Question 1

The diagram shows a test-tube containing clear jelly. A drop of blue ink is injected into the middle of the jelly.The blue colour of the ink spreads throughout the jelly.By which process does the blue ink spread through the jelly?

Accepted Answer

. The blue ink spreads through the jelly by diffusion.. Diffusion is the process by which molecules move from a region of higher concentration to a region of lower concentration.. In this diagram, when a drop of blue ink is added to the clear jelly, it occupies a particular position inside the jelly.. From there, every other point in the jelly is at a lower concentration of blue ink than the initial position occupied by it. So it starts moving to other points in the jelly, i.e, it diffuses throughout the jelly.. This diffusion continues till every point in the jelly is at a uniform concentration of the blue ink, hence turning the whole jelly blue.Therefore, the blue ink spreads through the jelly through diffusion.

Question 2

The apparatus shown was set up.Some hours later, the water in the beaker had turned blue, and the liquid in the glass tube had moved upwards.Which processes caused these changes?

Accepted Answer

Water turned blue - diffusion ; liquid in glass tube moved upwards - osmosis. Diffusion is the process by which molecules move from a region of higher concentration to a region of lower concentration. This diffusion continues till every point in the water is at a uniform concentration of the blue ink, hence turning the whole water blue.The concentration of water in the beaker is higher than the concentration of water in the glass tube. By the definition of osmosis, water always moves from an area of higher water concentration to lower water concentration through a semi-permeable membrane. Also, since the solvent molecules inside the tubing are less, water molecules from outside will rush in, leading to a rise in the water level. Therefore, liquid in glass tube moved upwards by the process of osmosis.

Question 3

The diagram shows a test-tube containing clear jelly. A drop of blue ink is injected into the middle of the jelly. The blue colour of the ink spreads throughout the jelly.By which process does the blue ink spread through the jelly?

Accepted Answer

Diffusion is the process by which molecules move from a region of higher concentration to a region of lower concentration.
In the given diagram, the point where blue ink is injected has higher concentration of blue ink as compared to all the other parts of the clear jelly. So the ink diffuses throughout the jelly and it continues till all the points in the jelly are at uniform concentration of blue ink.

Question 4

Which process describes osmosis?

Accepted Answer

Osmosis is the process where water molecules move across a partially permeable membrane from an area of lower solute concentration to an area of higher solute concentration. This movement of water helps balance the concentration of solutes on both sides of the membrane. It's a vital mechanism in cells and biological systems to regulate water levels and maintain proper function. Hence, the correct option is B.

Question 5

Which row describes active transport?

Accepted Answer

Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration using energy from respiration. Water crosses the cell membrane by passive transport.

Question 6

Which process uses energy released by respiration?

Accepted Answer

The energy released during respiration is stored in the form of Adenosine Triphosphate (ATP) and is used for carrying out various life processes needed for growth and development.The energy produced through respiration is used to support the formation of new cells, tissues, and structures during an organism's growth.Hence, the correct option is C.

Question 7

The cholera bacterium produces toxins that cause chloride ions to be secreted into the small intestine. How does this affect the water potential of blood in the intestinal capillaries and the intestinal contents?

Accepted Answer

The cholera bacterium produces toxins that cause chloride ions from blood capillaries to be secreted into the small intestine. As a result the water potential inside the intestine is reduced (solute potential high) as compared to that in blood capillary (solute potential low). If it enters the small intestine it can cause illness in the following way:- Bacteria attach to the wall of the small intestine.- They produce a toxin.- The toxin stimulates the cells lining the intestine to release chloride ions from inside the cells into the lumen of the intestine.- The chloride ions accumulate in the lumen of the small intestine and lower the water potential there.- Once the water potential is lower than that of the cells lining the intestine, water starts to move out of the cells into the intestine (by osmosis).- Large quantities of water are lost from the body in watery faeces.- The blood contains too little chloride ions and water.Therefore, the correct answer is water potential in the blood capillary is raised and water potential in the small intestine is lowered.

Question 8

Which process describes osmosis?

Accepted Answer

Osmosis is a process of movement of solvent molecules (like water) through a semi-permeable membrane from a region of lower solute concentration to higher solute concentration.

Question 9

Which row describes active transport?

Accepted Answer

Active Transport is defined as a process that involves the movement of molecules (such as ions) from a region of lower concentration to a region of higher concentration across a membrane against a gradient or an obstacle with the use of external energy. During the process of respiration, oxidation of high energy food substances takes place in the cells of the living organisms to produce energy. The energy released in the process is in the form of the chemical substance called ATP. This energy is used in active transport to facilitate movement of molecules against concentration gradient.Thus, row D describes active transport correctly.

Question 10

Fig. 1 shows a diagram of a palisade mesophyll cell before it has been placed in a concentrated sugar solution.Fig. 2 shows the same palisade mesophyll cell after it has been placed in a concentrated sugar solution for twenty minutes.(i) Describe the changes that have taken place in the cell between Fig.1 and Fig. 2. ............................................................................................................................................................................................................................................................................ (ii) Explain why the cell in Fig. 2 has changed. ............................................................................................................................................................................................................................................................................. (iii) Suggest how the cell in Fig. 2 could be treated so that it returned to its original appearance in Fig.1. .............................................................................................................................................................................................................................................................................

Accepted Answer

(i) Fig. 1 shows a diagram of a palisade mesophyll cell before it has been placed in a concentrated sugar solution. When the same cell is transferred to a concentrated sugar solution for twenty minutes (Fig. 2) the following observations are made-- Cell membrane is pulled away from cell wall. - Cell contents shrink and pull away from the cell wall.- Vacuole shrunk and became smaller.- Space between cell wall and cell membrane is filled by external solution.(ii) The sugar solution in Fig 2 is more concentrated. Plant cells have comparatively a lesser concentration of solutes in them. Now, according to osmosis the water always moves from its higher concentration to lower concentration. This means that water will move from inside of the cell towards the sugar solution via a semi-permeable membrane. This is known as exosmosis. This process is known as plasmolysis and the fully shrunken protoplast is called the plasmolyzed cell.(iii) In order to return the cell to its original from it must be kept in a dilute/less concentrated solution. By doing so, there will be movement of water molecules from outside to inside the cell. The  cell will swell up and regain its original shape and size.

Question 11

Three identical potato cylinders were used to investigate water movement in plant cells.Cranberry juice is a red fruit juice that contains natural sugars.Three test-tubes were set up as shown in figure and left for one hour.After one hour the potato cylinders were removed from test-tubes 1 to 3. The mass of each potato cylinder is recorded in the table.(i) Describe the results for test-tubes 1 and 2. test-tube 1 ......................................................................................................................... ........................................................................................................................................... test-tube 2 ......................................................................................................................... ........................................................................................................................................... (ii) Calculate the decrease in the mass of the potato cylinder in test-tube 3. ........................................................g  (iii) Explain why the potato cylinder lost mass in test-tube 3. ......................................................................................................................................  .....................................................................................................................................

Accepted Answer

(i) test-tube 1: Potato cylinder in test-tube 1 is increased in mass by 5g.    test-tube 2: Potato cylinder in test-tube 2 stayed the same.(ii) mass of the potato cylinder at the start/g in test-tube 3 = 25     mass of the potato cylinder after one hour/g in test-tube 3 = 19     The decrease in the mass of the potato cylinder in test-tube 3 = 25-19 = 6g(iii) The potato cylinder lost mass in test-tube 3 by the process of osmosis. Water moves out of the potato cylinder by osmosis because there is more water inside the potato than in the solution.

Question 12

Transmission of impulses relies on the flow of ions through the cell membranes of neurones down their concentration gradients. Active transport is responsible for maintaining the concentration gradients of ions across the membranes of neurones.Explain how ions are moved across membranes by active transport....................................................................................................................................................

Accepted Answer

Active transport is an energy-driven process where membrane proteins transport molecules across cells. In active transport this movement s achieved by acting against the formation of an equilibrium, typically by concentrating molecules depending on the various needs of the cell, e.g., ions, sugars, and amino acids. Active transport requires the use of energy (namely ATP) since it takes molecules from a lower to a higher concentration, i.e., against its concentration or electrochemical gradient. Conduction of nerve impulse occurs due to the presence of active and electronic potentials along the conductors. In particular, the sodium-potassium pump is required to maintain cell potentials and can be seen in neuronal action potentials.

Question 13

You are going to measure the distance moved by different concentrations of citric acid through agar. You are provided with a Petri dish labelled agar plate.The agar in the Petri dish contains Universal Indicator which will change colour in the presence of acid.You should use the safety equipment provided while you are carrying out the practical work.Step 1 Label three test-tubes A, B and C and place them in a test-tube rack.Step 2 Make three solutions, each containing a different concentration of citric acid, in the labelled test-tubes.            Use the volumes of 5% citric acid and distilled water shown in the table to make the solutions.Step 3 Turn the Petri dish over so the base side is up. Use a marker pen to draw three lines to divide the base into approximately equal sections. Label the sections A, B and C as shown in Fig.1.Step 4 Turn the Petri dish so the base side is down. Use a straw to cut a hole in the centre of each section of the agar in the Petri dish, as shown in Fig. 2.Hold the straw vertically and push through the agar to the bottom of the layer. As you remove the straw twist it slightly to pull out the agar. Squeeze the end of the straw  gently to push the agar you have removed onto a paper towel.Step 5 Use a pipette to transfer three drops of solution from test-tube A into the hole in the agar in section A of the Petri dish. Do not let the solution drip onto the surface of the agar.Step 6 Use a clean pipette to repeat step 5 for the solution in test-tube B and the hole in the agar in section B of the Petri dish.Step 7 Use a clean pipette to repeat step 5 for the solution in test-tube C and the hole in the agar in section C of the Petri dish. Step 8 Start the stop-clock and leave the Petri dish for five minutes.Step 9 After five minutes observe the appearance of the agar around each of the holes. Describe the appearance of the agar around the holes in A, B and C after five minutes. ...........................................................................................................................................

Accepted Answer

As the agar in petri dish contains universal indicator, it will change colour in the presence of acid. The appearance of the agar around the holes in A, B and C after five minutes will be pinkish in colour.

Question 14

You are going to measure the distance moved by different concentrations of citric acid through agar. You are provided with a Petri dish labelled agar plate.The agar in the Petri dish contains Universal Indicator which will change colour in the presence of acid.You should use the safety equipment provided while you are carrying out the practical work.Step 1 Label three test-tubes A, B and C and place them in a test-tube rack.Step 2 Make three solutions, each containing a different concentration of citric acid, in the labelled test-tubes.            Use the volumes of 5% citric acid and distilled water shown in the table to make the solutions.Step 3 Turn the Petri dish over so the base side is up. Use a marker pen to draw three lines to divide the base into approximately equal sections. Label the sections A, B and C as shown in Fig.1.Step 4 Turn the Petri dish so the base side is down. Use a straw to cut a hole in the centre of each section of the agar in the Petri dish, as shown in Fig.2. Hold the straw vertically and push through the agar to the bottom of the layer. As you remove the straw twist it slightly to pull out the agar. Squeeze the end of the straw gently to push the agar you have removed onto a paper towel. Step 5 Use a pipette to transfer three drops of solution from test-tube A into the hole in the agar in section A of the Petri dish. Do not let the solution drip onto the surface of the agar.Step 6 Use a clean pipette to repeat step 5 for the solution in test-tube B and the hole in the agar in section B of the Petri dish. Step 7 Use a clean pipette to repeat step 5 for the solution in test-tube C and the hole in the agar in section C of the Petri dish.Step 8 Start the stop-clock and leave the Petri dish for five minutes. Step 9 After five minutes observe the appearance of the agar around each of the holes.Step 10 Leave the Petri dish for a further 25 minutes. During this time, continue with the other questions.Step 11 After a total of 30 minutes use the ruler to measure the distance the citric acid has travelled from the edge of the hole in section A. You may need to use the hand lens.Step 12 Repeat step 11 for the holes in section B and section C of the Petri dish.  Describe how you decided where to measure the distance travelled by the citric acid solution............................................................................................................................................

Accepted Answer

In order to measure the distance travelled by the citric acid solution, we need to measure the distance between the edge of the hole (made from the straw in agar) to the edge of the red / yellow circle (formed because of the interaction of citric acid solution and universal indicator in agar).

Question 15

You are going to measure the distance moved by different concentrations of citric acid through agar.You are provided with a Petri dish labelled agar plate.The agar in the Petri dish contains Universal Indicator which will change colour in the presence of acid.You should use the safety equipment provided while you are carrying out the practical work.Step 1 Label three test-tubes A, B and C and place them in a test-tube rack.Step 2 Make three solutions, each containing a different concentration of citric acid, in the labelled test-tubes.Use the volumes of 5% citric acid and distilled water shown in the table to make the solutions.Step 3 Turn the Petri dish over so the base side is up. Use a marker pen to draw three lines to divide the base into approximately equal sections. Label the sections A, B and C as shown in Fig.1.Step 4 Turn the Petri dish so the base side is down. Use a straw to cut a hole in the centre of each section of the agar in the Petri dish, as shown in Fig.2.Hold the straw vertically and push through the agar to the bottom of the layer. As you remove the straw twist it slightly to pull out the agar. Squeeze the end of the straw gently to push the agar you have removed onto a paper towel.Step 5 Use a pipette to transfer three drops of solution from test-tube A into the hole in the agar in section A of the Petri dish.Do not let the solution drip onto the surface of the agar.Step 6 Use a clean pipette to repeat step 5 for the solution in test-tube B and the hole in the agar in section B of the Petri dish.Step 7 Use a clean pipette to repeat step 5 for the solution in test-tube C and the hole in the agar in section C of the Petri dish. Step 8 Start the stop-clock and leave the Petri dish for five minutes.Step 9 After five minutes observe the appearance of the agar around each of the holes.Step 10 Leave the Petri dish for a further 25 minutes. During this time, continue with the other questions.Step 11 After a total of 30 minutes use the ruler to measure the distance the citric acid has travelled from the edge of the hole in section A. You may need to use the hand lens.Step 12 Repeat step 11 for the holes in section B and section C of the Petri dish.State a conclusion for your results. ...........................................................................................................................................

Accepted Answer

In the above mentioned experiment, we can conclude that higher is the concentration of citric acid, further the citric acid solution moves which means, more is the concentration of citric acid in the solution, more distance will be covered by citric acid in the agar petri dish.

Question 16

You are going to measure the distance moved by different concentrations of citric acid through agar.You are provided with a Petri dish labelled agar plate.The agar in the Petri dish contains Universal Indicator which will change colour in the presence of acid.You should use the safety equipment provided while you are carrying out the practical work.Step 1 Label three test-tubes A, B and C and place them in a test-tube rack.Step 2 Make three solutions, each containing a different concentration of citric acid, in the labelled test-tubes.Use the volumes of 5% citric acid and distilled water shown in the table to make the solutions.Step 3 Turn the Petri dish over so the base side is up. Use a marker pen to draw three lines to divide the base into approximately equal sections. Label the sections A, B and C as shown in Fig.1.Step 4 Turn the Petri dish so the base side is down. Use a straw to cut a hole in the centre of each section of the agar in the Petri dish, as shown in Fig. 2.Hold the straw vertically and push through the agar to the bottom of the layer. As you remove the straw twist it slightly to pull out the agar. Squeeze the end of the straw gently to push the agar you have removed onto a paper towel.Step 5 Use a pipette to transfer three drops of solution from test-tube A into the hole in the agar in section A of the Petri dish.Do not let the solution drip onto the surface of the agar.Step 6 Use a clean pipette to repeat step 5 for the solution in test-tube B and the hole in the agar in section B of the Petri dish.Step 7 Use a clean pipette to repeat step 5 for the solution in test-tube C and the hole in the agar in section C of the Petri dish. Step 8 Start the stop-clock and leave the Petri dish for five minutes.Step 9 After five minutes observe the appearance of the agar around each of the holes.Step 10 Leave the Petri dish for a further 25 minutes. During this time, continue with the other questions.Step 11 After a total of 30 minutes use the ruler to measure the distance the citric acid has travelled from the edge of the hole in section A. You may need to use the hand lens.Step 12 Repeat step 11 for the holes in section B and section C of the Petri dish.The citric acid moves through the agar by diffusion. The diffusion coefficient is used to show the effect of concentration on diffusion. The formula to calculate the diffusion coefficient is: diffusion coefficient = time ( distance travelled )2​ Calculate the diffusion coefficient for a 10% solution of citric acid that travelled 14mm in 30 minutes. Give your answer to two significant figures. Space for working...................................... mm2 per minute

Accepted Answer

diffusion coefficient = time ( distance travelled )2​With this formula, taking the value of  distance travelled = 14mm and time =  30 minutes, we get: diffusion coefficient =30142​ = 30196​ = 6.5

Question 17

The diagram shows two cells. Which process can be carried out by only one of these cells?

Accepted Answer

The above given cell are of plant and animal.. The chemical reactions in the cell is controlled by both plant and animal cells. In plant and animal cells both, Cytoplasm is the site in the cell wherein multiple chemical reactions take place in the presence of enzymes.. The movement of substances into the cell is controlled by both plant and animal cells. Plasma membrane or cell membrane controls the movement of substances in and out of the cell.. Starch inside the cell is only synthesized by the plant cells. Starch is synthesized in the plastids—chloroplasts in leaves or specialized amyloplasts in the starch-storing tissues of staple crops.. . Glucose is used by both, plant and animal cells. Glucose is a potential source of energy for cells.

Question 18

A frog is an animal. A frog’s skin is permeable to oxygen and carbon dioxide.When a frog is swimming in pond water, in which directions will there be a net diffusion of oxygen and carbon dioxide?

Accepted Answer

Just beneath the frog's skin are networks of capillaries that carry off the oxygen to their cells and dump carbon dioxide into the air. Frogs have rather inefficient lungs. Without their gas-permeable skin, they would suffocate, depending solely on their lungs.Thus, there will be net inward diffusion of oxygen from the water into the frog and net outward diffusion of carbon dioxide from frog into water, making option A correct.

Question 19

Which process only involves the movement of water through the partially permeable membrane of a cell?

Accepted Answer

Osmosis refers to the process of diffusion of water (other solvents) from a region of low-solute concentration to a high solute concentration, through a membrane that is semi-permeable in nature . This solute or solvent transport is generally passive, which means that energy is not required for the osmosis process to take place.This takes place in order to equalize the concentrations of solute (or solvent) on both the sides of the cell membrane.

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