1. BEYOND DOMINANT & RECESSIVE ALLELES
INCOMPLETE DOMINANCE, CODOMINANCE, MULTIPLE ALLELES, POLYGENIC TRAITS
BEYOND DOMINANT & RECESSIVE ALLELES

2. DOMINANT? RECESSIVE? NEITHER?
NOT ALL GENES SHOW SIMPLE PATTERNS OF DOMINANT AND RECESSIVE ALLELES. IN MOST ORGANISMS, GENETICS IS MORE COMPLICATED, BECAUSE THE MAJORITY OF GENES HAVE MORE THAN TWO ALLELES. MANY IMPORTANT TRAITS ARE CONTROLLED BY MORE THAN ONE GENE.

3. INCOMPLETE DOMINANCE
PAIRS OF ALLELES PRODUCE A HETEROZYGOUS PHENOTYPE THAT RESULTS IN AN APPEARANCE IN BETWEEN THE PHENOTYPES OF THE TWO PARENTAL VARIETIES.
EXAMPLE: SNAPDRAGON COLOR

4. INCOMPLETE DOMINANCE
With incomplete dominance, a cross between organisms with two different phenotypes produces offspring with a third phenotype that is a blending of the parental traits. 

5. INCOMPLETE DOMINANCE - SNAPDRAGONS

6. INCOMPLETE DOMINANCE – Humans
HYPERCHOLESTEROLEMIA – DANGEROUSLY HIGH LEVELS OF CHOLESTEROL IN THE BLOOD
NORMAL INDIVIDUALS ARE HH
HETEROZYGOTES ARE Hh (1 IN 500)
HOMOZYGOUS INDIVIDUALS WITH hh (1 IN 1,000,000) 5X NORMAL CHOLESTEROL LEVEL

7. INCOMPLETE DOMINANCE PROBLEMS
YELLOW GUINEA PIGS CROSSED WITH WHITE GUINEA PIGS ALWAYS PRODUCE CREAM COLORED GUINEA PIGS.
A. COMPLETE THE PUNNETT SQUARE FOR THE CROSS BETWEEN A WHITE GUINEA PIG AND A YELLOW GUINEA PIG.
B. COMPLETE THE PUNNETT SQUARE FOR THE CROSS BETWEEN TWO CREAM-COLORED GUINEA PIGS.
C. GIVE THE PHENOTYPIC AND GENOTYPIC RATIO FOR THE OFFSPRING IN A AND B.

8. CODOMINANCE BOTH ALLELES CONTRIBUTE TO THE PHENOTYPE OF THE ORGANISM
EXAMPLE: IN HORSES, THE ALLELE FOR RED HAIR IS CODOMINANT WITH THE ALLELE FOR WHITE HAIR. HORSES WITH BOTH ALLELES ARE ROAN BECAUSE THEIR COATS ARE A MIXTURE OF BOTH RED AND WHITE HAIRS

9. CODOMINANCE
With codominance, a cross between organisms with two different phenotypes produces offspring with a third phenotype in which both of the parental traits appear together. 

10. CODOMINANCE PROBLEMS
1. WHEN A RED BULL IS BRED TO A WHITE COW, THE COLOR OF THE CALF IS ROAN.
A. COMPLETE THE PUNNETT SQUARE FOR THE CROSS BETWEEN A RED BULL AND A WHITE COW
B. GIVE THE GENOTYPIC RATIO AND PHENOTYPIC RATIO OF THE OFFSPRING.
C. A ROAN BULL AND A ROAN COW ARE BRED. GIVE THE GENOTYPIC RATIO AND PHENOTYPIC RATIO OF THIS CROSS.

11. MULTIPLE ALLELES MORE THAN TWO POSSIBLE ALLELES EXIST IN A POPULATION
EX. COAT COLOR IN RABBITS – DETERMINED BY A SINGLE GENE WITH AT LEAST 4 DIFFERENT ALLELES THAT CAN PRODUCE 4 POSSIBLE COAT COLORS
EX. HUMAN GENE FOR EYE COLOR

12. Multiple Alleles

13. Human Eye Color 3 known genes that
Explain typical patterns of green, brown, blue eye color inheritance
Hazel, grey and multiple shades of green, brown, blue are not explained – molecular basis unknown

14. Human Eye Color Genes Human Eye Color Genes:
EYCL1 (gey) green dominant over blue
Green/blue eye color,
EYCL2 (bey 1)
EYCL3 (bey 2) brown dominant over blue
Central Brown eye color gene

15. MULTIPLE ALLELES HUMAN BLOOD TYPES – CHROMOSOME 9
ONE GENE, THREE ALLELES, PRODUCE 4 PHENOTYPES
A PERSON’S BLOOD GROUP MAY BE O, A, B OR AB
THESE LETTERS REFER TO 2 CARBOHYDRATES THAT MAY BE FOUND ON THE SURFACE OF RED BLOOD CELLS (RBC)

16. ABO BLOOD GROUPS

17. Blood Types: What do they look like?

18. ABO BLOOD GROUPS Blood Type Genotypes ABO Enzymes Present
RBC Antigens Present
Serum Antibodies
"A"
AA, Ai
"H", "A"
A, H
anti-B
"B"
BB, Bi
"H", "B"
B, H
anti-A
"AB" AB
"H", "A", "B"
A, B, H
none
"O" ii
"H" H
anti-A, anti-B

19. ABO BLOOD GROUP PROBLEMS
COMPLETE A PUNNETT SQUARE FOR:
HOMOZYGOUS TYPE A x TYPE O
HETEROZYGOUS TYPE A x TYPE O
HOMOZYGOUS TYPE A x HOMOZYGOUS TYPE B
HETEROZYGOUS TYPE B x HOMOZYGOUS TYPE A
HETEROZYGOUS TYPE A x HETEROZYGOUS TYPE B
HETROZYGOUS TYPE B x TYPE O

20. Practice problems A TYPE AB MALE MARRIES A TYPE O FEMALE.
WHAT ARE THEIR GENOTYPES?
COMPLETE A PUNNETT SQUARE TO SHOW THE POSSIBLE GENOTYPES OF THEIR CHILDREN.

21. Practice problems
A TYPE A MALE MARRIES A TYPE B FEMALE. THEY HAVE TWO CHILDREN. ONE CHILD IS TYPE AB AND THE OTHER CHILD IS TYPE O.
WHAT ARE THE GENOTYPES OF THE PARENTS?
COMPLETE A PUNNETT SQUARE TO SHOW THE GENOTYPES OF THESE TWO CHILDREN AND FOR ANY FUTURE CHILDREN.

22. Practice problems
A TYPE A MALE IS MARIED TO A TYPE O FEMALE. USING PUNNETT SQUARES, EXPLAIN HOW IT WOULD BE POSSIBLE FOR THEM TO HAVE A:
TYPE A CHILD.
TYPE B CHILD.
TYPE O CHILD.

23. Practice problems
A TYPE AB MALE MARRIES A TYPE B FEMALE. USING PUNNETT SQUARES, EXPLAIN HOW IT WOULD BE POSSIBLE FOR THEM TO HAVE A:
TYPE AB CHILD
TYPE A CHILD
TYPE O CHILD
TYPE B CHILD

24. Practice problems
A TYPE AB MALE MARRIES A TYPE AB FEMALE. USING PUNNETT SQUARES, EXPLAIN HOW IT WOULD BE POSSIBLE FOR THEM TO HAVE A:
TYPE AB CHILD
TYPE A CHILD
TYPE B CHILD
TYPE O CHILD

25. Practice problems A TYPE O MALE MARRIES A TYPE O FEMALE.
WHAT ARE THE POSSIBLE GENOTYPES OF THEIR CHILDREN?

26. Practice problems
A TYPE A MALE, WHOSE FATHER IS TYPE O, MARRIES A TYPE O FEMALE. WHAT ARE THE POSSIBLE GENOTYPES OF THEIR CHILDREN?
THE MALE FROM THE PREVIOUS PROBLEM HAS A BROTHER WHO IS TYPE B. WHAT IS THEIR MOTHER’S BLOOD TYPE?
THE BROTHERS IN THE ABOVE PROBLEMS HAVE A SISTER WHOSE BLOOD TYPE IS O. IS THIS POSSIBLE?

27. POLYGENIC TRAITS TRAITS CONTROLLED BY TWO OR MORE GENES
EX. 3 GENES INVOLVED MAKING THE REDDISH-BROWN PIGMENT IN THE EYES OF FRUIT FLIES
EX. WIDE RANGE OF HUMAN SKIN COLOR DUE TO MORE THAN 4 DIFFERENT GENES CONTROLLING THE TRAIT

28. POLYGENIC TRAIT – COAT COLOR IN MICE

29. SEX-LINKED TRAITS
IN HUMANS, SEX IS DETERMINED BY THE 23RD PAIR OF CHROMOSOMES – THE SEX CHROMOSOMES!
THE SEX CHROMOSOMES ARE THE X AND THE Y CHROMOSOMES
XX – FEMALE
XY – MALE

30. SEX-LINKED TRAITS MOST SEX-LINKED GENES ARE X-LINKED GENES.
WHY? THE X CHROMOSOMES ARE LONGER AND CONTAINS THOUSANDS MORE GENES THAN THE Y CHROMOSOME.
X CHROMOSOME CONTAINS 1098 GENES.
Y CHROMOSOME CONTAINS 26 GENES.

31. KARYOTYPE

32. SEX-LINKED TRAITS
FOR EACH GENE EXCLUSIVELY ON THE X CHROMOSOME, THERE ARE TWO ALLELES OF EACH GENE
MALES, XY, HAVE ONLY ONE ALLELE
A MALE, XY, WITH A RECESSIVE ALLELE ON THE X CHROMOSOME, WILL ALWAYS EXHIBIT THAT RECESSIVE TRAIT BECAUSE THERE IS NO CORRESPONDING ALLELE ON THE Y CHROMOSOME.

33. EXAMPLES OF SEX-LINKED TRAITS
IN HUMANS:
Red-Green Color Blindness
Hemophilia
Duchenne Muscular Dystrophy
IN CATS
Calico coat color
IN FRUIT FLIES:
Eye Color

34. WORKING PUNNETT SQUARES
IDENTIFY EACH INDIVIDUAL AS MALE OR FEMALE ACCORDING TO THEIR SEX CHROMOSOMES
EX. XX or XY
THE LETTERS REPRESENTING THE TRAIT ARE PLACED AS SUPERSCRIPTS ABOVE THE X CHROMOSOME
EXAMPLE: XR or Xr

35. EYE COLOR IN FRUIT FLIES
THE GENE FOR EYE COLOR IS ON THE X CHROMOSOME.
THE ALLELE FOR RED EYES IS DOMINANT OVER WHITE EYES
PROBLEM:
IF A WHITE-EYED FEMALE FRUIT FLY IS MATED WITH A RED-EYED MALE, PREDICT THE POSSIBLE OFFSPRING

36. COAT COLOR IN CATS
Coat color in cats is an X-linked gene, with alleles for black and orange-brown
XBXB and XBY cats will have a black coat
XOXO and XOY will have an orange-brown coat
female cats with XBXO are Calico!!

37. Pedigrees
a diagram of a family history that shows relationships between family members and their status with respect to a particular hereditary condition.
The affected person through which the pedigree is discovered is indicated by an arrow on pedigrees.
Males are denoted by squares and females by circles.
A filled symbol indicates that an individual is affected by a specific condition.
Generations are represented on horizontal levels;
parents on one level and the children of those parents together on a level below them.

38. Pedigrees Autosomal Dominant

39. Autosomal Dominant
Dominant conditions are expressed in individuals who have just one copy of the mutant allele. The pedigree on the right illustrates the transmission of an autosomal dominant trait. Affected males and females have an equal probability of passing on the trait to offspring. Affected individual's have one normal copy of the gene and one mutant copy of the gene, thus each offspring has a 50% chance on inheriting the mutant allele. As shown in this pedigree, approximately half of the children of affected parents inherit the condition and half do not.

40. Pedigrees–Autosomal Recessive

41. Autosomal Recessive
Recessive conditions are clinically manifest only when an individual has two copies of the mutant allele. When just one copy of the mutant allele is present, an individual is a carrier of the mutation, but does not develop the condition. Females and males are affected equally by traits transmitted by autosomal recessive inheritance. When two carriers mate, each child has a 25% chance of being homozygous wild-type (unaffected); a 25% chance of being homozygous mutant (affected); or a 50% chance of being heterozygous (unaffected carrier).

42. Pedigrees-X-Linked Recessive

43. X-Linked Recessive
X-linked recessive traits are not clinically manifest when there is a normal copy of the gene. All X-linked recessive traits are fully evident in males because they only have one copy of the X chromosome, thus do not have a normal copy of the gene to compensate for the mutant copy. For that same reason, women are rarely affected by X-linked recessive diseases, however they are affected when they have two copies of the mutant allele. Because the gene is on the X chromosome there is no father to son transmission, but there is father to daughter and mother to daughter and son transmission. If a man is affected with an X-linked recessive condition, all his daughter will inherit one copy of the mutant allele from him.

44. Pedigrees-X-Linked Dominant

45. X-Linked Dominant
Because the gene is located on the X chromosome, there is no transmission from father to son, but there can be transmission from father to daughter (all daughters of an affected male will be affected since the father has only one X chromosome to transmit). Children of an affected woman have a 50% chance of inheriting the X chromosome with the mutant allele. X-linked dominant disorders are clinically manifest when only one copy of the mutant allele is present.

46. Famous Pedigree The PEDIGREE of
Nicholas (Nikolia) II ROMANOV (last CZAR) of RUSSIA
(Nicholas Alexandrovitch); Knight of the Garter; (since he possessed Russia, he may be considered the wealthiest man ever)