Codominance / Genetics

Hi can anyone help on these topics im really stuck, especially on codominance and multiple alelles i dont understand it at all

Monohybrid inheritance refers to the inheritance of a single gene with two alleles:
a. For two alleles, A and a, there are three possible genotypes (in a 1:2:1 ratio):
i. Homozygous dominant (AA) – dominant allele is expressed.
ii. Homozygous recessive (aa) – recessive allele is expressed.
iii. Heterozygous (Aa) – dominant allele is expressed.
b. There will be two phenotypes, in a 3:1 ratio.

Codominant alleles are ones that are both expressed in the phenotype, for example blood groups:
a. There are three alleles – IA and IB are dominant, and IO is recessive.
b. IA and IB code for A and B proteins on red blood cells, whereas IO codes for no relevant
proteins:
i. IOIO – blood group O.
ii. IAIA or IAIO – blood group A.
iii. IBIB or IBIO – blood group B.
iv. IAIB – blood group AB (codominance).

Dihybrid crosses involve two separate genes, each with two alleles, at the same time. The results
are organised in a Punnett square, as there will be 16 genotypes.
a. With parents of AABB and aabb, F1 will all be AaBb.
b. F2 will then give a 9:3:3:1 ratio of phenotypes, assuming the functions of the genes
(biochemically) are independent.

Epistasis is where the presence of one allele affects the expression of another, due to a biochemical
interaction – if two enzymes are produced by genes A and B respectively, then if A is not
expressed, then the pathway cannot proceed regardless of whether B can be produced or not:
a. In wild mice, two genes are needed for the agouti (grey/brown) colour:
i. If A and B are present, the mice will be agouti.
ii. For AAbb or Aabb, the mice will be black (due to lack of enzyme B).
iii. For aa, regardless of whether or not there is enzyme B, the mice will be albino,
as enzyme A is not present, and B depends on A.
b. This will result, in this case, in a 9:3:4 ratio of phenotypes.