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The Life Cycle of HIV Which of the following allows the AIDS virus, which contains RNA, to insert viral DNA into the DNA of a host T-cell after the AIDS virus enters the cell?

A. Receptor proteins located on the surface of the virus

B. The protein that makes up the capsid of the virus

C. Reverse transcriptase, an enzyme encoded by the virus

D. The phospholipids found on the envelope of the virus

Answer Explanation:

The AIDS virus (HIV virus) is Class VI or Group VI in the Baltimore Classification. Based on the Baltimore Classification, this specifically involves the reverse transcriptase enzyme and it is encoded by the virus.

This is the enzyme that will produce DNA molecules from the viral RNA. This is the mechanism involved where it will use the reverse transcriptase enzyme to insert viral DNA into the genetic material of the host cell.

Therefore, the Correct Answer is C.

More Questions on TEAS 7 Science

  • Q #1: Nitrogen gas is an extremely stable molecule because its structure contains which of the following?

    A. Resonance bonds

    B. Ionic bonds

    C. Triple covalent bonds

    D. Hydrogen bonds

    Answer Explanation

    The Lewis structure of nitrogen gas is shown below. Triple bonds are stronger than double or single bonds. Therefore, we can infer that the triple bond between the two nitrogen atoms makes nitrogen gas more stable.

  • Q #2: Which of these processes occurs in the mouth?

    A. Ingestion

    B. Mechanical digestion

    C. Chemical digestion

    D. All of the above

    Answer Explanation

    All of these processes occur in the mouth. Ingestion involves taking food into the mouth, mechanical digestion includes the physical breakdown of food through chewing and mixing with saliva, and chemical digestion begins with the action of enzymes in saliva, such as amylase, which starts to break down carbohydrates.

  • Q #3: An atom of lead (Pb) has a mass number of 207. How many neutrons are in the nucleus?

    A. 130

    B. 125

    C. 82

    D. 84

    Answer Explanation

    We find the number of neutrons as follows

    Number of neutrons

                = mass number – atomic number

                = 207 – 82

                = 125 neutrons