Download 3. Inheritance and hereditary

 3. Inheritance and hereditary
How do you determine how a phenotype is inherited? A specific phenotype is an
observable characteristic of an organism, and is encoded by the genotype, or genetic makeup,
of an organism. Specific phenotypes are usually studied with the goal of understanding how the
genes function within the organism to create its phenotype. For example, suppose you were
examining a population of nematodes and found a worm that was short and stout (dumpy)
instead of its usual body shape. Why are these worms dumpy? To start to understand this
question, you might begin by examining the inheritance pattern for this phenotype.
Understanding how this phenotype is inherited would allow you to learn things like how many
genes are responsible for the phenotype and how heritable this phenotype is. Once a gene is
pinpointed as being required for the inheritance of a phenotype, further studies on a gene can
be carried out. These types of studies include examining the function of that gene’s product
through genetic analysis (Chapter 4), investigating how that gene’s product interacts with other
gene products to determine phenotype (Chapter 5) and determining the physical location in the
genome (Chapter 9), which can tell you about the gene product encoded by the locus of
interest,
In this chapter, we discuss how to understand and analyze the pattern of inheritance of a
particular phenotype, starting with how to examine and understand the phenotype of interest.
We then discuss various modes of inheritance and the types of conclusions that can be made
as the inheritance pattern of phenotypes is observed. Next, we will discuss the mechanisms by
which inheritance patterns are governed, including the more traditional Mendelian inheritance
and the complications for genetic analysis from effects such as penetrance, expressivity, and
parent of origin effects. Then, we will discuss non-Mendelian inheritance, which includes
situations in which the observed phenotype is not directly predicted from the nuclear genotype.
Throughout the chapter, we will incorporate examples from key model organisms with
exemplary genetic characteristics to illustrate general principles that also apply to other
organisms, including new model organisms for which new technologies now or soon will allow
some aspects of genetic manipulation and study.
The general nomenclature used in this book is described in Box 3-1.
Understanding heritability of phenotypes
Phenotypes are observable characteristics or traits. Some phenotypes can be easily
visualized, such as a person’s eye color. Some phenotypes can require sophisticated research
tools to see, such as the localization of a particular protein within the cell (Fig. 1). The types of
phenotypes that can be studied include those that may be naturally occurring at easily
observable frequencies in populations, or those obtained in experimental organisms within the
lab (Chapter 7). Although the concept of a phenotype may seem deceptively simple, the proper
analysis of phenotypes can be complex.
Phenotypes can be defined by discrete or continuous variation
Before studying the heritability of a phenotype, exactly what is being measured as a
phenotype must be carefully defined. The classical examples of phenotypic analysis usually
involve discrete phenotypic variation, where an all-or-none trait is being discussed. This type
of variation can also be considered qualitative variation. Discrete phenotypic variation
comprises categories such as big versus small, blue eyes versus brown eyes, or alive versus
dead. On the other hand, a phenotype may involve continuous phenotypic variation, and is
usually described in a quantitative fashion. Continuous phenotypic variation is described as a
quantitative variable and involves measurements such as height, lifespan, the velocity of
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locomotion, or the resistance to stress (Fig. 2). These categories must be defined so that
phenotypes can be scored appropriately.
Phenotypes should be classified by a method that accurately and simply describes
variation
What determines whether a phenotype is treated as discrete or continuous? The
distinction depends on how the variation in a particular trait can be represented accurately, yet
as simply as possible. At one extreme, a trait whose variation is difficult to assign a numerical
value (for example, wrinkled versus smooth) is treated as discrete. On the other hand, a trait
readily assigned a numerical value (for example, height) can be treated as either discrete or
continuous depending on its pattern of variation. If the trait can be represented accurately as a
discrete trait, then the variation can be treated as a qualitative trait, which would simplify its
analysis.
For example, consider height as a phenotype you wish to study (Figure 3 [SGF-1025]). If
the heights of an organism cluster into discrete classes (in the simplest case, tall and short, with
no lengths in between), such variation can be treated as a qualitative, discrete phenotype,
simplifying its analysis. In this case, the organisms would be classified as either short or tall.
However, if the heights observed vary continuously from tall to short, then this variation is
probably best represented by describing individual phenotypes by their numerical values rather
than by partitioning them into categories, or bins.
For any qualitative phenotype, the description used to describe the discrete phenotype
implies an imposition of some threshold. Sometimes this threshold is intuitively obvious. For
example, the color of an organism can be scored as a qualitative phenotype. Whether an
organism is green or yellow is simply distinguished by the human eye, as we perceive colors
from different parts of the light spectrum as discrete colors. However, the color of an organism
can also be represented qualitatively by the wavelength of light reflected from the organism.
Although our eye perceives green and yellow as discrete, the distinction between green and
yellow represents two parts of the light spectrum (510 nm versus 570 nm, respectively) that are
not that far apart. The binning of these two parts of the light spectrum is essentially what our
eyes are doing when we decide whether an organism is either green or yellow.
Although our eye readily bins colors into discrete phenotypes, there are situations when
we might want to represent color as a continuous phenotype. For example, if your organism
displays phenotypes of “yellowish-green” or “greenish-yellow”, which are colors that lie in
between 510 and 570 nm of the light spectrum. Perhaps you suspect that multiple loci
contribute to the determination of the color of an organism, and an organism of the color from
535 nm is significantly distinct from an organism the color from 570 nm (Figure 4 [SGF-1199]). In
this case, the representation of color by the terms yellow and green would oversimplify the
phenotypes. Ultimately, this phenotypic oversimplification would be discovered when the
genetic models do not accurately predict inheritance patterns. In this chapter, we will focus on
methods used for phenotypes that show discrete phenotypic variation. The genetic dissection
of phenotypes showing continuous phenotypic variation will be discussed further in Chapter 6,
when we discuss the analysis of quantitative traits.
Determining patterns of inheritance for a phenotype
To understand how a phenotype is inherited, you will need to examine how a phenotype
behaves as it passes from the parents to its offspring. Often, the inheritance pattern over
several generations will need to be observed in order to understand the appropriate model of
inheritance.
Breeding crosses can be used to determine the pattern of inheritance
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Patterns of inheritance can be observed and determine by setting up specific breeding
crosses and examining the segregation of the phenotypes in the resulting progeny. This
strategy can be used to understand inheritance in model organisms, livestock, agricultural
organisms, or companion animals.
There are two main modes of breeding crosses: cross-fertilization and selffertilization. During cross-fertilization, organisms need to be crossed to another in order to
reproduce (i.e., Drosophila, humans, and most other common animal species) (Fig 5A [SGF1005]). Organisms that can self-fertilize include those that reproduce without being crossed to
another (Fig. 5B), such as C. elegans and some plant species. For the analysis of inheritance
pattern, microbes that undergo vegetative growth and create a new cell that is a copy of itself by
cell division (i.e., most bacteria like E. coli, as well as fungi such as the haploid S. cerevisiae)
are analyzed like self-fertilizing organisms. When these organisms grow vegetatively, no
introduction of genetic material is necessary to produce a new cell, much how a self-fertilizing
organisms propagates itself. Furthermore, when haploid microbes are mated to construct a
diploid (Fig. 5C [SGF-1042]), the creation of haploids from diploids occurs without introduction of
genetic material, thus behaving more like the self-fertilization process, as no additional crosses
are needed.
Pedigree analysis can be used to determine patterns of inheritance when breeding
crosses are not possible
In cases where crosses cannot be experimentally arranged (as in humans), pedigree
analysis is carried out to examine the pattern of inheritance. Pedigree analysis involves
examining different generations of individuals for the phenotype of interest, and using that
information to determine how that phenotype is expressed. Included in a pedigree is the
relation of each individual to the others within the pedigree. In human pedigrees, the sex of the
individual is usually stated as well. A simple example is shown in Fig. 6 [SGF-1002], where an
affected female mates with an unaffected male and generates two affected and two nonaffected progeny. Many fewer individuals are usually studied when pedigrees are used
compared to breeding crosses. Because of this, it can be difficult to narrow down a model of
inheritance for a particular phenotype. The more extensive a pedigree, the more likely you will
be able to narrow down the models of inheritance that fit that particular pedigree.
Simple versus complex models of inheritance
By examining inheritance patterns, we can construct models to explain how the
phenotype is inherited. When assessing patterns of inheritance, you usually approach the
problem by starting from one of two extreme alternative positions. At one extreme, the genetic
properties can be assumed to be simple (i.e., strict Mendelian inheritance, see below), with
more complicated hypotheses considered only when experimental observations conflict with the
simple model. At the other extreme, high complexity is anticipated from the outset.
Whether simple or complex models are considered depends on the particulars of the
analysis being conducted. If the alleles under study are present in a population that is otherwise
rather genetically homogeneous, the assumption that genetic properties are simple is most
expeditious. This is typically the case when studying laboratory-induced mutations in model
organisms, where the mutations are usually induced into a defined genetic background. In this
case, if it turns out that the genetic properties are not simple, additional genetic inter-breeding
experiments can be performed to dissect the complexity. However, for phenotypic variations
found in populations in which many genes are likely to vary, an approach involving few or no
assumptions about the nature of the underlying genetic properties is often employed. This is
typically the case for the examination of naturally occurring variations, like those studied in
humans. This approach recognizes the higher likelihood that variations in multiple genes
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underlie phenotypic variation within genetically heterogenous populations. Furthermore, in
many of these cases, the ability to test specific hypotheses through subsequent rounds of
interbreeding is limited, and thus it is better to make fewer assumptions. In these studies, data
are collected, and hypotheses subsequently formed to fit these data (e.g., Box 3-2).
Analyses involving simple models of inheritance usually involve discrete phenotypic
variation. For a simple model of inheritance, it is assumed that a single genetic locus is
responsible for the phenotypic effect and that all organisms of the appropriate genotype will
express the phenotype (i.e., Mendelian Inheritance, see below). These phenotypes are usually
assumed to be recessive to the allele found in the phenotypically wild-type strain. In addition,
the expression of the phenotype is assumed to be free of parental effect and not particularly
sensitive to environmental conditions.
However, there are many ways that inheritance does not follow the rules governing
simple inheritance, and a phenotype may display complex inheritance in many ways (Table 1).
Instead of the phenotype being due to the action of one locus, it is possible that multiple loci are
required for the phenotype. The phenotype may not be recessive to the allele found in wild type
strains. Furthermore, it is also possible that organisms that are genotypically similar may not
display the same phenotype, a phenomenon referred to as partial penetrance. The phenotype
may display differences depending on which parent it was inherited from (parent of origin effect)
and there may also be extracellular material transferred from the parent (i.e., maternal
inheritance) that may be involved. Finally, expression of the phenotype may be affected by
perturbations such that genotypically similar individuals may express the phenotype differently
depending on their environment. These more complex situations will be discussed later in this
chapter.
Like begetting like suggests genetic homogeneity
A true breeding phenotype is one where the organism and all its descendents display the
phenotype of interest (Fig. 7 [SGF-1044]). This phenomenon is observed at a population level.
For organisms that breed by crossing, a true breeding phenotype would be a phenotype that
exists in all members of a particular population where there is free crossing within that particular
population. For organisms that self-fertilize or reproduce by cell division, a true breeding
phenotype is one where an individual generates progeny of the same phenotype and each of
these progeny will generate descendents of that same phenotype.
Single cell colonies can display true breeding
True breeding phenotypes can be most simply observed in single celled organisms
growing vegetatively. A single cell that generates only cells of the same phenotype is inferred to
be true-breeding. Suppose that while studying a strain of yeast that normally forms creamcolored colonies, you isolate a colony containing a dark-pigmented sector (Fig. 8 [SGF-1014]). To
determine whether the dark pigmented phenotype is true breeding, you pick cells from the dark
colored part of the colony and use these cells to create many additional colonies, each founded
by a single dark-colored cell. Consider two different types of results you might obtain. First, the
simplest result you might obtain would be if cells that derive from the dark-colored colony all
give colonies that are dark colored. In this case, the dark-colored variant would appear to be
true breeding. A second type of result would be if cells that derive from the dark-colored colony
produce a mixture of dark and cream-colored colonies. This would suggest that the darkcolored variant was not true-breeding, as the dark-colored cells produce cells of varied
phenotype (Table 301).
Alleles represent different variants of a gene
The different variants of a given gene are referred to as alleles of that gene (see Box 3-1
for nomenclature conventions used in the text). For example, one diploid individual might
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contain two different alleles of gene A, namely a1 and a2. A diploid organism can be
homozygous for an allele, where the alleles at the gene are the same on both copies of the
chromosomes (i.e., for gene A, the organism is a1/a1). A heterozygous organism will have two
different alleles at the gene of interest (i.e., a1/a2) (Fig. 9 [SGF-1200]).
Sometimes diploid organisms are hemizygous at a particular locus and only have a
single copy of a gene. For example, if a gene were on a sex chromosome, the heterogametic
sex would be hemizygous. If the A locus were on the human X chromosome, a human male
would be hemizygous for the A locus and would be genotypically a1/O.
Mendelian Inheritance
Gregor Mendel demonstrated that variation within a single gene can underlie discrete
phenotypic variation. Thus, traits due to variation within a single gene are referred to as
Mendelian traits.
Simple Mendelian inheritance involves clear dominant-recessive relationships between
alleles
The simplest cases of Mendelian inheritance are those in which the multiple alleles of a
gene exhibit clear dominant-recessive relationships. In such a situation, a diploid animal will
express the phenotype associated with the dominant allele whenever at least one dominant
allele is present, while animals will express the phenotype associated with the recessive allele
only if the dominant allele is not present (Figure 10 [SGF-1201]). Thus, animals homozygous,
hemizygous or heterozygous for a dominant allele will show the dominant phenotype, while only
animals hemizygous or homozygous for recessive alleles will exhibit the recessive phenotype.
Often, though not always, wild-type phenotypes are associated with dominant alleles and
mutant phenotypes are associated with recessive alleles. The terms dominant and recessive
refer to the relationships of the various alleles within a genetic locus. More complicated
variations to the simple dominant-recessive relationship are discussed below under
complications to Mendelian inheritance.
Mendel’s laws describe simple patterns of inheritance
The two main genetic tenets from Mendel are the Law of Segregation and the Law of
Independent Assortment. The Law of Segregation is now understood to mean that when a
diploid organism produces gametes by meiosis, each allele is segregated into different gametes
(Fig. 11 [SGF-1003]). The Law of Independent Assortment says that this segregation of alleles
happens independently of other alleles (Figures 12 and13 [SGF-1004, 1013]). We now know this is
true for genetic loci that are not closely associated along a piece of DNA. These independently
assorting loci also known as unlinked loci, while those that that are closely associated along a
piece of DNA are considered linked (Chapter 9).
Together, Mendel’s two laws ensure that the segregation of genetic traits is intrinsically
probabilistic in nature: knowing the genotypes of the parents, you can predict the probability that
the offspring will have a particular genotype and phenotype. For a phenotype inherited in a
Mendelian fashion within a true-breeding population, the gene involved in the phenotype would
be homozygous. However, having a true-breeding strain does not always imply a homozygous
state. Sometimes, genetic situations can exist to maintain true-breeding phenotypes in a
heterozygous state (see Box 3-8); these types of situations can be useful for genetic analyses,
as we will see in later Chapters.
Mendelian traits display typical segregation patterns
One of the key features of Mendelian inheritance is the segregation of phenotypes in
precise ratios from a heterozygote, as this segregation pattern stems from Mendel’s laws (Box
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3-2). Whether a single gene determines a trait can be determined by examining the segregants
from a cross.
First consider the case in a diploid organism where we have two parents that are both
heterozygous at one locus, of genotype a1/a2. In this case, consider the a2 allele recessive to
the a1 allele such that only an organism homozygous for allele a2 displays the non-A phenotype.
For this type of cross, 1/4 of the progeny will homozygous for the a1 allele, 1/2 will be
heterozygous and have the genotype a1/a2, and 1/4 will be homozygous for the a2 allele (Fig. 14
[SGF-1055]). The ratio of progeny from this cross that will display the phenotype A to phenotype
non-A will therefore be 3:1.
As Mendel was studying peas, Mendel’s laws most simply apply to diploid organisms.
However, we sometimes speak of Mendelian inheritance in organisms that go through haploid
states in cases where inheritance follows similar rules as if the organism were always diploid.
One such example is the budding yeast S. cerevisiae, which can be grown as either a diploid or
a haploid. Rather than examining the segregation of a given trait by examining the phenotypes
of the diploid progeny created by crossing two diploid organisms, patterns of inheritance in
yeast are commonly studied by mating two haploid yeast cells to create a diploid yeast cells.
The diploid cell is then allowed to undergo meiosis through the process of sporulation, and the
meiotic products are scored (Fig. 15 [SGF-1202]). Because of this, if you were to start with a
diploid yeast cell that is heterozygous and of genotype a1/a2, the segregation pattern seen from
a yeast meiosis will be two haploid cells of phenotype A and two haploid cells of phenotype nonA.
What if more than a single locus is responsible for the phenotype of interest? This can
usually be determined from a simple cross. For a diploid organism, in the cross above, we see
that a 3:1 ratio would be expected, resulting in 3/4 of the offspring displaying the dominant
phenotype and 1/4 displaying the recessive phenotype. However, if two recessive loci were
required to produce the recessive phenotype, only 1 in 16 would display the recessive
phenotype given that there is only a 1/16 chance that both recessive alleles would be
homozygous in the offspring (Fig. 16 [SGF-1035]). For a number n genes, the expected proportion
of offspring displaying the recessive phenotype would be 1/4n, as each additional recessive
gene necessary for the phenotype contributes a factor of ¼.
Complications to Mendelian Inheritance
Traits that display simple Mendelian inheritance are relatively easy to analyze. These
traits are due to a single recessive allele that always displays a specific phenotype when the
recessive allele is homozygous. Many traits display this simple mode of inheritance. Others
that require more than one locus to display a phenotype can also display a relatively simple
mode of inheritance in the case where the phenotype is displayed only when all alleles involved
are homozygous, yielding offspring in an expected ratio (Fig. 16 [SGF-1035]). However, additional
factors can complicate the scoring of even a single gene trait including incomplete penetrance,
variable expressivity, incomplete dominance, parent of origin effect, and sex-linked inheritance.
In this next section, we will discuss these situations can that complicate the analysis of
inheritance patterns and will also provide examples of mechanisms that may underlie these
phenomena.
Incomplete penetrance complicates observing simple patterns
One factor that can complicate the scoring of a single gene trait is that not all organisms
of a given genotype will necessarily exhibit the phenotype, a phenomenon called incomplete
penetrance. Penetrance reflects the fraction of individual organisms of a given genotype that
display the phenotype (Fig. 17A [SGF-1015]).
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For example, consider a gene, B, important for viability. 100% of individuals who are
homozygous for b1 live. In this case, the allele b1 would be considered completely penetrant, as
all individuals of this genotype display the same phenotype. However, allele b2 is important for
viability, but behaves slightly differently from allele b1. Only 90% of individuals homozygous for
allele b2 are viable, while 10% of individuals homozygous for allele b2 are dead. The b2 allele is
considered incompletely penetrant (see Table 7 in Box 3-4).
When simply examining an individual for viability, it would be difficult to assess whether
an individual carries the b1 or b2 allele. To distinguish these alleles, you would need to examine
multiple members of a true breeding population, meaning, multiple members of a population
homozygous for the b1 allele and multiple member of a population homozygous for the b2 allele
(Figure 17A). If 90% of individuals in the population live, this would suggest the presence of
allele b2. On the other hand, if 100% live, this would suggest the presence of allele b1.
Examining multiple individuals of the same genotype will determine phenotype
penetrance
If you suspect the trait you are examining might have an incompletely penetrant
phenotype, the most important observation to make (if possible) is to examine the penetrance of
a homozygous individual. Since incomplete penetrance implies that individuals of the same
genotype can have distinct phenotypes, this also means that individuals with distinct
phenotypes can have the same genotype. Thus, analysis of multiple individuals with an
identical genotype will reveal incomplete penetrance: regardless of phenotype class, an
individual of the same genotype will segregate the same spectrum or proportion of phenotypes.
This is most readily seen in self-fertilizing organisms or haploid organisms (Figure 17B
[SGF-1015]). Consider allele c1, an incompletely penetrant allele of gene C where organisms who
are c1/ c1 will produce offspring that are 20% non-C phenotype and 80% C phenotype. In a
population of homozygous c1 individuals, an individual displaying either the non-C or C
phenotype will have offspring that are also 20% non-C phenotype and 80% C phenotype. Since
individuals from both phenotypic classes of the parents’ offspring generate descendents that
show the same spectrum of phenotypes as their parents’ and grandparents’ generations, we
can infer that the starting individual is homozygous for the c1 allele.
An example of incomplete penetrance can be seen when examining the C. elegans lin-3
gene (Figure 17C). When homozygous, the C. elegans lin-3(e1417) alelle results in 89% egglaying defective (Egl) hermaphrodites and 11% non-egg-laying defective (wild-type)
hermaphrodites. Thus, each homozygous lin-3(e1417) Egl hermaphrodite generates the same
distribution of Egl and wild-type progeny: 89% of the offspring will be Egl and 11% will be nonEgl, whether the hermaphrodite was of Egl or non-Egl phenotype. The lin-3 gene encodes the
major inductive signal necessary for induction of the vulva, which is required for egg-laying. At
the molecular level, the lin-3(e1417) allele appears to reduce the level of lin-3 activity to such a
degree that the probability that a vulva will form declines from 100% in wild-type to 11% in the
mutant.
Distinguishing incomplete penetrance from the effect of dominance involves examination
of the segregation pattern from crosses
How might you distinguish between the effects of incomplete penetrance or the effects of
dominance? Consider the example where you have two genetically identical individuals for
whom 75% of their offspring display phenotype and 25% display phenotype non-D (Fig. 18 [SGF1016]). Two hypotheses that could explain this segregation pattern are:
Hypothesis 1: The parents are heterozygous and both are genotypically d1/d2, where d1 is a
dominant allele. In this case, you would expect a 3:1 segregation of D phenotype: non-D
phenotype.
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Hypothesis 2: The parents are homozygous, and of genotype d1/d1, and the d1 allele is
incompletely penetrant such that only 75% organisms which are d1/d1 will display the D
phenotype.
One method to distinguish between these two possibilities is to breed two of the offspring
displaying the non-D phenotype together. If the non-D phenotype is due to a dominant, but
completely penetrant allele, 100% of the offspring should display the non-D phenotype.
However, if the non-D phenotype is due to incomplete penetrance, 25% of the offspring from
this cross will display the non-D phenotype.
Variable expressivity complicates observing simple patterns
The expressivity of a phenotype refers to the range of phenotypic severity that is
displayed in genetically identical organisms (Figure 20). For example, for a particular allele e1,
some individuals carrying this allele may have light blue eyes while other individuals have dark
blue eyes. The phenotypic range can be expressed in a discrete (i.e., light blue versus dark
blue) or in continuous (i.e., shades of blue, ranging from light to dark) terms.
A simple example of a phenotype with variable expressivity are temperature-sensitive
phenotypes. Geneticists have often isolated alleles of genes that show differing phenotypes at
different temperatures. These types of alleles have been particularly useful for genes that
otherwise would cause lethality when the gene function is completely removed. These
temperature sensitive alleles are also examples of alleles that display conditional inheritance,
which are alleles that display phenotypes only under particular conditions (e.g., Table 2).
The relationship of incomplete penetrance and variable expressivity is analogous to that
between continuous and discrete traits, as the two properties can be viewed as extremes along
a continuum. For incomplete penetrance, phenotypic variation is characterized as resulting in
multiple discrete classes of phenotype, while for variable expressivity, phenotypic variation
occurs over a more continuous spectrum with many gradations. Penetrance and expressivity
are both examples of genetically identical individuals that differ in phenotype. Box 3-4
discusses some general mechanisms how this might occurs.
Recessive and Dominant traits are not always absolute
In the simplest cases, the mutant allele is recessive to the wild-type allele. In this case,
the wild type allele is considered dominant, as in the heterozygote, the effect of the wild type
allele predominates (Fig. 21 [SGF-1204]). However, this is not always the case. Sometimes when
the heterozygous organism displays incomplete penetrance of the dominant phenotype, that
allele is said to be semi-dominant or partially dominant. Sometimes when effects of both alleles
are observed, the alleles are considered co-dominant (Table 3).
Dominant inheritance is most easily seen when two true-breeding strains that are each
homozygous for different alleles of a single gene are crossed to create a heterozygous
individual carrying two different alleles (Fig. 21 [SGF-1204]). If the alleles have a complete
dominant/recessive relationship, the allele whose phenotype is displayed would be the dominant
allele. When starting with a heterozygote, we can also determine which of the alleles it carries is
dominant by crossing it to a homozygous strain of the opposite phenotype (Fig. 22 [SGF-1205]).
What makes an allele dominant or recessive? Most recessive mutations disrupt some or
all of the function of a gene. Because the amount of gene product is either compensated due to
feedback within the cell or because the missing amount of gene product does not detectably
disrupt function, there is often no obvious immediate consequence of having only one functional
copy of a gene. The feedback control within cells is exquisite and takes care of a lot of the
noise inherent in any biological system. For example, quality control systems exist to get rid of
excess amounts of unbound subunits in multi-protein complexes such that proteins in
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complexes are stable and unbound subunits are marked for degradation. This is one kind of
mechanism that can compensate for the lack of two functional alleles, which is why alleles that
completely lack gene function can be recessive. The amount of gene product is in general,
proportional to the number of functional copies of the gene. So, a cell carrying a wild-type allele
and a complete loss-of-function allele would produce about 50% of the gene product compared
to cells that have two copies of the wild-type allele. In practice, this can sometimes exert a
biological effect. This phenomenon will be discussed further in Chapter 4.
For example, consider the case where gene H encodes an enzyme important for
production of pigment. Allele h1 specifies a functional enzyme while allele h2 specifies a nonfunctional enzyme. Allele h1 (which produces a functional enzyme) will likely be dominant. If
the exact level of enzyme is critical, then allele h1 may be semi-dominant. If allele h1 results in a
pigment being expressed in the tail and h2 results in a pigment being expressed in the belly,
they would be codominant, as now there would be functional allele in the tail as well as the belly
(Figure 23 [SGF-1206]).
Phenotypes can take time to develop [delete unless there are other mechanisms]
Sometimes there can be delay in the manifestation of a phenotype for a given genotype;
this is termed phenotypic lag. A phenotypic lag due to a stable protein being passed onto a
cell’s progeny is most often seen in single-celled organisms. The parent of origin effect is a
similar effect is seen in multicellular organisms and is described in the next section. One
mechanism for phenotypic lag involves genes that encode particularly stable proteins. In this
case, phenotypic lag is seen because while the progeny of that cell may inherit a gene that no
longer produces the gene product, the cell still has the gene product inherited from the mother
cell, and thus the phenotype of a cell lacking that gene product is not displayed. In this way, the
phenotype is not simply initially correlated with the genotype (Fig. 24 [SGF-1020]). If the protein is
present in sufficient quantities, particularly stable and also active in small quantities, there will
need to be enough cell divisions to occur until cells no longer inherit sufficient amounts of
protein so that the phenotype is displayed.
A clear-cut example of phenotypic lag involves the problems seen with telomere loss.
As discussed in Chapter 2 Box 2-1, telomeres have repeats that allow replication of the ends of
chromosomes. These repeats are lost each cycle of replication, but copies of these repeats are
added by the enzyme telomerase. In the absence of the telomerase enzyme, telomeres will
gradually shorten; however, cells often have kilobases of telomeric repeats and only a few
hundred bases will be lost each round of cell division. If an animal is missing the gene encoding
for the telomerase enzyme, the chromosome instability and other consequences of telomere
loss will not be manifested until the telomeric repeats are sufficiently shorted, leading to a
phenotypic lag. Indeed, in C. elegans, mutant lacking the telomerase enzyme will reproduce as
do wild type animals for several generations before becoming sterile. In Chapter 4, we will
discuss genetic mosaics in which perdurance of gene activity can be observed: an allele is lost
from a daughter cell but the effect of the gene continues for several cell generations.
One of the first examples of phenotypic lag involves the pep4 protease mutants in S.
cerevisiae. These mutants are defective in a set of vacuolar proteases (the yeast vacuole is
equivalent to the lysosome in higher eukaryotes). Proteinase A (the product of the PEP4 locus)
is necessary to cleave inactive pro-proteases to their active form. Mutant cells defective in pep4
gene nonetheless inherit enough active proteinase A and other proteinases (such as as
proteinase B), by cytoplasmic segregation even though they are unable to synthesize new
enzyme. After enough generations, the Pep4 protein has been diluted out. The yeast pep4-3
mutant is unable to cleave a carboxypeptidase Y substrate (N-acetylphenylalanine P-naphthyl
ester; APE) that can be in agar overlaid onto yeast colonies. The cleavage product, P-naphthol,
is treated with a chemical that results in an insoluble red dye deposited where there is enzyme
activity. Thus, colonies producing CPY stain red; those colonies lacking CPY activity are yellow.
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When a diploid heterozygous for a pep4-3 mutation undergoes meiosis and sporulation, two of
the spores are PEP4 (wild type) and two pep4-3. The PEP4 spores generate colonies that stain
red. The two pep4-3 spores generate mixed colonies, which stain with red and yellow regions.
Cells from yellow regions only generate yellow cells. The cells that are genotypically pep4-3 in
the red sectors are displaying phenotypic lag.
During each cell division, about half of the protein is lost from the cell. For this particular
protease, it can take tens of generations to manifest the phenotype! This is more the
exceptional case, as the stability of proteins are often regulated, including through the use of the
proteosome system of regulated protein destruction. Protein stability is highly regulated and
also differs among proteins.
Parent of origin effects arise from multiple causes
Parental effects are a causal effect of the parental genotype on the phenotype of the
offspring. Whenever parent of origin effects are seen, the genotype of the parent determines
the phenotype of the offspring, essentially overriding the genotype of the offspring. An effect
due to the genotype of the mother is called a maternal effect (Figure 25 [SGF-1207]). If the effect
is due to the genotype of the father, it is called a paternal effect.
Different types of maternal effects can be manifested. Sometimes a wild type copy of an
allele from a mother can lead to a situation where a recessive phenotype is not expressed in a
homozygous animal. Sometimes a mutant copy of an allele can lead to a situation where the
heterozygous individual from a homozygous mother will express a phenotype even though they
are heterozygous for a usually recessive allele (Figure 26 [SGF-1041]).
The situation with a homozygous mother’s genotype affecting the phenotype of her
offspring is illustrated in Figure 27 [SGF-1012]. In this case, allele s1 is recessive to the wild-type
allele (+). Allele s1 is a recessive allele, but also displays a maternal effect. Because allele s1 is
recessive, individuals that are s1/+ display the non-S phenotype. In generation II, a s1/+ mother
is crossed to an s1/+ father (generation I). The s1/ s1 progeny in generation II do not display the
S phenotype. However, when these s1/s1 females from generation II now mate to either a wild
type (+/+) male, all progeny in generation III now will display the S phenotype, regardless of
genotype. This would also be true if the s1/s1 females from generation II mated to a male that
was heterozygous (s1/+). However, if the s1/ s1males from generation II now are mated to a wild
type or heterozygous female, all progeny in generation III will be wild type, since the female was
genotypically wild type.
Maternal effects are more common than paternal effects because in many organisms,
the females produce oocytes, which contain much more than just the maternal DNA
contribution. In addition to the DNA, oocytes can contain large amounts of RNA and protein.
Thus, the mother’s genes, which are responsible for the contents of the egg, can have profound
effects on the newly formed organism.
Genomic imprinting can underlie some observed parental effects
Genomic imprinting is a heritable change that is not due to a change in DNA
sequence, but instead due to epigenetic marks on the DNA or chromatin. Genomic imprinting is
due to the parental genotype affecting the expression of particular alleles in the offspring and is
thus due to interplay between the parental and zygotic genomes.
In mammals, genomic imprinting can be caused by DNA methylation, where a methyl
group is covalently attached to a DNA base (Fig. 28A [SGF1053]). In this case, the methylation of
a particular region of DNA in one parent will decrease the expression of a particular allele in its
offspring, often leading to the lack of expression of the allele that is imprinted (Fig. 28B [SGF1209]). When cells divide in the offspring, the state DNA methylation can be inherited. Because
of this, the state of an imprinted allele can be stably maintained throughout the organism’s
development and its life (Fig. 28C [SGF-1210]). Because imprinting by DNA methylation results in
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the lack of expression of the allele, the offspring with the imprinted allele is in practice
hemizygous (Fig. 29 [SGF-1212]). The silencing of an allele by imprinting can lead to difficulties in
determining whether or not an allele behaves in a dominant or recessive fashion, as the
hemizygous state created by the imprinted allele means that the effect of an otherwise
recessive allele can now be exerted. Since only one active allele is inherited, imprinting is
referred to as a case of monoallelic inheritance. The inheritance of imprinting patterns can be
complex. In mice, during meiosis, the parental imprinting is removed and then reestablished in
the gamete according to the sexual genotype of the male or female.
Mouse Igf2 locus exemplifies imprinting
Genomic studies, discussed in Chapter 11, have revealed potentially hundreds of
imprinted loci in the mouse. For an individual locus, imprinting is perhaps best understood in
the mouse Igf2/H19 locus (Fig. 30 [SGF-1211]). igf2 encodes an insulin-like growth factor that is
necessary for growth of the embryo, and H19 is necessary to limit growth. Paternal imprinting
of the locus leads to increased expression of igf2 and decreased expression of H19. By
contrast, maternal imprinting of the locus leads to decreased expression of igf2 and increased
expression of H19.
The apparent dominance of either igf2 or H19 will depend upon the parent of origin. By
setting up reciprocal crosses, one can distinguish imprinting from dominance. In particular (Fig.
31 [SGF-1024]) if a3/a3 males are crossed with a1/a2 females and all progeny are of the a3associated phenotype, a3 appears dominant. However, if the cross is the reciprocal, with a3/a3
females crossed with a1/a2 males, then no progeny are of the a3-associated phenotype
Imprinting-like phenomena can also occur at a chromosomal level. Mammals have
evolved a mechanism to normalize the expression from the X chromosomes in males, who have
a single X chromosome, and females, who have two X chromosomes. This mechanism is called
dosage compensation. Mammalian dosage compensation occurs by preventing gene
expression from one of the two X chromosomes; this is also called X-inactivation. X-inactivation
occurs at random, and is stably maintained through somatic cell divisions. One consequence of
X-inactivation is that in a heterozygous individual (like all humans!), only one allele will be
expressed in a clone that has inactivated the other allele.
Phenotypic sex can affect the phenotypic expression
Another effect on phenotypic expression is an organism’s sexual phenotype. Some traits
are linked to sexual phenotype or limited to one sex (Fig. 32 [SGF-1032]). The simplest types of
sex-limited traits are secondary sex characteristics that are naturally sex limited, such as milk
production in females. These effects could be indirect, for example, deriving from secondary
sexual characteristics such as size. Some traits are genetically linked to sexual phenotype.
These are usually attributable to the mechanisms of sex determination.
An organism’s sex can be a striking phenotype (for sexually dimorphic species), and
thus it is often noted as it is a simple phenotype to observe in a pedigree or cross. Phenotypes
that co-segregate (are linked) to sexual phenotype are referred to as sex-linked. The nature of
the linkage depends upon the genetic determinants of sex.
Understanding chromosomal sex determination is necessary to understand the effect of
sex on phenotypic expression
To understand the segregation patterns of genes on sex chromosomes, you must first
know how chromosomal sex determination occurs in your organism of interest. Many organisms
specify sex based on chromosomal constitution such that there is often a heterogametic sex
that has two distinct sex chromosomes. The homogametic sex would have a pair of a single
sex chromosome. For example, in humans, males are the heterogametic sex and have a single
X and a single Y chromosome while females have two X chromosomes and no Y chromosome.
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In birds, the female is the heterogametic sex with one W chromosome and one Z chromosome
while the male has two W chromosomes. Sometimes the relative number of sex chromosomes
relative to can determine sex. In C. elegans, males have a single X chromosome while
hermaphrodites have two X chromosomes. In Drosophila melanogaster, sex determination also
depends on the ratio of X chromosomes to autosomes although there also is a Y chromosome.
The sex that has two distinct sex chromosomes is called the heterogametic sex (males
in humans and females in birds). Because the heterogametic sex has only a single of each
chromosome, a recessive allele can be expressed in the heterogametic sex because it is
hemizygous.
Chromosomal sex determination impacts the segregation patterns of genes on sex
chromosomes because recessive alleles can be expressed in the heterogametic sex since it is
hemizygous. This can affect the segregation patterns of genes. Instead of a 3:1 segregation for
a recessive allele, all males (or all females in birds) could display the recessive phenotype.
For example, in lovebirds (Agapornis roseicollis; small parrots), the standard color is
green and there is an autosomal recessive blue allele that confers blue feathers (Figure 33 [SGF1018]). In a blue/blue background, a Z linked lutino recessive allele confers yellow feathers. A
yellow male is homozygous for the blue allele and is homozygous for the W-linked recessive
lutino allele.
There are a wide variety of sex determination systems (Figure 34 [SGF-1036]). In animals,
there can be sex chromosomes such as the familiar XX-XY system of mammals, the WZ-WW
system of birds, or multiple sex chromosomes. Other less familiar sex determination systems
include maternal sex determination as in the carrion fly, or environmental sex determination in
which the sex of an individual depends on factors such as nutrition, population density or social
structure. In bacteria, either a cell has an F factor or not. In fungi there are one or more mating
type loci (Figure 35 [SGF-1208]). In a simple case, that of baker’s yeast, there is one mating type
locus and two sexes, a and α (alpha).
Non-Mendelian inheritance
Mendel’s great discoveries of segregation and independent assortment underlie the
most common modes of genetic inheritance. However, there are inherited characteristics that do
not obey Mendel’s laws and are thus considered non-Mendelian. While genes located on the
chromosomes in the nuclei of diploid cells underlie Mendelian inheritance, genes controlling
traits inherited in a non-Mendelian fashion reside on cytoplasmic organelles, cytoplasmic
elements (other than organelles), and plasmids. We need to understand non-Mendelian modes
of inheritance not only because it is crucial in its own right, but also because we need to rule out
other patterns of inheritance to infer Mendelian inheritance. The key characteristic of nonMendelian inheritance in eukaryotes is that the origin of the cytoplasm (and not the nuclear
DNA) determines the phenotype. Cytoplasmic inheritance can be due to organelles or any
other particles that can be passed on to the next cell generation and affect its phenotype (Fig.
36 [SGF-1017]).
Cytoplasmic inheritance can arise from variant organelle genomes
Eukaryotic cells have organelles, which include mitochondria and chloroplasts.
Mitochondria provide energy for the cell as well as carrying out some key biochemical reactions.
Chloroplasts carry out photosynthesis. Most importantly for genetic analysis, mitochondria and
chloroplasts have their own genomes. As these organelles and their genomes are partitioned at
mitosis and passed on to the next generation, they will have distinct patterns of inheritance (Fig.
36 [SGF-1017]). In most cases, one parent will provide the organelles for the offspring. This
parent is usually the female, as the oocyte contains the materials necessary in the cytoplasm for
the development of the zygote. There are diseases in humans that are due to mitochondrial
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DNA mutations. Typically in these cases, the transmission of the diseases states occurs only
via the female.
An example of a disease inherited through organellar inheritance is encephalomyopathy
(myoclonic epilepsy with ragged red fibers; abbreviated MERRF). MERRF results from a defect
in the ability of mutant mitochondria to synthesize proteins. This inability results from a mutation
of a Lysine-inserting tRNA gene encoded by the mitochondrial genome. A pedigree of a family
affected by MERRF is shown in Fig. 37 [SGF-1037]. All four progeny of the key individual studied in
a particular pedigree (the proband), individual I-1 display MERRF. The male II-4, who was
affected, had three normal children because these children received their mitochondria from
their unaffected mother. Mitochondrial inheritance is another mechanism by which a parental
effect can occur. However, unlike the cases described above, this type of inheritance would
involve a maternal effect of a non-Mendelian trait. Cytoplasmic inheritance can be similar to
dominant maternal effect inheritance, but in practice these can be distinguished (Box 3-7).
A typical cell also has many copies of the organellar genome, as there are many copies
of the mitochondria (and chloroplast, in cells that have them) within the cell. Having multiple
copies of the genome can affect the inheritance patterns and genome function. For example, a
cell can have a heterogeneous organelle set, with some organelles having a defective genome
that are carried along with other organelles containing a fully functional genome. The
organelles with the defective genome can function because of the presence of the other
organelles with a functional genome.
Cytoplasmic inheritance can also arise from other nucleic acid based entities
Similar to how organellar genomes can affect phenotype, any other genetically-based
entity that is carried within a cell or organism and is passed onto the organism’s progeny can
affect inheritance. Pathogens, symbionts, and cytoplasmically inherited viruses are examples of
these types of situations. These situations can be difficult to distinguish from an environmentally
based effect, but are important to keep in mind when complex inheritance patterns are seen.
For example, a double-stranded RNA virus in yeast called “killer element” causes yeast
cells that harbor this virus to secrete a toxin that kills cells that lack the killer element. This killer
element is cytoplasmically inherited, and thus displays properties similar to other
mitochondrially-inherited traits (Fig. 38 [SGF-1215]).
The popcorn mutant found in Drosophila is another example of this type of inheritance.
Flies that carry the popcorn mutant undergo brain degeneration (Fig. 39). However, the
popcorn gene was not found on any Drosophila chromosome. Instead, this mutation was found
on the genome of a mutant form of Wolbachia, a Rickettsial family microbe that is normally a
symbiont of insects.
A more complicated case of inheritance due to other genetic entities is the maternal
transfer of an infectious agent. The mouse mammary tumor virus (MMTV) is a retrovirus whose
genome can be integrated into the host genome and thus passed along to its offspring through
transmission of the genome. The presence of MMTV will predispose mice to mammary tumors.
However, passage through the genome is not always the mode of inheritance, as this virus can
also be transmitted from an infected mother to its offspring through its milk. Ingestion of cells
carrying this virus leads to the virus passing through the digestive system into the organism’s
immune cells, where it can infect some cells of the immune system and ultimately infect
mammary gland cells where it affects their proliferation. This unusual mode of transmission was
seen because adopted offspring (who are genetically dissimilar to the mother) were also
displaying a predisposition to mammary tumors when nursed by mothers infected with MMTV,
as the virus was being transmitted in the milk of the infected host to adoptive offspring through
nursing. HIV also passes through mother’s milk.
Box 3-9 describes a way to distinguish nuclear from cytoplasmic inheritance.
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Cytoplasmic inheritance can arise from protein states- prions
Prions describe a cytoplasmic inheritance agent that does not involve nucleic acids.
Instead, prions involve an altered conformation of a protein (designated as PrP*) that has the
ability to convert other normal conformations of that protein (designated as PrP) into a prion
state (Fig. 40 [SGF-1039, 1214]). This prion-forming ability is usually due to specific sequences
within the protein that allow for the protein in the prion state to form and create self-propagating
filaments from the normal protein. Prions include pathogenic agents for a variety of disease
such as human Cruetzfeldt-Jacob disease, bovine spongiform encephalopathy [a.k.a., mad cow
disease] and sheep scrapie as well as altered forms of a handful of yeast genes such as
SUP35, RNQ1, and URE3. Because the altered PrP* can convert the normally folded PrP
protein to the PrP* prion state, the transmission of the protein in a PrP* prion state can occur in a
cytoplasmic manner as cells divide. Once a cell inherits a protein in a prion state, this prion
protein can convert the normally folded proteins into the cell into a prion state. If the presence
of the prion now causes a phenotype, this phenotype has now been inherited in a cytoplasmic
fashion in a manner not involving nucleic acids. Prions apparently infect immune cells in the gut
and then spread into the brain.
Determining inheritance: the case of the Dumpy phenotype
How do you determine the inheritance pattern of a dumpy worm? First, you might want
to examine what the progeny of the dumpy worms are like: do they produce other dumpy
worms? Assuming that the dumpy phenotype is heritable, you might want to examine the
phenotype more carefully and see whether dumpy versus non- dumpy is a discrete phenotype
or a more continuous phenotype with different gradations of dumpiness. Since this is a
nematode, you can examine what happens when you cross worms from a dumpy culture with
worms from a non-dumpy culture: what phenotypes will segregate from the progeny of this
cross? Does this differ when you cross two dumpy strains together? Do your results suggest
simple Mendelian inheritance? Is it due to the inheritance of several loci that are required for
the phenotype to be expressed? If not, you will then have to consider the factors that can
complicate Mendelian inheritance and also non-Mendelian inheritance.
How would you examine a newly isolated phenotype: Let us consider an example from
C. elegans. Like Sydney Brenner, you pick a variant C. elegans that has a Dumpy (Dpy)
phenotype (Fig. 41): this animal appears shorter and squatter than the typical C. elegans worm.
First, you want to test whether this phenotype is heritable (Fig. 42 [SGF-1021-1]). You let
the worm you pick self-fertilize and find that the Dumpy phenotype breeds true: all its progeny
were similarly Dumpy. Since you are studying C. elegans, let’s assume this variation is due to
a simple model of inheritance; we will consider otherwise if the data suggests.
You can ask if this variant allele is recessive to the wild-type allele. You cross the Dpy
strain to a non-Dpy strain and all the F1 animals are non-Dpy (Fig. 43 [SGF-1021-2]). This
observation implies that the variant allele is recessive to the wild-type allele. Since you have
now created a heterozygous F1 hermaphrodite, you let an individual F1 hermaphrodite selffertilize, and find that about one-quarter of its progeny are Dpy, the remaining three-quarters
being non-Dumpy (Fig. 44 [SGF-1022]). These observations are consistent with the variant being
a single locus.
To test your hypothesis of the Dumpy phenotype being due to a single locus, you let
some of the F2 individuals self-fertilize. The F2 Dpy gives only Dpy progeny, consistent with
your original observation of it being recessive. You also observe that 2/3 of the non-Dpy F2
segregate both Dpy and non-Dpy progeny, while 1/3 of the non-Dpy F2 segregate only nonDpy progeny (Fig. 44 [SGF-1022]).
Why is there a 1:2 ratio of non-Dpy segregating F2 to individuals that segregate both
Dpy and non-Dpy progeny? Consider the genotype of the F1 used to generate these F2
individuals. The F1 is dpy/+ (non-Dpy) and produces ¼ dpy/ dpy (Dpy): ½ dpy /+ (non-Dpy): ¼
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+/+ non-Dpy progeny. Half of the non-Dpy progeny (which are ¾ of the total progeny are
heterozygous, and thus ½ / ¾ =2/3. In other words, we only look at the 2:1 part of the 1:2:1
ratio.
Now, suppose the Dumpy variant is dominant to the wild-type allele. You might first
notice this when you cross your original Dpy individual to the wild type animal and see that all
F1 individuals are Dpy (Fig. 45 [SGF-1023]). The segregation patterns from individual F1 Dpy
heterozygotes (dpy /+) will then yield one-quarter non-Dpy progeny and ¾ Dpy progeny. Of the
F2 Dpy individuals, 1/3 will be homozygous (dpy/dpy) and will only have Dpy progeny while 2/3
will be heterozygous (dpy /+) and will yield ¼ non-Dpy progeny and ¾ Dpy progeny.
Understanding how to analyze the inheritance patterns of a phenotype allows for more
mechanistic understanding of how the causative agent (usually a gene) affects the organism.
More generally, although most genes in most organisms are inherited in patterns consistent with
Mendel’s Laws, a fraction of the organism’s genes may display complicating factors such as
parent-of-origin effects and incomplete penetrance. Furthermore, phenotypes that are due to
non-Mendelian inheritance are ultimately also part of an organism. Thus, the understanding of
an individual comprises the organism and its associated symbionts and parasites. Because of
this, the phenotypic properties of an individual ultimately arises from the individual’s zygotic
genome, it’s maternal and paternal material, as well as the genotype of its associated
organisms.
Essential Concepts
Ø Phenotypic variation can discrete or continuous, either of which can allow
analysis of heredity
Ø True breeding phenotypes indicate genetic homogeneity
Ø Wild-type alleles are usually dominant to mutant alleles
Ø Single Mendelian inheritance is single locus segregation, clear dominance,
complete penetrance and no parental effects
Ø Models of inheritance are fit to data, statistically-tested and verified if possible
Ø Incomplete penetrance suggests an underlying stochastic process
Ø Traits can be co-dominant
Ø Phenotypes can take many cell generations to manifest
Ø Sex limited traits suggest sex linkage
Ø Sex determination is often based on chromosomal constitution
Ø Parent of origin effects have many and interesting causes
Ø Cytoplasmic inheritance has multiple causes including organelles, viruses, and
prions
References and recommended reading
Collinge, J. (2010). Prion strain mutation and selection. Science 328: 1111-1112.
SW Leibman and F Sherman Cytoplasmic Inheritance, In Landmark papers in yeast biology. P.
Linder, D. Shore and MN Hall, Eds. (2006).
Wrinkled Peas & White-Eyed Fruit Flies: The Molecular Basis of Two Classical Genetic Traits
Patrick Guilfoile The American Biology Teacher, Vol. 59, No. 2 (Feb., 1997), pp. 92-95 (article
consists of 4 pages) Published by: University of California Press on behalf of the National
Association of Biology Teachers Stable URL: http://www.jstor.org/pss/4450256
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East, E. M. (1935). Genetic reactions in Nicotinia. III. Dominance. Genetics 20: 443 S
Horvitz HR, Sulston JE. Isolation and genetic characterization of cell-lineage mutants of the
nematode Caenorhabditis elegans. Genetics. 1980 Oct;96(2):435-54. PubMed PMID: 7262539;
PubMed Central PMCID: PMC1214309.
Shoffner JM, Lott MT, Lezza AM, Seibel P, Ballinger SW, Wallace DC. Myoclonic epilepsy and
ragged-red fiber disease (MERRF) is associated with a mitochondrial DNA tRNA(Lys) mutation.
Cell. 1990 Jun 15;61(6):931-7. PubMed PMID: 2112427. Example of a mitochondrial mutation
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Systems Genetics Chapter 3 Figures and Tables, version 1
Figure 1. Pretty pictures of phenotypes.
Whole organism
Plant, animal
Cell
Behavioral
Figure 2. Examples of discrete vs continuous traits. Normal distributed phenotype,
real examples A. Discrete. B. continuous.
(blue vs brown eyes, different flower colors, agamous vs wild type flowers, etc.) ––
people of all different heights?
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Figure 3. [SGF-1025]. Continuous versus discrete traits.
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Figure 4. Colors as discrete versus continuous traits.
[SGF-1199]
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Figure 5. Typical mating systems. A. Cross-Fertilization. [SGF-1005 =A, B] B. Selffertilization. C. Yeast mating system [SGF-1042].
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Figure 6. Pedigrees. Black circles indicate females that display the particular trait of
interest (affected). Generations are indicated by roman numerals. Horizontal lines
between females and males indicate mating [SGF-1002]
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Table 1. Simple versus complex modes of inheritance
SIMPLE
COMPLEX
Discrete (as opposed to continuous)
Combination of additive and epistatic
variation
effects
One locus with all the effect
Multiple loci with a spectrum of effects
Complete penetrance
Partial penetrance
Recessive to the wild-type strain
Partial dominance
Phenotype immediately discernible
Delayed appearance of phenotype
Zygotic expression
Parental effects
Robust to environmental perturbations
Sensitivity to environment
Independent of paternal origin
Parent of origin effects (maternal effect)
Nuclear, chromosomal inheritance
Cytoplasmic inheritance
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Figure 7. True Breeding population [SGF-1044] Figure showing a true breeding
population for A) a crossing strain and B) a selfing strain and C) a microbial strain. C. A
single cell of genotype A1 gives rise to two progeny cells also of genotype A1. By
separating out the microbes to single cells and letting them grow, a colony derived from
a single parent cell is obtained.
D Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 8. Sectored colonies. [SGF-1014] Schematic of origin of sectored colonies.
Table 301: Results from the yeast sectoring experiment
Type of
result
I.
II.
Dark gives:
III.
Dark with
cream
sectors
Dark only
Dark
Cream gives: Dark true
breeding?
Cream only
Yes
Cream with
Yes
dark sectors
Cream with
No
dark sectors
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
Cream true
breeding?
Yes
No
No
8
Fig. 9. Definition of homozygous, heterozygous and hemizygous. [SGF-1200]
Fig. 10. Definition of dominant and recessive. [SGF-1201]
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Figure 11. Mendel’s law of Segregation. [SGF-1003] Segregation of chromosomes at
meiosis into individual gametes.
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Figure 12. Mendel’s Law of Independent Assortment. [SGF-1004] Each pair of
homologous chromosomes independently undergoes meiosis (replication and
disjunction). With two chromosomes each heterozygous, meiosis generates four
gamete types.
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Figure 13 Punnett square of two locus segregation. [SGF-1013] Each female
gamete, or male gamete can have either allele at either of the two loci. There are thus
four types of gametes and nine genotypes possible.
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Figure 14. Diploid segregation patterns for Mendelian traits [SGF-1055].
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Figure 15. Yeast-like alternation of haploid and diploid. [SGF-1202]
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Figure 16. For each independent locus necessary to confer the phenotype, there is a
factor of ¼, because of Mendel’s first Law. [SGF-1035]
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Figure 17. Penetrance. A. Incomplete penetrance: by contrast to wild-type
(penetrance v = 0), and a fully penetrant (v=1), incompletely penetrant phenotypes have
some wild-type and some mutant, for example, live worms or dead eggs. B. From a
cross of potentially true-breeding individuals, some of the progeny (proportion v) are of
Phenotypically R while others are not (proportion 1-v). Breeding of Phenotype R
individuals yields both R and non-R, and breeding of non-R also yields R and non-R
progeny in proportion v : 1-v. [SGF-1015]. C. lin-3(e1417) is incompletely penetrant.
[SGF-1226]
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C. C. elegans lin-3 example of incomplete penetrance. add Photographs.
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Figure 18. Incomplete penetrance versus dominant phenotype. Two Phenotype
D individuals are crossed, and among their progeny 75% are Phenotype D and 25% are
phenotype non-D. Progeny of phenotype D are crossed together and their progeny
examined. There are two simple possibilities. If again 75% are of phenotype D and
25% non-D, we infer that the D individuals are true breeding but of incomplete
penetrance (75% penetrance). On the other hand, if all the progeny are phenotype D,
then we infer that the original D parents were heterozygous for a dominant allele
conferring the D phenotype. Only 1/16 of the progeny of all the D F2 animals will be
non- D, they must be from two heterozygous parents ( ½ x ½ ), and inherit the + allele
from both ( ½ x ½ ) for 1/16. [SGF-1016]
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Fig. 19. [SGF-1230] Calculation for dominant crosses.
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Figure 20. Expressivity. A. C. elegans Multivulva phenotypes. top, wild type, nonMultivulva hermaphrodite; middle, partial Multivulva phenotype; bottom, strong
Mulitvulva phenotype. Blue arrow indicates functional vulva; red arrow indicateds
pseudo-vulva. Left-lateral views of L4 stage hermaphrodite larvae. Modified from Zhong
and Sternberg Science 2006. B. temperature sensitive yeast growth. The expressivity
of the ts mutant depends on temperature
INCLUDE WHOLE ANIMAL PICTURES
A.
B.
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Table 2. Conditional alleles
genotype f+ f+ f1 f1 f2 f2 f3
f3
type of allele
wild-­‐type wild-­‐type non-­‐conditional non-­‐conditional ts ts cs cs temperature 16 30 16 30 16 30 16 30 Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
phenotype alive alive dead dead alive dead dead alive 21
Figure 21. Using two true-breeding strains to determine allele relationships. Cross
of two homozygotes. [SGF-1204]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
22
Table 3: Dominance.
Complete
dominance
Co-dominance
Semi-dominance
Genotype
a1/a2
Phenotype
All organisms display A phenotype.
a1/a2
All organisms display an intermediate
phenotype compared to organisms that are
a1/a1 and a2/a2 or there is reduced penetrance
relative to the homozygotes
Only some organisms display the A
phenotype, a fraction will display the non-A
phenotype.
a1/a2
Table 3a: Numerical example of semi-dominance
Genotype Phenotype
a1/a1
100% display A phenotype
a1/a2
10% display A phenotype.
a2/a2
0% display the A phenotype,
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 22. [SGF-1205] Determining dominance starting with a heterozygote.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 23 [SGF-1206]. Codominance. h1 allele specifies tail pigmentation; h2 alleles
specifies belly pigmentation.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
25
Figure 24. Phenotypic lag. The a1 mutant allele becomes homozygous without
phenotypic effect for several generations, and then is stable. There is thus a lag in
expression of the R phenotype. In the pictorial example, the R phenotype is lack of
telomeres, which get shorter each generation in the absence of the a+ allele [SGF1020]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
26
Figure 25. Maternal versus Paternal Effects. [SGF-1207] Crosses that demonstrate
maternal effects or paternal effects. Alleles k1 and p1 are recessive (not shown) and
confer the K and P phenotypes, respectively. The progeny have different phenotypes
depending on which parent carries the allele.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
27
Figure 26. Maternal rescue: [SGF-1041] A. Maternal Rescue. Homozygous offspring
are rescued to a wild-type phenotype by the presence of the + allele in mother. B.
Dominant maternal effect. Genotypically wild-type offspring display the mutant
phenotype due to the dominant maternal effect.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 27. Pedigree displaying a recessive Maternal Effect trait. A. Pedigree. B.
Inferred genetic model. [SGF-1012]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
29
Figure 28. Methylation of DNA. A. [SGF-1053] Methylation of the base cytosine at the 5
position (5-methyl cytosine) is an important modification of DNA. Specific enzymes catalyze this
methylation reaction. B. Methylation can interfere with transcription [SGF-1209]. After DNA
replication, a maintenance DNA methylase recognizes the hemi-methylated DNA and
methylates the CpG.[SGF-1210]
A
B.
C.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
30
Figure 29. Imprinting affects transcriptional regulation; the simple view. A. An
enhancer and insulator are located 5’ to a coding regions. B. A female (non-imprinted
allele) has an active insulator, which blocks activation of transcription. C. A paternallyimprinted insulator element does not prevent transcriptional activation. [SGF-1212]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
31
Figure 30. [SGF-1211] Imprinting at the IGf2-H19 locus of the mouse. A. The Igf2
and H19 coding regions are nearby. The transcriptional enhancers for both promoters
are located 3’ to H19. B. Alternative looping allows the enhancers to interact with etiher
H19 (left) or Igf2 (right). C. A female (non-imprinted allele) binds CCTF complex and
favors looping to H19. D. A paternally-imprinted insulator element does not bind the
CCTF complex and allows the distal enhancer to activate transcription of Igf2.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
32
Figure 31. Female imprinting. Asterisk indicates an imprinted and thus not
expressed allele. In this case the maternal alleles are imprinted. [SGF-1024]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
33
Figure 32. Human pedigree demonstrating sex linkage. This pedigree demonstrates
sex-linkage in which only male grandchildren (III-3 and III-6) of an affected grandfather
(I-2) are affected, and then only half of those grandchildren. This pedigree is fully
consistent with X-linked recessive inheritance. [SGF-1032]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
34
Figure 3. Sex linkage [SGF-1018 ]. A. Sex linkage in lovebirds B. [Add pictures of
birds if available]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
35
Figure 34. Sex determination [SGF-1036] Many familiar sex determination systems
are based on asymmetry in the types of chromosomes (top) or the numbers of
chromosomes.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
36
Figure 35. Chromosomal sex determination in fungi. Cells with the MATa locus are
a-type mating cells; cells with the MATα locus are α cells. [SGF-1208]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
37
Fig. 36. Patterns seen with mitochondrial inheritance [SGF-1017]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
38
Figure 37. Simplified pedigree of mitochondrial disease MERRF. Data described in
Cell 1990. [SFG-1037]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
39
Figure 38 [SGF-1215] Killer element. A. One killer element has a double-spranded
RNA in a virus like particle. B. L RNA replicates and the RNA and particle segregate
at mitosis.
Figure 39. Flies that carry the popcorn mutant undergo brain degeneration.
photo
¿Add MMTV here – as a pedigree or picture showing inheritance?
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
40
Figure 40. Prions. [SGF-1039]. A. Abnormal Prion conformation converts normal
conformation. B. [SGF-1214] At cell division, the cytoplasmic PrP* is segregated
to both daughters, and converts newly synthesized PrP to PrP*.
B.
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 41 Picture of a Dpy worm
Figure 42 C. elegans genetics: recovering a mutant in the F2 of a mutagenized
grandparent. [SGF-1021-1]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
42
Figure 43. Demonstrating recessivity. [SGF-1021-2]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
43
Figure 44. [SGF-1022]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
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Figure 45. Demonstrating dominance. [SGF-1023]
Systems Genetics, Chapter 3 Figures & Tables v1, 4/1/13
45
Systems Genetics Chapter 3: BOXES
Box 3-1.
[SGB-321]
Nomenclature.
Genes are referred to by capital letters, italicized, e.g., A.
Alleles will be lower case, either just letters or letters plus numbers if more than one
allele is being discussed, e.g., a, a1, a2, a3. If the a1 allele referred to is the wild-type
allele, this allele sometimes represented with a “+”.
Phenotypes will be referred to by bold, capital letters, such as A. A wild-type
phenotype will be referred to as non-A, or sometimes wild type.
A protein product will be capital non-italicized. Thus, the protein product encoded by
gene A will be protein A.
When we discuss specific examples, we will use the approved nomenclature for that
organism. Organism nomenclature is described in the Boxes introducing the genetics of
that organism (see Boxes __ in Chapter 1).
The absence of a chromosome in a diploid animal will be indicated by an “O”. This
nomenclature is mostly relevant for discussion of sex determination. For example, XX
indicates and animal with two X chromosomes while XO would refer to an animal with
only one X chromosome.
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 1
Box 3-2 [SGB-322] Real data do not always yield clear results: using statistics to
determine significance
Sometimes your data do not yield a clear-cut interpretation. For example, if you
wanted to determine whether a single genetic locus is involved in a phenotype, you
could examine the segregation of progeny from heterozygous parents. If this single
locus involved an allele that behaved in a recessive fashion, you would predict a 3:1
segregation pattern of wild type to mutant, as predicted by Mendel’s laws. However,
sometimes you obtain data that does not obviously segregate as 3:1. One option is to
collect more data, as a small sample size is more subject to sampling error. (Imagine
that you flip a coin 4 times and it comes up heads each time; is it a rigged coin? No,
since four heads will come up once every 16 times.) Sometimes collecting more data is
not an option, and statistics can be used to determine the likelihood that your phenotype
is segregating as a single genetic locus.
Let’s consider the situation where you take a true breeding variant, with
phenotype A, and cross it to a wild-type strain. If we assume a simple Mendelian
model, the F1 organisms created by this cross are presumably heterozygous at the A
locus. Now, if you cross the F1 organisms to each other, you can examine the
phenotypes of the grandprogeny in the F2 generation. When you examine 100 grandprogeny, you see that 82 are phenotypically wild type (non-A) while 18 have the A
phenotype.
Is your observation consistent with the A phenotype segregating as a simple
Mendelian trait? The Chi-squared test for significance is used to determine whether the
data obtained are consistent with a hypothesis. This test will return the probability that
the observed values deviate from the expected values simply by chance. In other
words, this test will tell you what are the chances you will get the observed values
purely by the luck of the draw. Convention is that a hypothesis is rejected if there is less
than 1 in 20 (0.05) chance that the observed data fit the expectation. Note that for
performing a chi-squared analysis, the data are analyzed as the actual observation, and
are not converted into proportions or percentages. Because of this, the significance
value obtained does take into account whether few or many observations were made.
To calculate chi-squared:
χ 2 = Σover all categories of observation (Observed Number– Expected Number) 2
Expected Number
(The math symbol for sum is the capital Greek letter S: Σ.)
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 2
Figure 1/Box3-2 [SGF-1216]
Phenotype
A
non-A
Observed
18
82
Expected (hypothesis of 3:1)
25
75
Obs-Exp
7
7
(O-E)2 /e
49/25
49/75
We calculate χ2 = 2.613, P=0.11
For the example above, χ2 = (18-25)2 /25 + (82-75) 2 /75 = 49/25 + 49/75 = 2.613.
We look up in Table 6 the P value for one degree of freedom and find P=0.11. The
degrees of freedom are the number of independent variables, typically one less than the
number of classes. If you have two phenotypes with proportion f and q, then f + q = 1
since an individual can be either of the two phenotypes. Thus q = 1 - f, and f is the only
variable (degree of freedom). If a parameter is estimated from the data, then it does
not count as an independent variable. For this analysis, the degrees of freedom are
one less than the number of classes of phenotypes (or genotypes).
This result means that there is an 11% chance these observed data fit the
hypothesis of a 3:1 segregation (indicating a Mendelian trait) were to account for the
phenotype. From this result, we conclude that our observed data are consistent with
this phenotype being caused by a single gene.
Now, consider the same experiment described above. However, what if we
instead observed 15 individuals of phenotype A and 85 individuals of phenotype non-A?
Are these ratios consistent with the phenotype being caused by a single locus (giving a
3:1 segregation) or two loci (giving a 15:1 segregation)? First, we need to calculate the
expected ratios for each hypothesis and calculate the chi-squared values.
Phenotype
A
Non-A
Observed
15
85
Expected (hypothesis of 3:1)
25
75
χ2 = 5.3, P=0.02
Phenotype
A
Non-A
Observed
15
85
Expected (hypothesis of 15:1)
6.25
93.75 χ2 = 13.067, P=0.0003
From these observations, the chi-squared values indicate that our results are
consistent with neither hypothesis. This does not necessarily mean that both
hypotheses are incorrect. Instead, data inconsistent with neither hypothesis may be due
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 3
to an insufficient number of segregants observed, giving the chance that these ratio of
observed phenotypes do not fit the expected values.
How might you resolve this inconclusive situation? One solution might be to
examine more segregants from the heterozygote, to increase your numbers and see if
more data points make it so that you will now be consistent with one of these
observations.
Perhaps a better way to go is to generate a new hypothesis that explains the 5:1
ratio. See BOX 3-3.
A better statistical test for small numbers (and any numbers if you have sufficient
computer power) is Fisher’s Exact Test as the chi-squared is an approximation that
matches the real values more closely with large numbers than with small numbers (see
Appendix).
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 4
BOX 3-3. [SGB-323] What happens when your data do not fit simple Mendelian
rules: the 5:1 ratio.
Gregor Mendel was lucky when he picked the alleles of the traits he studied, as
they all behaved like 100% penetrant recessive alleles. If the alleles you are interested
do not behave in a simple Mendelian fashion, how do you proceed.
For example, let’s consider the case where you think a single genetic locus is
required for the trait in which you are interested. However, you do not see the 3:1
segregation pattern you might expect from Mendel’s laws, and instead observe a 5:1
ratio. How might you explain a 5:1 ratio? First, you need to make a hypothesis to test.
For example, one model that might explain a 5:1 ratio is that there is a single locus, with
incomplete penetrance.
Hypothesis: one-locus, incompletely penetrant recessive
genotype
proportion
penetrance fraction
from cross (f)
with
phenotype
+/+
¼
0
0
a1/+
½
0
0
a1/a1
¼
0.6
0.15
TOTAL
1
0.15
To obtain 0.15 of the progeny with the phenotype, when only the a1/a1 individuals
contribute with penetrance f, we need ¼ f = 0.15. This implies that f = 0.6.
If this is an experimentally manipuable organism, we can test this hypothesis
experimentally. To do so, we can verify whether this is the case by examining
organisms of a1/a1 genotype (a fraction of which will display the A phenotype). In
practice, you could take individuals of A phenotype, which are presumable homozygous
a1/a1, and examine their progeny. If this hypothesis is valid, you would expect 60% of
their progeny to display the A phenotype and 40% to display the non-A phenotype.
Let’s examine a more complicated hypothesis: two loci are involved, where one
allele is recessive and one allele is dominant.
Hypothesis: two loci, a1 is recessive; b1 is dominant
genotype
proportion from
penetrance (f)
fraction with A
cross
phenotype
a/a b/b
1/16
1.0
0.0625
a/a b/+
2/16
0.7
0.0875
a/a +/+
1/16
0
0
+/+ b/b
1/16
0
0
a/+ b/b
2/16
0
0
a/+ b/+
4/16
0
0
a/+ +/+
2/16
0
0
+/+ b/+
2/16
0
0
+/+ +/+
1/16
0
0
TOTAL
1
0.15
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 5
If an allele is segregating according to this model, we want 15% of the offspring
to display the A phenotype. The double homozygotes (a1/a1 b1/b1) contribute 0.0625, so
we need .0875 (0.15 – 0.0625 = 0.0875) with penetrance f, where 2/16 (0.125) of the
progeny are of this genotype. We thus have
0.125 f = 0.0875, and thus
f = 0.0875/0.125 = 0.7.
If we took phenotypically A animals and crossed each to genotypically a1/a1
b1/b1true-breeding parents, we would expect that
58% (0.0875/0.15) would be a1/a1 b1/+ x a1/a1 b1/b1 and thus give
0.5 a1/a1 b1/b1 of which all are A
0.5
0.5 a1/a1 1b/+ of which 0.7 are A
0.35
total
0.85 are of A phenotype
and 42% (0.0625/0.15) of the crosses would be a1/a1 b1/b1 x a1/a1 b1/b1 and thus give all
A progeny.
We could then compare our expectations by a statistical test). For example, if we
observed 12 crosses, with 6 crosses giving ~0.85 A progeny and 6 crosses giving all A
progeny, we could use Fisher’s Exact Test (see Appendix/FET) to compare 6:6 to our
expected 58:42 ratio, and find that it is not significantly different from expectation.
Table 6. Values of χ 2. chi-squared table (NIST)
See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda3674.htm
Upper critical values of chi-square distribution with degrees of freedom
Probability of exceeding the critical value
0.10
0.05
0.025
0.01
0.001
1
2.706
3.841
5.024
6.635
10.828
2
4.605
5.991
7.378
9.210
13.816
3
6.251
7.815
9.348
11.345
16.266
The chi-squared test cannot be used if the number of individual observations in any
class is <5b because it is a poor approximation to the real values of the distribution. For
small numbers it is easy to use Fisher’s Exact Test (see Appendix/FET or use
http://www.graphpad.com/quickcalcs/contingency1.cfm).
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 6
Box 3-4. [SGB-­‐324] Stochastic effects can play a role in the expression
of phenotypes
How might you get differences in penetrance and expressivity for a particular
allele for two individuals that carry an identical allele? One source of variability is
genetic: that is, there can be unknown genetic modifiers segregating in the population.
In this case, it can be argued that then the phenotype is not be due to a single genetic
locus, as the modifier is also involved in the phenotype. Such cases of multiple loci
determining phenotypes will be discussed in Chapters 5 and 6.
Let’s consider the simplest case where two genetically identical individuals show
differing penetrance and or expressivity for a gene. Although this can be a difficult
situation to obtain in populations without controlled breeding, this situation is usually the
case when working with genetically tractable model organisms.
The environment can be one source of variability. For example, for phenotypes
such as animal body weight and height, nutrition can affect how these phenotypes are
expressed. Similarly, temperature, humidity, and light can affect how genetically
identical plant might grow.
Stochastic, or random, effects also play a role in penetrance and expressivity.
These random effects can be due to many sources, which may include effects due to
the probabilities of basic biochemical reactions or effects due to the distribution of
cellular components during the cytoplasmic and membrane partitioning seen during cell
division processes.
One way to envision stochastic effects due to partitioning is to consider a case
where there are a limited number of particles. An example of this type of particle might
be an organelle such as a mitochondrion. In the extreme, when there is only one
particle, this single particle can segregate at random to either cell (Figure 1). If this
particular particle is required for viability, then one daughter cell at each cell division will
die.
Figure 1 [SGF-1007] illustrates the random segregation of a small number of
particles at a cell division. If there is one particle, either of the daughters will have the
particle, each receiving it 50% of the time (Figure 1A). A population would have equal
numbers (1:1 ratio) of cells with zero or one particle. If there are two particles, then ¼ of
the cells will have zero particles, ½ will have one particle and ¼ will have two particles
(Figure 1B). The one particle class is twice as frequent because there are two ways a
cell might have one particle. This can be seen if we arbitrarily color the particles
(purple and red). If there are three particles, then 1/8 of the cells will have none, 3/8 will
have one, 3/8 will have two and 1/8 will have three (Figure 1C). We can see how these
frequencies arise if we label the particles purple, red and and green. If there are four
particles, then 1/16 will have none, ¼ will have one, 3/8 will have two, ¼ will have three
and 1/16 will have all four (Figure 1D). Again, this can be seen if we color the particles.
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 7
Figure 1 [SGF-1007]. Binomial sampling.
If we now consider two particles that segregate randomly and independently,
there are three possible outcomes:
1. The left daughter has no particles while the one on the right has two.
2. Each daughter cell has a single particle. This particular case will
happen twice as often, because the left and right daughter cells look
equivalent in this situation compared to situations 1 and 3.
3. The left daughter cell has two particles while the one on the right has
none.
In this example, if you were to examine each daughter cell for these three possible
outcomes, you can see that there are two daughter cells that have 0 particles, four
daughter cells will have 1 particle, and two daughter cells will have 2 particles, giving a
ratio of 1:2:1 for cells with 0, 1, or 2 particles. If the particle were required for viability,
you would then end up with 1/4th (or 25%) of cells dying.
If we now consider three particles that segregate randomly and independently,
the ratio now will be 1:3:3:1 for cells that have 0, 1, 2, and 3 particles. In this case, 1/8th
(or 12.5%) of the cells will die. In general, 1 / 2N daughters dies.
To think about stochastic effects having to do with a biochemical function,
consider cases where gene activity is dependent on a threshold level. (Consider an
example of incompletely penetrant allele of a gene required for viability. This allele has
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 8
partial loss of gene activity, due to a mutation in the gene that causes most of the
proteins produced to fold incorrectly and be biochemically inactive; however, a small
fraction of the proteins will fold correctly and provide enzymatic activity. Furthermore,
only 1/5th of the activity of the enzyme (compared to wild type cells) is necessary in
order for cells to live. Thus, even though only a fraction of the enzymes produced by the
cell are active, 90% of the cells have enough enzyme activity to live and thus do not die.
If the allele were a complete loss of function, complete penetrance would be seen and
100% of the cells would be dead (Table 7).
Table 7. Incomplete penetrance due to reduction of gene activity
Activity of enzyme
viability
wild-type
100%
partial loss of function
90%
complete loss of function
0%
A threshold (shown in orange) splits a distribution of gene activities (blue) into part
above threshold (green) and a part below threshold (shown in red). As the threshold is
centered to the left of the peak of the distribution of gene activity, a greater proportion of
individuals are over threshold than are under. The interpretation of the threshold curve
is that the probability of an individual being green is proportional to the height of the
curve. The penetrance would be the ratio of the area under the red curve to the area
under the blue curve. (Figure 2 [SGF-1009-2])
Figure 2 Incomplete penetrance can arise from a threshold effect.
An alternative possibility for creating an incompletely penetrant phenotype can
involve loss of function alleles in distinct pathways of gene action that have the same
final result In this case, a complete loss of function of a gene can still create an
incompletely penetrant phenotype. If you consider a case where pathway A contributes
80% of the effect of a phenotype while pathway B contributes 20% of the effect (Figure
3 [SGF-112]). If a cell were to completely lose function in pathway A, there would only be
an 80% defect, as pathway B was still intact. Similarly, a complete loss of pathway B
would result in a less severe 20% defect, as pathway A is still providing 80% of the
activity. In particular, one pathway (pathway A) makes product Z from substance X.
Another pathway (pathway B) makes the same product Z from substance Y. A.
Pathway A makes 0.8 of the total product Z, while pathway B makes 0.2 of the total
product Z. B. Loss of pathway A results in only 0.2 of product Z begin produced (by
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 9
pathway B). C. Loss of pathway B results in only 0.8 of product Z begin produced (by
pathway A). D. Loss of both pathway A and pathway B results in complete lack of
product Z. If the phenotype is simply the amount of Z produced, then pathway A mutant
would display 80% penetrance as it lacks 80% of the wild-type activity.
Figure 3. [SGF-112]. Two contributing pathways creating incompletely penetrant
phenotypes.
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 10
Box 3-5 [SGB-325 ] Using pedigrees to analyze modes of
inheritance
Figure 1.
When controlled crosses are not possible, pedigree analysis can be used to
analyze modes of inheritance. Because for most pedigree analysis the sample size is
small, chi-squared analysis does not apply. Instead, you can calculate the probability
that the segregation pattern you see would fit your model by chance.
Consider the two pedigrees in Figure 1, both of which show hereditary patterns
consistent with the segregation of dominant alleles or alleles that display incomplete
penetrance.
The pedigree in A has two of four children of an affected mother displaying the
phenotype. There are several simple models with which this pedigree is consistent.
These data are consistent with dominant allele conferring the phenotype, with the
mother being heterozygous for the dominant allele. These data are also consistent with
her being homozygous for the dominant allele and the trait being about 50% penetrant.
Finally, these data are consistent with a recessive allele that is homozygous in the
mother and heterozygous in the father.
The pedigree shown in B has all four children of an affected mother displaying
the phenotype. One obvious explanation is that the phenotype is due to a dominant
allele inherited from the mother. However, is it possible that this segregation pattern is
due to a recessive allele that is homozygous in the mother and heterozygous in the
father? If this were the case, we would expect a 1:1 ratio in the progeny of affected
individuals to non-affected individuals.
Given the small sample size in this situation (4 progeny), what is the probability
of observing 4 affected individuals if the expected situation is 2 affected individuals and
2 non-affected individuals? To calculate this probability you would calculate 1/2N where
N is the number of affected individuals. For N=4, 1/24 is 1/16. In other words, there is
a 1/16 chance that you would obtain this result by chance. Because conventionally, an
hypothesis is rejected if there is a 1/20 probability of obtaining the data by chance, we
cannot reject the hypothesis here and conclude that the difference is not statistically
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 11
significant. Therefore, either hypothesis fits the data because the sample size is too
small.
On the other hand, if there were 5 progeny that all were affected (and thus) N=5,
the probability of obtaining this result by chance 1/32. In this case, since the probability
is less than 0.05 and thus significant, we would reject the hypothesis of the
heterozygous father. Similarly, with 7 progeny, all affected, the probability would be
1/128, which is <0.01 and thus highly significant.
How can one distinguish these possibilities? One way is to obtain a larger
sample. This larger sample could come from including additional generations in the
pedigree, or from additional families with the trait. If additional families are included,
you have to assume that it has the same genetic cause.
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 12
Box 3-7. [SGB-327 ] Can a dominant maternal effect mutation be distinguished
from cytoplasmic inheritance associated with mitochondria?
How can you distinguish cytoplasmic inheritance from a dominant, strict maternal
effect? Most simply, the mapping of the locus to a chromosome would indicate that it is
not due to cytoplasmic inheritance. Methods associated with the traditional mapping of
genetic loci will be discussed in Chapter 9.
Sometimes, mapping is not simple. Another method to distinguish a maternal
effect mutation from an effect due to mitochondrial inheritance is to set up crosses and
analyze their segregants. For example, if you were to take an affected female and
cross to an affected male, then take the females from this cross (who are all affected)
and cross it to an unaffected male, you can then look at the F2 female progeny and
examine their offspring. If the effect were due to a strict maternal effect, only ½ of the
F2 females will give rise to affected progeny. However, if the effect were due to a
mitochondrial marker (assuming a homogenous mitochondrial population), all F3
progeny of the F2 females will produce affected progeny. If the mitochondria were to
exist as a genetically diverse population, there may be a mixture of non-affected and
affected females that can be seen segregating in all generations.
Figure 1. A. dominant maternal effect. B. cytoplasmic inheritance. [SGF-1033].
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 13
Box 3-6
[SGB-326]
Balanced lethals and balancer chromosomes.
True-breeding strains do not necessarily have to be homozygous, as there can be
stable strains that are maintained in heterozygous state (Figure 1 [SGF-1019]). For
example, from a heterozygote + a1 lethal2/ lethal1 a2 +, one-half of the progeny will also
be this heterozygote, and ½ the progeny will die from either of the recessive lethals.
Such a situation is called balanced lethals because the recessive lethal mutations keep
each other in an otherwise precarious situation (see Chapter 13).
As balanced lethals allow for the maintenance of the heterozygous state of the A gene,
they can also be useful for strain maintenance and strain construction, particularly if the
homozygous state of one of the A alleles leads to a suboptimal condition such as
sterility, death, or slow growth. This concept was extended to the construction of
specialized balancer chromosomes used in model organisms to maintain a
heterozygous state. Balancer chromosomes are placed in trans to the chromosome
carrying your allele of interest, and will either suppress the recombination from the other
chromosome or allow you to detect that a recombination event occurred (see Chapter 9)
More generally, one can use marked chromosomes to keep track of alleles for
maintenance of strains, strain construction or mutagenesis (see Chapter 7). Figure 2
shows a simple generic balancer chromosome. This contains a dominant marker, for
example a GFP expressing transgene, to distinguish any individual that has the
balancer chromosome. A recessive lethal kills off the homozygous balancer, which
would also express GFP. An aberration such as a large inversion or a translocation that
suppresses recombination (see Chapter 14) prevents recombination.
Figure 1 Balanced lethals [SGF-1019]
Figure 2. Generic balancer chromosomes. [SGF-1217]
Systems Genetics, Chapter 3 Boxes V1, 4/1/13 14
Box 3-9. [SGB-329] Exchanging compartments between cells allows for
the distinction of nuclear from cytoplasmic effects.
Sometimes the physical mapping of the trait to either the nucleus or the
cytoplasm can be accomplished by the transfer of material from one cell to the other
Figure 1 [SGF-1218]. For example, by transf erring the nucleus from an affected cell to the
cytoplasm of an enucleated unaffected cell (and vice versa), you can determine which
cellular compartment is important for conveying the inheritance of the trait (Table 8). If
the trait followed the nucleus, the conclusion is that the nucleus (and presumably the
chromosomes) is important for inheritance.
Figure 1 [SGF-1218]. Nuclear transplant.
Table 8. Logic of nuclear transplant experiments.
genotype
phenotype
nucleus
cytoplasm
+
+
wild type
+
mutant
+
wild type
wild type
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