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B. INCOMPLETE DOMINANCE
INCOMPLETE DOMINANCE TRAITS
• OF THE GENES THAT CONTROL THE TRAIT, NEITHER ALLELE IS DOMINANT
OR RECESSIVE OVER THE OTHER.
• IN THE HETEROZYGOUS GENOTYPE, A MIXING OR BLENDING OF THE
ALLELES OCCURS.
• THIS CREATES A THIRD PHENOTYPE.
EXAMPLE
IN SNAPDRAGONS, A TYPE OF FLOWER, TWO ALLELES CONTROL THE
TRAIT OF FLOWER COLOR.
• ONE ALLELE (R) PRODUCES RED COLORED FLOWERS.
• ANOTHER ALLELES (W) PRODUCES WHITE FLOWERS.
• IN THE HETEROZYGOUS GENOTYPE, THE FLOWERS ARE PINK IN COLOR.
Possible Genotypes
RR
RW
WW
Possible Phenotypes
Red
Pink
White
PRACTICE PROBLEM
CROSS A RED FLOWERED SNAPDRAGON WITH A PINK FLOWERED
SNAPDRAGON.
1.
WHAT ARE THE PARENTS’ GENOTYPES?
2.
SHOW THE PUNNETT SQUARE FOR THIS CROSS.
3.
SHOW THE PROBABILITY FOR ALL GENOTYPES AND PHENOTYPES.
1.
PARENTS’ GENOTYPES = RR X RW
2.
PUNNETT SQUARE
R
R
W
R
RR
red
RW
pink
RR
red
RW
pink
3.
GENOTYPE PROBABILITY = RR 50 %; RW 50 %
4.
PHENOTYPE PROBABILITY = RED 50%; PINK 50%
MORE PRACTICE PROBLEMS (SPONGE BOB)
C. CO-DOMINANCE
OF THE GENES THAT CONTROL THE TRAIT, NEITHER ALLELE IS DOMINANT
OR RECESSIVE TO THE OTHER.
• IN THE HETEROZYGOUS GENOTYPE, BOTH OF THE ALLELES ARE
EXPRESSED AT THE SAME TIME. (CO-DOMINANT)
• THIS CREATES A THIRD PHENOTYPE
EXAMPLES: HUMAN ABO BLOOD TYPES
• IN HUMANS, 3 DIFFERENT ALLELES FOR BLOOD TYPE ARE POSSIBLE.
• A ALLELE = PRODUCES PROTEINS CALLED ANTIGEN A ON THE RED BLOOD
CELL
• B ALLELE = PRODUCES ANTIGEN B ON THE RBC’S.
• O ALLELE = PRODUCES NO ANTIGENS ON THE RBC’S
• A AND B ANTIGENS ARE CO-DOMINANT
• O IS RECESSIVE
• WHEN A AND B ALLELES ARE PRESENT = RBC’S HAVE BOTH A AND B
ANTIGENS
HUMAN ABO BLOOD TYPES
Possible genotypes
for blood type:
Possible blood
types:
Can receive blood
from:
Can donate blood
to:
AA
AO
Blood type A
AA, AO, OO
AA, AO, AB
BB
BO
Blood type B
BB, BO, OO
BB, BO, AB
Blood type AB
AA, AB, AO, BO, BB
OO
(Universal receiver)
AB
OO
AA, AO, AB, BB, BO,
OO
(universal donor)
AB
(least common)
OO
(most common)
Blood type O
PRACTICE BLOOD TYPE PROBLEMS
1.
MARY HAS BLOOD TYPE AB AND HER HUSBAND HAS BLOOD TYPE
O.
A.
WHAT ARE MARY'S AND HER HUSBAND’S GENOTYPES?
AB X OO
B.
SHOW THE PUNNETT SQUARE FOR THIS CROSS.
A
B
AO
BO
AO
BO
O
O
C. SHOW THE PROBABILITY FOR ALL GENOTYPES AND PHENOTYPES.
Genotypes: AO 50%; BO 50%
Phenotypes: Blood type A 50 %; Blood type B 50%
D. WHAT IS THE CHANCE THAT MARY WILL HAVE A BABY WITH THE SAME
BLOOD TYPE AS HER?
None 0%
2.
CROSS AN INDIVIDUAL WHO IS HETEROZYGOUS FOR BLOOD TYPE
A WITH AN INDIVIDUAL WHO IS HETEROZYGOUS FOR BLOOD TYPE
B.
A.
WHAT ARE THE PARENTS’ GENOTYPES?
B.
AO X BO
SHOW THE PUNNETT SQUARE FOR THIS CROSS.
A
B
O
O
AB
BO
AO
OO
C. SHOW THE PROBABILITY FOR ALL THE GENOTYPES AND
PHENOTYPES.
Genotypes: AB 25%, AO 25%, BO 25%, OO 25%
Phenotypes: Type AB 25%, Type A 25%, type B 25%, type O 25%
D. WHAT IS THE CHANCE THAT THESE 2 INDIVIDUALS WILL PRODUCE A
CHILD WITH A BLOOD TYPE THAT IS NOT THE SAME AS EITHER OF
THEIR BLOOD TYPES?
50 % (Type A and Type B are NOT the same as the parents’ blood types)
E.
MORE PRACTICE PROBLEMS ON WORKSHEET.
D. SEX-LINKED TRAITS (X-LINKED)
1.
MOST TRAITS ARE CARRIED ON CHROMOSOME PAIRS 1-22
(AUTOSOMES).
2.
A FEW TRAITS ARE CARRIED ON THE X SEX CHROMOSOME (BUT
NOT ON THE Y).
3.
THE RECESSIVE ALLELE FOR THESE TRAITS CAUSES A DISORDER.
4.
SYMBOLS FOR SEX-LINKED TRAITS:
XN = an X carrying the normal dominant allele (no disorder)
Xn = an X carrying the recessive allele (could have a disorder)
5.
SEX-LINKED TRAITS
Possible Genotypes
6.
Possible Phenotypes
XN XN
XN Xn
XnXn
Female with no disorder
Female with no disorder (carrier)
Female with the disorder
XN Y
XnY
Male with no disorder
Male with the disorder
DISORDERS CAUSED BY RECESSIVE SEX-LINKED GENES ON THE X
CHROMOSOME:
• HEMOPHILIA = BLOOD DOES NOT CLOT PROPERLY
• DUCHENNE’S MUSCULAR DYSTROPHY = SLOW DETERIORATION OF MUSCLE
TISSUE
• COLOR BLINDNESS = INABILITY TO SEE CERTAIN COLORS PROPERLY
PRACTICE PROBLEMS
7.
CROSS A WOMAN WHO IS HETEROZYGOUS FOR COLOR
BLINDNESS (BUT DOES NOT HAVE COLOR BLINDNESS WITH A MAN
WHO IS NOT COLOR BLIND.
A.
WHAT ARE THE PARENTS’ GENOTYPES?
B.
XN Xn X X N Y
SHOW THE PUNNETT SQUARE FOR THE CROSS.
XN
Xn
XN
Y
XN XN
XN Y
XN Xn
XnY
C. SHOW THE PROBABILITY FOR ALL GENOTYPES AND PHENOTYPES.
Genotypes: XNXN 25%; XNXn 25%; XNY 25%;
XnY 25%
Phenotypes: female no disorder 50%; male no disorder 25%;
male with color blindness 25%
D. WHAT IS THE CHANCE THESE PARENTS WILL HAVE A COLOR BLIND
SON?
25%
8.
CROSS A COLOR BLIND WOMAN WITH A MAN WHO IS ALSO
COLOR BLIND.
A.
WHAT ARE THE PARENTS’ GENOTYPES?
X nXn X XnY
B.
SHOW THE PUNNETT SQUARE FOR THIS CROSS.
Xn
Xn
Xn
Y
XnXn
X nX n
XnY
Xn Y
C.
SHOW THE PROBABILITY FOR ALL GENOTYPES AND PHENOTYPES.
Genotypes: XnXn 50 %; XnY 50%
Phenotypes: daughter with color blindness 50%; son with color blindness 50%
D.
WHAT IS THE CHANCE THESE PARENTS WILL HAVE A COLOR BLIND
DAUGHTER?
50%
E.
WHAT IS THE CHANCE THESE PARENTS WILL PRODUCE A COLOR BLIND
CHILD OF EITHER SEX?
100%
MORE PRACTICE PROBLEMS