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Alexxx [7]
3 years ago
12

What factors do you predict might influence the likelihood of mutations occurring in viruses?

Biology
1 answer:
Neporo4naja [7]3 years ago
3 0

The factors which might influence the likelihood of mutations occurring in

viruses is Spontaneous nucleic acid damage and replication inhibition.

Viruses are pathogenic micro-organisms which mutate rapidly than the

humans do. Its ability to mutate very rapidly inside the host cells is what

makes it unique and deadly.

Factors which might influence mutations in Virus include nucleic acid

damage and replication inhibition. When this occurs, the virus isn't able to

mutate thereby causing less damage to the host.

Read more on brainly.com/question/2472845

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A red blood cell placed in a hypertonic solution will shrink in a process called crenation. A red blood cell placed in a hypoton
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Answer:

The correct answers are A. "crenation", B. "hemolysis", C. "hemolysis", D. "crenation" and E. "neither will occur".

Explanation:

0.9% (m/v) NaCl or 5.0% (m/v) glucose are isotonic solutions at which the cells will not suffer any harmful consequence. A solution with a higher concentration than the isotonic conditions would result in the cells crenation, while a solution with a lower concentration would result in the cells hemolysis. Therefore the consequences of putting the red cells to the solutions stated in the question are as following:

A: 3.21% (m/v) NaCl Solution  = crenation (higher than 0.9% (m/v) NaCl)

B: 1.65% (m/v) glucose Solution  = hemolysis (lower than 5.0% (m/v) glucose)

C: distilled H2O Solution  = hemolysis (lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose)

D: 6.97% (m/v) glucose Solution  = crenation (higher than 5.0% (m/v) glucose)

E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl = neither will occur (equal to 5.0% (m/v) glucose and 0.9% (m/v) NaCl)

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