Transcript Incomplete Dominance/Codominance

Incomplete
Dominance
and
Codominance
Incomplete Dominance
F1
hybrids have an appearance
somewhat in between the phenotypes
of the two parental varieties.
Example: snapdragons (flower)
r
red (RR) x white (rr)
RR = red flower
rr = white flower
R
R
r
Incomplete Dominance
r
r
R Rr
Rr
R Rr
Rr
produces the
F1 generation
All Rr = pink
(heterozygous pink)
Incomplete Dominance
Codominance
Heterozygote
expresses both
alleles’ conditions
Ex.
A black rooster bred with a white
hen produces a black and white
checkered chicken.
Usually
uses both capital letters
(black= B, white=W , checkered
= BW)
Codominance practice: cross a
black rooster with a white hen
Key:
W
W
B = black
BW BW
W = white
BW = checkered
Results:
100% checkered
(BW)
B
B
BW BW
Now let’s cross a checkered hen
with a checkered rooster.
B
W
B
BB
BW
W
BW
WW
Results:
25% Black
50% Checkered
25% White
(1:2:1)
Codominance
Two
alleles are expressed (multiple
alleles) in heterozygous individuals.
Example: blood type
1.
2.
3.
4.
type
type
type
type
A
B
AB
O
=
=
=
=
IAIA or IAi
IBIB or IBi
IAIB
ii
Codominance Problem
Example:homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IA
i
IB
IAIB
IBi
IB
IAIB
IBi
1/2 = IAIB
1/2 = IBi
Another Codominance Problem
Example: male Type O (ii)
x
female type AB (IAIB)
IA
IB
i
IAi
IBi
i
IAi
IBi
1/2 = IAi
1/2 = IBi
Codominance
Question:
If a boy has a blood type O and
his sister has blood type AB,
what are the genotypes and
phenotypes of their parents?
boy - type O (ii) X
girl - type AB (IAIB)
Codominance
Answer:
IA
IB
i
i
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
Polygenic Inheritance
Trait
controlled by 2 or more
genes
May be on the same or different
chromosomes
Shows a range, intermediate is
most common phenotype
Upper and lower case letters
used
Many variations in skin color
Skin color: A polygenic trait
Determining # of Genes Involved in Skin Color
Expected
distribution- 4
genes
Number of individuals
Observed
distribution
of skin color
Expected
distribution1 gene
Light
Expected
distribution- 3
genes
Right
Range of skin color
Skin Color
Skin
color is actually due to 5 genes
Genotypes darkest to lightest:
AABBCCDDEE
would be darkest
skinned
AaBbCcDdEe would be medium
skinned
aabbccddee would be lightest skinned
Influence of external
environment
In
arctic foxes
temperature has
an effect on the
expression of
coat color. In
winter, fur is
white.
Arctic Fox in summer coloration:
influence of external
environment
Influence of external
environment
External influences can also be seen in leaves.
Leaves can have different sizes, thicknesses, and
shapes depending on the amount of light they
receive.
Influence of internal
environment
The internal
environments of males
and females are
different because of
hormones and
structural differences.
An organism’s age
can also affect gene
function.
Peacock (male) Peahen (female)
Blood type quick facts
Red
blood cells are called
erythrocytes
Proteins on their surfaces are
called antigens, controlled by
genes
Antigens make antibodies to
foreign substances, which
includes RBCs with different
antigens on their surface
Phenotype A
Surface molecule A
• The lA allele is
dominant to i, so
inheriting either the
lAi alleles or the lA lA
alleles from both
parents will give
you type A blood.
• Surface molecule
A is produced.
Phenotype B
• The lB allele is also
dominant to i.
• To have type B
blood, you must
inherit the lB allele
from one parent and
either another lB
allele or the i allele
from the other.
• Surface molecule B is
produced.
Surface molecule B
Phenotype AB
• The lA and lB alleles
are codominant.
• If you inherit the lA
allele from one parent
and the lB allele from
the other, your red
blood cells will produce
both surface molecules
and you will have type
AB blood.
Surface molecule B
Surface molecule A
Phenotype O
•No antigens produced
Check your chart!
Blood
Group
Antigens
Antibodies
Can
receive
from
Can give to
A
A
B
A or O
A or AB
B
B
A
B or O
B or AB
AB
A and B
None
AB
O
None
both
A, B, AB,
O
O
A, B, AB,
O
Possible genotypes for each
phenotype:
A
A
A
I I
=
or
B = IBIB or IBi
A
B
AB =I I
O
= ii
A
I i
Sex-linked Traits
Traits
(genes) located on the sex
chromosomes
Sex chromosomes are X and Y
XX genotype for females
XY genotype for males
Many sex-linked traits carried on
X chromosome
Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
Sex-linked Trait Problem
Example:
Eye color in fruit flies
(red-eyed male) x (white-eyed female)
XRY
x
X rX r
Remember: the Y chromosome in males
does not carry traits.
Xr
Xr
RR = red eyed
Rr = red eyed
rr = white eyed
XY = male
XX = female
XR
Y
Sex-linked Trait Solution:
Xr
XR
XR
Xr
Y
Xr Y
Xr
XR
Xr
Xr Y
50% red eyed
female
50% white eyed
male
Crosses
with sex-linked traits;
colorblindness and hemophilia
Female Carriers
Females
are carriers of sex-linked traits
if they have the heterozygous
genotype.
 Males cannot be carriers because
they only have one X chromosome.
Female parents who are carriers, pass
sex-linked traits to all children, but
males are usually the ones who express
the trait.
Pedigrees
Tay Sachs disease
Occurs
in people of Jewish
descent
Enzyme that breaks down lipids in
the brain is defective. Lipid
buildup kills brain cells.
Always results in death, usually by
age 5
Caused by a recessive allele
Pedigree for Tay Sachs
disease
carrier
carrier
Huntington’s Disease
Caused
by a rare dominant
allele
Doesn’t show up till age 30-50
Breaks down areas of the brain,
loss of control of all body
functions
No treatment
Pedigree for Huntington’s
Disease
normal
carrier
What if the trait is sex-linked?
What gives you a clue?
Is this sex-linked or not?
karyotypes
Autosomal
inheritance patterns;
sickle cell, cystic fibrosis,
huntingtons
Genetic Practice
Problems
Breed the P1 generation
tall
(TT) x dwarf (tt) pea plants
t
T
T
t
Solution:
tall (TT) vs. dwarf (tt) pea plants
t
t
T
Tt
Tt
produces the
F1 generation
T
Tt
Tt
All Tt = tall
(heterozygous tall)
Breed the F1 generation
tall
(Tt) vs. tall (Tt) pea plants
T
T
t
t
Solution:
tall (Tt) x tall (Tt) pea plants
T
t
T
TT
Tt
t
Tt
tt
produces the
F2 generation
1/4
1/2
1/4
1:2:1
3:1
(25%) = TT
(50%) = Tt
(25%) = tt
genotype
phenotype