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Chapter 13
Extensions of and
Deviations from Mendelian
Genetic Principles
Copyright © 2010 Pearson Education Inc.

Not all genes have only
two forms (alleles); many
have multiple alleles.
◦ a. Differences in the DNA
sequence of a gene result in
multiple alleles.
◦ b. No matter how many
alleles for the gene exist in
the multiple allelic series, a
diploid individual will have
only two alleles, one on each
homologous chromosome.

The number of possible genotypes in a multiple allelic
series depends on how many alleles are involved
◦ a. The formula n(n1+1)/2 calculates possible genotypes for n
alleles.
◦ b. Of the genotypes predicted by the formula, n are
homozygotes, and n(n 2-1)/2 are heterozygotes.


ABO blood groups result from a series of three
alleles (IA, IB, and i) that combine to produce four
phenotypes (A, B, AB, and O).
Both IA and IB are dominant to i, while IA and IB are
codominant to each other. The resulting
phenotypes are:
◦
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a. People
b.People
c. People
d.People
with
with
with
with
genotype
genotype
genotype
genotype
i/i are blood type O.
IA/IA or IA/i are blood type A.
IB/IB or IB/i are blood type B.
IA/IB are blood type AB.

ABO inheritance follows Mendelian principles. For example,
a type O individual’s genotype is i/i. Possible genotypes of
the parents could be:
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a. i/i and i/i (both blood type O).
b. IA/i and i/i (one type A, the other type O).
c. IA/i and IA/i (both type A).
d. IB/i and i/i (one type B, the other type O).
e. IB/i and IB/i (both type B).
f. IA/i and IB/i (one type A, the other type B).
Blood typing may be used in cases of disputed parentage.
Blood typing does not prove the identity of a parent. It can,
however, eliminate individuals who are not biological
parents of a particular child. Example:
◦ a. A child with blood type AB (IA/IB) could not have a parent with
type O (i/i).
◦ b. Blood-type data are not considered adequate legal proof for
parenthood in most states, and DNA fingerprinting is generally
used.

Cellular antigens are important
in blood transfusions, since
recipient antibodies may
respond to antigens on donor
cells. Recipients generally can
accept blood containing
antigens the recipient already
has..
◦ a. Type A blood can be transfused
into type A or AB recipients.
◦ b. Type B blood can be transfused
into type B or AB recipients.
◦ c. Type AB blood can be
transfused only into type AB
recipients, the only ones who will
recognize both antigens as self.
◦ d. Type O blood has neither
antigen, and so can be transfused
into a recipient of any blood type.

Summary of the relationship between the ABO alleles and RBC
antigens:
◦ a. The ABO locus produces RBC antigens by encoding
glycosyltransferases, which add sugars to existing polysaccharides on
membrane glycolipid molecules. These polysaccharides act as the antigen
in the ABO system.
◦ b. In most people, the glycolipid is the H antigen.
i. The IA gene product is a glycosyltransferase that adds a-Nacetylgalactosamine to the H antigen, converting it to the A antigen.
 ii. The IB gene product is a glycosyltransferase that adds galactose to its
polysaccharide, converting the H antigen to the B antigen.
 iii. Both enzymes are present in an IA/IB individual, and some H antigens will be
modified to the A antigen while others are modified to the B antigen.
 iv. Neither enzyme is present in an i/i individual and so the H antigen remains
unmodified.


Drosophila has over 100 mutant alleles at the eyecolor locus on the X chromosome. Homozygotes
for each allele have eyes of a distinct color in the
spectrum between white and red, depending on
how much pigment depositing function remains in
the encoded protein.


The eosin allele (we) is recessive
to wild type.
Eosin and white are mutations of
a single gene. The relationship
between these multiple alleles is:
◦ a. The allele red (wild-type) is
dominant to eosin and white.
◦ b. The eosin allele is recessive to red
but dominant to white.
◦ c. For example, in the cross of an
eosin-eyed (we/we) female with a
white-eyed male (w/Y), the F1 females
are all we/w. They have eosin eyes,
showing that we is dominant over w.

Next the eosin-eyed F1 females
(we/w) are crossed with redeyed males w+/Y.
◦ i. All female progeny are red-eyed
(w+/w or w+/we).
◦ ii.Male progeny are 1⁄2 eosin-eyed
(we/Y), and 1⁄2 white-eyed (w/Y).


Genes encode proteins, and changes in amino
acids of those proteins may change a phenotype.
Multiple alleles exist for many genes, because
there are many sites within a gene where
introduction of a mutation will alter the protein
product.
Consequences of multiple alleles in human genetic
disorders include:
◦ a. Variation in disease symptoms depending on the
patient’s allele(s).
◦ b.Complications in designing a single DNA-based test to
diagnose the disease or detect carriers.

Complete dominance and complete recessiveness
are two extremes in the range of dominance
possible between pairs of alleles. Many allelic pairs
are less extreme in their expression, showing
incomplete dominance or codominance.
◦ a. In incomplete dominance, a heterozygote’s phenotype
will be intermediate between the two possible
homozygous phenotypes.
◦ b.In codominance, the heterozygote shows the
phenotypes of both homozygotes.
◦ c. At the molecular level, these relationships between pairs
of alleles depend upon patterns of gene expression.



Incomplete dominance is an allelic
relationship where dominance is
only partial.
In a heterozygote, the recessive
allele is not expressed. The
dominant allele is unable to
produce the full phenotype as in
homozygous dominant individual.
The result is intermediate
phenotype.
Palomino horses when interbred,
the progeny are:
◦ a. 1⁄4 cremello (cream colored) with
genotype Ccr/Ccr.
◦ b. 1⁄2 palomino with genotype C/Ccr.
◦ c. 1⁄4 light chestnut (full coat color)
with genotype C/C.


Incomplete dominance often occurs in plants. An
example is flower color in snapdragons involving
two alleles, CR and CW. Red-flowered plants (CR/CR)
crossed with white-flowered ones (CW/CW) produce
all pink progeny (CR/CW).
The sickle-cell mutation in humans is another
example.

In codominance, the heterozygote’s
phenotype includes the phenotypes of both
homozygotes. Examples include:
◦ a. The ABO blood series, in which a heterozygous
IA/IB individual will express both antigens, resulting
in blood type AB.
◦ b.The human M-N blood group involves red blood
cell antigens that are less important in transfusions.
There are three types:
 i.
Type M, with genotype LM/LM.
 ii.
Type MN, with genotype LM/LN.
 iii.
Type N, with genotype LN/LN.

Current explanations involve levels of gene expression
for each allele in the pair.
◦ a. In codominance, both alleles make a product, producing a
combined phenotype.
◦ b. In incomplete dominance, the recessive allele is not
expressed, and the dominant allele produces only enough
product for an intermediate phenotype.
◦ c. By contrast, a completely dominant allele creates the full
phenotype by one of two methods:
 i. It produces half the amount of protein found in a homozygous
dominant individual, but that is sufficient to produce the full
phenotype. These genes are haplosufficient.
 Ii. Expression of the one active allele may be upregulated,
generating protein levels adequate to produce the full phenotype.
◦ d. Gain-of-function alleles may also show incomplete
dominance, with heterozygotes intermediate in phenotype
between wild-type and fully mutant homozygotes.

Some genes are required for life (essential genes),
and mutations in them (lethal alleles) may result in
death. Dominant lethal alleles result in death of
both homozygotes and heterozygotes, while
recessive lethal alleles cause death only when
homozygous.
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Yellow crossed with nonyellow results in a ratio
of 1 yellow : 1 nonyellow. This suggests yellow
is heterozygous.
Yellow mice never breed true, another indication
of heterozygosity. When yellow is bred with
yellow, the result is about 2 yellow : 1 nonyellow
(instead of the predicted 3:1).
Castle and Little (1910) proposed that yellow
homozygotes die in utero and are therefore
missing from the progeny. The yellow allele has
a dominant effect on coat color but also acts as
a recessive lethal allele.
Yellow is an allele of the agouti locus,
designated AY. shows the yellow X yellow cross.
i. The cross is AY/A 3 AY/A, and death of the
homozygous yellow animals (AY/AY) results in a 2:1
ratio.
◦
ii. When two heterozygotes are crossed and
produce a 2:1 ratio of progeny, a recessive lethal allele
is suspected.
◦
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Wild-type agouti mice express the agouti gene only during
hair development in the days after birth and when plucked
hair is being regenerated. Gene expression is seen in no
other tissues and at no other time.
Heterozygous mice (AY/A) express the AY allele at high
levels in all tissues during all developmental stages. Tissuespecific regulation appears to be lost in the AY allele.
The AY allele transcript RNA is 50% longer than that of the
wild-type allele (A1). This is because:
◦ (1)
The AY allele results from deletion of an upstream
sequence, removing the normal promoter of the agouti gene.
◦ (2)
The gene is transcribed from the promoter of an
upstream gene called Raly. The beginning of the sequence
encoding Raly is fused with the agouti gene, producing a longer
transcript.

Embryonic lethality of AY/AY mice probably results from lack
of Raly gene activity rather than from the defective agouti
gene.



Tay–Sachs disease, resulting from an inactive gene for
the enzyme hexosaminidase. Homozygous individuals
develop neurological symptoms before 1 year of age,
and usually die within the first 3–4 years of life.
Hemophilia, resulting from an X-linked recessive allele,
is lethal if untreated.
A dominant lethal gene causes Huntington disease,
characterized by progressing central nervous system
degeneration. The phenotype is not expressed until
individuals are in their 30s. Dominant lethal are rare,
since death before reproduction would eliminate the
gene from the pool.

Development of a multicellular organism from a zygote
is a series of generally irreversible phenotypic changes
resulting from interaction of the genome and the
environment. Four major processes are involved:
◦
◦
◦
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
a. Replication of genetic material.
b. Growth.
c. Differentiation of cells into types.
d. Arrangement of cell types into defined tissues and organs.
Internal and external environments interact with the
genes by controlling their expression and interacting
with their products.

Penetrance describes how completely
the presence of an allele corresponds
with the presence of a trait. It depends
on both the genotype (e.g., epistatic
genes) and the environment of the
individual.
◦ a. If all those carrying a dominant mutant
allele develop the mutant phenotype, the
allele is completely (100%) penetrant.
◦ b. If some individuals with the allele do
not show the phenotype, penetrance is
incomplete. If 80% of individuals with the
gene show the trait, the gene has 80%
penetrance.
◦ c. Human examples include:
 i. Brachydactyly involves abnormalities of the
fingers and shows 50–80% penetrance.
 Ii. Many cancer genes are thought to have
low penetrance, making them harder to
identify and characterize.

Expressivity describes variation
in expression of a gene or
genotype in individuals.
◦ a. Two individuals with the same
mutation may develop different
phenotypes due to variable
expressivity of that allele.
◦ b.Like penetrance, expressivity
depends on both genotype and
environment and may be constant or
variable.

Osteogenesis imperfecta, inherited as an
autosomal dominant with nearly 100% penetrance.
◦ i. Three traits are associated with the allele:
 (1) Blueness of the sclerae (whites of eyes).
 (2) Very fragile bones.
 (3) Deafness.
◦ ii. Osteogenesis imperfecta shows variable expressivity
because an individual with the allele may have one, two,
or all three of its symptoms, in any combination. Bone
fragility is also highly variable.

An example is neurofibromatosis.
◦ a. The allele is an autosomal dominant that shows
50–80% penetrance and variable expressivity.
◦ b. Individuals with the allele show a wide range of
phenotypes:
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 i. The mildest form of the disease is a few pigmented
areas on the skin (café-au-lait spots).
ii. More severe cases may include:
(1) Neurofibroma tumors of various sizes.
(2) High blood pressure.
(3) Speech impediments.
(4) Headaches.
(5) Large head.
(6) Short stature.
(7) Tumors of eye, brain, or spinal cord.
(8) Curvature of the spine.
Incomplete penetrance and variable
expressivity complicate medical genetics
and genetic counseling.

Age of onset is an effect of the individual’s internal
environment. Different genes are expressed at different
times during the life cycle, and programmed activation and
inactivation of genes influences many traits. Human
examples include:
◦ a. Pattern baldness, appearing in males aged 20–30 years.
◦ b. Duchenne muscular dystrophy, appearing in children aged 2–5
years.

Sex of the individual affects the expression of some
autosomal genes.
◦ a. Sex-limited traits appear in one sex but not the other. Examples
include:
 i. Milk production in dairy cattle, where both sexes have milk genes but
only females express them.
 ii. Horn formation in some sheep species, where only males express the
genes used to produce horns.
 iii. Facial hair distribution in humans.

Sexes show either a difference in
frequency of occurrence or an altered
relationship between genotype and
phenotype. Human examples include:
◦ i. Pattern baldness, controlled by an
autosomal gene that is dominant in males
and recessive in females.
 (1) The genotype b/b produces pattern
baldness in both men and women.
 (2) The genotype b+/b+ gives a nonbald
phenotype in both sexes.
 (3) The genotype b+/b will lead to the bald
phenotype in men and the nonbald phenotype
in women.
 (4) This gene shows variable expressivity in:
 (a)
 (b)
 (c)
Age of onset.
Site of baldness (crown or forehead).
Degree of hair loss.

Temperature may alter the activity of
enzymes so that they function normally
at one temperature but are
nonfunctional at another. An example
is fur color in Siamese cats.
◦ a. Siamese phenotype occurs in cats
homozygous for the recessive cS allele of the
C (albino) locus, which encodes a tyrosinase.
◦ b. Tyrosinase is required in the melanin
synthesis pathway. Activity of the tyrosinase
encoded by cS is temperature sensitive.
 i. Kittens are uniformly warm, and so stay light
colored.
 ii. As cats grow, extremities become cooler, and
tyrosinase becomes active, allowing melanin to
be made and fur on the points to become
darker.
 iii. Color on the points depends on other coat
color genes in the animal.

Chemicals can have significant effects. Example:
◦ a. Phenylketonuria (PKU) is an autosomal recessive defect
in metabolism of the amino acid phenylalanine. If the
defect is not treated by restricting phenylalanine in the
diet, severe mental retardation and other symptoms
result.

Some maternally derived phenotypes are
produced by the maternal nuclear genome
(maternal effect) rather than inherited as
extranuclear genes (maternal inheritance).
◦ a. Proteins and/or mRNA deposited in the oocyte
before fertilization direct early development in the
embryo.
◦ b.The genes encoding these products are on
nuclear chromosomes. No mtDNA is involved.

Shell coiling in the snail Limnaea peregra.
◦ a. Shell coiling is determined by a pair of nuclear alleles,
with the dominant D allele producing a dextral (right) coil
and the recessive d producing sinistral (left) coiling.
◦ b.The shell-coiling phenotype is always determined by
the mother’s genotype.
◦ c. In all crosses of true-breeding dextral and sinistral
snails, the F1s have the same genotype (D/d) but the
reciprocal crosses produce different phenotypes.

A dextral female (D/D) crossed
with a sinistral (d/d) male
produces a dextral F1 (D/d).
◦ (1) The F2 genotypes have a 1:2:1
ratio (D/D : D/d : d/d). All F2
snails, including those with
genotype d/d, have dextral shells.
◦ (2) Selfing the F2 produces an F3
that is 3⁄4 dextral and 1⁄4 sinistral.
The sinistral snails are the progeny
of F2 d/d mothers (who had dextral
shells).

A sinistral female (d/d) crossed
with a dextral male (D/D)
produces a sinistral F1 (D/d).
◦ (1) The F2 genotypes also have a 1:2:1
ratio (D/D : D/d : d/d). All F2 snails
have sinistral shells.
◦ (2) Selfing the F2 produces an F3 that
is all dextral, due to the D/d
genotype of the F2 mothers (who had
sinistral shells).