Incomplete dominanceCo-dominance

Multiple alleles

IB topic 4.3 (pages 103-104)

Review: Dominant/Recessive

• One allele is dominant over the other (capable of masking the recessive allele)

PP = purple pp = white Pp = purple

Review Problem: Dominant/Recessive

P

p

P p

pp

Pp

Pp

PP- PP (1); Pp (2); pp (1)- ratio 1:2:1

- purple (3); white (1)- ratio 3:1

GENOTYPES:

PHENOTYPES:

In pea plants, purple flowers (P) are dominant over white flowers (p) show the cross between two heterozygous plants.

Incomplete Dominance

• A third (new) phenotype appears in the heterozygous condition.

• Flower Color in 4 O’clocks

RR = red rr = white Rr = pink

Problem: Incomplete Dominance

• Show the cross between a pink and a white flower.

GENOTYPES:• - Rr (2); rr (2)• - ratio 1:1PHENOTYPES:• - pink (2); white (2)• - ratio 1:1

rrRr

rrRr

R r

r

r

Codominance

• The heterozygous condition, both alleles are expressed equally

• Sickle Cell Anemia in HumansNN = normal cells

SS = sickle cells NS = some of each

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:

Multiple Alleles

• There are more than two alleles for a trait

• Blood type in humans• Blood Types (4 different phenotypes)

• Type A, Type B, Type AB, Type O

• Blood Alleles• Pg. 104

Blood types

• The letters refer to 2 carbohydrates (A substance and B substance) that may be found on the surface of the red blood cells

• A person’s blood cells may have:• One type or the other (A or B)• Both (type AB)• Neither (type O)

Blood types continued

• The four blood groups result from various combinations of the three different alleles of one gene

• Genotypes: • IA (A carbohydrate)• IB (B carbohydrate)• i (giving rise to neither)

• Because each person carries 2 alleles, there are 6 possible genotypes

Blood types continued

• Both the IA and IB alleles are dominant to the i allele

• Therefore, • IA IA and IAi individuals have type A• IB IB and IBi individuals have type B • IA IB individuals have type AB (codominant) • ii individuals have type O

Why is this important?

• Matching compatible blood groups is critical for blood transfusions• A person’s blood produces specific proteins called

antibodies again foreign blood factors • If the donors blood has a factor (A or B) that is

foreign to the recipient, antibodies produced by the recipient bind to the foreign molecules and cause the donated blood cells to clump together

• BAD!!

Random facts

• O is the universal donor• AB is the universal recipient • + and – comes from the Rh test

• Another type of antigen • Key test for pregnant woman• If fetus and mother are Rh incompatible,

the mother can receive a vaccine

Problem: multiple alleles • Show the cross between a mother who

has type O blood and a father who has type AB blood.

- IAi (2); IBi (2) - ratio 1:1

- AO (2); BO (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.

-IAIB, IBi, IAi, ii - 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

Mix-up at the hospital• One busy night in an understaffed maternity unit, four

children were born about the same time. Then the babies were mixed up (by mistake); it was not certain which child belonged to which family. Fortunately, the children had different blood groups: A, B, AB, and O.

• The parents blood groups were also known:• Mr. and Mrs. Jones (A x B) • Mr. and Mrs. Gerber (O x O)• Mr. and Mrs. Lee (B x O)• Mr. and Mrs. Santiago (AB x O)

• The nurses were able to decide which child belonged to which family. Deduce how this was done.

Answer

• Jones: parents of the AB child• Lees: parents of the B child• Gerbers: parents of the O child• Santiagos: parents of the A child

How? • Jones

• A x B• AA x BB (AB), AO x BO (AB or O), AA x BO (AB or A), AO x BB (AB or B)• So, they could be the parents of any of the kids

• Lees• B x O

• BB x OO (B), BO x OO (B or O) • Parents of either the B or O children

• Gerbers• O x O

• OO x OO • Must be parents of the O child• Therefore, Lees must be parents of the B child

• Santiagos • AB x O

• AB x OO (A or B)• Can’t be parents of the B child b/c of the Gerbers, so they are the parents

of the A child• Therefore, the Jones are the parents of the AB child