---
title: "NCERT Solutions Class 9 Science Reproduction: How Life Continues"
url: https://www.swavid.com/science/class/9/chapter/reproduction-how-life-continues/ncert-solutions
dateModified: 2026-10-07T16:34:58+00:00
---

# NCERT Solutions Class 9 Science Reproduction: How Life Continues

This chapter contains various questions related to reproduction in plants and animals, including asexual and sexual methods, pollination, fertilisation, and human reproductive health. The exercises cover conceptual understanding, data analysis, and critical thinking based on the biological processes discussed.

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## Think It Over

### Question y

*3 marks · Short answer*

When does a farmer prefer asexual or sexual methods of reproduction for crops production?

**Solution**

1. A farmer prefers asexual methods of reproduction (such as vegetative propagation using cuttings, grafting, or layering) when they want to produce a large number of genetically identical crops quickly, preserve desirable characteristics, and ensure uniform crop quality.
2. A farmer prefers sexual methods of reproduction when they want to introduce genetic variations, develop new varieties with desired traits through plant breeding, or when crops naturally rely on seeds for wide-scale dispersal and growth.

**Answer:** Farmers prefer asexual methods for fast, uniform production of identical desirable crops, and sexual methods for creating new varieties through genetic variation.

> Common mistake: Stating that farmers only use one method for all crops without distinguishing between uniformity and variation.

### Question y

*3 marks · Short answer*

Why do you think most complex animals and flowering plants use sexual reproduction, while many simple organisms, like yeast and hydra mainly reproduce asexually?

**Solution**

1. Complex animals and flowering plants have intricate body designs and multicellular structures, so they use sexual reproduction to create genetic variations through the mixing of characteristics during gamete formation and fertilisation.
2. These variations help the offspring adapt better to changing environments and contribute to evolution over generations.
3. Simple organisms like yeast and hydra mainly reproduce asexually because it is a faster method involving a single parent, allowing them to rapidly increase their population when environmental conditions are favourable.

**Answer:** Complex organisms use sexual reproduction for generating adaptive variations, whereas simple organisms use asexual reproduction for rapid population growth from a single parent.

> Common mistake: Confusing the speed of asexual reproduction with the evolutionary advantage of variation provided by sexual reproduction.

## Activity 11.1: Let us explore

### Question 1

*Activity*

Interact with gardeners working in your school garden or farmers working in a field.

**Solution**

1. This is an activity-based question where students interact with gardeners or farmers to understand practical agricultural and horticultural methods.
2. Expected observation: Gardeners and farmers use natural and artificial vegetative propagation methods like cutting, grafting, and layering to cultivate desirable crops efficiently on a large scale.

**Answer:** Students should record their practical observations of gardener interactions in their notebooks.

### Question 2

*Activity*

Observe the techniques of cutting, grafting and layering followed by them. Discuss these techniques with them and record your observations in your notebook.

**Solution**

1. Observe the steps of cutting, grafting, and layering demonstrated or practiced by professionals in the field.
2. Expected observation: Cuttings are prepared from healthy shoots, grafted onto rooted plants to combine desirable features, and layered by burying flexible twigs to develop new roots.

**Answer:** Record the techniques of cutting, grafting, and layering with proper steps in the notebook.

### Question 1(i)

*Activity*

Does the gardener, scientist or horticulturist cut the overgrown branches of a plant at the end of its growing season? (Different plants have different growing seasons).

**Solution**

1. Observe whether gardeners or horticulturists prune overgrown branches at the end of the growing season.
2. Expected observation: Yes, overgrown or old branches are often trimmed at the end of the growing season to prepare healthy stem cuttings for growing new plants.

**Answer:** Yes, overgrown branches are cut at the end of the growing season.

### Question 1(ii)

*Activity*

Observe them prepare the cuttings from a plant for the purpose of growing new plants. Note the average length of the cuttings.

**Solution**

1. Observe the preparation of plant cuttings used for growing new plants.
2. Expected observation: The average length of a stem cutting typically ranges around 15 to 20 cm, containing active growing regions.

**Answer:** Note the average length of the cuttings as observed in the field.

### Question 1(iii)

*Activity*

Count the number of nodes and internodes on the cuttings.

**Solution**

1. Examine the prepared stem cuttings and count the structural divisions.
2. Expected observation: Cuttings generally possess multiple nodes (from where new shoots or roots sprout) and internodes.

**Answer:** Count and record the number of nodes and internodes present on the cuttings.

### Question 2

*Activity*

Collect the cuttings of the shoots in the morning for planting.

**Solution**

1. Collect shoot cuttings in the morning when the plant tissues are fully turgid with water.
2. Expected observation: Morning collection ensures high moisture content in the shoots, improving the success rate of planting and rooting.

**Answer:** Collect shoot cuttings in the morning for optimal planting results.

### Question 3

*Activity*

Remove leaves from the lower half of each cutting.

**Solution**

1. Removing leaves from the lower half of the cutting reduces water loss through transpiration.
2. It also prevents fungal infection or rotting when that part of the cutting is inserted into the soil.

**Answer:** Leaves are removed from the lower half to reduce water loss and prevent rotting.

### Question 4

*Activity*

Insert the cuttings up to approximately half of their length in the soil mixed with compost at an angle of about 45–60° from the soil surface (Fig. 11.2).

**Solution**

1. Inserting the cutting at an angle of about 45-60 degrees ensures that sufficient nodes remain in contact with the soil.
2. This facilitates proper absorption of water and nutrients from the soil mixed with compost to sprout new roots.

**Answer:** Cuttings are inserted at an angle to provide maximum contact with the soil for root development.

### Question 5

*Activity*

Water them regularly and observe the change, if any.

**Solution**

1. Regular watering keeps the soil moist, which is essential for the activation of cells and growth of new roots and shoots.
2. Observing the changes helps track the successful development of the cutting into a new independent plant.

**Answer:** Regular watering maintains soil moisture required for root and shoot growth.

## Grafting

### Question 1

*Activity*

For grafting, take a healthy rooted plant (Plant A) (Fig. 11.3a) (for example, a wild rose variety) and a healthy stem piece from another plant (Plant B) of other varieties (for example, a yellow rose plant and/or a pink rose plant) (Fig. 11.3c and Fig. 11.3d).

**Solution**

1. Select a healthy rooted plant (Plant A) to serve as the stock and a healthy stem piece from another variety (Plant B) to serve as the scion.
2. Observation: This step sets up the two plant parts of different varieties that will be joined together in grafting.

**Answer:** Preparation of stock (Plant A) and scion (Plant B) for grafting as shown in the figure in the textbook (Fig. 11.3a, c, d).

### Question 2

*Activity*

Create a wound or a slit on a twig of Plant A (Fig. 11.3b).

**Solution**

1. Create a wound or a slanting cut (slit) on the twig of the rooted Plant A.
2. Observation: This cut exposes the inner tissues of the stock plant to receive the scion.

**Answer:** A slit or wound is created on the twig of Plant A as shown in the figure in the textbook (Fig. 11.3b).

### Question 3

*Activity*

Insert and fit the cutting of stem of Plant B into the slit of stem of Plant A (Fig. 11.3e).

**Solution**

1. Take the stem cutting of Plant B and insert and fit it firmly into the slit made in the stem of Plant A.
2. Observation: Proper contact between the tissues of Plant A and Plant B allows them to grow together.

**Answer:** The stem cutting of Plant B is fitted into the slit of Plant A as shown in the figure in the textbook (Fig. 11.3e).

### Question 4

*Activity*

Protect the wound or slit by using a cotton cloth or by wrapping film to avoid pests entering the graft until it heals (Fig. 11.3f). Cut the other branches of Plant A.

**Solution**

1. Wrap the wound or slit using a cotton cloth or wrapping film to prevent pests from entering until the graft heals.
2. Cut off the other branches of Plant A so that the plant's energy is directed towards the graft.

**Answer:** The graft is protected with a wrapping film or cloth and other branches of Plant A are cut as shown in the figure in the textbook (Fig. 11.3f).

### Question 5

*Activity*

Water the plant regularly and observe the growth of Plant B along with Plant A.

**Solution**

1. Water the grafted plant regularly.
2. Observation: Observe the growth and development of Plant B along with Plant A as a single unified plant.

**Answer:** Regular watering leads to the successful growth of Plant B along with Plant A.

## Layering

### Question 1

*Activity*

For layering, select a flexible, thin twig of a tree or a shrub, such as a lemon and bury the middle part of the twig under the soil surface (Fig. 11.4).

**Solution**

1. This describes the first step of the layering technique in artificial vegetative propagation.
2. It shows that a flexible, thin twig of a plant like lemon is selected and its middle part is buried under the soil surface as illustrated in Fig. 11.4.

**Answer:** Selected twig is buried under the soil for layering.

### Question 2

*Activity*

Water it regularly and observe the growth of new leaves on the twig buried in the soil.

**Solution**

1. This step involves regular watering of the buried portion of the twig.
2. Regular watering maintains moisture and supports the growth of new leaves on the twig part buried in the soil.

**Answer:** Regular watering supports the growth of new leaves on the buried twig.

### Question 3

*Activity*

After 10–15 days, the roots will develop from the area of the twig buried in the soil.

**Solution**

1. This observation notes the developmental milestone in the layering process.
2. After 10 to 15 days, adventitious roots develop from the specific area of the twig that is buried under the soil.

**Answer:** Roots develop from the buried area of the twig after 10–15 days.

### Question 4

*Activity*

Once roots have developed, cut the twig from the parent plant, so that it can grow as a new plant.

**Solution**

1. This is the final step of vegetative propagation by layering.
2. Once roots are fully developed, the twig is cut away from the parent plant so that it can independently grow and develop as a new plant.

**Answer:** The rooted twig is cut from the parent plant to grow independently.

## Activity 11.2: Let us explore

### Question 1

*Activity*

Take 20 mL of sugar solution (1 g in 10 mL) in a test tube.

**Solution**

1. Take 20 mL of sugar solution (prepared by dissolving 1 g of sugar in 10 mL of water) in a test tube.

**Answer:** Sugar solution taken in a test tube.

### Question 2

*Activity*

Add a pinch of yeast granules to it and then place a cotton plug on the mouth of the test tube.

**Solution**

1. Add a pinch of yeast granules to the sugar solution and place a cotton plug on the mouth of the test tube.

**Answer:** Yeast added and test tube plugged with cotton.

### Question 3

*Activity*

Keep it undisturbed in a warm place to allow the yeast to become active.

**Solution**

1. Keep the test tube undisturbed in a warm place to allow the yeast to become active and multiply.

**Answer:** Test tube kept in a warm place.

### Question 4

*Activity*

After 1–2 hours, place a small drop of the yeast mixture from the test tube onto a glass slide and mount it with a coverslip.

**Solution**

1. After 1–2 hours, place a small drop of the yeast mixture from the test tube onto a glass slide and mount it with a coverslip.

**Answer:** Glass slide prepared with a drop of yeast mixture.

### Question 5

*Activity*

Observe the slide under a compound microscope at different magnifications and draw a diagram of what you observe.

**Solution**

1. Observe the slide under a compound microscope at different magnifications and draw a diagram of the observed yeast cells.

**Answer:** Slide observed under microscope and diagram drawn.

### Question Q

*3 marks · Short answer*

Do you observe any small, round outgrowths (buds) emerging from the parent yeast cells as shown in Fig. 11.6? Do these features indicate that the yeast is duplicating? How do these observations help you in understanding reproduction in yeast?

**Solution**

1. Small, round outgrowths known as buds are observed emerging from parent yeast cells.
2. These features indicate that the yeast is actively duplicating by asexual reproduction called budding.
3. This observation helps us understand that a single parent yeast cell can produce new individuals by forming small outgrowths that eventually separate.

**Answer:** Yeast reproduces asexually through budding, where small outgrowths emerge from parent cells and indicate duplication.

> Common mistake: Confusing budding in yeast with spore formation or sexual reproduction.

## Activity 11.3: Let us experiment

### Question 1

*Activity*

Take a small slice of bread or a roti and lightly moisten it with a few drops of water.

**Solution**

1. Moistening the bread slice or roti provides the necessary moisture required for fungal spores present in the air to germinate and grow.

**Answer:** Moistening the bread or roti provides moisture for mould growth.

### Question 2

*Activity*

Prepare a moist chamber using a plastic box or steel dabba. Place a thin layer of cotton in it, cover it with tissue paper and moisten it with pre-boiled water. Put the slice of bread or roti on the wet cotton bed covered with tissue paper.

**Solution**

1. Preparing a moist chamber using a plastic box or steel dabba with a wet cotton and tissue paper bed maintains high humidity, creating a suitable microclimate for the spores to thrive.

**Answer:** The moist chamber maintains humidity for spore germination.

### Question 3

*Activity*

Keep the moist chamber in a warm and dark place, away from direct sunlight (if the bread or roti starts drying, add a few drops of water to keep it moist).

**Solution**

1. Keeping the chamber in a warm (25–35 degrees Celsius) and dark place provides the optimum temperature and conditions required for rapid reproduction and growth of fungal spores.

**Answer:** Warm and dark conditions accelerate spore growth.

### Question 4

*Activity*

Observe the bread or roti every day for any changes. Record your observations without touching it directly.

**Solution**

1. Observing daily without direct touching helps record changes safely, as mould spores or mature colonies should not be inhaled or touched directly to avoid hygiene issues.

**Answer:** Daily observation tracks the progressive growth of mould safely.

### Question 5

*Activity*

After three days, observe the surface of the bread or roti carefully using a magnifying glass. Do you notice the growth of mould?

**Solution**

1. After three days, fluffy, fuzzy patches of fungal growth called mould become visible on the surface of the bread or roti when viewed through a magnifying glass.

**Answer:** Yes, fuzzy patches of mould are observed on the bread surface.

### Question 6

*Activity*

When enough mould grows and spreads on it, carefully take the box to the school laboratory.

**Solution**

1. Taking the container to the laboratory allows detailed microscopic examination of the mould hyphae, sporangia, and spores using stains like cotton blue under the guidance of a teacher.

**Answer:** The sample is taken to the laboratory for microscopic study.

### Question 7

*Activity*

With the help of a needle, carefully transfer a little mould onto a microscope slide. Under the guidance of your teacher, add cotton blue stain—a coloured dye to help see it better.

**Solution**

1. Transfer a small amount of mould from the bread slice onto a microscope slide using a needle.
2. Add a drop of cotton blue stain to the mould to highlight its structures clearly under the microscope.

**Answer:** Mould transferred to the slide and stained with cotton blue dye for microscopic observation.

### Question 8

*Activity*

Observe the mould under the microscope and draw its diagram based on your observations.

**Solution**

1. Observe the stained slide of mould under the compound microscope.
2. Identify thread-like structures called hyphae and round sac-like structures containing spores.
3. Draw a clear, labeled diagram based on the observed structures.

**Answer:** Observed thread-like hyphae and spore-containing sacs, and drew the corresponding diagram.

### Question 9

*Activity*

Compare the diagram you have drawn with Fig. 11.8 and share your observations with your classmates.

**Solution**

1. Compare the drawn diagram of the observed mould with Fig. 11.8 showing Rhizopus and Aspergillus.
2. Note the presence of hyphae, sporangia (sacs), and spores in both.
3. Discuss and share the observations with classmates.

**Answer:** Compared the drawn diagram with Fig. 11.8, confirming the presence of hyphae and spore-containing sacs similar to Rhizopus.

## Activity 11.4: Let us explore

### Question 1

*Activity*

Take three pairs of beads of different colours (Fig. 11.9), each pair representing two contrasting characters on different chromatids of different chromosomes, such as: Pair 1 (green): One light green bead and one dark green bead representing blonde and black hair colour, respectively on different chromatids of chromosome 1. Pair 2 (blue): One light blue bead and one dark blue bead representing straight and curly hair, respectively on different chromatids of chromosome 2. Pair 3 (red): One light red bead and one dark red bead representing brown and black eye colour, respectively on different chromatids of chromosome 3.

**Solution**

1. This activity demonstrates how different pairs of chromosomes segregate independently during gamete formation by meiosis.
2. Each pair of beads represents contrasting characters on different chromatids of three different chromosome pairs.

**Answer:** The activity models the independent assortment and segregation of alleles during meiosis.

### Question 2

*Activity*

Make a combination from it by randomly picking one bead from each pair.

**Solution**

1. Randomly picking one bead from each of the three pairs simulates the separation of chromosomes into gametes during meiosis.
2. This random mixing of characteristics from different chromosome pairs produces multiple unique genetic combinations.

**Answer:** A unique combination of three traits is formed by randomly selecting one bead from each pair.

### Question 3

*Activity*

Write your combination as ‘light green, light blue, light red’.

**Solution**

1. Select one bead randomly from each of the three pairs representing different chromosomes.
2. Combine the chosen character traits into a sequence representing a possible gamete combination.
3. Write the sequence of traits as a combination, for example, 'light green, light blue, light red'.

**Answer:** The combination is written as a sequence of three chosen traits, such as 'light green, light blue, light red'.

### Question 4

*Activity*

How many combinations can you make with just these three pairs of characters? Each time you make a combination using beads, you will get either the same combination or a different one. With just three pairs of characters, eight combinations are possible.

**Solution**

1. Each of the three pairs of characters has 2 alternative forms (beads of different colours).
2. The total number of unique combinations is calculated by multiplying the number of options for each pair: $2 \times 2 \times 2$.
3. This gives a total of 8 possible combinations.

**Answer:** Eight combinations are possible with three pairs of characters.

### Question 5

*Activity*

Imagine how many combinations are possible with 23 pairs of chromosomes, each carrying genetic information for many characters.

**Solution**

1. Each pair of chromosomes can segregate in 2 different ways during gamete formation by meiosis.
2. With 23 pairs of chromosomes, the total number of possible chromosome combinations in gametes is calculated as $2^{23}$.
3. This results in $8,388,608$ ($8.4$ million) possible combinations from a single parent's gametes.
4. When fertilization combines gametes from two parents, trillions of unique genetic combinations are possible, creating vast variation among siblings.

**Answer:** $2^{23}$ (or $8,388,608$) possible chromosome combinations

> Common mistake: Multiplying 23 by 2 instead of raising 2 to the power of 23 ($2^{23}$).

## Activity 11.5: Let us explore

### Question 1

*Activity*

Collect different types of flowers from your surroundings.

**Solution**

1. Collect different types of flowers from the surroundings to study their floral parts and structure.

**Answer:** Different types of flowers are collected from the surroundings for observation.

### Question 2

*Activity*

Carefully observe each part of the flowers you have collected, starting from the outer whorl to the inner one.

**Solution**

1. Observe each part of the collected flowers carefully, starting from the outermost whorl to the innermost whorl.

**Answer:** Floral parts are observed sequentially from the outer to the inner whorl.

### Question 3

*Activity*

Record the presence of various floral parts in the different flowers that you collected in Table 11.1.

**Solution**

1. Record the presence or absence of sepals, petals, stamens, and pistils in the different flowers in Table 11.1.

**Answer:** The presence of various floral parts is recorded in Table 11.1.

### Question 4

*Activity*

Analyse the function of each part of the flower based on visible characters.

**Solution**

1. Analyse the function of each floral part based on its visible characters such as color, structure, and position.

**Answer:** The function of each part of the flower is analysed based on its visible characters.

### Question 5

*Activity*

Cut a transverse and a longitudinal section of the ovary (swollen base of the pistil) and observe it under a dissecting microscope.

**Solution**

1. Cut a transverse and a longitudinal section of the ovary, which is the swollen base of the pistil.
2. Observe the sections under a dissecting microscope to examine internal structures like ovules.

**Answer:** Transverse and longitudinal sections of the ovary are cut and observed under a dissecting microscope.

### Question 6

*Activity*

Record any other feature(s) in Table 11.1.

**Solution**

1. Record any other unique features observed in the flowers or their parts in Table 11.1.

**Answer:** Any additional features observed are recorded in Table 11.1.

### Question 7

*Activity*

Draw a diagram of the structure you observed under the microscope.

**Solution**

1. Take a transverse and longitudinal section of the ovary (the swollen base of the pistil) as instructed in Activity 11.5.
2. Observe the sections under a dissecting microscope to view the internal structure.
3. Diagram: Draw a neat, labeled diagram showing the ovary containing ovules and egg cells, similar to the structure shown in Fig. 11.11 in the textbook.

**Answer:** A labeled diagram of the transverse and longitudinal sections of the ovary showing ovules.

## Activity 11.6: Let us investigate

### Question 1

*Activity*

Identify sweet pea (matar) or garden pea plants in a garden or a nearby field.

**Solution**

1. This is an activity step requiring the identification of sweet pea or garden pea plants in a nearby field or garden.
2. Expected observation: Locating healthy pea plants to conduct the experimental setup for studying pollination.

**Answer:** Identification of pea plants for the investigation.

### Question 2

*Activity*

Select two juvenile (less developed) flower bud and three freshly blossomed flowers on the same pea plant.

**Solution**

1. This is an activity step involving the selection of two juvenile flower buds and three freshly blossomed flowers on the same pea plant.
2. Expected observation: Choosing flowers and buds at appropriate developmental stages for the experiment.

**Answer:** Selection of specific flower buds and blossomed flowers on the same plant.

### Question 3

*Activity*

Carefully remove the stamens from one of the two selected flowers buds and one of the three selected flowers.

**Solution**

1. This is an activity step where stamens are carefully removed (emasculation) from one of the selected flower buds and one of the selected blossomed flowers.
2. Expected observation: Removal of male reproductive parts to prevent self-pollination in the chosen flowers.

**Answer:** Removal of stamens from selected flower buds and flowers.

### Question 4

*Activity*

Take muslin cloth bags and loosely wrap them around the flower bud, the flower bud of which stamens are removed, the flower of which stamens are removed and a freshly blossomed flower (Fig. 11.12).

**Solution**

1. This is an activity step involving the loose wrapping of muslin cloth bags around the selected flower buds and flowers as shown in Fig. 11.12.
2. Expected observation: Enclosing the flowers in muslin cloth bags to prevent unwanted transfer of pollen.

**Answer:** Wrapping specific flower buds and flowers with muslin cloth bags.

### Question 5

*Activity*

Leave one freshly blossomed flower uncovered (without muslin cloth bag).

**Solution**

1. This is an activity step where one freshly blossomed flower is left completely uncovered without any muslin cloth bag.
2. Expected observation: Leaving a control flower exposed to natural pollination by external agents.

**Answer:** Leaving one freshly blossomed flower uncovered as a control.

### Question 6

*Activity*

Observe them regularly and notice the development of fruits in place of the flowers that were not covered with muslin cloth. Allow them to grow for a few more days.

**Solution**

1. This is an activity step involving regular observation of the experimental set-up to check for the development of fruits.
2. Expected observation: Fruits form in place of flowers in all treatments except in the flower bud where stamens were removed and it remained covered, demonstrating that pollen transfer is necessary for fruit formation.

**Answer:** Observation of fruit development across different treatments.

### Question 7

*Activity*

Once the pods are fully developed in the flowers without muslin cloth, remove the muslin cloth from all the wrapped flowers and observe them.

**Solution**

1. Remove the muslin cloth bags from all wrapped flower buds and flowers after the pods have fully developed in the uncovered flowers.
2. Observe whether pods and seeds have formed or not in each of the treated flowers.

**Answer:** The muslin cloth is removed to check for fruit and seed formation across different experimental conditions.

### Question 8

*Activity*

Note your observations in Table 11.2.

**Solution**

1. Examine the presence or absence of fruit formation in each of the five treatments listed in Table 11.2.
2. Record 'Yes' for treatments where fruits are formed and 'No' where they are not formed.

**Answer:** Observations show that fruits form in all treatments except where stamens were removed from the flower bud.

## Activity 11.7: Let us find out

### Question 1

*3 marks · Short answer*

Compare and analyse the two strategies in terms of (Table 11.3)— Pollen to seed ratio, Efficiency of pollination and seed formation

**Part (i)**

1. Wind-pollinated grasses release $5,00,000\text{--}10,00,000$ pollen grains per flower to form only $50\text{--}200$ seeds, resulting in a very high pollen-to-seed ratio.
2. Insect-pollinated plants release $20,000\text{--}40,000$ pollen grains per flower and form $800\text{--}1,000$ seeds, resulting in a much lower pollen-to-seed ratio.

Answer (i): Wind-pollinated plants have a vastly higher pollen-to-seed ratio than insect-pollinated plants.

**Part (ii)**

1. Wind pollination relies on random air currents, making it less efficient with heavy wastage of pollen grains.
2. Insect pollination uses specific animal vectors, making it highly efficient with a greater proportion of pollen grains successfully reaching the target stigma and forming seeds.

Answer (ii): Insect-pollinated plants show higher efficiency of pollination and seed formation compared to wind-pollinated plants.

**Answer:** Wind-pollinated grasses have a very high pollen-to-seed ratio and low efficiency, whereas insect-pollinated plants have a lower pollen-to-seed ratio and much higher efficiency.

> Common mistake: Confusing pollen-to-seed ratio with pollination efficiency.

### Question 2

*3 marks · Short answer*

Explain why producing a very large number of pollen grains can still be an effective pollination strategy.

**Solution**

1. Wind pollination relies on random air currents to transport pollen grains from one flower to another.
2. Since wind blows in all directions and much pollen is lost or fails to reach a compatible stigma, a vast majority of pollen grains do not achieve pollination.
3. Producing millions of pollen grains ensures that even with massive wastage, at least a sufficient number successfully reach the stigmas of other flowers to guarantee seed formation and species survival.

**Answer:** Producing a very large number of pollen grains compensates for the high random loss during wind transport, ensuring successful pollination and seed formation.

> Common mistake: Stating that all pollen grains are used up in pollination.

## Pause and Ponder

### Question 1

*3 marks · Short answer*

In a china-rose (hibiscus or gudhal) plant, a pollen tube grows and continues through the style after pollen lands on the stigma. Which process is about to happen next?

**Solution**

1. The pollen tube grows down through the style and carries the male gamete towards the ovule in the ovary.
2. Upon reaching the ovule, the male gamete fuses with the egg cell (female gamete).
3. This process of fusion of gametes is called fertilisation, which marks the beginning of a new life.

**Answer:** Fertilisation (fusion of the male gamete with the egg cell inside the ovule) is about to happen next.

> Common mistake: Confusing pollination with fertilisation.

### Question 2

*3 marks · Short answer*

Look at the pictures (Fig. 11.16) of calotropis (madar) seeds and dandelion seeds given below. Can you guess what kind of seed dispersal these seeds are adapted for?

**Solution**

1. Calotropis (madar) and dandelion seeds have specialized structures such as hair-like tufts or lightweight wings.
2. These structures make the seeds lightweight so they can easily float through air currents.
3. Therefore, these seeds are adapted for seed dispersal by wind.

**Answer:** Seed dispersal by wind (anemochory).

> Common mistake: Writing dispersal by animals instead of wind.

### Question 3

*3 marks · Short answer*

A farmer plants two varieties of maize side by side, but notices that seeds form only when pollen from one variety reaches the stigma of the other. What type of pollination is this?

**Solution**

1. Pollination involves the transfer of pollen grains from the anther to the stigma of a flower.
2. When pollen is transferred from the flower of one plant to the stigma of another plant of the same type, it is called cross-pollination.
3. Since seeds form only when pollen reaches the stigma of the other variety, this represents cross-pollination.

**Answer:** Cross-pollination.

> Common mistake: Writing self-pollination.

### Question 4

*3 marks · Short answer*

Why do animals with external fertilisation generally produce more eggs than animals with internal fertilisation?

**Solution**

1. In external fertilisation, gametes and eggs are released into the external environment such as water.
2. A large number of these eggs are destroyed by water currents or eaten by other animals.
3. To compensate for these heavy losses and ensure the survival of the species, animals with external fertilisation produce a very large number of eggs.

**Answer:** Animals with external fertilisation produce more eggs to compensate for the high loss of eggs due to water currents and predators in the external environment.

> Common mistake: Failing to mention the destruction of eggs by predators and water currents.

### Question 5

*3 marks · Short answer*

In animals, which fertilisation method the gametes are more protected?

**Solution**

1. In external fertilisation, gametes are released into water where they are exposed to environmental hazards and predators.
2. In internal fertilisation, fertilisation takes place inside the body of the female.
3. Therefore, gametes and the resulting zygote are much more protected in internal fertilisation.

**Answer:** Internal fertilisation.

> Common mistake: Stating external fertilisation.

### Question 6

*3 marks · Short answer*

Ravi suddenly notices that he is growing taller rapidly, his shoulders are broadening, and his voice cracks. What stage of life is he entering?

**Solution**

1. Physical changes such as rapid increase in height, broadening of shoulders, and cracking of the voice occur during puberty.
2. Puberty is the period during adolescence when reproductive organs mature and secondary sexual characters develop in boys and girls.
3. Thus, Ravi is entering the stage of adolescence or puberty.

**Answer:** Adolescence (or puberty).

> Common mistake: Writing adulthood instead of adolescence/puberty.

### Question 7

*3 marks · Short answer*

Rina’s period occurs every 28 days. Her last period was on the 5th of March. On which day is she most likely to get her next period?

**Solution**

1. State the duration of the menstrual cycle and the date of the last period: cycle length is 28 days, and the last period started on 5th March.
2. Calculate the remaining days in March: March has 31 days, so the number of days left after 5th March is $31 - 5 = 26$ days.
3. Calculate the date of the next period in April: the remaining days for the 28-day cycle are $28 - 26 = 2$ days into April, making the expected date 2nd April.

**Answer:** 2nd April

> Common mistake: Forgetting that March has 31 days and incorrectly adding 28 directly to 5 without considering month lengths.

### Question 8

*3 marks · Short answer*

A human zygote has just formed. How many chromosomes does it have?

**Solution**

1. State the concept: A human zygote is formed by the fusion of a male gamete (sperm) and a female gamete (egg).
2. Recall the chromosome number: Each human gamete is haploid and contains 23 chromosomes.
3. Determine the total: The resulting zygote is diploid and contains 23 + 23 = 46 chromosomes.

**Answer:** 46 chromosomes (23 pairs)

> Common mistake: Writing 23 chromosomes instead of 46.

### Question 9

*3 marks · Short answer*

What protective devices can be used during sexual activity to reduce the spread of STIs?

**Solution**

1. Identify the barrier method: Condoms are the primary protective devices used during sexual activity.
2. Explain their function: They act as barriers that prevent the exchange of bodily fluids between partners.
3. State the outcome: This significantly reduces the transmission of Sexually Transmitted Infections (STIs) and helps prevent unwanted pregnancy.

**Answer:** Condoms

> Common mistake: Confusing oral contraceptive pills with protective devices against STIs.

### Question 10

*3 marks · Short answer*

If a couple uses oral contraceptive pills but not condoms, which risks remain and why?

**Solution**

1. State the effect of oral pills: Oral contraceptive pills alter hormone levels to prevent the release of eggs and avoid pregnancy.
2. Identify the remaining risk: Since pills only prevent pregnancy, they do not provide any physical barrier against bodily fluids.
3. Conclusion: The risk of contracting or spreading Sexually Transmitted Infections (STIs) remains entirely unprotected.

**Answer:** The risk of contracting Sexually Transmitted Infections (STIs) remains unprotected because pills only prevent pregnancy and do not act as barriers against infections.

> Common mistake: Assuming that preventing pregnancy also protects against infections.

### Question 11

*3 marks · Short answer*

In many animals, the young ones can walk or find food soon after birth but human babies are completely dependent on adults for a long time. What might be some advantages and disadvantages of this for humans as a species?

**Solution**

1. State the advantage: Extended dependency allows for a longer period of brain development, learning, and acquiring complex social and survival skills.
2. State the disadvantage: It places a heavy burden and responsibility on adult parents for care, protection, and provisioning over a long time.
3. Conclusion: These factors shape human social structures and evolutionary success as a species.

**Answer:** Advantages include advanced brain development and learning of complex skills, while disadvantages include prolonged vulnerability and heavy burden of care on adults.

> Common mistake: Listing only advantages or only disadvantages instead of both.

## Revise, Reflect, Refine

### Question 1

*1 mark · MCQ*

A flower’s anthers are removed before it matures. Later, pollen from another plant of the same species is dusted onto its stigma and seeds are produced. Which process has been ensured here?

- Self-pollination
- Cross-pollination
- Fertilisation
- Tissue culture

**Solution**

1. 1. When pollen is transferred from the anther of a flower of one plant to the stigma of a flower of another plant of the same species, it is called cross-pollination.
2. 2. Since the anthers of the flower were removed and pollen from another plant was dusted, cross-pollination is ensured.

**Answer:** (ii) Cross-pollination

> Common mistake: Confusing cross-pollination with self-pollination.

### Question 2

*3 marks · Short answer*

Arrange the following stages of sexual reproduction in plants in the correct order: (i) Pollen germination on stigma (ii) Fertilisation (iii) Pollination (iv) Formation of zygote

**Solution**

1. 1. Step (iii) Pollination occurs first, involving the transfer of pollen grains from the anther to the stigma.
2. 2. Step (i) Pollen germination on stigma follows, where the pollen grain produces a pollen tube.
3. 3. Step (ii) Fertilisation takes place when the male gamete fuses with the egg cell inside the ovule, leading to Step (iv) Formation of zygote.

**Answer:** (iii), (i), (ii), (iv)

> Common mistake: Writing fertilisation before pollen germination.

### Question 3

*1 mark · Assertion and reason*

Assertion (A): The zygote formed after fertilisation immediately attaches to the uterus wall. Reason (R): The uterus wall is always prepared to receive the zygote.

- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. 1. Before ovulation and fertilisation, the inner lining of the uterus becomes thick and rich with blood vessels to receive and nourish the developing zygote.
2. 2. Therefore, Assertion (A) is false because the zygote undergoes mitotic divisions while travelling to the uterus before implanting, but Reason (R) is true as the uterus wall prepares itself.

**Answer:** (iv) A is false, but R is true.

> Common mistake: Assuming the zygote attaches immediately without undergoing cleavage divisions during transport.

### Question 4

*3 marks · Short answer*

Why does asexual reproduction produce offsprings that are genetically identical to the parent?

**Solution**

1. 1. Asexual reproduction involves only a single parent.
2. 2. The central process behind asexual reproduction is mitosis, a type of cell division that produces two daughter cells having the same number of chromosomes identical to the parent cell.
3. 3. Consequently, the offspring produced are genetically identical to the parent and are called clones.

**Answer:** Asexual reproduction involves only a single parent and relies on mitosis, producing genetically identical offspring called clones.

> Common mistake: Forgetting to mention mitosis and single parent involvement.

### Question 5

*3 marks · Short answer*

Explain why the menstrual cycle stops during pregnancy.

**Solution**

1. 1. During pregnancy, if fertilisation occurs, the zygote develops into an embryo which gets implanted in the thick inner lining of the uterus.
2. 2. The inner lining of the uterus is maintained and kept rich in blood vessels to nourish the developing embryo throughout pregnancy.
3. 3. As a result, the uterine lining does not shed, causing the menstrual cycle to stop during pregnancy.

**Answer:** The menstrual cycle stops during pregnancy because the thick uterine lining is required to nourish the developing embryo and is not shed.

> Common mistake: Stating that ovulation continues during pregnancy.

### Question 6

*3 marks · Short answer*

Why are flowers that bloom at night white or light in colour as compared to flowers that bloom during the day?

**Solution**

1. 1. Flowers that bloom at night are usually pollinated by nocturnal animals such as moths and bats.
2. 2. White or light-coloured flowers are easily visible in the dark or dim light to attract these nocturnal pollinators.
3. 3. This is an adaptive strategy of plants to ensure successful pollination in the absence of sunlight.

**Answer:** Flowers that bloom at night are white or light in colour to make them easily visible in the dark to attract nocturnal pollinators.

> Common mistake: Confusing night-blooming flower adaptations with day-pollinated flower features like bright colours for insects.

### Question 7

*3 marks · Short answer*

Why do vegetatively propagated plants tend to be more vulnerable to diseases than sexually reproduced plants?

**Solution**

1. Vegetatively propagated plants are clones, meaning they are genetically identical to the parent plant.
2. Because they lack genetic variation, they do not have different traits that might provide resistance to specific diseases.
3. Therefore, if a disease affects one plant, it is likely to affect the entire population, making them more vulnerable.

**Answer:** Vegetatively propagated plants are genetically identical clones, lacking the genetic variation needed to resist diseases, which makes them more vulnerable than sexually reproduced plants.

> Common mistake: Students often forget to mention that lack of genetic variation is the primary reason for vulnerability.

### Question 8

*3 marks · Short answer*

If all flowers in a type of plant were only capable of self-pollination, how would it affect the genetic diversity over several generations? Explain.

**Solution**

1. Self-pollination involves the transfer of pollen within the same flower or plant.
2. This process does not involve the mixing of genetic material from two different individuals.
3. Over several generations, this leads to a reduction in genetic diversity as the offspring remain genetically similar to the parent.

**Answer:** Continuous self-pollination reduces genetic diversity over generations because it prevents the mixing of genetic material from different individuals.

> Common mistake: Confusing self-pollination with asexual reproduction; while both reduce diversity, they are different processes.

### Question 9

*3 marks · Short answer*

A farmer wants to produce a large number of genetically identical plants quickly. Suggest suitable reproduction methods and explain why they are effective.

**Solution**

1. Suitable methods include vegetative propagation techniques like cutting, grafting, layering, or tissue culture.
2. These methods are effective because they involve only one parent and produce genetically identical offspring (clones).
3. This allows farmers to maintain desirable traits and produce large numbers of identical crops quickly.

**Answer:** Vegetative propagation methods like cutting, grafting, or tissue culture are suitable as they produce genetically identical clones, ensuring the preservation of desirable traits on a large scale.

> Common mistake: Suggesting sexual reproduction, which would lead to variations and not identical plants.

### Question 10

*3 marks · Short answer*

Suresh prepares slides with pollen grains in different sugar concentrations (0%, 2.5%, 5%, 7.5%, 10%) to study the germination of pollen. (i) What are the different hypotheses which can be tested using this set-up? (ii) What parameters should be kept the same in this set-up?

**Part (i)**

1. The hypothesis is that the concentration of sugar solution affects the rate or percentage of pollen grain germination.

Answer (i): Sugar concentration affects pollen germination.

**Part (ii)**

1. Parameters to keep constant include temperature, time duration for observation, type of pollen grains used, and light conditions.

Answer (ii): Temperature, time, type of pollen, and light conditions.

**Answer:** The experiment tests the effect of sugar concentration on pollen germination.

> Common mistake: Forgetting to mention that the type of pollen must be the same for a fair test.

### Question 11

*3 marks · Short answer*

Look at the picture given below and think in line with the given prompts and find out which type(s) of pollination might have been followed in these flowers— Tomato, Wheat, Papaya

**Solution**

1. Tomato: Self-pollination, as the stamens cover the stigma within the same flower.
2. Wheat: Wind pollination, as the flowers open after pollination and are adapted for wind-borne pollen.
3. Papaya: Cross-pollination, as male and female flowers are borne on different trees.

**Answer:** Tomato: Self-pollination; Wheat: Wind pollination; Papaya: Cross-pollination.

> Common mistake: Assuming all flowers use the same pollination method.

### Question 12

*4 marks · Case-based*

In the lower Himalayan region of northern India, apples are an important cash crop that contribute significantly to farmer’s livelihoods. The fruit yield in apple cultivation is declining continuously, associated with climate change and a significant decline in the population of natural pollinators. A researcher-farmer group set up two experimental apple orchards at two distinct locations: Places A and B. In apple orchards at Place A, they allowed natural pollinators to pollinate the flowers of the apple. In apple orchards at Place B, they applied mixed farming techniques of beekeeping. Along with honey, the farmer yielded apples. The yield of apples is depicted in Fig. 11.24, in terms of fruit setting (number of fruits/the total number of corresponding fruit-bearing branches) and fruit drop (premature falling of developing fruits) in the two types of experimental places of apple orchards. (i) What are the hypotheses the researcher-farmers group has thought of for this investigation? (ii) What are the different parameters in the experiment? (iii) Compare and analyse the data of two experimental orchards Places A and B, in terms of high yields of apple fruits. (iv) Based on your analysis, what do you infer from the data?

**Part (i)**

1. The hypothesis is that the presence of additional pollinators (bees) increases the pollination efficiency, leading to higher fruit set and lower fruit drop.

Answer (i): Additional pollinators increase fruit set and decrease fruit drop.

**Part (ii)**

1. The parameters include the location of the orchard, the variety of apple trees, the presence or absence of beekeeping, and the environmental conditions.

Answer (ii): Location, tree variety, presence of bees, and environmental conditions.

**Part (iii)**

1. Place B (with beekeeping) shows a significantly higher percentage of fruit set and a lower percentage of fruit drop compared to Place A (natural pollination).

Answer (iii): Place B has higher fruit set and lower fruit drop than Place A.

**Part (iv)**

1. The data indicates that introducing managed pollinators like bees improves the reproductive success of apple trees, resulting in higher yields.

Answer (iv): Managed pollinators improve fruit yield and reduce fruit drop.

**Answer:** Beekeeping increases fruit set and reduces fruit drop in apple orchards.

> Common mistake: Misinterpreting the bar chart by confusing fruit set with fruit drop.

### Question 13

*3 marks · Short answer*

A student claims, “In humans, ovulation always happens on day 14 of the menstrual cycle”. Critically examine this claim and state whether the claim is correct or not. Give at least two reasons for your answer.

**Solution**

1. The student's claim is incorrect because the length of the menstrual cycle is not fixed at 28 days for all individuals.
2. The menstrual cycle typically ranges from 21 to 35 days, and ovulation occurs approximately 14 days before the start of the next menstrual period.
3. Consequently, in cycles that are shorter or longer than 28 days, the day of ovulation will shift accordingly rather than always occurring on day 14.

**Answer:** The claim is incorrect because the menstrual cycle length varies between 21–35 days, and ovulation occurs about 14 days before the next period, not necessarily on day 14.

> Common mistake: Assuming the menstrual cycle is always exactly 28 days long for every female.

## Frequently asked questions

### How many questions are there in the NCERT Solutions for Class 9 Science Chapter 11 Reproduction: How Life Continues?

This chapter for the 2026-27 session based on the new NCERT book contains a total of 81 questions across various interactive sections. You can find step-by-step solutions for all of them in the free PDF available right on this SwaVid page.

### Which specific topics and activities are covered in these Class 9 Science Chapter 11 solutions?

The solutions cover diverse concepts such as vegetative propagation techniques like stem cuttings, grafting, and layering, along with budding in yeast and mould structures in Activity 11.2 and 11.3. They also include meiosis bead models, floral part functions, pollination investigations, and human reproduction topics explored in Pause and Ponder.

### What is the hardest question type in this chapter and how should students approach it?

Assertion-reason and case-based questions found in the Revise, Reflect, Refine section are often considered the most challenging because they require a deep conceptual understanding rather than rote memorization. Students should approach them by first verifying the individual truth value of the assertion and the reason, and then checking if the reason correctly explains the assertion.

### How can I write answers to score full marks in Class 9 Science exams for this reproduction chapter?

To secure full marks, structure your answers clearly by using scientific terminology like nodes, internodes, scion, stock, and chromosome segregation. You should also refer to the detailed explanations and expert tips provided in SwaVid's free PDF solutions on this page.

### Is the free PDF for Class 9 Science Chapter 11 available for download?

Yes, the comprehensive free PDF containing complete solutions for all activities, Think It Over questions, and exercises is available on this SwaVid page for the 2026-27 session. It is aligned with the new NCERT book to help you revise efficiently.

## Related pages

- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student