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saw5 [17]
3 years ago
9

A viral infection in which the virus can remain latent in the host cell for weeks to years is described as a(n) _________ infect

ion, and may lead to cancer if host DNA is altered.
Biology
1 answer:
alina1380 [7]3 years ago
4 0

Answer:

Persistent infection

Explanation:

Viruses have the ability to remain in the host cell in a dormant state or latent state for many years. This is a part of reproduction cycle of a virus called the lysogenic cycle. This causes persistent infection in the host because the virus remains in the host cell and cause infection.

This persistent infection can cause cancer in the host because the viral genome incorporates in the host cell genome and take controls on its gene machinery. For example, HPV can incorporate itself in the host genome and can alter it which can cause cancer. So the correct answer is persistent infection.

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Which of the following is an example of specific immunity?
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Specific immunity is also called as Acquired immunity and Anamnestic response

2nd option is the answer

lymphocytes recognize, attack and destroy a pathogen

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Which type of neuron stimulates muscles to contract?
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Answer:

Motor neurons stimulate muscles to contract answer:C

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Contrary to the common opinion that most of the elderly are infirm, only about _____ percent of the over-64 population live in n
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Answer: 5 percent.

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3 years ago
You will be given a stock solution of BSA protein with a BSA concentration of 2 mg/mL. (Here, "mgimL is the concentration unit i
Korolek [52]

Answer: See table in the attachment.

Explanation: 1) The <u>Parallel</u> <u>Dilution</u> <u>Formula</u> is given by:

C_{1}.V_{1} = C_{2}.V_{2}

where:

C is concentration

V is volume

For the given data, index 1 represents the concentration and volume of the stock solution and index 2, the concentration and volume of the diluted solution. For example:

V_{1} = \frac{C_{2}.V_{2}}{C_{1}}

V_{1} = \frac{0.1.100}{2} = 5 μL

V_{1} = \frac{0.15.100}{2} = 7.5 μL

V_{1} = \frac{0.2.100}{2} = 10 μL

V_{1} = \frac{0.25.100}{2} = 12.5 μL

V_{1} = \frac{0.3.100}{2} = 15 μL

V_{1} = \frac{0.35.100}{2} = 17.5 μL

V_{1} = \frac{0.2.100}{2} = 20 μL

Substituing C_{2} for each concentration given and find each volume needed.

The results is shown in the second column of the table.

2) The stock and total volume is determined. So, to calculate the volume of water needed:

V_{H_{2}O} = V_{total} - V_{stock}

V_{H_{2}O} = 100 - 5 = 95 μL

V_{H_{2}O} = 100 - 7.5 = 92.5 μL

V_{H_{2}O} = 100 - 10 = 90 μL

V_{H_{2}O} = 100 - 12.5 = 87.5 μL

V_{H_{2}O} = 100 - 15 = 85 μL

V_{H_{2}O} = 100 - 17.5 82.5 μL

V_{H_{2}O} = 100 - 20 = 80 μL

As shown in the third column of the table.

Download pdf
3 0
3 years ago
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