MODIFICATIONS TO
MENDEL’S CLASSIC
RATIOS
LEARNING OBJECTIVES
1. DISTINGUISH MENDELIAN FROM NON-MENDELIAN MODES OF
INHERITANCE
2. DESCRIBE SOME CASES OF NON-MENDELIAN GENETIC TRAITS
RECALL: MENDEL’S CLASSIC RATIO
COMMON OUTCOME FROM
SUCH CROSSES.
• HAVE 3 GENOTYPES
(TT, TT, & TT)
• IN A 1:2:1 GENOTYPIC
RATIO
• HAVE 2 DIFFERENT
PHENOTYPES (TALL &
SHORT) IN A 3:1
PHENOTYPIC RATIO
INCOMPLETE DOMINANCE
CO-DOMINANCE
MULTIPLE ALLELES
• SNAPDRAGONS
http://faculty.pnc.edu/pwilkin/incompdominance.jpg
http://www.dobermann-review.com/info/genetics/mendels_genetic_laws/Gregor%20Mendel.jpg
INCOMPLETE DOMINANCE
• NEITHER ALLELE IS COMPLETELY DOMINANT OVER THE OTHER
ALLELE.
• A HETEROZYGOUS PHENOTYPE
 A MIXTURE OR BLENDING OF THE TWO
FOUR-O’ CLOCK
FLOWERS
• INCOMPLETE DOMINANCE
• NEITHER RED (R) OR WHITE (W) IS DOMINANT
WHEN A HOMOZYGOUS RED FLOWER
(RR)
MIX WITH A HOMOZYGOUS WHITE
FLOWER
(WW), THE ALLELES BLEND IN THE
HYBRID
(RW) TO PRODUCE PINK FLOWERS
ANDALUSIAN
CHICKENS
• INCOMPLETE DOMINANCE
• NEITHER BLACK (B) OR
WHITE (W) ARE DOMINANT
THE OFFSPRING OF A BLACK
FEATHERED
CHICKEN (BB) AND A WHITE
FEATHERED
CHICKEN (WW) ARE BLUE (BW) –
BLUE HEN!!
The heterozygous genotype produces a
phenotype that falls in between the dominant
trait and the recessive trait.
Incomplete Dominance
Example 1:
In flowers, petal color demonstrates incomplete dominance. Red
results when a flower has homozygous dominant alleles for the
trait. White results when a flower has homozygous recessive
alleles for the trait. A flower that is heterozygous for this trait will
be pink. Predict the genotypic and phenotypic ratios of the
offspring for the following:
Rr Rr
Rr Rr
Genotypic Ratio:
100% Rr
Phenotypic Ratio:
100% pink flowers
1. Red and white RR x rr
R R
r
r
Example 2:
In flowers, petal color demonstrates incomplete dominance. Red
results when a flower has homozygous dominant alleles for the
trait. White results when a flower has homozygous recessive
alleles for the trait. A flower that is heterozygous for this trait will
be pink. Predict the genotypic and phenotypic ratios of the
offspring for the following:
RR RR
Rr Rr
Genotypic Ratio:
50% RR 50% Rr
Phenotypic Ratio:
50% Red flowers, 50%
Pink flowers
2. Red and pink RR x Rr
R R
R
r
CO-DOMINANCE
• TWO EQUALLY DOMINANT ALLELES ARE EXPRESSED AT
THE SAME TIME.
• HETEROZYGOUS PHENOTYPE WILL HAVE BOTH
PHENOTYPES VISIBLE
SHORTHORN
CATTLE
• CO- DOMINANCE
• HOMOZYGOUS RED (RR)
• HOMOZYGOUS WHITE (WW)
THE OFFSPRING OF WILL HAVE
BOTH RED AND WHITE HAIRS (RW)
THE OFFSPRING ARE
HETEROZYGOUS AND CALLED
“ROAN”
http://search.vadlo.com/b/q?rel=2&keys=Dominance+Incomplete+Dominance+Codominance+PPT
Roan Horse
SICKLE- CELL
ANEMIA
• CO- DOMINANCE
• CAUSED BY AN ABNORMAL
HEMOGLOBIN, THE PROTEIN THAT
RED BLOOD CELLS USE TO CARRY
OXYGEN
NORMAL HEMOGLOBIN IS (RR)
SICKLE CELL SHAPED BLOOD CELLS (SS)
PEOPLE WHO ARE CARRIERS
(HETEROZYGOUS)
FOR THE DISEASE THERE IS A MIXTURE
OF BOTH
Co-dominance
Condition in which both alleles of a gene are
expressed in heterozygous offspring.
Example 1:
In cows black coat color and white coat color are co-
dominant. Heterozygous offspring will be spotted.
1. Black cow and White cow BB x WW
BW BW
BW BW
B B
W
W
Genotypic Ratio:
100% BW
Phenotypic Ratio:
100% Spotted cow
Example 2:
In cows black coat color and white coat color are co-
dominant. Heterozygous offspring will be spotted.
2. Black cow and Spotted cow BB x BW
BB BB
BW BW
B B
B
W
Genotypic Ratio:
50% BB
50% BW
Phenotypic Ratio:
50% Black cow
50% Spotted cow
Example 3:
In cows black coat color and white coat color are co-
dominant. Heterozygous offspring will be spotted.
3. Both Spotted cows BW x BW
BB BW
BW WW
B W
B
W
Genotypic Ratio:
25 % BB
50% BW
25% WW
Phenotypic Ratio:
25% Black cow
50% Spotted cow
25% White cow
PROBLEM: CODOMINANCE
• SHOW THE CROSS BETWEEN AN INDIVIDUAL WITH SICKLE-CELL
ANEMIA AND ANOTHER WHO IS A CARRIER BUT NOT SICK.
N S
S
S
NS
NS
SS
SS
- NS (2) SS (2)
- ratio 1:1
- carrier (2); sick (2)
- ratio 1:1
GENOTYPES:
PHENOTYPES:
• WHEN THERE IS MORE THAN 2 ALLELES POSSIBLE FOR A
GIVEN GENE.
• ALLOWS FOR A LARGER NUMBER OF GENETIC AND
PHENOTYPIC POSSIBILITIES.
Multiple Alleles
Example 4:
In blood typing, the gene for type A and the gene for
type B are co-dominant. The gene for type O is
recessive.
Blood Type (phenotype) Genotype
A AA or AO
B BB or BO
AB AB
O OO
MULTIPLE ALLELES • HUMAN BLOOD TYPE
IS AN EXAMPLE OF
BOTH CO-
DOMINANCE AND A
TRAIT WITH
MULTIPLE ALLELES.
• AB = UNIVERSAL
ACCEPTOR
• O = UNIVERSAL
DONOR
BLOOD
TYPE
GENOTYPE CAN RECIVE
BLOOD FROM
A IAIA , IAi A, O
B IBIB , IBi B, O
AB IAIB A, B, AB, O
O ii O
http://www.pennmedicine.org/health_info/images/19450.jpg
Determine the possible blood types of the offspring
when:
1. Mother is type O, Father is type A (homozygous)
AO AO
AO AO
O O
A
A
Genotypic Ratio:
100% AO
Phenotypic Ratio:
100% Type A
Blood
Determine the possible blood types of the
offspring when:
AA AB
AO BO
A B
A
O
Genotypic Ratio:
25 % AA 25% AO
25% AB 25% BO
Phenotypic Ratio:
50% Type A Blood
25% Type AB Blood
25% Type B Blood
2. Mother is type AB, Father is Type A (heterozygous)
Determine the possible blood types of the
offspring when:
AB BO
AO OO
A O
B
O
Genotypic Ratio:
25 % OO 25% AO
25% AB 25% BO
Phenotypic Ratio:
25%Type O Blood
25% Type A Blood
25% Type AB Blood
25% Type B Blood
3. Mother is type A (heterozygous), Father is type B
(heterozygous)
RULES FOR BLOOD TYPE
• A AND B ARE CO-DOMINANT
• AA = TYPE A
• BB = TYPE B
• AB = TYPE AB
• A AND B ARE DOMINANT OVER O
• AO = TYPE A
• BO = TYPE B
• OO = TYPE O
http://duongchan.files.wordpress.com/2007/05/abobloodsystem.jpg
http://image.wistatutor.com/content/feed/tvcs/blood_type5B15D.jpg
http://www.biologycorner.com/anatomy/blood/images/bloodtypes.jpg
RHESUS FACTOR (RH) FACTOR
• THE RHESUS FACTOR IS AN ANTIGEN, OR MORE SPECIFICALLY A
PROTEIN, THAT EXISTS ON THE SURFACE OF RED BLOOD CELLS.
• IF A PERSON HAS EITHER TWO (+) GENES FOR RH OR ONE (+)
AND ONE (-) RH GENE, THEY WILL TEST RH(+). A PERSON WILL
BE NEGATIVE ONLY IF THEY HAVE 2 (-).
RHESUS FACTOR
• THE RHESUS FACTOR GETS
ITS NAME FROM
EXPERIMENTS CONDUCTED
IN 1937 BY SCIENTISTS KARL
LANDSTEINER AND
ALEXANDER S. WEINER.
• INVOLVED RABBITS WHICH
WHEN INJECTED WITH THE
RHESUS MONKEY’S RED
BLOOD CELLS, PRODUCED
AN ANTIGEN PRESENT IN
THE RED BLOOD CELLS OF
MANY HUMANS http://content9.clipmarks.com/blog_cache/latimesblogs.latimes.com/img/E0C83714-56E5-4757-B9F1-604FCE5643C3
• IT IS VERY IMPORTANT IN TERMS OF
BABIES: E.G. AN RH(-)
MOTHER MAY MAKE ANTIBODY
AGAINST AN RH(+) FETUS IF THE BABY
GETS A (+) GENE FROM ITS FATHER
• THE ABO AND RH GENES ARE ONLY
TWO OF MANY BLOOD ANTIGENS
THAT ARE PRESENT ON HUMAN RED
CELLS AND MUST BE MATCHED UP
FOR SUCCESSFUL BLOOD
TRANSFUSIONS.
RELEVANCE OF RH FACTOR & ABO
TYPING?
MULTIFACTORIAL
• REFERRING TO CONTROL OF THE EXPRESSION OF A TRAIT BY
SEVERAL GENES AND ENVIRONMENTAL FACTORS.
• MANY MULTIFACTORIAL TRAITS SHOW CONTINUOUS
DISTRIBUTION.
MULTIFACTORIAL
• E.G. HUMAN HEIGHT
• USUALLY BETWEEN
120CM AND 200CM,
HOWEVER THESE GENES
CANNOT BE FULLY
EXPRESSED WITHOUT
ALL THE NECESSARY
NUTRIENTS FROM A
HEALTHY DIET http://www.babble.com/CS/blogs/droolicious/menace-height.jpg
PRACTICE QUESTIONS
1. IN A CERTAIN CASE A WOMAN’S BLOOD TYPE WAS TESTED
TO BE AB. SHE MARRIED AND HER HUSBANDS BLOOD TYPE
WAS TYPE A. THEIR CHILDREN HAVE BLOOD TYPES A, AB,
AND B.
WHAT ARE THE GENOTYPES OF THE PARENTS? WHAT ARE
THE GENOTYPIC RATIOS OF THE CHILDREN?
2. IN A CERTAIN BREED OF COW THE GENE FOR RED FUR, R, IS
CO-DOMINANT WITH THAT OF WHITE FUR, W. WHAT WOULD
BE THE PHENOTYPIC & GENOTYPIC RATIOS OF THE OFFSPRING
IF YOU BREED A RED COW AND A WHITE BULL? WHAT WOULD
THEY BE IF YOU BREED A RED & WHITE COW WITH A RED &
WHITE BULL?
3. A ROOSTER WITH GREY FEATHERS IS MATED WITH A HEN OF
THE SAME PHENOTYPE. AMONG THEIR OFFSPRING 15 CHICKS
ARE GREY, 6 ARE BLACK AND 8 ARE WHITE.
A. WHAT IS THE SIMPLEST EXPLANATION FOR THE
INHERITANCE OF THESE COLORS IN CHICKENS?
B. WHAT OFFSPRING WOULD YOU EXPECT FROM THE MATING
OF A GREY ROOSTER AND A BLACK HEN?
4. PROBLEM: MULTIPLE ALLELES
• SHOW THE CROSS BETWEEN A MOTHER WHO HAS
TYPE O BLOOD AND A FATHER WHO HAS TYPE AB
BLOOD.
- AO (2) BO (2)
- ratio 1:1
- type A (2); type B (2)
- ratio 1:1
GENOTYPES:
PHENOTYPES:
O O
A
B
AO
BO
AO
BO
PROBLEM: MULTIPLE ALLELES
• SHOW THE CROSS BETWEEN A MOTHER WHO
IS HETEROZYGOUS FOR TYPE B BLOOD AND A
FATHER WHO IS HETEROZYGOUS FOR TYPE A
BLOOD.
-AB (1); BO (1);
AO (1); OO (1)
- ratio 1:1:1:1
-type AB (1); type B (1)
type A (1); type O (1)
- ratio 1:1:1:1
GENOTYPES:
PHENOTYPES:
A O
B
O
AB
OO
BO
AO

Lesson4_Modifications to Mendel’s Classic Ratios

  • 1.
  • 2.
    LEARNING OBJECTIVES 1. DISTINGUISHMENDELIAN FROM NON-MENDELIAN MODES OF INHERITANCE 2. DESCRIBE SOME CASES OF NON-MENDELIAN GENETIC TRAITS
  • 3.
    RECALL: MENDEL’S CLASSICRATIO COMMON OUTCOME FROM SUCH CROSSES. • HAVE 3 GENOTYPES (TT, TT, & TT) • IN A 1:2:1 GENOTYPIC RATIO • HAVE 2 DIFFERENT PHENOTYPES (TALL & SHORT) IN A 3:1 PHENOTYPIC RATIO
  • 4.
  • 5.
  • 6.
    INCOMPLETE DOMINANCE • NEITHERALLELE IS COMPLETELY DOMINANT OVER THE OTHER ALLELE. • A HETEROZYGOUS PHENOTYPE  A MIXTURE OR BLENDING OF THE TWO
  • 7.
    FOUR-O’ CLOCK FLOWERS • INCOMPLETEDOMINANCE • NEITHER RED (R) OR WHITE (W) IS DOMINANT WHEN A HOMOZYGOUS RED FLOWER (RR) MIX WITH A HOMOZYGOUS WHITE FLOWER (WW), THE ALLELES BLEND IN THE HYBRID (RW) TO PRODUCE PINK FLOWERS
  • 8.
    ANDALUSIAN CHICKENS • INCOMPLETE DOMINANCE •NEITHER BLACK (B) OR WHITE (W) ARE DOMINANT THE OFFSPRING OF A BLACK FEATHERED CHICKEN (BB) AND A WHITE FEATHERED CHICKEN (WW) ARE BLUE (BW) – BLUE HEN!!
  • 9.
    The heterozygous genotypeproduces a phenotype that falls in between the dominant trait and the recessive trait. Incomplete Dominance
  • 10.
    Example 1: In flowers,petal color demonstrates incomplete dominance. Red results when a flower has homozygous dominant alleles for the trait. White results when a flower has homozygous recessive alleles for the trait. A flower that is heterozygous for this trait will be pink. Predict the genotypic and phenotypic ratios of the offspring for the following: Rr Rr Rr Rr Genotypic Ratio: 100% Rr Phenotypic Ratio: 100% pink flowers 1. Red and white RR x rr R R r r
  • 11.
    Example 2: In flowers,petal color demonstrates incomplete dominance. Red results when a flower has homozygous dominant alleles for the trait. White results when a flower has homozygous recessive alleles for the trait. A flower that is heterozygous for this trait will be pink. Predict the genotypic and phenotypic ratios of the offspring for the following: RR RR Rr Rr Genotypic Ratio: 50% RR 50% Rr Phenotypic Ratio: 50% Red flowers, 50% Pink flowers 2. Red and pink RR x Rr R R R r
  • 12.
    CO-DOMINANCE • TWO EQUALLYDOMINANT ALLELES ARE EXPRESSED AT THE SAME TIME. • HETEROZYGOUS PHENOTYPE WILL HAVE BOTH PHENOTYPES VISIBLE
  • 13.
    SHORTHORN CATTLE • CO- DOMINANCE •HOMOZYGOUS RED (RR) • HOMOZYGOUS WHITE (WW) THE OFFSPRING OF WILL HAVE BOTH RED AND WHITE HAIRS (RW) THE OFFSPRING ARE HETEROZYGOUS AND CALLED “ROAN”
  • 14.
  • 15.
    SICKLE- CELL ANEMIA • CO-DOMINANCE • CAUSED BY AN ABNORMAL HEMOGLOBIN, THE PROTEIN THAT RED BLOOD CELLS USE TO CARRY OXYGEN NORMAL HEMOGLOBIN IS (RR) SICKLE CELL SHAPED BLOOD CELLS (SS) PEOPLE WHO ARE CARRIERS (HETEROZYGOUS) FOR THE DISEASE THERE IS A MIXTURE OF BOTH
  • 16.
    Co-dominance Condition in whichboth alleles of a gene are expressed in heterozygous offspring.
  • 17.
    Example 1: In cowsblack coat color and white coat color are co- dominant. Heterozygous offspring will be spotted. 1. Black cow and White cow BB x WW BW BW BW BW B B W W Genotypic Ratio: 100% BW Phenotypic Ratio: 100% Spotted cow
  • 18.
    Example 2: In cowsblack coat color and white coat color are co- dominant. Heterozygous offspring will be spotted. 2. Black cow and Spotted cow BB x BW BB BB BW BW B B B W Genotypic Ratio: 50% BB 50% BW Phenotypic Ratio: 50% Black cow 50% Spotted cow
  • 19.
    Example 3: In cowsblack coat color and white coat color are co- dominant. Heterozygous offspring will be spotted. 3. Both Spotted cows BW x BW BB BW BW WW B W B W Genotypic Ratio: 25 % BB 50% BW 25% WW Phenotypic Ratio: 25% Black cow 50% Spotted cow 25% White cow
  • 20.
    PROBLEM: CODOMINANCE • SHOWTHE CROSS BETWEEN AN INDIVIDUAL WITH SICKLE-CELL ANEMIA AND ANOTHER WHO IS A CARRIER BUT NOT SICK. N S S S NS NS SS SS - NS (2) SS (2) - ratio 1:1 - carrier (2); sick (2) - ratio 1:1 GENOTYPES: PHENOTYPES:
  • 21.
    • WHEN THEREIS MORE THAN 2 ALLELES POSSIBLE FOR A GIVEN GENE. • ALLOWS FOR A LARGER NUMBER OF GENETIC AND PHENOTYPIC POSSIBILITIES. Multiple Alleles
  • 22.
    Example 4: In bloodtyping, the gene for type A and the gene for type B are co-dominant. The gene for type O is recessive. Blood Type (phenotype) Genotype A AA or AO B BB or BO AB AB O OO
  • 23.
    MULTIPLE ALLELES •HUMAN BLOOD TYPE IS AN EXAMPLE OF BOTH CO- DOMINANCE AND A TRAIT WITH MULTIPLE ALLELES. • AB = UNIVERSAL ACCEPTOR • O = UNIVERSAL DONOR BLOOD TYPE GENOTYPE CAN RECIVE BLOOD FROM A IAIA , IAi A, O B IBIB , IBi B, O AB IAIB A, B, AB, O O ii O
  • 24.
  • 25.
    Determine the possibleblood types of the offspring when: 1. Mother is type O, Father is type A (homozygous) AO AO AO AO O O A A Genotypic Ratio: 100% AO Phenotypic Ratio: 100% Type A Blood
  • 26.
    Determine the possibleblood types of the offspring when: AA AB AO BO A B A O Genotypic Ratio: 25 % AA 25% AO 25% AB 25% BO Phenotypic Ratio: 50% Type A Blood 25% Type AB Blood 25% Type B Blood 2. Mother is type AB, Father is Type A (heterozygous)
  • 27.
    Determine the possibleblood types of the offspring when: AB BO AO OO A O B O Genotypic Ratio: 25 % OO 25% AO 25% AB 25% BO Phenotypic Ratio: 25%Type O Blood 25% Type A Blood 25% Type AB Blood 25% Type B Blood 3. Mother is type A (heterozygous), Father is type B (heterozygous)
  • 28.
    RULES FOR BLOODTYPE • A AND B ARE CO-DOMINANT • AA = TYPE A • BB = TYPE B • AB = TYPE AB • A AND B ARE DOMINANT OVER O • AO = TYPE A • BO = TYPE B • OO = TYPE O
  • 29.
  • 30.
  • 31.
  • 32.
    RHESUS FACTOR (RH)FACTOR • THE RHESUS FACTOR IS AN ANTIGEN, OR MORE SPECIFICALLY A PROTEIN, THAT EXISTS ON THE SURFACE OF RED BLOOD CELLS. • IF A PERSON HAS EITHER TWO (+) GENES FOR RH OR ONE (+) AND ONE (-) RH GENE, THEY WILL TEST RH(+). A PERSON WILL BE NEGATIVE ONLY IF THEY HAVE 2 (-).
  • 33.
    RHESUS FACTOR • THERHESUS FACTOR GETS ITS NAME FROM EXPERIMENTS CONDUCTED IN 1937 BY SCIENTISTS KARL LANDSTEINER AND ALEXANDER S. WEINER. • INVOLVED RABBITS WHICH WHEN INJECTED WITH THE RHESUS MONKEY’S RED BLOOD CELLS, PRODUCED AN ANTIGEN PRESENT IN THE RED BLOOD CELLS OF MANY HUMANS http://content9.clipmarks.com/blog_cache/latimesblogs.latimes.com/img/E0C83714-56E5-4757-B9F1-604FCE5643C3
  • 34.
    • IT ISVERY IMPORTANT IN TERMS OF BABIES: E.G. AN RH(-) MOTHER MAY MAKE ANTIBODY AGAINST AN RH(+) FETUS IF THE BABY GETS A (+) GENE FROM ITS FATHER • THE ABO AND RH GENES ARE ONLY TWO OF MANY BLOOD ANTIGENS THAT ARE PRESENT ON HUMAN RED CELLS AND MUST BE MATCHED UP FOR SUCCESSFUL BLOOD TRANSFUSIONS. RELEVANCE OF RH FACTOR & ABO TYPING?
  • 35.
    MULTIFACTORIAL • REFERRING TOCONTROL OF THE EXPRESSION OF A TRAIT BY SEVERAL GENES AND ENVIRONMENTAL FACTORS. • MANY MULTIFACTORIAL TRAITS SHOW CONTINUOUS DISTRIBUTION.
  • 36.
    MULTIFACTORIAL • E.G. HUMANHEIGHT • USUALLY BETWEEN 120CM AND 200CM, HOWEVER THESE GENES CANNOT BE FULLY EXPRESSED WITHOUT ALL THE NECESSARY NUTRIENTS FROM A HEALTHY DIET http://www.babble.com/CS/blogs/droolicious/menace-height.jpg
  • 37.
    PRACTICE QUESTIONS 1. INA CERTAIN CASE A WOMAN’S BLOOD TYPE WAS TESTED TO BE AB. SHE MARRIED AND HER HUSBANDS BLOOD TYPE WAS TYPE A. THEIR CHILDREN HAVE BLOOD TYPES A, AB, AND B. WHAT ARE THE GENOTYPES OF THE PARENTS? WHAT ARE THE GENOTYPIC RATIOS OF THE CHILDREN?
  • 38.
    2. IN ACERTAIN BREED OF COW THE GENE FOR RED FUR, R, IS CO-DOMINANT WITH THAT OF WHITE FUR, W. WHAT WOULD BE THE PHENOTYPIC & GENOTYPIC RATIOS OF THE OFFSPRING IF YOU BREED A RED COW AND A WHITE BULL? WHAT WOULD THEY BE IF YOU BREED A RED & WHITE COW WITH A RED & WHITE BULL?
  • 39.
    3. A ROOSTERWITH GREY FEATHERS IS MATED WITH A HEN OF THE SAME PHENOTYPE. AMONG THEIR OFFSPRING 15 CHICKS ARE GREY, 6 ARE BLACK AND 8 ARE WHITE. A. WHAT IS THE SIMPLEST EXPLANATION FOR THE INHERITANCE OF THESE COLORS IN CHICKENS? B. WHAT OFFSPRING WOULD YOU EXPECT FROM THE MATING OF A GREY ROOSTER AND A BLACK HEN?
  • 40.
    4. PROBLEM: MULTIPLEALLELES • SHOW THE CROSS BETWEEN A MOTHER WHO HAS TYPE O BLOOD AND A FATHER WHO HAS TYPE AB BLOOD. - AO (2) BO (2) - ratio 1:1 - type A (2); type B (2) - ratio 1:1 GENOTYPES: PHENOTYPES: O O A B AO BO AO BO
  • 41.
    PROBLEM: MULTIPLE ALLELES •SHOW THE CROSS BETWEEN A MOTHER WHO IS HETEROZYGOUS FOR TYPE B BLOOD AND A FATHER WHO IS HETEROZYGOUS FOR TYPE A BLOOD. -AB (1); BO (1); AO (1); OO (1) - ratio 1:1:1:1 -type AB (1); type B (1) type A (1); type O (1) - ratio 1:1:1:1 GENOTYPES: PHENOTYPES: A O B O AB OO BO AO