Class 11

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.

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.

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:

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.

(c) Differences between Pectoral and Pelvic Girdle

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

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 12 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 12 Thermodynamics

PSEB 11th Class Physics Guide Thermodynamics Textbook Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10⁴ J/g?
Solution:
Water is flowing at a rate of 3.0 liter/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
Rise in temperature, ΔT = T₂ -T₁
= 77-27 = 50°C

Heat of combustion = 4 x 10⁴ J/g
Specific heat of water, C = 4.2 J g^-1 °C^-1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mC ΔT
= 3000 x 4.2 x 50
= 6.3 x 10⁵ J/min
∴ Rate of consumption = \(\frac{6.3 \times 10^{5}}{4 \times 10^{4}}\) = 15.75 g/min

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol^-1K^-1).
Solution:
Mass of nitrogen, m = 2.0 x 10^-2 kg = 20g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂,M =28
Universal gas constant, R = 8.3 J mol^-1K^-1
Number of moles, n = \(\frac{m}{M}\)
= \(\frac{2.0 \times 10^{-2} \times 10^{3}}{28}\) = 0.714
Molar specific heat at constant pressure for nitrogen,
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) x 8.3
= 29.05J mol^-1 K^-1

The total amount of heat to be supplied is given by the relation
ΔQ = nC_pΔT
= 0.714 x 29.05 x 45 = 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J

Question 3.
Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂) / 2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂) / 2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution:
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4 For an adiabatic process,
we have
P₁V₁^γ = P₂V₂^γ
The final volume is compressed to half of its initial volume.

Hence, the pressure increases by a factor of 2.639.

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:
The work done (W) on the system while the gas changes from state A to state Bis 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
ΔQ = 0
ΔW = -22.3 J (since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW

where, ΔU = Change in the internal energy of the gas .
ΔU = ΔQ – ΔW = -(-22.3 J)
ΔU = +22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is
ΔQ =9.35cal = 9.35 x 4.19 = 39.1765J
Heat absorbed, ΔQ = ΔU + ΔW
ΔW = ΔQ – ΔU = 39.1765 – 22.3 – = 16.8765J
Therefore, 16.88 J of work is done by the system.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock.
A contains a gas at standard temperature and pressure.
B is completely evacuated. The entire system is thermally insulated.
The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P -V – T surface?
Solution:
(a) 0.5 atm
The volume available to the gas is doused as soon as the stopcock between cylinders A and B is opened? Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5atm.

(b) Zero
The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(c) Zero
Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) No
The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Work done by the steam engine per minute, W = 5.4 x 10⁸ J
Heat supplied from the boiler, H = 3.6 x 10⁹ J
Efficiency of the engine = \(\frac{\text { Output energy }}{\text { Input energy }}\)
∴ η = \(\frac{W}{H}=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
Hence, the percentage efficiency of the engine is 15%.
Amount of heat wasted= 3.6 x 10⁹ – 5.4 x 10⁸
= 30.6 x 10⁸ = 3.06 x 10⁹ J
Therefore, the amount of heat wasted per minute is 3.06 x 10⁹J.

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
Heat is supplied to the system at a rate of 100 W.
∴ Heat supplied, ΔQ = 100 J/s
The system performs at a rate of 75 J/s.
∴ Work done, ΔW = 75 J/s

From the first law of thermodynamics, we have
ΔQ = ΔU + ΔW
where ΔU = Rate of change in internal energy
ΔU = ΔQ – ΔW = 100 – 75 = 25 J/s = 25W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure given below.

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Solution:
Total work done by the gas from D to E to F = Area of ΔDEF
Area of ΔDEF = \(\frac{1}{2}\) DF x EF
where, DF = Change in pressure
=600 N/m²
= 300N/m² = 300N/m²
FE = Change in volume
5.0 m³ – 2.0 m³ = 3.0m³
Area of ADEF = \(\frac{1}{2} \) x 300 x 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, Calculate the coefficient of performance.
Solution:
Temperature inside the refrigerator, T₁ = 9°C = 273 + 9 = 282 K
Room temperature, T₂ = 36°C = 273+36
Coefficient of performance = \(\frac{T_{1}}{T_{2}-T_{1}}\)
= \(\frac{282}{309-282}=\frac{282}{27}\)
309-282 = 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 15 Plant Growth and Development Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

PSEB 11th Class Biology Guide Plant Growth and Development Textbook Questions and Answers

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:
Growth: It is an irreversible permanent increase in size of an organ or its parts or even of an individual cell.
Determinate Growth: Although growth in most of the plant parts is unlimited, certain parts grow up to a certain level and then stop growing. This kind of growth is known as determinate growth.

Development: It is the process of whole series of changes which an organism goes through during its life cycle.
Meristem: The cells of which the capacity to divide and self-perpetuate.

Growth Rate: The increased growth per unit time is termed as growth rate. Thus, rate of growth can be expressed mathematically. An organism or a part of the organism can produce more cells in a variety of ways. The growth rate shows an increase that may be arithmetic or geometrical.

Differentiation: The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation.

Dedifferentiation: The living differentiated cells, that by now have lost the capacity to divide can regain the capacity of division under certain conditions. This phenomenon is termed as dedifferentiation. For example, formation of meristems; interfascicular cambium, and cork cambium from fully differentiated parenchyma cells. Redifferentiation: While undergoing dedifferentiation plant cells once again lose their capacity to divide and become mature. This process is called redifferentiation.

Question 2.
Why is not anyone parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Anyone parameter is not good enough to demonstrate growth throughout the life of a flowering plant because the plants exhibit different types of growth during different stages of their life cycle. In the seedling stage, they are in state of active cell division (i.e., mitotic divisions), then they undergo active cell enlargement stage during growing stage. In the reproductive or flowering stage of their life cycle, they exhibit reductional divisions. Finally, after the formation of various organs, they undergo cell differentiation or get matured.

Question 3.
Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Answer:
(a) Arithmetic Growth: In arithmetic growth, following mitotic cell division only One daughter cell continues to divide while the other differentiates and matures. Example is a root elongating at a constant rate.

(b) Geometric Growth: In geometrical growth, in most systems, the initial growth is Size of slow (lag phase), and it organ increases rapidly thereafter at an exponential rate (log or exponential phase). Here, both the progeny cells following mitotic cell division retain the ability to divide and continue to do so.

(c) Sigmoid Growth Curve: If we plot the parameter of growth against time, we get a typical sigmoid or S-curve. A sigmoid curve is a characteristic of living organism growing in a natural environment. It is typical for all cells, tissues and organs of a plant.

(d) Absolute and Relative Growth Rates: The measurement and the comparison of total growth per unit time is called the absolute growth rate. And the growth of the given system per unit time expressed on a common basis, e. g., per unit initial parameter is called the relative growth rate.

Question 4.
List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions, and agricultural/ horticultural applications of any one of them.
Answer:
Five main groups of natural plant growth regulators are auxins, gibberellins, ethylene, cytokinins and abscisic acid.
Auxins
(i) Discovery: Auxins was first isolated from human urine. They are generally produced by the growing apices of the stems and roots. Auxins like IAA (indole acetic acid) and indole butyric acid (IBA) have been isolated from plants. NAA (naphthalene acetic acid) and 2, 4-D (2, 4-dichloro phenoxy acetic) are synthetic auxins.

(ii) Physiological Function: They help to initiate rooting in stem cuttings, an application widely used for plant propagation. Auxins promote flowering, i. e., in pineapples. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.

(iii) Agricultural/Horticultural Applications: Auxins also induce parthenocarpy, e. g., in tomatoes. They are widely used as herbicides 2, 4-D, widely used to kill dicotyledonous seeds, does not affect mature monocotyledonous plants. It is used to prepare seed-free lawns by gardeners. Auxin also controls xylem differentiation and helps in cell division.

Question 5.
What do you understand by photoperiodism and vernalization? Describe their significance.
Answer:
1. Photoperiodism: The response of plants to periods of day/night is termed as photoperiodism. The site of perception of light/dark duration are the leaves.

Significance: The significance of photoperiodism is in regulating flowering in plants. Flowering is an important step towards seed formation and seeds are responsible for continuing the generation of a plant.

2. Vernalisation: There are plants in which flowering is either quantitatively or qualitatively dependent on exposure to low temperatures. This phenomenon is termed as vernalization. Vernalisation refers especially to the promotion of flowering by a period of low temperature.

Significance: Vernalisation prevents precocious reproductive development late in the growing season. This enables the plant to have sufficient time to reach maturity.

Question 6.
Why is abscisic acid also known as stress hormone?
Answer:
Abscisic acid acts as a general plant growth inhibitor and an inhibitor of plant metabolism. ABA inhibits seed germination. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone.

Question 7.
‘Both growth and differentiation in higher plants are open Comments.
Answer:
Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

Question 8.
‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Answer:
Petkus winter rye (Secale cereal) gives responses of low temperature at very young seedlings or even at seed stage. If winter rye is shown in the spring, the seeds germinate and produce vegetative plants in the following summer. In this case, the period of vegetative growth is extended and flowering occurs only in the next summer when the cold requirements is fulfilled during winters. The same variety, if grown in early autumn produces flowers in the following summer.

Question 9.
Which one of the plant growth regulators would you use if you are asked to:
(a) induce rooting in a twig
(b) quickly ripen a fruit.
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Answer:
(a) Cytokinins
(b) Ethylene
(c) Cytokinins
(d) Auxin
(e) Gibberellins
(f) Abscisic acid.

Question 10.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Answer:
No, a defoliated plant do not respond to the photoperiodic cycle. Because leaves of a plant are the sites of light perception for the induction of flowering.

Question 11.
What would be expected to happen if:
(a) GA₃ is applied to rice seedlings.
(b) dividing cells stop differentiating.
(c) a rotten fruit gets mixed with unripe fruits.
(d) you forget to add cytokinin to the culture medium.
Answer:
(a) The rice seedlings show extraordinary elongation of stem and leaf sheaths.
(b) Tissue and organ differentiation will not take place.
(c) Unripe fruits will also get rotten due to the ethylene hormone secreted by rotten fruit.
(d) No root and shoot formation will take place.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 7 Structural Organisation in Animals Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals

PSEB 11th Class Biology Guide Structural Organisation in Animals Textbook Questions and Answers

Question 1.
Answer in one word or one line:
(i) Give the common name of Periplanata americana.
(ii) How many spermathecae are found in earthworm?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of cockroach?
(v) Where do you find Malpighian tubules?
Answer:
(i) American cockroach
(ii) 4 pairs
(iii) Under the 4th, 6th abdominal terga
(iv) 10 segments in adults
(v) At the junction of midgut and hindgut in cockroach.

Question 2.
Answer the following:
(a) What is the function of nephridia?
(b) How many types of nephridia are found in earthworm based on their location?
Answer:
(a) The Function of Nephridia: The nephridia regulate the volume and composition of the body fluids. The nephridium starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular parts of the nephridium, which delivers the wastes through a pore to the surface in the body wall in the digestive tube.

(b) Based on their location, there are following three types of nephridia in earthworm:

• Septal nephridia.
• Pharyngeal nephridia.
• Integumentary nephridia.

Question 3.
Draw a labelled diagram of the reproductive organs of an earthworm.
Answer:

Question 4.
Draw a labelled diagram of alimentary canal of a cockroach.
Answer:

Question 5.
Distinguish between the following:
(a) Prostomium and peristomium.
(b) Septal nephridium and pharyngeal nephridium.
Answer:
(a) Prostomium and Peristomium: The first segment of earthworm with a ventral mouth is known as peristomium. Prostomium is a dorsal, lobe which is present on the ventral mouth.

(b) Septal Nephridia and Pharyngeal Nephridia: Septal nephridia are present on both the sides of intersegmental septa of segment 15 to the ‘ last that open into intestine.
The pharyngeal nephridia are closed (no nephrostome) nephridia present as three paired groups (of about 100) in 4th, 5th and 6th segments.

Question 6.
What are the cellular components of blood?
Answer:
The cellular components of blood are red blood cells, white blood cells and platelets.

Question 7.
What are the following and where do you find them in animal body (a) Chondrocytes (b) Axons (c) Ciliated epithelium.
Answer:
(a) Chondrocytes: These are the matrix secreting cells of the cartilage. These are found in the cartilage of connecting tissue.

(b) Axon: It is a long fibre, the distal end of which is branched. Each branch terminates as a bulb like structure called synaptic knob. The axon transmit nerve impulses away from the cell body.

(c) Ciliated Epithelium: If the columnar or cuboidal cells of columnar and cuboidal epithelium bear cilia on their free surface they are called ciliated epithelium.

Question 8.
Describe various types of epithelial tissues with the help of labelled diagrams.
Answer:
Epithelial Tissues: Epithelial tissues provide covering to the inner and outer lining of various organs. There are following two types of epithelial tissues :
1. Simple epithelium: Simple epithelium is composed of a single layer of cells. It functions as a lining for body cavities, ducts and tubes.

2. Compound epithelium: The compound epithelium consists of two or more cell layers. It has protective function as it does in our skin. It covers the dry surface of the skin, the moist surface of buccal cavity, pharynx, inner lining of ducts of salivary glands and of pancreatic ducts.
On the basis of structural modifications of the cells, simple epithelium is further divided into three types. These are :
(a) Squamous epithelium: The squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in functions like forming a diffusion boundary.

(b) Cuboidal epithelium: The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in the ducts of glands and tubular parts of nephrons in kidneys. Its main functions are secretion and absorption.

(c) Columnar epithelium: The columnar epithelium is composed of a single layer of tall and slender cells. They are found in the lining of stomach and intestine and help in absorption and secretion.
(i) When the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium organs like bronchioles and fallopian tubules.
(ii) Some of the columnar or cuboidal cells get specialized for secretion are called glandular epithelium. They are unicellular and multicellular.

Question 9.
Distinguish between:
(a) Simple epithelium and compound epithelium
(b) Cardiac muscle and striated muscle
(c) Dense regular and dense irregular connective tissue
(d) Adipose tissue and blood tissue
(e) Simple gland and compound gland
Answer:
(a) Differences between Simple and Compound Epithelium

Simple Epithelium	Compound Epithelium
1. It is composed of a single layer of	It consists of two or more cell layers.
2. It functions as a lining for body cavities, ducts and tubes.	It is protective in function like our skin.

(b) Differences between Cardiac and Striated Muscle

Cardiac Muscle	Striated Muscle
1. It occurs only in the wall of heart.	It occurs in the body wall, limb, tongue, pharynx, etc.
2. They are short and cylindrical with truncate ends.	They are long and cylindrical with blunt ends.
3. They have nerve supply from brain and autonomous nerve system	They have nerve supply from central nervous system.

(c) Differences between Dense Regular and Dense Irregular Connective Tissues

Dense Regular Connective Tissue	Dense Irregular Connective Tissue
Collagen fibres are present in rows between many parallel bundle of fibres. Examble: Tendons	Fibroblasts and many fibre are present that are oriented differently. Examble: Cartilage, bones and blood.

(d) Differences between Adipose Tissue and Blood Tissue

Adipose Tissue	Blood Tissue
1. It is a soft gel like connective tissue.	It is a fluid connective tissue.
2. It is partitioned into lobules by septa.	There are no partitions.
3. It is a storage tissue.	It is a transport tissue.
4. Matrix is secreted by the cells.	Matrix is not secreted by the cells.
5. It contain fibres.	Fibres are not conspicuous.
6. Adipocytes contain fat droplets.	No cell of the tissue contains fat droplets.

(e) Differences between Simple and Compound Gland

Simple Gland	Compound Gland
1. These glands have single unbranched duct.	These glands have branched system of ducts.
2. These may be simple tubular glands, simple coiled tubular glands and simple alveolar glands.	These may be compound tubular glands, compound alveolar glands

Question 10.
Mark the odd one in each series:
(a) Areolar tissue, blood, neuron, tendon
(b) RBC, WBC, platelets, cartilage
(c) Exocrine, endocrine, salivary gland; ligament
(d) Maxilla, mandible, labrum, antennae
(e) Protonema, mesothorax, metathorax, coxa
Answer:
(a) Neuron,
(b) Cartilage,
(c) Ligament,
(d) Antennae,
(e) Protonema.

Question 11.
Match the terms in column I with those in column II.

Column I	Column II
A. Compound epithelium	1. Alimentary canal
B. Compound eye	2. Cockroach
C. Septal nephridia	3. Skin
D. Open circulatory system	4. Mosaic vision
E. Typhlosole	5. Earthworm
F. Osteocytes	6. Phallomere
G. Genitalia	7. Bone

Answer:

Column I	Column II
A. Compound epithelium	3. Skin
B. Compound eye	4. Mosaic vision
C. Septal nephridia	5. Earthworm
D. Open circulatory system	2. Cockroach
E. Typhlosole	1. Alimentary canal
F. Osteocytes	7. Bone
G. Genitalia	6. Phallomere

Question 12.
Mention briefly about the circulatory system of earthworm.
Answer:
Pheretima exhibits a closed type of blood vascular system, consisting of blood vessels, capillaries and heart. Due to closed circulatory system, blood is confined to the heart and blood vessels. Contractions keep blood circulating in one direction. Smaller blood vessels supply the gut, nerve cord, and the body wall. Blood glands are present on the 4th, 5th and 6th segments. They produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature. Earthworms lack specialised breathing devices. Respiratory exchange occurs through moist body surface into their blood stream.

Question 13.
Draw a neat diagram of digestive system of frog.
Answer:

Question 14.
Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm
Answer:
(a) Functions of Ureters in Frog: In male frog, two ureters emerge from the kidneys. The ureters act as urinogenital duct which opens into the cloaca. Thus, the ureters carry both sperms and excretory wastes to the cloaca. In female frog, the ureters and oviduct open separately in the cloaca. The ureters in frog, thus acts as carrier of sperms and ova.

(b) Functions of Malpighian Tubules of Cockroach: Excretion is carried out by Malpighian tubules. Each tubule is lined by glandular cells. They absorb excretory waste products and converts them into uric acid which is excreted out through the hindgut.

(c) Functions of Body Wall of Earthworm: The body wall of earthworm has five layers – cuticle, epidermis, circular muscle layer, longitudinal muscle layer, peritoneum.

• Cuticle is a non-cellular elastic layer.
• The columnar cells of body wall provide support and therefore, are also known as supporting cells.
• Epidermis also has gland cells, receptor cells and basal cells.
• The glandular cell secrete mucus and thus, keep the skin moist, this
  help in cutaneous respiration.
• The last layer of the body wall is the outer membrane of the coelom called coelomic epithelium. The various muscle layers of the body wall provide strength and rigidity.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 16 Digestion and Absorption Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

PSEB 11th Class Biology Guide Digestion and Absorption Textbook Questions and Answers

Question 1.
Choose the correct answer among the following:
(i) Gastric juice contains
(a) pepsin, lipase and rennin
(b) trypsin, lipase and rennin
(c) trypsin, pepsin and lipase
(d) trypsin, pepsin and rennin

(ii) Succus entericus is the name given to
(a) a junction between ileum and large intestine
(b) intestinal juice
(c) swelling in the gut
(d) appendix
Answer:
(i) (a) Pepsin, lipase, and rennin
(ii) (b) Intestinal juice.

Question 2.
Match column I with column II.

Column I	Column II
A. Bilirubin and biliverdin	1. Parotid
B. Hydrolysis of starch	2. Bile
C. Digestion of fat	3. Lipases
D. Salivary gland	4. Amylases

Answer:

Column I	Column II
A. Bilirubin and biliverdin	2. Bile
B. Hydrolysis of starch	4. Amylases
C. Digestion of fat	3. Lipases
D. Salivary gland	1. Parotid

Question 3.
Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form?
(c) What are the basic layers of the wall of alimentary canal?
(d) How does bile help in the digestion of fats?
Answer:
(a) The mucosa layer of alimentary canal forms small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance. These modifications increase the surface area enormously.
Villi are supplied with the network of capillaries and large lymph vessel called the lacteal mucosal.

(b) The inactive form of enzyme pepsinogen is activated by Rd.

(c) The wall of alimentary canal from esophagus to rectum possesses four layers namely serosa, muscularis, sub-mucosa and mucosa. Serosa is the outermost layer, followed by muscularis, sub-mucosa and mucosa.

(d) Bile salts help in emulsification of lipids and activate the lipases.

Question 4.
State the role of pancreatic juice in digestion of proteins.
Answer:
The pancreatic juice contains inactive enzymes trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases, and nucleases. Trypsinogen is aëtivated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Proteins, proteases and peptones (partially hydrolyzed proteins) in the chyme reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice as given below:

Question 5.
Describe the process of digestion of protein in stomach.
Answer:
The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme. The pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).

HCl provides the acidic pH (pH 1.8) optimal for pepsins. Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins.

Question 6.
Give the dental formula of human beings.
Answer:
The dental formula of human beings is
\(\frac{2123}{2123} \times 2\).

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Answer:
Bile is yellowish-green alkaline solution with 89-98% water, having no digestive enzymes. The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol, and phospholipids but no enzymes. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases.

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?
Answer:
Chymotrypsin is the active form of chymotrypsinogen. It is activated by trypsin. It curdles milk. Nucleases like DNA ase and RNAase and pancreatic lipase are other enzymes secreted by the pancreas.

Question 9.
How are polysaccharides and disaccharides digested?
Answer:
The chemical process of digestion of carbohydrates is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About 30 percent of starch is hydrolyzed here by this enzyme (optimum pH 6.8) into a disaccharide-maltose. Further, carbohydrates in the chyme are hydrolyzed by pancreatic amylase into disaccharides.

Question 10.
What would happen if HCl were not secreted in the stomach?
Answer:
The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from excoriation by the highly concentrated hydrochloric acid.

HCl provides the acidic pH (pH 1.8) optimal for pepsins. Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins. Small amount of lipases are also secreted by gastric glands.

Question 11.
How does butter in your food get digested and absorbed in the body?
Answer:
Bile helps in emulsification of fats, i. e., breaking down of the fats into very small micelles. Bile also activates lipases.

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Answer:
Digestion of Protein in Stomach: The proenzyme pepsinogen, on exposure to HCl, gets converted into active enzyme pepsin.

Pepsin always outs in acidic medium (pH 1.8). In infants, main proteins are digested by rennin.

Digestion of Protein in Small Intestine: Pancreatic juice contains proenzyme trypsinogen. It is activated by enterokinase, secreted by intestinal mucosa, into active trypsin. Trypsin acts in alkaline medium.

The dipeptides are changed into amino acids by the enzyme succus enterics (intestinal juice).

Question 13.
Explain the term ‘the codont’ and ‘diphyodont’.
Answer:
Each tooth is embedded in a socket of jaw bone. This type of attachment is called thecodont. The majority of mammals including human beings forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Answer:
An adult human has 32 permanent teeth, which are of four different types (heterodont dentition), i.e., incisors (I), canine (C), premolars (Pm), and molars (M), and their number are 4, 2, 4, 6 respectively.

Question 15.
What are the functions of liver?
Answer:
Liver is the largest gland in human body which is mainly responsible for the digestion of food.
Role of liver in digestion of food :

• Its hepatic cells secrete bile juice which passes through the hepatic duct into the gall bladder.
• It has its major role in digestion and processing of proteins.
• Bile secreted by it is mainly responsible for digestion of fats for easy absorption in the body.
• It also responsible for the removal of toxins from blood.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 6 Anatomy of Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

PSEB 11th Class Biology Guide Anatomy of Flowering Plants Textbook Questions and Answers

Question 1.
State the location and function of different types of meristems.
Answer:
A meristematic tissue represents a group of cells that have retained the power of division throughout the life of an individual. The meristematic tissues are of three types, i.e., apical, intercalary and lateral meristem.

• Apical Meristem: These meristems are present at the apices of shoot and roots of the plants. Apical meristems are responsible for the increase in length of all the primary tissues.
• Intercalary Meristem: It is the meristem that occurs between the mature tissues. They occur in grasses and regenerate parts removed by the grazing herbivores. It contributes to the formation of the primary plant body.
• Lateral Meristem: It occurs in the mature regions of roots arid shoots of many plants, particularly that produce woody axis. These meristems are responsible for producing the secondary tissues.

Question 2.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Answer:
Yes, cork cambium forms tissues that form cork. As the stem continues to increase in girth another meristematic tissue called cork cambium or phellogen develops in cortex region of stem. The phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem. The inner cells differentiate into secondary cortex or phelloderm. Cork is impervious to water due to suberin and provides protection to underlying tissues.

Question 3.
Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Answer:
Secondary growth in stems of woody angiosperms occur by two types of cambia, i.e., vascular cambium and cork cambium.
(i) Vascular Cambium: Certain cells of medullary rays become meristematic to form interfascicular cambium. The fascicular cambium and the interfascicular cambium join to form a complete ring called cambial ring. The cells of the cambial ring undergo mitotic divisions and produce secondary phloem on its outer side and secondary xylem on its inner side.

At places, vascular cambium possesses ray initials. They form vascular rays, phloem rays in secondary phloem and wood rays in secondary xylem.

As new secondary phloem becomes functional, the previous older phloem gets crushed. Secondary xylem or wood persists. As a result wood grows

with age in the form of annual rings. In each annual ring, there is wide or broader spring or early wood or spring wood and narrow autumn or late wood.

In old stems, the central part of wood becomes non-functional and dark coloured due to tyloses and deposit of resins, gums, tannins. It is called r duramen or heartwood. The outer, functional wood is called sapwood.

(ii) Cork Cambium: As the stem continues to increase in girth due to the activity of vascular cambium the outer cortical and epidermal layers get broken and need to be replaced to provide new protective cell layers. In this way, cork cambium or phellogen develops in the cortex region. Phellogen cuts of cells on both sides.

The outer cells differentiate into cork or phellem while, the inner cells ; differentiate into secondary cortex or phelloderm. Due to the activity of cork cambium, pressure builds up on remaining layers peripheral to
phellogen and ultimately these layers die and slough off. At places, aerating pores called lenticels develop, which have loosely arranged , complementary cells.

Significance of Secondary Growth

• It replaces old non-functional tissues.
• It provides fire proof, insect proof and insulating cover around the older plant parts.
• Commercial cork is a product of secondary growth.
• Wood is the product of secondary growth.

Question 4.
Draw illustrations to bring out the anatomical difference between:
(a) Monocot root and Dicot root
(b) Monocot stem and Dicot stem
Answer:

(a) Anatomical Differences between Monocot Root and Dicot Root
(i) Anatomy of Monocot Root
(a) The structure of epidermis, cortex, endodermis and pericycle of a monocot root resembles exactly those of a dicot root.
(b) Vascular bundles are radial, and polyarch.
(c) Pith is large and well-developed; it is composed of parenchyma cells

(ii) Anatomy of dicot root

T.S of Dicot root
(a) Epidermis is single layered and many cells bear root hairs; cuticle is absent.
(b) Cortex is made of several layers of parenchyma cells.
(c) Endodermis consists of a single-layer of cells. The cells have a deposition of suberin, in the form of casparian strips, on their radial and tangential walls.
(d) Pericycle comprises a few layers of specialised parenchyma cells inner to the endodermis.
(e) Vascular bundles are radial and may range between two and six, though commonly there are four groups; i.e., tetrarch; xylem is exarch.
(f) Pith is very small and made of parenchyma cells.

(b) Anatomical Differences between Monocot Stem and Dicot stem
(i) Anatomy of monocot stem
(a) Epidermis is single layered and trichomes are absent; cuticle is present on its outer surface.
(b) Hypodermis consists of two or three layers of sclerenchyma cells.
(c) Ground tissue is parenchymatous and is not differentiated into cortex or pith.
(d) Vascular bundles are many and scattered in the ground tissue; they vary in size.

T.S. of Monocot Stem

• Each vascular bundle is surrounded by sclerenchymatous bundle sheath.
• The vascular bundles are conjoint and closed; xylem is endarch and characteristically a protoxylem lacuna is present.

(ii) Anatomy of dicot stem
(a) Epidermis is the outermost layer of cells; externally it is covered with a cuticle and may bear trichomes and a few stomata.
(b) Hypodermis consists of a few layers of collenchyma cells, just below the epidermis.
(c) Cortex consists of parenchyma cells.
(d) Endodermis is single layered and the cells are rich in starch grains and hence it is also referred to as starch sheath.
(e) Pericyde occurs inner to the endodermis, above the phloem of vascular bundles in the form of semi-lunar patches (hence also referred to as bundle caps); it is composed of sclerenchyma.
T.S. of Dicot Stem

(f) Vascular bundles are characteristically arranged in the form of a ring.

• Each vascular bundle is conjoint and open with intra-fascicular cambium; xylem is endarch.
  (g) Medullary rays are the few layers of radially placed parenchyma cells, in between the vascular bundles.
  (h) Pith is composed of parenchyma cells.

Question 5.
Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give reasons.
Answer:
Transverse section of a monocot stem possess following characters:

• Dumbbell-shaped guard cells in stomata in epidermis.
• Sclerenchymatous hypodermis.
• No concentric arrangement of internal tissues.
• Unifrom ground tissue showing no tissue differentiation.
• More than 8 scattered vascular bundles.
• Bundle sheath is present.
• No secondary growth normally.
• Xylem vessels arranged in Y-shaped manner.
• Protoxylem cavity usually present in vascular tissues.

Transverse section of a dicot stem possess following characters:

• Kidney-shaped guard cells in stomata present in epidermis.
• Collenchymatous hypodermis.
• Concentric arrangement of internal tissues.
• Differentiation of ground tissue into cortex, endodermis, pericycle and pith.
• The vascular bundles are arranged in a ring.
• Conjoint, collateral and open vascular bundles.
• Without bundle sheath.
• Secondary growth takes place.
• Xylem vessels arranged in rows.

[Note: For figures refer to Q.No. 4]

Question 6.
The transverse section of a plant material shows the following anatomical features:
(a) The vascular bundles are conjoint, scattered and surrounded by a sclerenchymatous bundle sheaths.
(b) Phloem parenchyma is absent. What will you identify it as?
Answer:
It is a transverse section of monocotyledonous stem.

Question 7.
Why are xylem and phloem called complex tissues?
Answer:
Xylem and phloem are made up of more than one type of cells that is why they are called as complex tissues :
(i) Xylem is composed of four different kinds of elements, namely-tracheids, vessels, xylem fibres and xylem parenchyma. Tracheids are dead tube-like cells which are thick walled, vessels are made up of large number of tube cells placed end to end. Xylem fibres are thick walled cells that maybe septate and aseptate. Xylem parenchyma is living and thin walled cells.

(ii) Phloem is composed of sieve tube elements, companion cells, phloem parenchyma and phloem fibres. Sieve tube elements are tube-like cells, whereas phloem parenchyma are living cells and phloem fibres are thick walled lignified cells.

Question 8.
What is stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Answer:
The minute pores present in the epidermis are known as stomata. The stomata may be surrounded by either bean-shaped (in dicots) or by dumb-bell-shaped (in monocots) guard cells. The guard cells in turn are surrounded by other epidermal cells, which are known as subsidiary or

accessory cells. The stomatal aperture, guard cells, accessory cells constitute the stomatal apparatus.

Question 9.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
The three basic; tissue systems in flowering plants are as follows:

• Epidermal Tissue System: The tissue related to this system are epidermis, cuticle and wax, stomata and trichomes.
• Ground Tissue System: It consists of cortex, endodermis, pericycle, medullary rays, pith and ground tissue of leaves.
• Vascular Tissue System: It contains conducting tissues like xylem and phloem.

Question 10.
How is the study of plant anatomy useful to us?
Answer:
Plant anatomy is the study of internal structure of living organisms.

• It describes the tissues involved in assimilation of food and its storage, transportation of water, i.e., xylem tissue, transportation of minerals. i.e., phloem and those involved in providing mechanical support to the plant,
• Study of internal structure of plants helps to understand their adaptations of diverse environments.
• The study of plant anatomy also help in understanding the functional organisation of higher plants.

Question 11.
What is periderm? How does periderm formation take place in
the dicot stems?
Answer:
Phellogen, phellem and phelloderm are collectively called as periderm. Phellogen develops usually in the cortex region. Phellogen is a couple of layers thick. Phellogen cuts off cells on both sides. The outer cells 1 differentiate into cork or phellem, while the inner cells differentiate into secondary cortex or phelloderm. All these together form periderm.

Question 12.
Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.
Answer:
The vertical section of a dorsiventral leaf through the lamina shows three ; main parts namely, epidermis, mesophyll and vascular system. The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata. The tissue between the upper and the lower epidermis is called the mesophyll. Mesophyll, which possesses chloroplasts and carry out photosynthesis, is made up of parenchyma. It has two types of cells – the palisade parenchyma and the spongy parenchyma.

The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other. The oval or round and loosely arranged spongy parenchyma is situated below the palisade cells and extends to the lower epidermis. There are numerous large spaces and air cavities between these cells. Vascular system includes vascular bundles, which can be seen in the veins and the midrib. The size of the vascular bundles are dependent on the size of the veins. The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick walled bundle sheath cells.
T.S. of Dorsiventral Leaf

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 17 Breathing and Exchange of Gases Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases

PSEB 11th Class Biology Guide Breathing and Exchange of Gases Textbook Questions and Answers

Question 1.
Define vital capacity. What is its significance?
Answer:
Vital Capacity (VC): The maximum volume of air a person can breathe in after a forced expiration is called vital capacity. Vital capacity is higher in athletes and singers. Vital capacity shows the strength of our inspiration and expiration.

Question 2.
State the volume of air remaining in the lungs after a normal breathing.
Answer:
The volume of air remaining in the lungs even after a forcible expiration averages 1100 ml to 1200 ml.

Question 3.
Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
Answer:
Alveoli are the primaty sites of gas exchange in the respiratory system. Exchange of gases occur between blood and these tissues. O₂ and CO₂ are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient. The diffusion membrane for gas exchange is made up of three major layers.
These layers are :

1. Squamous epithelium of alveoli.
2. Endothelium of alveolar capillaries.
3. Basement substance in between them. Its total thickness is much less than a millimetre. Therefore, all the factors in our body are favourable for diffusion of O₂ from alveoli to tissues and that of CO₂ from tissues to alveoli.

Question 4.
What are the major transport mechanisms for CO₂? Explain.
Answer:
Transport of Carbon Dioxide: CO₂ in gaseous form diffuses out of the cells into capillaries, where it is transported in following ways :
(i) Transport in dissolved form: About 7% CO₂ is carried in dissolved form through the plasma because of its high solubility.
(ii) Transport as bicarbonate: The largest fraction (about 70%) is carried in plasma as bicarbonate ions (HCO₃). At the tissues site, where pCO₂ is high due to catabolism, CO₂ diffuses into the blood (RBCs and plasma) and forms HCO₃^– and H^–.

This reaction is faster in RBCs because they contain an enzyme carbonic anhydrase. Hydrogen ion released during the reaction bind to Hb, triggering the Bohr effect.
At the alveolar site, where pCO₂ is low, the reaction proceeds in opposite direction forming CO₂ and H₂O. Thus, CO₂ trapped as bicarbonate at tissue level and transported to alveoli is released as CO₂.

(iii) Transport as carbaminohaemoglobin: Nearly 20-25% CO₂ is carried by haemoglobin as carbaminohaemoglobin, CO₂ entering the blood combines with the NH₂ group of the reduced Hb.

The reaction releases oxygen from oxyhaemoglobin.
Factors affecting the binding of CO₂ and Hb are as follows:

• Partial pressure of CO₂.
• Partial pressure of O₂ (major factor).

In tissues, pCO₂ is high and pO₂ is low, more binding of CO₂ occurs while, in the alveoli, pCO₂ is low and pO₂ is high, dissociation of CO₂ from HbCO₂ takes place, i.e., CO₂, which is bound to Hb from the tissues is delivered at the alveoli.

Question 5.
What will be the pO₂ and pCO₂, in the atmospheric air compared to those in the alveolar air?
(i) pO₂ lesser and pCO₂ higher
(ii) pO₂ higher and pCO₂ lesser
(iii) pO₂ higher and pCO₂ higher
(iv) pO₂ lesser and pCO₂ lesser
Answer:
(i) In the alveolar tissues, where low pO₂, high pCO₂, high H⁺ concentration, these conditions are favourable for dissociation of oxygen from the oxyhaemoglobin.

(ii) When there is high pO₂, low pCO₂, less H⁺ concentration and lesser temperature, the factors are all favourable for formation of oxyhaemoglobin.

(iii) When pO₂ is high in the alveoli and pCO₂ is high in the tissues then the oxygen diffuses into the blood and combines with oxygen forming oxyhaemoglobin and CO₂ diffuses out.

(iv) When pO₂ is low in the alveoli and pCO₂ is low in the tissues then these conditions are favourable for dissociation of oxygen from the oxyhaemoglobin.

Question 6.
Explain the process of inspiration under normal conditions.
Answer:
Inspiration is the process during which atmospheric air is drawn in. Inspiration is initiated by the contraction of diaphragm, which increases the volume of thoracic chamber in the anteroposterior axis. The contraction of external intercostal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis.

The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intrapulmonary pressure to less than the atmospheric pressure, which forces the air from outside to move into the lungs, i. e., inspiration. On an average, a healthy human breathes 12-16 times/minute.

Question 7.
How is respiration regulated?
Answer:
Respiratory rhythm centre is primarily responsible for regulation of respiration. This centre is present in the medulla.
Pneumotaxic centre, present in the pons region, also coordinates respiration. Apart from them, receptors associated with aortic arch and carotid artery, can also recognize changes in CO₂ and H⁺ concentration and send signal to the rhythm centre for proper action.

Question 8.
What is the effect of pCO₂ on oxygen transport?
Answer:
Partial pressure of CO₂ (pCO₂) can interfere the binding of oxygen with haemoglobin, i.e., to form oxyhaemoglobin.
(i) In the alveoli, where there is high pO₂ and low pCO₂, less H⁺ concentration and low temperature, more formation of oxyhaemoglobin occur.

(ii) In the tissues, where low pO₂, high pCO₂, high H⁺ concentration and high temperature exist, the conditions are responsible for dissociation of oxygen from the oxyhaemoglobin.

Question 9.
What happens to the respiratory process in a man going up a hill?
Answer:
At hills, the pressure of air falls and the person cannot get enough oxygen in the lungs for diffusion in blood. Due to deficiency of oxygen, the person feels breathlessness, headache, dizziness, nausea, mental fatigue and a bluish colour on the skin, nails and lips.

Question 10.
What is the site of gaseous exchange in an insect?
Answer:
The actual site of gaseous exchange in an insect is tracheoles and tracheolar end cells.

Question 11.
Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
A sigmoid curve is obtained when percentage saturation of haemoglobin with O₂ is plotted against the pO₂.
This curve is called the oxygen dissociation curve and is highly useful in studying the effect of factors like pCO₂, H⁺ concentration, etc., on binding of O₂ with haemoglobin.

In the alveoli, where there is high pO₂, low pCO₂, lesser H⁺ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low pO₂, high pCO₂, high H⁺ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. This dearly indicates that O₂ gets bound to haemoglobin in the lung surface and gets dissociated at the tissues. Every 100 ml of oxygenated blood can deliver around 5 ml of O₂ to the tissues under normal physiological conditions.

Question 12.
Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
Answer:
Hypoxia is the shortage of oxygen supply to the blood due to :
(a) normal shortage in air
(b) oxygen deficiency on high mountains (mountain sickness), anaemia and phytotoxicity or poisoning of electron transport system.

Question 13.
Distinguish between:
(a) IRV and ERV
(b) Inspiratory Capacity and Expiratory Capacity
(c) Vital Capacity and Total Lung Capacity
Answer:
(a) IRV and ERV Inspiratory Reserve Volume (IRV): Additional volume of air, a person can inspire by a forcible inspiration. This is about 2500-3000 mL. Expiratory Reserve Volume (ERV): Additional volume of air, a person can expire by a forcible expiration. This is about 1000-1100 mL.

(b) Inspiratory Capacity and Expiratory Capacity Inspiratory Capacity (IC): Total volume of air a person can inspire after a normal expiration. This includes tidal volume and inspiratory reserve volume (TV+IRV).
Expiratory Capacity (EC): Total volume of air a person can expire after a normal inspiration. This includes tidal volume and expiratory reserve volume (TV+ERV)

(c) Vital Capacity and Total Lung Capacity
Vital Capacity (VC): The maximum volume of air, a person can breathe in after a forced expiration. This includes ERV, TV and IRV or the maximum volume of air a person can breathe out after a forced inspiration.
Total Lung Capacity (TLC): Total volume of air accommodated in the lungs at the end of a forced inspiration. This includes RV, ERV, TV and IRV or vital capacity + residual volume.

Question 14.
What is tidal volume? Find out the tidal volume (approximate value) for a healthy human in an hour.
Answer:
Tidal Volume (TV): Volume of air inspired or expired during a normal respiration is called tidal volume. It is about 500 mL., i.e., a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 18 Body Fluids and Circulation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation

PSEB 11th Class Biology Guide Body Fluids and Circulation Textbook Questions and Answers

Question 1.
Name the components of the formed elements in the blood and mention one major function of each of them.
Answer:
(a) Erythrocytes: They are also known as Red Blood Cells (RBC). They are the most abundant of all the cells in blood. A healthy’adult man has, on an average, 5 millions to 5.5 millions of RBCs mm ^-3 of blood. RBCs are formed in the red bone marrow in the adults.
RBCs are devoid of nucleus in most of the mammals and are biconcave in shape. They have a red coloured, iron containing complex protein called haemoglobin. A healthy individual has 12-16 gms of haemoglobin in every 100 ml of blood.
these molecules play a significant role in transport of respiratory gases. RBCs have an average life span of 120 days after which they are destroyed in the spleen. Hence, spleen is also known as the graveyard of RBCs.

(b) Leucocytes: They are also known as White Blood Cells (WBC) as they are colourless due to the lack of haemoglobin. They are nucleated and are relatively lesser in number which averages 6000-8000 mm^-3 of blood. Leucocytes are generally short-lived.

There are two main categories of WBCs :
1. Granulocytes, e.g., neutrophils, eosinophils and basophils
2. Agranulocytes. e.g., lymphocytes and monocytes.
Neutrophils are the most abundant cells (60-65 per cent) of the total WBCs and basophils are the least (0.5-1 per cent) among them. Neutrophils and monocytes (6-8 per cent) are phagocytic cells which destroy foreign organisms entering the body.

Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions. Eosinophils (2-3 per cent) resist infections and are also associated with allergic reactions. Lymphocytes (20-25 per cent) are of two major types- ‘B’ and T forms. Both B and T lymphocytes are responsible for immune responses of the body.

(c) Platelets: Platelets or thrombocytes, are involved in the coagulation or clotting of blood. A reduction in their number can lead to clotting disorders, which will lead to excessive loss of blood from the body.

Question 2.
What is the importance of plasma proteins?
Answer:
Fibrinogen, globulins and albumins are the major plasma proteins. Fibrinogens are needed for clotting or coagulation of blood. Globulins primarily are involved in defense mechanisms of the body and the albumins help in osmotic balance.

Question 3.
Match column I with column II.

Column I	Column II
A. Eosinophils	1. Coagulation
B. RBC	2. Universal recipient
C. AB group	3. Resist infections
D. Platelets	4. Contraction of heart
E. Systole	5. Gas transport

Answer:

Column I	Column II
A. Eosinophils	3. Resist infections
B. RBC	5. Gas transport
C. AB group	2. Universal recipient
D. Platelets	1. Coagulation
E. Systole	4. Contraction of heart

Question 4.
Why do we consider blood as a connective tissue?
Answer:
Blood is a mobile connective tissue derived from mesoderm which consists of fibre-free fluid matrix, plasma and other cells. It regularly circulates in the body, takes part in the transport of materials.

Question 5.
What is the difference between blood and lymph?
Answer:
Differences between Blood and Lymph

Blood	Lymph
It is red in colour due to the presence of haemoglobin in red cells.	It is colourless as red blood cells are absent.
It consists of plasma, RBC, WBC and platelets.	It consists of plasma and less number of WBC.
Glucose concentration is low.	Glucose concentration is higher than blood.
Clotting of blood is a fast process.	Clotting of lymph is comparatively slow.
It transports materials from one organ to other.	It transports materials from tissue cells into the blood.
Flow of blood is fast.	Lymph flows very slowly.
Its plasma has more proteins, calcium and phosphorus.	Its plasma has less protein, calcium and phosphorus.
It moves away from the heart and towards the heart.	It moves in one direction, i. e., from tissues to sub-clavians.

Question 6.
What is meant by double circulation? What is its significance?
Answer:
Double Circulation: In double circulation, the blood passes twice through the heart during one complete cycle. Double circulation is carried out by two ways :
1. Pulmonary circulation,
2. Systemic circulation

Significance: In birds and mammals, oxygenated and deoxygenated blood received by the left and right atria respectively passes on to the ventricles of the same sides. The ventricles pump it out without mixing up, i.e., two separate circulatory pathways are present in these organisms. This is the importance of double circulation.

Question 7.
Write the differences between:
(a) Blood and lymph
(b) Open and closed system of circulation
(c) Systole and diastole
(d) P-wave and T-wave
Answer:
(a)

Blood	Lymph
It is red in colour due to the presence of haemoglobin in red cells.	It is colourless as red blood cells are absent.
It consists of plasma, RBC, WBC and platelets.	It consists of plasma and less number of WBC.
Glucose concentration is low.	Glucose concentration is higher than blood.
Clotting of blood is a fast process.	Clotting of lymph is comparatively slow.
It transports materials from one organ to other.	It transports materials from tissue cells into the blood.
Flow of blood is fast.	Lymph flows very slowly.
Its plasma has more proteins, calcium and phosphorus.	Its plasma has less protein, calcium and phosphorus.
It moves away from the heart and towards the heart.	It moves in one direction, i. e., from tissues to sub-clavians.

(b) Differences between Open and Closed Circulatory Systems

Open Circulatory System	Closed Circulatory System
1. It is present in arthropods and molluscs.	It is present in annelids and chordates.
2. Blood pumped by heart passes through large vessels into open spaces or body cavities called sinuses.	Blood pumped by the heart is circulated through a loosed network of blood vessels.
3. Flow of blood is not regulated precisely.	It is more advantageous as the blood flow is more precisely regulated.

(c) Differences between Systole and Diastole

Systole	Diastole
1. The contraction of the muscles of auricles and ventricles is called systole.	It is the relaxation of atria and ventricle muscle.
2. It increases the ventricular pressure causing the closure of tricuspid and bicuspid valves due to attempted backflow of blood into atria.	The ventricular pressure falls causing the closure of semilunar valves which prevent backflow of blood into the ventricle.
3. Systolic pressure is higher and occurs during ventricular contraction.	Diastolic pressure is lower and occurs during ventricular expansion.

(d) Differences between P-wave and T-wave

P-wave	T-wave
The P-wave represents the electrical excitation (or depolarisation) of the arrÍa, which leads to the contraction of both the arria.	The T-wave represents the return of the ventricles from excited to normal state (repolarisation). The end of the T-wave marks the end of systole.

Question 8.
Describe the evolutionary change in the pattern of heart among the vertebrates.
Answer:
The heart among the vertebrates shows different patterns of evolution. Different groups of animals have evolved different methods for this transport. All vertebrates possess a muscular chambered heart.
Fishes have a 2-chambered heart with an atrium and a ventricle.
Amphibians and the reptiles (except crocodiles) have a 3-chambered heart with two atria and a single ventricle.
In crocodiles, birds and mammals possess a 4-chambered heart with two atria and two ventricles.

In fishes, the heart pumps out deoxygenated blood which is oxygenated by the gills and supplied to the body parts from where deoxygenated blood is returned to the heart.

In amphibians and reptiles, the left atrium receives oxygenated blood from the gills/lungs/skin and the right atrium gets the deoxygenated blood from other body parts. However, they get mixed up in the single ventricle which pumps out mixed blood.

In birds and mammals, oxygenated and deoxygenated blood received by the left and right atria respectively passes on to the ventricles of the same sides. The ventricles pump it out without any mixing up, i. e., two separate circulatory pathways are present in these organisms, hence, these animals have double circulation.

Question 9.
Why do we call our heart myogenic?
Answer:
Heart is myogenic in origin because the cardiac impulse is initiated in our heart muscles.

Question 10.
The sino-atrial node is called the pacemaker of our heart. Why?
Answer:
The sino-atrial node of heart is responsible for initiating and maintaining the rhythmic activity, therefore it is known as pacemaker of the heart.

Question 11.
What is the significance of atrioventricular node and atrioventricular bundle in the functioning of heart?
Answer:
Atrioventricular Node (AVN): It is the mass of tissue present in the lower-left corner of the right atrium close to the atrioventricular septum. It is stimulated by the impulses that sweep over the atrial myocardium. It is too capable of initiating impulses that cause contraction but at slower rate than SA node.

Atrioventricular Bundle (AV Bundle): It is a bundle of nodal fibres, which continues from AVN and passes through the atria-ventricular septa to emerge on the top of interventricular septum. The AV bundle, bundle branches and Purkinje fibres convey impulses of contraction from the AV node to the apex of the myocardium. Here the wave of ventricular contraction begins, then sweeps upwards and outwards, pumping blood into the pulmonary artery and the aorta.
This nodal musculature has the ability to generate action potentials without any external stimuli.

Question 12.
Define a cardiac cycle and the cardiac output.
Answer:
Cardiac Cycle: The sequential event in the heart which is cyclically repeated is called the cardiac cycle. It consists of systole and diastole of both the atria and ventricles.

Cardiac Output: It is the volume of blood pumped out by each ventricle per minute and averages 5000 mL or 5 L in a healthy individual. The body has the ability to alter the stroke volume as well as the heart rate and thereby the cardiac output. For example, the cardiac output of an athlete will be much higher than that of an ordinary man.

Question 13.
Explain heart sounds.
Answer:
During each cardiac cycle, two prominent sounds are produced which can be easily heard through a stethoscope. The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves, whereas the second heart sound (dup) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance.

Question 14.
Draw a standard ECG and explain the different segments in it.
Answer:
Electrocardiograph (ECG): ECG is a graphical representation of the electrical activity of the heart during a cardiac cycle. A patient is connected to the machine with three electrical leads (one to each wrist and to the left ankle) that continuously monitor the heart activity. For a detailed evaluation of the heart’s function, multiple leads are attached to the chest region.

Each peak in, the ECG is identified with a letter from P to T that corresponds to a specific electrical activity of the heart. The P-wave represents the electrical excitation (or depolarization) of the atria, which leads to the contraction of both the atria. The QRS complex represents the depolarization of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.

• The T-wave represents the return of the ventricles from excited to normal state (repolarisation).
• The end of the T-wave marks the end of systole.
• Obviously, by counting the number of QRS complexes that occur in a given time period, one can determine the heartbeat rate of an individual.
• Since the ECGs obtained from different individuals have roughly the same shape for a given lead configuration, any deviation from this shape indicates a possible abnormality or disease.
• Hence, it is of a great clinical significance.

Punjab State Board PSEB 11th Class Biology Book Solutions Morphology of Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Morphology of Flowering Plants

PSEB 11th Class Biology Guide Morphology of Flowering Plants Textbook Questions and Answers

Question 1.
What is meant by modification of root? What type of modification of root is found in the:
(a) Banyan tree
(b) Turnip
(c) Mangrove trees
Answer:
Modification of Root: Roots in some plants change their shape and
structure and become modified to perform functions, other than absorption and conduction of water and minerals. The roots are modified for water, absorption, support, storage of food and respiration.
(a) A banyan tree have hanging roots known as prop roots.
(b) The roots of turnip get modified to become swollen and store food.
(c) The roots of mangrove trees get modified to grow vertically upwards and help to get oxygen for respiration. These are known as pneumatophores.

Question 2.
Justify the following statements on the basis of external features:
(a) Underground parts of a plant are not always roots.
(b) Flower is a modified shoot.
Answer:
(a) Underground parts of a plant are not always roots, they are subterranean stems which do not have root hairs and root cap. Have terminal bud, nodes and internodes. Have leaves on the nodes.
Most of the underground stems such as sucker, rhizome, corm, tubers, bulb, etc., store food, form aerial shoots.
(b) Flower is a modified shoot because:

• It possess nodes and internodes.
• It may develop in the axil of small leaf-like structure called bract.
• Flowers get modified into bulbils or fleshy buds in some plants.
• Anatomically the pedicel and thalamus of a flower resemble that of stem.
• The vascular supply of different organs of flower resemble that of normal leaves.
• In the flower of Degeneria, the stamens are expanded like leaves and the carpels appear like folded leaves.

Question 3.
How is a pinnately compound leaf different from a palmately compound leaf?
Answer:
In pinnately compound leaf, the number of leaflets are present on a common axis, the rachis, which represents the midrib of the leaf as in neem. In case of a palmately compound leaf, the leaflets are attached at a common point, i e., at the tip of petiole as in silk cotton.

Question 4.
Explain with suitable examples the different types of phyllotaxy.
Answer:
Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. This is usually of three types—alternate, opposite and whorled.

• In alternate phyllotaxy, a single leaf arises at each node in alternate manner, as in China rose, mustard and sunflower plants.
• In opposite type of phyllotaxy, a pair of leaves arise at each node and lie opposite to each other as in Calotropis and guava plants.
• If more than two leaves arise at each node and form, a whorl, it is called as whorled, as in Alstonia.

Question 5.
Define the following terms:
(i) Aestivation
(ii) Placentation
(iii) Actinomorphic
(iv) Zygomorphic
(v) Superior ovary
(vi) Perigynous flower
(vii) Epipetalous stamen
Answer:
(i) Aestivation: The mode of arrangement of sepals or petals in relation to one another in a flower bud is called aestivation.
(ii) Placentation: The pattern by which the ovules are attached in an ovary is called placentation.

(iii) Actinomorphic: A flower having radial symmetry. The parts of each whorl are similar in size and shape. The flower can be divided in two equal halves along more than one median longitudinal plane.

(iv) Zygomorphic: A flower having bilateral symmetry. The parts of one or more whorls are dissimilar. The flower can be divided into two equal halves in only one vertical plane.

(v) Superior ovary: The ovary is called superior when it is borne above the point attachment of perianth and stamens on the thalamus.

(vi) Perigynous flower: It is the condition in which gynoecium of a flower is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level.

(vii) Epipetalous stamen: Stamens adhere to the petals by their filaments so, appear to arise from them.

Question 6.
Differentiate between:
(a) Racemose and cymose inflorescence
(b) Fibrous root and adventitious root
(c) Apocarpous and syncarpous ovary
Answer:
(a) Differences between Racemose and Cymose Inflorescence.

Racemose	Cymose
1. This is further divided into (a) Raceme (b) Catkin (c) Spike (d) Spadix (e) Corymb (f) Umbel or capitutum	It is further divided into (a) Monochasial cyme (b) Dichasial Cyme (c) Polychasial Cyme.
2. Branches develop indefinitely and further branches arise laterally in acropetal manner.	The branches arise from terminal buds and stop growing after some time Lateral branches grow much vigorously and spread like a dome.

(b) Differences between Fibrous Root and Adventitious Root

Fibrous Root	Adventitious Root
In monocotyledonous plants, the primary root is short lived and is replaced by a latge number of roots. These roots originate from the base of the fibrous root system say, for example in wheat plants.	In some plants, say for example, in grass and banyan tree there are roots arising from parts of the plant other than the radicle. These are called adventitious roots.

(c) Differences between Apocarpous Ovaiy and Syncarpous Ovary

Apocarpous Ovary	Syncarpous Ovary
When more than one carpel is present, they may be free (as in lotus and rose) and are called apocarpous ovary.	They are termed syncarpous ovary when fused, as in mustard and tomato. After fertilisation, the ovules develop into seeds and like ovary matures into a fruit.

Question 7.
Draw the labelled diagram of the following:
(a) gram seed
(b) V.S. of maize seed
Answer:
(a) Gram Seed

(b) V.S. of maize seed

Question 8.
Describe modifications of stem with suitable examples. [NCERT]
Answer:
Modifications of stem are as follows:

• Tendrils help plants to climb on the support, e. g., Cucumber.
• Thorns are woody, pointed, straight structures to protect plants from browsing animals, e. g., Bougainvillea.
• The plants in arid regions modify their stems into flattened (Opuntia) or fleshy cylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.
• Underground stems of some plants such as grass and snawberry, etc., spread to new riches and when older parts die, new plants are formed.
• In Pistia and Eichhornia, a lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots is found.
• Stolons or runners help in vegetative propagation in jasmine and grass, respectively.

Question 9.
Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagram after studying them.
Answer:
(i) Fabaceae: This family was earlier called Papilionoideae, a subfamily of family Leguminosae.
(a) Habit: Trees, shrubs, herbs, climbers, etc.
(b) Root System: Tap root system with root nodules, that harbour nitrogen fixing bacterium.
(c) Leaves: Leaves are alternate, simple or pinnately compound, pulvinate, and stipulate; venation reticulate.
(d) Inflorescence: Racemose usually, a raceme.
(e) Flowers: Bracteate, bracteolate, bisexual, zygomorphic, hypogynous, and pentamerous.
(f) Calyx: Five sepals, gamosepalous, irregular, odd sepal anterior (characteristic feature of the family) and valvate aestivation.

(g) Corolla: Corolla consists of five petals, polypetalous, characteristically papilionaceous, with an odd posterior large petal called standard or vexillum, a pair of lateral petals, called wing or alae and two anterior keel or carina, which enclose the essential organs; aestivation is vexillary.
(h) Androecium: Ten stamens, diadelphous, [(9) + 1] and anthers dithecous.
(i) Gynoecium: Ovary is superior, monocarpellary, unilocular with many ovules on marginal placenta; style single, curved or bent at right angles to the ovary.
(j) Fruits and Seeds: Characteristically a legume/pod and seeds are non-endospermic.
(k) Floral Formula:
(l) Economic Importance: Plants of this family yield pulses, edible oil, dye, fodder, fibres and wood; some yield products of medicinal value,

(ii) Solanaceae (Potato family)
(a) Habit: Plants are mostly herbs or shrubs or small trees; stem is erect, cylindrical, branched (cymose type); stem is underground in potato CSolarium tuberosum).
(b) Leaves: Simple, alternate, exstipulate with reticulate venation.
(c) Inflorescence: Axillary or extra-axillary cymose, or solitary.
(d) Flowers: Bisexual, actinomorphic, hypogynous and pentamerous.
(e) Calyx: Five sepals, gamosepalous, persistant and valvate aestivation.
(f) Corolla: Five petals, gamopetalous, valvate or imbricate, rotate/wheel-shaped.
(g) Androecium: Five stamens, epipetalous and alternating with the petals.
(h) Gynoecium: Bicarpellary, syncarpous, superior with many ovules on swollen axile placenta; carpels are obliquely placed.
(i) Fruits and Seeds: A berry (tomato and brinjal) or a capsule; seeds are endospermic.
(j) Floral Formula :
(k) Economic Importance: Many plants are used as source of food (vegetables), spice, medicines of fumigatory; some are ornamental plants.

Question 10.
Describe the various types of placentations found in flowering plants.
Answer:
Types of Placentations: The arrangement of ovules within the ovary is known as placentation. The placentations are of different types – marginal, axile, parietal, free central and basal.

Marginal placentation: In this placentation, the placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows as in pea.

Axile placentation: In this placentation, the placenta is axile and the ovules are attached to it in a multilocular ovary as in China rose, tomato, etc.

Parietal placentation: In this placentation, the ovules develop on the inner wall of the ovary or on peripheral part. Ovary is one chambered but it becomes two chambered due to the formation of a false septum known as replam, e.g., mustard.

Free central placentation: In this type of placentation, the ovules are present on the central axis of ovary and septa are absent as in Dianthus and primrose.

Basal placentation: In this placentation the placenta develops at the base of ovary and a single ovule is attached to it, as in sunflower.

Question 11.
What is a flower? Describe the parts of a typical angiospermic flower.
Answer:
Flower: It is a condensed modified reproductive shoot found in angiosperms. It often develops in the axile of a small leaf-like structure called bract. The stalk of the flower is called pedicel. The tip of the pedicel or the base of flower has a broad highly condensed multinodal region called thalamus.
A flower has following four floral structures:

• Calyx: It is made up of sepals. These are green in colour and help in photosynthesis.
• Corolla: It is the brightly coloured part containing petals.
• Androecium: It is the male reproductive part which consists of stamens. A stamen has a long filament and terminal anther. The anther produces the pollen grains.
• Gynoecium: It is the female reproductive part which consists of
  carpels. A carpel has three parts, i.e., style, stigma and ovary. The ovary bears the ovules.

Question 12.
How do the various leaf modifications help plants?
Answer:
Leaf Modifications in Plants
(i) In some plants, the leaf and leaf parts get modified to form green, long, thin unbranched and sensitive thread-like structures called tendrils. The tendrils coil around the plant and provide support to the plant in climbing. Tendrils are present in pea, garden Nasturtium, Clematis, Smilax, etc.

(ii) In some plants, the leaves get modified to form curved stiff claw like hooks to help the plant in clinging to the support. Leaflet hooks are present in Bignonia.

(iii) In case of Acacia and Zizyphus, the leaves get modified to form vasculated, hard, stiff and pointed structures.

(iv) In case of Acacia longifolia, the expanded petiole gets modified and perform the function of photosynthesis in absence of lamina.

(v) In plants such as Nepenthes, the lamina is modified to form large pitcher. It is used for storing water and for digesting insect protein.

(vi) In case of Utricularia, the leaf segments are modified into small bladders, to trap small animals.

Question 13.
Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Answer:
The arrangement of flowers on the floral axis is termed as inflorescence. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescences are defined – racemose and cymose. In racemose type of inflorescence the main axis continues to grow, the flowers are borne laterally in an acropetal succession.

In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipetal order, as depicted in figure.

Question 14.
Write the floral formula of a actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five f free stamens and two united carpels with superior ovary and axile placentation.
Answer:
Floral formula

Question 15.
Describe the arrangement of floral members in relation to their ‘ insertion on thalamus.
Answer:
A flower is a condensed specialised reproductive shoot found in angiosperms. The stalk of the flower is known as pedicel. The tip of the pedicel or the base of the flower has a broad highly condensed multinodal region called thalamus. The floral parts of a flower are present on the thalamus. Starting from below they are green sepals or calyx, coloured petals or corolla, stamens or androecium and carpels or gynoecium.