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PSEB 11th Class Biology Solutions Chapter 13 Photosynthesis in Higher Plants

August 1, 2024July 31, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 13 Photosynthesis in Higher Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants

PSEB 11th Class Biology Guide Photosynthesis in Higher Plants Textbook Questions and Answers

Question 1.
By looking at a plant externally can you tell whether a plant is C₃ or C₄? Why and how?
Answer:
Usually plants growing in dry conditions use C₄-pathways. It cannot be said conclusively, if the plant is a C₃ or C₄ by looking at external appearance.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
The particularly large cells around the vascular bundles of the C₄– pathway plants are called bundle sheath cells and the leaves, which have such anatomy are said to have ‘Kranz’ anatomy.
‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.
The bundle sheath cells may form several layers around the vascular bundles they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

Question 3.
Even though a very few cells in a C₄ plant carry out the biosynthetic-Calvin pathway, yet they are highly productive. Can you discuss why?
Answer:
The productivity of a plant is measured by the rate at which it photosynthesizes. The amount of carbon dioxide present in a plant is directly proportional to the rate of photosynthesis. C₄ plants have a mechanism for increasing the concentration of carbon dioxide. In C₄ plants, the Calvin cycle occurs in the bundle-sheath cells.

The C₄ compound (malic acid) from the mesophyll cells is broken down in the bundle sheath cells. As a result, CO₂ is released. The increase in CO₂ ensures that the enzyme RuBisCo does not act as an oxygenase, but as a carboxylase.
This prevents photorespiration and increases the rate of photosynthesis. Thus, C₄ plants are highly productive.

Question 4.
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ -plants?
Answer:
RuBisCO has a much greater affinity for CO₂ than for O₂. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme. In C₄ -plants some O₂ does bind to RuBisCO and hence, CO₂ fixation is decreased. Here the RuBP instead of being converted to two molecules of PGA binds with O₂ to form one molecule and phosphoglycolate in a pathway called photorespiration.

In the photorespiratory pathway, there is neither synthesis of sugars, nor of ATP. Rather it results in the release of CO₂ with the utilization of ATP. in the photorespiratory pathway, there is no synthesis of ATP or NADPH. Therefore, photorespiration is a wasteful process.

In C₄-plants, photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site. This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂, this results in increasing the intracellular concentration of CO₂. In turn, this ensures that the RuBisCO functions as a carboxylase minimizing the oxygenase activity.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll-b but lacked chlorophyll a, would it carry out
photosynthesis? Then why do plants have chlorophyll-b and other accessory pigments?
Answer:
‘Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll-b, xanthophylls and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to chlorophyll-a. Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect chlorophyll-a from photo-oxidation.

Question 6.
Why is the colour ola leaf kept n the dark frequently yellow or pale green? Which pigment do you think is more stable?
Answer:
This is due to the interconversion of pigments, i.e., change of green chlorophyll pigment into yellow-colored carotenoids. The carotene pigment is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer:
The leaves of the same plant on the sunny side are dark green as compare it with the leaves on the sunny side due to more chlorophyll pigment.

Question 8.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:
(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
PSEB 11th Class Biology Solutions Chapter 13 Photosynthesis in Higher Plants 1
Answer:
(a) Points K-C of the curve, the rate did not increase with an increase in its concentration because under these conditions, light becomes limiting factor.

(b) The rate of photosynthesis shows proportionate increase upto a certain CO₂ concentration (In region A of the curve), beyond which the rate again hcomes constant, not showing any increase by increasing CO₂ concentration.

(c) lithe light inrensiry is doubled, i.e., the plants are exposed to 2 units of light, CO₂ concentration again becomes limiting factor beyond this concentration (Points C and D represent on the curve.)

Question 9.
Give comparison between the following:
(a) C₃ and C₄ pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C₃ and C₄ plants
Answer:
(a) Comparison between C₃ and C₄ pathways

C₃ Pathway	C₄ Pathway
The primary acceptor of CO₂ is RuBP, a 5 carbon compound.	The primary acceptor of CO₂ is PEP, a 3 carbon compound.
It operated under low concentration of CO₂ in mesophyll cells.	It can operate under very low CO₂ concentration in mesophyll cells.
CO₂ once fixed is not released back.	CO₂ once fixed is released back in bundle sheath cells.
Fixation of one molecule of CO₂ needs 3ATP and 2 NADPH₂ molecules. It requires 18ATP for the synthesis of one molecule of glucose.	C₄ pathway requires 30 ATP for the synthesis of one molecule of glucose.
C₃ -cycle operates in all categories of plants.	It operates in only C₄-plants.

(b) Comparison between cyclic and non-cyclic photophosphorylation

Cyclic Photophosphorylation	Non-cyclic Photophosphorylation
It occurs in photosystem-I in stromal or intergranal thylakoids.	It is carried out by both PS-I and PS-II in the granal thylakoids.
It is not connected to photolysis of water so no oxygen is evolved.	It is connected with photolysis of water, so oxygen is evolved in it.
It is activated by light of 700 nm wavelength.	It occurs in 680 nm as well as 700 nm wavelength.
It generates ATP only there is no formation of NADPH₂.	It produces both ATP as well as NADPH₂.
Chlorophyll does not receive any electrons from donor.	The source of electrons is photolysis of water.
This system does not take part in photosynthesis except in bacteria.	This system is connected with CO₂ fixation and is dominant in green plants.

C₃ Leaves	C₄ Leaves
Bundle-sheath cells are absent.	Bundle-sheath cells are present.
RuBisCo is present in the mesophyll cells.	RuBisCo is present in the bundle sheath cells.
The first stable compound produced is 3-phosphoglycerate, a three-carbon compound.	The first stable compound produced is oxaloacetic acid; a four-carbon compound.
‘ Photorespiration occurs.	‘ Photorespirarion does not occur.

PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants

August 1, 2024July 31, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants

Very short answer type questions

Question 1.
Define translocation.
Answer:
Transport over longer distances proceeds through the vascular system (the xylem and the phloem) and known as translocation.

Question 2.
Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both. [NCERT Exemplar]
Answer:
Pressure and concentration gradient.

Question 3.
The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because. [NCERT Exemplar]
Answer:
The cell wall is freely permeable to water and other substances in the solution but the plasma membrane is selectively permeable.

Question 4.
Why does rate of transport reach maximum or becomes saturated in facilitated diffusion?
Answer:
The transport rate reach a maximum because all the transport proteins are occupied/saturated.

Question 5.
Imbibition is considered a method of diffusion. Comment.
Answer:
Imbibition is considered as a method of diffusion because the movement of water occurs along the concentration gradient during this process.

Question 6.
Give one basic difference between antiport and uniport.
Answer:
In antiport both the molecules cross the membrane in opposite directions whereas, in uniport molecules moves across a membrane independent of any other passing molecule.

Question 7.
Mention two .factors on which net direction of molecules and rate of osmosis depends.
Answer:
The two factors responsible are:

Pressure gradient
Concentration gradient.

Question 8.
A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after sometime the plant dies. This may be due to which factor? [NCERT Exemplar]
Answer:
As urea content make the soil hypertonic in nature, therefore, the plant dies due to exosmosis.

Question 9.
The endodermis is impervious to water. Comment.
Answer:
The inner boundary of the cortex, i.e., endodermis is impervious to water because of a band of suberised matrix called the casparian strip.

Question 10.
Identify the vascular tissue responsible for translocation of organic and inorganic substances from leaves to other parts of the plant.
Answer:
Phloem is responsible for this type of translocation.

Question 11.
How do root hairs increase the absorption of water by plants?
Answer:
Root hairs increase the surface area of roots. This helps in making contact with larger volume of water. Thus, the presence of root hairs helps in absorption by plants.

Question 12.
It is seen that the number of stomata are greater on the lower surface of the leaf than the upper surface. Why is it so ?
Answer:
Stomata are present in greater number on the lower surface because if more number of stomata will be present on the upper surface, it would lead to great amount of water loss through transpiration. Thus, to avoid the excessive transpiration, stomata are present in greater number on lower surface of the leaf.

Question 13.
Elucidate the channels of food transport in plants.
Answer:
The channels of food transport are sieve tubes and sieve cells of phloem.

Question 14.
How are companion cells helpful to sieve tubes?
Answer:
The companion cells are connected to the sieve tubes by plasmodesmata and provide them with proteins, ATP and other nutrients.

Short answer type questions

Question 1.
Define facilitated diffusion.
Answer:
Membrane proteins provide sites at which movement of certain molecules takes place. These molecules have hydrophilic moiety and hence it is difficult for them to cross a membrane. They- need assistance of membrane proteins to cross the membrane. This is called facilitated diffusion.

Question 2.
Give a comparison table of different transport mechanisms.
Answer:
Comparison of Different Transport Mechanisms
PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants 1

Question 3.
Photosynthesis needs constant supply of water. But transpiration can hamper this supply. How do plants of desert area manage to get sufficient water in spite of faster transpiration?
Answer:
Desert plants have a different mechanism of photosynthesis and it is called C₄ pathways. The evolution of the C₄ photosynthetic system is probably one of the strategies for maximising the availability of CO₂ while minimising water loss. C₄ plants are twice as efficient as C₃ plants in terms of fixing carbon (making sugar). However, a C 4 plant loses only half as much water as a C₃ plant for the same amount of CO₂ fixed.

Question 4.
What is the significance of transpiration?
Ans. Significance of Transpiration

Transpiration pull facilitates movement of water from roots.
Transpiration supplies water for photosynthesis.
It pulls minerals from soil.
Helps in cooling of plants.
Maintains shape of plant cells.

Question 5.
Describe the movement of water in leaves.
Answer:
Evaporation from the leaf sets up a pressure gradient between the outside air and the air spaces of the leaf. The gradient is transmitted into the photosynthetic cells and on the water-filled xylem in the leaf vein. This facilitates movement of water from xylem to the guard cells of stomata.

Long answer type questions

Question 1.
Observe the given figure and answer the following questions:
PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants 2
(i) State the nature of solution marked (1).
(ii) What process has been depicted in figures C to D?
(iii) In which figures turgor pressure will be zero?
(iv) In which figures wall pressure will he positive?
Answer:
(i) The solution marked as (1) will he hypertonic (more concentrated) due to which cell shrinks.
(ii) From C to D, figure is showing the process of deplasmolysis as shrinked cell has again regained its original shape.
(iii) Turgor pressure will be zero in figure B and C because cell is in a flaccid condition.
(iv) Wall pressure will be positive in figure A and D because in these figure cell wall is exerting ‘equal and opposite pressure against the expanding protoplasm.

Question 2.
Minerals are present in the soil in sufficient amount. [NCERT Exemplar]
(i) Do plants need to adjust the types of solute that reach xylem.
(ii) Which molecules help to adjust this?
Answer:
(i) An analysis of the xylem exudates shows that though some of the nitrogen travels as inorganic ions, much of it is carried in the organic form as amino acids and related compounds. Similarly, small amount of P and | S are carried as organic compounds. In addition, small amount of exchange of materials does take place between xylem and phloem.

(ii) Mineral ions are frequently remobilised, particularly from older, sensecing parts. Older dying leaves export much of their mineral content to younger leaves. Similarly, before leaf fall in deciduous plants, minerals are removed to other parts. Elements most readily mobilised are phosphorus, sulphur, nitrogen and potassium. Some elements that are structural components like calcium are not remobilised.

PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination

Very short answer type questions

Question 1.
Write the mode of excretion performed by Xenopus.
Answer:
Dual excretion (mainly ammonotelism and partly ureotelism).

Question 2.
In which of the organism antennary glands are found as excretory organ?
Answer:
Crustaceans.

Question 3.
What is the excretory product from the kidney of reptiles? [NCERT Exemplar]
Answer:
Uric acid.

Question 4.
Give the name of vessel of peritubular capillaries that runs parallel to the loop of Henle.
Answer:
Vasa recta.

Question 5.
Give the name of the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 6.
Give the name of cells that are responsible for the formation of filtration slits or slit pores.
Answer:
Podocytes (epithelial cells of Bowman’s capsule).

Question 7.
In which part of excretory system of mammals you can first use the term urine?
Answer:
The filtrate formed by the process of ultrafiltration in the Bowman’s capsule is called glomerular filtrate or primary urine.

Question 8.
What is the ratio of the concentrated filtrate to that of the initial filtrate?
Answer:
The concentrated urine (filtrate) is nearly four times concentrated than the initial filtrate formed.

Question 9.
What will be the effect on the amount of urine released when water is abundant in the body tissues in case of vertebrates?
Answer:
Vertebrates excrete large quantities of dilute urine when water is abundant in the body tissues and vice-versa.

Question 10.
What is the pH of urine? [NCERT Exemplar]
Answer:
It ranges from 4.5-8.2, average pH is 6.0.

Question 11.
What are the two substances responsible for causing the gradient for increasing hyperosmolarity of medullary interstitium?
Answer:
NaCl and urea.

Question 12.
Give the name of the main component that play an important role in the counter-current mechanism.
Answer:
Henle’s loop and vasa recta.

Short answer type questions

Question 1.
Describe the structure of human kidney.
Answer:

Shape and Size of Kidney: Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g.
Structure of Kidney: Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter.
Inner Structure: Inner to the hilum is a broad funnel-shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and
an inner medulla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces. The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Question 2.
What is the function of proximal convoluted tubules in the kidney?
Answer:
Function of Proximal Convoluted Tubule (PCT): PCT is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption. Nearly all of the essential nutrients, and 70-80 percent of electrolytes and water are reabsorbed by this segment. PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO from it.

Question 3.
What is the function of Henle’s loop?
Answer:
Function of Henle’s Loop: Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 4.
What is the function of distal convoluted tubule?
Answer:
Function of Distal Convoluted Tubule (DCT): Conditional reabsorption of Na⁺ and water takes place in this segment. DCT is also capable of reabsorption of HCO and selective secretion of hydrogen and potassium ions and NH₃ to maintain the pH and sodium-potassium balance in blood.

Question 5.
What is the function of collecting duct?
Answer:
Collecting Duct

Large amounts of water could be reabsorbed from this region.
This segment also allows the transport of small amounts of urea into the medullary interstitium, to maintain the osmolarity.
It also plays a role in maintaining pH and ionic balance of the body fluids by selective section of K⁺ and H⁺ ions.

Question 6.
How does reabsorption take place in the excretory system in human?
Answer:
Reabsorption: A comparison of the volume of the filtrate formed per day (180 litres per day) with that of the urine released (1.5 litres), suggest that nearly 99 per cent of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. The tubular epithelial cells in different segments of nephron perform this either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc., in the filtrate are reabsorbed actively whereas the nitrogenous wastes are absorbed by passive transport. Reabsorption of water also occurs passively in the initial segments of the nephron. During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.

Question 7.
Which gland releases ADH? What is the role of ADH in excretion?
Answer:
Hypothalamus releases ADH. ADH facilitates water reabsorption from latter part of tubules. This prevents diuresis. Excess fluid loss, through urine is called diuresis.

Question 8.
Write a short note on-disorders of the excretory system.
Answer:
Disorders of the Excretory System
The disorders related to kidneys are :

Uremia
Renal calculi and
Glomerulonephritis

Hemodialysis is the process of removal of nitrogenous wastes from the blood of a uremia patient.
Kidney transplantation is the ultimate method of correcting urinary failure, in which a functioning kidney from a suitable donor is transplanted.

Hemodialysis

Blood from the artery of an uremia patient is taken, cooled to 0°C and mixed with an anticoagulant like heparin.
It is put into the cellophane tubes of the artificial kidney, where cellophane is permeable to micromolecules, but not to macromolecules like plasma protein.
Outside the cellophane tube is the dialysing fluid, which has the same composition as that of plasma except the nitrogenous molecules like urea, uric acid, creatine, etc.
Hence, the nitrogenous molecules from within the cellophane tubes flow into the dialysing fluid, following concentration gradient, (dialysis)
The blood coming out of the artificial kidney is warmed to body temperature, mixed with antiheparin and restored to a vein of the patient.

Long answer type questions

Question 1.
The glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain. [NCERT Exemplar]
Answer:

The gradient of increasing hyperosmolarity of medullary interstitium is maintained by a counter current mechanism and the proximity between the Henle’s loop and vasa recta.
This gradient is mainly caused by NaCl and urea. The transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism.
NaCl is transported by the ascending limb of Henle’s loop, which is exchanged with the descending limb of vasa-recta. NaCl is returned to the medullary interstitium by the ascending part of vasa recta.
But, contrarily, the water diffuses into the blood of ascending limb of vasa recta and is carried away into the general blood circulation.
Permeability to urea is found only in the deeper parts of thin ascending limbs of Henle’s loops and collecting ducts. Urea diffuses out of the collective ducts and enters into the thin ascending limb.
A certain amount of urea recycled in this way is trapped in medullary interstitium by the collecting tubule. This mechanism helps in the maintenance of a concentration gradient in the medullary interstitium.
Presence of such gradient helps in an easy passage of water from the collecting tubule, resulting in the formation of concentrated urine (filtrate) i.e., nearly four times concentrated than the initial filtrate formed.

PSEB 11th Class Biology Solutions Chapter 19 Excretory Products and their Elimination

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 19 Excretory Products and their Elimination Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination

PSEB 11th Class Biology Guide Excretory Products and their Elimination Textbook Questions and Answers

Question 1.
Define Glomerular Filtration Rate (GFR).
Answer:
Glomerular Filtration Rate (GFR) is the amount of the filtrate formed by the kidneys per minute. GFR in a healthy individual is approximately 125 mt/mm, i.e., 180 L day.

Question 2.
Explain the autoregulatory mechanism of GFR.
Answer:
Regulation of GFR: The kidneys have built-in mechanisms for the regulation o glomerular filtration rate. One such efficient mechanism is carried out by the juxtaglomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.

Question 3.
Indicate whether the following statements are true or false:
(a) Micturition is carried out by a reflex.
(b) Al)H helps In water elimination, making the urine hypotonia.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.
Answer:
(a) True,
(b) False,
(C) True,
(d) True,
(e) True.

Question 4.
Give a brief account of the counter-current mechanism.
Answer:
Counter-current Mechanism

The counter-current mechanism takes place in Henle’s loop and vasa recta.
The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus, forms a counter-current.
The flow of blood through the two limbs of vasa recta is also in a counter-current pattern.
NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.
NaCl is returned to the interstitium by the ascending portion of vasa recta.
Similarly small amount of urea enters the thin segment of the ascending limb of Henle’ loop, which is transported back to the interstitium by the collecting tubule.
This transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter-current mechanism,
It helps to maintain a concentration gradient in the medullary interstitium, which facilitates an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).

Question 5.
Describe the foie of liver, lungs and skin in excretion.
Answer:
Liver, lungs and skin also play an important role in the process of excretion. Role of the Liver: Liver is the largest gland in vertebrates. It helps in the excretion of cholesterol, steroid hormones, vitamins, drugs, and other waste materials through bile. Urea is formed in the liver by the ornithine cycle. Ammonia-a toxic substance is quickly changed into urea in the liver and thence eliminated from the body. Liver also changes the decomposed haemoglobin pigment into bile pigments called bilirubin and biliverdin.

Role of the Lungs: Lungs help in the removing waste materials such as carbon dioxide from the body.
Role of the Skin: Skin has many glands which help in excreting waste products through pores. It has two types of glands-sweat and sebaceous glands.

Sweat glands are highly vascular and tubular glands that separate the waste products from the blood and excrete them in the form of sweat. Sweat excretes excess salt and water from the body.
Sebaceous glands are branched glands that secrete an oily secretion called sebum.

Question 6.
Explain micturition.
Answer:
Micturition: The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex. An adult human excretes on an average 1 to 1.5 L of urine per day. The urine formed is a light yellow coloured watery fluid which is slightly acidic (pH 6.0) and has a characteristic odour.

Question 7.
Match the items of Column-I with those of Column-II:

Column-I	Column-II
(a) Ammonotelism	(i) Birds
(b) Bowman’s capsule	(ii) Water reabsorption
(c) Micturition	(iii) Bony fish
(d) Uricotelism	(iv) Urinary bladder
(e) ADH	(v) Renal tubule

Answer:

Column-I	Column-II
(a) Ammonotelism	(iii) Bony fish
(b) Bowman’s capsule	(v) Renal tubule
(c) Micturition	(iv) Urinary bladder
(d) Uricotelism	(i) Birds
(e) ADH	(ii) Water reabsorption

Question 8.
What is meant by the term osmoregulation?
Answer:
It is the phenomenon of regulation of change in the concentration of body fluids according to the concentration of external environment. Le., most marine invertebrates, some freshwater invertebrates, hagfish (a vertebrate).

Question 9.
Terrestrial animals are generally either republic or uricotelic, not ammonotelic, why?
Answer:
Terrestrial adaptation requires the production of less toxic nitrogenous wastes like urea and uric acid for the conservation of water. Aquatic animals excret atnmonia which requires large amounts of water to dissolve. This huge quantity of water is easily available to such animals from surroundings.

However, for terrestrial animals, such a huge quantity of water is not available, hence, they modify their original water product NH3 to comparatively less toxic products like urea and uric acid which requires less amount of water for their excretion.

Question 10.
What is the significance of juxtaglomerular apparatus (JGA) in Sidney function?
Answer:
The juxtaglomerular apparatus (JGA) plays an important role in monitoring and regulation of kidney functioning by hormonal feedback, mechanism, involving the hypothalamus and to a certain extent, the heart.

Question 11.
Name the following:
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Answer:
(a) Flarworms,
(b) Columns of Bertini,
(c) Vasa recta

Question 12.
Fill in the gaps:
(a) Ascending limb of Henle’s loop is ………………………… to water whereas the descending limb is …………………….. to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ……………………………. .
(c) Dialysis fluid contains all the constituents as in plasma except …………………………… .
(d) A healthy adult human excretes (on an average) ………………………… gm of urea/day.
Answer:
(a) impermeable, permeable;
(b) vasopressin or ADH.;
(c) the nitrogenous wastes;
(d) 25-30.

PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom

Very short answer type questions

Question 1.
What are coelomates? Give two examples.
Answer:
Animals possessing coelom are called coelomates, e.g., Annelids, molluscs, arthropods, etc.

Question 2.
Body cavity is the cavity present between body wall and gut wall. In some animals the body cavity is not lined by mesoderm. Such animals are called: [NCERT Exemplar]
Answer:
Pseudocoelomates.

Question 3.
In some animal groups, the body is found to be divided into compartments with atleast some organs/organ repeated. Name this characteristic feature. [NCERT Exemplar]
Answer:
The segmentation that simultaneously divides body both externally and internally is called metamerism.

Question 4.
Name the group which lack digestive tract.
Answer:
Porifera.

Question 5.
Name an animal having canal system and spicules. [NCERT Exemplar]
Answer:
Scypha (sycon)/Leucosolenia.

Question 6.
Give an example of animal which exhibit alternation of generation. [NCERT Exemplar]
Answer:
Obelia, shows alternation of asexual and sexual phases.

Question 7.
Name the animal which exhibits the phenomenon of bioluminescence. Mention the phylum to which it belongs.
[NCERT Exemplar]
Answer:
Pleurobrachia, phylum – Ctenophora.

Question 8.
How is mesoglea different in ctenophores that in cnidarians?
Answer:
The mesoglea in ctenophores also contains amoebocytes, elastic fibres and muscle cells.

Question 9.
What is the role of radula in Mollusca? [NCERT Exemplar]
Answer:
It is the rasping organ that help in feeding.

Question 10.
Identify the animal in which adults exhibit radial symmetry and larvae exhibit bilateral symmetry. [NCERT Exemplar]
Answer:
In Echinodermata, the adults are radially symmetrical and larvae are bilaterally symmetrical.

Question 11.
Why are urochordates called tunicates?
Answer:
This is because the soft body of urochordates is surrounded by a thick test or tunic, often transparent or translucent.

Question 12.
In which fishes cartilaginous skeleton is present?
Answer:
Chondrichthyes.

Question 13.
Which amphibians show branchial (gills) respiration?
Answer:
Young ones of most of the amphibians which are aquatic show branchial respiration.

Question 14.
What is the importance of pneumatic bones and air sacs in Aves? [NCERT Exemplar]
Answer:

Pneumatic bones are light but strong, the feature which helps in flight,
Air sacs increase the efficiency of respiration and provide buoyancy to the animal.

Short answer type questions

Question 1.
Define tissue level of organization in the Kingdom Animalia. Give some examples of organisms showing tissue level of organization.
Answer:
In tissue level of organization there are specified groups of cells to carry out specific functions. This is a start of division of labour in the Animal Kingdom. Examples: Aurelia and Hydra.

Question 2.
What are diploblastic and triploblastic organisation?
Answer:
Animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm, are called diploblastic animals, e.g., coelenterates. An undifferentiated layer, mesoglea, is present in between the ectoderm and the endoderm. Those animals in which the developing embryo has a third germinal layer, mesoderm, in between the ectoderm and endoderm, are called triploblasdc animals (platyhelminthes to chordates).

Question 3.
Write two characters on the basis of which you can say that Coelenterata is more evolved than Porifera.
Answer:
(a) Porifera shows cellular level of organization, while Coelenterata shows tissue level of organisation.
(b) All members of Porifera are sessile, i.e., they are attached to the substratum, while some members of Coelenterata are motile, showing further improvement.

Question 4.
On the basis of which characters you can say that Aschelminthes are more advanced compared to Platyhelminthes?
Answer:
(a) Platyhelminthes are acoelomate,while Aschelminthes are pseudocoelomates. This indicates development of mesoderm.
(b)In Platyhelminthes both sexes are on the same animal, while in Aschelminthes there is segregation of sexes. This shows another point of evolution.

Question 5.
Give a description of the water vascular system in Echinodermata..
Answer:
Water Vascular System in Echinodermata: The water vascular system is a hydraulic system used by echinoderms, such as starfish and sea urchins, for locomotion, food and waste transportation, and respiration. The system is composed of canals connecting numerous tube feet. Echinoderms move by alternately contracting muscles that force water into the tube feet, causing them to extend and push against the ground, then relaxing to allow the feet to retract.

Long answer type questions

Question 1.
Differentiate between polyp and medusa. [NCERT Exemplar]
Answer:
Differences between polyp and medusa are as follows :

Polyp	Medusa
1. It is a fixed zooid.	It is free swimming.
2. It is asexual.	It is sexual.
3. It is cylindrical in outline.	Medusa is umbrella-shaped.
4. Tentacles found at upper end of manubrium.	Tentacles occur along the margin of umbrella.
5. Mouth is circular and terminal over upright manubrium.	Mouth is four sided, lies at the lower end of hanging manubrium.
6. Velum and sense organs are absent.	Medusa have a-circular velum and eight sense organs or statocysts.

Question 2.
Differentiate between Chordates and Non-Chordates.
Answer:
Differences between chordates and non-chordates are given below:

Chordates	Non-Chordates
•» Notocord is present at stages in some stages of development.	Notochord is absent.
•» Central nervous system is dorsal, hollow, single and non-ganglionated.	Central nervous system is ventral, solid, double and ganglionated.
•» Gill slits present on lateral side of pharynx in sum stages of throughout life.	Gill slits are absent.
•» Tail is present in some stages and throughout life.	Tail generally absent.
•» Heart is ventral.	Heart is dorsal.
•» Haemoglobin is present in RBCs.	It is present in plasma.

Question 3.
Which features make mammals as most successful and dominant animals?
Answer:
Features which make mammals as dominant and successful animals are as follows:

The presence of an insulating and protective hairy exoskeleton
They are warm blooded so have high rate of metabolism.
They are viviparous animals and show placentation and intrauterine development which increases the chances for survival of young ones.
They show high degree of parental care.
They have more developed hearing efficiency due to the presence of pinna, three ear-ossicles and coiled cochlea in the ear.
They are able to speak through language.
They have good power of learning due to the presence of more developed brain.

PSEB 11th Class Biology Solutions Chapter 4 Animal Kingdom

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 4 Animal Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 4 Animal Kingdom

PSEB 11th Class Biology Guide Animal Kingdom Textbook Questions and Answers

Question 1.
What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
Answer:
The major difficulties in the classification of animals are on the following lines :

Some show cellular level of organisation, some have tissue level and even some have organ system level of organisation.
Regarding symmetry, some are radially symmetrical, while some have bilateral symmetry.
Some have open circulatory system, while others have closed type.
Regarding digestion, some animals have extracellular digestion, while others have intracellular digestion.
In case of body cavity, some have true coelom while others are pseudocoelomates.
Regarding reproduction, some have only asexual reproduction, while others reproduce both by sexual and asexual means.
So, these are difficulties that zoologists face in the classification of animals.

Question 2.
If you are given a specimen, what are the steps that you would follow to classify it?
Answer:
If I am given an animal specimen, then I will classify it on the basis of fundamental features which are common to all animal types inspite of the presence of some major differences in the structure and form of animals. The features taken into consideration during classification of animal are as follows:

The type of arrangement of cells.
Body symmetry.
Nature of coelom.
Pattern of digestive system.
Type of circulatory system.
Type of methods of reproduction.

Question 3.
How useful is the study of the nature of body cavity and coelom in the classification of animals?
Answer:
Presence or absence of a cavity between the body wall and the gut wall is very important in classification. The body cavity which is lined by, mesoderm, is called coelom.

Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates.
In some animals, the body cavity is not mesoderm, instead the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., Aschelminthes.
The animals in which the body cavity is absent are known as acoelomates, e.g. , Platyhelminthes.

Question 4.
Distinguish between intracellular and extracellular digestion?
Answer:
Differences between intracellular and extracellular digestion are as follows:
(i) Intracellular Digestion: It occurs inside the living cells with the help of lysosomal enzymes. Food particle is taken in through endocytosis. It forms a phagosome which fuses with a lysosome. The digested material pass into the cytoplasm. The undigested matter is throw out by exocytosis. It occurs in Amoeba, Paramecium, etc.,

(ii) Extracellular Digestion: In case of coelentrates digestion occurs in gastrovascular cavity. This cavity has gland cells which secret digestive enzymes over the food. The partially digested fragmented food particles are ingested by nutritive cells. It occurs in Hydra, Aurelia, etc.

Question 5.
What is the difference between direct and indirect development?
Answer:
In direct development the embryo directly develops into an adult, while in indirect development there is an intermediate larval stage. Certain members of arthropods show larval stage of development.

Question 6.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
Hooks and suckers are present in the parasitic forms. They are parasitic flatworms commonly called flukes. The body is unsegmented leaf like, which is covered by a thick living tegument. There is no epidermis. The mouth is anterior and is armed with suckers for attachment in the host. Life history includes larval stage and involves more than one hosts.
Examples : Fasciola (the liver fluke), Schistosoma (the blood fluke.)

Question 7.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods constitute the largest group of the animal kingdom. It is estimated that the Arthropoda population of the world is approximately a billion (10¹⁸) individuals, in terms of species diversity, number of individuals and geographical distribution. It is the most successful phylum on the Earth that have ever existed. Arthropods are equipped with jointed appendages, which are variously adapted for walking, swimming, feeding, sensory reception and defence. The appendages of abdomen are associated with locomotion, reproduction and in some cases with defence as well.

The appendages of head are related to defence, whereas those of thorax are mainly associated with locomotion. These features are responsible for its large diversity.

Question 8.
Water vascular system is the characteristic of which group of the following?
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
Echinodermata have the water vascular system.

Question 9.
‘All vertebrates are chordates but all chordates are not vertebrates’.’Justify the statement.
Answer:
Notochord is a characteristic feature of all chordates. The members of sub-phylum – Vertebrata possess notochord during the embryonic stage. But in adults the notochord is replaced by a cartilaginous or bony vertebral column. Whereas in member of other Sub-phyla of Chordata the notochord remain as such. The urochordate and cephalochordates retain the notochord during their entire life cycle. Thus, the absence of notochord in adult vertebrates suggest that all vertebrates are chordates but all chordates are not vertebrates.

Question 10.
How important is the presence of air bladder in Pisces?
Answer:
In fishes, air bladder regulates buoyancy and helps in floating in water. If it is absent, animals need to swim constantly to avoid sinking.

Question 11.
What are the modifications that are observed in birds that help them fly?
Answer:
Flight adaptations in birds are as follows :

Boat-shaped body helps to propel through the air easily.
Feathery covering of body to reduce the friction of air.
Holding the twigs automatically by hindlimbs.
Extremely powerful muscles that enables the wings to work during flight.
Bones are light, hollow and provide more space for muscle attachment. Presence of pneumatic bones which reduce the weight of body and help in flight.
The first four thoracic vertebrae are fused to form a furculum for walking of the wings.
Lungs are solid and elastic and have associated air sacs.
The power of accomodation of eyes is well developed due to the presence of comb-like structure pecten.
A single left ovary and oviduct to reduce the body weight.

Question 12.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
The number of eggs or young ones produced by an oviparous or viviparous mother cannot be equal. An oviparous mother gives rise to more number of eggs as some of them die during hatching and as they have to pass through a large number of developmental stages before becoming an adult. On the other hand, a viviparous mother gives rise to fewer number of young ones because there are less chances of their death. Moreover, they did not have to pass through any larval stage.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Annelida
(c) Aschelminthes
(d) Arthropoda
Answer:
Segmentation in the body is first observed in Annelida. This phenomenon is known as metamerism.

Question 14.
Match the following:

A. Operculum	1. Ctenophora
B. Parapodia	2. Mollusca
C. Scales	3. Porifera
D. Comb plates	4. Reptilia
E. Radula	5. Annelida
F. Hairs	6. Cyclostomata and Chondrichthyes
G. Choanocytes	7. Mammalia
H. Gill slits	8. Osteichthyes

Answer:

A. Operculum	8. Osteichthyes
B. Parapodia	5. Annelida
C. Scales	4. Reptilia
D. Comb plates	1. Ctenophora
E. Radula	2. Mollusca
F. Hairs	7. Mammalia
G. Choanocytes	3. Porifera
H. Gill slits	6. Cyclostomata and Chondrichthyes

Question 15.
Prepare a list of some animals that are found parasitic on human heings.
Answer:
A list of parasitic animals on human beings:

Parasite	In Part of Human Body
Leishmania donovani	Blood
Trichomonas vaginalis	Vagina of human female
Plasmodium vivax	Blood
Taenia solium	Intestine
Ascaris lumbricoides	Small intestine
Wuchereria bancrofti	Lymphatic and muscular system
Loa loa	Eyes
Fasciola hepatica	Liver and bile ducts
Entamoeba histolytica	Intestine
Trypanosoma gambiense	Blood

PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom

Very short answer type questions

Question 1.
Name the types of classification of plants.
Answer:
Artificial, natural and phylogenetic.

Question 2.
Which system indicates evolutionary as well as genetic relationships among organisms?
Answer:
Phylogenetic system of classification.

Question 3.
What is cytotaxonomy?
Answer:
Cytotaxonomy is a method of classification. It is based on cytological structure and their relatedness.

Question 4.
Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
[NCERT Exemplar]
Answer:
Phaeophyceae.

Question 5.
Holdfast, stipe and frond constitutes the plant body in case of
[NCERT Exemplar]
(a) Rhodophyceae
(b) Chlorophyceae
(c) Phaeophyceae
(d) All of these
Answer:
The lamina of Phaeophyceae members large sized body with differentiation of holdfast, stipe and lamina.

Question 6.
The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorptions. What is the equivalent of the roots in the less developed lower plants? [NCERT Exemplar]
Answer:
Rhizoids.

Question 7.
Most algal genera show haplontic life cycle. Name an alga which is (i) Haplodiplontic (ii) Diplontic [NCERT Exemplar]
Answer:
Haplodiplontic: Ulva, Dictyota.
Diplontic: Fucus, Cladophora, Glomerata.

Question 8.
How are mosses considered ecologically important?
Answer:
Mosses, along with lichens are the first organism to colonise rocks, hence are ecologically important.

Question 9.
A prothallus is
(a) a structure in pteridophyte formed before the thallus develops
(b) a sporophytic free-living structure formed in pteridophyte
(c) a gametophytic free-living structure formed in pteridophytes
(d) a primitive structure formed after fertilisation in pteridophyte
[NCERT Exemplar]
Answer:
Gametophyte is a free-living small thalloid structure called prothallus. In most ferns, the prothallus is green and autotrophic.

Question 10.
Where are seeds located in gymnosperm?
Answer:
Seeds lie naked or exposed on the surface of megasporophyll.

Question 11.
The embryo sac of an angiosperm is made up of:
(i) 8 cells
(ii) 7 cells and 8 nuclei
(iii) 8 nuclei
(iv) 7 cells and 7 nuclei [NCERT Exemplar]
Answer:
(ii) Embryo sac of angiosperm develops up to 8 nucleate state prior to fertilisation. There is a three celled egg apparatus, three antipodal cells and two polar nuclei.

Question 12.
What is alternation of generations?
Answer:
Alernation of generations is regular switch over from gamete bearing haploid gametophyte to haploid spore producing diploid sporophyte.

Short answer type questions

Question 1.
What are the main differences among Chlorophyceae, Phaeophyceae and Rhodophyceae?
Answer:
PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom 1

Question 2.
What is the general structure of bryophytes?
Answer:
Structure of Bryophytes: It is thallus-like and prostrate or erect, and attached to the substratum by unicellular or multicellular rhizoids. They lack true roots, stem or leaves. They may possess root-like, or stem-like structures.

Question 3.
What is the general structure of pteridophytes?
Answer:

The main plant body is a sporophyte which is differentiated into true root, stem and leaves. These organs possess well-differentiated vascular tissues.
The leaves in pteridophyta are small (microphylls) as in Selaginella or large (macrophylls) as in ferns.
The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases sporophylls may form distinct compact structures called strobili or cones (Selaginella, Equisetum).

Question 4.
Write short notes on:
(a) Importance of carbon fixation by algae.
(b) Importance of Gymnosperms
(c) Importance of Angiosperms
(d) Medicinal use of algae
Answer:
(a) About 50% of carbon fixation is done by algae. This enables majority of sea organisms to get the required food.
(b) Gymnosperms are mainly used as decorative plants. Certain paints are
prepared from Gymnosperm plants.
(c) Angiosperms are the major providers of food-grains to the mankind.
(d) Spirullina is made by algae and is used as a nutritional supplement.

Long answer type questions

Question 1.
Algae are known to reproduce asexually by a variety of spores under different environmental condition. Name these spores and the conditions under which they are produced. [NCERT Exemplar]
Answer:
Zoospores: Flagellate spores, under favourable conditions.
Aplanospores: Non-flagellate, thin-walled spores under approaching unfavourable conditions.
Hypnospores: Thick-walled, resting spores in unfavourable conditions.
Akinetes: Thin-walled and thick-walled spores formed from whole cells in unfavourable conditions. .
Autospores: Spores which look exactly like parent cell formed under favourable conditions.

Question 2.
Write about habit and habitat of algae.
Answer:
Habit and Habitat of Algae: Algae are chlorophyll-bearing, simple, thalloid, autotrophic and largely aquatic (both fresh water and marine) organisms. They occur in a variety of other habitats : moist stones, soils and wood. Some of them also occur in association with fungi (lichen) and animals (e.g., on sloth bear).

Question 3.
What are the differences betweenpinus and cycas?
Answer:

Pinus	Cycas
1. Roots are micorrhizal.	1. Roots are not micorrhizal.
2. Stems are branched.	2. Stems are unbranched.
3. Male and female strobili are on same tree.	3. They are on different trees.

PSEB 11th Class Biology Solutions Chapter 3 Plant Kingdom

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 3 Plant Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 3 Plant Kingdom

PSEB 11th Class Biology Guide Plant Kingdom Textbook Questions and Answers

Question 1.
What is the basis of classification of algae?
Answer:
Basis of classification of algae are as follows:

Kinds of pigments.
Nature of reserve food.
Kinds, number and points of insertion of flagella of motile cells.
Presence or absence of organised nucleus in the cell.

Question 2.
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Answer:
Reduction division in the life cycle of a liverwort, a moss, a fern and a gymnosperm take place during the production of spores from spore mother cells. In case of an angiosperm, the reduction division occurs during pollen grain formation from anthers and during production of embryo sac from ovule.

Question 3.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Answer:
Three groups of plants that bear archegonia are bryophytes, pteridophytes and gymnosperms.
Life Cycle of a Pteridophyte: The life cycle of a pteridophyte consists of two morphologically distinct phases:
1. The gametophytic phase
2. The sporophytic phase.
These two phases come one after another in the life cycle of a pteridophyte. This phenomenon is called alternation of generation. The gametophyte is haploid with single set of chromosomes. It produces male sex organs antheridia and female sex organs archegonia.

The antheridia may be embedded or projecting type. Each antheridium has single layered sterile jacket enclosing a mass of androcytes.
The androcytes are flask-shaped, sessile or shortly stalked and differentiated into globular venter and tubular neck.
The archegonium contains large egg, which is non-mo tile.
The antherozoids after liberation from antheridium, reaches up to the archegonium fuses with the egg and forms a diploid structure known as zygotes.
The diploid zygote is the first cell of sporophytic generation. It is retained inside the archegonium and forms the embryo.
The embryo grows and develop to form sporophyte which is differentiated into roots, stem and leaves.
At maturity the plant bears sporangia, which encloses spore mother cells.
Each spore mother cell gives rise to four haploid spores which are usually arranged in tetrads.
The sporophytic generation ends with the production of spores.
Each spore is the first cell of gametophytic generation. It germinates to produce gametophyte and completes its life cycle.

PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification 1

Question 4.
Mention the ploidy of the following:
Protonemal cell of a moss, primary endosperm nucleus in dicot, leaf cell of a moss, prothallus cell of a fern, gemma cell in Marchantia, meristem cell of monocot, ovum of a liverwort and zygote of a fern.
Answer:
Protonemal cell of a moss – haploid
Primary endosperm nucleus in dicot – triploid
Leaf cell of a moss – haploid
Prothallus cell of a fern – haploid
Gemma cell in Marchantia – haploid
Meristem cell of monocot – diploid
Ovum of a liverwort – haploid
Zygote of a fern – diploid

Question 5.
Write a note on economic importance of algae and gymnosperms.
Answer:
Economic Importance of Algae

Red algae provides food, fodder and commercial products. Porphyra tenera is rich in protein, carbohydrates and vitamin-A, B, E and C.
Corallina has vermifuge properties.
Agar-agar a gelatin substance used as solidifying agent in culture media is obtained from Gelidium and Gracilaria algae. Funori is a glue used as adhesive and in sizing textiles, papers, etc. Chondrus is most widely used in sea weed in Europe.
Mucilage extracted from Chondrus is used in sampoos, shoe polish and creams.
Carrageenin is a sulphated polysaccharide obtained from cell wall of Chondrus crispus and Gigartina and is used in confectionary, bakery, jelly, creams, etc.

Economic Importance of Gymnosperms

Gymnosperms hold soil particles and thus check soil erosion.
Many gymnosperms are grown in gardens as ornamental plants, i.e., Cycas, Thiya, Araucaria, Taxus, Agathis, Maiden hair tree, etc.
Sago is a kind of starch obtained from cortex and pith of stem and seeds of Cycas. Roasted seeds of Pinus geradiana (chilgoza) are used as dry fruit.
Paper pulp is obtained from wood of Picea (spruce), Gnetum, Pinus (pine) and Larix (larck).
The wood of Juniperus virginiana (red cedar) is used to make pencils, holders and cigar boxes. Wood of Taxus is heaviest amongst soft woods and is used for making bows for archery.
Dry leaves of Cycas are used to make baskets and brooms. Needles of Pinus in making fibre board. Electric and telephone poles are made of stem of conifers.
Essential oils are obtained from Juniperus, Tsugo, Picea, Abies, Cedrus, etc. Resins are obtained from many species of Pinus.

Question 6.
Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Answer:
Both gymnosperms and angiosperms bear seeds, but they are yet classified separately. Because, in case of gymnosperms the seeds are naked, i.e., the seeds are not produced inside the fruit but in case of angiosperms the seeds are enclosed inside the fruit.

Question 7.
What is heterospory? Briefly comment on its significance. Give two examples.
Answer:
Heterospory is the phenomenon in which a plant produces two types of spores, namely microspores and megaspores.
Heterospory is significant in the following ways:

Microspores give rise to male gametophyte and megaspores give rise to female gametophyte.
Female gametophyte is retained on the parent plant. The development of zygote takes place within the female gametophyte.
This leads to formation of seeds.
Examples: All gymnosperms and all angiosperms, Pinus, Gnetum, neem, peepal, etc.

Question 8.
Explain briefly the following terms with suitable examples:
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Diplontic
(v) Sporophyll
(vi) Isogamy.
Answer:
(i) Protonema: It is the juvenile stage of a moss. It results from the germinating meiospore. When fully grown, it consists of a slender green, branching system of filaments called the protonema.

(ii) Antheridium: The male sex organ of bryophyte and pteridophyte is known as antheridium. It has a single-layered sterile jacket enclosing in a large number of androcytes. The androcytes’ metamorphose into flagellated motile antherozoids.

(iii) Archegonium: The female sex organ of bryophytes, which is multicellular and differentiated into neck and venter. The neck consists of neck canal cells and venter contains the venter canal cells and egg.

(iv) Diplontic: A kind of life cycle in which the sporophyte is the dominant, photosynthetic, independent phase of the plant and alternate with haploid gametophytic phase is known as diplontic life cycle.

(v) Sporophyll: The sporangium bearing structure in case of Selaginella is known as sporophyll.

(vi) Isogamy: It is the process of fusion between two similar gametes, i.e., Chlamydomonas.

Question 9.
Differentiate between the following:
(i) red algae and brown algae,
(ii) liverworts and moss,
(iii) homosporous and heterosporous pteridophytes.
(iv) syngamy and triple fusion.
Answer:
(i) Differences between Red Algae and Brown Algae

Red Algae

Brown Algae

1. It belongs to the It belongs to the class-Rhodophyceae

It belongs to the It belongs to the class Phaeophyceae.

2. It is red in colour due to the presence of pigments chlorophyll-a, c and phycoerythrin.

Example: Stylolema, Rhodela.

It is brown in colour due to the presence

Example: Sargassum, Microcystis.

(ii) Differences between Liverworts and Moss

Liverwort	Moss
1. These are the member of class-Hepaticopsida of bryophyta.	These belongs to class-Bryop’sida of bryophyta.
2. Thallus is dorsoventrally flattened and lobed liver like	Thallus is leafy and radially symmetrical.
3. Rhizoids are unicellular.	Rhizoids are multicellular
4. Elaters are present in capsule to assist dispersal of spores.	Elaters are absent, but peristome teeth are present in the capsule to assist dispersal of spores.

(iii) Differences between Homosporous and Heterosporous Pteridophytes

Homosporous Pteridophyte

Heterosporous Pteridophyte

Pteridophytes, which produce only one kind of spores.

Example: Lycopodium

These produce two kinds of spores, i.e., large megaspore and smaller microspore.

Example: Selaginella

(iv) Differences between Syngamy and Triple Fusion

Syngamy	Triple Fusion
It is the act of fusion of one male gamete with the egg cell to form zygote.	The act of fusion of second male gamete with secondary nucleus to form triploid enddsperm is called triple endosperm is called triple fusion.

Question 10.
How would you distinguish monocots from dicots?
Answer:

Dicotyledons (Dicots)	Monocotyledons (Monocots)
•» Tap root system	Fibrous root system
•» Two cotyledons	One cotyledon
•» Reticulate Venation	Parallel venation
•» Tetramerous or Pentamerous flowers	Trimerous flowers

Question 11.
Match the followings (column I with column II)

Column I	Column II
(a) Chlamydomonas	(i) Moss
(b) Cycas	(ii) Pteridophyte
(c) Selaginella	(iii) Algae
(d) Sphagnum	(iv) Gymnosperm

Answer:

Column I	Column II
(a) Chlamydomonas	(iii) Algae
(b) Cycas	(iv) Gymnosperm
(c) Selaginella	(ii) Pteridophyte
(d) Sphagnum	(i) Moss

Question 12.
Describe the important characteristics of gymnosperms.
Answer:
Characteristics of gymnosperms are as follows :

Naked-seeded plants, i.e., their ovules are exposed and not enclosed in ovaries. Hence, the seeds are naked without fruits.
Tap root system is present. They show symbiotic as speciation with fungi
to form mycorrhizae or with N₂-fixing cyanobacteria to form colloidal roots as in Cycas.
Leaves are large and needle-shaped.
Vascular tissues are well developed.
Gymnosperms are heterosporous.
Pollination by wind and deposited in ovules.
Fertilisation occurs in archegonia.
Retention of female gametophyte inside the ovule and the ovules on the sporophytic plant for complete development is responsible for the development of seed habit.

PSEB 11th Class Biology Solutions Chapter 20 Locomotion and Movement

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 20 Locomotion and Movement Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

PSEB 11th Class Biology Guide Locomotion and Movement Textbook Questions and Answers

Question 1.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:
PSEB 11th Class Biology Solutions Chapter 20 Locomotion and Movement 1

Question 2.
Define sliding-filament theory of muscle contraction.
Answer:
The sliding-filament, theory states that the contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Answer:
Mechanism of Muscle Contraction:

The mechanism of muscle contraction is explained by the sliding filament theory.
This theory states that contraction of a muscle fibre is due to the sliding of the thin (actin) filaments over the thick (myosin) filaments.
Muscle contraction is initiated by a neural signal from the central nervous system through a motor neuron.
When the neural signal reaches the neuromuscular junction, it releases a neurotransmitter, i.e., acetylcholine, which generates an action potential in the sarcolemma.
This spreads through the muscle fibre and causes the release of Ca⁺⁺ ions from the sarcoplasmic reticulum into the sarcoplasm.
The Ca⁺⁺ ions bind to the subunit of troponin and brings about conformational changes; this removes the masking of the active site for myosin.
The myosin head binds to the active site on actin to form a cross-bridge; this utilises energy from the hydrolysis of ATP.
This pulls the actin filaments towards the centre of A-band.
As a result, the Z-lines limiting the sarcomere are pulled closer together, causing a shortening of the sarcomere or contraction of muscle.
Thus, during muscle contraction, the length of A band remains unchanged, while that of I-band decreases.
The myosin goes back to its relaxed state.
A new ATP binds and the cross-bridge is broken and the actin filaments slide out of A-band.
The cycle of cross bridge-formation and cross bridge breakage continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic reticulum which leads to the masking of the active site on F-actin.

Question 4.
Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.
Answer:
(a) True
(b) False, H-zone represents thick filaments
(c) True
(d) False, there are 12 pairs of ribs in man.
(e) True

Question 5.
Write the difference between:
(a) Actin and Myosin
(b) Red and White Muscles
(c) Pectoral and Pelvic Girdle
Answer:
(a) Differences between Actin and Myosin Filament

Actin Filaments	Myosin Filaments
1. These are found in I-band.	These are found in A-band.
2. These are thin.	These are thick.
3. Cross bridges (heads) are absent.	Cross bridges (heads) are present.
4. It is a globular protein with low molecular weight.	It is a heavy molecular weight polymerised protein.

(b) Differences between Red and White Muscles

Red Muscles	White Muscles
1. In some muscles, myoglobin content is high, which gives a reddish colour to them, such muscles are called red muscles.	Some muscles possess very less quantity of myoglobin, so they appear whitish called as white muscles.
2. These contain plenty of mitochondria.	These have less number of mitochondria but amount of sarcoplasmic reticulum is high.
3. These are called aerobic muscles.	They depend on anaerobic process of energy.

Pectoral Girdle	Pelvic Girdle
1. It helps in the articulation of upper limbs.	It helps in the articulation of lower limbs.
2. It is situated in the pectoral region of the body.	It is situated in the pelvic region of the body.
3. Each half of pectoral girdle is formed of a clavicle and a scapula.	Pelvic girdle consists of two coxal bones.
4. Scapula is a large triangular flat bone and clavicle is a long slender bone.	Each coxal bone is formed of three bones, ilium, ischium and pubis.
5. An expanded process, acromion from scapula forms a depression called glenoid cavity, which articulates with the head of humerus to form shoulder joint.	Ilium, ischium and pubis fuse at a point to form a cavity called acetabulum to which the thigh bone articulates.

Question 6.
Match Column-I with Column-II

Column-I	Column-II
(a) Smooth muscle	(i) Myoglobin
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(iii) Sutures
(d) Skull	(iv) Involuntary

Answer:

Column-I	Column-II
(a) Smooth muscle	(iv) Involuntary
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(i) Myoglobin
(d) Skull	(iii) Sutures

Question 7.
What are the different types of movements exhibited by the cells of human body?
Answer:
Cells of the human body exhibit three main types of movements-amoeboid, ciliary and muscular.
(i) Amoeboid Movement: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

(ii) Ciliary Movement: Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

(iii) Muscular Movement: Movement of our limbs, jaws, tongue, etc., require muscular movement. Locomotion requires a perfect coordinated activity of muscular, skeletal and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Answer:

Skeletal Muscle	Cardiac Muscle
1. The, cells of skeletal muscles are unbranched.	1. The cells of cardiac muscles are branched.
2. Intercalated disks are absent.	2. The cells are joined with one another by intercalated disks that help in coordination or synchronization of the heartbeat.
3. Alternate light and dark bands are present.	3. Faint bands are present.
4. They are voluntary muscles.	4. They are involuntary muscles.
5. They contract rapidly and get fatigued in a short span of time.	5. They contract rapidly but do not get fatigued easily.
6. They are present in body parts such as the legs, tongue, hands, etc.	6. These muscles are present in the heart and control the contraction and relaxation of the heart.

Question 9.
Name the type of joint between the following:
(i) Atlas/Axis
(ii) Carpal/Metacarpal of thumb
(iii) Between phalanges
(iv) Femur/Acetabulum
(v) Between cranial bones
(vi) Between pubic bones in the pelvic girdle.
Answer:
(i) Pivot joint
(ii) Saddle joint
(iii) Hinge joint
(iv) Ball and socket joint
(v) Fibrous joint
(vi) Cartilagenous joint

Question 10.
Fill in the blank spaces.
(a) All mammals (except a few) have ………………………………. cervical vertebra.
(b) The number of phalanges in each limb of human is ……………………………………
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ………………………. and ………………..
(d) In a muscle fibre Ca²⁺ is stored in ………………………..
(e) ………………….. and ……………………………….. pairs of ribs are called floating ribs.
(f) The human cranium is made of …………………………. bones.
Answer:
(a) seven
(b) fourteen.
(c) troponin and tropomyosin
(d) sarcoplasm
(e) 11 th; 12th
(f) eight

PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics

Very Short Answer Type Questions

Question 1.
Can a system be heated and its temperature remain constant? (NCERT Exemplar)
Answer:
If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

Question 2.
Air pressure in a car tyre increases during driving. Explain. (NCERT Exemplar)
Answer:

During driving, temperature of the gas increases while its volume remains constant.
So, according to Charles’ law, at constant V, p ∝ T.
Therefore, pressure of gas increases.

Question 3.
Write conditions for an isothermal process.
Answer:
The conditions for an isothermal process are :

The walls should be diathermic.
The process should be quasi-static.

Question 4.
Why air quickly leaking out of a balloon becomes cooler?
Answer:
Leaking of air is adiabatic expansion and adiabatic expansion produces cooling.

Question 5.
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer:
Here, heat removed is less than the heat supplied and hence the room become hotter.

Question 6.
Is reversible process is possible in nature?
Answer:
A reversible process is never possible in nature because of dissipative forces and condition for a quasi-static process is not practically possible.

Question 7.
On what factors, the efficiency of a Carnot engine depends?
Answer:
On the temperature of source of heat and the sink.

Question 8.
Which thermodynamic law put restrictions on the complete conversion of heat into work?
Answer:
According to second law of thermodynamics, heat energy cannot converted into work completely.

Short Answer Type Questions

Question 1.
What are the limitations of the first law of thermodynamics?
Answer:
Following are the limitations of the first law of thermodynamics :

It does not tell us about the direction of flow of heat.
It fails to explain why heat cannot be spontaneously converted into work.

Question 2.
Two bodies at different temperatures T₁ and T₂ are brought in contact.
Under what condition, they settle to mean temperature? (after they attain equilibrium)
Answer:
Let m₁ and m₂ are masses of bodies with specific heats s₁ and s₂, then if their temperature after they are in thermal equilibrium is T.
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 1
Then, if > T₁> T₂ and assuming no heat loss.
Heat lost by hot body = heat gained by cold body
m₁s₁(T₁-T)=m₂s₂(T-T₂)
⇒ \(\frac{m_{1} s_{1} T_{1}+m_{2} s_{2} T_{2}}{m_{1} s_{1}+m_{2} s_{2}}\) = T[equilibrium temperature]
So for, bodies to settle down to mean temperature,
m₁ = m₂ and s₁ = s₂
means bodies have same specific heat and have equal masses.
Then, T = \(\frac{T_{1}+T_{2}}{2}\) [mean temperature]

Question 3.
When ice melts, then change in internal energy, is greater than the heat supplied, why?
Solution:
When ice melts, volume of water formed is less than that of ice. So, surroundings (environment) does work on the system (ice). And by first law,
ΔQ = ΔW+ΔU
⇒ ΔU = ΔQ-ΔW
(ΔW = negative as work is done on the system)
⇒ ΔU>ΔQ

Question 4.
Calculate the work done for adiabatic expansion of a gas.
Solution:
Consider (say µ mole) an ideal gas, which is undergoing an adiabatic expansion.
Let the gas expands by an infinitesimally small volume dV, at pressure p, then the infinitesimally small work done given by
dW = pdV
The net work done from an initial volume V₁ to final volume V₂ is given by
W= ∫v₁v² pdV
For an adiabatic process, pV^γ = constant = K
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 2
For an adiabatic process, K = p₁V^γ = p₂V^γ
For an ideal gas, p₁V₁ = μRT₁ and p₂V₂ = μRT₂.
So, we have
W = \(\frac{1}{(1-\gamma)}\left[\mu R T_{2}-\mu R T_{1}\right]=\frac{\mu R}{(\gamma-1)}\left[T_{1}-T_{2}\right]\)

Question 5.
What is a heat engine? What is the best way to increase efficiency of a heat engine? Is it possible to design a thermal engine that has 100% efficiency?
Solution:
A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, η = \(\frac{\text { Work output }}{\text { Heat input }}=1-\frac{T_{2}}{T_{1}}\)
where, T₂ = temperature of sink
T₁ = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ =0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

Question 6.
An ideal engine works between temperatures T₁ and T₂. It derives an ideal refrigerator that works between temperatures T₃ and T₄. Find the ratio Q₃/Q₁ in terms of T₁, T₂, T_3, and T₄.
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 3
Solution:
W = work done by engine = Q₁ – Q₂
and W = work done supplied to refrigerator = Q₃ -Q₄
Q₁ – Q₂ =Q₃ – Q₄
Dividing by Q₁, on both sides of the above equation, we get

Question 7.
Under what condition, an ideal Carnot engine has 100% efficiency?
Solution:
Efficiency of a Carnot engine is given by η = \(\left(1-\frac{T_{2}}{T_{1}}\right)\)
where, T₂ = temperature of sink
and T₁ = temperature of sink source
So for η = 1 or 100%, T₂ = 0 K or heat is rejected into a sink at 0 K temperature.

Question 8.
Draw p-V diagram of a Carnot cycle.
Solution:
p-V diagram for Carnot cycle
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 5

Long Answer Type Questions

Question 1.
A cycle followed by a machine (made of one mole of a perfect gas in a cylinder with a piston) is shown in figure
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 6
A to B: volume constant B to C: adiabatic
C to D: volume constant D to A : adiabatic
V_C = V_D = 2 V_A = 2 V_B
(i) In which part of the cycle, heat is supplied to the machine from outside?
(ii) In which part of the cycle, heat is being given to the surrounding by the machine?
(iii) What is the work done by the machine in one cycle? Write your answer in terms of P_A’ P_B’ V_A.
(iv) What is the efficiency of the machine?
Take γ = \(\frac{5}{3}\) for the gas and C_V =R for one mole.
Solution:
(i) A to B because T_B > T_A, as p ∝ T [ ∴ V = constant]
(ii) C to D because T_C>T_D, as P ∝ T [∴ V=constant]
(iii) W_AB = \(\int_{B}^{C} \) pdV=O and W_CD =0 [∴ V= constant]
Similarly,
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 7

(iv) Heat supplied during process A to B
dQ_AB = dU_AB
Q_AB = \(\frac{3}{2} n R\left(T_{B}-T_{A}\right)=\frac{3}{2}\left(p_{B}-p_{A}\right) V_{A}\)
∴ Efficiency = \(=\frac{\text { Net work done }}{\text { Heat supplied }}=\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\)

Question 2.
Explain with the suitable example that a reversible process must be carried slowly and a fast process is necessarily irreversible.
Answer:
A reversible process must pass through equilibrium states which are very close to each other so that when process is reversed, it passes back through these equilibrium states. Then, it is again decompressed or it passes through same equilibrium states, system can be restored to its initial state without any change in surroundings. e.g., If a gas is compressed as shown But a reversible process can proceed without reaching equilibrium in intermediate states.
PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics 9