Bihar Board Class 10 Science Solutions Chapter 6 Life Processes

Bihar Board Class 10th Science Book Solutions विज्ञान Chapter 6 Life Processes – NCERT पर आधारित Text Book Questions and Answers Notes, pdf, Summary, व्याख्या, वर्णन में बहुत सरल भाषा का प्रयोग किया गया है.

Bihar Board Class 10 Science Solutions Chapter 6 Life Processes

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans ?
In multicellular organisms, all the cells are not in direct contact with surrounding environment. Therefore, simple \ diffusion will not meet the oxygen requirement of all the cells,

Question 2.
What criteria do we use to decide whether something is alive ?
Living organisms must keep repairing and maintaining 1 their structures. All these structures are made up of molecules. ” They must be capable of moving molecules around, all the time. So, molecular (invisible) movement is necessary to identify whether something is alive.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 3.
What are outside raw materials used by ah organism ?

Food as a source of supplying energy and materials.
Oxygen for the breakdown of the food to obtain energy.
Water for proper digestion of food and other functions inside the body.
Question 4.
What processes would you consider essential for maintaining life?
There are various life processes which are essential for maintaining life.
Some of them are :

Intext Questions (Page 101)

Question 1.
What are the differences between autotrophic . nutrition and heterotrophic nutrition ?
Autotrophic nutrition : When the organisms like y green plants prepare their own food by CO2 and water in presence of chlorophyll and sunlight, then they are called autotrophs. The process is called photosynthesis.
Heterotrophic nutrition : When organisms do not prepare their own food but depend on others for their food, then they are called heterotrophs, e.g., fungi and human beings.

Question 2.
Where do the plants get each of the raw materials required for photosynthesis ?
Plants need the following things for photosynthesis.

Carbon dioxide : Plants get CO2 from atmosphere through stomata.
Water : Plants absorb water from soil through roots and transport to leaves.
Question 3.
What is the role of acid in our stomach ?
Roles of acid in the stomach are as follows :
(i) It makes an acidic medium in the stomach which is it necessary for activation of pepsin enzyme.
(ii) It kills the germs present in the food,

Question 4.
What is the function of digestive enzymes ?
The food materials are very complex in nature. Digestive ; enzymes help to break these complex molecules into smaller molecules so that they can be absorbed by the walls of small intestine.

Question 5.
How is the small intestine designed to absorb digested food ?
Small intestine is designed to provide more and more area for absorption of digested food and transfer it into the blood for circulation throughout the body. The inner lining of small intestine has a large number of fingerrlike projections called
villi. These villi provide a large surface area for absorption of food.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 1

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Intext Questions (Page 105)

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtain oxygen for respiration ?
The organisms that live in water use oxygen dissolved in surrounding water. Since air dissolved in water has fairly low concentration of oxygen, the aquatic organisms has much faster rate of breathing. Terrestrial organisms take oxygen from the oxygen-rich atmosphere through respiratory organs. So, they have much less breathing rate than aquatic organisms.

Question 2.
What are the differenfcways in which glucose is oxidised to provide energy in various organisms ?
First step of breakdown of glucose (six carbon molecules) takes place in the cytoplasm of cells of all organisms. This process yields a three carbon molecule compound called pyruvate. Further breakdown of pyruvate takes place in different manners in different organisms.

Anaerobic respiration : This process takes place in absence of oxygen, e.g., in yeast during fermentation. In this case pyruvate is converted into ethanol and carbon dioxide.
Aerobic respiration : In aerobic respiration, breakdown of pyruvate takes place in presence of oxygen to give rise three molecules of carbon dioxide and water. The release of energy in aerobic respiration is much more than anaerobic respiration.
Lack of oxygen : Sometimes, when there is lack of oxygen, especially during vigorous activity, in our muscles, pyruvate is converted into lactic acid(three carbon molecule , compounds). Formation of lactic acid in muscles causes cramp.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 2
Question 3.
How are oxygen and carbon dioxide transported in human beings?
(a) Transport of oxygen : The respiratory pigments (haemoglobin) present in the red blood cells takes up the oxygen from the air to the lungs. They carry the oxygen to tissues which . are deficient in oxygen.
(b) Transport of carbon dioxide : Carbon dioxide is more soluble in water. Hence, it is mostly transported from body tissues in the dissolved form in our blood plasma to lungs where it diffuses from blood to air in the lungs and then expelled out through nostrils.

Question 4.
How are lungs designed in human beings to maximize the area for exchange of gases ?
In the lungs, the air passage (windpipe) divides into smaller tubes, called bronchi in turn form bronchioles. The bronchioles which terminate in balloon-like structures, called alveoli. The alveoli present in the lungs provide maximum surface for exchange of gases. The alveoli have very thin walls and contain an extensive network of blood vessels to facilitate exchange of gases.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 3
Intext Questions (Page 110)

Question 1.
What are the components of the transport system in human beings ? What are the functions of these components ? .
Transport system in human beings consists of heart, blood and blood vessels.

Functions :
(i) Heart: Heart is a pumping organ to push and pull blood around the body. It receives the deoxygenated blood from various parts of the body and pumps oxygenated blood throughout the body.
(ii) Blood : It is a fluid connective tissue. It consists of: (a) plasma, (b) RBC, (c) WBC and (d) blood platelets. Plasma transports food, CO2 and nitrogenous wastes in dissolved form. RBC transports respiratory gases and hormones. WBC protects the body from infections and platelets prevent the loss of blood at the time of injury by forming blood clots.
(iii) Blood vessels : There is a network of vessels. They help in the circulation of blood throughout the body.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds ?
It is necessary to separate the oxygenated and deoxygenated blood to maintain efficient supply of oxygen into the body. This system is essential in animals that have high energy need. Mammals and birds require constant supply of oxygen to obtain energy.’

Question 3.
What are the components of the transport system in highly organised plants ?
The main components of transport system, in plants are :
(i) Xylem, (ii) Phloem.
(i) Xylem : It consists of vessels and tracheids. Xylem helps to conduct water and minerals from soil to the leaves.
(ii) Phloem : It consists of sieve tubes and companion cell. Phloem helps to transport food materials from leaves to various parts of the plant. This process is called translocation.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 4.
How are water’and minerals transported in plants ?
Water and minerals are transported through xylem cells from soil to the leaves. The xylem cells of roots, stem and leaves are interconnected to form a conducting channel. The root cells take ions from the soil. This creates a difference between the concentration of ions of roots and soil. Therefore, there is a steady movement of water into xylem. An osmotic pressure is formed and water and minerals are transported from one cell to the other cell due to osmosis. The continuous loss of water takes place due to transpiration. Thus, the continuous movement of water and mineral molecules remain and transportation of water and minerals takes place.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 5
Fig. 6.18. Movement of water during transpiration in a tree.

Question 5.
How is the food transported in plants ?
The prepared food is transported in the plants through phloem to the storage organs of roots, fruits, seeds and growing parts. This process is called translocation. This function is done by sieve tubes and companion cells. The movement of food particles takes place upward and downward.

Mechanism of translocation is an active process that utilizes energy. Materials are transferred from leaf cells or from the site of storage into phloem tissue. For this, energy is required which is provided by the ATP molecules. This energy increases the osmotic pressure, as a result, water from outside moves into the phloem. This pressure maintains the movement of food through all the parts of plants.

Intext Questions (Page 112)

Question 1.
Describe the structure and functioning of nephron.
Structure of nephron : Nephron is the filtration unit of kidney. It consists of a tubule which is connected with collecting duct at one end and a cup-shaped structure at the other end.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 6

This cup-shaped structure is called Bowman’s capsule. Every Bowman’s capsule contains a cluster of capillaries, called glomerulus, within the cup-shaped structure. The blood enters into glomerulus through afferent arteriole of renal artery and leaves it through efferent arteriole.

Functioning of nephron :
(i) Filtration : Filtration of blood takes place in Bowman’s capsule from the capillaries of glomerulus. The filtrate passes into the tubular part of the nephron. This filtrate contains glucose, amino acids, urea, uric acid, salts and a major amount of water.

(ii) Reabsorption : As the filtrate flows along the tubule useful substances such as glucose, amino acids, salts and water are selectively reabsorbed into the blood by capillaries surrounding the nephron tubule.
The amount of water reabsorbed depends on the need of the body and also on the amount of wastes to be excreted.

(iii) Urine : The filtrate which remains after reabsorption is called urine. Urine contains dissolved nitrogenous waste, i.e., urea and uric acid, excess salts and water. Urine is collected from nephrons by the collecting duct to carry it to the ureter.

Question 2.
What are the methods used by plants to get rid of excretory products ?
To get rid of excretory products, plants use the following ways :

Many waste products are stored in vacuoles of the cells. Plant cells have comparatively large vacuoles.
Some waste products are stored in the leaves. They are removed as the leaves fall off.
Some waste products such as resins and gums are stored, especially in non-functional old xylem.
Some waste products such as tannins, resins, gums are stored in bark, thereby removed as peeled off.
Plants also excrete some waste substances through roots into the soil around them.
Question 3.
How is the amount of urine produced regulated ?
The amount of urine largely depends on the amount of water reabsorbed. The amount of water reabsorbed by the nephron tubule depends on :
(i) How much water in excess is in the body removed ?
When water is abundant in the body tissues, large quantities of dilute urine is excreted out. When water is less in quantities in the body tissues, a small quantity of concentrated urine is excreted.

(ii) How much dissolved wastes, especially nitrogenous wastes like urea and uric acid and salts to be excreted from the body ?
When there is more quantity of dissolved wastes in the body, more quantity of water is required to excrete them. So, the amount of urine produced increases.

Bihar Board Class 10 Science Chapter 6 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) transpiration
(c) excretion

Question 2.
The xylem in plants are responsible for –
(а) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen
(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
(d) all of the above

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in –
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus
(b) mitochondria

Question 5.
How are fats digested in our body ? Where does this process take place ?
(i) Digestion of fat takes place in the small intestine.
(ii) Digestion of fat : The fats are present in the form of large globules in the small intestine. Fat-digesting enzymes are not able toact upon large globules efficiently. Bile juice secreted by the liver is poured in the intestine along with pancreatic juice. The bile salts present in the bile juice emulsify the large globules of fats.

So, by emulsification large globules break down into fine globules to provide larger surface area to act upon by the enzymes. Lipase enzyme present in the pancreatic juice causes breakdown of emulsified fats. Glands present in the wall of small intestine secrete intestinal juice which contains lipase enzyme that converts fats into fatty acids and glycerol.

Small intestine : Region of digestion of fats :
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 7
Question 6.
What is the role of saliva in the digestion of food?
(i) The saliva contains salivary amylase enzyme that breaks down starch to sugars like maltose.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 8
(ii) The saliva moistens the food that helps in chewing and b’eaking down the big pieces of food into smaller ones, so that, salivary amylase can digest the starch efficiently.

Question 7.
What, are the necessary canditions for autotrophic nutrition and what are its by-products?
Necessary conditions for autotrophic nutrition:
(i) (a) Presence of chlorophyll in the living cells.
(b) Provision of supply of water to green parts or cells of the plant either through roots or by surrounding environment.
(c) Availability of sufficient of sunlight to provide light energy required to carry out photosynthesis.
(d) Sufficient supply of carbon dioxide which is one of the important component for the formation of carbohydrates during

Question 8.
What are the diffòrences between aerobic and anaerobic respirations ?Name some organisms that use
the anaerobic respiration.
Answer: (i)

Aerobic respiration :

Aerobic respiration takes place in the presence of free oxygen.
In aerobic respiration, complete oxidation of glucose takes place.
End products of aerobic respiration are CO2, water and energy C6H12O6 + 6O2 → 6H2O + 6CO2 + energy (686 kcal).
Large amount of energy is released, 38 molecules of ATP per glucose molecule.
First step of aerobic respiration (glycolysis) takes place in cytoplasm while the next step takes place in mitochondria.
Anaerobic respiration :

It takes place in the absence of free oxygen.
In anaerobic respiration, the glucose rholecule is incompletely broken down.
End products of anaerobic respiration are ethyl alcohol (or lactic acid), CO2 and a little energy.
C6H12O6 →2CO2 + 2C2H5OH + energy (58 kcal)
Small amount of energy is released, i.e., 2 ATP molecules per glucose molecule.
Complete anaerobic respiration occurs in cytoplasm.
(ii) Organisms which use anaerobic respiration are yeasts, bacteria and parasites like tapeworm (Taenia), Ascaris, etc.

Question 9.
How are the alveoli designed to maximize the exchange of gases ?
(i) The alveoli are thin walled and richly supplied with a network of blood vessels to facilitate exchange of gases between blood and the air filled in alveoli.
(ii) Alveoli have balloon-like structure that provide maximum surface area for the exchange gases.

Question 10.
What would be the conse-quences of deficiency of haemoglobin in our bodies ?
The average haemoglobin content of blood, irrespective of sex, is 14.5 g per 100 mL. If haemoglobin content reduces in blood, its oxygen carrying capacity decreases. So, the person shows symptoms of deficient oxygen such as breathlessness, often one of the first indications of iron deficiency is anaemia.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 11.
Describe double circulation in human beings. Why is it necessary ?
In human beings, the blood goes through the heart twice during each cycle, i.e., the blood passes through the human heart two times to supply once to the whole body. So, it is called double circulation of blood. The double circulation of blood includes :
(i) Systemic circulation and
(ii) Pulmonary circulation
(i) Systemic circulation : It supplies oxygenated blood from left auricle to left ventricle, thereby pumped to various body parts. The deoxygenated blood is collected from the various body organs by the veins to pour into vena cava and finally into the right-atrium (auricle). Right atrium transfers the blood into the right ventricle.

(ii) Pulmonary circulation : The deoxygenated blood is pushed by the right ventricle into the lungs for oxygenation. The oxygenated blood is brought back to left atrium of the human heart. From left atrium, the oxygenated blood is pushed into the left ventricle. The left ventricle pumps oxygenated blood into aorta for systemic circulation.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 9
Necessity of double circulation : The right side and the left side of the human heart are useful to keep deoxygenated and oxygenated blood from mixing. This type of separation of oxygenated and deoxygenated blood ensures a highly efficient supply of oxygen to the body. This is useful it case of humans because it constantly gives maintain their body temperature.

Question 12.
What are the differences between the transport of materials in xylem and phloem ?
Transport of materials :
Xylem :

Xylem conducts water and dissolved mineral from roots to leaves and other part.
In xylem, the transport of material takes pláce through vessels and tracheids which are dead tissues.
In xylem, upward movement of water and dissoved minerals is mainly achieved by transpiration pull. It is created by the evaporation of water molecules from the cells of a leaf.
Movement of water is achieved by simple physicalforces. There is no expenditure of energy. So, ATP molecules are not required.
Phicem :

Phloem conducts prepared food material from leaves to other parts of plant in dissolved form.
In phloem, transport of material takes place through sieve tubes .with the help of companion cells, which are living cells.
In translocation, material is transferred into phloem tissue using energy from ATP. This increases the osmotic pressure that moves the material in the phloem to the tissues which have less pressure.
The translocation in phloem is an active process and requires energy. This energy is taken from ATP.
Question 13.
Compare the functioning of alveoli in the lungs and nephron in the kidneys with respect to their structure and functioning.
Comparison between alveoli and nephron :
Alveoli :

Alveoli hâve thin walled balloon-like structure. Surface is fine and delicate.
Alveoli are supplied with extensive network of thin walled blood vessels, i.e., capillaries for exchange of gases.
Alveoli increàse surface area for diffusion of CO2 from blood to air and 02 from air to blood.
Alveoli only provide surface for exchange of gases in the lungs.
Alveoli are very small and a large number of them are present ineach lung.
Nephron :

Nephrons have thin walled, cup-shaped structure attached with thin walled tubule.
Bowman’s capsule is supplied with a cluster of capillaries, called glomerulus for filtration. A network of blood vessels is present around the tubular part of nephron for reabsorption of useful substances and water.
Nephrons also increase surface area for filtration of blood and reabsorption of useful substances and water from filtrate leaving behind urine.
Tubular part of nephron also carries the urine to collecting duct.
A large number of nephrons, the basic filtration unit, are present in each kidney.
Bihar Board Class 10 Science Chapter 6 Life Processes Textbook Activities
Activity 6.1 (Page 96)

Take a potted plant with variegated leaves – for example, money plant or crotons.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 10
Keep the plant in a dark room for three days so that all th6 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.
Question 1.
What happens to the colour of the leaf ? What is the colour of the solution ?

The colour of the leaf disappears and it becomes colourless because chlorophyll dissolves in alcohol. The colour of the solution becomes green.
Now dip the leaf in a dilute solution of iodine for a few minutes.
Take out the leaf and rinse off the iodine solution.
Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning (Fig. 6.5).
Question 2.
What can you conclude about the presence of starch in various areas of the leaf ?
The areas of leaf which become dark’blue or black due to iodine solution show the presence of starch while those which remain colourless show that no starch formation took place. This activity indicates that chlorophyll is essential for photosynthesis.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Activity 6.2 (Page 97)

Take two healthy potted plants which are nearly the same size.
Keep them in a dark room for three days.
Cover both plants with separate bell-jars as shown in Fig. 6.6.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 11
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.
Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
Keep the – plants in sunlight for about two hours.
Pluck a’Teaf 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 sarftfe amount of starch?
No, both the leaves do not show same amount of starch because both get different amount of C02 due to the presence of KOH Iri one setup.

Question 2.
What can you conclude from this activity ?
This activity indicates that amount of carbon dioxide is an essential factor for the photosynthesis.

Activity 6.3 (Page 99)

Take 1 mL starch solution (1%) in two test tubes (A and B).
Add 1 mL- saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
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 ?
In test tube B, colour is changed because it contains starch while in test tube A, starch is converted into sugar, so no colour change is observed.

Question 2.
What does this indicate about the presence or absence of starch in the two test tubes ?
It indicates that test tube B contains starch while test tube A does not.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 3.
What does this tell us about the action of saliva oil starch ?
It tells us that saliva acts on the stanch and converts it in other substance (maltose sugar).

Activity 6.4 (Page 101)

Take some freshly prepared lime water in a test tube.
Blow air through this lime water.
Note how long it takes for the lime water to turn milky.
Use a syringe or pichkari to pass air through some fresh lime water taken in another test tube (Fig. 6.10).
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 12
Question 1.
Blow air through lime water and note how long it takes for the lime water to turn milky.
Do it yourself.

Question 2.
How long it takes for the lime water to turn jnilky?
Do it yourself.

Question 3.
What does this tell us about the amount of carbon dioxide in the air that we breathe out ?
The lime water in first case [Fig. 6.10 (a)] takes more time to turn milky than that in the second case [Fig. 6.10 (£>)]. This indicates that the exhaled air contains more carbon dioxide than fresh, air. Exhaled air turns lime water milky quicker than normal air. This shows that exhaled air contains more carbon dioxide.

Activity 6.5 (Page 101)

Take some fruit juice or sugar solution and add some yeast to this. Take this mixture in a test tube fitted with a one- holed cork.
Fit the cork with a bent glass tube. Dip the free end of the glass tube into a test tube containing freshly prepared lime water.

Question 1.
What change is observed in the lime water and how long does it take for this change to occur ?
Lime water becomes milky because carbon dioxide is formed by mixing yeast in sugar and alcohol is also formed. The time should be noted by students themselves on the basis of observation.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 2.
What does this tell us about the products of fermentation ?
This tells us that carbon dioxide is one of the products with other products (alcohol).

Activity 6.6 (Page 103)

Question 1.
Observe fish in an aquarium. They open and close their mouths and the gill-slits (or the operculum which covers the gill-slits) behind their eyes also open and close. Are the timings of the opening and closing of the mouth and the gill slits coordinated in some manner ?
Yes. They open and close alternately.

Question 2.
Count the number of times the fish opens and closes its mouth in one minute.’
Opening and closing of mouth varies from fish to fish and size of the fish species, etc. Students are advised to do it themselves.

Question 3.
Compare this to the number of times you breathe in and out a minutes.
Fish breathe much faster than us because less amount of oxygen is available in water than air.

Activity 6.7 (Page 105)

Question 1.
A health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
The normal range of haemoglobin content in human beings is 13.8 to 17.2g/dl (gram per deci litre) is men and 12.1 to 15.1 g/dl in women.

Question 2.
Is it same for children and adults ?
No. In adult male as mentioned in previous answer, it is 13.8 to 17.2 g/dl while in children, it is 11 to 16 g/dl.

Question 3.
Is there any different in the haemoglobin levels for men and women ?
Yes, The difference is mentioned in the previous answers.

Question 4.
Visit a veterinary clinic in your locality. Find out what is the normal range of haemoglobin content in a animal like the buffalo or cow.
The normal range of haemoglobin content in cattles is

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 5.
Is this content different in calves, male and female animals ?
Yes. The haemoglobin content in calves is higher than that of adult cattles. .

Question 6.
Compare the difference seen in male and female human beings and animals.
The haemoglobin levels are as following :
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 4
Question 7.
How would the difference, if any, be explained ?
We know that haemoglobin is essential for the transportation of O2 and CO2. Clearly, as adult men do more work than that of women or children, the amount of haemoglobin is more in men. Similarly, due to the nature and variety of work human being does the haemoglobin content is more than that of cattles.

Activity 6.8 (Page 105)

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.
Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
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 ?
Yes. The water droplets are seen in the plastic sheet in the pot which has plant. These water droplets are formed due to condensation of water vapour released by the plant in the process of transpiration. While, there is no water drops in other pot which has stick.

Bihar Board Class 10 Science Chapter 6 Life Processes NCERT Exemplar Problems
Short Answer Type Questions

Question 1.
Name the following:
(a) The process in plants that links light energy with chemical energy
(b) Organisms that can prepare their own food
(c) The cell organelle where photosynthesis occurs
(d) Cells that surround a stomatal pore
(e) Organisms that cannot prepare their own food
(f) An enzyme secreted from gastric glands in stomach that acts on proteins.
(a) Photosynthesis
(b) Autotrophs
(c) Chioroplast
(d) Guard Cells
(e) Heterotrophs
(f) Pepsin

Question 2.
“All plants give out oxygen during day and carbon dioxide during night”. Do you agree with this statement? Give reason.
During day time, as the rate of photosynthesis is more than the rate of respiration, the net result is evolution of oxygen. At night there is no photosynthesis, so they give out carbon dioxide due to respiration.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 3.
How do the guard cells regulate opening and closing of stomatal pores?
The swelling of guard cells due to absorption of water causes opening of stomatal pores while shrinking of guard cells closes the pores. Opening and closing of stomata occurs due to turgor changes in guard cells. When guard cells are turgid, stomatal pore is open while in flaccid conditions, the stomatal aperture closes.

Question 4.
Two green plants are kept separately in oxygen free containers, one in the dark and the other in continuous light. Which one will live longer? Give reasons.
Plant kept in continuous light will live longer, because it will be able to produce oxygen required for its respiration by the process of photosynthesis.

Question 5.
If a plant is releasing carbon dioxide and taking in oxygen during the day, does it mean that there is no photosynthesis occurring? Justify your answer.
Release of CO2 and intake of O2 gives evidence that either photosynthesis is not taking place or its rate is too low. Normally during day time, the rate of photosynthesis is much more than the rate of respiration. So, CO2 produced during respiration is used up for photosynthesis hence CO2 is not released.

Question 6.
Why do fishes die when taken out of water?
Fishes respire with the help of gills. Gills are richly supplied with blood capillaries and can readily absorb oxygen dissolved in water. Since fishes cannot absorb gaseous oxygen they die soon after they are taken out of water.

Question 7.
Differentiate between an autotroph and a heterotroph.
Autotroph :

Orgasms that prepare their own food.
They have chlorophyll.
Heterotroph :

Organisms that are dependent on other organisms for food.
They lack chlorophyll.
Question 8.
Is ‘nutrition’ a necessity for an organism? Discuss.
Food is required for the following purposes :
(a) It provides energy for the various metabolic processes in the body.
(b) It is essential for the growth of new cells and repair or replacement of worn-out cells.
(c) It is needed to develop resistance against various diseases.
Thus, nutrition is a necessity for an organism.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 9.
What would happen if green plants disappear from earth?
Green plants are the sources of energy for all organisms. If all green plants disappear from the earth, all the herbivores will die due to starvation and so will the carnivores that are dependent on them.

Question 10.
Leaves of a healthy potted plant were coated with vaseline. Will this plant remain healthy for long? Give reasons for your answer.
This plant will not remain healthy for a long time because.
(a) it will not get oxygen for respiration.
(b) it will not get carbon dioxide for photosynthesis.
(c) Upward movement of water and minerals would be hampered due to lack of transpiration. .

Question 11.
How does aerobic respiration differ from anaerobic respiration?
Aerobic respiration :

Oxygen is utilised for the break-down of respira-tory substrate.
It takes place in cytoplasm (glyco-lysis) and inside mitochondria (Krebs cycle).
End products are carbon dioxide and water.
More energy is released.
Anaerobic respiration :

Oxygen is not required.
It takes place in cytoplasm only.
End products are lactic acid or ethanol and carbon dioxide.
Less energy is released.
Question 12.
Match the words of Column (A) with that of
Column (B)
Column (A) Column (B)
(a) Phloem – (i) Excretion
(b) Nephron – (ii) Translocation of food
(c) Veins – (iii) Clotting of olood
(d) Platelets – (iv) Deoxygenated blood
(b) → (i),
(c) → (iv),
(d) → (iii).

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 13.
Differentiate between an artery and a vein.
Artery :

Have thick elastic, muscular walls.
Lumen is narrow.
Vein :

Have thin, non-elastic, walls.
Lumen is wide.
Question 14.
What are the adaptations of leaf for photosynthesis?
(a) Leaves provide large surface area for maximum light absorption.
(b) Leaves are arranged at right angles to the light source in a way that causes overlapping.
(c) The extensive network of veins enables quick transport of substances to and from the mesophyll. cells.
(d) Presence of numerous stomata for gaseous exchange.
(e) The chloroplasts are more in number on the upper surface of leaves.

Question 15.
Why is small intestine in herbivores longer than in carnivores?
Digestion of cellulose takes a longer time. Hence, herbivores eating grass need a longer small intestine to allow complete digestion of cellulose. Carnivorous animals cannot digest cellulose, hence they have a shorter intestine.

Question 16.
What will happen if mucus is not secreted by the gastric glands?
Gastric glands in stomach release hydrochloric acid, enzyme pepsin and mucus. Mucus protects the inner lining of stomach from the action of hydrochloric acid and enzyme pepsin. If mucus is not released, it will lead to erosion of inner lining of stomach, leading to acidity and ulcers.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 17.
What is the significance of emulsification of fats?
Fats are present in food in the form of large globules which makes it difficult for enzymes to act on them. Bile salts present in bile break them down mechanically into smaller globules which increases the efficiency of fat digesting enzymes.

Question 18.
What causes movement of food inside the alimentary canal?
The wall of alimentary canal contains muscle layers. Rhythmic contraction and relaxation of these muscles pushes the food forward. This is called peristalsis, which occurs all along the gut.

Question 19.
Why does absorption of digested food occur mainly in the small intestine? ‘
Maximum absorption occurs in small intestine because
(a) digestion is completed in small intestine.
(b) inner lining of small intestine is provided with villi which increases the surface area for absorption.
(c) wall of intestine is richly supplied with blood vessels
(which take the absorbed food to each and every cell of the body). ‘

Question 20.
Match Group (A) with Group (B)
(а) Autotrophic nutrition – (i) Leech
(b) Heterotrophic nutrition – (ii) Paramecium
(c) Parasitic nutrition – (iii) Deer
(d) Digestion in food vacuoles – (iv) Green Plant
(a) → (iv)
(b) → (iii)
(e) → (i)
(d) → (ii)

Question 21.
Why is the rate of breathing in aquatic organisms much faster than in terrestrial organisms?
Aquatic organisms like fishes obtain oxygen from water present in dissolved state through their gills. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 22.
Why is blood circulation in human heart called double circulation?
The blood circulation in human heart is called double circulation because the blood passes through the heart twice in one complete cycle of the body – once through the right half in the form of deoxygenated blood and once through the left half in the form of oxygenated blood.

Question 23.
What is the advantage of having four chambered heart?
In four chambered heart, left half is completely separated from right half by septa. This prevents oxygenated and deoxygenated blood from mixing. This allows a highly , efficient supply of oxygenated blood to all parts of the body. This is useful in animals that have high energy needs, such as birds and mammals.

Question 24.
Mention the Ynajor events during photosynthesis.
The major events during photosynthesis are –
(a) absorption of light energy by chlorophyll.
(b) conversion of light energy to chemical energy.
(c) splitting of H2O into H2, O2 and e
(d) reduction of CO2 to carbohydrates.

Question 25.
In each of the following situations what happens to the rate of photosynthesis?
(а) Cloudy days
(b) No rainfall in the area
(c) Good manuring in the area
(d) Stomata get blocked due to dust
(a) Decreases
(b) Decreases
(c) Increases
(d) Decreases

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 26.
Name the energy currency in the living organisms. When and where is it produced?
Adenosine triphosphate (ATP) is the energy currency in the living organisms. It is produced in mitochondria during respiration in living organisms. In plants, during photosynthesis too some molecules of ATP are produced in chloroplasts.

Question 27.
What is common for cuscuta, ticks and leeches?
All are parasites, they derive nutrition from plants or animals without killing them

Question 28.
Explain the role of mouth in digestion of food.
(a) Food is crushed into small pieces by the teeth.
(b) It mixes with saliva and the enzyme amylase (found in saliva) breaks down starch into sugars.
(c) Tongue helps in thorough mixing of food with saliva.

Question 29.
What are the functions of gastric glands present in the wall of the stomach?
(a) Production of pepsin enzyme that digests proteins
(b) Secretion of mucus for protection of inner lining of stomach.

Question 30.
Match the terms in Column (A) with those in Column (B)
(а) Trypsin – (i) Pancreas
(b) Amylase – (ii) Liver
(c) Bile – (iii) Gastric glands
(d) Pepsin – (iv) Saliva
(a) → (i)
(b) → (iv)
(c) → (ii)
(d) → (iii)

Question 31.
Name the correct substrates for the following enzymes :
(a) Trypsin
(b) Amylase
(c) Pepsin
(d) Lipase
(a) Protein
(b) Starch
(c) Protein
(d) Fats

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 32.
Why do veins have thin walls as compared to arteries?
Arteries carry blood from the heart to various organs of the body under high pressure so thev have think and election walls. Veins collect the blood from different organs and bring it back to the heart. The blood is no longer under pressure so the walls are thin with valves to ensure that blood flows only in one direction.

Question 33.
What will happen if platelets were absent in the blood?
The blood has platelet cells which circulate around the body and plug the leaks in blood carrying tubes by helping to clot the blood at the points of injury. In the absence of platelets, the process of clotting would be affected.

Question 34.
Plants have low energy needs as compared to , animals. Explain.
Plants do not move. In a large plant body there are many dead cells like schlerenchyma as a result it requires less energy as compared to animals.

Question 35.
Why and how does water enter continuously into the root xylem?
Cells of root are in close contact with soil and so actively take up ions. The ion-concentration increases inside the root and hence osmotic pressure increases the movement of water from the soil into the root which occurs continuously. .

Question 36.
Why is transpiration important for plants ?
Transpiration is important because
(a) it helps in absorption and upward movement of water and minerals from roots to leaves.
(b) it prevents the plant parts from heating up.

Question 37.
How do leaves of plants help in excretion?
Many plants store waste materials in the vacuoles of ‘ mesophyll cells and epidermal cells. When old leaves fall, the waste materials are excreted along with the leaves.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Long Answer Type Questions

Question 38.
Explain the process of nutrition in Amoeba.
Ans. The process of nutrition in Amoeba : First of all, amoeba takes in food using temporary finger-like extensions of the cell surface which fur over the food particle forming a food- vacuole. Inside the food-vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. The remaining undigested material is moved to the surface of the cell and thrown out of the body.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 13
(a) Amoeba approaching the food particle.
(b) Extensions of plasma membrane forming pseudopodia to surround the food particle.
(c) Engulfed food particle.
(d) Food vacuole containing food particle in the cytoplasm of amoeba.

Question 39.
Describe the alimentary canal of man.
The human alimentary canal is basically a lpng tube extending from the mouth to the anus. This tube has different parts. Let’s discuss different parts of the human alimentary canal alPng with their functions :

Mouth (Buccal cavity) : The mouth (buccal cavity) contains teeth, tongue and salivary glands. When food is taken in the mouth, teeth chew the food and saliva secreted by the salivary glands gets mixed with it. The saliva contains an enzyme called salivary amylase that breaks down starch which is a complex molecule to give sugar. The food is mixed thoroughly with saliva and moved around the mouth while chewing by the muscular tongue.

Oesophagus : From the mouth, the food is taken to the stomach through the food-pipe or oesophagus. The rhythmic contraction and expansion of the oesophagus help in pushing the food forward. No digestion of food occurs in oesophagus but food reaches from mouth to stomach through it.

Stomach : The stomach is a large organ which expands when food enters it. The muscular- walls of the stomach help in mixing the food thoroughly with more digestive juices. These digestion functions are taken care of by the gastric glands present in the wall of the stomach. These release hydrochloric acidj a protean digesting enzyme called pepsin and mucus. The exit of food from the stomach is regulated by a sphincter muscle which releases it in small amounts into the small intestine.

Small Intestine : From the stomach, the food now enters the small intestine. This is the longest part of the alimentary canal which is fitted into a compact space because of extensive coiling. The small intestine is the site of the complete digestion of carbohydrates, proteins and fats. The digested food is taken up by the walls of the intestine.

Large Intestine : The unabsorbed food is sent into the large intestine where water is absorbed by its walls from the material. The rest of the material is removed from the body via the anus.

Question 40.
Explain the process of breathing in man.
In human beings, air is taken into the body through the nostrils. The air passing through the nostrils is filtered by fine hairs that line the passage. The passage is also lined with mucus which helps in this process. From Ivre, the air passes through the throat and into the lungs. Rings of cartilage are present in the throat. These ensure that the air-passage does not collapse.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 14
Within the lungs, the passage divides into smaller and smaller tubes which finally terminate in balloon like structures which are called alveoli. The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood-vessels.

When we breathe in, we lift our ribs and flatten our diaphragm, and the chest cavity becomes larger as a result. Because of this, air is sucked into the lungs and fills the expanded alveoli. The blood brings carbon dioxide from the rest of the body for release into the alveoli and the oxygen in the alveolar air is taken up by blood in the alveolar blood vessels to be transported to all the cells in the body.

During the breathing cycle, when air is taken in and let out, the lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for carbon dioxide to be released.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes.

Question 41.
Explain the importance of soil for plant growth.
Soil anchors the plants. Without soil plants cannot grow. For plants, it is the nearest and richest source of raw materials like nitrogen, phosphorus and other minerals. Roots of the plants absorb these minerals and water from the soil and transport them to other parts of the plants. Soil also makes oxygen available for respiration of root cells as air remains trapped in small empty spaces present in the soil.

It can happen because soil is porous. Besides, soil is a favourite habitat of many microbes. In the soil, most of the microbes live close to the roots of plants. This area near the roots is called the ‘rhizosphere’ which is the thin layer of soil that sticks to the roots. The rhizosphere is a huge habitat in the soil because plants have so many root fibres. Some microbes that live in rhizosphere have very close relationships with plants. The association of microbes and roots of plants often benefits both organisms.

The plant gives the microbes food, such as sugars and amino acids and the microbes give the plants minerals, some vitamins, nitrogen and some amino acids. This type of association in plants and microbes in which both the organisms get benefitted is called symbiotic association.

Thus, from the above discussion, it is clear that soil provides anchoring, water, minerals and oxygen for roots to breathe which help in growth of plants. Also it facilitates the symbiotic association of plant and microbes.
From all this, it can be concluded that soil is very important for plant growth.

Question 42.
Draw the diagram of alimentary canal of man and label the following parts :
Mouth, Oesophagus, Stomach, Intestine
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 15
Question 43.
How do carbohydrates, proteins and fats get digested in human beings?
We eat various types of food that is broken down into simple molecules and is digested. Let’s discuss how and where carbohydrates, proteins and fats get digested in our body. In the buccal cavity (mouth), the saliva, secreted by the salivary glands, breaks down starch which is a complex molecule to give sugar. The enzyme called salivary amylase contained in the saliva helps in breaking down of starch. From the mouth, the food is taken to the stomach through the food-pipe or oesophagus. In the stomach proteins of the food

gets digested as the muscular walls of the stomach secrete digestive juices containing protein digesting enzyme, pepsin.
From the stomach the food is sent into the small intestine. The small intestine is the site of complete digestion of carbohydrates, proteins, and fats. It receives the secretions of liver and pancreas for this purpose. The food coming from the stomach is acidic and has to made alkaline for the pancreatic enzymes to act.

Bile juice from the liver accomplishes this in addition to acting on fats. Fats are present in the intestine in the form of large globules which makes it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The pancreas secretes pancreatic juice which contains enzymes like trypsin for digesting proteins and lipase for breaking down emulsified fats. The walls of the small intestine contain glands which secrete intestinal juice.

The enzymes presentin it finally convert the proteins to amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol. The digested food is taken up by the walls of the intestine. This is how carbohydrates, proteins and fats get digested in our body.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes
Question 44.
Explain the mechanism of photosynthesis.
Photosynthesis is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This material is taken in the form of carbon dioxide and water which is converted into carbohydrates in the presence of sunlight and chlorophyll. Carbohydrates are utilised for providing energy to the plant. The carbohydrates which are not used immediately are stored in the form of starch, which serves as the internal energy reserve to be used as and when required by the plant.

Let’s now see what actually happens during the process of photosynthesis. The following events occur during this process :
First of all, absorption of light energy by chlorophyll takes place. Then light energy is converted into the chemical energy and splitting of water molecules into hydrogen and oxygen occurs. Finally carbon dioxide reduces to carbohydrates. In the form of a chemical equation, photosynthesis can be represented as follows:
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 16
Question 45.
Explain the three pathways of breakdown of glucose in living organisms.
In living organisms, the food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to break dowil glucose completely into carbon dioxide and water, some use other pathways that do not involve oxygen. In all cases, the first step is the breakdown of glucose, a six-carbon molecule, into a three-carbon molecule called pyruvate. This process takes place in the cytoplasm. Further, the pyruvate may be converted into ethanol, and carbon dioxide.

This process takes place in yeast during fermentation. Since this process takes place in the absence of air (oxygen), it is called anaerobic respiration. Break down of pyruvate using oxygen takes place in the mitochondria. This process breaks up the ‘ three-carbon pyruvate molecule to give three molecules of carbon dioxide. The other product is water. Since this process takes place in the presence of air (oxygen), it is called aerobic respiration.

The release of energy in this aerobic process is a lot greater than in the anaerobic process. Sometimes, when there is a lack of oxygen in our muscle cells, another pathway for the breakdown of pyruvate is taken. Here the pyruvate is converted into lactic acid which is also a three-carbon molecule.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 17
Question 46.
Explain the mechanism of flow of oxygenated and deoxygenated blood through the human heart.
Describe the flow of blood through the heart of human beings.
The mechanism of flow of oxy-genated and deoxygenated blood through the human heart can be explained in the following manner :
The muscles of all the four chambers of heart are relaxed (= joint diastole). The deoxygenated blood from the anterior and posterior vena cava pour into the right atrium (= auricle) and the oxygenated blood from the pulmonary veins (coming from lungs) pour into the left atrium.
Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes – 18
Now, the right and left atria contract (= atrial systole). The , right atrium pumps the deoxygenated blood into the right
ventricle through the right atrio-ventricular aperture. Simultaneously, the left atrium pumps the oxygenated blood into the left ventricle through left atrio-ventricular aperture.

Next, the two ventricles contract (= ventricular systole). The . deoxygenated blood from the right ventricle flows to the lungs
through pulmonary artery, while the oxygenated blood from the left ventricle flows to all parts of the body through the aorta, which is the largest artery in human body.

Bihar Board Class 10 Science Solutions Chapter Chapter 6 Life Processes

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