Cell: The Unit of LifeHard
Question
If in a cell water potential is -30 bar and osmotic potential is -72 bar then what is the value of the turgidity developed in the form of TP in the cell?
Options
A.102 bar
B.-102 bar
C.-42 bar
D.42 bar
Solution
ΨW = ΨS + ΨP [NCERT (E), (H) Pg. # 180]
Water potential (ΨW) = - 30 bar
Osmotic potential (ΨS) = - 72 bar ΨP = ΨW - ΨS
ΨP = - 30 - (-72) = - 30 + 72 = 42 bar
ΨP = TP s o TP = 42 bar
Water potential (ΨW) = - 30 bar
Osmotic potential (ΨS) = - 72 bar ΨP = ΨW - ΨS
ΨP = - 30 - (-72) = - 30 + 72 = 42 bar
ΨP = TP s o TP = 42 bar
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