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Chloroplast is the name of the green plastid.

(1) It carries oxygen to the cells. 

(2) It forms clots to plug leaking vessels, to stop bleeding.

The answer is (C) Haemoglobin

The answer is (D) Complete digestion of fat

It is the systemic arterial blood pressure, which the blood flowing through the arteries exerts on their walls.

The answer is (A) 120 nm

1. (c) nearby

2. (a) accepted

3. (a) deteriorate

4. (c) disinterest

5. (b) few

1. When Jay saw the Himalayas, he developed an attraction for the mountains.

2. Walking on tender grass in the morning is very refreshing.

3. The enemy struck the guards in the stillness of the night.

4. The sun is gradually setting at the horizon.

1. The princess was so frightened that she stared at the devil.

2. The wise old man could sense that his son was in some trouble.

3. The prince aimed an arrow at the cruel giant and killed him.

4. As the light grew faint, the children ran towards their houses.

Mass of the body, m = 25kg 

Momentum, p = 125 kg.m/s 

p = mv 

v = p/m = 5 m/s 

Velocity of the body is 5 m/s

Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps. 

Glycolysis consists of two major phases: 

1. Preparatory phase (1-5 steps). 

2. Payoff phase (6-10 steps). 

1. Preparatory phase: 

a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed. 

b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other. 

c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate. 

d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed. 

e. Preparatory phase of glycolysis ends. 

2. Payoff phase: 

a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP. 

b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

(i) Alstonia: Whorled phyllotaxy 

(ii) Guava: Opposite and decussate

Morphology of flowering plants deals with the study of the shape and structure of plants. The flowering plants also are referred to as angiosperms. they seem during a sort of sizes, shapes, and forms. The morphological structures of flowering plants include the roots, the stem, leaves, flowers. The roots form the underground a part of the plant and absorb water and nutrients from the soil. The stem grows above the soil. The leaves contain chlorophyll which helps within the synthesis of food. They also bear pores for transpiration. We have MCQ Questions for class 11 with Answers to assist students understands the concept alright.

Let start practice of the Class 11 Biology MCQ Questions of Morphology of Flowering Plants with Answers and revise the chapters in proper manner. These MCQ Questions for class 11 BIology are designed according to the Latest Exam Pattern and Syllabus. Class 11 Biology MCQ Questions of Morphology of Flowering Plants with Answers are helpful board exams and many more competitive examinations. These MCQ Questions for Class 11 with answers clear for a speedy correction of the Chapter.

Practice MCQ Questions for class 11 Biology Chapter-Wise

1. Artabotrys is a hook climber in which the hooks are modified

(a) inflorescence axis
(b) petiole
(c) roots
(d) stipules

2. Exceptional roots of Cuscuta are

(a) haustorial
(b) coralloid
(c) mycorrhizal
(d) all of the above.

3. The coloured part of a Bougainvillea flower is the

(a) corolla
(b) calyx
(c) bracts
(d) androecium.

4.  Replum is found in the ovary of

(a) Brassicaceae
(b) Malvaceae
(c) Liliaceae
(d) Asteraceae.

5. One seeded winged fruit is

(a) Nut
(b) Samara
(c) Cypsela
(d) Achene

6. A fruit developed from a condensed inflorescence is

(a) Composite fruit
(b) Simple fruit
(c) Aggregate fruit
(d) Etaerio

7. Stem modified to perform the function of a leaf and having many internodes is called as

(a) phylloclade
(b) cladode
(c) offset
(d) phyllode.

8. Keel is the characteristic features of:

(a) Tulip
(b) Indigofera
(c) Aloe
(d) Tomato

9. A typical lower with superior ovary and other floral parts inferior is called:  

(a) Polygamous
(b) Hypogynous
(c) Perigynous
(d) Epigynous

10. Placenta and pericarp are edible portions in

(a) Potato
(b) Banana
(c) Tomato
(d) Apple

11. Dry indehiscent single-seeded fruit formed by the bi-carpellary syncarpous inferior ovary is

(a) Cremocarp
(b) Berry
(c) Caryopsis
(d) Cypsela

12. Placentation in tomato and lemon is

(a) Parietal
(b) Free central
(c) Marginal
(d) Axile

13. Pineapple (ananas) fruit develops from

(a) a multipistillate syncarpous flower
(b) a cluster of compactly borne flowers on a common axis
(c) a multilocular monocarpellary flower
(d) a unilocular polycarpellary flower

14. The aleurone layer in maize grain is especially rich in

(a) auxins
(b) proteins
(c) starch
(d) lipids

15. What would be the number of chromosomes of the aleurone cells of a plant with 42 chromosomes in its root tip cells?

(a) 42
(b) 63
(c) 84
(d) 21

16. A large globular root that tapers sharply at the lower end is called

(a) fusiform
(b) napiform
(c) conical
(d) tuberous

17. A short, green, flattened branch resembling a leaf arising from the axil of a reduced scale leaf is called

(a) phylloclade
(b) cladode
(c) phyllode
(d) stipule

18. Sweet potato is a modification of

(a) Primary root
(b) leaf
(c) underground root
(d) Adventitious root.

19. Winged petiole is found in

(a) citrus
(b) acacia
(c) radish
(d) peepal

20. Velamen is found in

(a) only aerial roots of orchids
(b) leaves of Ficus elastica
(c) aerial and terrestrial roots of orchids
(d) roots of screwpine

21. Maize grain is a fruit known as

(a) caryopsis
(b) achene
(c) cypsela
(d) legume

22. Geocarpic fruit is 

(a) potato 
(b) groundnut
(c) onion
(d) garlic

23. Free central placentation is found in 

(a) Dianthus
(b) Argemone
(c) Brassica
(d) Citrus

24. Heterospory and seed habit are often discussed in relation to a structure called

(a) spathe
(b) bract
(c) petiole
(d) ligule

25. The morphological nature of the edible part of a coconut is

(a) Cotyledon
(b) Perisperm
(c) Pericarp
(d) Endosperm

Answer:

1. Answer:(a) inflorescence axis

Explanation: Floral stalks or pedicels of Artabotrys produce a curbed hook. This hook is the modification of the inflorescence axis that helps the plant in climbing.

2. Answer: (a) haustorial

Explanation: Haustorial root is present in (1) Zea mays (2) Cactus (3) Cuscuta (4) Monotropa. The root is an underground part of the plant that absorbs water and minerals from the soil and anchors the plant firmly. Sucking or Haustorial roots – These roots are found in parasitic plants.

3. Answer:(c) bracts

Explanation: The color of the Bougainvillea flower depends upon the color of its bracts. These bracts are the specialized leaf that is modified. They may even form the petals or sepals of the flowers giving them different colors.

4. Answer: (a) Brassicaceae

Explanation::Replum is present in the ovary of flower of mustard. A false septun called replum develops between the two parietal placentae in mustard and other members of family Brassicaceae

5. Answer: (b) Samara

Explanation:A samara is a winged achene, a type of fruit in which a flattened wing of fibrous, papery tissue develops from the ovary wall. A samara is a simple dry fruit, and is indehiscent (not opening along a seam).

6. Answer: (a) Composite fruit

Explanation: Such fruit is often referred to as fruitlessness is a Composite fruit. There are composite fruits of two kinds.

7. Answer: (a) phylloclade

Explanation:Phylloclade is a modification of stem because the stem modifies into a green fleshy leaf life structure having distinct nodes and internodes.

8. Answer: (b) Indigofera

Explanation: Keel is petal is vexillary aestivation, which is a characteristic of family - Fabaceae, i.e. Indigofera tomato belongs to the family - Solanaceae, tulip and while aloe to Liliaceae. Hence, the keel is the characteristic feature of the flower of a family - Papilionaceae.

9. Answer: (b) Hypogynous

Explanation:A typical flower with superior ovary and other floral parts inferior is. A typical flower that has an ovary placed superior along with the other floral organs is known as hypogynous. The other floral organs are also attached under the ovary to the receptacle.

10. Answer: (c) Tomato

Explanation:The placenta can be defined as an outgrowth of a parenchymatous tissue in the inner wall of the ovary to which the ovule or ovules remain attached. Thus, both placenta and pericarp in tomatoes are edible.

11.Answer: (d) Cypsela

Explanation: Cypsela is dry, indehiscent, single-seeded fruit develops from an unilocular, single ovulate inferior ovary of bicarpellary, syncarpous, gynoecium possessing basal placentation. The fruit wall develops from pericarp and thalamus and is thin and remains attached to the seed at one point, eg, Helianthus.

12.Answer: (d) Axile

Explanation: Axile type of placentation is seen in lemon and tomato. In this type of placentation, the ovary is partitioned into two or more chambers by septa. The placenta is formed in the central region where all the septa intersect or meet and therefore an axle column bearing ovules is formed.

13.Answer: (b) a cluster of compactly borne flowers on a common axis

Explanation:Pineapple fruit develops from a cluster of compactly borne flowers on a common axis. It develops from spike, spadix, or catkin. In such cases, the flowers remain with their succulent petals, the axis bearing them grows and becomes fleshy or woody.

14. Answer: (b) proteins

Explanation: The aleurone layer of maize grain is especially rich in proteins. Aleurone is a protein found in protein granules of mature seeds and tubers. The aleurone layer is the outermost layer is the outermost layer of the endosperm.

15.Answer: (b) 63

Explanation: The number of chromosomes in the diploid cell of the root tip as given is 42, which means one set is of 21 chromosomes. So the cells of the aleurone layer of the given plant will have 63 chromosomes. 

16. Answer: (b) napiform

Explanation: Napiform root- The root is nearly globular or spherical in shape. The basal portion of the root is much swollen which suddenly tapers towards the apex giving a top-shaped appearance.

17. Answer: (b) cladode

Explanation: Cladodes (also called cladophylls or phylloclades) are shoot systems in which leaves do not develop; rather, the stems become flattened and assume the photosynthetic functions of the plant. In asparagus (Asparagus officinalis; Asparagaceae), the scales found on the asparagus spears are the true leaves.

18.Answer: (d) Adventitious root

Explanation: Sweet potato is a changed adventitious root. Few plants are modified by their root systems and execute specialized features. There may be fleshy and swollen transformed roots Modified roots, in the form of water and starch, serve as a storage portion to store the food required for the plant.

19. Answer: (a) citrus

Explanation:Plants like Citrus and Dionaea have winged petiole in them. This also gives a characteristic foliage arrangement to the plant. Outgrowth present on each side of the petiole in some species is called stipules.

20. Answer: (a) only aerial roots of orchids

Explanation:Velamen tissue is found in the aerial roots of some orchids or vanda. Vanda is a genus in the orchid family orchidaceae. Velamen is a spongy, multiple epidermis that covers the roots of some epiphytic or semi-epiphytic plants, such as orchid species.

21.Answer: (a) caryopsis

Explanation:Caryopsis is a term used for small, dry, indehiscent, single-seeded fruits that are developed from a monocarpellary ovary. Here the pericarp is fused with the seed coat. Therefore, the maize grain is a fruit known as a caryopsis.

22. Answer: (b) groundnut

Explanation:Geocarpic fruit is groundnut.Geocarpy refers to an extremely rare means of plant reproduction. Here the plants produce via diaspores within the soil. This has evolved as an effective means of ensuring a suitable environment for the plant's offspring.

23.Answer: (a) Dianthus

Explanation:Free central : It is found in bicarpellary to multicarpellary syncarpous ovary. Due to degradation of false septum unilocular condition is formed and ovules are arranged on the central axis. E.g.: Dianthus, Primula (primroses).

24. Answer: (d) ligule

Explanation: In early heterosporous plants, magaspores were released form the parent. But in seed plants, these are retained and fertilised to become seed. This habit is seen in Selaginella which bears a small multicellular scale-like structure called ligule at the base of leaf on adaxial side.

25.Answer: (d) Endosperm

Explanation: The morphological nature of the edible part of coconut is. The edible part of coconut is endosperrn. Coconut water is free nuclear endospenn and white kernel is the cellular endosperm.

Click here to practice Morphology Of Flowering Plants MCQ Questions for Class 11 

Morphology is the branch of science or biology that studies the external features of an organism. 

Morphology provides distinguishing features of stem type, leaf type, inflorescence, flower structure, fruit type etc., in classification.

Any alteration in the structure or function of an organism or any of its part that results from natural selection and by which the organism becomes better fitted to survive and multiply in its environment.

The root is underground part of the plant and develops from elongation of radicle of the embryo.

Various types of root :

(i) Tap root : Originates from radicle. Dicotyledonous plants,

e.g., gram, pea, mango. 

(ii)  Fibrous root : Originates from base of the stem Monocotyledonous plants, e.g., wheat, paddy, grasses.

(iii) Adventitious root : Originates from parts of the plant other than radicle. Banyan tree (Prop roots) Maize (Still roots) Rhizophora (Respiratory roots)

Important functions of root are: 

(i) Fixation : The root fixes the plant to the soil i.e, ground. 

(ii) Absorption : Root absorb nutrients and water from the soil. 

(iii) Conduction : It conducts absorbed materials from the soil to the aerial parts of the plant.

(iv) Special Functions : In addition to the above functions, some adventitious roots perform different function i.e., in Cuscuta (a parasitic plant) they absorb food from the host's body, in banyan, the prop roots provide support to the plant, in maize, Rhizophora, they support the plant; in Tinospora the green roots perform the function of photosynthesis; in some plants they get swollen and perform as storage organs for the plant; other perform the function of vegetative reproduction Some roots perform the functions of storage of food reproduction, climbing, giving the support to plant.

In some aquatic insectivorous plants like Utricularia, some of the leaf segments of much dissected leaves are modified to form small bladders. 

Function : These bladders have sensitive hair, bristles, trap valve and glands for trapping and digesting small insects.

The primary functions of a leaf are : 

(i) Synthesis of organic food through photosynthesis. 

(ii) Leaves possess minute pores called stomata. The stomata help in the exchange of gases necessary for photosynthesis and respiration. 

(iii) Leaves are the main seat of loss of water called force for the ascent of sap and keeps the temperature low in summer.

The fruit which develops only from ovary is called as true fruit or eucarp. e.g. Mango. The fruit which develops from other than ovary floral part is called as false fruit or pseudocarp

Differences between simple and compound leaf:

The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.

Iodine is the active ingredient in the produc tion of thyroxin hormone. If there is insufficient quantity of iodine in food, the thyroxin hormone is not produced. As a result, thyroid gland increases in size which is Visible as a swelling in the neck. This condition is known as goitre.

Leaves show different types of modification as follows: 

1. Leaf spines: 

Sometimes entire leaf is modified into spines (Opuntia) or margin of leaf becomes spiny (Agave) or stipule modifies into spine (Acacia, Zizyphus) to check the rate of transpiration and to protect plant from grazing. 

2. Leaf tendril: 

In some weak stems, leaf, leaflet or other part modifies to produce thin, green, wiry, coiled structure called as leaf tendril. It helps in climbing and provides additional support. 

3. Leaf hooks: 

In plants like Bignonia unguis-cati (Cat’s nail) the terminal three leaflet get modified into three! stiff curve and pointed hooks used to cling over the bark of tree. 

4. Phyllode: 

When petiole of leaf becomes flat, green and leaf like it is called as phyllode. In Acacia auriculoformis the normal leaf is bipinnately compound and falls off soon. The petiole modifies itself into phyllode. It is a xerophytic adaptation.

The thymus gland is located in the mediastinum in front of the heart. This secretes a hormone thymosin which stimulates the lymphocytes to destroy invading microorganisms and antigens. It degenerates on ageing, decreasing the immunity in old age.

1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert. 

2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration. 

3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

1. Figure ‘a’ shows fruit of tomato. 

• It is a simple fruit as it develops from a single flower with bicarpellary syncarpous gynoecium. 
• It is a berry, because it has fleshy endocarp and many seeds. 

2. Figure ‘b’ shows fruit of Custard apple. 

• It is an aggregate fruit, because it develops from a single flower with polycarpellary, apocarpous gynoecium. 
• Here, the ovary of each carpel gives rise to a part of the fruit called fruitlet. Hence, it is called an aggregation of fruitlets. 
• Custard apple can be further described as Etaerio of berries.

3. Figure ‘c’ shows fruit of pineapple. 

• It is a composite fruit, because it develops from a complete inflorescence. 
• Pineapple can be further described as Sorosis, as it develops from catkin type of inflorescence.

4. Figure ‘d’ shows fruit of milkweed. 

• It is a simple dehiscent dry fruit. 
• It has many seeds. When pericarp becomes dry and thin, it breaks open by one ventral suture

The endocrine gland present in the human body are:

(i) Pineal gland

(ii) Hypothalamus

(iii) Pituitary

(iv) Thyroid

(v) Parathyroid

(vi) Thymus

(vii) Pancreas

(viii) Adrenal glands

(ix) Testes (in males)

(x) Ovaries (in females). Pancreas, testes and ovaries function as exocrine glands.

Iodine is essential for the production of thyroxin by the thyroid gland. The people living in hilly regions receive insufficient iodine in their diet, due to the fact that they drink iodine deficient water from rivers and streams fed by melting snow. Even their diet has little or no iodine. When there is a deficiency of iodine in the diet the thyroid gland enlarges in order to compensate for this deficiency resulting in goitre.

Functions of pituitary gland:

(i) Regulates growth.

(ii) Influences the thyroid gland, the adrenals and gonads.

(iii) Can produce changes in the skin colour of many amphibians.

(iv) Influence migration and nest buildings in birds.

(v) Controls and stimulates the secretion of other endocrine glands.

Exophthalmic goitre or Grave’s disease is caused by overactivity of the thyroid gland. The thyroid increases in size and leads to an increased metabolic rate, a high rate of heart beat and wasting away of the tissues of the body.

The apparent symptoms are:

(a) Goitre (Swelling in the neck). 

(b) Protuberance of the eyeballs.

(c) The patient feels tired, nervous and restless.

Osteoporosis is a disease caused due to excessive secretion of parathormone. In this, the bones become weak, elastic and bent.

The correct answer is (A) cytoplasm

The correct answer is : (d) 1

The correct answer is : (c) -5

Give the function f: x → x² – 5x + 6. 

(i) f(-1) = (-1)² – 5(1) + 6 = 1 + 5 + 6 = 12 

(ii) f(2a) = (2a)² – 5(2a) + 6 = 4a² – 10a + 6 

(iii) f(2) = 2² – 5(2) + 6 = 4 – 10 + 6 = 0 

(iv) f(x – 1) = (x – 1)² – 5(x – 1) + 6 

= x² – 2x + 1 – 5x + 5 + 6 

= x² – 7x + 12

Correct option is (A) 10³

\(1\,cm^3=10^{-6}m^3\)   \((\because\) 1 cm \(=\frac1{100}m\) \(\Rightarrow 1cm^3=\frac1{(10^2)^3}m^3=10^{-6}m^3)\)

\(=10^{-6}\times(10^3)^3mm^3\)   \((\because\) 1 m = 1000 mm \(=10^3mm)\)

\(=10^{9-6}mm^3\) \(=10^3mm^3\)

Correct option is  A) 10³

We have gof(x) = g((3x + 4)/(5x - 7)) = (7((3x + 4)/(5x - 7)) + 4)/(5((7x + 4)/(5x - 7)) - 3)

= (21x + 28 + 20x - 28)/(15x + 20 - 15x + 21) = 41x/41 = x

Similarly, fog(x) = f((7x + 4)/(5x - 3)) = (3(7x + 4)/(5x - 3)) + 4)/(5(7x + 4)/(5x - 3)) - 7)

= (21x + 12 + 20x - 12)/(35x + 20 - 35x + 21) = 41x/41 = x

Thus, gof (x) = x, ∀ x ∈ B and fog (x) = x, ∀ x ∈ A, which implies that gof = I_B and fog = I_A.

f is not one-one or injective as f (1) = f (-1) ⇒ 1 = -1 

f is not on to or surjective as f(x) = 2 ⇒ x = √2∉ Z 

Let ‘a’ be edge if a cube, so edge = 1m or 100 cm

Volume of cube of 100 cm edge = (side)³

= (100)³

= 1000000 cm³

Volume of cubes of 10cm edge = 10³ = 1000 cm³

Now,

Number of required cubes = (Volume of cube of 100 cm edge)/(Volume of cubes of 10cm edge)

= 1000000/1000000

= 1000

Answer!!

f is not injective as f(x₁) = f(x₂) ⇒ x₁= ± x₂ 

Range off does not contain any negative real. 

Hence not surjective.

Correct option is (C) 1

\(1\,m^3\) = 1000 liters

= 1 kilo liters

Correct option is  C) 1

Answer is (D)

We have [x]² – 5[x] + 6 = 0 => [(x – 3)([x] – 2) = 0

=> [x] = 2 , 3 .

For [x] = 2, x ∈ [2, 3)

For [x] = 3, x ∈ [3,4)

x ∈ [2, 3) u [3,4)

Or x ∈ [2,4)

Let x₁, x₂ ∈ N

f(x₁) = f(x₂) ⇒ x³₁ = x³₂

hence one-one

Range of f= {1,8,27,…… } ≠ N

f is not surjective.

Given: [x]² – 5 [x] + 6 = 0, where [ . ] denote the greatest integer function

To find: range of x

Explanation: we have

[x]² – 5 [x] + 6 = 0

We will split the middle term, we get

⇒ [x]² – 3 [x] -2[x] + 6 = 0

⇒ [x]([x]–3)-2([x]-3) = 0

⇒ ([x]-3)([x]-2) = 0

⇒ [x]-3 = 0 or [x]-2 = 0

⇒ [x] = 3 or [x] = 2

⇒ [x] = 2, 3

For [x] = 2, x ∈ [2,3)

For [x] = 3, x ∈ [3,4)

[x] ∈ [2,3) ∪ [3,4)

So, x ∈ [2,4]

Hence the correct answer is option (C).

Volume of a cube is 729 cm³ (given)

we know, Volume of the cube = (side)³

(side)³ = 729

or Side = 9

Each side measure of a cube is 9 cm.

Now,

Total surface area of the cube = 6(side)²

= 6 × 9²

= 6 × 81

= 486

Total surface area of the cube is 486 cm².

Let x₁, x₂ ∈ Z 

f is one-one or injective 

Range of f = {0, ±1, ±8, ±27,…………….. } ≠Z 

f is not surjective