1. Lecture 5: Major Genes, Polygenes, and QTLs

2. Major genes --- genes that have a significant effect on the phenotype
Polygenes --- a general term of the genes of small
effect that influence a trait
QTL, quantitative trait locus --- a particular gene
underlying the trait.
Usually used when a gene underlying a trait is
mapped to a particular chromosomal region
Candidate gene --- a particular known gene that is of
interest as being a potential candidate for contributing
to the variation in a trait
Mendelizing allele. The allele has a sufficiently large
effect that its impact is obvious when looking at phenotype

3. Major Genes Major morphological mutations of classical
genetics that arose by spontaneous or induced mutation
Genes of large effect have been found selected lines
pygmy, obese, dwarf and hg alleles in mice
booroola F in sheep
halothane sensitivity in pigs
Major genes tend to be deleterious and are at very low frequencies in unselected populations, and contribute little to Var(A)

4. Genes for Genetic modification of muscling
“Natural” mutations in the myostatin gene in cattle

5. “Natural” mutation in the callipyge - gene in sheep

6. “Booroola” gene in sheep increasing ovulation rate
Merino Sheep

7. Major genes for mouse body size
The mutations ob or db cause deficiencies in
leptin production, or leptin receptor deficiencies

8. Major Genes and Isoalleles
What is the genetic basis for quantitative variation?
Honest answer --- don’t know.
One hypothesis: isoalleles. A locus that has an allele of
major effect may also have alleles of much smaller effect
(isoalleles) that influence the trait of interest.
Structural vs. regulatory changes
Structural: change in an amino acid sequence
Regulatory: change affecting gene regulation
General assumption: regulatory changes are likely more
important

9. Cis vs. trans effects
Cis effect --- regulatory change only affects gene
(tightly) linked on the same chromosome
Trans effect --- a diffusible factor that can influence
regulation of unlinked genes
Trans-acting locus. This locus influences genes on
other chromosomes and non-adjacent sites on the same
chromosome
Cis-acting locus. The allele influences
The regulation of a gene on the same
DNA molecule

10. TRANS-modifiers
MASTER modifiers
CIS-modifiers
Genomic location of genes on array
Genomic location of mRNA level modifiers

11. Polygenic Mutation For “normal” genes (i.e., those with large effects)
simply giving a mutation rate is sufficient
(e.g. the rate at which an dwarfing allele
appears)
For alleles contributing to quantitative variation,
we must account for both the rate at which mutants
appear as well as the phenotypic effect of each
Mutational variance, Vm or s2m - the amount of new additive
genetic variance introduced by mutation each generation
Typically Vm is on the order of 10-3 VE

12. Simple Tests for the Presence of Major Genes
Simple Visual tests:
• Phenotypes fall into discrete classes
• Multimodality --- distribution has several modes (peaks)
Simple statistical tests
• Fit to a mixture model (LR test)
p(z) = pr(QQ)p(z|QQ) +pr(Qq)p(z|Qq) + pr(qq)p(z|qq)
• Heterogeneity of within-family variances
Select and backcross

13. ( ) Mixture Models p ( z ) = X P r ' ; π æ
The distribution of trait value z is the weighted sum of n
underlying distributions p ( z ) = n X i 1 P r
The probability that a random individual is from
class i
The distributions of phenotypes conditional of
the individual belonging to class i
The component distributions are typically assumed normal p ( z ) = n X i 1 P r ' ; π æ 2 ( ) ' z ; π i æ 2 = 1 p e x ∑ ∏
Normal with mean m and variance s2
Typically assume common variances -> 2n-1 parameters
3n-1 parameters:
n-1 mixture proportions, n means, n variances

14. ` ( z ) = ; ¢ Y In quantitative genetics, the underlying classes are
typically different genotypes (e.g. QQ vs. Qq) although
we could also model different environments in the same
fashion
Likelihood function for an individual under a mixture model ` ( z j ) = P r Q p + q ' ; π æ 2
Mixture proportions follow from Hardy-Weinberg,
e.g. Pr(QQ) = pQ* pQ
Likelihood function for a random sample of m individuals ` ( z ) = 1 ; 2 m Y j
. . .

15. Likelihood Ratio test for Mixtures
Null hypothesis: A single normal distribution is
adequate to fit the data. The maximum of the
likelihood function under the null hypothesis is m a x ` ( z 1 ; 2 ) = S e p X j @ A … X S 2 = 1 m ( z i )
The LR follows a chi-square distribution with n-2 df, where
n-1 = number of fitted parameters for the mixture
The LR test for a significantly better fit under a mixture
is given by 2 ln (max { likelihood under mixture}/max l0 )

16. Complex Segregation Analysis
A significant fit to a mixture only suggests the possibility
of a major gene.
A much more formal demonstration of a major gene is
given by the likelihood-based method of Complex
Segregation Analysis (CSA)
Testing the fit of a mixture model requires a sample of
random individuals from the population.
CSA requires a pedigree of individuals. CSA uses
likelihood to formally test for the transmission of
A major gene in the pedigree

17. Building the likelihood for CSA
Start with a mixture model
Difference is that the mixing proportions are not the
same for each individual, but rather are a function of
its parental (presumed) genotypes ` ( z i j g f ; m ) = 3 X o 1 P r ' π æ 2
Transmission Probability of an offspring having genotype
go given the parental genotypes are gf, gm.
Phenotypic value of individual j in family i
Major-locus genotypes of parents
Phenotypic variance conditioned on major-locus genotype
Mean of genotype go
Sum is over all possible genotypes, indexed by go =1,2,3 ( g o = 3 j f 1 ; m 2 ) q Q
Example: code qq=3, Qq=2,QQ=1
Sum over all possible parental genotypes
Likelihood for family i ` ( z i ) = 3 X g f 1 m j ; ` ( z i j g f ; m ) = n Y 1
Conditional family likelihood

18. Transmission Probabilities
Explicitly model the transmission probabilities P r ( q j g f ; m ) = 1 ø Q + -
Probability that the father transmits Q
Probability that the mother transmits Q
Formal CSA test of a major gene (three steps):
• Significantly better overall fit of a mixture model compared
with a single normal
• Failure to reject the hypothesis of Mendelian segregation :
tQQ = 1, , tQq = 1/2, tqq = 0
• Rejection of the hypothesis of equal transmission for all genotypes (tQQ = tQq = tqq )

19. CSA Modification: Common Family Effects
Families can share a common environmental effect
Expected value for go genotype, family i is mgo + ci
Likelihood conditioned on common family effect ci ` ( z i j g f ; m c ) = n Y 1 2 4 3 X o P r ' π + æ 5 ` ( z i j g f ; m ) = Z 1 c ' æ 2 d
Unconditional likelihood (average over all c --- assumed
Normal with mean zero and variance sc2
Likelihood function with no major gene, but family effects ` ( z i ) = Z 1 j c ' ; æ 2 d 4 n Y π + 3 5

20. Maps and Mapping Functions
The unit of genetic distance between two markers is
the recombination frequency, c
If the phase of a parent is AB/ab, then 1-c is the
frequency of “parental” gametes (e.g., AB and ab),
while c is the frequency of “nonparental” gametes
(e.g.. Ab and aB).
A parental gamete results from an EVEN number of
crossovers, e.g., 0, 2, 4, etc.
For a nonparental (also called a recombinant) gamete,
need an ODD number of crossovers between A & b
e.g., 1, 3, 5, etc.

21. Hence, simply using the frequency of “recombinant”
(i.e. nonparental) gametes UNDERESTIMATES
the m number of crossovers, with E[m] > c
In particular, c = Prob(odd number of crossovers)
Mapping functions attempt to estimate the expected
number of crossovers m from observed recombination
frequencies c
When considering two linked loci, the phenomena
of interference must be taken into account
The presence of a crossover in one interval typically
decreases the likelihood of a nearby crossover

22. c = + ° 2 ( 1 ± ) Suppose the order of the genes is A-B-C.
If there is no interference (i.e., crossovers occur
independently of each other) then
Even number in A-B, odd number in B-C
Probability(odd number of crossovers btw A and C)
Odd number of crossovers btw A & B and even
number between B & C c A C = B ( 1 ) + 2
We need to assume independence of crossovers in
order to multiply these two probabilities
When interference is present, we can write this as
Interference parameter c A C = B + 2 ( 1 )
d = 1 --> complete interference: The presence of
a crossover eliminates nearby crossovers
= 0 --> No interference. Crossovers occur
independently of each other

23. m = ° l n ( 1 2 c ) Mapping functions. Moving from c to m
Haldane’s mapping function (gives Haldane map
distances)
Assume the number k of crossovers in a region
follows a Poisson distribution with parameter m
This makes the assumption of NO INTERFERENCE
Pr(Poisson = k) = lk Exp[-l]/k!
l = expected number of successes c = 1 X k p ( m ; 2 + ) e ! -
Odd number
Prob(Odd number of crossovers) m = l n ( 1 2 c )
This gives the estimated Haldane distance as
Usually reported in units of Morgans or Centimorgans (Cm)
One morgan --> m = One Cm --> m = 0.01

24. Linkage disequilibrium mapping
Idea is to use a random sample of individuals from
the population rather than a large pedigree.
Ironically, in the right settings this approach has
more power for fine mapping than pedigree analysis.
Why?
Key is the expected number of recombinants.
in a pedigree, Prob(no recombinants) in n
individuals is (1-c)n
LD mapping uses the historical recombinants in
a sample. Prob(no recomb) = (1-c)2t, where t =
Time back to most recent common ancestor

25. Expected number of recombinants in a sample of
n sibs is cn
Expected number of recombinants in a sample of
n random individuals with a time t back to the
MRCA (most recent common ancestor) is 2cnt
Hence, if t is large, many more expected recombinants
in random sample and hence more power for very
fine mapping (i.e. c < 0.01)
Because so many expected recombinants, only works
with c very small

26. Fine-mapping genes
Suppose an allele causing a large effect on the trait
arose as a single mutation in a closed population
New mutation arises on
red chromosome
Initially, the new mutation is
largely associated with the
red haplotype
Hence, markers that define the red haplotype are
likely to be associated (i.e. in LD) with the mutant allele

27. This linkage disequilibrium decays slowly with time if
c is small
Let p = Prob(mutation associated with original haplotype)
p =(1-c)t
Thus if we can estimate p and t, we can solve for c,
c = 1- p 1/t

28. Diastrophic dysplasis (DTD) association with
CSF1R marker locus alleles
Allele
Normal
DTD-bearing
1-1
4 (3.3%)
144 (94.7%)
1-2
28 (22.7%)
1 (0.7%)
2-1
7 (5.7%)
0 (0%)
2-2
84 (68.3%)
7 (4.6%)
Most frequent allele type varies between
normal and DTD-bearing haplotypes
Hence, allele 1-1 appears to be on the original haplotype
in which the DTD mutation arose --> p = 0.947
c = 1- p 1/t = /100
100 generations to MRCA used
for Finnish population
Gives c = between marker and DTD. Best
Estimate from pedigrees is c = (1.2cM)

29. Candidate Loci and the TDT
Often try to map genes by using case/control contrasts,
also called association mapping.
The frequencies of marker alleles are measured in both a
case sample -- showing the trait (or extreme values)
control sample -- not showing the trait
The idea is that if the marker is in tight linkage, we might
expect LD between it and the particular DNA site causing
the trait variation.
Problem with case-control approach: Population
Stratification can given false positives.

30. When population being sampled actually consists of
several distinct subpopulations we have lumped together,
marker alleles may provide information as to which group
an individual belongs. If there are other risk factors in
a group, this can create a false association btw marker and
trait
Example. The Gm marker was thought (for biological
reasons) to be an excellent candidate gene for
diabetes in the high-risk population of Pima indians
in the American Southwest. Initially a very strong
association was observed:
Gm+
Total
% with diabetes
Present
293 8%
Absent
4,627
29%
The association was re-examined in a population of Pima
that were 7/8th (or more) full heritage:
Problem: freq(Gm+) in Caucasians (lower-risk diabetes
Population) is 67%, Gm+ rare in full-blooded Pima
Gm+
Total
% with diabetes
Present 17
59%
Absent
1,764
60%

31. Transmission-disequilibrium test (TDT)
The TDT accounts for population structure. It requires
sets of relatives and compares the number of times a
marker allele is transmitted (T) versus not-transmitted
(NT) from a marker heterozygote parent to affected
offspring.
Under the hypothesis of no linkage, these values
should be equal, resulting in a chi-square test for
lack of fit: 2 t d = ( T N ) +

32. Scan for type I diabetes in Humans. Marker locus
D2S152
Allele T NT c2 p
228 81 45
10.29
0.001
230 59 73
1.48
0.223
240 36 24
2.30
0.121 2 = ( 8 1 4 5 ) + : 9