Diffusion and Osmosis + Water Potential,, How Plants life...........The differences...???

Diffusion, Osmosis & Water Potential Quiz
 

1. What is the water potential (Ψ_w) of a beaker of pure water _0 Mpa_. Pressure potential?  _0 Mpa_ Solute potential? _Negative of P_s value (- X Mpa)_.
    
2. Calculate the osmotic potential (Ψ_s) of a 2.4 molal sucrose solution at 24^oC. Show your work.

As known ; Molality of sucrose (m)      = 2,4 molal

                   Temperature (T)                 = 24ºC = 24 + 273 = 297ºK

                   Ionization Constanta (i)    = 1,0 for non-electrolit

                   Gases Constanta (R)        = 0,0083 ltr Mpa/mol

Asked       ; Osmotic potential (Ψ_s) . . . . . .?

Solving      ; Ψ_s = -m . i . R . T

                        = -(2,4 molal) . (1,0) . (0,0083 ltr Mpa/mol) . (297ºK)

                        = -5,91624 Mpa           

3. Calculate the water potential (Ψ_w) of a 2.4 molal sucrose solution at 24^oC. Show your work.
   As known ; Molality of sucrose (m)      = 2,4 molal

                   Temperature (T)                 = 24ºC = 24 + 273 = 297ºK

                   Ionization Constanta (i)    = 1,0 for non-electrolit

                   Gas Constanta (R)            = 0,0083 ltr Mpa/mol

Asked       ; Water potential (Ψ_w) . . . . . .?

Solving                 ; Ψ_w  = Ψ_s + Ψ_p

                                                = (-m . i . R . T) + 0 (unknown)

                        = (-(2,4 molal) . (1,0) . (0,0083 ltr Mpa/mol) . (297ºK)) + 0

                        = -5,91624 Mpa

4. Assume that a plant cell with a water potential of -1.0 MPa is placed in a beaker containing a sucrose solution that has a water potential of -4.0 MPa. Further, assume the temperature of the solution is 24 C. Thus:

a.	The plant cell will become:	a. larger     b. smaller   c. not change
b.	The weight of the plant cell will:	a. increase   b. decrease   c. not change
c.	The concentration of the sucrose solution in the beaker will:	a. increase   b. decrease  c. not change
d.	The turgidity of the plant cell will:	a. increase   b. decrease  c. not change
e.	The osmotic potential of the sucrose solution will become:	a. more negative  b.  less negative
f.	There will be a net movement of water from the:	a. cell to the solution  b. solution to the cell
g.	After a few hours the cell is removed. A drop of sucrose (-4.0 MPa) placed in the solution will	a. float    b. sink    c. hover & disperse
h.	The refractive index of the sucrose solution will:	a. increase  b. decrease  c. not change
i.	The cell will likely plasmolyze:	a. false  b. true

5. Calculate the pressure (MPa) in a cell if Ψ_w = -1.7 MPa and Ψ_s = -4.2 MPa

As known         ; Water potential (Ψ_w) = -1,7 MPa

  Osmotic potential (Ψ_s) = -4.2 MPa

            Asked              ; Ψ_p . . . . . .?

            Solved             ; Ψ_w  = Ψ_p + Ψ_s

                                                  Ψ_p  = Ψ_w – Ψ_s

                                                                    = -1,7 MPa – (-4,2 Mpa)

                                            = 2,5 MPa

6. If Ψ_s for a tissue is -0.80 MPa, what do you predict the freezing point depression of expressed sap will be? Show your work.

As known         ; Osmotic potential of tissue (Ψ_{s tissue})                     = -0.80 MPa

                          Osmotic potential of 1 molal solution (Ψ_{s solution)}  = -2.27 MPa

                          Lower freezing point (fp₂)                                   = 1.86ºC

Asked              ; fp₁. . . . . ?

Solved             ; Ψ_{s tissue} / fp₁=  Ψ_s2 / fp₂

   (-0.80 Mpa)/ fp1         = -2.27MPa/ 1.86 ^oC

            fp1        = (-2.27MPa/ 1.86 ^oC) . (0.80 MPa)

                                                            fp1       = (-4,22 MPa/^oC) . (0.80 MPa)

            fp1       = -3.38ºC

Osmometer Question:  The diagram represents a simple osmometer.  The membrane is freely permeable to water by not solute (sucrose).  The membrane is inelastic.  Inside the membrane is a solution of 0.1 molar sucrose. 

There will be a net movement of water _a_the membrane sac.         a.  into        b.  out of The water level in the tube will: _b_         a.  fall        b.  rise        c.  not change The pressure in the sac will:_b_         a.  decrease    b. increase        c.  not change Sketch a graph plotting water potential in the sac vs. time The graph of condition : water potential in the sac vs. the time           Water potential                                                                                                                                                                                time   Sketch a graph plotting osmotic potential in the sac vs. time -    The graph plotting of osmotic potential in the sac vs. time     Ψs   Sketch a graph plotting pressure in the sac vs. time. Graph plotting the pressure in the sac vs. time        Ψp

Some Multiple Choice Questions: (circle all answers that apply)

1. A cell whose internal osmotic concentration is 0.3 osmoles/liter is placed in a solution that is 0.5 osmoles/liter.  The solution is: b.

a.  Isoosmotic to the cell               d.  Isotonic to the cell
b.  Hypoosmotic to the cell           e.  Hypertonic to the cell
c.  Hyperosmotic to the cell      f.  Hypotonic to the cell

2.   A cell is placed in a solution and swells.  The solution is: c

a.  Isoosmotic to the cell               d.  Isotonic to the cell
b.  Hypoosmotic to the cell           e.  Hypertonic to the cell
c.  Hyperosmotic to the cell      f.  Hypotonic to the cell

3.  A cell is placed in a solution and the cell swells.  The solution is probably, but not necessarily: c
        a.  isoosmotic to the cell
        b.  hypoosmotic to the cell
        c.  hyperosmotic to the cell.

Short Answer Questions:   Using water potential terminology, explain why:

1. Over-fertilizing your lawn will cause it to “burn” (the leaves wilt, become dry and fall off).

yes, it could be. Over-fertilizing could cause  it to “burn”. it will be happen when we give in to excessive fertilizer plant or over-fertilizing, known as that the fertilizer have higher concentration than the ground. When fertilizer absorbed into the ground, there will be differences in the contrast between the internal of roots with the external environment. This triggered the osmosis of the soil hypertonic environment into hypotonic of the roots (the water molecules of roots would be through out to the ground then the molecules of fertilizer would be absorbed fastly by roots to the all part of plant). The higher concentration of fertilizer will cause dehydrated of tissue in these plants, decreased the photosynthesis and other metabolic processes in the body of these plants. The leaves will wilt, become dry then it will be fall off

2. Strawberries become juicy when you add sugar to them.

The causes of juicy condition of strawberries is the difference concentration of the two. The strawberries as the lower concentration then the sugars as higher concentration. When they put into one place, the waters molecules will be moved to the sugars that from the hypotonic to hypertonic. Then the molecules of sugars will be moved in to the strawberries that from the hypertonic to the hypotonic. It also make the strawberries sweet.

3. Cooks put wilted lettuce in cold water.

When we do this, the lettuce will be more fresh than before. Means, this is the osmotic process when the water molecules in low temperature moved to the internal tissue of lettuce and fill them. The waters molecules will moved from the low concentration to the higher concentration of lettuce tissue.

Sequence Question:  Number the following tissues in sequence from the one with the highest water potential (least negative = 1) to the lowest (most negative = 9) water potential.

9	air (50% RH)	1	soil (saturated)
4	Endodermis	3	Cortex
5	Pericycle	8	palisade cell
6	root xylem	2	root hair
7	leaf xylem