Question

You now are given a normal sample of muscle tissue which contains many individual cells. The tissue is minced and digested with enzymes to liberate individual muscle cells. You are given a solution of Dextrose in half-normal saline (NaCl) which you are told is isosmotic but you are not told the concentration of Dextrose or NaCl but they are both present in the solution. In this case Dextrose is permeable to the cell. You place the isolated muscle cells in this solution. Assume all of the solutes inside the cell are not permeable to the cell membrane. What happens to the cells when you place them in this solution and it has reached equilibrium? (5 Points) a. The cells remain the same size b. The cells have increased in size; they swell c. The cells have decreased in size; they shrink d. There is not enough information given in the problem to determine what happens

Answer 1

• The solution of dextrose and sodium chloride is isosmotic means that there are 5% dextrose and 0.9% NaCl.
• This 0.9% NaCl is isosmotic with the muscle cell so that NaCl movement will not occur, but the dextrose content is more so that the glucose from the dextrose enters the cell and the movement of the solute to the cell is followed by the movement of the water from the exterior so that the volume of the cell increase and it swells up.
• But the glucose in the cell undergoes an aerobic respiration process so that CO2 and water are released from the glucose. So due to decrease in solute in the cell and the water make it hypotonic result in the diffusion of water out of it so that cell remains in the same size. So the answer is option A.
• And option B is wrong due to the above reason.
• And option C is wrong as the solution is hypertonic to the cell so that solute movement into it followed the water of the surrounding area make it hypotonic.