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murzikaleks [220]
3 years ago
10

The diagram to the right represents the lysogenic cycle. What is occurring at step D? The provirus is created. The provirus is r

eplicated. The provirus leaves the host cell’s nucleic acid. The host cell divides. The steps of the lysogenic cycle are shown. In Step A, attachment and entry occurs. In step B, the provirus is formed. In step C, the cell begins to divide. In step D, the provirus leaves. In step E, the virus is replicated and assembled. In step F, lysis and release of the virus occurs.

Biology
2 answers:
dem82 [27]3 years ago
5 0

Answer: The Answer is C. The provirus leaves the host cell 's nucleic acid

Explanation:

qaws [65]3 years ago
3 0

Answer:

<u>Lysogenic Cycle -</u> "Viruses can replicate it DNA and grow in  number by attacking the host cell and using the resources of the host cell to form new and more viruses by going through the process of lysognenic cycle."

There are two methods or ways for a virus to replicate its DNA and produce more virus bodies like itself which are,

  • Lysogenic Cycle
  • Lytic Cycle

Explanation:

The lysogenic cycle can be explained by the given steps below:

  • <u> </u><u>Step A- Absorption:</u>The Bacteriophage attacks the host cell. As it sticks to the surface of its host.
  • <u>Step B- Penetration</u>: The bacteriophage starts the process by penetrating the surface of the host cell and releasing its DNA into its body.
  • <u>Step C- Prophage formation:</u> The cells DNA takes the shape of the viral genome which is yet to be connected to the host cell's genome for replication.
  • <u>Step D- Spontaneous</u>: The prophage formed is induced into the genome of the host cells genome and becomes a part of that for its replication and nourishment purpose.
  • <u>Step E-Replication and Maturation: </u>The virus genome is replicated and the new bodes formed are grown into a more matured form by using the resources of the host cell.
  • <u>Step F-Releasing:</u> The grown viral bodies breaks the membrane and spreads into the outside environment to attack more host cell and increase there number with the same process.

-Now specifically at step D of the cycle the spontaneous induction of the the prophage into the host cell genome by cutting of or lysis of some portion of the host genome and rearrangement of itself genomic code to match up according the host's genomic code.

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Iron-55 has a half-life of 3 years. How much of a 40 gram sample will remain after 12 years? A. 0 grams B. 4 grams C. 2.5 grams
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Answer:

\boxed {\boxed {\sf C. \ 2.5 \ grams }}

Explanation:

We are asked to find how much of a 40 gram sample remains after 12 years.

Iron-55 has a half-life of 3 years. Therefore, after 12 years, 4 half-lives have been completed.

  • 12 years/3 years = 4 half-lives

Every time a half-life is completed, half of the sample's mass decays. Remember we start with a 40 gram sample.

  • 1 half- life: 40 g / 2 = 20 g
  • 2 half-lives: 20 g / 2= 10 g
  • 3 half-lives: 10 g / 2 = 5 g
  • 4 half-lives: 5 g / 2 = 2.5 g

There is also a formula that can be used to solve this problem.

A= A_o(\frac {1}{2})^{\frac{t}{hl}

Where A₀ is the initial amount, t is the time, and hl is the half-life.

We know 40 grams is the inital amount, 12 years is the time, and 3 years is the halflife.

  • A₀= 40 g
  • t= 12
  • hl= 3

A= 40 \ g (\frac{1}{2})^\frac{12}{3}

A= 40 \ g (\frac{1}{2})^4

A= 40 \ g * 0.0625

\bold {A= 2.5 \ g}

After 12 years, <u>2.5 grams </u> of Iron-55 will remain.

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