A typical virus life cycle includes the attachment of virus to host cell, entry, replication or latency (depending on the type of life cycle), and exit. If the virus immediately replicates and lyses the cell, it undergoes a lytic cycle. If the virus incorporates itself into the host cell’s genome and becomes latent, then it undergoes lysogenic cycle. Note that a virus in lysogenic cycle never exits the cell; to exit the cell, the virus must become reactivated by environmental cues. Once activated, the virus enters the lytic cycle and causes the eventual lysis of the cell. Therefore, the only common theme between lytic and lysogenic cycles is the mode of entry.

Lysogenic cycle of a virus involves the incorporation of the viral genome into the host cell’s genome. Upon incorporation, the virus becomes a provirus and remains latent until activated. The only remnant of the virus found in provirus is its genome, or its nucleic acid content. Note that viruses can have DNA or RNA. If it has RNA, virus uses an enzyme called reverse transcriptase to convert the RNA to DNA. The DNA can now incorporate itself into the host cell’s genome (host cell genome is made up of DNA only).

One of the key characteristics of a provirus is that it does not cause any symptoms such as fever, headaches, etc. It can only cause infection and symptoms after reactivation, which can occur due to environmental cues.

Cell rupture, or cell lysis, is caused by viruses in lytic cycle. In a lytic cycle, the virus replicates and exits the host cell, causes an infection, and elicits an immune response.

The question states that the virus is known to cause a deadly disease. It also states that the rat did not develop any symptoms for the entire observatory period (2 weeks). One possible explanation for this result is that the virus enters the cell and becomes latent. It incorporates itself into the host cell's genome and causes disease once re-activated. This reactivation can occur years after the primary infection; therefore, a patient infected with this virus can have no symptoms for years. Recall that a virus that enters the cell and becomes latent is undergoing the lysogenic cycle; therefore, this virus must be undergoing this life cycle.

The question states that the rat is immunosuppressed. This means that the rat does not have the ability to activate the immune system (B cells, T cells, macrophages, etc.) to fight infections. Also, remember that even a completely immuno-competent rat will not produce an immune response to this virus. This is because the immune system does not recognize a latent virus in the lysogenic cycle and cannot, therefore, produce an immune response.

The correct answer is that cells must be actively replicating. Retroviral gene integration into host genomes can only occur when the host is actively replicating. Pantropic retroviruses are able to infect many different cell types from many species. Additionally, the presence of a lipopolysaccharide layer indicates that the target cell is a bacterium. Bacteriophages, not retroviruses, infect bacteria. 

Temperate phages a so named because after infecting a bacterial cell, they can enter either the lysogenic or lytic cycle. Other phages cannot enter the lysogenic cycle and will always enter the lytic one. Since they have the choice to either hide out in the cell unnoticed or immediately replicate and lyse the cell lysogenic phages are refereed to as temperate.

During assembly, newly synthesized viral proteins and the replicated viral genomes associate with one another to form new virions. The build up of new virions causes rupture and release. Replication describes the production of viral mRNAs which are translated using host machinery. Penetration describes the entry of the original virus into the cell. 

The only answer that properly explains a 2:1 ratio is if the homozygous dominant phenotype is lethal. In our punnett square, this would give a two-thirds chance for a heterozygous offspring and a one-third chance for a recessive offspring. We know the recessive phenotype is not lethal because homozygous recessive offspring were produced.

Parents: Aa x Aa

Offspring genotypes: AA, Aa, Aa, aa

Offspring phenotypes: lethal, dominant, dominant, recessive (only three live offspring produced)

Offspring ratio: 2 dominant to 1 recessive

There is also no evidence that codominance or incomplete dominance is present. If black and white spotted offspring or gray offspring were produced then these theories would have merit.

This problem requires a standard dihybrid cross. The crossed genotypes are AaBb x AaBb. This results in a phenotypic ratio of 9 dominant for both traits, 3 dominant for a single trait, 3 dominant for the other trait, and 1 recessive for both traits. In this cross, it will result in: 9 AxBx, 3 Axbb, 3 aaBx, and 1 aabb.

Since we know that the genes are both capable of making the red coloration we actually need to add together all of the choices that contain at least a single dominant allele. Essentially, AxBx, Axbb, and aaBx all show the exact same phenotype. This leaves us with a 15:1 ratio of red to white flowers. 

We are not given the mother's full genotype in the question; she could reasonably carry two A alleles, or an A allele and a recessive O allele. We know that the father must carry one copy of the A allele and one copy of the B allele.

Two punnett squares can answer this question, corresponding to the two possible maternal genotypes: one crossing AA x AB and the other crossing AO x AB. From the first cross there is a 50% chance of blood type A versus 50% chance of blood type AB (half AA and half AB). The second cross shows that there is a potential chance of 50% for type A, 25% for type AB, and 25% for type B (one AA, one OA, one AB, and one OB).

Based on these possibilities, the child could have blood type A, B, or AB. The child cannot have blood type O.

There are nine distinct genotypes present after a standard dihybrid cross. This question can easily be answered by setting up a Punnett square (AaBb x AaBb) and counting the number of unique genotypes present after doing the cross. The numbers also conveniently work out that however many offspring display the dominant phenotype is equal to the number to of genotypes present (this is true for monohybrid and trihybrid crosses as well).