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.

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.

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.

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 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.

Antibiotic resistance genes are often found on plasmids, which are small DNA molecules which are easily transfered to other prokaryotic cells.

Antibiotic resistance genes are often found on plasmids, which are small DNA molecules which are easily transfered to other prokaryotic cells.

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.

Conjugation is a type of genetic recombination that requires one bacterium to have the F-plasmid in order to create a sex pilus. This sex pilus will connect with another bacterium and allow DNA to pass between the bacteria.

Transduction is the transfer of genetic information to a bacterium via a vector, such as a bacteriophage. Transformation occurs when a bacterial cell receives genetic material from its surrounding environment. Binary fission does not involve recombination, and is the term for bacterial cell division that results in two identical offspring from a single parental cell.