BioNotesICSE Class X Biology Practice Paper 2 - Question Paper
Practice question paper (80 marks, 2 hours) for ICSE Class X Biology, Science Paper 3, 2027 pattern.
ISC Class XII Biology Practice Paper 1 - Question Paper
Practice question paper (70 marks, 3 hours) for ISC Class XII Biology Paper 1 Theory, 2027 pattern.
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Online
Marking scheme and model answers for ISC Class XII Biology Practice Paper 1, 2027 pattern.
Marks are awarded point-wise, in the manner of CISCE step-marking. Alternative correct answers should be accepted.
(i) (b) ensure seed set even in the absence of pollinators
(ii) (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(iii) (b) 360
(G = 20% = 240; C = 240; A + T = 1200 − 480 = 720; A = T = 360)
(iv) (c) Bacillus thuringiensis – biofertiliser (it is a biocontrol agent)
(v) (c) A forest with a single large tree supporting many herbivorous insects
(vi) (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(vii) (b) variable number of tandem repeats (VNTRs)
(viii) Non-albuminous (exalbuminous) seed; example: pea / bean / groundnut / gram.
(ix) (a) IgA [½]; (b) IgE [½].
(x) (any one) Inhibit ovulation; inhibit implantation; alter the quality of the cervical mucus to retard the entry of sperm.
(xi) Final population = 2000 + (births + immigration) − (deaths + emigration) = 2000 + (40 + 10) − (20 + 5) = 2025.
(xii) Grass 10,000 J → Grasshopper 1000 J → Frog 100 J → Snake 10 J.
(xiii) RNA polymerase.
(xiv) 5′–AUG CCU AAG–3′.
(xv) Natality = 60 / 1500 = 0.04 per individual per year.
(xvi) (a) Adenosine Deaminase [½]; (b) Sutton and Boveri [½].
(xvii) Taq polymerase is thermostable, so it remains active during the high-temperature denaturation step and synthesises the new DNA strand (extension, ~72 °C) using the primers.
(xviii) (a) The large, feathery stigma helps to trap the light, non-sticky pollen grains carried by wind [1].
(b) Males have only one X chromosome (hemizygous), so a single recessive allele is expressed; a female needs two recessive alleles (homozygous recessive) to show the disorder [1].
(any two, 1 mark each, with example/explanation)
(i) ³²P batch — the radioactive phosphorus of DNA was found inside the bacterial cells [1].
(ii) DNA, and not protein, is the genetic material [1].
(i) Incomplete dominance [1].
(ii) 1 red : 2 pink : 1 white [1].
(i) Primary lymphoid organ: thymus / bone marrow [½]; secondary lymphoid organ: spleen / lymph nodes / MALT [½].
(ii) In the thymus [1].
(i) (a) Adenosine deaminase (ADA) [1].
(b) Lymphocytes from the patient's blood are grown in culture; a functional ADA gene (cDNA) is introduced into them using a retroviral vector and the cells are returned to the patient [1]. (For permanent cure, the gene is introduced into the bone marrow/early embryonic cells.)
OR
(ii) (a) Parents: I^A i × I^B i. Possible children: I^A I^B (AB), I^A i (A), I^B i (B), ii (O) [1].
(b) Probability of blood group O = ¼ (25%) [1].
(i) Methane (CH₄), ammonia (NH₃), hydrogen (H₂) and water vapour [1] (at ~800 °C with electric discharge).
(ii) Amino acids (simple organic compounds) [1].
| In situ conservation | Ex situ conservation |
|---|---|
| Conservation of organisms in their natural habitat | Conservation outside the natural habitat |
| Example: national parks, wildlife sanctuaries, biosphere reserves, sacred groves | Example: zoos, botanical gardens, seed banks, cryopreservation |
(1 mark for each correct pair of difference and example.)
1 mark for neat and correct diagram; 2 marks for any four correct labels.
(i) Mature embryo sac (7-celled, 8-nucleate):
www.bionotes.inLabels: Antipodal cells (3, chalazal end), Polar nuclei (central cell), Egg cell, Synergids, Filiform apparatus, Micropylar end, Chalazal end.
OR
(ii) Human sperm:
www.bionotes.inLabels: Head, Acrosome, Nucleus, Middle piece, Mitochondria, Centriole, Tail (axoneme).
(i) Denaturation (~94 °C) → Annealing (~54 °C) → Extension (~72 °C) [1].
(ii) Primers are short single-stranded oligonucleotides complementary to the ends of the target DNA; they anneal to the template and provide a free 3′-OH end from which DNA polymerase begins synthesis [1].
(iii) 2⁵ = 32 copies [1].
(i) p = 1 − q = 1 − 0.3 = 0.7 [1].
(ii) 2pq = 2 × 0.7 × 0.3 = 0.42 [1].
(iii) p² = 0.49; 0.49 × 1000 = 490 individuals [1].
(i) NPP is the amount of organic matter (biomass) available to consumers, i.e. Gross Primary Productivity minus the energy lost in respiration (NPP = GPP − R) [1].
(ii) NPP = GPP − R = GPP − 0.4 GPP = 0.6 GPP = 1200 ⇒ GPP = 2000 g/m²/year [1]; R = 0.4 × 2000 = 800 g/m²/year [1].
1 mark for neat and correct diagram; 2 marks for any four correct labels.
www.bionotes.inLabels: Two heavy chains, Two light chains, Disulphide bonds, Variable region, Constant region, Antigen-binding sites, Hinge region.
(i) Heavy nitrogen (¹⁵N) and light nitrogen (¹⁴N); E. coli [1].
(ii) After one generation (one round of replication in ¹⁴N medium), all the DNA was of intermediate (hybrid, ¹⁵N/¹⁴N) density [1].
(iii) DNA replication is semi-conservative — each daughter DNA has one old (parental) strand and one newly synthesised strand [1].
www.bionotes.in(i) Human Immunodeficiency Virus (HIV); mainly infects helper T-lymphocytes (CD4⁺ cells) (also macrophages) [1].
(ii) Reverse transcriptase [1].
(iii) ELISA (Enzyme-Linked Immunosorbent Assay) [1].
(a) Man: X^c Y [½]; Woman: X^C X^c (carrier — she received X^c from her colour-blind father) [½].
(b) [2]
| X^C | X^c | |
|---|---|---|
| X^c | X^C X^c (carrier daughter) | X^c X^c (colour-blind daughter) |
| Y | X^C Y (normal son) | X^c Y (colour-blind son) |
(c) Probability of a colour-blind daughter = ¼ (25%) [1].
(d) Probability of a colour-blind son = ¼ (25%) [1].
(a) [2]
| True fruit | False fruit |
|---|---|
| Develops only from the ovary | Other floral parts (e.g. thalamus, calyx, peduncle) also contribute to the fruit |
| Example: mango / tomato / coconut | Example: apple / pear / strawberry / cashew |
(b) Double fertilisation: one pollen tube enters the embryo sac and releases two male gametes [½]. One male gamete fuses with the egg cell (syngamy) to form the diploid zygote [½]. The other male gamete fuses with the two polar nuclei (triple fusion) to form the triploid primary endosperm nucleus [1].
(c) Perisperm is the persistent remnant of the nucellus in a seed (e.g. black pepper, beet) [1].
(i) (1 mark for each phase + 1 mark for correct hormonal interrelationship)
www.bionotes.inOR
(ii) Translation in prokaryotes:
www.bionotes.inDiagrams in the marking scheme are from the BioNotes diagram bank. Question–answer design by the BioNotes teaching team; syllabus alignment: CISCE ISC 2027 Biology (863) Class XII.
ICSE Class X Biology Practice Paper 2 - Question Paper
Practice question paper (80 marks, 2 hours) for ICSE Class X Biology, Science Paper 3, 2027 pattern.
ISC Class XII Biology Practice Paper 1 - Question Paper
Practice question paper (70 marks, 3 hours) for ISC Class XII Biology Paper 1 Theory, 2027 pattern.