Biology
Board Question Paper : July 2019 Solutions
SECTION – A
Q.1.
_______ drug is used for patients who have undergone surgery.
Answer:
(C) Morphine
Q.2.
Name the process by which all the three types of non-genetic RNAs are produced on DNA template.
Answer:
(B) Transcription
Q.3.
Which of the animal groups show uricotelism?
Answer:
(C) Land snail, bird, lizard
Q.4.
Approximately how many eggs are produced by a normal healthy human female up to the age of 25 years if the age of menarche is 12 years _______.
Answer:
(A) 169
Calculation: (25 years - 12 years) = 13 reproductive years. 13 years × 13 eggs/year = 169 eggs.
Calculation: (25 years - 12 years) = 13 reproductive years. 13 years × 13 eggs/year = 169 eggs.
Q.5.
Name the process in which a tumour successfully spreads to the other parts of the body, grows and destroys healthy tissues.
Answer:
Metastasis
Q.6.
What is humification?
Answer:
Humification is the process of decomposition of detritus to form a dark coloured, amorphous, colloidal substance called humus. It is highly resistant to microbial action and undergoes decomposition at an extremely slow rate.
Q.7.
Name the sexually transmitted disease caused by Treponema pallidum.
Answer:
Syphilis
Q.8.
Which is the process that removes introns from RNA?
Answer:
Splicing (or RNA Splicing)
HSC Biology
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SECTION – B
Q.9.
Define fermentation. Write the names of substrate of alcoholic and lactic acid fermentation.
Answer:
Definition: Fermentation is the anaerobic breakdown of glucose or other organic nutrients into simpler compounds like alcohol or lactic acid along with the release of a small amount of energy.
Substrate: The primary substrate for both alcoholic and lactic acid fermentation is Glucose.
Substrate: The primary substrate for both alcoholic and lactic acid fermentation is Glucose.
Q.10.
Complete the following chart and rewrite it:
| Genotype | Phenotype |
|---|---|
| \(I^A I^A\) or \(I^A i\) | Blood Group A |
| \(I^B I^B\) or \(I^B i\) | B |
| \(I^A I^B\) | Blood Group AB |
| ii | O |
Answer:
The completed chart is filled above in bold.
Q.11.
Your friend wants to start a business of Apiculture. Enlist the equipment he would need.
Answer:
Equipment needed for Apiculture includes:
- Bee hive (Artificial wooden box) with frames and foundation sheets
- Hive stand
- Smoker (to calm bees)
- Bee veil / Protective clothing (gloves, overall)
- Queen Excluder
- Honey Extractor
- Uncapping knife
- Brush
Q.12.
Give the role of:
i. Tissue plasminogen activator (TPA)
ii. Tissue growth factor-Beta (TGF-β) in Gene therapy.
i. Tissue plasminogen activator (TPA)
ii. Tissue growth factor-Beta (TGF-β) in Gene therapy.
Answer:
i. Tissue plasminogen activator (TPA): It is an enzyme used as a thrombolytic agent (clot buster) to dissolve blood clots in blood vessels, useful in treating heart attacks and strokes.
ii. Tissue growth factor-Beta (TGF-β): It is used to stimulate cell growth, new blood vessel formation (angiogenesis), and promotes wound healing and tissue repair.
ii. Tissue growth factor-Beta (TGF-β): It is used to stimulate cell growth, new blood vessel formation (angiogenesis), and promotes wound healing and tissue repair.
Q.13.
Match the following and rewrite it:
Answer:
- i. Invertase — b. Saccharomyces cerevisiae
- ii. Lipase — d. Rhizopus spp. (or Candida)
- iii. Cellulase — a. Trichoderma konigi
- iv. Pectinase — c. Sclerotinia libertinia
Q.14.
(Diagram should show the Clover leaf structure with: 3' Acceptor arm with CCA end, 5' end, TΨC loop, DHU loop, Anticodon loop with anticodon triplet, and Variable arm.)
Sketch and label hairpin model of tRNA.
(Diagram should show the Clover leaf structure with: 3' Acceptor arm with CCA end, 5' end, TΨC loop, DHU loop, Anticodon loop with anticodon triplet, and Variable arm.)
Answer:
See diagram description above.
Q.15.
[Image References from Question Paper]
Identify and write the names of given diagrams A, B, C and D.
[Image References from Question Paper]
Answer:
A: Eosinophil (Acidophil)
B: Neutrophil
C: Lymphocyte
D: Monocyte
B: Neutrophil
C: Lymphocyte
D: Monocyte
OR
Dilip and Mohsin measured their blood pressure. Dilip’s B.P. is 120/80 mmHg and Mohsin’s B.P. is 160/100 mmHg. Who is suffering from hypertension? What are its causes?
Dilip and Mohsin measured their blood pressure. Dilip’s B.P. is 120/80 mmHg and Mohsin’s B.P. is 160/100 mmHg. Who is suffering from hypertension? What are its causes?
Answer:
Who is suffering: Mohsin is suffering from hypertension (High Blood Pressure).
Causes: Stress, obesity, high dietary salt intake, smoking, alcoholism, arteriosclerosis (hardening of arteries), kidney disorders, or genetic factors.
Causes: Stress, obesity, high dietary salt intake, smoking, alcoholism, arteriosclerosis (hardening of arteries), kidney disorders, or genetic factors.
Q.16.
Give the functions of Kidney.
Answer:
Functions of Kidney:
- Excretion: Removal of nitrogenous metabolic wastes like urea and uric acid.
- Osmoregulation: Maintenance of water and salt balance (homeostasis) in the body.
- pH Regulation: Maintaining the acid-base balance of blood.
- Secretion: Produces hormones like Erythropoietin (for RBC formation) and Renin (for BP regulation).
Q.17.
Give the location of following valves within human heart:
i. Eustachian valve
ii. Thebesian valve
iii. Bicuspid valve
iv. Tricuspid valve
i. Eustachian valve
ii. Thebesian valve
iii. Bicuspid valve
iv. Tricuspid valve
Answer:
i. Eustachian valve: Guards the opening of the Inferior Vena Cava in the Right Atrium.
ii. Thebesian valve: Guards the opening of the Coronary Sinus in the Right Atrium.
iii. Bicuspid (Mitral) valve: Located between the Left Atrium and Left Ventricle.
iv. Tricuspid valve: Located between the Right Atrium and Right Ventricle.
ii. Thebesian valve: Guards the opening of the Coronary Sinus in the Right Atrium.
iii. Bicuspid (Mitral) valve: Located between the Left Atrium and Left Ventricle.
iv. Tricuspid valve: Located between the Right Atrium and Right Ventricle.
Q.18.
Define Green House Gases. Give any two examples.
Answer:
Definition: Greenhouse gases are atmospheric gases that absorb infrared radiation (heat) emitted from the Earth's surface and re-radiate it back, trapping heat in the atmosphere and causing the greenhouse effect.
Examples: Carbon dioxide (\(CO_2\)), Methane (\(CH_4\)), Chlorofluorocarbons (CFCs), Nitrous oxide (\(N_2O\)).
Examples: Carbon dioxide (\(CO_2\)), Methane (\(CH_4\)), Chlorofluorocarbons (CFCs), Nitrous oxide (\(N_2O\)).
SECTION – C
Q.19.
Explain Homologous and Analogous organs with example.
Answer:
1. Homologous Organs:
- Organs that have the same fundamental structure and embryonic origin but perform different functions.
- They indicate divergent evolution and common ancestry.
- Example: Forelimbs of Human (for grasping), Whale (for swimming), Bat (for flying), and Cheetah (for running). All share the same bone structure (humerus, radius, ulna, etc.) but differ in function.
- Organs that have different structural details and embryonic origins but perform the same function.
- They indicate convergent evolution.
- Example: Wings of a Butterfly (chitinous membrane) and Wings of a Bird (feathers and bones). Both are used for flight but are structurally different.
Q.20.
A homozygous tall pea plant is crossed with its homozygous recessive parent. Find out the genotypic and phenotypic ratio with the help of Punnet square method.
Answer:
Parents: Homozygous Tall (TT) × Homozygous Dwarf (tt)
Gametes: (T) and (t)
Punnett Square (F1 Generation):
F1 Result: All offspring are Heterozygous Tall (Tt).
(Note: If the question implies finding the F2 ratio by selfing the F1 generation, which is standard for such questions):
Selfing F1 (Tt × Tt):
F2 Phenotypic Ratio: 3 Tall : 1 Dwarf
F2 Genotypic Ratio: 1 Homozygous Tall (TT) : 2 Heterozygous Tall (Tt) : 1 Homozygous Dwarf (tt)
Gametes: (T) and (t)
Punnett Square (F1 Generation):
| ♂ \ ♀ | t | t |
|---|---|---|
| T | Tt (Tall) | Tt (Tall) |
| T | Tt (Tall) | Tt (Tall) |
F1 Result: All offspring are Heterozygous Tall (Tt).
(Note: If the question implies finding the F2 ratio by selfing the F1 generation, which is standard for such questions):
Selfing F1 (Tt × Tt):
F2 Phenotypic Ratio: 3 Tall : 1 Dwarf
F2 Genotypic Ratio: 1 Homozygous Tall (TT) : 2 Heterozygous Tall (Tt) : 1 Homozygous Dwarf (tt)
Q.21.
(Diagram should show: Afferent arteriole, Efferent arteriole, Glomerulus (capillary knot), Bowman’s capsule (Cup shape), Parietal layer, Visceral layer, Capsular space, and PCT.)
Sketch and label the structure of Malpighian body and explain the structure of Bowman’s capsule.
(Diagram should show: Afferent arteriole, Efferent arteriole, Glomerulus (capillary knot), Bowman’s capsule (Cup shape), Parietal layer, Visceral layer, Capsular space, and PCT.)
Answer:
Structure of Bowman’s Capsule:
- It is a double-walled, cup-shaped structure located at the beginning of the nephron.
- It consists of an outer Parietal layer composed of simple squamous epithelium and an inner Visceral layer composed of specialized cells called Podocytes.
- Podocytes have foot-like processes (pedicels) that wrap around glomerular capillaries, leaving slit pores for filtration.
- The space between the two layers is called the Capsular space or urinary space, which receives the glomerular filtrate.
Q.22.
[Image References from Question Paper]
Write down the names of missing intermediate compounds in a sequence in the given diagrammatic representation of Kreb’s cycle.
[Image References from Question Paper]
Answer:
Based on the sequence of Kreb's cycle:
- Acetyl Co-A (Enters the cycle to combine with Oxaloacetate)
- Oxaloacetate (Combines with Acetyl Co-A to form Citrate)
- Oxalosuccinate (Intermediate between Isocitrate and \(\alpha\)-Ketoglutarate)
- Succinyl Co-A (Formed from \(\alpha\)-Ketoglutarate)
- Succinate (Formed from Succinyl Co-A)
- Malate (Formed from Fumarate)
Q.23.
Define jumping genes. Classify them on the basis of their mechanism.
Answer:
Definition: Jumping genes, or Transposons, are DNA sequences that have the ability to move (transpose) from one location to another within the genome.
Classification based on mechanism:
Classification based on mechanism:
- Retrotransposons (Class I): They move via a "copy and paste" mechanism. The DNA is transcribed into RNA, and then reverse-transcribed back into DNA which is inserted at a new location.
- DNA Transposons (Class II): They move via a "cut and paste" mechanism. The DNA segment is excised from its original position and inserted directly into a new location.
Q.24.
[Image References from Question Paper]
Identify A, B, C in the given diagram and give their functions.
[Image References from Question Paper]
Answer:
A: Acrosome - It contains hydrolytic enzymes (hyaluronidase) that help the sperm penetrate the egg during fertilization.
B: Mitochondria (Nebenkern) / Middle Piece - It provides energy (ATP) for the movement of the sperm.
C: Tail (Flagellum) - It provides motility to the sperm, allowing it to swim towards the ovum.
B: Mitochondria (Nebenkern) / Middle Piece - It provides energy (ATP) for the movement of the sperm.
C: Tail (Flagellum) - It provides motility to the sperm, allowing it to swim towards the ovum.
OR
Explain various mechanical methods of birth control.
Explain various mechanical methods of birth control.
Answer:
Mechanical methods (Barrier methods) prevent the physical meeting of sperm and ovum:
- Condom (Nirodh): A rubber/latex sheath worn over the penis (male) or inside the vagina (female) to collect semen and prevent it from entering the uterus. It also protects against STDs.
- Diaphragm / Cervical Cap: Rubber domes inserted into the female reproductive tract to cover the cervix, blocking sperm entry.
- Intrauterine Devices (IUDs): Devices like Lippes loop, Cu-T, or Multiload 375 inserted into the uterus. They increase phagocytosis of sperm and suppress sperm motility/fertilizing capacity. (Sometimes classified separately, but mechanically block implantation).
Q.25.
Identify disorders developed in the given genotypes and give two symptoms of each:
i. 44 + XO
ii. 44 + XXY
i. 44 + XO
ii. 44 + XXY
Answer:
i. Genotype 44 + XO: Turner’s Syndrome
Symptoms:
ii. Genotype 44 + XXY: Klinefelter’s Syndrome
Symptoms:
Symptoms:
- Sterile female with rudimentary ovaries.
- Short stature, webbed neck, and shield-shaped chest.
ii. Genotype 44 + XXY: Klinefelter’s Syndrome
Symptoms:
- Sterile male with under-developed testes (Hypogonadism).
- Development of breast in males (Gynecomastia) and sparse body hair.
Q.26.
Name the interaction in:
i. Lichen
ii. Sucker fish and shark
iii. A protozoan living in the digestive tract of a flea living on a dog.
i. Lichen
ii. Sucker fish and shark
iii. A protozoan living in the digestive tract of a flea living on a dog.
Answer:
i. Lichen: Mutualism (Symbiotic relationship between Algae and Fungi).
ii. Sucker fish and Shark: Commensalism (Sucker fish benefits from transport/food, Shark is unaffected).
iii. Protozoan in flea on dog: Hyperparasitism (The protozoan is a parasite of the flea, which is itself a parasite of the dog).
ii. Sucker fish and Shark: Commensalism (Sucker fish benefits from transport/food, Shark is unaffected).
iii. Protozoan in flea on dog: Hyperparasitism (The protozoan is a parasite of the flea, which is itself a parasite of the dog).
Q.27.
Given an account of various steps involved in tissue culture.
Answer:
The steps involved in tissue culture are:
- Explant Selection: Selecting a healthy plant part (shoot tip, leaf, etc.) to be cultured.
- Sterilization: Sterilizing the glassware, nutrient medium, and surface sterilization of the explant to prevent microbial contamination.
- Inoculation: Transferring the explant onto the nutrient medium in aseptic conditions (Laminar air flow).
- Incubation: Keeping the culture in a controlled environment (light, temperature, humidity) to allow growth.
- Callus Formation: The explant cells divide to form an undifferentiated mass of cells called Callus.
- Organogenesis: Differentiation of callus into roots and shoots by altering the ratio of hormones (Auxins and Cytokinins).
- Hardening: Gradual acclimatization of plantlets to the natural environment.
- Transfer: Planting the hardened plantlets in the field.
SECTION – D
Q.28.
(Diagram should show Mesophyll cell and Bundle Sheath cell interaction: CO2 fixed by PEP to OAA -> Malate -> Transport to BS cell -> Decarboxylation to release CO2 -> Calvin Cycle -> Pyruvate -> Transport back to Mesophyll -> Regeneration of PEP.)
Give the diagrammatic representation of HSK-pathway and answer the following questions:
i. Why is photorespiration avoided in C4 pathways?
ii. Give any two examples of C4 plants.
iii. Name the CO2 acceptor in mesophyll cells during HSK pathway.
i. Why is photorespiration avoided in C4 pathways?
ii. Give any two examples of C4 plants.
iii. Name the CO2 acceptor in mesophyll cells during HSK pathway.
(Diagram should show Mesophyll cell and Bundle Sheath cell interaction: CO2 fixed by PEP to OAA -> Malate -> Transport to BS cell -> Decarboxylation to release CO2 -> Calvin Cycle -> Pyruvate -> Transport back to Mesophyll -> Regeneration of PEP.)
Answer:
i. Avoidance of Photorespiration: In C4 plants, the enzyme RuBisCO is present only in the bundle sheath cells. The C4 mechanism concentrates \(CO_2\) in these cells (by decarboxylation of malate), ensuring a high \(CO_2\) to \(O_2\) ratio. This prevents RuBisCO from acting as an oxygenase, thus avoiding photorespiration.
ii. Examples: Maize (Corn), Sugarcane, Sorghum.
iii. CO2 Acceptor: Phosphoenolpyruvate (PEP).
ii. Examples: Maize (Corn), Sugarcane, Sorghum.
iii. CO2 Acceptor: Phosphoenolpyruvate (PEP).
OR
Identify and explain with the help of diagrammatic representation, type of photophosphorylation in which P700 (PS II) and P680 (PS I) both are involved.
Identify and explain with the help of diagrammatic representation, type of photophosphorylation in which P700 (PS II) and P680 (PS I) both are involved.
Answer:
Identification: The process involves both Photosystem II (P680) and Photosystem I (P700), so it is Non-Cyclic Photophosphorylation (Z-Scheme).
Explanation:
Explanation:
- Light hits PS II (P680), exciting electrons which are accepted by a primary acceptor.
- Electrons flow down an electron transport chain (Plastoquinone -> Cytochrome b6f -> Plastocyanin) to PS I.
- During this flow, ATP is synthesized from ADP + Pi.
- Simultaneously, photolysis of water occurs at PS II to replace lost electrons, releasing \(O_2\).
- Light hits PS I (P700), exciting electrons which reduce NADP+ to NADPH via Ferredoxin.
- Result: Synthesis of ATP and NADPH, and release of Oxygen.
Q.29.
Give reasons:
i. Pituitary gland was formerly called as ‘master endocrine gland’.
ii. Oxytocin is ‘birth hormone’.
iii. People living in hilly region are advised to use iodised salt.
iv. Old age persons show weakened immune response.
v. Pancreas is a dual gland.
i. Pituitary gland was formerly called as ‘master endocrine gland’.
ii. Oxytocin is ‘birth hormone’.
iii. People living in hilly region are advised to use iodised salt.
iv. Old age persons show weakened immune response.
v. Pancreas is a dual gland.
Answer:
i. The Pituitary gland secretes hormones that control and regulate the secretions of many other endocrine glands (like Thyroid, Adrenal cortex, Gonads). Hence, it was called the master gland (though it is itself controlled by the Hypothalamus).
ii. Oxytocin stimulates vigorous contraction of the smooth muscles of the uterus at the end of pregnancy, causing expulsion of the fetus (parturition). Hence, it is called the birth hormone.
iii. The soil in hilly regions is often deficient in iodine, leading to iodine-deficient food. Iodine is essential for Thyroxine synthesis. To prevent Goiter (thyroid enlargement) caused by this deficiency, iodised salt is advised.
iv. The Thymus gland, which acts as a training school for T-lymphocytes, degenerates with age. This leads to a decrease in the production of T-cells and cell-mediated immunity, causing a weakened immune response in old age.
v. Pancreas acts as both an exocrine gland (secreting digestive enzymes like trypsin, lipase via ducts) and an endocrine gland (Islets of Langerhans secreting hormones Insulin and Glucagon into the blood). Hence, it is a dual (heterocrine) gland.
ii. Oxytocin stimulates vigorous contraction of the smooth muscles of the uterus at the end of pregnancy, causing expulsion of the fetus (parturition). Hence, it is called the birth hormone.
iii. The soil in hilly regions is often deficient in iodine, leading to iodine-deficient food. Iodine is essential for Thyroxine synthesis. To prevent Goiter (thyroid enlargement) caused by this deficiency, iodised salt is advised.
iv. The Thymus gland, which acts as a training school for T-lymphocytes, degenerates with age. This leads to a decrease in the production of T-cells and cell-mediated immunity, causing a weakened immune response in old age.
v. Pancreas acts as both an exocrine gland (secreting digestive enzymes like trypsin, lipase via ducts) and an endocrine gland (Islets of Langerhans secreting hormones Insulin and Glucagon into the blood). Hence, it is a dual (heterocrine) gland.
OR
Describe functional areas of cerebrum with the help of neat and labelled diagram.
Describe functional areas of cerebrum with the help of neat and labelled diagram.
Answer:
Functional Areas of Cerebrum:
[Image: Diagram of Lateral view of Human Brain showing lobes and functional areas]
- Frontal Lobe: Contains the Motor area (controls voluntary movements), Broca’s area (motor speech area), and areas for intellect, memory, and will-power.
- Parietal Lobe: Contains the Somatosensory area (perception of touch, pain, pressure, temperature, taste).
- Temporal Lobe: Contains the Auditory area (hearing), Olfactory area (smell), and Wernicke’s area (understanding speech).
- Occipital Lobe: Contains the Visual area (sight/vision).
[Image: Diagram of Lateral view of Human Brain showing lobes and functional areas]
Q.30.
(Diagram of Inverted ovule showing Funiculus, Hilum, Outer/Inner Integuments, Micropyle, Nucellus, Chalaza, and Embryo Sac with Egg apparatus, Polar nuclei, Antipodals.)
Define pollination. Explain different types of self and cross pollination with suitable examples.
Answer:
Definition: Pollination is the transfer of pollen grains from the anther to the stigma of the same or a different flower.
1. Self Pollination: Transfer of pollen to the stigma of the same flower or another flower on the same plant.
1. Self Pollination: Transfer of pollen to the stigma of the same flower or another flower on the same plant.
- Autogamy: Pollination within the same flower. E.g., Pea.
- Geitonogamy: Pollination between different flowers of the same plant. E.g., Cucurbits.
- Anemophily (Wind): E.g., Maize, Wheat.
- Hydrophily (Water): E.g., Vallisneria, Zostera.
- Entomophily (Insects): E.g., Rose, Jasmine.
- Ornithophily (Birds): E.g., Butea, Bombax.
OR
Sketch and label the V.S. of anatropous ovule and answer the following questions:
i. How many mitotic divisions are required to produce embryo sac?
ii. Which part of ovule is converted into seed coat?
iii. Which part provides the passage for entry of pollen tube during fertilization?
Sketch and label the V.S. of anatropous ovule and answer the following questions:
i. How many mitotic divisions are required to produce embryo sac?
ii. Which part of ovule is converted into seed coat?
iii. Which part provides the passage for entry of pollen tube during fertilization?
(Diagram of Inverted ovule showing Funiculus, Hilum, Outer/Inner Integuments, Micropyle, Nucellus, Chalaza, and Embryo Sac with Egg apparatus, Polar nuclei, Antipodals.)
Answer:
i. Mitotic Divisions: After meiosis forms the functional megaspore, 3 successive free nuclear mitotic divisions occur to produce the 8-nucleate, 7-celled embryo sac.
ii. Part converted to seed coat: The Integuments (Outer integument forms Testa, Inner integument forms Tegmen).
iii. Passage for pollen tube: The Micropyle.
ii. Part converted to seed coat: The Integuments (Outer integument forms Testa, Inner integument forms Tegmen).
iii. Passage for pollen tube: The Micropyle.