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Page No 262:

Question 1:

In the sketch of the stomatal apparatus given alongside, which one of the following is missing?


(1) Cell membranes of the cells
(2) Cell walls of the cells
(3) Nuclei in the guard cells
(4) Chloroplasts in the guard cells
 

Answer:

The nuclei in the guard cells is missing from the sketch of the stomatal apparatus given above.

Hence, the correct answer is option 3.

Page No 262:

Question 2:

A student focussed the leaf epidermal peel under the low power of microscope but could not see all the parts. He should :
(1) use the coarse adjustment knob again to focus the slide.
(2) use the fine adjustment knob to increase magnification.
(3) focus under high power using coarse adjustment knob.
(4) focus under high power using fine adjustment knob.

Answer:

In order to view all the parts of a leaf epidermal peel, the student should focus it under high power using fine adjustment knob. Cells are microscopic structures which can be clearly observed only under high power. It also enables us to observe various other components of cells.

Hence, the correct answer is option 4.

Page No 262:

Question 3:

The temporary mount of the leaf epidermal peel which looked pinkish red under the microscope was :
(1) stained in acetocarmine and mounted in glycerine
(2) stained in iodine and mounted in water
(3) stained in safranin and mounted in glycerine
(4) stained in methylene blue and mounted in water

Answer:

The temporary mount of leaf is stained with safranine and mounted with glycerine.
Safranine is a dye which can be taken up by cells and gives them a pink colour. The cell as well as the background are both transparent and it is difficult to visualise the cells as such. Staining imparts colour to the cells or its components and enhances its contrast and makes it easier to see them under the microscope.

Hence, the correct answer is option 3.

Page No 262:

Question 4:

A well stained leaf peel mount when observed under the high power of a microscope shows nuclei in :
(1) only epidermal cells
(2) only guard cells
(3) guard cells and epidermal cells
(4) guard cells, epidermal cells and stoma

Answer:

 Only guard cells and epidermal cells will be visible under the high power of a microscope.

Hence, the correct answer is option 3.

Page No 262:

Question 5:

Which structure out of I, II, III and IV marked in the given diagram of the epidermal peel of leaf should be labelled as stoma?



(1) I
(2) II
(3) III
(4) IV

Answer:

Stoma/stomatal pore is an opening surrounded by guard cells. Structure III marked in the given diagram represents stoma.

Hence, the correct answer is option 3.

Page No 262:

Question 6:

A student had drawn the diagram of stomata in a hurry, as shown alongside. He could not be given full marks as he :



(1) forgot to draw nuclei in guard cells and also to label the diagram.
(2) did not draw nuclei in guard cells and other cells.
(3) should have drawn nuclei and chloroplasts in guard cells and nuclei in all epidermal cells.
(4) did not label the stoma in the correct position.

Answer:

 The student could not get full marks for his diagram because he should have drawn nuclei and chloroplasts in guard cells and nuclei in all epidermal cells.

Hence, the correct answer is option 3.

Page No 262:

Question 7:

Bottles containing the undermentioned liquids were available in the laboratory. Which liquid did a student use for putting a drop on the slide before placing the coverslip while preparing the temporary mount of leaf epidermal peel?
(1) Water
(2) Safranin
(3) Glycerine
(4) Iodine

Answer:

The student must have put a drop of glycerine on the slide before placing the coverslip on the slide. Glycerine is a viscous liquid which prevents the specimen from drying out and keeps it moist for longer duration. Also glycerine tends to refract light due to its refractive nature. As a result of which, the image appears clearer under the microscope. 

Hence, the correct answer is option 3.

Page No 262:

Question 8:

To prepare a good temporary mount of the petunia leaf peel showing many stomata, the student has to get the peel from the :
(1) tip of the leaf
(2) upper surface of the leaf
(3) lower surface of the leaf
(4) point of attachement of leaf to its petiole

Answer:

To prepare a good temporary mount, the student should take the peel from the lower surface of the leaf.
Stomata are structures which help in gaseous exchange and are usually scattered on the lower surfaces of leaf.

Hence, the correct answer is option 3.



Page No 263:

Question 9:

While preparing a good temporary mount of leaf peel to observe stomata, care should be taken to avoid:
(1) adding glycerine to the slide
(2) using water to wash the slide
(3) having air bubbles in the slide
(4) staining the peel with safranin

Answer:

A good temporary mount of leaf should be free of any air bubbles. These air bubbles lower the viewing resolution. Thus, all precautions should be taken to avoid having air bubbles in the slide. 

Hence, the correct answer is option 3.

Page No 263:

Question 10:

In the sketch of stomatal apparatus given alongside, the parts I, II, III and IV were labelled differently by four students. The correct labelling is shown in :
 

 
(1) (I) guard cells, (II) stoma, (III) starch granule, (IV) nucleus
(2) (I) cytoplasm, (II) nucleus, (III) stoma, (IV) chloroplast
(3) (I) guard cells, (II) starch, (III) nucleus, (IV) stoma
(4) (I) cytoplasm, (II) chloroplast, (III) stoma, (IV) nucleus

Answer:

The correct labelling of the above diagram is -
(I) cytoplasm, (II) nucleus, (III) stoma, (IV) chloroplast

Hence, the correct answer is option 2.

Page No 263:

Question 11:

In the slide of an epidermal peel, the parts which appear pink coloured after staining with safranin are :
(1) stomata only
(2) nuclei only
(3) cell membrane and cytoplasm
(4) all parts in the peel

Answer:

Safranin is a stain which stains all the parts of the epidermal peel pink.

Hence, the correct answer is option 4.

Page No 263:

Question 12:

Four students, A, B, C and D, make the records given below, for the parts marked X and Y in this diagram.


 

Student X Y
A
B
C
D
Stoma
Guard cell
Epidermal cell
Stoma
Guard cell
Stoma
Stoma
Epidermal cell

The correct record, out of these, is that of student:
 
(1) A
(2) B
(3) C
(4) D

Answer:

The correct record is made by student A as X represents stoma whereas Y represents guard cells.
Stoma also known as the stomatal pore is an opening which is surrounded by a pair of guard cells.
Guard cells are bean shaped cells which regulate the opening and closing of the stomatal pore.

Hence, the correct answer is option 1.

Page No 263:

Question 13:

The steps involved in making a slide of epidermal peel of leaf are given as follows :
I. Pull out a thin peel from the lower surface of the leaf
II. Place a drop of glycerine on the slide
III. Stain the peel in safranin
IV. Place the stained peel on the glycerine
V. Remove the extra stain by washing with water
VI. Place the coverslip over the peel
Which one is the correct sequence of steps to be followed?
(1) I, II, III, IV, V, VI
(2) I, III, V, II, IV, VI
(3) I, III, IV, II, V, VI
(4) I, II, IV, III,V, VI

Answer:

The correct sequence of steps for making a slide of epidermal peel are -
I. Pull out a thin peel from the lower surface of the leaf
III. Stain the peel in safranin
V. Remove the extra stain by washing with water
II. Place a drop of glycerine on the slide
IV. Place the stained peel on the glycerine
VI. Place the coverslip over the peel

I, III, V, II, IV, VI

Hence, the correct answer is option 2.

Page No 263:

Question 14:

Given alongside is a sketch of a leaf partially covered with black paper and which is to be used in the experiment to show that light is compulsory for the process of photosynthesis. At the end of the experiment, which one of the leaf parts labelled I, II and III will become blue black when dipped in iodine solution?
 

 
(1) I only
(2) II only
(3) I and III
(4) II and III

Answer:

According to the above sketch, parts I and III will become blue black when dipped in iodine. It is because the starch will be synthesized in parts I and III and whereas it will not be synthesized in part II.

Hence, the correct answer is option 3.

Page No 263:

Question 15:

Before testing the leaf for starch at the end of the experiment, ‘light is necessary for photosynthesis’, the experimental leaf should be boiled in:
 
(1) Water
(2) Alcohol
(3) KOH solution
(4) Hydrochloric acid

Answer:

In the above mentioned experiment, the leaf should be boiled with alcohol to remove chlorophyll.

Hence, the correct answer is option 2.

Page No 263:

Question 16:

Given below are the steps to be followed for performing ‘starch test’ on a green leaf.
(A) Boil the leaf in alcohol
(B) Boil the leaf in water
(C) Dip the leaf in iodine solution
(D) Wash the leaf in water
Which one of the following sequences should the students follow to get the correct result?
(1) (A), (D), (B), (C)
(2) (D), (A), (B), (C)
(3) (B), (D), (A), (C)
(4) (B), (A), (D), (C)

Answer:

The following sequence should be followed by students to obtain the correct result in starch experiment -

(B) Boil the leaf in water - to break down the cell membrane of leaf cells and making them more permeable to iodine solution.
(A) Boil the leaf in alcohol - to remove the cholorphyll which interferes in test for starch
(D) Wash the leaf in water - to soften it and remove any chlorophyll molecules attacahed to it
(C) Dip the leaf in iodine solution - to check the presence of starch

(B), (A), (D), (C)

Hence, the correct answer is option 4.



Page No 264:

Question 17:

The steps necessary for setting up the experiment, ‘To demonstrate that light is necessary for photosynthesis’ are not given here in proper sequence :
I. Keep the potted plant in sunlight for 3 to 4 hours.
II. Keep the potted plant in darkness for about 48 hours.
III. Cover a leaf of the plant with a strip of black paper.
IV. Pluck the leaf and test it for starch.
The correct sequence of steps is :
(1) I, III, IV, II
(2) I, IV, III, II
(3) II, IV, III, I
(4) II, III, I, IV

Answer:

The following sequence should be followed for setting up the experiment "To demonstrate that light is necessary for photosynthesis" -

II. Keep the potted plant in darkness for about 48 hours. - This is done to destarch the leafs so that no starch is present in the beginning of the expriment itself.
III. Cover a leaf of the plant with a strip of black paper. - This is done to prevent the synthesis of starch in that region.
I. Keep the potted plant in sunlight for 3 to 4 hours. - This is done so that photosynthesis can occur and synthesis of starch can take place.
IV. Pluck the leaf and test it for starch. - The leaf is tested for the presence of starch.

 II, III, I, IV

Hence, the correct answer is option 4.

Page No 264:

Question 18:

On completion of the experiment to demonstrate that ‘Light is necessary for photosynthesis’, four students reported the inference as follows. Identify the correct inference:
(1) Part of the leaf covered with strip can only undergo photosynthesis.
(2) Uncovered part of the leaf cannot synthesise starch
(3) Photosynthesis takes place only in the presence of sunlight
(4) Light is necessary for the synthesis of starch in green plants.

Answer:

 The correct inference for the above experiment is light is necessary for the synthesis of starch in green plants. Green plants cannot synthesise starch in the absence of light. Important requirement for photosynthesis to occur in plants is chlorophyll, sunlight and water.

Hence, the correct answer is option 4.

Page No 264:

Question 19:

In the experiment to show that light is necessary for photosynthesis, the plucked leaf is boiled in ethanol and then washed with water. After this, it is tested for the presence of a carbohydrate by a chemical which is :
(1) salt solution
(2) sugar solution
(3) iodine solution
(4) starch solution

Answer:

The chemical used for testing the leaf is iodine solution. It is used to test the presence of starch in leafs.

Hence, the correct answer is option 3.

Page No 264:

Question 20:

In the experiment to show that light is necessary for photosynthesis, some of the steps of the experiment deal with the following activities:
I. Starch reacts with iodine and gives blue-black colour
II. Chlorophyll is dissolved in ethanol
III. Hot water makes leaf tissue soft
The correct sequence of these steps is :
(1) I–II–III
(2) II–III–I
(3) III–II–I
(4) I–III–II

Answer:

The correct sequence of steps for the above mentioned experiment is II - III - I.

II. Chlorophyll is dissolved in ethanol
III. Hot water makes leaf tissue soft
I. Starch reacts with iodine and gives blue-black colour

Hence, the correct answer is option 2.

Page No 264:

Question 21:

A portion of each of four destarched leaves of a plant was covered with paper strips of various kinds. The plant was exposed to sunlight for 5 hours. Thereafter, the strips were removed and the leaves tested for starch in the covered portion. Which one out of the four leaves gave the starch test in the covered portion?
(1) that covered with black paper strip
(2) that covered with green paper strip
(3) that covered with white paper strip
(4) that covered with transparent paper strip

Answer:

The leaf covered with transparent paper strip will give a positive starch test as the light will be easily absorbed by leafs.

Hence, the correct answer is option 4.

Page No 264:

Question 22:

In order to destarch the leaves for an experiment to show that sunlight is necessary for photosynthesis, the:
(1) leaves are kept in alcohol and boiled in a water bath
(2) leaves are soaked in iodine for two hours
(3) plant with the leaves is kept in dark room for 48 hours.
(4) plant with the leaves is exposed to light of a lamp, a night before the experiment

Answer:

In order to destarch the leaves for the experiment to show that sunlight is necessary for photosynthesis, the plants with leaves is kept in dark room for 48 hours.

Hence, the correct answer is option 3.

Page No 264:

Question 23:

Which one of the following is the correct combination of relevant materials required for setting up an experiment to show that light in necessary for photosynthesis?
(1) destarched leaves, strips of black paper, starch solution, and iodine crystals
(2) a potted plant, strips of coloured paper, starch solution, iodine and potassium iodide
(3) destarched leaves, strips of black paper, starch solution, and potassium iodide
(4) destarched leaves, strips of black paper, and iodide solution

Answer:

 The correct combination of materials for the above mentioned experiment are -
destarched leaves - to check the synthesis of starch
strips of black paper - to cover the leafs
iodine solution - to check the presence of starch

Hence, the correct answer is option 4.

Page No 264:

Question 24:

A portion of destarched leaf of a potted plant was covered with a black strip of paper. The plant was then exposed to sunlight for six hours and then tested for starch. It was observed that:
(1) both covered and uncovered parts of leaf turned blue-black
(2) both covered and uncovered parts of leaf turned yellow-brown
(3) only the uncovered part of leaf turned blue-black
(4) only the covered part of leaf turned blue-black

Answer:

In the above case, only the uncovered part of leaf turned blue black. Although the leaf had been destarched earlier, keeping it exposed to sunlight led to the synthesis of starch. But this process would not occur in parts which had been covered.

Hence, the correct answer is option 3.



Page No 265:

Question 25:

A star-shaped figure was cut in the black paper strip used for covering the leaf of a destarched plant used for demonstrating that light is necessary for photosynthesis. At the end of the experiment when the leaf was tested for starch with iodine, the star-shaped figure on the leaf was found to be:


(1) colourless
(2) green in colour
(3) brown in colour
(4) blue-black in colour

Answer:

The star shaped figure would be found to be blue black in colour indicating the presence of starch in that area.

Hence, the correct answer is option 4.

Page No 265:

Question 26:

In an experiment to test the presence of starch in a leaf, the leaf is boiled in alcohol for a few minutes using a water bath. This is an essential step in the experiment because alcohol:
(1) softens the leaf
(2) disinfects the leaf
(3) allows iodine to enter the leaf
(4) dissolves chlorophyll from leaf

Answer:

Boiling the leaf in alcohol before testing the presence of starch in a leaf is an essential step. It removes the chlorophyll from leaves.

Hence, the correct answer is option 4.

Page No 265:

Question 27:

For the experiment ‘Light is necessary for photosynthesis’, the potted plant was first kept in darkness for a day. This is to:
(1) deactivate chloroplasts
(2) destarch leaves
(3) activate chloroplasts
(4) prepare leaves for photosynthesis

Answer:

For the experiment ‘Light is necessary for photosynthesis’, the potted plant was first kept in darkness for a day to destarch the leaves.

Hence, the correct answer is option 4.

Page No 265:

Question 28:

When asked to set up an experiment to show that ‘light is necessary for photosynthesis’, a student ran to the school garden and set up the experiment by using a plant growing in the school garden. The experiment failed. His classmates made the following suggestions to get success in the experiment:
Student A : Safranin should be used instead of iodine
Student B : The leaf should not be boiled in alcohol to remove chlorophyll before testing for starch
Student C : Transparent paper should be used instead of black paper strip.
Student D : The leaf should be destarched before starting the experiment
The correct suggestion is given by :
(1) student A
(2) student B
(3) student C
(4) student D

Answer:

The correct suggestion was given by student D. The leaf should be destarched before performing the experiment. The leafs need to be destarched so that we can check during the experiment that starch synthesis occurs or not in the region covered with black strip.
 
Hence, the correct answer is option 4.

Page No 265:

Question 29:

The steps taken for setting up an experiment to demonstrate that ‘light is necessary for photosynthesis’, were as follows :
• A strip of black paper was clipped on the leaf of a potted plant
• The plant was kept in the sun for four hours
• The strip was removed and the leaf was placed in boiling alcohol in water bath
• The leaf was washed and tested for starch
The result was not as expected. Identify the step which had been missed out :
(1) The plant was kept in the dark for 24 hours before starting the experiment
(2) The leaf was boiled in water after placing it in boiling alcohol
(3) The leaf was sprinkled with water before placing it in black paper strip
(4) A transparent strip was used to cover the black paper strip

Answer:

In the above experiment, the step which requires to keep the plant in dark for 24 hrs was missing before starting the experiment.
It is important to keep the plant in dark before starting the experiment so that all the leafs can be destarched.

Hence, the correct answer is option 1.

Page No 265:

Question 30:

Out of the following figures, choose the one showing the correct procedure for removing chlorophyll from the leaf in the experiment ‘light is necessary for photosynthesis’:


The correct procedure is :
(1) A
(2) B
(3) C
(4) D

Answer:

The figure in option C shows the correct figure for the experiment 'light is necessary for photosynthesis'. The correct method of destarching the leafs is by boiling the leaf in ethanol kept in a water bath. Ethanol cannot be directly heated as it is a flammable substance.

Hence, the correct answer is option 3.



Page No 266:

Question 31:

A leaf of destarched healthy potted plant was covered by black paper strip as shown in Figure I, and kept in sunlight in the morning. In the evening, it was plucked off and tested for the presence of starch by using iodine solution.


The observation is matched with the Figure No.:
(1) II
(2) III
(3) IV
(4) V

Answer:

The observation in figure II correlates with the observation. The covered part shows yellow colour whereas the exposed parts show blue black colour on testing with iodine. Presence of blue black colour indicates the synthesis of starch in that region.

Hence, the correct answer is option 1.

Page No 266:

Question 32:

In an experiment to test the presence of starch in a leaf, the plucked leaf is first boiled in water for a few minutes. This is an important step in the experiment because it :
(1) converts glucose made in the leaf into starch
(2) dissolves chlorophyll present in the green leaf
(3) extracts starch to destarch the leaf
(4) makes the leaf more permeable to iodine solution

Answer:

In the above mentioned experiment, the plucked leaf is first boiled in water to extract the starch and destarch the leaf.

Hence, the correct answer is option 3.

Page No 266:

Question 33:

Why is some KOH placed in a small test tube in the flask with germinating seeds in the experiment to demonstrate occurrence of respiration in germinating seeds?
(1) To provide oxygen required by the seeds for respiration.
(2) To absorb carbon dioxide and create partial vacuum in the flask.
(3) To absorb water from the seeds to make them dry.
(4) To make the air present in the flask alkaline.

Answer:

KOH is placed in the small tube to absorb carbon dioxide and create partial vacuum in the flask. This absorption of CO2 by KOH creates partial vacuum in the flask resulting in increase in the water level in the bent tube.

Hence, the correct answer is option 2.

Page No 266:

Question 34:

Which one of the following is the correct set of three precautions for setting up the experiment to demonstrate that carbon dioxide is evolved during respiration?
(1) Air tight set up; delivery tube dips in water in beaker; flask has seeds which have just germinated.
(2) Thread holding KOH test tube; air tight flask; delivery tube above surface of water in the beaker.
(3) Germinated seeds under water in the flask; experimental set up not air tight; delivery tube above water level.
(4) Delivery tube touching bottom of water; KOH test tube held by a thick wire; seeds covered by water.

Answer:

The precautions which should be taken -
Air tight set up - an air tight set up is required, so that the CO2 produced during respiration does not escape out
Delivery tube dips in water in beaker - the delivery tube needs to touch the water, so that the water can rise in the tube when partial vacuum is created in the flask.
Flask has seeds which have just germinated - germinated seeds will produce CO2 during the experiment which is an important requirement for this experiment.

Hence, the correct answer is option 1.

Page No 266:

Question 35:

The experimental set up shown here to demonstrate that ‘CO2 is given out during respiration’, did not yield expected results because :



(1) the flask was not air tight
(2) there was no KOH in test tube in the flask
(3) the delivery tube was dipped in water
(4) the germinating seeds were not immersed in water

Answer:

The above experiment did not yield the desired results as there was no KOH tube in the flask. KOH absorbs CO2 and creates partial vacuum in the flask which leads to increase in the water level in the delivery tube. The increase in the water level in the bent tube is an indication that 
COis produced during respiration.

The correct set up is



Hence, the correct answer is option 2.

Page No 266:

Question 36:

The following experimental set-ups were kept in the laboratory to show that ‘CO2 is given out during respiration’.



After two hours, students observed that water rises in the delivery tube:
(1) only in set up (A)
(2) only in set up (B)
(3) in both (A) and (B)
(4) neither in set up (A) nor in set up (B)

Answer:

In this experiment, the water will rise only in set up A. In set up A, the flask is made air tight with the help of a cork whereas in set up B, a cotton plug is used. It does not create an air tight condition as a result of which  water would not rise in set up B.

Hence, the correct answer is option 1.



Page No 267:

Question 37:

In the experiment to show that CO2 is given out during respiration by germinating seeds, the student uses :
(1) Lime water
(2) Alcohol
(3) KOH solution
(4) Iodine solution

Answer:

In the experiment to show that COis produced during respiration by germinating seeds, KOH solution is used. KOH solution is used to absorb the CO2 produced during respiration which results in the creation of vacuum in the flask

Hence, the correct answer is option 3.

Page No 267:

Question 38:

The most appropriate reason for taking germinating seeds in the experiment to show that carbon dioxide is produced during respiration, is :
(1) germinating seeds create high temperature
(2) germinating seeds are easy to handle
(3) germinating seeds are living things
(4) germinating seeds are in dormant to state

Answer:

The most appropriate reason to take germinating seeds in this experiment is that germinating seeds are living things. Only living things show the process of respiration. Use of seeds for this experiment does not lead to any ethical issues and are also easier to work with as compared to animals.

Hence, the correct answer is option 3.

Page No 267:

Question 39:

Before setting up an experiment to show that seeds release carbon dioxide during respiration, the seeds should be:
(1) dried completely
(2) boiled to make them soft
(3) soaked in potassium hydroxide solution
(4) kept moist till they germinate

Answer:

In the above mentioned experiment, the seeds need to be kept moist till they germinate. If the seeds are not kept moist, they would dehydrate and eventually die.

Hence, the correct answer is option 4.

Page No 267:

Question 40:

Which of the following precautions are to be taken for a successful run of the experiment to show that carbon dioxide is given out during respiration?
A. Cork should be air-tight
B. Seeds in the flash should be totally dry
C. A small tube with freshly prepared KOH solution should be placed in the flask
D. The end of the delivery tube should be above water level
The correct answer is :
(1) A and B
(2) A and C
(3) A, B and C
(4) A, B and D

Answer:

The precautions to be taken in the above experiment are -
A. Cork should be air-tight so that the carbon dioxide produced during the process does not escape out.
C. A small tube with freshly prepared KOH solution should be placed in the flask so that it can absorb the carbon dioxide produced and create a vacuum in the flask. This vacuum is necessary for the water level to rise in the bent tube.

Hence, the correct answer is option 2.

Page No 267:

Question 41:

The chemical required in the experiment to show that carbon dioxide gas is released during respiration is:
(1) potassium bicarbonate
(2) potassium dichromate
(3) potassium permanganate
(4) potassium hydroxide

Answer:

The chemical required in the experiment to show that carbon dioxide gas is released during respiration is potassium hydroxide. Potassium hydroxide absorbs the carbon dioxide produced during the process of respiration and creates a vacuum in the flask.

Hence, the correct answer is option 4.

Page No 267:

Question 42:

In the experiment to demonstrate that CO2 is given out during respiration, what would you observe in the delivery tube dipped in water?
(1) Water level rises in the delivery tube
(2) Water turns milky and rises in the delivery tube
(3) Water turns milky but does not rise in the delivery tube
(4) Water level in the delivery tube remains unchanged

Answer:

 In the experiment to demonstrate that CO2 is given out during respiration, it should be observed that water level rises in the delivery tube.

Hence, the correct answer is option 1.

Page No 267:

Question 43:

An experimental set-up to demonstrate respiration in germinating seeds is shown here. It is observed that water from the beaker has not risen into the delivery tube (bent tube). This is because :



(1) the set-up is airtight
(2) the beaker has coloured water
(3) carbon dioxide is not being absorbed
(4) no oxygen is available to seeds for respiration

Answer:

In the above experiment, it is observed that water from the beaker has not risen into the delivery tube which means that carbon dioxide is not being absorbed.
In the small tube, KCl is present instead of KOH which is required for absorbing CO2 and creating partial vacuum.

Hence, the correct answer is option 3.

Page No 267:

Question 44:

In the experiment to show that carbon dioxide is given out during respiration in humans, the student uses :
(1) lime water
(2) alcohol
(3) potassium hydroxide solution
(4) iodine solution

Answer:

In the above mentioned experiment, the student uses lime water which turns milky when CO2  passes through it.

Hence, the correct answer is option 1.

Page No 267:

Question 45:

An experimental set-up is given here to demonstrate that CO2 is given out during respiration. Four students made the following observations marked as I, II, III and IV.



I. Level of water remained the same in both the beaker and the delivery tube
II. Level of water increased in the delivery tube
III. Level of water reduced in both the beaker and the delivery tube
IV. Water ascends into the delivery tube and back flows into the beaker.
Which one of the above is the correct observation?
(1) I
(2) II
(3) III
(4) IV

Answer:

 The correct observation is that the level of water increased in the delivery tube.

Hence, the correct answer is option 2.



Page No 268:

Question 46:

The given slides A and B were identified by four students I, II, III and IV as stated below:


 

Slide A Slide B
I. Binary fission in Amoeba Daughter cells of Amoeba
II. Budding in yeast Buds of yeast
III. Binary fission in Amoeba Buds of yeast
IV. Budding in yeast Daughter cells of Amoeba

Of the above mentioned identification of slides A and B, which one is correct?

(1) I
(2) II
(3) III
(4) IV

Answer:

The observation made by student I is correct as slide A shows binary fission in Amoeba and slide B shows daughter cells of Amoeba.

Hence, the correct answer is option 1.

Page No 268:

Question 47:

The correct diagram showing an Amoeba undergoing binary fission is :



(1) I
(2) II
(3) III
(4) IV

Answer:

The correct diagram showing an Amoeba undergoing binary fission is diagram III.

The process of binary fission begins with the division of the nucleus into two, which is then followed by the division of the amoebic body into 2 daughter cells. Each daughter cell contains a nucleus.

Hence, the correct answer is option 3.

Page No 268:

Question 48:

The figure given here shows:



(1) Amoeba undergoing binary fission
(2) Yeast undergoing binary fission
(3) Yeast undergoing budding
(4) Amoeba undergoing budding

Answer:

The above figure shows Yeast undergoing budding.

Hence, the correct answer is option 3.

Page No 268:

Question 49:

Four stages of binary fission in Amoeba are shown below. The stage at which nuclear fission and cytokinesis are observed is, stage:




(1) I
(2) II
(3) III
(4) IV

Answer:

The stage at which nuclear fission and cytokinesis are observed in Amoeba is stage II.

The process of binary fission begins with the division of the nucleus into two (cytokinesis), which is then followed by the division of the amoebic body into 2 daughter cells. Each daughter cell contains a nucleus.

Hence, the correct answer is option 2.

Page No 268:

Question 50:

In the slides showing binary fission in Amoeba and budding in yeast, the correct observations are :
(1) the daughter cells of Amoeba and the bud of yeast are smaller than their respective parental cells 
(2) the daughter cells of Amoeba and the bud of yeast are of the same size as their respective parental cells
(3) the daughter cells of Amoeba are bigger than the parent but the bud of yeast is smaller than the parent
(4) the daughter cells of Amoeba are smaller than the parent but the bud of yeast is larger than the parent

Answer:

The correct observations are - the daughter cells of Amoeba and the bud of yeast are smaller than their respective parental cells. Since, in both the cases, the daughter cells are a result of division of the parent cell, their sizes are comparatively smaller to the parent cell.

Hence, the correct answer is option 1.



Page No 269:

Question 51:

In the figure of budding in yeast given here, the structures A, B, C and D should be labelled respectively as :



(1) nucleus of bud, bud, yeast, nucleus
(2) dividing nucleus of bud, bud, yeast, nucleus
(3) nucleus of bud, bud, yeast, dividing nucleus of yeast
(4) dividing nucleus of yeast, yeast, bud, nucleus of bud

Answer:

The structures A, B, C and D should be labelled respectively as nucleus of bud, bud, yeast, nucleus.

Hence, the correct answer is option 1.

Page No 269:

Question 52:

The diagram given alongside illustrates :



(1) bud formation in yeast
(2) binary fission in amoeba
(3) formation of daughter cells in yeast
(4) formation of pseudopodia in amoeba

Answer:

The diagram shows binary fission in Amoeba

The process of binary fission begins with the division of the nucleus into two, which is then followed by the division of the amoebic body into 2 daughter cells. Each daughter cell contains a nucleus.



Hence, the correct answer is option 2.

Page No 269:

Question 53:

Which stage out of those marked I, II, III and IV is showing the binary fission in Amoeba?



(1) I
(2) II
(3) III
(4) IV

Answer:

Stage IV shows binary fission in Amoeba.

Hence, the correct answer is option 4.

Page No 269:

Question 54:

The following figures illustrate the binary fission in Amoeba in an incorrect sequence :




The correct sequence is :

(1) I, III, IV, II
(2) II, III, IV, I
(3) IV, III, II, I
(4) III, IV, II, I

Answer:

The correct sequence of binary fission in amoeba is II, III, IV, I.

The process of binary fission begins with the division of the nucleus into two, which is then followed by the division of the amoebic body into 2 daughter cells. Each daughter cell contains a nucleus.

Hence, the correct answer is option 2.

Page No 269:

Question 55:

Which one of the following sketches does not illustrate budding in yeast?
 

 
(1) I
(2) II
(3) III
(4) IV

Answer:

Sketch II does not represent budding in Yeast. It represents a stage of binary fission in Amoeba.

Hence, the correct answer is option 2.

Page No 269:

Question 56:

Slides A and B show stages of asexual reproduction in two different organisms:
 



The slides A and B are depicting :
(1) binary fission in both Amoeba and Yeast
(2) budding in both Amoeba and Yeast
(3) binary fission in yeast and budding in Amoeba
(4) binary fission in Amoeba and budding in Yeast

Answer:

Slides A and B represent binary fission in Amoeba and budding in Yeast respectively.

Hence, the correct answer is option 4.



Page No 270:

Question 57:

The given slides A and B were identified by four students I, II, III and IV as sated below:


 

Slide A Slide B
I. Binary fission in Amoeba
II. Budding in Yeast
III. Binary fission in Amoeba
IV. Budding in Yeast
Daughter cells of Amoeba
Buds of Yeast
Buds of Yeast
Daughter cell in Amoeba

Of the above mentioned identifications of slides A and B, which one is correct?

(1) I
(2) II
(3) III
(4) IV

Answer:

 Identification I is correct as slide A shows binary fission in Amoeba and slide B shows daughter cells of Amoeba.

Hence, the correct answer is option 1.

Page No 270:

Question 58:

Each of the three beakers A, B and C contained 50 mL of distilled water. A student placed five raisins in each beaker. The raisins for each beaker weighed the same. The beakers were kept at room temperature. The raisins were removed from beaker A after 10 minutes, from beaker B after 20 minutes and from beaker C after one hour. On calculating the percentage of water absorbed by the raisins, it was found that:
(1) maximum absorption of water by raisins was in beaker C
(2) maximum absorption of water by raisins was in beaker B
(3) minimum absorption of water by raisins was in beaker C
(4) absorption of water was equal in raisins of all the three beakers

Answer:

The maximum absorption of water by raisins was in beaker C as they were placed in the beaker for the longest duration.

Hence, the correct answer is option 1.

Page No 270:

Question 59:

The following data was obtained on performing an experiment for determining the percentage of water absorbed by raisins:
 

Mass of water in the beaker = 50 g
Mass of dry raisins = 20 g
Mass of raisins after soaking in water = 30 g
Mass of water left in the beaker after the experiment = 40 g

The percentage of water absorbed by raisins will be:

(1) 10%
(2) 25%
(3) 45%
(4) 50%

Answer:

The precentage of water absorbed is calculated as -

Final weight-Initial weightInitial weight×100

30 -2020 ×100 = 100020                           = 50%

Hence, the correct answer is option 4.

Page No 270:

Question 60:

At the end of the experiment, 'to determine the percentage of water absorbed by raisins', the raisins are wiped just before weighing. This is to ensure that:

(1) hands do not get wet
(2) the raisins lose water before weighing
(3) the weighing scale does not get wet
(4) only water absorbed by raisins is weighed

Answer:

In the above experiment, raisins are wiped just before weighing so that only the water absorbed by raisins is weighed.

Hence, the correct answer is option 4.

Page No 270:

Question 61:

A student soaked 5 g of raisins in beaker (A) containing 25 mL of ice-chilled water and another 5 g of raisins in beaker (B) containing 25 mL of tap water at room temperature. After one hour the student observed that:
(1) water absorbed by raisins in beaker (A) was more than that absorbed by raisins of beaker (B).
(2) water absorbed by raisins in beaker (B) was more than that absorbed by raisins of beaker (A).
(3) the amount of water absorbed by the raisins of both beakers (A) and (B) was equal.
(4) no water was absorbed by raisins in either of the beakers (A) and (B).

Answer:

It will be observed that the water absorbed by raisins in beaker (B) was more than that absorbed by raisins of beaker (A). It is because rate of imbibition is directly proportional to the temperature. Higher the temperature of water, higher will be the rate of imbibition of water.

Hence, the correct answer is option 2.

Page No 270:

Question 62:

Raisins are soaked in water for determining the percentage of water absorbed by raisins. The formula used by a student for calculating the percentage of water absorbed, is:

(1) Initial weight-Final weightInitial weight×100

(2) Final weight-Initial weightInitial weight×100

(3) Final weight-Initial weightFinal weight×100

(4) Initial weight-Final weightFinal weight×100
 

Answer:

The correct formula for calculating the percentage of water absorbed, is:

Final weight-Initial weightInitial weight×100

Hence, the correct answer is option 2.

Page No 270:

Question 63:

A student soaked 10 g of raisins in 50 mL of distilled water in two beakers A and B each. She maintained beaker A at 25°C and beaker B at 50°C. After an hour, the percentage of water absorbed will be:
(1) the same in both A and B
(2) more in A than in B
(3) more in B than in A
(4) exactly twice in B than in A

Answer:

The percentage of water absorbed will be more in B than in A. It is because rate of imbibition is directly proportional to the temperature. Higher the temperature of water, higher will be the rate of imbibition of water.

Hence, the correct answer is option 3.



Page No 271:

Question 64:

A Student dissolved 5 g of sugar in 100 mL of distilled water in beaker A. He dissolved 100 g of sugar in 100 mL of distilled water in beaker B. Then he dropped a few raisins of equal weight in each beaker. After two hours he found the raisins in A swollen and those in B shrunken. The inference drown is that:
(1) sugar concentration of raisins is lower than that of solution A and higher than that solution B.
(2) sugar concentration of raisins is higher that that of solution A and lower that that of solution B.
(3) in B the cell membrane of raisins was damaged resulting in bleaching.
(4) in A the permeability of water of the cell membrane of raisins was enhanced

Answer:

In the first case, the sugar concentration of raisins was less as compared to the solution in the beaker. As a result, water moves into the raisins causing them to swell.

Hence, the correct answer is option 2.

Page No 271:

Question 65:

Which of the following set of materials is required to set up an experiment to determine the percentage of water absorbed by raisins ?

(1) raisins, breaker of water, filter paper, petri dish, weight box, balance
(2) raisins, petri dish, beaker, filter paper, weight box, balance
(3) raisins, beaker of water, blotting paper, physical balance, weight box
(4) raisins, beaker, blotting paper, petri dish, weight box, balance

Answer:

The following set of materials is required to set up an experiment to determine the percentage of water absorbed by raisins - raisins, beaker of water, blotting paper, physical balance, weight box.

Hence, the correct answer is option 3.

Page No 271:

Question 66:

An experiment was set up to determine the percentage of water absorbed by raisins.If the mass of dry raisins was 40 g, and the mass of wet raisins was 45 g,then the percentage of water absorbed would be :

(1) 45 g40 g×100

(2) 40 g45 g×100

(3) (45-40)g40g×100

(4) (45-40)g45 g×100

Answer:

Percentage of water absorbed by raisins  = W2-W1W1×100

The percentage of water absorbed in the above case would be  = (45-40)g40 g×100

Hence, the correct answer is option 3.

Page No 271:

Question 67:

5 dry raisins were placed in each of the two beakers containing 50 mL of water. After four hours, the raisins were taken out and wiped. For calculating the percentage of water absorbed by raisins, the raisins should have been weighed:

(1) only before placing in water

(2) only after four hours of their being in water

(3) both before and after placing in water

(4) before and at intervals of every hours

Answer:

The formula for calculating  the percentage of water absorbed = Final weight-Initial weightInitial weight×100
So, for calculating the percentage of water absorbed by raisins, they should have been weighed both before and after placing in water.

Hence, the correct answer is option 3.

Page No 271:

Question 68:

Raisins swell up after being placed in a beaker containing water for some time because :

(1) the concentration of water in the cell sap is higher than the water  in the beaker

(2) the concentration of water in the cell sap is lower than the water in the beaker

(3) the concentration of water in the cell sap is the same as that of water in the beaker

(4) water inside the raisins passed out of them when placed in a beaker of water

Answer:

As the concentration of water inside the cell sap is lower, the water from beaker will move inside the cell sap resulting in the swelling of raisins. This process is termed as osmosis.

Hence, the correct answer is option 2.

Page No 271:

Question 69:

A group of students performed an experiment to determine the percentage of water absorbed by raisins.The initial weight of raisins is 5 grams and final weight is 8 grams. The  percentage of water absorbed will be :

(1) 62.5

(2) 160

(3) 60

(4) 20

Answer:

The  percentage of water absorbed will be  = Final weight-Initial weightInitial weight×100

                                                                      = 8-55×100 =60%

Hence, the correct answer is option 3.

Page No 271:

Question 70:

If the weight of dry raisins is W1 and that of soaked raisins is W2 ,then the correct equation for calculating the percentage of water absorbed by raisins will be :

(1) W1-W2

(2) W2-W1

(3) W2-W1W1×100

(4) W1-W2W2×100
 

Answer:

The correct equation for calculating the percentage of water absorbed by raisins will be 

W2-W1W1×100

where, W1 = Initial weight of raisins
           W2 = Final weight of raisins

Hence, the correct answer is option 3.

Page No 271:

Question 71:

In the experiment to show that carbon dioxide is produced during respiration by germinating seeds, the alkali solution kept in a small test-tube absorbs:

(1) only O2 gas

(2) only CO2 gas

(3) both Oand  CO2 gases

(4) neither Onor CO2 gases

Answer:

In the above mentioned experiment, the alkali (KOH) kept in a small test tube absorbs only CO2 gas and creates partial vacuum in the flask.

Hence, the correct answer is option 2.

Page No 271:

Question 72:

During the preparation of slide, a drop of glycerine is used so that:

(1) material sticks on the  slide

(2) bacteria may not attack the material

(3) material may not dry up

(4) visibility of  material through the microscope may improve

Answer:

During the preparation of slide, glycerine is used to prevent the drying up of material used as a specimen. Glycerine is a viscous liquid which prevents the specimen from drying out, so that it can be clearly viewed under the microscope.

Hence, the correct answer is option 4.



Page No 272:

Question 73:

During the preparation of a temporary mount of leaf peel, the excess glycerine is removed by:

(1) dipping slide in water

(2) a blotting paper

(3) a cotton cloth

(4) tilting the slide

Answer:

The excess of glycerine is removed by using a blotting paper. Excess of glycerine, interferes with observation of the specimen under the microscope.

Hence, the correct answer is option 2. 

Page No 272:

Question 74:

The apparatus required to perform the experiment to show the evolution of carbon dioxide during respiration includes:

(1) flat-bottom flask,rubber cork , small glass tube, water,KOH solution,delivery tube ,germinating seeds,thread,vaseline, small beaker

(2) round-bottom flask,rubber cork, boiling tube,water,NaOH solution,  delivery tube ,dry seeds, thread, vaseline, small beaker

(3) measuring-flask, rubber cork,small glass tube,water,Na2COsolution,delivery tube,germinating seeds, thread,vaseline,small beaker

(4) flat-bottom flask, rubber cork, small glass tube,water,KOH solution, delivery tube, dry seeds,thread, vaseline,small beaker

Answer:

For the above mentioned experiment, the following apparatus is required -
flat-bottom flask, rubber cork (to close the flask), small glass tube (to keep KOH), water, KOH solution (to absorb CO2), delivery tube, germinating seeds, thread (to hang the KOH tube), vaseline (to seal the flask and rubber crok), small beaker

Hence, the correct answer is option 1.
 

Page No 272:

Question 75:

In the experiment to demonstrate that starch is made as food by the process of photosynthesis, the plucked leaf is first boiled in water for about 3 to 5 minutes in order to :

(1) remove chlorophyll from leaf cells

(2) break down the cell membranes of leaf cells 

(3) soften the brittle leaf

(4) convert starch into glucose so that is can be tested easily

Answer:

In the above mentioned experiment, the plucked leaf is first boiled in water for about 3 to 5 minutes in order to remove the chlorophyll from leaf cells. It is done to destarch the leafs.

Hence, the correct answer is option 1.



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