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.

This question can be solved by making a punnett square. The genotypes are given: AaBb x aabb.

The potential gametes the AaBb dog can produce are AB, Ab, aB, and ab. The aabb dog can only produce one gamete: ab.

Putting these gametes in our punnett square we can see that we end up with the following potential offspring: AaBb, Aabb, aaBb, and aabb.

Each of these possible offspring carries a different phenotype. AaBb will carry both dominant traits and be black with long ears. Aabb will be yellow with long ears. aaBb will be black with short ears. Finally, aabb will be yellow with short ears. Each of these gametes is produced in the same ratio, making these phenotypes exist in a 1:1:1:1 probability.

Punnett squares give information about the potential genotype ratios of offspring possible from the cross of two members of the parental generation. The letters represent alleles of various genes, but do not give any information about the allele frequencies. To get information about the allele frequencies, more information about the size and make-up of the population would be needed. The actual cross is between potential gametes produced by the parental generation. Each square shows the potential offspring from these potential gametes.

The shortcut for this problem involves the standard phenotypic ratios for a dihybrid cross. Nine offspring will show both dominant traits. Three will show one dominant trait and the other recessive trait. Three will show the inverse phenotypes, with the opposite dominant trait and recessive trait combination. One offspring will show both recessive traits. Based on these ratios, we can see that three of the sixteen offspring will show the dominant smooth trait and the recessive green phenotype.

We can also solve by using a dihybrid punnett square. The cross described will be AaBb x AaBb, in which the A alleles signify color and the B alleles signify shape.

Consider the color of the peas. In order to have green peas, two recessive alleles must combine in the next generation. According to a punnett square where both sides are heterozygous for the trait, there is only a one in four chance of this taking place. Since smooth is the dominant shape for the peas, a punnett square where each side is heterozygous shows a three in four chance that pea plants will have this shape. By multiplying these two probabilities, we determine that three out of sixteen pea plants will have smooth, green peas.

The three main modes of genetic transfer for prokaryotes are transformation, transduction, and conjugation. Transformation occurs when a bacterium picks up a piece of genetic material from its external environment and incorporates it into its own genome. Transduction is genetic transfer using a bacteriophage as a vector. Conjugation is direct gene transfer via sex pili.

The IIR bacteria became deadly when exposed to the remains of the IIIS bacteria. This means that the IIR bacteria managed to receive genetic material from the environment and incorporate it into their genome. This is an example of transformation, a process that results in genetic recombination. In this case, the recombination made the formerly harmless bacteria deadly in mice.

Transduction is the process by which new genetic information is introduced to a bacterium via a vector, such as a bacteriophage. Conjugation is the transfer of genetic material between bacteria via a sex pilus. Binary fission is not a means of recombination; rather, the parent cell divides to produce two identical copies of itself.