statistical commonplace that the
  interpretation of a body of data requires a
  knowledge of how it was obtained Nevertheless,
  in human genetics especially, statistical methods
  are sometimes put forward, and their respective
  claims advocated with entire disregard of the
  conditions of ascertainment.

17
Truncated binomial method

• Consider Dd x Dd families with X observed
  affected individuals out of s total offspring.
• X is binomial random variable with parameters s
  and expected fraction affected of p ¼ for a
  rare recessive disorder.
• Assume all families with Xgt0 are ascertained.
• We wish to compute
• which can be plugged into the maximum
  likelihood framework already described.

18
Population genetics as another tool to establish
the mode of inheritance.

• The field of population genetics is concerned
  with the distribution patterns of alleles and the
  factors that alter or maintain their frequencies
• One might expect a detrimental allele to
  disappear from the population over time, but
  population genetic analysis shows that this is
  not usually the case
• The main measure is the allele frequency (or gene
  frequency), i.e., the frequency with which an
  allele is present in the population.
• Keep in mind that this measure is not the same as
  the frequency of different genotypes, which
  involve two alleles in combination.

19
Population Genetics 101Measuring Genetic
Variation

• Genotype frequencies depend only on allele
  frequencies
• For recessive disorders, the genotype BB may
  result in selective disadvantage, but this is
  balanced by the occurrence of new mutations (see
  next slide). This is the Hardy-Weinberg
  equilibrium (HWE).
• pA frequency of allele A
• pB frequency of allele B
• P(A/A) pA2 P(A/B) pB2 P(A/B) 2pApB
• pA pB 1
• (pA pB)2 1 ? pA2 2pApB pB2 1

20
A
a
New mutations
Elimination by disease
21
Population Genetics (continued)

• When the allele frequency is known, the expected
  genotype frequencies can be determined.
• For instance, if the frequency p of allele A is
  60 p 0.6q 0.4Frequency of AA
  0.36Frequency of Aa 0.48Frequency of aa
  0.16
• Conversely, when genotype frequencies are known,
  the allele frequencies can be estimated. For
  instance, for a recessive disorder one knows the
  frequency of disease which provides an estimate
  of q2. From this, all other genotype frequencies
  can be estimated.
• In the case of linkage analysis, the
  demonstration of Hardy-Weinberg equilibrium is
  very strong evidence for a genetic basis for a
  trait.

22
Which loci make good markers?

• Clearly, they must be polymorphic (q gt 1).
• However, we also want high heterozygosity, since
  it is heterozygous matings that are the most
  informative.
• Observed vs. expected heterozygosity
• Ho Observed fraction of heterozygous
  individuals
• He Expected fraction based on allele
  frequencies
• The frequency f(X) of allele X is the fraction of
  times it occurs over all loci (2 per individual)
• He 1 the probability of homozygosity
• 1 f2(X) f2(Y) for all alleles
  (X,Y,)

23
Example 10 Unique Genotypes(in bp lengths of
microsatellite)
Ho 0.30 He 0.69
H 1 high diversity H 0 asexual
mitotic reproduction Ho ltlt He indicates
selective pressure or non-random mating
24
Components of the genetic model

• Components of the genetic model include
  inheritance pattern (dominant vs. recessive,
  sex-linked vs. autosomal), trait allele frequency
  (a common or rare disease?), and the frequency of
  new mutation at the trait locus.
• Another important component of the genetic model
  is the penetrance of the trait allele. Knowing
  the penetrance of the disease allele is crucial
  because it specifies the probability that an
  unaffected individual is unaffected because he's
  a non-gene carrier or because he's a
  non-penetrant gene carrier. The frequency of
  phenocopies is an important component, too.
• Rough estimates of the disease allele frequency
  and penetrance can often be obtained from the
  literature or from computer databases, such as
  Online Mendelian Inheritance in Man
  (http//www3.ncbi.nlm.nih.gov/Omim/). Estimates
  of the rate of phenocopies and new mutation are
  frequently guesses, included as a nuisance
  parameter in some cases to allow for the fact
  that these can exist.
• Linkage analysis is relatively robust to modest
  misspecification of the disease allele frequency
  and penetrance, but misspecification of whether
  the disease is dominant or recessive can lead to
  incorrect conclusions of linkage or non-linkage.

25
Steps to linkage analysis

• In pedigrees in which the genetic model is known,
  linkage analysis can be broken down into five
  steps
• State the components of the genetic model.
• Assign underlying disease genotypes given
  information in the genetic model.
• Determine putative linkage phase.
• Score the meiotic events as recombinant or
  non-recombinant.
• Calculate and interpret LOD scores.
• Let's take a look at each of these steps in
  detail.

26
State the components of the model

• In this example, the disease allele will be
  assumed to be rare and to function in an
  autosomal dominant fashion with complete
  penetrance, and the disease locus will be assumed
  to have two alleles
• N (for normal or wild-type)
• A (for affected or disease)

27
Assign underlying disease genotypes

• The assumption of complete penetrance of the
  disease allele allows all unaffected individuals
  in the pedigree to be assigned a disease genotype
  of NN. Since the disease allele is assumed rare,
  the disease genotype for affected individuals can
  be assigned as AN.

28
Determine putative phase

• Individual II-1 has inherited the disease trait
  together with marker allele 2 from his affected
  father. Thus, the A allele at the disease locus
  and the 2 allele at the marker locus were
  inherited in the gamete transmitted to II-1.
  Once the putative linkage phase (the disease
  allele "segregates" with marker allele 2) has
  been established, this phase can be tested in
  subsequent generations.

29
Score the meiotic events asrecombinant (R) or
non-recombinant (NR)

• There are four possible gametes from the
  affected parent II-1 N1, N2, A1, and A2. Based
  on the putative linkage phase assigned in step 3,
  gametes A2 and N1 are non-recombinant.

30
Calculate LOD scores

• In this example, the highest LOD score is -0.09
  at q 0.40. At no value of q is the lod score
  positive, let alone gt3.0, so this pedigree has no
  evidence in favor of linkage between the disease
  and marker loci.

31
Deviations from Mendelian segregation

• The trait is not governed by the alleles of a
  single locus
• Ascertainment bias heterozygous parents with no
  diseased children are not sampled
• Differential survival some genotypes do not
  survive
• Phenocopies the disease can also be caused by
  environmental factors
• Incomplete penetrance the genotype does not
  always lead to disease

For each of these possibilities, is the
segregation ratio increased or decreased?