Chapter 22 - Quantitative Genetics
These notes are provided to help direct your study from the textbook. They are not designed to explain all aspects of the material in great detail; that is what class time and the textbook is for. If you were to study only these notes, you would not learn enough genetics to do well in the course.
Quantitative traits exhibit continuous variation in their phenotypic
expression, e.g. height, weight, IQ, or skin color in humans.
These traits are considered to have phenotypic expressions that are due
to the additive influence of three or more genes that govern the same
trait. These complexes of genes are referred to as multiple genes or
polygenes (Figure 18.1).
Refer back to a monohybrid cross in which the alleles are incompletely
dominant (figure 22.1 and 22.2 and table 22.1).
(red) RR x rr (white)
F1 x F1 Rr (pink)
F2 RR Rr rr -- genotype
(red) (pink) (white) -- phenotype
1 2 1 -- genotypic ratio
2 1 0 -- units of red color
Dihybrid, incomplete dominance (Figure 18.3 and Figure 18.5).
(red) AABB x aabb (white)
F1 AaBb (pink)
F1 x F1 red reddish pink pink whitish pink white
1 4 6 4 1
1 AABB (red, 4 additive alleles)
2 AaBB, 2 AABb (reddish pink, 3 additive alleles)
1 aaBB, 4 AaBb, 1 AAbb (pink, 2 additive alleles)
2 aaBb, 2 Aabb (whitish pink, 1 additive alleles)
1 aabb (white, 0 additive alleles)
AABB AaBB aaBB AABb AaBb aaBb aabb Aabb AAbb
1 2 1 2 4 2 1 2 1
3 genes (table 22.1 and figure 22.5)
AABBCC x aabbcc
F2 yields a phenotypic ratio of . . .
red pink white
1 6 15 20 15 6 1
This produces a curve in which most individuals are pink with a very
small number of individuals that are either red or white. With more
genes, the curve becomes smoother until it approaches the standard bell
Skin color in humans is a quantitative trait. The children (F1)
of black-skinned and white-skinned parents have skin color that is
intermediate between the parents.
By examining the children (F2) from a F1 x F1 marriage,
the F2s are on average similar to the F1s, but with greater variance.
Remember that in a 4-hybrid cross, the completely homozygous recessive
individual should occur 1 out of (24)2 times (Table 2.3).
In marriages between individuals who are F1s, the most extreme
phenotypes (black or white) occur about 1 in every 256 births.
By solving the equation
----- = ------
for n, we can see that n equals 4. Thus, the data are consistent with a
model of four loci, each segregating two alleles. At each locus, one
allele adds a measure of color whereas the other does not. The additive
genetic effects at a locus are what determines that an offspring will,
in general, have a phenotypic value intermediate between its parents.
Additive Genetic Variance
Ultimately, we would like to determine the effect of genes on the
expression of traits in a population. This influence will be due to
genetics and to the environment. In evaluating the effects of genes, we
need to examine such parameters as the heritability of a trait and a
population's response to selection. Both of these phenomenon depend
upon an understanding of the additive genetic variance and the
additive genetic variance depends upon the concept of allele
frequencies. As response to selection and heritability are population
phenomenon, we need to introduce a little population genetics.
Frequencies of alleles in a population
The phenotypic distribution of flower color among a sample of 300 flowers.
Red flowers (RR) = 120 f(RR) = 120/300 = .40
Pink flowers (Rr) = 150 f(Rr) = 150/300 = .50
White flowers (rr)= 30 f(rr) = 30/300 = .10
# of r alleles
f(r) = -----------------
total # of alleles
(It is easier to think in terms of alleles rather than genotypes.)
2(120) + 150 304
p = f(R) = ------------- = --- = .65
2(30) + 150 96
q = f(r) = ------------ = --- = .35
As there are only two alleles in this example, p + q = 1 and q = 1 - p.
(See pages 671 - 672)
Phenotypic Mean and Variance of a Population
What is the mean flower color of the population above?
mean(xbar) = sigma (sum across all classes of) fi*xi
where fi is the frequency of the ith class and xi is the phenotypic
value of ith class. This equation is derived from equation 22.1.
For the population above, where red is assigned a phenotypic value
of 1, pink is assigned a value of 0, and white is assigned a value
of -1, the equation for the mean phenotypic value becomes
mean(xbar) = .40*1 + .50*0 + .10*(-1) = .30, which means that the average
color of a flower is reddish-pink, though no flower in the population
will have this exact color.
Now we need to derive a more general formula for the mean and then one
for the variance.
In our 1 locus problem, the mean phenotype in the population is a weighted
average of the three phenotypic values times the genotypic frequencies.
The genotypic frequencies in a population are determined by the allele
frequencies. The relationship is called the Hardy-Weinberg relationship,
which we will cover more thoroughly later.
RR Rr rr -- genotype
p2 2pq q2 --expected genotypic ratio
a 0 -a -- phenotypic value
where p2 is the frequency of the red flowers, 2pq is the frequency
of the pink flowers, q2 is the frequency of the white flowers,
and a is a unit of red color produced by a single additive allele
mean (xbar) x = p2(a) + 2pq(0) + q2(-a)
remember that p + q = 1 and simplify the equation to
mean (xbar) = 2ap - a
variance V = sigma fi(xi - xbar)2
This equation is derived from equation 18.2.
V = p2[a - (2ap - a)]2 + 2pq[0 - (2ap - a)]2 + q2[-a - (2ap - a)]2
remember that p + q = 1 and simplify the equation to
V = 2pqa2
Note that V is directly proportional to p*q, and p*q is a maximum when
p = q = .5
We now have an equation to determine the additive genetic variance for a
population, if we know the allelic frequencies in the population and the
contribution to the phenotype of each additive allele. For the vast
majority of cases, the inheritance of traits is too complicated to use
this formula to determine the additive genetic variance and we must use
other methods. However, it is important to understand that the additive
genetic variance is the variance among the phenotypes in a population
that is due to the additive effects of the alleles affecting the trait
Partitioning the Phenotypic Variance
The total phenotypic variance can be partitioned (split) into that due
to genetics and that due to the environment.
VP = VG + VE + VGE (equation 22.11)
VP = total phenotypic variance in the population
VG = contribution to the variance from genetics
VE = contribution to the variance from the environment
VGE = contribution to the variance from a genotype by environment interaction
The genetic variance VG can be further partitioned into the additive
genetic variance, the dominance genetic variance, and the epistatic (genic interaction)
genetic variance, where
VG = Va (additive) + Vd (dominance) + Ve (epistatic)
VP = Va + Vd + Ve + VE + VGE (equation 22.13)
heritability in the narrow sense = hN2 = Va/VP
Heritability in the narrow sense represents the proportion of the total
phenotypic variance in a population that is due to the additive genetic
variance. One of several ways to estimate hN2 is to plot the offspring
phenotypic value versus the midparent phenotypic value for a large
number of families in a population. The slope of the regression line
through the data is hN2. (Figure 22.17).
Another way to determine hN2 is to calculate the correlation
for a phenotypic trait bwtween pairs of dizygotic twins and then divide
that by the expected correlation coeficient based on the degree of
relatedness. That ratio is an estimate of hN2.
If we set up careful experiments using regression against other relatives,
we can estimate Vd and Ve. However, these estimates require
large sample sizes if we are striving for precision.
Response to Selection
Practically any character you could examine or measure of a species
would vary within a population. This includes not only morphological
traits, but physiological traits (heart rate), behavioral traits
(aggression), biochemical traits (resistance to toxins), etc.
In addition, many of the characters are inherited in some fashion.
That is hN2 is positive and there is a positive Va.
Because of this, for most characters that have been examined, the mean
of a population can be shifted via selection (selective breeding
regimens) (figures 22.21 and 22.22). For a character that has no additive genetic
variance (Va = 2pqa2 = 0), a change due to selection is not possible.
Selection can only cause a change in a population when Va > 0,
and the rate and extent of change is proportional to Va.
hN2 = ----------------- = --------
Va + Vd + Ve + VE VP
The response to selection is directly proportional to hN2, where
R = hN2 * S, where
S is called the selection differential and
R is the response to selection
If we have a population whose mean is Ybar before selection and we
select a few individual with one extreme phenotype for breeders for the
next generation, then the mean of the offspring (Ybaro) minus Ybar
is the response to selection. The mean of the breeders (Ybarb) minus Ybar
is the selection differential. R/S equals hN2
Monozygotic twins result from one fertilized egg splitting into two
embryos Pages 141 - 145. Monozygotic rates are relatively constant when considering race
and maternal age.
Dizygotic twins result from two eggs being fertilized. They occur
because of multiple ovulations and are more common in older women.
This may be under some genetic control as it tends to run in families.
Also, some ethnic groups have higher rates.
Monozygotic twins reared together are genetically identical (G) and have
the same familial environment (E2) and they only differ on their
individual environments (E1). If you plot the phenotypic value
of one twin versus the other twin, the correlation between them is
reduced from 1.0 due to the effects of their different individual
Monozygotic twins reared apart are genetically identical (G), but have
different familial and individual environments (E1 + E2).
The correlation between monozygotic twins reared apart is due to the
common genotype that they share. Thus that correlation is G.
The difference between the correlations for monozygotic twins reared apart
and monozygotic twins reared together is E2, the familial environment.
In one study on intelligence, the correlation between monozygotic twins
reared together was 0.87, which means that E1 is equal to 0.13.
The correlation between monozygotic twins reared apart was 0.69, which
means that G is equal to 0.69. The difference between the two
correlation coefficients (0.87 - 0.69) is 0.18, which is equal to E2.
Thus genetics plays a major role in determining the variance in
intelligence in this population.
Multiple gene hypothesis has 6 assumptions concerning the behavior
of the alleles and their expression. These six assumptions basically
summarize the behavior of multiple loci and their effects on a
1) no allelic pairs exhibit dominance of one allele over another
2) each contributing allele in a series has an equal effect
3) the effect of each contributing allele is additive
4) no genetic interaction, or epistasis, occurs among the alleles
at different loci in a polygenic series
5) no genetic linkage is exhibited between the genes in a polygenic
series, and therefore they assort independently
6) there are no environmental effects
The last two assumptions are violated most often.
The multiple gene hypothesis forms a satisfactory place from which to
start when analyzing a quantitative trait.
Last update on 24 October 2004
Provide comments to Dwight Moore at email@example.com
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