HBSE 10th Class Science Solutions Chapter 6 Life Processes

Haryana State Board HBSE 10th Class Science Solutions Chapter 6 Life Processes Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 6 Life Processes

HBSE 10th Class Science Life Processes Textbook Questions and Answers

Question 1.
The kidneys in human beings are a part of the system for ……….
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) excretion

Question 2.
The xylem in plants is responsible for ……….
(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen
Answer:
(a) transport of water

Question 3.
The autotrophic mode of nutrition requires ……….
(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above
Answer:
(d) all of the above.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in ……….
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus
Answer:
(b) mitochondria

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
1. Fat digestion occurs in small intestine.

2. First of all the large tat molecules gets converted into small molecules. This helps the fat-digesting enzymes to act on the fat molecules efficiently.

3. The bile juice which comes from the gall bladder (produced by liver) contains bile salts. These bile salts emulsify the fat molecules. Here, the surface tension of fat molecules decrease and they become smaller

4. As the surface area of these smaller molecule is more, it helps in effective action of enzymes.

5. The enzyme lipase which does this task comes from pancreas as well as intestinal glands. This lipase converts the fats into fatty acids and glycerol and this is how fats get digested.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
Human digestive system:
1. The function of digestive system is to break down larger molecules of food into smaller forms so that they can be absorbed easily into the body.
The human digestive system consists of the alimentary canal and its associated glands.

2. The main organs of the human digestive system are:
Mouth, oesophagus, stomach, small intestine and large intestine.

The main glands associated with digestion are:
Salivary gland, liver, and pancreas.

Salivary gland:
1. The salivary gland secretes saliva in our mouth. It is a watery liquid and so it wets (lubricates) the food in mouth. It is easy to swallow wet food.
2. The salivary gland also secretes an enzyme called amylase. Amylase breaks the complex molecule called starch present in the food into sugar.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are Its by-products?
Answer:
(a) Necessary conditions:

Presence of chlorophyll in the plants
Absorption of light energy
Splitting of water molecules
Reduction of carbon dioxide to carbohydrates

(b) By-products:
Oxygen is the by-product and carbohydrate (glucose) is the main product of the autotrophic nutrition.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:
1. Yeast, fungi. endoparasites (tape-worm, ascans), some bacteria are examples of some organisms who show anaerobic mode of respiration.

Aerobic respiration

Anaerobic respiration

1. It takes place in presence of oxygen.

2. End products are CO₂ and water.

3. It takes place in cytoplasm and mitochondria.

4. Aerobic respiration produces a considerable amount of energy.

1. It takes place in absence of oxygen.

2. End products are ethanol or lactic acid.

3. It takes place only in cytoplasm.

4. Anaerobic respiration produces quite Less energy.

Question 9.
How are the alveoli designed to maximize the exchange of gases?
Answer:
1. Alveoli are balloon-like structures. They can be also compared with a bunch of grapes. Their spacious structure facilitates the gaseous exchange with increased surface area.

2. Additionally, they are rich in blood supply and are thin-walled which helps in maximizing the exchange of gases.

Question 10.
What would be the consequences of deficiency of hemoglobin In our bodies?
Answer :
The desirable value of hemoglobin in males is 13-16 gibo mL blood and in females 12-15 gibo mL blood.
1. It hemoglobin is less, the oxygen-carrying capacity of blood decreases. It leads to lack of supply of oxygen to the body organs.

2. So, the heart has to pump the blood rapidly. As a result, a person gets tired soon and breathes faster than normal.

Question 11.
Describe double circulation of blood in human beings. Why Is it necessary?
Answer:
Double circulation:
In higher animals, the blood goes twice in the heart during each cardiac cycle. Hence, it is known as double circulation. The two circulations are called:

Systemic circulation and
Pulmonary circulation.

The systemic circulation occurs between the body cells (except lungs) and the heart while in pulmonary circulation, the circulation occurs only between the heart and the lungs.

Need of double circulation:
The need of oxygen is more in higher animals. In double circulation, the oxygenated and deoxygenated bloods do not mix. So, the body organs can get pure blood. It is helpful to maintain their metabolic needs.

Question 12.
What are the differences between the transport of materials In xylem and phloem?
Answer:

Xylem	Phloem
1. Xylem transports water and mineral salts in the plants. 2. Its transport route is from roots to leaves. 3. Here, transportation occurs in upward direction only. 4. Xylem contains tracheids and vessels.	1. Phloem transports organic food materials in the plants. 2. Its transport route is from leaves to various plant organs. 3. Here, translocation occurs in both upwards as well as downward direction and also in lateral direction. – in any and all directions. 4. Poem contains sieve tubes and sieve cells.

Question 13.
Compare the functioning of alveoli In the lungs and nephrons In the kidneys with respect to their structure and functioning.
Answer:

Functions of alveoli

Functions of nephron

1. It is the unit of respiratory system

2. It is thin-walled balloon-like structure

3. It increases the surface area for respiratory gas exchange

1. It is the unit of excretory system

2. It is thin-walled having cup shaped structure

3. It filters blood and formulates urine

HBSE 10th Class Science Life Processes InText Activity Questions and Answers

Textbook Page no – 95

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multi-cellular organisms like humans?
Answer:
1. In multicellular organisms such as humans, all the cells of, body are not in direct contact with surrounding environment.

2. Moreover, the size of human body is quite large as well as complex. So, each and every cell cannot perform gaseous exchange with environment by diffusion.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
1. Movement is one of the basic criterions of life to decide if something is alive.

2. Animals show visible movement called locomotion, while plants show invisible movement of biomolecules with themselves.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Chief raw materials used by an organism from outside are:
(a) Food (Carbon based food source):
Constituents: Carbohydrates. proteins, fats etc.
Chief function: Providing energy, photosynthesis in plants
(b) Oxygen: For respiration
(c) Water:
Constituents: H₂O and many water soluble minerals and salts.
Chief function: Universal medium for various metabolic processes in the body.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
Processes essential for maintaining life include —

Nutrition,
Respiration,
Transportation,
Excretion,
Control and Coordination,
Movement,
Reproduction.

Textbook Page no – 101.

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic nutrition	Heterotrophic nutrition
1. In autotrophic nutrition, organisms produce their own food using water, carbon dioxide and sun light. 2. In this nutrition, organisms obtain energy by producing carbohydrates with the help of carbon dioxide, water and sunlight. 3. Autotrophic nutrition has no further classification. Example: All green plants and some bacteria.	1. In heterotrophic nutrition, organisms derive energy by digesting organic substances obtained from plants and animals. 2. In this nutrition, the organisms first eat the food, then digest it into simpler forms and finally obtain energy. 3. Heterotrophic nutrition can be classified into (A) Saprophytic, (B) Parasitic and (C) Hoiozoic nutrition. Example: Herbivores, carnivores and omnivores.

Question 2.
Where do plants get each of the raw materials required for photosynthesis?
Answer:
Plants need water. CO₂, chlorophyll and sunlight for photosynthesis. They get these raw material from the following sources:

Water: Plants (Terrestrial plants) absorb water from the soil through roots.
CO₂ : Plants get CO₂ from the environment through the stomata.
Chlorophyll: Chlorophyll is a pigment found in the organelle called chloroplast of the cells of green parts of the plants.
Sunlight: The sun gives sunlight.

Question 3.
What is the role of the acid In our stomach?
Answer:
Stomach:

Stomach is a large organ which expands when the food enters it. Moreover, the stomach releases certain juices that help in digestion.
The muscular walls of the stomach churn the food thoroughly with these digestive juices.
During this process, the food is broken down into smaller pieces and is converted into a semi solid paste.
The wall of stomach contains three tubular glands which secrete gastric juice.

The gastric juice contains:

Dilute hydrochloric acid,
An enzyme called pepsin and
Mucus

(a) Dilute hydrochloric acid:

Since the stomach releases dilute hydrochloric acid, the digestive juices are acidic in nature.
The presence of acid enables the enzyme pepsin to digest protein present in the food.
Therefore, the function of hydrochloric acid is to create an acidic medium in the stomach.
It also kills the bacteria that enter the stomach through food.

(b) Pepsin: Pepsin is an enzyme that helps in digesting protein and converting food into smaller molecules.

Question 4.
What is the function of digestive enzymes?
Answer:
1. The digestive enzymes play an important role in digestion.
2. The food we eat contains complex macromolecules.
3. The digestive enzymes break down these complex molecules into smaller and simpler molecules.
4. These molecules are then easy to absorb by intestine.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
1. The process of digestion of food is almost completed in small intestine.
2. The small intestine is designed to provide maximum surface area for absorption of digested food.
3. To increase the surface area, the inner lining of the small intestine has numerous finger-like projections called ýilli.
4. These villi have a rich blood supply which helps to absorb food effectively.

Textbook Page no – 105

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
In aquatic environment. i.e. in water, the dissolved O₂ content is quite low compared to the oxygen-rich atmosphere of the terrestrial animal. So. the aquatic organisms have to breath at a much faster rate compared to the terrestrial organisms.

Question 2.
What are the different ways in which glucose Is oxidized to provide energy in various organisms?
Answer:
1. The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. The process of converting food into energy varies among organisms.

2. In any case, the first step is to break-down glucose which is a 6-carbon molecule, into two 3-carbon molecule called pyruvate. This process takes place in the cytoplasm.

Further conversion of pyruvate:

1. In absence of oxygen (anaerobic respiration):

Conversion of pyruvate in absence of oxygen is called anaerobic respiration.
In the absence of oxygen, the conversion of pyruvate may take place in two ways. They are-

(a) The pyruvate may be converted into ethanol and carbon dioxide.
This process takes place in yeast during fermentation.

(b) When our muscles lack oxygen, the pyruvate gets converted into lactic acid which is also a 3-carbon molecule.
When we over-stress our muscles they build up lactic acid which then results in cramps.

(2) In presence of oxygen (aerobic respiration):

The conversion of pyruvate in the presence of oxygen is called aerobic respiration.
In aerobic respiration, pyruvate breaks down in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. Water is also formed in this process.
Aerobic process releases much greater energy as compared to aerobic process.
The energy released during cellular respiration is immediately used to synthesize a molecule called ATP which is used to fuel all other activities in the, cell.

Question 3.
How is oxygen and carbondioxide transported in human beings?
Answer:
(a) Transport of oxygen:

In human body, hemoglobin has a very high affinity for oxygen.
Hemoglobin is present in the red blood cells. It carries oxygen from the respiratory surface of the lungs to the body cells.

(b) Transport of CO₂:
Answer:
Carbon dioxide is a respiratory gas to be exhaled out of the body. CO₂ is more soluble in water than oxygen. So, majority of CO₂ is transported in the dissolved form in blood plasma.

Question 4.
How are the lungs designed in human beings to maximize the area for exchange of gases?
Answer:
1. Lungs have balloon-like structure.
2. The windpipe gets divided into left and right bronchi.
3. The bronchi enters into the lungs and further divides into narrower tubes called bronchioles.
4. The bronchioles terminate into balloon-like structure known as alveoli.
5. These alveoli provide a large surface area for maximum exchange of respiratory gases.

Textbook Page no – 110

Question 1.
What are the components of the transport system In human beings? What are the functions of these components?
Answer:
The transport system of human body contains three main components. They are –
(i) Heart,
(ii) Blood and
(iii) Blood vessels and
(iv) Lymph

(i) Functions of heart:

The human heart is a muscular pump which keeps blood flowing in the body till the person ¡s alive.
The heart receives purified (oxygenated) blood from the lungs and pumps it towards the body cells.
The heart receives deoxygenated blood from the body parts and sends it towards the lungs for purification.

(ii) Function of various blood cells:

Red blood cells (RBCs): They contain hemoglobin. Hence, they transport O₂ from lungs to the body cells.
White blood cells (WBCs): They play an important role in providing immunity to the body. They protect the body from micro-organisms.
Blood platelets: When any blood vessel gets cut, the platelets clot the blood and prevent bleeding.
Blood plasma: Blood plasma transport nutrients, excretory substances, CO₂, hormones, enzymes, etc.

(iii) Functions of blood vessels:

Arteries carry blood from heart to the different organs of the body.
Exchange of materials between blood and its surrounding takes place through these capillaries.
Veins collect blood from different parts of the body and bring it back to the heart.

(iv) Functions of lymph: Lymph carries digested and absorbed fat from intestine and drain excess fluid from intercellular spaces back into blood.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
1. Human heart (and also birds) require high amount of energy. They need to maintain their body temperature constantly. So, they need constant energy supply and hence constant supply of oxygen.

2. The septum separates the right side and the left side of te heart. As a result, the oxygenated blood and deoxygenated blood remain in separate chambers and do not mix.

3. Such a separation enables a very efficient supply of oxygen to the body cells.

Question 3.
What are the components of the transport system in highly organized plants?
Answer:
Xylem tissue and phloem tissue are the components of the transport system in highly organized plants.

Question 4.
How are water and minerals transported In plants?
Answer:
Transport of water in higher plants:

Higher plants possess xylem for transporting water to all its parts.
Xylem contains tracheids and vessels.
Xylem tissues of all the organs of a plant are connected end to end with each other to form a network of conducting tubes.

Process of transport:
(a) Through diffusion:

The cells of the roots are in direct contact with the soil and they take up ions from the soil.
Due to this, a difference is created between the concentration of ions in the roots and that of ions in the soil.
The ions present in the soil water are at higher concentrations and so the water moves up from soil to the roots through osmosis.
This water movement creates a water column under which water is steadily pushed upwards.

(b) Through conduction (transpiration):

In higher plants transporting water till the highest point of the plant with this system is inefficient. As a result, higher plants take help of a process called transpiration.
The transpiration or evaporation of water from the leaves create a suction which pulls up water from the xylem vessels.
Thus the process of transpiration helps in the upward movement of water from roots to the leaves through the stem.
Since the stomata are open during the daytime, the transpiration pull becomes a major driving force in the movement of water in the xylem.

Transpiration:

The loss of water in the form of water vapour from the aerial parts of the plant is known as transpiration.
Plants absorb the water through the roots and transport it to all its parts.
This absorption is in fact driven by transpiration that occurs through the tiny pores on the leaves called stomata.
Transpiration creates a suction which pulls the water from the xylem cells of roots.
Thus, we can say transpiration plays an important role in making it possible for all the parts of the plant to receive water.
Transpiration also helps in regulating temperature.

Textbook Page no – 112.

Question 1.
Describe the structure and functioning of nephrons.
Answer:
Nephron:

Nephron is the main functional unit of kidney.
Each human kidney possesses about 10 lakh nephrons.

Structure of nephron:
Bowman’s capsule:

Each nephron has a double walled cup-shaped bag at its upper end which is called Bowman’s capsule.
The Bowman’s capsule contains a mass (bundle) of blood capillaries which is called glomerulus.

Tubule:

1. The lower end of the Bowman’s capsule is called tubule.
2. The part of tubule which is near the Bowman’s capsule is quite small and is called the neck.
3. After the neck, the tubule becomes very narrow and coiled.
4. This region of tubule consists of

A proximal convoluted tubule,
A Henle’s loop and
A distal convoluted tubule.

5. The posterior end of nephron is known as collecting tube.
6. Collecting tubule opens into renal pelvis.
7. The renal pelvis opens into ureter.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Excretion in Plants:

For plants, oxygen is one of the end products of photosynthesis and can also be considered as a waste product.
Plants emit this oxygen into atmosphere through diffusion process.
Plants get rid of excess water through transpiration.
For other waste products, many of the plant tissues contain dead cells in themselves.
Some of the material is also lost by falling leaves.
In the plant cell, vacuoles are the excretory organelle.
Around the root system in the soil, plant excretes some of the waste products.
Peeling of the bark is also the example of plant excretion.
Resins and gums are excretions of plants.

Question 3.
How is the amount of urine produced In the body, regulated?
Answer:
The amount of urine formed in the body depends on —

Amount of excess water present in the body
Amount of nitrogenous wastes produced in the body
Amount of re-absorption in kidneys
Certain hormones like ADH (Anti Diuretic Hormone)
Surrounding temperature

Activities

Activity 1.

Aim: To demonstrate that chlorophyll is essential for photosynthesis.

Procedure:

Take a potted plant with variegated leaves — for example, money-plant or croton.
Keep the plant in a dark room for three days so that all the starch gets used up.
Now keep the plant in sunlight for about six hours.
Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
Dip the leaf in boiling water for a few minutes.
After this, immerse it in a beaker containing alcohol.
Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
Now, dip the leaf in a dilute solution of iodine for a few minutes, Fig
Take out the leaf and rinse off the iodine solution.

Question 1.
What happens to the colour of the leaf? What is the colour of the solution?
Answer:
The leaf becomes colourless because on boiling the leaf, it loses the pigmentation. The water becomes green in colour due to presence of chlorophyll that was present in the leaf.

Question 2.
Observe the colour of the leaf and compare this with the tracing of the leaf done In the beginning.
Answer:
The leaf shows blue-black and colourless patches.

Question 3.
What can you conclude about the presence of starch in various areas of the leaf?
Answer:
The green patched areas become blue-black when they are tested with iodine. While the chlorophyll-

Activity 2.

Aim: To demonstrate that carbon dioxide is necessary for photosynthesis.

1. Take two healthy potted plants which are nearly the same size.
2. Keep them in a dark room for three days.
3. Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one
of the plants. The potassium hydroxide is used to absorb carbon dioxide.
4. Cover both plants with separate bell-jars as shown in figure 2.
5. Use Vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
6. Keep the plants in sunlight for about two hours.
7. Pluck a leaf from each plant and check for the presence of starch as in the above activity.

Question 1.
Do both the leaves show the presence of the same amount of starch?
Answer:
The leaves of plant ‘a’ show absence of starch, while the leaves of plant ‘b’ show presence of starch.

Question 2.
What can you conclude from this activity?
Answer:
1. Starch is not formed in the leaf of the plant kept under the bell-jar along with potassium hydroxide.
This is so because potassium hydroxide absorbs carbon dioxide from air. Hence, the plant does not get carbon dioxide and it can not undergo photosynthesis.
2. From the above activity, we can conclude that a good amount of CO₂ is necessary to carry out photosynthesis at proper rate.

Activity 3.

Aim: To check the effect of saliva on starch.
1. Take 1 mL starch solution (1%) in two test tubes (A and B).
2. Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
3. Now, add a few drops of dilute iodine solution to the test tubes.

Question 1.
In which test tube do you observe a colour change?
Answer:
The solution of test tube B will turn bluish black.

Question 2.
What does this indicate about the presence or absence of starch in the two test tubes?
Answer:
It shows that starch is present in test tube ‘B’ while test tube ‘A’ does not have starch.

Question 3.
What does this tell us about the action of saliva on starch?
Answer:
The result shows that the saliva of test tube ‘A’ acts upon starch. So, starch gets converted into sugar and the solution does not show reaction to iodine test.

Activity 7.

Aim: To find out the haemoglobin content in human beings and in animals such as buffalo and cow.

Question 1.
Visit a health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
Answer:
It is 13 — 18 g/dL (Note: dL = decilitre which means 1/10th of 1 litre or say loo mL)

Question 2.
Is It the same for children and adults?
Answer:
No, it is 12.5 g/dL for children

Question 3.
Is there any difference in the haemoglobin levels for men and women?
Answer:
In males, it is 13 — 18 g/dL and in females, it is 12 -10 g/dL.

Question 4.
Visit a veterinary clinic In your locality. Find out what is the normal range of hemoglobin content in an animal like buffalo or cow.
Answer:
It Is 10.4 — 16.4 g/dL in such animals.

Question 5.
Is this content different in calves, male and female animals?
Answer:
Yes, the haemoglobin level of calves is more than the adult male and female animals.

Question 6.
How would the difference, if any, be explained?
Answer:
Hemoglobin content depends on the personal health, genetic map and environment.

Activity 8.

Aim : To demonstrate physiological process of transpiration in plants.

1. Take two small pots of approximately the same size and having the same amount of soil. One should have a plant in it. Place a stick of the same height as the plant in the other pot.
2. Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
3. Cover both sets, one with the plant and the other with the stick, with plastic sheets and place in bright sunlight for half an hour.

Question 1.
Do you observe any difference in the two cases?
Answer:
The plant which is covered with a plastic sheet, shows some water droplets inside the plastic sheet. This demonstrates that there is water loss from the aerial parts of a plant. This process is called transpiration.

Haryana Board 10th Class Science Solutions Chapter 6 Life Processes

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