Question 1

Question 1(a)Name the following:(i) The cell body of a nerve cell.(ii) The waxy layer on the epidermis of the leaf meant to reduce transpiration.(iii) A non-biodegradable pesticide.(iv) The physical expression of genes in an individual.(v) Knot-like mass of blood capillaries inside the bowman's capsule. [5]

Accepted Answer

(i) Cyton(ii) Cuticle(iii) DDT(iv) Phenotype(v) Glomerulus

Question 2

Question 1(b)State the exact location of the following:(i) Chloroplast(ii) Incus(iii) Corpus callosum(iv) Guard cells(v) Pulmonary semilunar valve [5]

Accepted Answer

(i) Chloroplast are present in mesophyll cells located between the upper epidermis and the lower epidermis (i.e. in palisade cells and spongy cells) of leaves. They are also found in the guard cells of stomata and in the outer layers of young green stems.(ii) Incus is located between Malleus and Stapes of ear ossicle which is found in the middle ear.(iii) Corpus callosum is located between the two cerebral hemispheres.(iv) Guard cells are located in Stomata.(v) Pulmonary semilunar valves are located at the opening of the right ventricle into the pulmonary artery.

Question 3

Question 1(g)The figure given below represents an experiment to demonstrate a particular aspect of photosynthesis. The alphabet 'A' represents a certain condition inside the flask.(i) What is the aim of the experiment ?(ii) Identify the special condition inside the flask.(iii) Name an alternative chemical that can be used instead of KOH.(iv) In what manner do the leaves 1 and 2 differ at the end of the starch test? [5]

Accepted Answer

(i) The aim of the experiment is to show that plants require Carbon dioxide (CO2) for photosynthesis.(ii) There is no Carbon dioxide (CO2) present inside the flask because KOH absorbs it all.The leaf inside the flask do not turn blue black when tested with iodine as the leaf inside the flask does not get atmospheric Carbon dioxide (CO2) because KOH absorbs all the Carbon dioxide (CO2) in the flask.(iii) An alternative chemical that can be used instead of KOH is NaOH (Sodium hydroxide).(iv) Leaf 1 does not turn blue black at the end of starch test because due to lack of Carbon dioxide (CO2) inside the flask, it is not able to perform photosynthesis and produce starch. Leaf 2 on the other hand, gets atmospheric Carbon dioxide (CO2)​. Hence, it is able to perform photosynthesis as it turns blue black after the starch test.

Question 4

Question 3(a)Draw a diagram of the human eye as seen in a vertical section and label the parts which suits the following descriptions relating to the:(i) photosensitive layer of the eye.(ii) strucutre which is responsible for holding the eye lens in its position.(iii) structure which maintains the shape of the eye ball and the area of no vision.(iv) anterior chamber seen in front of the eye lens.(v) outer most transparent layer seen in front of the eye ball. [5]

Accepted Answer

Labelled diagram of the human eye is given below:

Question 5

Question 3(b)Differentiate between the following pairs on the basis of what is mentioned in brackets :(i) Photolysis and Photophosphorylation. (Definition)(ii) Bicuspid valve and Tricuspid valve. (Function).(iii) Vasectomy and Tubectomy. (Explain)(iv) Cerebrum and Spinal cord. (Arrangement of nerve cells)(v) Bowman's capsule and Malpighian capsule. (Parts included) [5]

Accepted Answer

(i) Difference between Photolysis and Photophosphorylation. (Definition)PhotolysisPhotophosphorylationChlorophyll gets activated by absorbing light energy. The absorbed energy is used in splitting the water molecule (H2O) into its two components (Hydrogen and Oxygen) and releasing electrons. This reaction is known as Photolysis.Formation of ATP from ADP and inorganic phosphate by the utilization of energy released by electrons emitted due to photolysis of water. This process is called Photophosphorylation.(ii) Difference between Bicuspid valve and Tricuspid valve (Function).Bicuspid valveTricuspid valvePrevents the reverse flow of blood from left ventricle to left atriumPrevents the reverse flow of blood from right ventricle to right atrium(iii) Difference between Vasectomy and Tubectomy. (Explain)VasectomyTubectomyIt is the process in which vas deferens from each testes is ligated and a small piece between the two ligatures is removed. It is contraceptive practice in males.Tubectomy is the process which involves cutting or ligating of fallopian tubes by nylon thread to close the passage of egg. It is contraceptive practice in females.(iv) Difference between Cerebrum and Spinal cord. (Arrangement of nerve cells)CerebrumSpinal cordThe grey matter containing cytons lies in the cortex (outer region) while the white matter containing axons lies in the medullary region (inner region).The grey matter containing cytons lies in the medullary region (inner side) while the white matter containing axons lies in the cortex i.e. (outer region).(v) Difference between Bowman's capsule and Malpighian capsule. (Parts included)Bowman's capsuleMalpighian capsuleBowman's capsule is a thin walled (single-cell thick epithelium) cup, containing the glomerulus.The Bowman's capsule along with the knot-like mass of blood capillaries called glomerulus together are known as Malpighian capsule.

Question 6

Question 4(a)Given below is a schematic diagram showing Mendel's Experiment on sweet pea plants having axial flowers with round seeds (AARR) and Terminal flowers with wrinkled seeds (aarr). Study the same and answer the questions that follow:(i) Give the phenotype of F1 progeny.(ii) Give the phenotypes of F2 progeny produced upon by the self-pollination of progeny.(iii) Give the phenotypic ratio of F2 progeny.(iv) Name and explain the law induced by Mendel on the basis of the above observation. [5]

Accepted Answer

(i) The phenotype of F1 progeny will be Axial Round as it is dominant phenotype.(ii) The phenotypes of F2 progeny will be Axial Round, Axial Wrinkled, Terminal Round and Terminal Wrinkled as in F2 generation, recessive trait is also expressed.(iii) Phenotypic ratio of F2 progeny will be 9:3:3:1.(iv) Above observation are based on Law of Independent Assortment. This law states that factors or genes controlling different characters assort independently without influencing each other during the formation of gametes.

Question 7

Question 4(b)Complete the following table by filling in the blanks from 1 to 10 with appropriate terms. [5]S.No.GlandSecretionFunction/Effect on body1Thyroid1223Vasopressin4356Promotes glucose utilization by the body cells4Lacrimal gland785Adrenal medulla910

Accepted Answer

S.No.GlandSecretionFunction / Effect on body1Thyroid(1) Thyroxine(2) Promotes tissue metabolism, growth and differentiation.2(3) Posterior pituitaryVasopressin(4) Increases reabsorption of water from kidneys.3(5) Pancreas(6) InsulinPromotes glucose utilization by the body cells.4Lacrimal gland(7) Tears(8) Lubricates eye surface and kills germs5Adrenal medulla(9) Adrenaline(10) Prepares body for any emergency

Question 8

Question 5(b)Give biological reasons for the following :(i) The wall of the ventricle is thicker than the auricles.(ii) The renal cortex has a dotted appearance.(iii) Wooden frames of doors get jammed during the monsoon season.(iv) Throat infections can lead to ear infections.(v) The hand automatically shows the direction to turn a cycle without thinking. [5]

Accepted Answer

(i) Ventricles have thicker walls than auricles because they have to pump blood to different organs and the pressure with which the blood flows through them is more than the auricles. Therefore, the walls are thicker in ventricles than in auricles.(ii) As the cortex region contains numerous nephrons or kidney tubules, therefore, it shows a dotted appearance.(iii) The wooden frames absorb moisture from the atmosphere due to imbibition and swell up. This causes them to get jammed during the monsoon season.(iv) The Eustachian tube is a part of the middle ear. It also connects the cavity of the middle ear to the throat. Hence, throat infections can lead to ear infections.(v) It occurs due to conditioned reflexes. These reflexes are learned responses to stimuli acquired individually during the life of an organism.

Question 9

Question 6(a)The figure given below shows the epidermal cells of an onion bulb. This cell was then transferred to a drop of sugar solution.(i) Draw a well labelled diagram of the epidermal cell as it would appear after immersion in a strong sugar solution.(ii) What scientific term is used for the changes as shown in (i) above ?(iii) What should be done to restore the cell back to its original condition ?(iv) Give the scientific term for the recovery of the cell as a result of the step taken in (iii) above.(v) Define the term osmosis. [5]

Accepted Answer

(i) Diagram of the plasmolysed epidermal cell after immersing it in a strong sugar solution is shown below:(ii) Plasmolysis is technical term used for the above change.(iii) To restore the cell back to its original condition, it must be placed in hypotonic solution immediately after it gets plasmolysed.(iv) Deplasmolysis is technical term used for recovery of the cell as a result of the step taken in (iii) above.(v) Osmosis is the movement of water molecules from their region of higher concentration (dilute solution or with a lower solute concentration) to their region of lower concentration (concentrated solution or with a higher solute concentration) through a semi permeable membrane.

Question 10

Question 6(b)Briefly explain the following terms:(i) Genes.(ii) Cytokinesis in plant cells.(iii) Guttation.(iv) Diabetes insipidus.(v) Disinfectants. [5]

Accepted Answer

(i) Genes are specific sequences of nucleotides on a chromosome that encode particular proteins which express in the form of some particular feature of the body. They are the units of heredity which are transferred from parents to offsprings and are responsible for some specific characteristics of the offspring.(ii) Cytokinesis is the division of cytoplasm. In plant cells, a cell plate appearing at the centre of the middle grows up to the periphery, splitting the cytoplasm into two and thus producing two new cells.(iii) Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure. It occurs early mornings or late nights.(iv) Deficiency of Vasopressin (Anti-diuretic hormone or ADH) causes Diabetes Insipidus in which urination is frequent and copious, resulting in loss of water from the body and the person becomes thirsty.(v) These are the chemicals that kill microbes once they come in contact with them, they are too strong to be used on the body. Example — Phenol and Cresol.

Question 11

Question 7(a)(i) Draw a well labelled diagram to show the anaphase stage of mitosis in a plant cell having four chromosomes.(ii) State any two harmful effects of acid rain.(iii) Expand the following biological abbreviations :NADPACT [5]Answer(i) Diagram showing the anaphase stage of mitosis in a plant cell having four chromosomes is shown below:(ii) Two harmful effects of acid rain are:Damage to vegetation by pollution of the soil.Fish and other aquatic organisms are harmed due to increased acidity of water in lakes and rivers.(iii) The full forms are mentioned below:NADP — Nicotinamide Adenine Dinucleotide PhosphateACT — Adrenocorticotropin.

Accepted Answer

Answer: [5], NADP, ACT

Question 12

Question 7(b)(i) List any two major activities of the Red Cross.(ii) Write any two major reasons for the population explosion in the world.(iii) Write the names of four nitrogenous bases in a DNA molecule. [5]

Accepted Answer

(i) Two major activities of the Red Cross are:To extend relief and help to the victims of any calamity — fire, flood, famine, earthquakes etc.To procure and supply blood for the victims of war and other calamities.(ii) Two major reasons for the population explosion in the world are:Better health care at all age groups and fewer deaths due to better medical aids.Large scale immunization against communicable diseases.(iii) The four nitrogenous bases in the DNA molecule are Guanine, Thymine, Adenine and Cytosine.

Solved 2013 Question Paper ICSE Class 10 Biology

Section I 40 Marks

Section Ii 40 Marks