Prepare for NEET Biology Transport in Plants with MCQs & PYQs on NEET.GUIDE. Access free practice, previous year questions, and expert guidance to understand xylem, phloem, and movement of water.
Two identical plant cells are placed in separate solutions. Cell A is placed in a 1M sucrose solution, and Cell B is placed in a 1M NaCl solution. Which cell will experience greater plasmolysis, assuming ideal conditions and complete dissociation of NaCl?
Cell A
Cell B
Both cells will plasmolyze equally
Neither cell will plasmolyze
A plant cell is placed in a solution, and its volume initially increases. After a while, the volume increase stops. Which of the following BEST explains why the volume stopped increasing?
The cell membrane reached its maximum stretching capacity.
All available water in the solution was absorbed by the cell.
The cell wall became impermeable to water.
Turgor pressure reached equilibrium with the external solution's osmotic pressure.
Certain integral membrane proteins facilitate the diffusion of molecules across the membrane without directly binding to the transported molecule. These proteins create a continuous aqueous pathway through the lipid bilayer. Which of the following is the BEST example of such facilitated diffusion?
GLUT transporters moving glucose
Sodium-potassium pump maintaining ion gradients
Aquaporins transporting water
Chloride channels gated by voltage
Which of the following statements regarding facilitated diffusion is INCORRECT?
It requires transport proteins.
It is a passive process and does not require energy.
It can move molecules against their concentration gradient.
Transport proteins involved can become saturated.
Imagine a plant cell undergoing plasmolysis. If the external solution's water potential is further decreased, which of the following is the MOST likely consequence?
Deplasmolysis and return to a turgid state.
Incipient plasmolysis with the cell membrane just touching the cell wall.
No further change in the cell's state.
Increased plasmolysis with further shrinkage of the protoplast.
A dry wooden block and a dry gelatine block of equal mass are placed in water. The gelatine block imbibes considerably more water than the wooden block. This difference is primarily attributed to:
The higher density of wood compared to gelatine.
The greater number of hydrophilic colloids in gelatine compared to wood.
The presence of lignin in wood, which repels water.
The lower osmotic potential of gelatine compared to wood.
Seeds of certain desert plants exhibit delayed germination due to the presence of inhibitors. How does imbibition contribute to overcoming this dormancy?
Imbibition increases the seed's internal temperature, denaturing the inhibitors.
Imbibition activates enzymes within the seed that break down the inhibitors.
Imbibition allows water to enter the seed and leach out the inhibitors, promoting germination.
Imbibition increases the seed's turgor pressure, rupturing the seed coat and releasing the inhibitors.
If dry wooden blocks are placed in a solution with a solute potential of -0.5 MPa, and the initial water potential of the wood is -2.0 MPa, which of the following will occur?
Water will move from the solution into the wooden blocks.
Water will move from the wooden blocks into the solution.
No net movement of water will occur.
The wooden blocks will dissolve in the solution.
Which of the following scenarios would result in the LEAST efficient water absorption by a plant's roots?
Slightly acidic soil with optimal nutrient availability
Well-aerated soil with moderate water content
Low soil temperature coupled with high humidity
High soil salinity coupled with low soil oxygen levels
A plant with a mutation that disables aquaporins in its root cells would likely exhibit:
Complete cessation of water uptake.
Enhanced water uptake due to increased reliance on the apoplast pathway.
Reduced rate of water uptake but not complete cessation.
No change in water uptake as aquaporins are primarily involved in transpiration.
