1. Incomplete dominance Co-dominance Multiple alleles IB topic 4.3 (pages 103-104)

2. Review: Dominant/Recessive One allele is dominant over the other (capable of masking the recessive allele) PP = purple pp = white Pp = purple

3. Review Problem: Dominant/Recessive PpPp P p 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.

4. Incomplete Dominance A third (new) phenotype appears in the heterozygous condition. Flower Color in 4 O’clocks RR = red rr = whiteRr = pink

5. Problem: Incomplete Dominance Show the cross between a pink and a white flower. GENOTYPES: - Rr (2); rr (2) - ratio 1:1 PHENOTYPES: - pink (2); white (2) - ratio 1:1 rrRr rr Rr rrrr

6. Codominance The heterozygous condition, both alleles are expressed equally Sickle Cell Anemia in Humans NN = normal cells SS = sickle cells NS = some of each

7. Codominance Show the cross between an individual with sickle-cell anemia and another who is a carrier but not sick. N S SSSS SS - NS (2) SS (2) - ratio 1:1 - carrier (2); sick (2) - ratio 1:1 GENOTYPES: PHENOTYPES:

8. 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

9. 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)

10. Blood types continued The four blood groups result from various combinations of the three different alleles of one gene Genotypes: I A (A carbohydrate) I B (B carbohydrate) i (giving rise to neither) Because each person carries 2 alleles, there are 6 possible genotypes

11. Blood types continued Both the I A and I B alleles are dominant to the i allele Therefore, I A I A and I A i individuals have type A I B I B and I B i individuals have type B I A I B individuals have type AB (codominant) ii individuals have type O

12. 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!!

13. 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

14. Problem: multiple alleles Show the cross between a mother who has type O blood and a father who has type AB blood. - I A i (2); I B i (2) - ratio 1:1 - AO (2); BO (2) - ratio 1:1 GENOTYPES: PHENOTYPES: O O ABAB AO BO AO BO

15. 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. -I A I B, I B i, I A i, 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 BOBO AB OO BO AO

16. 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.

17. 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

18. 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