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Cell Biology - Viruses (HL)

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Viruses are non-living infectious agents that lack a metabolism and cannot reproduce independently, requiring a host cell's machinery.

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The basic structure of a virus consists of a nucleic acid core (genetic material) surrounded by a protein coat called a capsid. Some viruses also possess an external lipid envelope derived from the host cell membrane.

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Viral genomes are highly diverse; they can consist of either DNA or RNA, which may be single-stranded (ssss) or double-stranded (dsds).

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The Lytic Cycle involves the immediate replication of the virus within the host, leading to the lysis (bursting) of the cell and release of new virions. The stages are: Attachment, Penetration, Biosynthesis, Maturation, and Release.

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The Lysogenic Cycle involves the integration of the viral DNA into the host genome, where it is referred to as a prophage. The viral DNA is replicated along with the host DNA without destroying the cell, until an environmental trigger induces the lytic cycle.

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Retroviruses, such as HIV, contain ssRNAssRNA and use the enzyme reverse transcriptase to synthesize DNADNA from their RNARNA template, which then integrates into the host's chromosomes.

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There are three main hypotheses for the origin of viruses: 1. The Regressive (Reduction) Hypothesis (evolved from free-living cells), 2. The Progressive (Escape) Hypothesis (evolved from mobile genetic elements like transposons), and 3. The Virus-First Hypothesis (evolved before or alongside the first cells).

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Viruses show high rates of mutation, especially RNARNA viruses, because RNARNA polymerase lacks the proofreading mechanisms found in DNADNA polymerases.

📐Formulae

Magnification=Size of ImageActual Size of ObjectMagnification = \frac{\text{Size of Image}}{\text{Actual Size of Object}}

A=IMA = \frac{I}{M}

1 nm=10−3 μm=10−6 mm=10−9 m1\text{ nm} = 10^{-3}\text{ }\mu\text{m} = 10^{-6}\text{ mm} = 10^{-9}\text{ m}

💡Examples

Problem 1:

A transmission electron micrograph (TEM) shows a spherical virus with a diameter of 25 mm25\text{ mm}. If the magnification of the image is ×250,000\times 250,000, calculate the actual diameter of the virus in nanometers (nmnm).

Solution:

Using the formula A=IMA = \frac{I}{M}: A=25 mm250,000A = \frac{25\text{ mm}}{250,000} A=0.0001 mmA = 0.0001\text{ mm} To convert mmmm to nmnm, multiply by 10610^6: 0.0001×1,000,000=100 nm0.0001 \times 1,000,000 = 100\text{ nm}

Explanation:

The actual size is determined by dividing the measured image size by the magnification factor, then converting units to the appropriate scale (viruses are typically measured in nmnm).

Problem 2:

Contrast the genetic material of the T4T_4 bacteriophage and the Human Immunodeficiency Virus (HIV).

Solution:

The T4T_4 bacteriophage contains double-stranded DNA (dsDNAdsDNA), whereas HIV contains single-stranded RNA (ssRNAssRNA).

Explanation:

Viruses are classified based on their nucleic acid type. Bacteriophages often use DNA, while many animal viruses like retroviruses use RNA and require reverse transcription for integration.

Problem 3:

Explain the role of reverse transcriptase in the life cycle of a retrovirus.

Solution:

Reverse transcriptase is an enzyme that catalyzes the reaction: RNA→DNARNA \rightarrow DNA. This allows the viral RNARNA genome to be converted into a DNADNA sequence that can be integrated into the host cell's DNADNA by the enzyme integrase.

Explanation:

In retroviruses, the flow of genetic information is reversed from the standard biological dogma (DNA→RNA→ProteinDNA \rightarrow RNA \rightarrow Protein), allowing the virus to become a permanent part of the host's genetic makeup.