Incomplete Dominance and Codominance Incomplete Dominance F 1

Incomplete Dominance and Codominance

Incomplete Dominance F 1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example: snapdragons (flower) r red (RR) x white (rr) RR = red flower rr = white flower R R r

Incomplete Dominance r r R Rr Rr produces the F 1 generation All Rr = pink (heterozygous pink)

Incomplete Dominance

Codominance Heterozygote expresses both alleles’ conditions Ex. A black rooster bred with a white hen produces a black and white checkered chicken. Usually uses both capital letters (black= B, white=W , checkered = BW)

Codominance practice: cross a black rooster with a white hen Key: W B = black BW BW W = white BW = checkered Results: 100% checkered (BW) W B B BW BW

Now let’s cross a checkered hen with a checkered rooster. B W B BB BW WW Results: 25% Black 50% Checkered 25% White (1: 2: 1)

Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood type 1. 2. 3. 4. type A B AB O = = IAIA or IAi IBIB or IBi I AI B ii

Codominance Problem Example: homozygous male Type B (IBIB) x heterozygous female Type A (IAi) IA i IB I AI B I Bi 1/2 = IAIB 1/2 = IBi

Another Codominance Problem Example: male Type O (ii) x female type AB (IAIB) IA IB i I Ai I Bi 1/2 = IAi 1/2 = IBi

Codominance Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents? boy - type O (ii) X girl - type AB (IAIB)

Codominance Answer: IA IB i i I AI B ii Parents: genotypes = IAi and IBi phenotypes = A and B

Polygenic Inheritance Trait controlled by 2 or more genes May be on the same or different chromosomes Shows a range, intermediate is most common phenotype Upper and lower case letters used

Many variations in skin color

Skin color: A polygenic trait Determining # of Genes Involved in Skin Color Expected distribution- 4 genes Number of individuals Observed distribution of skin color Expected distribution 1 gene Light Expected distribution- 3 genes Right Range of skin color

Skin Color Skin color is actually due to 5 genes Genotypes darkest to lightest: AABBCCDDEE would be darkest skinned Aa. Bb. Cc. Dd. Ee would be medium skinned aabbccddee would be lightest skinned

Influence of external environment In arctic foxes temperature has an effect on the expression of coat color. In winter, fur is white.

Arctic Fox in summer coloration: influence of external environment

Influence of external environment External influences can also be seen in leaves. Leaves can have different sizes, thicknesses, and shapes depending on the amount of light they receive.

Influence of internal environment The internal environments of males and females are different because of hormones and structural differences. An organism’s age can also affect gene function.

Peacock (male) Peahen (female)

Blood type quick facts Red blood cells are called erythrocytes Proteins on their surfaces are called antigens, controlled by genes Antigens make antibodies to foreign substances, which includes RBCs with different antigens on their surface 4 phenotypes: A, B, AB, O

Phenotype A Surface molecule A • The l. A allele is dominant to i, so inheriting either the l. Ai alleles or the l. A alleles from both parents will give you type A blood. • Surface molecule A is produced.

Phenotype B • The l. B allele is also dominant to i. • To have type B blood, you must inherit the l. B allele from one parent and either another l. B allele or the i allele from the other. • Surface molecule B is produced. Surface molecule B

Phenotype AB • The l. A and l. B alleles are codominant. • If you inherit the l. A allele from one parent and the l. B allele from the other, your red blood cells will produce both surface molecules and you will have type AB blood. Surface molecule B Surface molecule A

Phenotype O • No antigens produced

Check your chart! Blood Group Antigens Antibodies Can receive from Can give to A A B A or O A or AB B B A B or O B or AB AB A and B None AB O None both A, B, AB, O O A, B, AB, O

Possible genotypes for each phenotype: A A A I I A I i = or B = IBIB or IBi A B AB = I I O = ii

Sex-linked Traits (genes) located on the sex chromosomes Sex chromosomes are X and Y XX genotype for females XY genotype for males Many sex-linked traits carried on X chromosome

Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male

Sex-linked Trait Problem Example: Eye color in fruit flies (red-eyed male) x (white-eyed female) X RY x X r Remember: the Y chromosome in males does not carry traits. Xr Xr RR = red eyed Rr = red eyed rr = white eyed XY = male XX = female XR Y

Sex-linked Trait Solution: Xr XR XR Xr Y Xr XR Xr Xr Y 50% red eyed female 50% white eyed male

Crosses with sex-linked traits; colorblindness and hemophilia

Female Carriers Females are carriers of sex-linked traits if they have the heterozygous genotype. Males cannot be carriers because they only have one X chromosome. Female parents who are carriers, pass sex-linked traits to all children, but males are usually the ones who express the trait.

Pedigrees

Tay Sachs disease Occurs in people of Jewish descent Enzyme that breaks down lipids in the brain is defective. Lipid buildup kills brain cells. Always results in death, usually by age 5 Caused by a recessive allele

Pedigree for Tay Sachs disease carrier

Huntington’s Disease Caused by a rare dominant allele Doesn’t show up till age 30 -50 Breaks down areas of the brain, loss of control of all body functions No treatment

Pedigree for Huntington’s Disease normal carrier

What if the trait is sex-linked? What gives you a clue?

Is this sex-linked or not?

karyotypes

Autosomal inheritance patterns; sickle cell, cystic fibrosis, huntingtons

Genetic Practice Problems

Breed the P 1 generation tall (TT) x dwarf (tt) pea plants t T T t

Solution: tall (TT) vs. dwarf (tt) pea plants t t T Tt Tt produces the F 1 generation T Tt Tt All Tt = tall (heterozygous tall)

Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants T T t t

Solution: tall (Tt) x tall (Tt) pea plants T t T TT Tt tt produces the F 2 generation 1/4 1/2 1/4 1: 2: 1 3: 1 (25%) = TT (50%) = Tt (25%) = tt genotype phenotype