Interphase is the portion of the cell life cycle where the cell is preparing for division. This portion of the life cycle includes growth, as well as reproducing organelles and proteins. The genome is also replicated in interphase. Interphase is composed of the G1, S, and G2 phases. G1 and G2 correspond to periods of growth, protein synthesis, and organelle replication. The S phase is the period of DNA replication.

The M phase is mitosis, and is the portion of the cell cycle during which the cell is undergoing division. As a result, it is not part of interphase.

When the cell is not preparing to divide any time soon, it will spend its time in the G₀ phase. This phase is distinct from the actual cell cycle, and can prevent the cell from actively growing and preparing for division. Cells arrested in the G₀ phase are considered quiescent.

Interphase includes S phase plus the two gap phases . The M phase is the remainder of the cell cycle, made up of mitosis (and meiosis in germ cells) and cytokinesis. Cytokinesis (physical cell division) occurs at the end of M phase, but before interphase.

Cells progress through the cell cycle by controlled expression and degradation of cyclin proteins. Cyclin-dependent kinases (CDKs) are always present in the cell and are not degraded after progression to a new stage of the cell cycle. Cyclins bind their respective CDKs to activate them. This activation causes a chain of events that allow the cell to progress to the next phase of the cell cycle. Afterwards, cyclins are ubiquinated and degraded until they are needed again.

During anaphase, sister chromatids are pulled apart towards opposite poles. The cell becomes longer and the cleavage furrow (contractile ring) forms. This marks the beginning of cytokinesis. The process completes during telophase, producing two new daughter cells. Cytokinesis must be preceeded by karyokinesis (physical movement of the chromosomes).

Since colorblindness is a recessive disease, all copies of the X-chromosome must have the diseased allele in order for the person to be colorblind. Daughters have two copies of the X-chromosome: one from the mother and the other from the father. Males only have one copy of the X-chromosome (from the mother) and a Y-chromosome from the father.

Since we know that the father has normal vision, he does NOT carry the colorblind allele. Since the daughters for this couple can only potentially receive one colorblind allele (from the mother), all of their daughters will have normal vision. This means that there is a zero percent chance for colorblindness in their daughters.

The cross would look like this, taking X^b as the colorblind allele:

Parents: XX^b x XY

Offspring: XX or XX^b (normal daughters), XY (normal son), YX^b (colorblind son)

The chance of a colorblind daughter will be zero, but the chance of a colorblind son will be 50%.

X-linked recessive inheritance dictates that expression of themutant phenotype will only occur if the individual is homozygous for the mutation on the X-chromosomes. Therefore, a female must have inherited two mutant X-chromosomes to have hemophilia A, while a male only requires one mutant X-chromosome to have the disorder. By virtue of the father not having hemophilia A, we know the daughter is inheriting at least one wild-type X-chromosome, and therefore there is zero chance she will be homozygous and have hemophilia A.

Incomplete dominance indicates that there is no dominant allele. In these cases, the phenotype associated with inheriting one copy of each allele (the heterozygotes, Bb) is often a blending of the phenotypes associated with homozygosity of each allele. As such, a genotype of BB will result in black hair, bb will produce white hair, and Bb will result in grey hair.

The incorrect answers are too limited in scope to be cases of incomplete dominance. The correct answer identifies that there will be three unique phenotypes. 

Because the disorder is autosomal dominant, the statement "If an individual does not have the disorder, they can still pass on the mutant gene if one of their parents has the disorder" must be false.

If the indivdual in question does not have the disorder, that means they did not inherit ANY copies of the mutant gene, and therefore cannot pass it on.

For autosomal dominant disorders, the individual only needs to inherent a single copy of a mutated allele to then show symptoms of that disorder. If it were recessive, both alleles would have to be mutant. X-linked recessive is incompletely correct for males since they only have one X-chromosome, and incorrect for females since 2 copies of the X-chromosome are needed, and thus 2 copies of the allele. Complex inheritance describes situations beyond a single gene, and Mendelian inheritance is not a specific method of inheritance. Note that Y-linked disorders are passed from father to son, and since males only have one copy of the Y-chromosome, if there is a genetic mutation on the Y-chromosome, the individual will be affected.