Transport in Plants Test

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Transport in Plants Test - 81

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Question 1
1 / -0

Read the passage and answer the following question:
In order to prevent changes in volume due to osmosis, unfertilized frog oocytes were bathed in Frog Ringer's solution. The Frog Ringer's solution was supplemented with radiolabeled amino acids. A sample of oocytes was taken every 30 minutes and assayed for radioactivity. At time 1, the oocytes were fertilized. Samples were taken at five-minute intervals after fertilization and assayed for radioactivity. The results are presented in the graph.
Based on the results of the experiment, the process most likely occurring after fertilization is

A
DNA replication.

B
Organogenesis.

C
Protein synthesis.

D
RNA transcription.

E
Endocytosis.
Question 2
1 / -0

Compare the statement A and B.
Statement A : To counteract the increase in turgor pressure in plant cells, the cell wall produces an equal and opposite pressure, ie, wall pressure.
Statement B : When plant cells undergo endosmosis, they swell but do not burst.
Select the correct description.

A
Statement A is wrong and B is correct

B
Both the statement A and B are correct and A is not the reason for B

C
Both the statement A and B are correct and A is the reason for B

D
Statement A is correct and B is wrong

Solution

In a hypoosmotic solution, a plant cell will take up water and become turgid. As long as the plasma membrane is intact, the turgor pressure of a cell is equal and opposite to the wall pressure. This is the main source of structural support in herbaceous plants. Plant cells are enclosed by a rigid cell wall. When the plant cell is placed in a hypotonic solution, it takes up water by osmosis and starts to swell, but the cell wall prevents it from bursting. The plant cell is said to have become turgid i.e., swollen and hard. The pressure inside the cell rises until this internal pressure is equal to the pressure outside. This liquid or hydrostatic pressure called as the turgor pressure prevents further net intake of water.
Question 3
1 / -0

Read the passage and answer the following question:
In order to prevent changes in volume due to osmosis, unfertilized frog oocyte was bathed in Frog Ringer's solution. The Frog Ringer's solution was supplemented with radiolabeled amino acids. A sample of oocytes was taken every 30 minutes and assayed for radioactivity. At time 1, the oocytes were fertilized. Samples were taken at five-minute intervals after fertilization and assayed for radioactivity. The results are presented in the graph.
How could a control be created for this experiment?

A
Immerse the oocytes in non-radiolabeled Frog Ringer's solution prior to fertilization.

B
Sample the oocytes at five-minute intervals prior to fertilization.

C
Sample the oocytes at 30-minute intervals after fertilization.

D
Divide the oocytes into two groups, immerse one group in non-radiolabeled Frog Ringer's solution and one group in radiolabeled solution.

E
Divide the oocytes into two groups, fertilize one group, and simulate fertilization in the other, but do not actually introduce sperm into the oocytes.
Question 4
1 / -0

A demonstration of chemiosmosis by Peter Mitchell allowed ................. to move down their concentration gradient by using two different ................ .

A
Potassium ions, light frequencies

B
Hydrogen ions, light frequencies

C
Potassium ions, pH solutions

D
Hydrogen ions, pH solutions

E
Potassium ions, electromagnetic frequencies

Solution

Peter Mitchell demonstrated the chemiosmotic mechanism by first immersing mitochondria in a medium at pH 8. All mitochondrial compartments equilibrate at pH 8 and since mitochondria can not set their own proton gradient using the electron transport chain, they starved. This was followed by placing the same mitochondria in an acidic medium with a much higher concentration of protons. Protons move quickly through the freely permeable outer membrane into the intermembrane space and thereby lowering the pH of this compartment. The established proton gradient across the inner mitochondrial membrane allows the flow of protons from high concentration to low concentration through specialized channels in the inner mitochondrial membrane with simultaneous ATP synthesis. Thus, the correct answer is D.
Question 5
1 / -0

What will happen if cells are placed in distilled water?

A
The cells remain unchanged.

B
Water moves from inside of the cell to the outside.

C
Water moves from outside of the cell to the inside.

D
The cells shrink.

E
The cells dissolve.
Solution
- Solute concentration is higher inside the cell as compared to that in the distilled water i.e., distilled water is hypotonic and the cell's interior is hypertonic.
- Water will move from outside to the inside of the cell through the semi-permeable cell membrane and will cause the cell to swell.
Therefore, the correct answer is option C.
Question 6
1 / -0

Read the passage and answer the following question:
In order to prevent changes in volume due to osmosis, unfertilized frog oocyte was bathed in Frog Ringer's solution. The Frog Ringer's solution was supplemented with radiolabeled amino acids. A sample of oocytes was taken every 30 minutes and assayed for radioactivity. At time 1, the oocytes were fertilized. Samples were taken at five-minute intervals after fertilization and assayed for radioactivity. The results are presented in the graph.
The radioactivity measured after fertilization was taken up by

A
Being incorporated into DNA.

B
Being incorporated into proteins.

C
Attaching to the cell membrane.

D
Endocytosis.

E
Associating with carbohydrates.

Solution

Amino acids are monomer units that join together by peptide linkage during protein synthesis. In the given experiment, radiolabelled amino acid was taken up by zygote which would have been incorporated in proteins. DNA and carbohydrates are polymer of nucleotides and simple sugars respectively; none of these monomers have amino acids.
Therefore, the correct answer is option B.
Question 7
1 / -0

The process responsible for the movement of water from soil into the roots of plant is

A
Evaporation of water from the leaves.

B
Capillary action.

C
Both A and B.

D
Osmosis.

E
Active transport.

Solution

When soil moisture is high, movement of water and dissolved minerals from soil into root hair and finally to the root xylem cells build up root pressure which in turn pushes the water up the xylem vessels. The increased soil moisture and the high root pressure causes increased rate of water absorption which in turn forces water droplets to come out from stomata and hydathodes. Lenticular and stomatal transpiration creates transpiration pull which in turn carry the water and minerals up in stem thereby decreasing root pressure.
Therefore, the correct answer is option A.
Question 8
1 / -0

Read the passage and answer the following question.
Wet mounts of three living samples of elodea cells are prepared for viewing under the light microscope. Each slide is mounted with a different solution and viewed after 5 minutes.
Sample A - Elodea + 5 drops of 10% NaCl
Sample B - Elodea + 5 drops of isotonic saline
Sample C - Elodea + 5 drops of distilled water
What would happen to the cells in sample A?

A
Exhibit turgor pressure

B
Undergo lysis

C
Swell and burst

D
Exhibit plasmolysis

E
Remain unchanged

Solution

Plasmolysis is the process in which cells lose water in a hypertonic solution. It is cell shrinking due to loss of water from a cell. Water leaves the cell because the elodea is in a hypertonic environment with a lower concentration of water outside the cell than inside the cell.
Therefore, the correct answer is option D.
Question 9
1 / -0

Cell A and cell B are adjacent plant cells. In cell A, $${ \psi }_{ S }=-20$$ bars and $${ \psi }_{ P }=8$$ bars. In cell 'B', $${ \psi }_{ S }=-12$$ bars and $${ \psi }_{ P }=2$$. Then,

A
Water moves from cell B to cell A.

B
Equal amount of water is simultaneously exchanged between cell A and cell B.

C
Water moves from cell A to cell B.

D
There is no movement of water between cell A and cell B.

Solution

For cell A, Ψs = -20 bars and Ψp = 8 bars
Water potential Ψw = Ψs + Ψp = -20 + 8 = -12 bars
For cell B, Ψs = -12 bars and Ψp = 2 bars
Water potential Ψw = Ψs + Ψp = -12 + 2 = -10 bars
Ψw for cell A < Ψw for cell B
Since water moves from a region of higher water potential to a region of low water potential, therefore, water will move from cell B to cell A.
(Note: The more negative is the value, the lower is the water potential.)
So, the correct answer is A.
Question 10
1 / -0

An animal cell when placed in a 0.9% solute solution, the cell will

A
Remain unchanged.

B
Swell and burst.

C
Shrivel.

D
Swell and divide.

E
Release solute by exocytosis.

Solution

Animal cells do not have a cell wall, but they do have a plasma membrane. The plasma membrane is semi-permeable and performs an important task of keeping the cell alive by maintaining an isotonic environment. The cells will not gain or lose water if placed in isotonic solution. The normal saline solution which is 0.9% NaCl is isotonic with blood. Hence, the animal cells remain unchanged.

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Transport in Plants Test - 81

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Transport in Plants Test - 81

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SHARING IS CARING

Question 1

DNA replication.

Organogenesis.

Protein synthesis.

RNA transcription.

Endocytosis.

Question 2

Statement A is wrong and B is correct

Both the statement A and B are correct and A is not the reason for B

Both the statement A and B are correct and A is the reason for B

Statement A is correct and B is wrong

Solution

Question 3

Immerse the oocytes in non-radiolabeled Frog Ringer's solution prior to fertilization.

Sample the oocytes at five-minute intervals prior to fertilization.

Sample the oocytes at 30-minute intervals after fertilization.

Divide the oocytes into two groups, immerse one group in non-radiolabeled Frog Ringer's solution and one group in radiolabeled solution.

Divide the oocytes into two groups, fertilize one group, and simulate fertilization in the other, but do not actually introduce sperm into the oocytes.

Question 4

Potassium ions, light frequencies

Hydrogen ions, light frequencies

Potassium ions, pH solutions

Hydrogen ions, pH solutions

Potassium ions, electromagnetic frequencies

Solution

Question 5

The cells remain unchanged.

Water moves from inside of the cell to the outside.

Water moves from outside of the cell to the inside.

The cells shrink.

The cells dissolve.

Solution

Question 6

Being incorporated into DNA.

Being incorporated into proteins.

Attaching to the cell membrane.

Endocytosis.

Associating with carbohydrates.

Solution

Question 7

Evaporation of water from the leaves.

Capillary action.

Both A and B.

Osmosis.

Active transport.

Solution

Question 8

Exhibit turgor pressure

Undergo lysis

Swell and burst

Exhibit plasmolysis

Remain unchanged

Solution

Question 9

Water moves from cell B to cell A.

Equal amount of water is simultaneously exchanged between cell A and cell B.

Water moves from cell A to cell B.

There is no movement of water between cell A and cell B.

Solution

Question 10

Remain unchanged.

Swell and burst.

Shrivel.

Swell and divide.

Release solute by exocytosis.

Solution

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