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Genetics Chapter 5
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Variations on Dominance
Mendelian genetics :
• dominant allele requires only 1 copy to determine a
phenotype ; masks expression of recessive allele
• recessive allele requires 2 copies to determine a
phenotype ; expression masked by dominant allele
• crossing 2 monohybrid heterozygotes when
dominance is complete gives a 3:1 phenotypic ratio
sometimes this simple dominant-recessive relationship
does not hold, and a heterozygous genotype does not
always express a dominant phenotype
Incomplete Dominance in Four-
o’clock Plants (Mirabilis jalapa)
Incomplete dominance : heterozygotes exhibit a
phenotype intermediate to the phenotypes
expressed by the two homozygotes
F2 Phenotypic ratio 1 : 2 : 1
Neither allele is
dominant to the
other : use R1,R2
instead of R, r
Incomplete Dominance
R1 : specifies red pigment
R2 : specifies no color
• R1, R2 heterozygotes : lighter red because they have only 1 allele that produces color
• Mendelian patterns of inheritance are still followed (1:2:1 ratio in F2), but phenotypes
do not follow patterns of strict dominance
The ABO Blood Classification
Erythrocytes expressing one (or both) of the ABO
antigens have an antibody against the other antigens
in their serum
Codominance
Codominance : the heterozygote expresses
phenotypes of both homozygotes simultaneously
The ABO Blood Classification
Codominance
IA IA x IB IB
F1 : 100% IAIB
(type A) (type B)
(type AB)
Phenotype includes both A and B
simultaneously : codominance
F2 : ?
Pedigree Analysis of the Inheritance
of Tay-Sachs Disease
Analysis of the pedigree indicates that Tay-Sachs is inherited as an autosomal
recessive trait
at the level of viability, shows Mendelian genetics of complete dominance
(only homozygous recessive is not viable)
this disease causes severe neurodegeneration before age 3
homozygous normal and heterozygotes do not have any symptoms
Hexosaminidase-A in normal and Tay-Sachs
individuals at the level of enzyme activity
The heterozygote individuals possess less activity than the homozygote
dominant individual. The homozygous recessive individual exhibits no activity.
at the level of enzyme activity, shows incomplete dominance
Figure 5.2b
Hexosaminidase-A in normal and Tay-Sachs
individuals at the level of protein structure
One variant in Tay-Sachs disease produces a protein smaller than the wild-type
protein.
•Afflicted individuals (aa) produce only the smaller protein
•Heterozygote individuals (Aa) produce both proteins
•Homozygous normal (AA) produce the larger protein
at the level of protein structure, shows codominance
Multiple Alleles
• we have considered genes that have only 2 alleles
• a gene can have more than 2 alleles
- diploid organism can have only 2 alleles, but many
different alleles of any given gene can exist in a
population polymorphic alleles
- in fact, multiple alleles are the rule rather than the
exception
The ABO Blood Classification
3 alleles : IA, IB, and i
IA, IB : glocosyltransferase enzymes that make A and B
structures on a sugar molecule (H structure) on RBCs
i : nonfunctional enzyme (recessive to IA and IB)
The ABO Blood Classification
Complete dominance (Mendelian) :
i is recessive to IA and IB
- IA and/or IB will modify the H product and
mask the fact that the i allele is present
Codominance :
IA and IB can be expressed simultaneously
Fur Color in the Mouse : Controlled by
Several Genes Including Agouti • over 25 different agouti alleles have been identified
• these alleles either exhibit a unique phenotype or have
different genetic interactions with other agouti alleles
Consider 4 of these alleles : A, Aw, at, and a
Heterozygous mice reveal that the Aw allele
is dominant to the other three agouti alleles
(Wild type)
(recessive)
(“white-bellied” allele ; dominant)
(“black&tan” allele)
Phenotypes of the Different Allelic
Combinations of Mouse Agouti Gene
Wild-type agouti allele (A)
-Dark gray fur color
-Deposition of yellow and black
pigments
White-bellied agouti allele (Aw)
-Dark gray fur on the back, white
or cream on the belly dominant
Nonagouti allele, a, is recessive to
other alleles
Phenotypes of the Different Allelic
Combinations of Mouse Agouti Gene
These alleles show a hierarchy of dominance :
Aw is dominant to A
Aw and A are dominant to a
. . . and complex patterns of dominance :
A is dominant to at on the back
at is dominant to A on the belly
Testing Allelism
* There may be many different variations of the
2 alleles that code for a gene (such as with
the gene for mouse fur color)
* Also, mutations in different genes may cause
similar or identical phenotypes
To study how phenotypes are inherited, need to
know which mutations are different
versions of the same gene (alleles) and which
are in completely different genes
Partial map of Drosophila chromosomes
ruby and scarlet are
2 different mutations
of 2 different genes on
different chromosomes,
but they both produce
a similar eye color
Complementation test in flies: determine
whether 2 recessive mutations are alleles
•Cross flies with homozygous recessive mutations, examine progeny :
1) wild-type phenotype : mutations complement each other
mutations are in different genes
arrangement of the two recessive mutations on different
chromosomes is called the trans configuration (each homolog contains 1
mutation)
wingless wingless
mutant gene 1, wt gene 2
wt gene 1, mutant gene 2
wild-type
: mutation
Noncomplementation (1)
2) none of the F1 progeny exhibit the wild-type phenotype:
mutations do not complement each other
mutations are in different alleles of the same gene
wingless wingless
mutant gene 1, wt gene 2
mutant gene 1, wt gene 2
wingless
Noncomplementation (2)
• Cross flies with nonidentical recessive mutations, examine
progeny
- F1 progeny : small, crinkled wings (not wild-type trait)
these mutations do not complement each other and must be
alleles
wingless Small, wrinkled wings
Very Small, wrinkled wings
mutant gene 1, wt gene 2
mutant gene 1, wt gene 2
(cis configuration is a control for test
to make sure a wt can be produced)
Parents : i) both mutants on same chromosome i) 1 mutant on 1 chromosome
ii) wild-type ii) other mutant on other chromosome
The Cis-Trans Test • examine a large # of mutants and place them in complementation groups
** identifies : 1) mutations in separate genes and 2) independent alleles of a
single gene
(trans configuration determines if
mutations are allelic or not)
The Cis-Trans Test
can use this test to determine how many genes control
a biochemical pathway
- try to mutate every gene in a pathway, then use cis-trans test
to determine how many distinct genes code for proteins that
affect the same mutant phenotype
helps to treat genetic diseases
Lethal Alleles
• cause a deviation from expected 3:1 phenoptypic
ratio in a monohybrid cross
• caused by certain genotypes that result in death
and are not seen in the progeny
• these alleles can still follow standard Mendelian
dominant-recessive relationships but will skew
Mendelian phenotypic ratios
Relationship of yellow (AY) allele to the
wild-type agouti (A) allele
• Homozygous agouti X yellow : always produces ½ yellow and
½ agouti mice (expected 1:1 ratio for a monohybrid test cross)
because the agouti mice are known to be homozygous,
yellow mice must be heterozygous with yellow dominant to
agouti
(1)
2 important crosses :
? we already know agouti
mice are genotypically
AA
•Yellow X yellow 2/3 yellow, 1/3 agouti
•Expected results phenotypic ratio 3:1 yellow to agouti
(monohybrid cross of heterozygotes)
•Possible explanations?
Relationship of yellow (AY) allele to the
wild-type agouti (A) allele
(2)
Examination of pregnant mice revealed 25% of the
developing embryos were dead
•Homozygous yellow allele is lethal: AyAy
•2/3 of the surviving mice will be yellow (heterozygous)
•1/3 will be agouti (homozygous)
Relationship of yellow (AY) allele to the
wild-type agouti (A) allele
AY allele is dominant for coat color, recessive for lethality
Pleitropy : when a single mutation causes multiple
phenotypes
Relationship of yellow (AY) allele to the
wild-type agouti (A) allele
Other Lethal Mutations • Dominant lethal mutations
Can only exist if :
i. Lethality is only expressed after sexual maturity
- example : Huntington disease (onset : after age 40)
ii. Incomplete penetrance : a phenotype is not
expressed for a certain genotype
• Deleterious mutations
- reduce viability without always causing death ;
depends on environment
- example : temperature sensitive mutations in flies
What about when a specific genotype
does not express the expected
phenotype ?
3 possibilities :
i. incomplete penetrance
ii. variable expressivity
iii. 1 gene affects the expression of
another gene
Incomplete Penetrance: A Percentage of
individuals with a genotype do not express
the expected phenotype
Only 60% of the individuals are expressing the expected
phenotype ; it is 60% penetrant
• Most genotypes are 100% penetrant
(all are homozygous recessive genotype)
Variable Expressivity : the range of
phenotypes associated with a specific
genotype is increased
(all are homozygous recessive genotype)
All express a mutant phenotype, but there is a range of
these phenotypes (light red to complete white)
- these phenotypes are associated with a single
mutant allele (caused by different expression levels?)
Incomplete Penetrance and Variable
Expressivity : Polydactyly
•Autosomal dominant trait
•Extreme manifestations characterized
by an extra digit on each hand and one
or two extra toes on each foot
•Variable expressivity:
•Extra toes or extra fingers
•Portion of an extra digit – thumb
•Partial extra little toe
incomplete penetrance and variable expressivity : affected by genetics and
environmental factors
Genotypic Interactions
Mendelian Inheritance :
Cross between 2 independent traits should
always produce 9:3:3:1 phenotypic ratio
- 9/16 have both dominant phenotypes
- 1/16 has both recessive phenotypes
Variations in phenotypes or in ratios
indicates that 2 genes may be interacting
•Cross Rose-combed chicken X pea-
combed chicken (or vice versa)
•F1 offspring are walnut-combed
•Cross F1 Heterozygous walnut-
combed chickens
•F2 progeny – 9:3:3:1
indicates 2 genes are involved
•9/16 walnut-combed fowl
•3/16 rose-combed fowl
•3/16 pea-combed fowl
•1/16 single-combed fowl
prediction of genotypes ?
2 Genes Contribute to a Single Phenotype: 1) mixed phenotypes : genetic interactions in the
combs of chickens
Interpretation:
•Dominant alleles for each gene: (R-)
(P-) : walnut-combed
•Dominant rose allele (R-) and
homozygous pea (pp) – rose-combed
•Dominant pea allele (P-) and
homozygous rose (rr) – pea-combed
•Homozygous recessive (pprr) for
both alleles – single-combed fowl
2 genes affect 1 phenotype
2 Genes Contribute to a Single Phenotype: 1) mixed phenotypes : genetic interactions in the
combs of chickens
•Crossing 2 white corn varieties yields all
purple kernels in F1
•F1 cross yields 9/16 purple
7/16 white ???
-Must be dealing with 2 traits, each with 2
alleles because ratios are in sixteenths
-9:7 is variation of 9:3:3:1 ratio
-genotypic possibilities in Punnet square :
purple color appears when at least one
dominant allele for both genes is present
white color appears when one or both
genes only have recessive alleles
2 Genes Contribute to a Single Phenotype: 2) complementary gene action : genetic interactions in
the color of corn kernels
2 Genes Contribute to a Single Phenotype: 2) complementary gene action : genetic interactions in
the color of corn kernels
• a dominant allele for 2 different genes must be present to
produce a phenotype (purple)
• without this allele at either or both genes, white color results
can order these genes in a biochemical pathway
Colorless precursor colorless intermediate purple product
Gene A
catalyzes
Gene B
catalyzes
Shepherd’s purse plant (Capsella bursa-pastoris)
Indicates heart-shaped is dominant to narrow;
should be a simple monohybrid cross
???
- expect 3:1, 1:2:1, or 2:1 for monohybrid
F2 ratio
Ratio is modification of 9:3:3:1 2 genes are involved
2 Genes Contribute to a Single Phenotype: 3) duplicate gene action : 2 genes produce a
phenotype ; both genes appear equivalent
15:1 ratio indicates that the dominant allele (for either the A or B gene) is
sufficient to produce the heart-shaped phenotype.
•the enzymes encoded by the A and B alleles are equivalent in the biochemical
pathway :
• 2 different genes encode identical enzymes
one of these enzymes can have a slightly different function ;
creates gene families with similar function (as with -globin family
composed of 5 genes, each expressed at different times in development)
2 Genes Contribute to a Single Phenotype: 3) duplicate gene action : 2 genes produce a
phenotype ; both genes appear equivalent
•Crossing a pure-breeding black
mouse with a pure-breeding
albino mouse produce 100%
agouti mice
•The albino phenotype results
from the absence of pigment in
the hair
2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene (epistatic
gene) masks the phenotype produced by 2nd gene
(hypostatic gene)
Coat Color in Mice: Black, Albino, or Agouti
* * Indicated by modified 9:3:3:1 ratio with only 3 phenotypes
F2 9:3:4 ratio is a modification of 9:3:3:1 recessive epistasis
F2 progeny:
•9/16 agouti A-C-
•3/16 black aaC-
•4/16 albino
(3/16 A-cc and 1/16 aacc)
2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene
masks the phenotype produced by 2nd gene
any genotype with cc is albino and masks phenotype of the A
gene (agouti or black) ; without this, the A gene expresses (A>a)
Color of Summer Squash Can Be White, Yellow, or Green
P1: pure-breeding white X
pure-breeding green
•F1 progeny : all white
squash
•F1 is self-crossed
2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene
masks the phenotype produced by 2nd gene
2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene
masks the phenotype produced by 2nd gene
F2 12:3:1 ratio is a modification of 9:3:3:1 dominant epistasis
dominant A allele masks phenotype of B gene
Defects in the Biochemical Pathway Producing
Melanin Results in Albino Mice
Mouse coat color :
Both the agouti
and black
phenotypes need
melanin in order to
be expressed
Albinism is caused by defects pathway
for melanin production :
The Epistatic Gene Functions Early in the Pathway
Resulting in the Albino Phenotype for cc Mice
The recessive c allele is responsible for the albino pigment by blocking
the pathway ; its product is required by next gene (A) in making the
yellow coat color
In biochemical pathways involving epistasis :
epistatic gene functions earlier than hypostatic gene
genetic analysis complements a biochemical
analysis
Suppression of the Vestigial
(vg) Wing Phenotype in
Drosophila :
In this case the su(vg)
recessive mutation “pushes”
the vestigial phenotype
toward the wild-type
recessive suppressor
2 Genes Contribute to a Single Phenotype: 5) suppression : one gene “pushes” the mutant
phenotype of a 2nd gene towards wild-type
phenotype
classes:
+ +
+ - -
- - - -
- - + -
Suppressors :
1) May have no other phenotype than to suppress
another gene’s mutant phenotype
dihybrids produce only 2 phenotypes
2) Only make the second gene’s mutant phenotype
more like wild-type (no effect on wild-type allele)
3) Can be either dominant or recessive, may suppress a
dominant or recessive allele
4) May have both dominant or recessive phenotypes
A suppressor gene may
encode a protein that
specifically interacts with
another protein :
Two genes (m and su) encode
different proteins that interact
to convert a substrate to a
product and produce the wild-
type phenotype
* finding these suppressor mutations
is a powerful way to identify which
gene product forms a protein
complex with another gene product
Possible Mechanisms of Suppression:
Other mechanisms (nonspecific; can act
on several other genes):
1) increase or decrease expression
level of gene
2) affect translation of an mRNA
- prevent a stop in translation
induced by nonsense mutation
Possible Mechanisms of Suppression:
Modified 9:3:3:1 ratio in F2 generation
2 genes are interacting in the
phenotype
epistasis : often only 3 phenotypes
9:3:4
12:3:1
suppression : often only 2 phenotypes
13:3
Example of epigenetics :
mosaicism in human females
If the female is heterozygous for a sex-linked gene, she is mosaic
(composed of two or more genetically distinct tissues or cell types)
- some tissues express dominant allele, other tissues express recessive
allele
Random inactivation of X chromosome during embryonic
development in female mammals -some cells (and developing tissues) have paternal X chromosome
inactivated, others have maternal X chromosome inactivated
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