👉 Download Biology Question Paper PDF - SET 10
PART A: BOTANY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
✅ Rhodophyceae
PS II: ……………….
✅ Red light of 680 nm
(a) ABA (b) NAA (c) IAA (d) 2, 4-D
✅ (a) ABA
✅ Interkinesis
(a) Pasteur (b) Ivanowsky
(c) Beijerinck (d) Stanley
✅ (c) Beijerinck
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
(a) Actinomorphic flowers and Zygomorphic flowers.
(b) Trimerous flowers and Tetramerous flowers.
✅ Answer
(a) Actinomorphic flowers (radial symmetry): Here, a flower can be divided into 2 equal radial halves in any radial plane.Zygomorphic (bilateral symmetry): Here, a flower can be divided into two similar halves only in a particular vertical plane.
(b) Trimerous flowers: Here, floral appendages are multiple of 3.
Tetramerous flowers: Here, floral appendages are multiple of 4.
✅ Answer
A= Chloroplasts B= LeucoplastsC= Amyloplasts D= Aleuroplasts
✅ Answer
A= Aleurone layer B= EndospermC= Scutellum D= Radicle
a) Photorespiration is a wasteful process.
b) Photorespiration does not occur in C4 plants.
✅ Answer
a) In Photorespiration, there is no synthesis of sugars, ATP and NADPH. So Photorespiration is a wasteful process.b) Photorespiration does not occur in C4 plants because they can increase CO2 concentration at the enzyme site. This takes place when C4 acid from the mesophyll is broken down in the bundle cells to release CO2. This minimises the oxygenase activity of RuBisCO.
✅ Answer
• Energy production is limited. Less than 7% of the energy in glucose is released and not all of it is trapped as high energy bonds of ATP.• Hazardous products (acid or alcohol) are formed.
✅ Answer
The sexual cycle involves the following three steps:(i) Fusion of protoplasm between two motile or non-motile gametes called plasmogamy.
(ii) Fusion of two nuclei called karyogamy.
(iii) Meiosis in zygote resulting in haploid spores
✅ Answer
Vascular bundles of dicot stem: Large in number. Conjoint & open. Their ring arrangement is a characteristic of dicot stem. Protoxylem is endarch.Vascular bundles of monocot stem: Many & Scattered. Conjoint & closed. Peripheral vascular bundles are smaller than central ones.
✅ Answer
(a) (X) = Metacentric (Y) = Submetacentric(b) Some chromosomes have non-staining secondary constrictions at a constant location. It is called satellite.
(b) How does the shape of guard cells of dicot plants differ from that of monocot plants?
✅ Answer
(a) Stomata regulate the transpiration and gaseous exchange.(b) In dicot plants, guard cells of dicot plants are bean-shaped.
In monocot plants (grasses) they are dumbbell shaped.
(b) How temperature affects the rate of photosynthesis?
✅ Answer
(a) “If a biochemical process is affected by more than one factor, its rate is determined by the factor nearest to its minimal value: it is the factor which directly affects the process if its quantity is changed.”(b) Dark reactions, being enzymatic, are temperature controlled.
The C4 plants respond to higher temperatures and show higher rate of photosynthesis.
C3 plants have a much lower temperature optimum.
III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
✅ Answer
Prophase I.Features: It is typically longer and more complex. It is subdivided into phases - Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.
The processes such as synapsis, crossing over, chiasmata formation etc. occur during Prophase I.
• Splitting of water
• Utilization of ATP and NADPH
• Biosynthesis of sugars
• Release of oxygen
↓ Mitochondrial matrix
3CO2 + ……. + ……. + ……. + …….
(i) 1 NADH (ii) 1 FADH2
✅ Answer
(a) 3CO2 + 4NADH+ + 4H+ + FADH2 + ATP(b) (i) 3 ATP (ii) 2 ATP
(a) Respiratory climactic
(b) Bolting
(c) Heterophylly
✅ Answer
(a) Ethylene enhances the respiration rate during ripening of the fruits. This rise in rate of respiration is called respiratory climactic.(b) Bolting is the internode elongation just prior to flowering.
(c) Heterophylly is the presence of different types of leaves at different phases of life or due to different environments.
PART B: ZOOLOGY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
Mangifera: Genus
indica: ……………….
✅ Species
i) P-wave represents auricular repolarization
ii) P-wave represents ventricular repolarization
iii) P-wave represents auricular depolarization
iv) P-wave represents ventricular depolarization
✅ iii) P-wave represents auricular depolarization.
✅ Ichthyophis
✅ Lyases
✅ Columns of Bertini
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
Urinary bladder
Cloacal aperture
Ureters
Kidneys
Cloaca
✅ Answer
Kidneys → Ureters → Urinary bladder → Cloaca → Cloacal aperture.✅ Answer
i. Primary metabolites. E.g., amino acids, sugars etc.ii. Secondary metabolites. E.g., pigments, alkaloids etc.
a) Adjacent vertebrae
b) Skull bones
c) Atlas and axis
d) Carpal & metacarpal of thumb
✅ Answer
a) Cartilaginous (Slightly movable) jointsb) Fibrous (immovable) joints
c) Pivot joint
d) Saddle joint
✅ Answer
Considering the basic functions, I agree to this statement. Malpighian tubules in cockroaches serve to excrete metabolic waste and maintain osmotic balance, similar to the kidneys in higher organisms.b) How the labelled part A differs from B?
✅ Answer
a) (a) Coelomate(b) Pseudocoelomate
(c) Acoelomate.
b) A is true coelom.
B is false coelom (pseudocoelom).
✅ Answer
a) A= Adenine B= Uracilb) Nucleotide of Adenine is Adenylic acid.
Nucleotide of Uracil is Uridylic acid.
✅ Answer
A= Renin B= Angiotensin IC= Aldosterone D= Rise in GFR
a) Identify the animal and name the class to which this animal belongs.
b) Write any two characteristic features of this class.
✅ Answer
a) Animal = Chelone Class = Reptilia.b) Dry & cornified skin, Epidermal scales or scutes, Crawling mode of locomotion.
✅ Answer
Actin, Myosin, Troponin and Tropomyosin.III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
a) IRV and ERV
b) IC and EC
c) RV and FRC
✅ Answer
a) Inspiratory reserve volume (IRV): Additional volume of air that can inspire by a forcible inspiration.Expiratory reserve volume (ERV): Additional volume of air that can expire by a forcible expiration.
b) Inspiratory capacity (IC): Total volume of air inspired after a normal expiration (TV + IRV).
Expiratory capacity (EC): Total volume of air expired after a normal inspiration (TV + ERV).
c) Residual volume (RV): Volume of air remaining in lungs after a forcible expiration.
Functional residual capacity (FRC): Volume of air remaining in the lungs after a normal expiration (ERV + RV).
b) Name the ions involved in this process.
c) How first stage is maintained?
✅ Answer
a) (1) Stage of resting membrane potential (RMP).(2) Stage of Depolarisation.
b) Na+ and K+ ions.
c) It is maintained by the active transport of ions by the Na-K pump. It transports 3 Na+ outwards for 2 K+ into the cell. Thus, outer surface becomes positive charge and inner surface becomes negative charge.
(i) Diabetes mellitusa) Specify which hormone deficiency causes these disorders.
(ii) Dwarfism
(iii) Diabetes insipidus
(iv) Cretinism
b) Name the diseases that are caused due to the overproduction of the hormones related to (ii) and (iv).
✅ Answer
(a) (i) insulin (ii) growth hormone (GH)(iii) ADH (iv) thyroxine
(b) Overproduction of GH causes Gigantism/ Acromegaly.
Overproduction of thyroxine causes Exophthalmic goitre.
a) Identify the phylum.
b) Write any four other characteristics of this phylum.
c) Give one example of this phylum.
✅ Answer
(a) Cnidaria (Coelenterata)(b) i. Tentacles with cnidoblasts.
ii. Gastrovascular cavity (coelenteron) present.
iii. Polyp and medusa forms are present.
iv. Mouth is present on hypostome.
(c) Hydra.