Incomplete dominance Codominance Multiple alleles IB topic 4

Incomplete dominance Co-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 In pea plants, purple flowers (P) are dominant over white flowers (p) show the cross between two heterozygous plants. GENOTYPES: - PP (1); Pp (2); pp (1) - ratio 1: 2: 1 PHENOTYPES: - purple (3); white (1) - ratio 3: 1 P p P PP Pp pp

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

Codominance • • The heterozygous condition, both alleles are expressed equally Sickle Cell Anemia in Humans NN = 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 GENOTYPES: • - NS (2) SS (2) - ratio 1: 1 PHENOTYPES: - carrier (2); sick (2) - ratio 1: 1 S NS SS

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. O O GENOTYPES: • - IAi (2); IBi (2) - ratio 1: 1 PHENOTYPES: - AO (2); BO (2) - ratio 1: 1 A AO AO B BO 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. GENOTYPES: -IAIB, IBi, IAi, ii - ratio 1: 1: 1: 1 A B PHENOTYPES: -type AB (1); type B (1) O type A (1); type O (1) - ratio 1: 1: 1: 1 O AB BO AO OO

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 • Ax. 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 • Bx. O • BB x OO (B), BO x OO (B or O) • Parents of either the B or O children Gerbers • Ox. O • OO x OO • Must be parents of the O child • Therefore, Lees must be parents of the B child Santiagos • 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