Skip to main content

10th Biology Solved Test Paper - Heredity and Evolution with practice paper

 Chapter 9 Heredity and Evolution solved Questions for slow lerner      
Very short answer type Question
1) Name the scientist who proposed the theory of natural selection.
Ans- Charles Darwin.
2) What is micro-evolution?
Ans-Evolution on a relatively small scale
3) Write the expended form of DNA.
Ans-Doxyribo Nucleic Acid
4) Name two human traits which show variation.
Ans- Colour of eyes and height of an individual.
5) What type of reproduction give rise to more number of successful variation?
Ans- Sexual reproduction.
6) What is the basis of evolution?
Ans-Selection of variants by environmental factors forms the basis of evolution.
7) Older fossils are found closer to the surface-True or False.
Ans- False.
8) What will be the sex of a child who inherit Y Chromosome from his/her father?
Ans- The sex of the child will be boy.
9) If a trait A exists in 10% of 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?
Ans- Trait B
10) What are fossils?
Ans- Fossils are all the preserved traces or remains of living organisms of geological past.
Short Answer Type Questions (2 Marks )
1) Differentiate between homologous organs and analogous organs.Give one example of each.
Ans- Homologous organs have similar basic structure.These may look different and may perform different functions.e.g.- fore limb of man and flipper of a whale.
Analogous organs have different basic structure.These look alike and perform same functions.e.g.- wings of birds and wings of insects.
2) How the study of fossil is considered significant?
Ans- The study of fossil is considered significant because-
        (i) Fossils provide direct evidence of past life.
        (ii) These provide convincing proof of organic evolution.
3)  How will you determine the age of fossils?
Ans- The age of fossils can be determined by two ways-
        (i) If we dig into the earth and start finding fossils, it can be assumed that the fissils closer to the surface are more recent to those found in deeper layers.
        (ii) By detecting the ratio of different isotopes of the same element in the fossil materials.
4) How does the creation of variations in a species ensure survival?
Ans- The creation of variations in a species enable them to adopt according to the changes and the new needs. Thus, they will enable the survival of the species.
5) Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Ans- No, the wing oe a butterfly and the wing of a bat cannot be considered homologous organs they have a common function for flyig but their origin and structure  are not common . So they are analogous organ.
6) Will geographical isolation be a major factor in  the speciation of a self-pollinating plant species? Why or why not?
Ans- No, because, new genes do not enter in the population and no species is formed.Hence, there is a little variation possible.
7) Explain the importance of fossils in deciding evolutionary relationships.
Ans- Fossils and their study is useful in knowing about the species which are no longer alive. They provide evidence and missing links between two classes. They are helpful in forming a sequence of organisms in the pathway of evolution.
Thus, fossils have an importance in deciding evolutionary relationships.
8) How does Mendel’s experiment show that traits may be dominant or recessive?      
Ans- Mendel selected and crossed tall pea plant with dwarf pea plant . The F1 progeny had Tt genotype but the plant was tall. This shows that T is dominant and t is recessive.
(9) Dead remains of two organisms A and B were buried. Later , only B's fossils were found , but not A's. Give reasons
Ans: Animal A must not have got preserved or must have have got fully decomposed by the process of anaerobic bacteria. The other reason could be that it must have moved away to some other places by the action of wind or rains etc.
10) Do all variations in a particular species have equal chances of survival in an environment?
Ans: No, all variations in a particular species do not have equal chances for survival. The variations which are beneficial for the species, that is, which help in the better survival of the species have a better survival as compared to the other variations.

Short Answer Type Questions (3Marks)
1) How is the sex of a new born individual determined?
Ans- Sex is deternined at the time of fertilisation. A human male  has XY sex chromosome and produces two types of  sperms, either with X- chromosome or with Y- chromosome.
A human female has XX- chromosome and produce ova of one type. If the sperm containing X chromosome fuses with ovum , the individual will be girl and the sperm containing Y chromosome fuses with ovum, the individual will be boy.


2) What is sex chromosome? Name the two types of sex chromosomes. Mention the chromosomes present in male and female .
Ans-  The chromosome which determine the sex of a person are called sex chromosome.
The two types of sex chromosomes are X chromosome and Y chromosome.
Sex chromosomes in male -- XY 
Sex chromosomes in females -- XX.
3) Explain with an example how traits get expressed?
 Ans- Gene is made up of a segment of DNA which provide information for protein. Protein is needed for biochemical process. If the proteins work efficiently,the traits get expressed in better way.
For erxample, the plant height depends onthe amount of growth hormone released by protein. If the protein is efficient more growth hormone will be released and plant will be taller.But tf the protein in inefficient the plant will be shorter.  Thus, genes control the traits.
4)     Describe the human evolution.
Ans- All human beings belong to a single species that evolved in Africa.
Few thosand years ago some ancestors left Africa while other stayed back.
The residents spread across Africa and the migrants spread acrossthe planet from Africa to West Asia , Central Asia,
Eurasia, South asia, Indonesia,  Australia and America.
They went forwards and backwards with groups separating from each other, or sometime comong together.
Like all other vspecies, they  were also living their lives to the best of thiir ability.
5) Discuss law of segregation or purity of gametes.      
Ans- Both factors or traits are present in pairs and during gamet formation, the two factors segregate and one gamete is pure for one character only.After fertilization one trait from male and other trait from female fuses to restore the paired feature in zygote.
6) What is speciation? How it occurs?
Ans- Origin of new species from existing one, is called speciation. It nay occur due to the following reasons-         Due to mutation  Due to natural selection.
7) Why are traits acquired during the lifetime of an individual not inherited?
Ans- Only the traits which are passed on to the DNA of germ cells are inheritedto the progeny of the organisms reproducing sexually. Traits acquired during lifetime do not pass on to the DNA of germ cells and they are not inherited.
8) Who was Mendel? Why was he called the father of ganetics?
Ans- Gregor Johann Mendel was an Australian geneticist. he sowed the garden pea and fond flowers of different colours.He formulated the law of heredity by his experiments. His law gave the base for futher study of genetics. Threrfore, he was called the father of Genetics.
9).(a) Why Mendel selected garden pea plant for his experiments. List a reason
(b)What are recessive and dominant genes?
Ans- (a) Mendel selected garden pea plant for his experiments as these can be self pollinated.
 (b) Recessive genes are not able to express but dominating genes express
10) What is genetics? State any two factors that could lead to the rise of new species.
Ans: Genetics is a field of biology that deals with the study of genes, heredity and variation of different organisms.

Two factors that could give rise to new species  includes:
1. Geographical isolation of individuals of a particular population into two or more groups.
2. Genetic drift.

11) When is a recessive trait capable of expressing itself? Write its expression with respect to height of plant (Genotype).

Ans: A recessive trait can be defined as the trait which is expressed only when the recessive alleles are present in a homozygous condition, which means that the individual has both the alleles as recessive alleles.
In pea plant, short plant height is the recessive trait. It is expressed when the individual plant has both the alleles as 't'. So the genotype of a plant with short height will be tt.
12) When a pesticide is sprayed on a population of insects,all insects do not get killed but few of them survive. Give reason.

Ans When a pesticide is sprayed on a field most of the insects are killed because they are sensitive to the pesticide. A few of the insects survive because they have variations which made them resistant against the pesticide, such that the pesticide has no effect on these insects.

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca