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3. The Genetics of a Single Locus: Qualitative Traits
Key Concepts________________________________________________________________
•
With complete dominance, the phenotype of the heterozygote is the same as the phenotype of the
dominant homozygote.
•
With incomplete dominance, the phenotype of the heterozygote is between the phenotypes of both
homozygotes.
•
With co-dominance, the heterozygote phenotype shows the characteristics of both homozygote
phenotypes.
•
The allele frequency is the frequency of the homozygote plus half the frequency of the heterozygote.
•
With random mating, genotype frequencies are given by the Hardy-Weinberg equilibrium frequencies.
•
In animal breeding, single locus genetics of qualitative traits is important in connection with monogenic
diseases that occur in animal breeding populations.
1. Modes of expression
This chapter deals with the genetics of a single locus that affects a qualitative trait. Remember that
qualitative traits are not measured on an ordered scale, but instead qualitative traits are classified
according to a number of discrete classes. As an example of a qualitative trait, we will use the color
of garden peas, which was first used by Gregor Mendel when he discovered what we now call
"Mendelian Inheritance".
Consider a locus with two alleles, denoted by A and a, where A is the allele for yellow pea color and
a the allele for green pea color. At this locus, there are three possible genotypes, both homozygotes,
AA and aa, and the heterozygote Aa. The phenotype that corresponds to a particular genotype
depends on the mode of expression. The simplest mode of expression is described by the term
complete dominance. With complete dominance, the heterozygote has the same phenotype as the
dominant homozygote. If one allele is completely dominant, the other allele is recessive. In peas,
yellow is completely dominant over green, i.e. green is recessive. Thus heterozygote peas are
yellow, and the relationship between genotypes and phenotypes is:
AA, Aa
:
yellow
aa
:
green
In most cases, the mode of expression is more complex. Often, the phenotype of the heterozygote is
somewhere in between the phenotypes of both homozygotes, which is called incomplete dominance.
An example of incomplete dominance is flower color in four-o'clocks. (Four-o'clocks are plants native
to tropical America, Griffiths et al., 1999, p. 110). In four-o'clocks, the relationship between genotypes
and phenotypes is:
CC
:
red flowers
_
Cc
cc
:
:
pink flowers
white flowers
Finally, there is the situation where the heterozygote expresses both the phenotype of the one
homozygote and also the phenotype of the other homozygote, which is called co-dominance. An
example of co-dominance are the human blood groups A, B and AB. The blood group of a person is
determined by the presence of a unique antigen, which is deposited on the surface of the red blood
cells. The relationship between the genotype (the alleles) and the phenotype (the antigen or blood
group) is:
AA
:
only antigen A, thus bloodgroup A
AB
:
both antigen A and B, thus blood group AB
BB
:
only antigen B, thus blood group B.
Thus the AB heterozygote has both the A-antigen and the B-antigen, i.e. the phenotypes of both
homozygotes. In classical animal breeding, co-dominance is not very important. In molecular
genetics, however, co-dominance is important with respect to genetic markers. A "co-dominant
marker" is a marker for which we can distinguish the heterozygote from both homozygotes, because
the heterozygote expresses both marker alleles.
2. Genes at the population level
In the previous section, we have looked at the expression of genes within individuals. The properties
of a whole population, however, do not depend only on the expression of genes within individual
animals, but also on the frequency and distribution of alleles within the population.
First of all, it is important to distinguish between the outcome of a specific cross versus the genotype
frequencies in a whole population. In a mono-hybrid cross, such as Aa × Aa, the proportions of the
different genotypes are ¼AA, ½Aa and ¼aa. Mono-hybrid crosses are common in plant breeding. In
animal breeding, however, we deal with the allele and genotype frequency in a population, which
differ from ¼ : ½ : ¼. Thus, for whole populations, one can NOT use the mono-hybrid ratios.
Consider a locus with two alleles A and a, in a population. The frequency of the A-allele in the
population is denoted by p(A) or simply by p, and the frequency of the a-allele by q. Because the
allele frequency is a proportion, p and q always sum to one.
p(A) = frequency of A
q(a) = frequency of a
p+q=1
(3.1)
The frequency of a particular allele is the proportion of all alleles in the population that are of this
type. For example, if we have a population of 1000 individuals, so that there are 2000 alleles, and
when there are 600 A-alleles in this population, the frequency of the A-allele equals p(A) = 600/2000
= 0.3.
In reality, we cannot observe alleles directly, but sometimes we can observe the genotypes of
animals. In that case, we have to estimate the allele frequency from the genotype frequency. The
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The Genetics of a Single Locus: Qualitative Triats
genotype frequencies are the proportions of the AA, Aa and aa genotypes, and are denoted by fAA,
fAa and faa. The relationship between allele frequencies and genotype frequencies is:
p(A) = fAA + ½fAa = frequency of the A-allele
q(a) = faa + ½fAa = frequency of the a-allele
(3.2)
The logic behind Equation 3.2 is that the heterozygote contains only 50% A-alleles, so that to
calculate p(A), we have to count the heterozygote only half.
Example 3.1: Consider a population of 1000 animals for which we want to calculate the allele
frequencies for a single locus that shows incomplete dominance (such as flower color in fouro'clocks). Because there is incomplete dominance, we can distinguish the heterozygote from both
homozygotes. Counting the genotypes in the population resulted in 150 individuals with genotype
AA, 100 individuals with Aa, and 750 individuals with aa, which gives genotype frequencies of fAA =
150/1000 = 0.15, fAa = 0.1 and faa = 0.75. The allele frequencies in this population are: p(A) = fAA +
½fAa = 0.15 + 0.05 = 0.2, and q(a) = faa + ½fAa = 0.75 + 0.05 = 0.8. Note that p + q = 1, as it should
be. Instead of using Equation 3.2, the allele frequencies can also be determined by simple counting.
In the population there are 150×2 + 100 = 400 A-alleles, and 750×2 + 100 = 1600 a-alleles, and in
total there are 2000 alleles. Thus the allele frequencies are: p(A) = 400/2000 = 0.2 and q(a) =
1600/2000 = 0.8.
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The frequency of an allele is the proportion of alleles in the population that are of this type.
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In a randomly mating population, there is a specific relationship between the allele frequencies and
the genotype frequencies. Consider a population with allele frequencies p and q. For this population,
a proportion p of the sperm and eggs will carry the A-allele and a proportion q will carry the a-allele.
Random union of sperm and eggs, therefore, will produce a proportion p2 of offspring with genotype
AA, a proportion 2pq of offspring with genotype Aa and a proportion q2 of offspring with genotype aa.
Thus in a randomly mating population, the genotype frequencies are given by:
fAA : fAa : faa
=
p2 : 2pq : q2
(3.3)
The relationship fAA : fAa : faa = p2 : 2pq : q2 is called Hardy-Weinberg equilibrium. Figure 3.1 shows a
graphical explanation of the Hardy-Weinberg equilibrium. Equation 3.3 can be used in two ways.
First, if you know the allele frequency p of a population, and you know that this is a randomly mating
population, then you can calculate the expected genotype frequencies fAA, fAa, and faa as p2, 2pq and
q2. Second, for traits that show complete dominance, you cannot estimate p and q from the genotype
frequencies, because you cannot distinguish the heterozygote. However, if you know that it is a
randomly mating population, then you know also that faa = q2, where q is the frequency of the
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recessive allele. With a completely dominant trait, you can observe the recessive phenotype and thus
determine faa and, with random mating, estimate q as: q = √faa and p = 1 – q.
Figure 3.1 The Hardy-Weinberg equilibrium frequencies that result from random mating. The frequencies of A and a
among both eggs and sperm are p and q (=1 – p), respectively. The total frequencies of the zygote genotypes are p2
for AA, 2pq for Aa and q2 for aa. The frequency of the A-allele in the zygotes is the frequency of AA plus one half the
frequency of Aa, or p2 + pq = p(p + q) = p (Griffiths et al., 1999).
Example 3.2: Consider again the population of example 3.1, where we had 150 individuals with
genotype AA, 100 individuals with Aa, and 750 individuals with aa. The question is whether this
population is in Hardy-Weinberg equilibrium? To solve this question, first you have to derive the
genotype frequencies that you would expect when there is Hardy-Weinberg equilibrium, and second,
compare those expected genotype frequencies to the observed genotype frequencies. The expected
genotype frequencies with Hardy-Weinberg equilibrium are given by fAA : fAa : faa = p2 : 2pq : q2. From
example 3.1, we know that p = 0.2 and q = 0.8, so that the expected genotype frequencies are fAA :
fAa : faa = 0.04 : 0.32 : 0.64. (Note that fAA + fAa + faa = 1, which is always the case.) From example 3.1,
the observed genotype frequencies are: 0.15 : 0.1 : 0.75, which clearly deviates from the expected
values, i.e., 0.15 : 0.1 : 0.75 ≠ p2 : 2pq : q2. This population, therefore, is not in Hardy-Weinberg
equilibrium. In this example, the difference between observed and expected genotype frequencies is
quite large and the result is clear. When the difference between the observed and the expected
genotype frequencies is small, you can use a Chi-squared statistical test to see whether it is
significant.
There are two common mistakes that you should not make on the exam. First, the fact that p2 + 2pq
+ q2 = 1 does NOT mean that there is HW-equilibrium; p2 + 2pq + q2 is always equal to one, also
when there is no HW-equilibrium. Second, when you are asked to check whether a population is in
HW-equilibrium, you CANNOT calculate the allele frequency as q = √faa; you can only use q = √faa
when you already know that there is HW-equilibrium.
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The Genetics of a Single Locus: Qualitative Triats
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With random mating, genotype frequencies are given by: fAA : fAa : faa = p2 : 2pq : q2
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3. Monogenic recessive diseases
In most cases in animal breeding, we deal with quantitative traits that are determined by many loci.
However, there are qualitative traits that are determined by a single locus only. Monogenic heritable
diseases are an example of such traits.
In most cases, monogenic diseases are recessive. This is not a coincidence, but can be understood
easily from an evolutionary perspective. First consider a monogenic dominant disease. With a
dominant disease, there are two types of animals that express the disease, the heterozygote Dd and
the homozygote DD. The other homozygote, dd, does not express the disease. Suppose that the
disease occurs early in life and is reasonably detrimental to the animal. In that case, animals that are
either Dd or DD, i.e. the animals that show the disease, will have reduced fertility and viability, and
consequently those animals will contribute fewer offspring to the next generation than animals that
are free of the disease. In the next generation, therefore, there will be a relative increase of the
number of offspring of dd parents, whereas the number of offspring of either Dd or DD parents will be
decreased. As a consequence, the frequency of the D-allele will decrease over generations, i.e. there
is natural selection against the disease. Consider, for example, a lethal disease, so that the Dd and
DD animals die before reproductive age. In this case, the only individuals that contribute offspring are
the dd individuals, so that the disease is eradicated within a single generation.
Next, consider a recessive disease. With a recessive disease, only the dd individuals express the
disease, whereas the DD and Dd individuals are healthy. When the disease is lethal and occurs
before reproductive age, the dd individuals will not contribute any offspring, but the Dd heterozygote
will contribute offspring. Thus the offspring generation will still contain the d-allele, i.e. the disease is
"hidden" in the heterozygote. Natural selection against recessive diseases, therefore, is less efficient
than natural selection against dominant diseases. With recessive diseases, the heterozygotes are
often called "carriers of the disease", or simply "carriers', because they carry the allele but do not
show the disease itself.
The efficiency of natural selection against recessive diseases depends on the allele frequency.
Assuming that there is random mating, so that genotype frequencies are in Hardy-Weinberg
equilibrium, a proportion 2pq of the population will be heterozygous and a proportion q2 will be
homozygous for the disease. In most cases, natural selection will have reduced the frequency of the
disease, so that q is much smaller than p, i.e. the allele that causes the disease is less frequent than
the other allele. Consider the case where q is close to zero, so that p is close to one. In that case, the
relative frequencies of the carrier and the homozygote are 2pq : q2 ≈ 2 : q. Because only half of the
alleles of heterozygotes are d-alleles, we have to count the heterozygote only half, so that the ratio
becomes 1 : q. For example, when q = 0.01, around 99% of the disease alleles will be hidden in the
heterozygote and only 1% is expressed in homozygotes. The fact that homozygous animals don’t
contribute offspring to the next generation will hardly reduce the frequency of the disease allele,
because for every single allele in a homozygous individual there are 99 alleles in heterozygous
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individuals that are passed on to the next generation. The general equation for the change of allele
frequency of a lethal recessive allele is given by:
q t +1 =
qt
1 + qt
(3.4)
where qt is the allele frequency at generation t. Figure 3.2 illustrates that the decrease of allele
frequency becomes increasingly slow when the allele frequency reaches low values.
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(Natural) selection against lethal recessive alleles is relatively inefficient.
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qt
0
0.2
0.4
0.6
0.8
0
(q t +1 -q t )/q t
-0.1
-0.2
-0.3
-0.4
-0.5
-0.6
Figure 3.2 Relative change of allele frequency, (qt+1–qt)/qt, of a recessive
lethal allele. Note that the allele frequency hardly changes when it is close
to zero.
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The Genetics of a Single Locus: Qualitative Triats
Box 3.1 BLAD
BLAD (Bovine Leukocyte Adhesion Deficiency) is an inherited disease in Holstein-Friesian dairy cattle. BLAD
is an autosomal recessive and lethal disease. A similar disease exists in humans. The symptoms are reduced
resistance against infectious diseases and leukocytosis with more than 30,000 cells per micro-liter of blood.
In cattle, the calves die at an early age due to bacterial infection. In humans, the disease can be cured by
bone marrow transplants. Animals that show the disease have a reduced expression of Beta 2 integrin on the
cell surface of leukocytes. As a consequence, the leukocytes can no longer migrate to the center of an
Box 3.1
infection. The genetic cause of the disease is a point mutation in the CD18 gene. The bovine mutation occurs
in a 250 amino acid region of CD18, where several mutations causing human LAD have been found also.
Carriers of the disease can be identified by means of a molecular genetic marker.
In the Holstein-Friesian dairy cattle population, the frequency of BLAD has increased substantially due to a
single bull; Carlin M Ivanhoe Bell (born in 1973). This bull was a very successful bull sire, but was also a
carrier of the BLAD allele. Due to the heavy use of Bell, BLAD was discovered in 1987 and subsequently a
genetic marker test has been developed to identify carriers. By using genetic markers, selection against
carrier animals has been very successful. At this moment, the disease is almost extinct in Holstein-Friesian
dairy cattle.
Carlin M Ivanhoe Bell
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