Question

A red blood cell placed in a hypertonic solution will shrink in a process called crenation. A red blood cell placed in a hypotonic solution will swell and potentially burst in a process called hemolysis.To prevent crenation or hemolysis, a cell must be placed in an isotonic solution such as 0.9% (m/v) NaCl or 5.0% (m/v) glucose. This does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes.A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur.Solution
A: 3.21% (m/v) NaClSolution
B: 1.65% (m/v) glucoseSolution
C: distilled H2OSolution
D: 6.97% (m/v) glucoseSolution
E: 5.0% (m/v) glucose and 0.9%(m/v) NaClDrag each solution to the appropriate bin.

Answer 1

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)