CLASS 9 Science CHAPTER 15 Improvement in Food Resources (NCERT Solution)

CLASS 9 Science                                                                                                 
CHAPTER 15             
Improvement in Food Resources (NCERT Solution) 

Intext Questions
On Page 204

Question 1: What do we get from cereals, pulses, fruits and vegetables?
Answer: The things we get from cereals, pulses, fruits and vegetables are as follows:
Cereals(wheat, rice, maize, etc.): are the sources of carbohydrates which provide energy.
Pulses(pea, gram and soybean, etc.): are source of proteins.
Vegetables and fruits: provide us vitamins, minerals, carbohydrates, proteins and fats.

On Page 205

Question 1: How do biotic and abiotic factors affect crop production?
Answer: Factors which affect crop production are:
Biotic factors which cause loss of grains are rodents, pests, insects, etc. and abiotic factors, such as temperature, humidity, moisture, etc.
Affects of both biotic and abiotic factors on crop production are in the following ways:
Weight loss
Infestation of insects
Poor germination ability
Discoloration

Question 2: What are the desirable agronomic characteristics for crop improvement?
Answer: Desirable agronomic characteristics are:
Desirable characters for fodder crops are tallness and profuse branching
Dwarfness is desired in cereals, so that fewer nutrients are consumed by these crops.

On Page 206

Question 1: What are the macronutrients and why are they called macronutrients?
Answer: Macronutrients are essential elements which are required by plants in major quantities.
Some of the main macronutrients are:
(i) N, P, S which are found in proteins.
(ii) Ca is found in cell wall.
(iii) Mg is a part of chlorophyll.

Question 2: How do plants get nutrients?
Answer: Soil is the main source of nutrients for plants. Plants absorb the dissolved nutrients by the roots from the soil. This water absorbed by the roots is transported by the xylem tissue throughout the plant body.

On Page 207

Question 1: Compare the use of manure and fertilisers in maintaining soil fertility.
Answer: Use of manure in maintaining soil quality:
(i) Manures are the very rich source of organic matter (humus) for the soil. Humus helps to restore water retention capacity of sandy soil and drainage in clayey soil.
(ii) Manures are the sources of soil organisms like soil friendly bacteria.

Use of fertilisers on soil quality
(i) Excess use of fertilisers cause to dryness of soil and hence the rate of soil erosion increases.
(ii) Continuous use of fertilisers decreases the organic matter which reduce the porosity of the soil and the plant roots do not get sufficient oxygen.

On Page 208

Question 1: Which of the following conditions will give the most benefits? Why?
(i) Farmers use high-quality seeds, do not adopt irrigation or use fertilisers.
(ii) Farmers use ordinary seeds, adopt irrigation, use fertilisers and use crop protection measures.
(iii) Farmers use quality seeds, adopt irrigation use fertilizer and use crop protection measures.
Answer: The (iii) third option is the best option which provides the best conditions to get most benefits. For this, farmers use quality seeds, adopt irrigation, use fertilizers and use crop protection measures. This is because the use of only quality seeds is not sufficient until they are properly irrigated, enriched with fertilizers and protected from biotic factors.

On Page 209

Question 1: Why should preventing measures and biological control methods be preferred for protecting crops?
Answer: In plants, diseases are caused by Pathogens. To remove pathogens, some preventive measure and biological control methods are used which are :
Minimise pollution without affecting the soil quality.

Question 2: What factors may be responsible for the losses of grains during storage?
Answer: Following are the factors responsible for loss of grains during storage:
(i) Abiotic factors like humidity and temperature.
(ii) Biotic factors like insects, rodents, birds, mites and bacteria.

On Page 210

Question 1: Which method is commonly used for improving cattle breeds and why? How is cross breeding useful in animals?
Answer: To improve the cattle breed breeds, we generally use the cross breeding method. It is a process in which a cross is made between indigenous varieties of cattle by exotic breeds to get a cross breed which is high-yielding. During cross breeding, the desired characters taken into considerations are the offsprings should be high Yielding, should have early maturity and should be resistant to diseases and climatic conditions.

On Page 211

Question 1: Discuss the implications of the following statement It is interesting to note that poultry is India's most efficient converter of low fibre food stuff (which is unfit for human consumption) into highly nutritious animal protein food.
Answer: The poultry birds are efficient converters of agricultural by products, particularly cheaper fibrous wastes into high quality meat and in providing egg, feathers and nutrient rich manure. So, the given statement is correctly said for the poultry birds.

Question 2: What management practices are common in dairy and poultry farming?
Answer: Well designed and hygenic shelter.
Proper food is required to get good yield of egg and meat.
Complete protection from diseases causing agents like virus, bacteria or fungi.

Question 3: What are the differences between broilers and layers and in their management?
Answer: The poultry bird giving meat are called broilers and the egg laying birds are called layers.
The ration required for broilers should be rich in protein with sufficient fat. The food should also have high vitamin-A and K. Layers require enough space and lighting.

On Page 213

Question 1: How are fish obtained?
Answer: Fish are obtained by two ways:
(i) Capture fisheries from natural resources
(ii) Fish farming by culture for commercial purposes.

Question 2: What is the advantage of composite fish culture?
Answer: Combination of five or six fish species in a single fish pond is known as composite fish culture. Basis of selection of species is food habits so that they do not compete for food among themselves. As a result, the food available in all parts of the pond is utilized without competing with each other. This increases the fish yield from the pond.

Question 3: What are the desirable characters of the varieties suitable for honey production?
Answer: Desirable characters in varieties for honey production are:
(a) Capacity to collect a large amount of honey.
(b)Should stay in beehive for a longer time.
(c) Should have good breeding capacity.

Question 4: What is pasturage and how is it related to honey producton?
Answer: Flowers available for nectar and pollen collection is known as pasturage . The quality and taste of honey depends on adequate quantity of pasturage and flowers available.

Exercises

Question 1: Explain anyone method of crop production which ensures high yield.
Answer: Inter cropping is a method used for a high yielding crop production. In this method, two or more crops are grown simultaneously on the same field in definite pattern. A few rows of one crop alternate with a few rows of second crop.
Example: soybean, maize or finger millet (bajara) and caw pea (labia).
The selected crops should have different nutrient requirements. This ensures maximum utilisatian of the nutrients supplied. It also prevents pests and diseases from spreading to all the plants belonging to one
crop in a field. This method gives a better crop yield.

Question 2: Why are manure and fertilizers used in fields?
Answer: The essential nutrients of the soil are full filled by manures and fertilizers supply. Therefore, they help in good vegetative growth, giving rise to healthy plants that rise in high crap production.

Question 3: What are the advantages of inter-cropping and crop rotation?
Answer: Advantages of inter-cropping
Maintain soil fertility.
Save time and labour.
Increases productivity per unit area.
Both crops can be easily harvested and threshed separately.
Advantages of crop rotation
Improves soil fertilitiy.
Reduces pest infestation and diseases. Helps in weed control.
Avoids depletion of a particular nutrient from soil.

Question 4: What is genetic manipulation? How is it useful in agricultural practices?
Answer: Transferring desirable genes from one plant to another plant far the production of varieties with desirable characters is known as genetic manipulation.
Example: Profuse branching in fodder craps, high yielding varieties in maize wheat, etc.
Uses in agricultural practices:
Better adaptability to adverse environmental conditions.
Contain desirable features.
Helps in increasing yield and quality.
Maturation period is shorter.

Question 5: How do storage grain losses occur?
Answer: The main reason for the losses of storage grain is abiotic and biotic factors.
The abiotic factors: Moisture and temperature
The biotic factors: Insects, rodents, birds, mites and bacteria.

Question 6: How do good animal husbandry practices benefit farmers?
Answer: Benefits of good farming practices:
(i) It provides improved breeds of domestic animals.
(ii)It increase the production of products like milk, egg and meat.
(iii) Proper shelter, feeding, care and protection against disesase help
the farmers to improve their economic conditions.

Question 7: What are the benefits of cattle farming?
Answer: Benefits of cattle farming
(i) Milk production is increased.
(ii) Good quality meat, fibre and skin is obtained.
(iii) Goad breed of draught animals can be obtained.

Question 8: For increasing production, what is common in poultry, fisheries and bee keeping?
Answer: Cross breeding is a common practice between poultry, fisheries
and bee keeping for increasing production.

Question 9: How do you differentiate between capture fishing, mariculture and aquaculture?
Answer: Differences between capture fishing, mariculture and aquaculture
Fishing
1.Fish are obtained from natural resources, like ponds, canals, rivers,
2.I Locating fish is easy and can be captured by using fishing nets.

Mariculture
1.A method of marine fish culture in the open sea.
2.Fish can be located with the help of satellites and echosounders.
These can be caught by many kinds of fishing nets using fishing boats.

Aquaculture
1.Production of fish from freshwater and brackish water resources.
2.Can be located easily and caught using fishing nets.

( From www.ncrtsolutions.in )

CLASS 9 Science CHAPTER 2 Is Matter around Us Pure (NCERT Solution)

CLASS 9 Science                                                                                                 
CHAPTER 2                                                                                                                           
Is Matter around Us Pure (NCERT Solution)

Intext Questions
On Page 15

Question 1: What is meant by a pure substance?
Answer: Substance having single type of particles is known as pure substance.
For example: Hydrogen, Water etc., are pure.
Note: All elements and compounds are considered to be pure.

Question 2: List the points of differences between homogeneous and heterogeneous mixtures.
Answer: Homogeneous mixture
Its constituent’s particles cannot be seen easily.
There are no visible boundaries of separation in a homogeneous mixture.
Its constituents cannot be easily separated.
Examples: Alloys, solution of salt in water etc.

Heterogeneous mixture
Its constituent particles can be seen easily.
Have visible boundaries of separation between the constituents.
Its constituents can be separated by simple methods.
Examples: Mixture of sand and common salt, mixture of sand and water etc.

On Page 18

Question 1: Differentiate between homogeneous and heterogeneous mixture with example.
Answer: Homogeneous mixture
Its constituent’s particles cannot be seen easily.
There are no visible boundaries of separation in a homogeneous mixture.
Its constituents cannot be easily separated.
Examples: Alloys, solution of salt in water etc.

Heterogeneous mixture
Its constituent particles can be seen easily.
Have visible boundaries of separation between the constituents.
Its constituents can be separated by simple methods.
Examples: Mixture of sand and common salt, mixture of sand and water etc.

Questions 2: How are sol, solution and suspension different from each other?
Answer: Solution or true solution is homogeneous.
NO Tyndall effect.
Solute particles cannot be filtered by using a filter paper
True solution is transparent.
Examples: Sea water, alloys, solution of lemon juice in water etc.
Sol (colloidal solution) :
Sol or colloidal solution is heterogeneous.

Tyndall effect.
Cannot be separated by ordinary filter paper.
It may be transparent or translucent.
Examples: Milk of magnesia, cough syrup, mist, fog, clouds, smoke, mud etc.

Suspension :
Suspension is also heterogeneous.
Tyndall effect.
It may be transparent or translucent.
Separated easily by filter paper.(because of large particles)
Examples: Mixture of sand in water, mixture of chalk in water.

Question 3: To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Answer: Mass of sodium chloride (solute) = 36 g
Mass of water (solvent) = 100 g
We know that, mass of solution = mass of solute + mass of solvent
= 36 g+ 100 g= 136 g
Concentration (mass percentage) of the solution 
      Mass of Solute
= ---------------------- x 100
     Mass of Solution
      36g
= --------- x 100 = 26.47%
    136 g

On Page 24

Question 1: How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Answer: Simple distillation is the method which can separate the mixture of kerosene and petrol (b.p. differ by more than 25°C).
 Method: In a distillation flask, a mixture of kerosene and petrol is taken. The mixture is heated slowly and the temperature is recorded with the help of thermometer. Petrol (b.p. = 70° C to 1200 ° C) vaporizes first and the temperature becomes constant for some time (till all petrol evaporates from the mixture). Vapours of petrol are condensed and collected in another container while the kerosene remains in the distillation flask. As soon as the temperature starts’ rising again, the heating is stopped and both the components are collected separately.

Question 2: Name the technique to separate
(i) Butter from curd
(ii) Salt from sea water
(iii) Camphor from salt
Answer: (i) By using centrifugation method, butter can be separated from curd.
(ii) By using evaporation method, salt from sea water can be separated. Water vaporises on evaporation leaving behind the salt.
(iii) Camphor from salt can be separated by sublimation method. On subliming camphor will be converted into vapour leaving behind the salt.

Question 3: What types of mixtures are separated by the technique of crystallisation?
Answer: Crystallisation method can be used for the purification of those mixtures whichccontain insoluble and/or soluble impurities.
Have crystalline nature cannot be separated by filtration as some impurities are soluble.

Question 4: Classify the following as chemical or physical changes
(a) Cutting of trees,
(b) Melting of butter in a pan,
(c) Rusting of almirah,
(d) Boiling of water to form steam,
(e) Passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,
(f) Dissolving common salt in water,
(g) Making a fruit salad with raw fruits, and
(h) Burning of paper and wood
Answer: Physical Change :
Cutting of trees
Melting of butter in a pan
Boiling of water to form steam
Dissolving common salt in water
Making a fruit salad with raw fruits

Chemical Change :
Rusting of almirah.
Passing of electric current, through water and the water breaking down
into hydrogen and oxygen gases.
Burning of paper and wood.

Question 5: Try segregating the things around you as pure substances or mixtures.
(a) Wood
(b) Coal
(c) Milk
(d) Sugar
(e) Common salt
(f) Soap
(g) Soil
(h) Rubber
Answer: (a) Mixture
(b) Mixture
(c) Mixture
(d) Pure substance
(e) Pure substance
(f) Compound/mixture
(g) Mixture
(h) Pure substance

Exercises
Question 1: Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Answer: (a) Evaporation
(b) Sublimation
(c) Filtration
(d) Chromatography.
(e) centrifugal machine or churning the curd by hand.
(f) Decantation
(g) Filtration.
(h) Magnetic Separation.
(i) Winnowing.
(j) Coagulation and decantation:

Question 2: Write the steps you would use for making tea. Use the words solution, solvent ,solute, dissolve, soluble, insoluble, filtrate and residue.
Solution: Method of preparation of tea
(i) Take some water (solvent) in a pan and heat it.
(ii) Add some sugar (solute) and boil to dissolve the sugar completely the
obtained homogeneous mixture is called solution.
(iii) Add tea leaves (or tea) in the solution and boil the mixture.
(iv) Now add milk and boil again.
(v) Filter the mixture through the tea stainer and collect the filtrate or soluble substances, i.e., tea in a cup. The insoluble tea leaves left behind as residue in the 8 trainer.

Question 3: Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer: (a) Mass of potassium nitrate needed to produce its saturated solution
in 100 g of water at 313 K = 62 g
Mass of potassium nitrate needed to produce its saturated solution in 50 g of water at 313K
       62 
= ------- x 50g = 31g
     100
(b) Crystals of potassium chloride are formed. This happens as solubility of solid decreases with decreasing the temperature.

(c) Solubility of each salt at 293 K
Potassium nitrate 32 g per          100 g water
Sodium chloride 36 g per           100 g water
Potassium chloride 35 g per       100 g water
Ammonium chloride 37 g per    100 g water
Note: Solubility of a solid is that amount in gram which can be dissolved in 100 g of water (solvent) to make saturated solution at a particular temperature.
Ammonium chloride has the maximum solubility (37 g per 100 g of water) at 293 K.

(d) Solubility of a (solid) salt decreases with decrease in temperature while it increases with rise in temperature.

Question 4: Explain the following giving examples.
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension
Answer: (a) Saturated solution: A solution in which no more amount of solute can be dissolved at a particular temperature is called saturated solution. Example: when sugar is dissolved repeatedly in a given amount of water, a condition is reached at which further dissolution of sugar is not possible in that amount of water at room temperature.

(b) Pure substance: A substance made up of single type of particles (atoms and/or molecules) is called pure substance. All elements and compounds are said to be pure. Example: water, sugar etc.

(c) Colloid: A heterogeneous mixture in which the solute particle size is too small to be seen with the naked eye, but is big enough to scatter light is known as Colloid. There are two phases in colloidal solution Dispersed phase: solute particles are said to be dispersed phase Dispersion medium: the medium in which solute particles are spread is called the dispersion medium. Example: Milk, clouds etc., are the example of colloid.

(d) Suspension: A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of suspension are visible to the naked eye. Example: Mixture of sand, Water and Muddy water etc.

Question 5: Classify each of the following as a homogeneous or heterogeneous mixture. Soda water, wood, air, soil, vinegar, filtered tea.
Answer: Homogeneous mixtures: Air, soda water, vinegar, filtered tea.
Heterogeneous mixtures: Wood, soil,

Question 6: How would you confirm that a colorless liquid given to you is
pure water?
Answer: If the given colorless liquid boils at 100°C sharp, it is pure water, otherwise not.

Question 7: Which of the following materials fall in the category of a "pure substance"?
(a) Ice (b) Milk (c) Iron
(d) Hydrochloric acid (e) Calcium oxide (f) Mercury
(g) Brick (h)Wood (i) Air
Answer: Ice, iron, calcium oxide, mercury are pure substance as they have definite composition.
Milk is a colloid, so it is a heterogeneous mixture.
Hydrochloric acid is also a mixture of hydrogen chloride gas and water.

Question 8: Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Answer: Sea water, air and soda water: Homogeneous mixture
Coal, Soil: Heterogeneous solution.

Question 9: Which of the following will show "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Answer: Milk and starch solution will show "Tyndall effect" as both of these are colloids.

Question 10: Classify the following into elements, compounds and
mixtures.
(a) Sodium (b) Soil (c) Sugar solution
(d) Silver (e) Calcium carbonate (f) Tin
(g) Silicon (h) Coal (i) Air
(j) Soap (k) Methane (l) Carbon dioxide
(m) Blood
Answer: Elements : Sodium, silver, tin and silicon
Compounds : Calcium carbonate, methane, and carbon dioxide
Mixtures : Soil, sugar solution, coal, air, soap and blood.

Question 11: Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle
Answer: Growth of a plant, rusting of iron, cooking of food, digestion of food, burning of a candle are chemical changes, because here the chemical composition of substance changes.

( From www.ncrtsolutions.in )

CLASS 10 Science CHAPTER 9 Heredity And Evolution (NCERT Solution)

CLASS 10 Science                                                                                                    CHAPTER 9                                                                                                             
Heredity  And Evolution (NCERT Solution)

Question 1: If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer: In asexual reproduction, the reproducing cells produce a copy of their DNA through
some chemical reactions. However, this copying of DNA is not accurate and therefore, the newly
formed DNA has some variations.


It can be easily observed in the above figure that in asexual reproduction, very few variations are
allowed. Therefore, if a trait is present in only 10% of the population, it is more likely that the trait
has arisen recently. Hence, it can be concluded that trait B that exists in 60% of the same population has arisen earlier than trait A.

Question 2: How does the creation of variations in a species promote survival?
Answer: Sometimes for a species, the environmental conditions change so drastically that their
survival becomes difficult. For example, if the temperature of water increases suddenly, most of
the bacteria living in that water would die. Only few variants resistant to heat would be able to
survive. If these variants were not there, then the entire species of bacteria would have been
destroyed. Thus, these variants help in the survival of the species. However, not all variations are useful. Therefore, these are not necessarily beneficial for the individual organisms.

Question 3: How do Mendel’s experiments show that traits may be dominant or recessive?
Answer: Mendel selected true breeding tall (TT) and dwarf (tt) pea plants. Then, he crossed these
two plants. The seeds formed after fertilization were grown and these plants that were formed
represent the first filial or F1 generation. All the F1 plants obtained were tall.


Then, Mendel self-pollinated the F1 plants and observed that all plants obtained in the
F2 generation were not tall. Instead, one-fourth of the F2 plants were short.


From this experiment, Mendel concluded that the F1 tall plants were not true breeding. They were
carrying traits of both short height and tall height. They appeared tall only because the tall trait is
dominant over the dwarf trait.

Question 4: How do Mendel’s experiments show that traits are inherited independently?
Answer: Mendel crossed pea plants having round green seeds (RRyy) with pea plants having
wrinkled yellow seeds (rrYY).


.
Since the F1 plants are formed after crossing pea plants having green round seeds and pea plants
having yellow wrinkled seeds, F1 generation will have both these characters in them. However, as
we know that yellow seed colour and round seeds are dominant characters, therefore, the F1 plants
will have yellow round seeds.
Then this F1 progeny was self-pollinated and the F2 progeny was found to have yellow round
seeds, green round seeds, yellow wrinkled seeds, and green wrinkled seeds in the ratio of 9:3:3:1.

In the above cross, more than two factors are involved, and these are independently inherited.

Question 4: A man with blood group A marries a woman with blood group O and their daughter
has blood group O. Is this information enough to tell you which of the traits − blood group A or O
− is dominant? Why or why not?
Answer: No. This information is not sufficient to determine which of the traits − blood group A
or O − is dominant. This is because we do not know about the blood group of all the progeny.
Blood group A can be genotypically AA or AO. Hence, the information is incomplete to draw any
such conclusion.

Question 5: How is the sex of the child determined in human beings?
Answer: In human beings, the females have two X chromosomes and the males have one X and
one Y chromosome. Therefore, the females are XX and the males are XY.
The gametes, as we know, receive half of the chromosomes. The male gametes have 22 autosomes
and either X or Y sex chromosome.
Type of male gametes: 22+X OR 22+ Y.
However, since the females have XX sex chromosomes, their gametes can only have X sex
chromosome.
Type of female gamete: 22+ X


Sex determination in humans
Thus, the mother provides only X chromosomes. The sex of the baby is determined by the type of
male gamete (X or Y) that fuses with the X chromosome of the female.

Question 6: What are the different ways in which individuals with a particular trait may increase
in a population?
Answer: Individuals with a particular trait may increase in a population as a result of the following:
(i) Natural selection: When that trait offers some survival advantage.
(ii) Genetic drift: When some genes governing that trait become common in a population.
(iii) When that trait gets acquired during the individual’s lifetime.

Question 7: Why are traits acquired during the life-time of an individual not inherited?
Answer: This happens because an acquired trait involves change in non-reproductive tissues
(somatic cells) which cannot be passed on to germ cells or the progeny. Therefore, these traits
cannot be inherited.

Question 8: Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer: Small numbers of tigers means that fewer variations in terms of genes are available. This means that when these tigers reproduce, there are less chances of producing progeny with some useful
variations. Hence, it is a cause of worry from the point of view of genetics.

Question 9: What factors could lead to the rise of a new species?
Answer: Natural selection, genetic drift and acquisition of traits during the life time of an
individual can give rise to new species. ill geographical isolation be a major factor in the
speciation of a self-pollinating plant species? Why or why not? Geographical isolation can prevent
the transfer of pollens among different plants. However, since the plants are self-pollinating,
which means that the pollens are transferred from the anther of one flower to the stigma of the
same flower or of another flower of the same plant, geographical isolation cannot prevent
speciation in this case.

Question 10: Will geographical isolation be a major factor in the speciation of an organism that
reproduces asexually? Why or why not?
Answer: Geographical isolation prevents gene flow between populations of a species whereas asexual reproduction generally involves only one individual. In an asexually reproducing organism, variations can occur only when the copying of DNA is not accurate. Therefore, geographical isolation cannot prevent the formation of new species in an asexually reproducing organism.

Question 11: Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer: The presence of feathers in dinosaurs and birds indicates that they are evolutionarily related. Dinosaurs had feathers not for flying but instead these feathers provided insulation to these warm-blooded animals. However, the feathers in birds are used for flight. This proves that reptiles and birds are closely related and that the evolution of wings started in reptiles.

Question 12: Can the wing of a butterfly and the wing of a bat be considered homologous organs?
Why or why not?
Answer: The wing of a butterfly and the wing of a bat are similar in function. They help the butterfly and the bat in flying. Since they perform similar function, they are analogous organs and not homologous.

Question 13: What are fossils? What do they tell us about the process of evolution?
Answer: Fossils are the remains of organisms that once existed on earth. They represent the
ancestors of plants and animals that are alive today. They provide evidences of evolution by
revealing the characteristics of the past organism and the changes that have occurred in these
organisms to give rise to the present organisms.

Question 14: Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer: A species is a group of organisms that are capable of interbreeding to produce a fertile
offspring. Skin colour, looks, and size are all variety of features present in human beings. These
features are generally environmentally controlled. Various human races are formed based on these
features. However, there is no biological basis to this concept of races. Therefore, all human beings are a single species as humans of different colour, size, and looks are capable of reproduction and can produce a fertile offspring.

Question 15: In evolutionary terms, can we say which among bacteria, spiders, fish and
chimpanzees have a ‘better’ body design? Why or why not?
Answer: Evolution cannot always be equated with progress or better body designs. Evolution simply creates more complex body designs. However, this does not mean that the simple body designs are inefficient. In fact, bacteria having a simple body design are still the most cosmopolitan organisms found on earth. They can survive hot springs, deep sea, and even freezing environment. Therefore, bacteria, spiders, fish, and chimpanzees are all different branches of evolution.

Question 16: A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers
with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer: (c) The genetic make-up of the tall parent can be depicted as TtWW Since all the progeny bore violet flowers, it means that the tall plant having violet flowers has WW genotype for violet flower colour. Since the progeny is both tall and short, the parent plant was not a pure tall plant. Its genotype
must be Tt. Therefore, the cross involved in the given question is
TtWw × ttww
         ↓
TtWw − ttww
Therefore, half the progeny is tall, but all of them have violet flowers.
Question 17: An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer: (b)An example of homologous organs is our teeth and an elephant’s tusks.

Question 18: In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer: (a) In evolutionary terms, we have more in common with a Chinese school boy.

Question 19: A study found that children with light- coloured eyes are likely to have parents with
light- coloured eyes. On this basis, can we say anything about whether the light eye colour trait is
dominant or recessive? Why or why not?
Answer: Let us assume that children with light- coloured eyes can either have LL or Ll or ll genotype. If the children have LL genotype, then their parents will also be of LL genotype.
LL × LL
     ↓
LL
If the children with light-coloured eyes have ll genotype, then their parents will also have ll
genotype.
ll×ll
 ↓
ll
Therefore, it cannot be concluded whether light eye colour is dominant or recessive.

Question 20: How are the areas of study − evolution and classification − interlinked?
Answer: Classification involves grouping of organism into a formal system based on similarities
in internal and external structure or evolutionary history.
Two species are more closely related if they have more characteristics in common. And if two species are more closely related, then it means they have a more recent ancestor.
For example, in a family, a brother and sister are closely related and they have a recent common
ancestor i.e., their parents. A brother and his cousin are also related but less than the sister and her
brother. This is because the brother and his cousin have a common ancestor i.e., their grandparents
in the second generation whereas the parents were from the first generation.
With subsequent generations, the variations make organisms more different than their ancestors.

Question 21: Explain the terms analogous and homologous organs with examples.
Answer: Homologous organs are similar in origin (or are embryologically similar) but perform
different functions. For example, the forelimbs of humans and the wings of birds look different
externally but their skeletal structure is similar. It means that their origin is similar (as wings in
birds are modifications of forearm) but functions are different - the wings help in flight whereas
human forearm helps in various activities.

Analogous organs, on the other hand, have different origin but perform similar functions. For example, the wings of a bird and a bat are similar in function but this similarity does not mean that these animals are more closely related. If we carefully look at these structures, then we will find that the wings of a bat are just the folds of skin that are stretched between its fingers whereas the wings of birds are present all along the arm. Therefore, these organs are analogous organs.

Question 22: Outline a project which aims to find the dominant coat colour in dogs.
Answer: Dogs have a variety of genes that govern coat colour. There are at least eleven identified
gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.
A dog inherits one gene from each of its parents. The dominant gene gets expressed in the
phenotype. For example, in the B series, a dog can be genetically black or brown.
Let us assume that one parent is homozygous black (BB), while the other parent is homozygous
brown (bb)
bb BB
B B
bBbBb
bBbBb
In this case, all the offsprings will be heterozygous (Bb). Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.
If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50 %
heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.
B b
BBBBb
bBbBb

Question 23: Explain the importance of fossils in deciding evolutionary relationships.
Answer: Fossils are the remains of the organism that once existed on earth. They represent the
ancestors of the plants and animals that are alive today. They provide evidences of evolution by
revealing the characteristics of the past organisms and the changes that have occurred in these
organisms to give rise to the present organisms. Let us explain the importance of fossils in
deciding evolutionary history with the help of the following example.
Around 100 million years ago, some invertebrates died and were buried in the soil in that area.
More sediment accumulated on top of it turning it into sedimentary rock.
At the same place, millions of years later, some dinosaurs died and their bodies were buried on top
of the sedimentary rock. The mud containing dinosaurs also turned into a rock.
Then, millions of years later, some horse-like creatures died in that area and got fossilized in rocks
above the dinosaur fossils.
Some time later, due to soil erosion or floods in that area, the rocks containing horse-like fossils
are exposed.
If that area is excavated deeper, then the dinosaur and invertebrates fossils can also be found.
Thus, by digging that area, scientists can easily predict that horse-like animals evolved later than
the dinosaurs and the invertebrates.
Thus, the above example suggests that the fossils found closer to the surface of the earth are more
recent ones than the fossils present in deeper layers.www.ncrtsolutions.in


Question 24: What evidence do we have for the origin of life from inanimate matter?
Answer: A British scientist, J.B.S. Haldane, suggested that life originated from simple inorganic
molecules. He believed that when the earth was formed, it was a hot gaseous mass containing
elements such as nitrogen, oxygen, carbon, hydrogen, etc. These elements combined to form molecules like water (H2O), carbon dioxide (CO2), methane (CH4), ammonia (NH3), etc.
After the formation of water, slowly the earth surface cooled and the inorganic molecules interacted with one another in water to form simple organic molecules such as sugars, fatty acids, amino acids, etc. The energy for these reactions was provided by solar radiations, lightning, volcanic eruptions, etc.
This was proved by the experiment of Stanley L. Miller and Harold C. Urey in 1953. They took a mixture of water (H2O), methane (CH4), ammonia (NH3), and hydrogen gas (H2) in a chamber and sparks were passed through this mixture using two electrodes. After one week, 15 % of the carbon from methane was converted into amino acids, sugars, etc. These organic molecules are polymerized and assembled to form protein molecules that gave rise to life on earth.
Miller and Urey experiment


Question 25: Explain how sexual reproduction gives rise to more viable variations than asexual
reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer: In sexual reproduction, two individuals having different variations combine their DNA
to give rise to a new individual. Therefore, sexual reproduction allows more variations, whereas in
asexual reproduction, chance variations can only occur when the copying of DNA is not accurate.
Additionally, asexual reproduction allows very less variations because if there are more variations,
then the resultant DNA will not be able to survive inside the inherited cellular apparatus.
However, in sexual reproduction, more variations are allowed and the resultant DNA is also able
to survive, thus making the variations viable.
Variation and Evolution: Variants help the species to survive in all the conditions. Environmental
conditions such as heat, light, pests, and food availability can change suddenly at only one place.
At that time, only those variants resistant to these conditions would be able to survive. This will
slowly lead to the evolution of a better adapted species. Thus, variation helps in the evolution of
sexually reproducing organisms.

Question 26: How is the equal genetic contribution of male and female parents ensured in the
progeny?
Answer: In human beings, every somatic cell of the body contains 23 pairs of chromosomes. Out
of these 23 pairs, the first 22 pairs are known as autosomes and the remaining one pair is known
as sex chromosomes represented as X and Y.
Females have two X chromosomes and males have one X and one Y chromosome. The gamete receives half of the chromosomes. Therefore, the male gametes have 22 autosomes and either X or Y chromosome.
The female gamete, on the other hand, has 22 autosomes and X chromosome.
During reproduction, the male and female gametes fuse and thus the progeny receives 22 autosomes and one X or Y chromosome from male parent and 22 autosomes and one X chromosome from the female parent.


Question 27: Only variations that confer an advantage to an individual organism will survive in a
population. Do you agree with this statement? Why or why not?
Answer: In species, variations that offer survival advantages are naturally selected. Individuals
adjust to their environments with the help of these selected variations and consequently these
variations are passed on to their progeny. Evolution of organisms occurs as a result of this natural
selection. However, there can be some other variations, which do not offer any survival advantage and arise only accidentally. Such variations in small populations can change the frequency of some genes
even if they are not important for survival. This accidental change in the frequency of genes in small populations is referred to as genetic drift. Thus, genetic drift provides diversity (variations) without any survival advantage.

( From www.ncrtsolutions.in )

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