
 
 
 
 
 
 
 
 


9
8
7
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:
 R= 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.
E ( X )   kl pk pl
k
l
 For the sake of simplicity let us normalize E(x)=0, and divide:
kl     k   l   kl ;   k pk  0

k
kl
pk  0
k
E ( X i X j )   2   mn kl p(m, n | k , l ) pk pl
k
l
m
n
    7 ij  ( k   l   kl ) 2 pk pl   8ij  ( k   l   kl )( k   m   km ) pk pl pm
2
k
l
k
l
m
 9ij  ( k   l   kl )( m   n   mn ) pk pl pm pn
k
l
m
n




  2   7 ij  2  k 2 pk    kl 2 pk pl    8ij  2  k 2 pk 
k
l
 k

 k

1
1

  2  2   7 ij  8ij  2  k 2 pk   7 ij   kl 2 pk pl
4
2
 k
k
l
  2  2ij a2   7 ij d2
cov( X i , X j )  2ij a2   7 ij d2
R
Relative Type
Risk Ratio
M
Identical Twin
sa/2K +2sd/2K2
S
Sibling
sa2/2K +2sd2/4K2
1
First Degree
sa2/2K2
2
Second Degree
sa2/4K2
3
Third Degree
sa2/8K2
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 2 will be used to indicate observational components of variance

Mean genotypic value is equal to the mean phenotypic value

E(Gj) = E(Pj )= Pj

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
Genotyp
e
A2A2
-a


A1A2
0
d
Genotypic
value
The
mean of environmental deviations is zero Ej = Pj – Gj
A1A1
+a
, ,(E = 0)

The correlation between genotypic values and environmental deviations for a
population of subjects is zero (GE = .00)
Genotype Frequency Value Freq x Value
A1A1
p2
+a
p2a
A1A2
2pq
D
2pqd
2
A2A2
q
-a
-q2a
Sum = a(p - q) + 2dpq

Recall that p2 - q2 = (p + q)(p - q) = p - q

P = a(p - q) + 2dpq
If multiple genes contribute to a phenotype in an additive way

 P =  a i (pi - qi ) + 2d i pi q i
i

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
G
Minus pop.
mean

pa + qd
-[a(p-q) +
2dpq]
q[(a + d(q-p)]
-qa + pd
-[a(p-q) +
2dpq]
-p[a+d(q-p)]
A1A1 A1A2 A2A2
+a
d
-a
A1
p
A2
q
p

q
Thus, the average effect for each allele also can be calculated for A1 and A2
in the following manner (Falconer, 1989):
1  pa  qd - [a( p - q)  2dpq]  pa  qd - ap  aq - 2 dpq

 q (d + a  2dp )  q ( a  d (q  p ))
 2  - p[a  d ( q - p)]
  1   2  q( a  d (q  p))  p[a  d (q - p)]  a  d (q - p)

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 + d(q - p)

When gene frequency is greater  is greater
q = 0.10 q = 0.40
+0.24
+1.44
1 =
-2.16
-2.16
2 =
2.40
3.60
 = 1 - 2
 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 2P attributable to 2A 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:
21 = 2q

A1A2:
 1 +  2 = (q - p)

A2A2:

Mean breeding value under HWC equilibrium is zero

2p2q + 2pq(q - p) - 2q2p2pqa(p + q - p - q) = 0
2 2 = -2p
2q
+
ad
0
(q p)
}
0

2p
-a
Genotypic values
Breeding values


A2A2
A1A2
A1A1
q2
2pq
p2
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
AA
Aa
aa
BB
Bb
bb
Locus
B
Locus
A

locus A shows an association with the trait

locus B appears unrelated
AA
Aa
aa
BB
Bb
bb
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
number of recombinant gametes
total gametes

Recombination fraction ( ) 

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.

 = 0 : no recombination => complete linkage

 < 0.5 : partial linkage

 = 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:
L( |n1,n2,n3,n4)=1/2 {[(1- )/2]n1+n2 ( /2)n3+n4 + ( /2)

n1+n2
[(1- )/2]
n3+n4
}
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)]

indicates the relative probabilities of finding that the sample is H1 or H0.
Thus, a lod score of 3 means that the probability of finding that the sample is
H1 is 1000 times greater than of finding that it is H0
("lod = logarithm of the odds").

Test For Linkage

The decision thresholds of the test are usually set at -2 and +3, so that if:

Z(  1) > 3 H0 is rejected, and linkage is accepted.

Z(  1) < -2 linkage of H1 is rejected.

-2 < Z(  1) < 3 it is impossible to decide between H0 and H1. It is necessary
to go on accumulating information.

For the thresholds chosen, -2 and +3, we can show that:

The first degree error, (False negative) < 10-3

The second degree error (false positive), < 10-2

The reliability, 1-

Significance of results

but a whole set of values between 0 and 1/2, with a step of various size (0.01
or 0.05).




-2



wn, the sample is not
sufficiently informative.


Criticism
The proposed test has the advantage of being very simple, and of providing
protection against falsely concluding linkage.

However, some criticisms can be leveled, not only against the criteria chosen ,
but also against the entire principle of using a sequential procedure .

The number of families typed is, indeed, rarely chosen in the light of the test
results.
Estimation Of The Recombination Fraction

If the test, on a sample of the family, has demonstrated linkage between the A
and B loci, then one may want to estimate the recombination fraction for these
loci.

The estimated value of
probability of observing linkage in the sample is greatest.
Recombination Fraction For A Disease Locus
And A Marker Locus

Let us assume we are dealing with a disease carried by a single gene,
determined by an allele, g0, located at a locus G (g0: harmful allele, G0:
normal allele).

We would like to be able to situate locus G relative to a marker locus T, which
is known to occupy a given locus on the genome. To do this, we can use
families with one or several individuals affected and in which the genotype of
each member of the family is known with regard to the marker T.

In order to be able to use the lod scores method described above, what is
needed is to be able to extrapolate from the phenotype of the individuals
(affected, not affected) to their genotype at locus G (or their genotypical
probability at locus G)