Risk Ratios and Genetic Model Discrimination

f / D9 / D8 / D7 / Relation
¼ / 0 / 1 / 0 / Parent-Offspring
1/8 / ½ / ½ / 0 / Half Sibling
¼ / ¼ / ½ / 1/4 / Full Sibling
1/16 / ¾ / ¼ / 0 / First Cousins
1/8 / 9/16 / 6/16 / 1/16 / Double First Cousins
1/64 / 15/16 / 1/16 / 0 / Second Cousins
1/8 / ½ / ½ / 0 / Uncle-Nephew

Risk Ratios and Genetic Model Discrimination.

· Let us assume that each person in the population is assigned a factor of X=1 if he/she is affected by a condition and X=0 otherwise.

· The Prevalence of the condition is K=E(X).

· Given two non-inbred relatives i and j and given that i is affected, what is the probability that J is affected?

· KR=P(Xj=1|Xi=1(

· P(Xj=1,Xi=1) = P(Xj=1|Xi=1(P(Xi=1) = KRK = E(XiXj)

· P(Xj=1|Xi=1) = E(XiXj)/K = (cov(Xi,Xj)+K2)/K = cov(Xi,Xj)/K+K

· This result simply represents the fact that the extra risk for j results from the covariance of X between i and j.

· The risk ratio can thus be defined as:

· lR= cov(Xi,Xj)/K2

· Let us compute this covariance, and following it the risk ratio.

· Let us assume that a given property is defined by a single gene with multiple alleles.

· For the sake of simplicity let us normalize E(x)=0, and divide:

Risk Ratio / Relative Type / R
sa2/K2+ sd2/K2 / Identical Twin / M
sa2/2K2+ sd2/4K2 / Sibling / S
sa2/2K2 / First Degree / 1
sa2/4K2 / Second Degree / 2
sa2/8K2 / Third Degree / 3

Phenotype vs Genotype

· A phenotype (P) is composed of genotypic values (G) and environmental deviations (E): P = G + E

· Whether we focus on mean, variance, or covariance, inference always comes from the measurement of the phenotype

· A distinction will be made:

o V will be used to indicate inferred components of variance

o s2 will be used to indicate observational components of variance

· Mean genotypic value is equal to the mean phenotypic value

· E(Gj) = E(Pj )= mPj

· Genotypic values are expressed as deviations from the mid-homozygote point

· Consider two alleles, A1 and A2, at a single locus.

· The two homozygous classes, A1A1 and A2A2, are assigned genotypic values +a and -a, respectively.

· Assume that the A1 allele increases the value of a phenotype while the A2 allele decreases the value.

· The heterozygous class, A1A2, is assigned a genotypic value of d

· Zero is midpoint between the two genotypic values of A1A1 and A2A2; d is measured as a deviation from this midpoint

· The mean of environmental deviations is zero Ej = Pj – Gj , ,(mE = 0)

· The correlation between genotypic values and environmental deviations for a population of subjects is zero (rGE = .00)

Genotype / Frequency / Value / Freq x Value
A1A1 / p2 / +a / p2a
A1A2 / 2pq / D / 2pqd
A2A2 / q2 / -a / -q2a
Sum = / a(p - q) + 2dpq

· Recall that p2 - q2 = (p + q)(p - q) = p - q

· mP = a(p - q) + 2dpq

If multiple genes contribute to a phenotype in an additive way

· Average effect of a particular gene (allele) is the mean deviation from the population mean of individuals which received that gene from one parent (assuming the gene transmitted from the other parent having come at random from the population)

Gamete / Frequ. & Values / mG / Minus pop. mean / a
A1A1
+a / A1A2
d / A2A2
-a
A1 / p / q / pa + qd / -[a(p-q) + 2dpq] / q[(a + d(q-p)]
A2 / p / q / -qa + pd / -[a(p-q) + 2dpq] / -p[a+d(q-p)]

· Thus, the average effect for each allele also can be calculated for A1 and A2 in the following manner (Falconer, 1989):

· Assume two alleles at a locus

· Select A2 genes at random from population; p in A1A2 and q in A2A2

· A1A2 to A1A1 corresponds to a change of d to +a, i.e., (a - d); A2A2 to A1A2 corresponds to a change of -a to d, (d + a)

· On average, p(a - d) plus q(d + a) or

· a = a + d(q - p)

· When gene frequency is greater a is greater

q = 0.10 / q = 0.40
a1 = / +0.24 / +1.44
a2 = / -2.16 / -2.16
a = a1 - a2 / 2.40 / 3.60

· The average effects of the parents’ genes determine the mean genotypic value of its progeny

· Average effect can not be measured (gene substitution), while breeding value can

· Breeding value: Value of individual compared to mean value of its progeny

· Mate with a number of random partners; breeding value equals twice the mean deviation of the progeny from the population mean (provides only half the genes)

· Breeding value is interpretable only when we know in which population the individual is to be mated

· Breeding values are referred to as “additive genotype”; variation due to additive effects of genes

· A symbolizes the breeding value of an individual

· Proportion of s2P attributable to s2A is called heritability (h2)

· G = A + D

· Statistically speaking, within-locus interaction

· Non-additive, within-locus effect

· A parent can not individually transmit dominance effects; it requires the gametic contribution of both parents

· Genotype: Breeding value

· A1A1: 2a1 = 2q a

· A1A2: a 1 + a 2 = (q - p) a

· A2A2: 2 a 2 = -2p a

· Mean breeding value under HWC equilibrium is zero

· 2p2q a + 2pq(q - p) a - 2q2p =2pqa(p + q - p - q) = 0

· Regression of genotypic value on gene dosage yields the genotypic values predicted by gene dosage average effect of an allele that which “breeds true”

· If there is dominance, this prediction of genotypic values from gene dosage will be slightly off. dominance is deviation from the regression line

Epitasis - Separate analysis

· locus A shows an association with the trait

· locus B appears unrelated

LINKAGE ANALYSIS

Recombination Fraction

· During synapsis, crossing-over may occur between any two non-sister chromatids

· If there are allelic differences at the site of crossing over, the genetic result is recombination

· Genes on the same chromosome are connected physically (syntenic)

· At least a thousand to several thousand for each human chromosome

Recombination fraction

· Prophase of the first meiotic cell division

· Homologous pairing of chromosomes (synapsis)

· Each chromosome consists of two fully formed chromatids joined at the centromere; four chromatids for each pair of chromosomes

· Each chromatid represents a separate DNA molecule

· If two syntenic genes are close enough that a crossover occurs between them less than once per meiosis, on the average, the two genes are genetically linked

· Recombination fraction (RF or ) is 1/2 the frequency of crossovers (a single crossover involves two of four chromatids in a synapsed pair of chromosomes)

· If two loci are so far apart that on average there is at least one crossover between them in every meiosis, then = 50%, the loci are unlinked

· can not be greater than 50%

· Morton et al. (1982) used cytological preps of spermatocytes and reported an average of 52 crossovers per male meiosis

· Recombination fraction is expressed in map units or centiMorgans (cM)

· 1 mu = 1 cM; of 1% over small distances

· One crossover on average implies a genetic map length of 50 cM

· If two loci are separated by a distance such that an average of one crossover occurs between them in every meitotic cell, then those loci are 50 cM apart

· 52 crossovers implies a total genetic map length of 2600 cM in humans; thus, 1 cM equals approximately 1 megabase of sequence

· Not additive over long distances due to multiple crossovers (positive or negative interference); mapping functions have been developed to address this phenomenon

· Linkage describes the phenomenon whereby allele at neighbouring loci are close to one another on the same chromosome, they will be transmitted together more frequently than chance.

· q = 0 : no recombination => complete linkage

· q < 0.5 : partial linkage

· q = 0.5 : no linkage

Linkage Analysis

· For a couple of which the genotypes at the A and B are known, the probability of observing the genotypes of the offspring depends on the value of 

· Let us assume the following crossing:

· Therefore, such a couple can have 4 types of offspring

· There are two possible situations:

· The alleles A1 and B1 may be on the same chromosome within the pair, in which case A1 and B1 are said to be "coupled";

· They may be on different chromosomes, in which case A1 and B1 are said to be in a state of "repulsion".

· Assuming that there is gamete equilibrium at the A and B loci, in parent 1 there is a probability of 1/2 that alleles A1 and B1 will be coupled, and a probability of 1/2 that they will be in repulsion.

· (1) A1 and B1 are coupled,

· The probability that parent (1) provides the gametes A1B1 and A2B2 is (1-)/2 and the probability that this parent provides gametes A1B2 and A2B1 is /2. The probability that the couple will have child of type (1) or (2) is (1-)/2, and that of their having a type (3) or type (4) child is /2.

· The probability of finding n1 children of type (1), n2 of type (2), n3 of type (3) and n4 of type (4) is therefore

· [(1- )/2]n1+n2 x (/2)n3+n4

· (2) A1 and B1 are in a state of repulsion

· The probability that parent (1) provides the gametes A1B2 and A2B1 is (1-)/2 and the probability that this parent provides gametes A1B1 and A2B2 is /2.

· The probability of the previous observation is therefore: (/2)n1+n2 x[(1-)/2]n3+n4

· With no additional information about the A1 and B1 phase, and assuming that the alleles at the A and B loci are in a state of coupling equilibrium, the probability of finding n1, n2, n3 and n4 children in categories (1), (2), (3), (4) is: p(n1,n2,n3,n4/)=1/2{[(1 -)/2]n1+n2 x (/2)n3+n4 + (/2) n1+n2 x [(1-)/2] n3+n4}

· So the liklihood of for an observation n1, n2, n3, n4 can be written:

· In the special case: number of children n= 1,regardless of the category to which this child belongs :L(q) = 1/2 [(1-)/2] + 1/2 [/2] = 1/4

· The likelihood of this observation for the family does not depend on . We can say that such a family is not informative for .

· An "informative family" is a family for which the liklihood is a variable function of .

· One essential condition for a family to be informative is, therefore, that it has more than one child. Furthermore, at least one of the parents must be heterozygotic.

· Definition: if one of the parents is doubly heterozygotic and the other is

o A double homozygote, we have a backcross

o A single homozygote, we have a simple backcross

o A double heterozygote, we have a double intercross

Definition Of The "Lod Score" Of A Family

· Take a family of which we know the genotypes at the A and B loci of each of the members.

· Let L() be the likelihood of a recombination fraction 0 and < 1/2

· L(1/2) is the likelihood of = 1/2, that is of independent segregation into A and B.

· The lod score of the family in is:

· Z() = log10 [L()/L(1/2)]

· Z can be taken to be a function of defined over the range [0,1/2].

· The likelihood of a value of for a sample of independent families is the product of the likelihoods of each family, and so the lod score of the whole sample will be the sum of the lod scores of each family.

· Several methods have been proposed to detect linkage: "U scores", were suggested by Bernstein in 1931, "the sib pair test" by Penrose in 1935, "likelihood ratios" by Haldane and Smith in 1947, "the lod score method" proposed by Morton in 1955 (1). Morton’s method is the one most commonly used at present.

· The test procedure in the lod score method is sequential (Wald, 1947 (2)). Information, i.e. the number of families in the sample, is accumulated until it is possible to decide between the hypotheses H0 and H1 :

· H0 : genetic independence = 1/2

· H1: linkage of 1 0 < 1 < 1/2

· The lod score of the 1 sample

· Z(1) = log10 [L(1)/L(l/2)]