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NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development

NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development

Edited By Irshad Anwar | Updated on Jul 15, 2025 11:40 AM IST

The NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development explain how plants grow, change, and respond to their surroundings. It also includes the different stages of plant growth. It also explains how both internal and external factors, like temperature, light, and nutrients, affect development. With the help of these answers, students will learn about special chemicals called plant hormones, such as auxin, gibberellins, etc. Students can also download the Plant Growth and Development new NCERT PDF for easy and quick access.

NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah Aakash | ALLEN

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This Story also Contains
  1. Download PDF of NCERT Solutions for Class 11 Biology Chapter 13
  2. Class 11 Biology NCERT Chapter 13 (Exercise Questions)
  3. Approach to Solve Questions of Class 11 Biology Chapter 13 Question Answer
  4. Important Question from Class 11 Biology NCERT Chapter 13
  5. What Extra Should Students Study Beyond the NCERT for NEET?
  6. NCERT Solutions for Class 11 Biology: Chapter-wise
NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development
NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development

The solutions allow students to understand difficult topics like seed dormancy, vernalisation, and photoperiodism. The Class 11 Biology chapter 13 solutions are prepared by subject experts in simple and easy language, along with well-labeled diagrams wherever required. This chapter is very important for students preparing for school exams as well as for competitive exams like NEET. Referring to the NCERT solutions helps students to build a strong foundation in the subject of Biology, making it easier for students to perform well in the exams.

Also, Read

Download PDF of NCERT Solutions for Class 11 Biology Chapter 13

Students can download the PDF of the Plant Growth and Development chapter from the link below. This PDF can be used offline, making it easy to revise anytime and anywhere. The NCERT Class 11 Solutions help students understand difficult concepts quickly without any confusion.

Download PDF

Class 11 Biology NCERT Chapter 13 (Exercise Questions)

The solved exercise questions of the chapter Plant Growth and Development are given below. Each question is answered in a clear, step-by-step manner to help students grasp key concepts easily. Students can use these Plant Growth and Development NCERT Solutions for strong conceptual clarity.

Question 1. Define growth, differentiation, development, dedifferentiation, development, redifferentiation, determinate growth, meristem and growth rate.

Solution:

Growth:

It is a permanent, irreversible increase in the size of an organ, one of its components, or even a single cell. Metabolic processes that are taking place as a result of the energy support growth.

Differentiation:

A localized qualitative change in the size, biochemistry, structure, and function of cells, tissues, or organs, such as mesophyll, a leaf, a fibre, a vessel, a trachea, or a sieve tube, is known as differentiation. As a result, both the form and the physiological activity change. It leads to specialization in certain functions.

Development:

Development refers to all the changes an organism experiences over the course of its existence.

Dedifferentiation:

Plants that have lost the ability to divide can regain it under specific circumstances. Dedifferentiation is the term used to describe this phenomenon. Example: meristem formation.

Redifferentiation:

It is referred to as redifferentiation when cells produced by dedifferentiation mature to carry out particular functions but once more lose the ability to divide.

Determinate growth:

Determinate growth is the capacity of a cell, tissue, or organism to grow for a specific amount of time. The majority of plants grow indefinitely; however, some plants reach a certain size before ceasing to expand.

Meristem:

The term "meristem" refers to plant tissue made up of undifferentiated cells (meristematic cells).

Growth rate:

A growth rate is an increase in growth per unit of time.

Question 2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?

Solution:

An increase in protoplasm production leads to growth. The parameters used to measure protoplasmic growth include changes in height, weight, number of cells, fresh tissue sample, length, area, volume, etc. As a result, it is challenging to identify a single growth parameter that characterizes a flowering plant's development over its lifetime.

Question 3. Describe briefly.

(a) Arithmetic growth

(b) Geometric growth

(c) Sigmoid growth curve

(d) Absolute and relative growth rates

Solution:

(a) Arithmetic growth

Only one daughter cell divides during arithmetic growth, while the other differentiates and matures. A root extending at a constant rate is the most basic illustration of arithmetic growth.

 Arithmetic growth

It can be mathematically expressed as:

Lt = Lo + rt

Lt = length of time ‘t’

Lo = length at the time ‘zero’

r = growth rate/elongation per unit of time

(b) Geometric growth

The initial phase of geometric growth is slow (lag phase), and the subsequent exponential or logarithmic phase sees a rapid increase. Here, both of the progeny cells that result from mitotic cell division are still able to divide. Due to the insufficient nutrient supply, the growth slows down and enters a stationary phase. In geometric growth, the number grows in a multiplicative pattern.

Geometric growth

If we plot the growth parameter against time for geometric growth, we obtain a typical sigmoid or S-curve.

Exponential growth can be expressed as:

W1 = Wo ert

W1 = final size

Wo = initial size of the period;

r = growth rate

t = time growth

e = base of natural logarithms

'r' stands for relative growth and the efficiency index, which measures how well plants can produce new plant materials. Wo's initial size determines the final size of W1, thus.

(c) Sigmoid growth curve

Plotting growth against time results in an S-shaped graph that has four main parts: a slow lag phase, exponential phase or rapid phase, stage of diminishing growth, and stationary phase.

Sigmoid growth curve

(d) Absolute and relative growth rates

The net growth per unit of time is known as the absolute growth rate. The growth rate per unit of time per unit of initial growth is known as the relative growth rate.

Question 4. List five main groups of natural plant growth regulators. Write a note on the discovery, physiological functions and agricultural/horticultural applications of any one of them.

Solution:

Plant growth regulators are the intercellular intrinsic factors (chemical substances) that are responsible for the growth and development of plants.

The following are the top five categories of natural plant growth regulators (PGR):

  • Auxins

  • Gibberellins

  • Cytokinins

  • Abscisic acid

  • Ethylene

These PGRs are synthesized in various plant parts and regulate various developmental and differentiation processes that occur throughout a plant's life cycle.

Gibberellins

Discovery:

  • One of the first gibberellins to be found, designated as GA1, GA2, and GA3, is gibberellic acid, which is acidic.

  • More than 100 gibberellins have been identified in a variety of different sources, including fungi and other higher plants.

Physiological functions:

  • Plants respond physiologically to gibberellins in a variety of ways.

  • They have the ability to increase the length of the axis, which can result in an increase in the length of grape stalks.

  • Fruits like apples elongate and acquire a better shape as a result of their presence.

  • They are in charge of putting off the senescence process.

Agricultural/horticultural applications:

  • The fruits can be left on the tree for longer to increase the market period because the senescence process is delayed.

  • Gibberellic acid, or GA3, is a substance that is used to accelerate the malting procedure in the brewing industry.

  • By spraying the sugarcane crop with gibberellins, which lengthen the stem and increase yield by up to 20 tonnes per acre because sugarcane stems store carbohydrates as sugar, the crop is able to store more carbohydrates as sugar.

  • Spraying GAs on young conifers can shorten their maturation period and promote early seed production.

  • Additionally, it encourages the bolting process in plants like cabbage and beets. The elongation of the internode that occurs just before flowering is known as bolting.

Question 5. Why is abscisic acid also known as stress hormone?

Solution:

Abscisic acid is responsible for stimulating the closure of stomata in the epidermis and raising the tolerance of plants to different types of stresses. This is why it is also known as the stress hormone. In order to ensure that seeds germinate under favourable conditions, abscisic acid is in charge of promoting seed dormancy. This makes it easier for seeds to withstand desiccation and induces dormancy in plants near the end of the growing season, which encourages the abscission of fruits, leaves, and flowers.

Question 6. ‘Both growth and differentiation in higher plants are open.’ Comment.

Solution:

Due to the presence of meristems at specific locations of their bodies, higher plants have the ability to retain the capacity to have indefinite growth throughout their life span. Because of these meristems, the cells have the ability to divide and grow on their own. This explains why the growth in higher plants is open. Following several rounds of cell division, some of these cells go through differentiation. Therefore, differentiation is also open.

Question 7. ‘Both a short-day plant and a long-day plant can produce flowers simultaneously in a given place.’ Explain.

Solution:

In a few plants, flowering depends on the relative durations of light and dark periods. Under the condition that they receive enough photoperiod, both long-day and short-day plants can bloom in the same location.

Question 8. Which one of the plant growth regulators would you use if you were 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.

Solution:

The plant growth regulators for the related events are listed below:

(a) Induce rooting in a twig. – Auxins

(b) Quickly ripen a fruit. – Ethylene

(c) Delay leaf senescence. – Cytokinins

(d) Induce growth in axillary buds. – Cytokinins

(e) ‘Bolt’ a rosette plant. – Gibberellins

(f) Induce immediate stomatal closure in leaves. – Abscisic acid

Question 9. Would a defoliated plant respond to the photoperiodic cycle? Why?

Solution:

No, a defoliated plant will not respond to the photoperiodic cycle. This is due to the fact that the leaves serve as the sites where dark or light duration is perceived. Therefore, plants would not respond to light if leaves were not present.

Question 10. 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.

Solution:

(a) If GA3 is applied to rice seedlings:

The rice seedlings will show internode-elongation, and hence, an increase in height will be observed.

(b) If dividing cells stop differentiating:

The various plant parts, including the stem and leaves, will not form if the dividing cells stop differentiating.

(c) If rotten fruit gets mixed with unripe fruit:

When unripe fruits are combined with rotten fruits, the unripe fruits ripen more quickly due to the ethylene produced by the rotten fruits, which is a plant growth regulator.

(d) If you forget to add cytokinin to the culture medium:

The processes of cell division, differentiation, and growth will be muted and slowed down if cytokinin is not added to the culture medium.

NCERT Solutions for Class 11: Subject-wise

Approach to Solve Questions of Class 11 Biology Chapter 13 Question Answer

To solve Plant Growth and Development questions efficiently, students can follow the steps given below:

  1. Understand key concepts like growth phases and plant hormones.

  2. Underline words like development, growth, and meristem.

  3. Revise the functions of auxins, gibberellins, cytokinins, ethylene, and ABA given in the Plant Growth and Development Class 11 Solutions.

  4. Outline the seed germination and growth stages of a plant, time and again.

  5. Refer to the NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development.

  6. Underline key processes like vernalization and photoperiodism.

  7. Practice the previous year's question papers and refer to the NCERT Solutions Class 11 Biology for revision before the exam.

Check the NCERT Books and NCERT Syllabus here:

Important Question from Class 11 Biology NCERT Chapter 13

Given below is a key question that will improve the students' understanding of the topic.

Question: Growth can be measured in various ways. Which of these can be used as parameters to measure growth?

Options:

1. Increase in cell number

2. Increase in cell size

3. Increase in length and weight

4. All the above

Answer: The correct answer is option 4) All the above

Explanation: Plants can continue to develop for the duration of their lives. This is because specific parts of their bodies contain meristems, which have the capacity to divide and reproduce themselves. The open form of growth refers to the process by which the meristem's activity continuously adds new cells to the plant body. The growth can be in lateral or apex regions and can cause an enlargement in height or weight; all of these changes are collectively termed growth.

What Extra Should Students Study Beyond the NCERT for NEET?

Here is a table for all the important topics from the chapter. These topics are highly useful for solving advanced questions in NEET. Students can also refer to Plant Growth and Development Class 11 NCERT Solutions for a better understanding of the basics.

NCERT Solutions for Class 11 Biology: Chapter-wise

NCERT solutions for all chapters are given below:-


Frequently Asked Questions (FAQs)

1. What are the phases of growth in plants as per Chapter 13 NCERT?

According to the NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development, plant growth occurs in three main phases: meristematic, elongation, and maturation. 

2. How do plant hormones regulate growth and development?

Plant hormones, also known as phytohormones, are chemical messengers that regulate plant growth and development by influencing processes like cell division, elongation, and differentiation, as well as responses to environmental cues.

3. How do auxins, gibberellins, and cytokinins influence plant growth?

Auxins, gibberellins, and cytokinins are plant hormones that play important roles in growth and development. Auxins promote cell elongation and root formation, gibberellins help in stem elongation and seed germination, while cytokinins promote cell division and shoot formation. As explained in the NCERT Solutions for Class 11 Biology Chapter 13 Plant Growth and Development, these hormones work together to regulate various stages of plant life.

4. Why is secondary growth important in plants?

Secondary growth is important for plants, particularly woody ones, as it increases girth, providing structural support and enabling the transport of water and nutrients to support a larger number of leaves. This concept is well covered in the Plant Growth and Development questions and answers PDF download resources.

5. What is the role of abscisic acid in plant growth regulation?

Abscisic acid is a plant hormone. It is also called a stress hormone because it shows different responses to stress conditions. It leads to seed dormancy and ensures seed germination only when favourable conditions return. It also causes the closing of stomata when plants don’t get enough water.

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Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

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be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

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Molality

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Weight fraction of solute

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Fraction of solute present in water

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Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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twice that in 60 g carbon

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6.023 × 1022

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half that in 8 g He

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558.5 × 6.023 × 1023

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less than 3

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more than 3 but less than 6

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more than 6 but less than 9

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more than 9

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