Consider solutions A, B, C, D and E: A - 150 mM NaCl B - 100 mM glucose + 100 mM

Question

Consider solutions A, B, C, D and E: A - 150 mM NaCl B - 100 mM glucose + 100 mM NaCl C - 100 mM durg X (a small non-polar molecule) + 150 mM NaCl D - 150 mM MgCl_2 E - 300 mM fructose Based on the background information above, red blood cells placed in solution A A lose water and have a crenated appearance B take in water but still remain intact C lose water and later burst D take in water and burst E remain a normal shape and not have any change in cell volume Based on the above background information, which of the following statements is regarding the comparison of the tonicity of solutions B, C, D and E with solution A? with solution A. A. all of the other solutions listed are isotonic B solutions B, C and E are hypertonic and solution D is isotonic C Solution D is hypertonic and all of the other solutions are isotonic D all of the other solutions listed are hypotonic E Solutions C and D are hypertonic and solutions B and E are isotonic

Explanation / Answer

Answer:

1)The red blood cell placed in solution A would remain a normal shape and not have any change in cell volume because they will neither gain and did not loss anything because they are in isotonic condition.

2) All the other solution listed are isotonic.

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