The best choice is that the individual has a mutation that accounts for his/her green eyes. This mutation would yield a new phenotype that was not a result of direct inheritance from the parent generation.

Codominance cannot only occur in one individual offspring; rather it would occur in some form in all offspring with respect to the given trait. Independent assortment always occurs in cases of typical Mendelian genetics. 

If either parent were homozygous dominant, all offspring would exhibit the dominant phenotype and the second phenotype would not be expressed.

Homozygous parent cross: RR x r_

Offspring: all offspring inherit a dominant R allele and will express the dominant phenotype.

If two heterozygotes are crossed, all three genotypes will be expressed, so all possible phenotypes will be expressed.

Both heterozygous: Rr x Rr

Offspring: 1 RR, 2 Rr, 1 rr with both dominant and recessive phenotypes expressed in a 3:1 ratio.

If one heterozygote and one homozygote are crossed the offspring will show only two genotypes, but both possible phenotypes.

One heterozygote: Rr x rr

Offspring: half Rr and half rr, with half showing dominant phenotype and half showing recessive phenotype.

The second and third examples give rise to both colors of offspring, regardless of which allele is dominant. The first example only shows the dominant phenotype, can cannot be possible for this question.

Biologists use a diagram called a Punnett square to aid them in predicting traits that will be inherited by offspring. In this case both parents are heterozygous, meaning that they both carry one dominant allele (W) and one recessive allele (w). The Punnett square for this cross would look like this:

Because the W allele is dominant, squares with a W will produce the large wing trait. Three of the four possible genotypes from this cross carry the dominant allele, meaning that 75% of the offspring will have large wings. If the parents produce sixteen offspring, then twelve of them will show the large wing phenotype.

The trait that is expressed in heterozygotes is the dominant allele. Thus, the normal plant must be homozygous recessive. If dwarfism was a recessive trait, it would not be expressed in the heterozygotes.

A genotype is the genetic makeup on an organism, while a phenotype is the composite of an organism’s observable traits. In short, the genotype determines the phenotype. The phenotype can also include any other observable traits, such as morphology, development, biochemical or physiological properties, phenology, behavior, and products of behavior. Environmental factors can also affect the phenotype of an organism (in addition to the inherited genotype).

Both of the parent's genotypes are Kk. This means that the offspring can be either KK, Kk, or kk.

Parental genotype: 

Offspring probability:

Once you properly set up your punnet square of a dihybrid cross, you should obtain a phenotypic ratio of 9:3:3:1. There will be 9 plants with dark blue leaves/yellow flowers, 3 plants with light blue leaves/yellow flowers, 3 plants with dark blue leaves/orange flowers, and 1 plant with light blue leaves/orange flowers.

For better visualization of this explanation we can use eye color as an example. BB and BB will be brown eyes and bb will be blue eyes. 

In a monohybrid cross you are dealing with only one gene. We can use "Bb" as an example for the gene as it is a hybrid of two alleles. BB & Bb are dominant and bb is recessive. 

Setting up a punnet square of Bb x Bb we will get 4 results: BB, Bb, Bb, and bb. 

The phenotype is what we see: eye color. Phenotype will be expressed the same for BB and Bb, therefore we have 3 that will express the dominant phenotype and 1 that will express the recessive phenotype; 3:1.

The genotype is the actual gene that creates the phenotype. So for this we have 3 different genes that arise from the monohybrid cross: BB, Bb, and bb. We get 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive; 1:2:1.

The genotype for Sharon is rr, because blonde is a recessive trait therefore in order to have blonde hair she must be homozygous recessive. Her husband is Rr, because it states that he has brown hair, which is dominant, in addition to being heterozygous. When drawing out a punnet square, you will find the offspring will be Rr, Rr, rr and rr. Therefore, their daughter has 50% chance of having brown hair and 50% chance of having blonde hair.

Since Molly is autosomal recessive (ss) and Fred is autosomal dominant (SS), all their children will be heterozygotes, or carriers and they will thus all be protected against malaria.