NEB Class 11 Science Biology Question Paper 2078 Nepal
This is the official NEB Class 11 (Science stream) Biology question paper for 2078, as set in the Model questions examination. It carries 75 full marks and a time allowance of 180 minutes, across 22 questions. On Kekkei you can attempt this Biology past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your NEB Class 11 Biology exam or solving previous years' question papers, this 2078 paper is a great way to practise under real exam conditions.
| Level | NEB Class 11 |
|---|---|
| Stream | Science |
| Subject | Biology |
| Year | 2078 BS |
| Exam session | Model questions |
| Full marks | 75 |
| Time allowed | 180 minutes |
| Questions | 22, all with step-by-step solutions |
Group 'A'
Circle the correct one from given alternatives.
Identify the correct bonding between nitrogen bases in DNA molecule from given pair.
A = T
In DNA, adenine pairs with thymine (A=T) via two hydrogen bonds. Answer: (d) A = T.
Miller and Urey designed an experiment to recreate the environment of primitive earth to support the theory of biochemical origin of life. Which composition did he use to produce simple amino acids?
Ammonia, Methane, Hydrogen, and water vapour
The Miller–Urey experiment used a mixture of ammonia, methane, hydrogen and water vapour. Answer: (b).
Which habitat is needed for xerophytic plants to survive?
habitat with dry condition
Xerophytes are adapted to dry conditions (scarcity of water). The best answer is a habitat with dry condition. Answer: (a).
"Red rust is one of the destructive diseases in tea plants which results adverse effect on tea yield. It is caused by a type of algae." Which discipline of Biology is related to explain the above statement?
Phycology
Since red rust of tea is caused by an alga (Cephaleuros), the relevant discipline is phycology (the study of algae). Answer: (d) Phycology.
Which of the following is a practice of in-situ conservation of biodiversity?
Wildlife sanctuary
In-situ conservation protects species in their natural habitat — e.g. a wildlife sanctuary. (Botanical garden, zoo and seed bank are ex-situ.) Answer: (b) Wildlife sanctuary.
Which one statement of the following tells the favor of prop root?
It has the function of support
Prop roots (e.g. in banyan) grow down from branches to the ground and give mechanical support to the heavy plant. Answer: (a) It has the function of support.
Forelimbs (wings) of Dragon fly, Pteranodon, Hawk and Bat are given. What type of organs are these?

Analogous
These wings perform the same function (flight) but have different structural origins/designs, so they are analogous organs (showing convergent evolution). Answer: (c) Analogous.
Which is the nearest ancestor of modern man?
Homo sapiens neanderthalensis
The nearest (most recent) ancestor among the options is Homo sapiens neanderthalensis. Answer: (a).
In the given figure, X denotes an organ of the frog which produces a particular sound. What does that sound indicate?

Sex appeal
X is the vocal sac of a male frog. The croaking sound produced is the mating call used to attract females, indicating sex appeal. Answer: (d) Sex appeal.
Trichocysts is an important organelle in Paramecium. Which activities does it associate for?
Defense, offence, and adhesion
Trichocysts are used by Paramecium for defence, offence (capturing prey) and anchorage/adhesion. Answer: (a) Defense, offence, and adhesion.
IUCN noticed the one-horned rhino restricted in Chitwan national park of Nepal as a high-risk group. Its population has declined due to habitat loss, excessive poaching, low rate of breeding, human encroachment and other factors. According to this description, in which risk group does the one-horned rhino of Nepal fall?
Endangered
Given the high risk and population decline described, the one-horned rhino is classified as Endangered. Answer: (c) Endangered. (IUCN lists the greater one-horned rhino as Vulnerable globally, but per the description of high risk the intended answer here is Endangered/critically endangered.)
Group 'B'
Give short answer to the following questions.
Write any five characteristics of Chlorophyceae.
Characteristics of Chlorophyceae (green algae): (1) they contain chlorophyll a and b, giving a grass-green colour; (2) the cell wall is made of an inner cellulose and outer pectose layer; (3) the reserve food is starch (stored in pyrenoids); (4) they are mostly aquatic (fresh water), some terrestrial; (5) the body is unicellular to filamentous/colonial (e.g. Chlamydomonas, Spirogyra, Volvox, Ulothrix); (6) motile cells/gametes bear two equal whiplash flagella; reproduction is vegetative, asexual (zoospores) and sexual.
'Gymnosperm is an advanced group than Pteridophyta'. Justify this statement with any five points.
OR
Compare between simple fruit and multiple fruit in two points. How can fruit play an important role for continuity of generation in plant life?
Gymnosperm advanced over Pteridophyta (any five): (1) Gymnosperms are seed-bearing (seed habit) whereas pteridophytes reproduce by spores; (2) gymnosperms show heterospory leading to seed formation; (3) pollination by wind — no water needed for fertilisation, unlike pteridophytes which need water for swimming sperms; (4) development of pollen tube (siphonogamy); (5) well-developed vascular tissue and secondary growth; (6) reduced, dependent gametophyte; (7) ovule and naked seed present.
OR — Simple vs multiple fruit (two points): A simple fruit develops from a single ovary of a single flower (e.g. mango); a multiple fruit develops from the whole inflorescence (many flowers fusing together, e.g. jackfruit, pineapple). Fruit aids continuity of generation by protecting the developing seeds and aiding their dispersal (by wind, water, animals), ensuring the species spreads and reproduces.
How can ecological factors influence in pond ecosystem? Discuss it with suitable examples with respect to its structural and functional aspects.
A pond is a self-sufficient freshwater ecosystem. Structural aspects: abiotic factors (light, temperature, dissolved O₂/CO₂, pH, nutrients, water) and biotic components — producers (phytoplankton, algae, aquatic plants), consumers (zooplankton, insects, fish — primary/secondary/tertiary) and decomposers (bacteria, fungi). Functional aspects: food chains/webs, energy flow (sunlight → producers → consumers), nutrient (biogeochemical) cycling, and productivity. Influence of ecological factors: e.g. light intensity controls photosynthesis (productivity); temperature affects metabolic rates and dissolved oxygen; nutrient levels affect algal growth (eutrophication if excessive); dissolved oxygen limits animal survival. Thus abiotic factors regulate the structure and functioning of the pond.
Explain any five ways how microorganisms play an important role in the field of medicine.
Roles of microorganisms in medicine (any five): (1) production of antibiotics (e.g. penicillin from Penicillium, streptomycin from Streptomyces) to treat bacterial infections; (2) production of vaccines (using attenuated/killed microbes) to immunise against diseases; (3) production of vitamins (e.g. B12, riboflavin) and other supplements; (4) production of enzymes and hormones (e.g. insulin via genetically engineered bacteria); (5) production of serums/antitoxins; (6) use in medical research and diagnosis; (7) production of organic acids and probiotics. (Any five.)
The excessive and unbalanced use of pesticides is a prominent environmental issue in Nepal. Mention any three consequences of overuse of pesticides in your locality and suggest an awareness campaign that you would follow in two points.
Three consequences of pesticide overuse: (1) soil and water pollution (contamination of drinking water and rivers); (2) harm to non-target organisms — bees, birds, fish, beneficial insects — and loss of biodiversity; (3) bioaccumulation/biomagnification in the food chain causing health hazards (cancer, poisoning) in humans; pesticide residues in food; development of pest resistance.
Awareness campaign (two points): (1) organise community education programmes/posters/street drama on safe use and alternatives (integrated pest management, organic/bio-pesticides); (2) train farmers on correct dosage, protective measures and promotion of organic farming, and lobby for regulation of harmful pesticides.
What is transverse binary fission? Write its process in Paramecium.
Transverse binary fission: An asexual reproduction in which the parent body divides into two equal daughter individuals by a division plane that runs transverse (at right angles) to the longitudinal axis.
Process in Paramecium: (1) The micronucleus divides first by mitosis (amitosis of macronucleus); the macronucleus elongates and divides amitotically. (2) A transverse constriction (furrow) appears at the middle of the body, deepening across the transverse axis. (3) The two contractile vacuoles and other organelles are distributed; new oral grooves form. (4) The body finally divides transversely into two daughter Paramecia, each with a micronucleus and macronucleus, which then grow to full size. It occurs in favourable conditions.
How can pancreas of a frog play an important role in digestion? Explain it in reference to physiology of intestinal digestion.
OR
Describe with suitable diagram, how copulation and cocoon formation takes place in earthworm.
Pancreas in frog digestion: The pancreas is a mixed gland; its exocrine part secretes pancreatic juice into the duodenum (via the bile-pancreatic duct). Pancreatic juice contains enzymes — trypsin (proteins → peptides/amino acids), amylase (starch → maltose), and lipase (fats → fatty acids + glycerol) — and it is alkaline, neutralising the acidic chyme. Thus in intestinal digestion the pancreas provides the major digestive enzymes that complete the breakdown of carbohydrates, proteins and fats in the small intestine. (The endocrine part secretes insulin/glucagon for blood-sugar regulation.)
OR — Earthworm copulation & cocoon formation: Earthworms are hermaphrodite but cross-fertilise. During copulation two worms come together with their ventral surfaces in opposite directions, held by the clitellum and genital setae; they exchange sperm, which is stored in each other's spermathecae. After separation, the clitellum secretes a mucous cocoon (a girdle) which slides forward over the body; as it passes, it receives ova from the female pores and stored sperm from the spermathecae, where fertilisation occurs. The cocoon then slips off the anterior end, its ends seal, and it is deposited in moist soil where the young develop.
How does Darwinism explain the theory of organic evolution? Clarify your answer.
Darwinism (natural selection) explains organic evolution through: (1) overproduction — organisms produce far more offspring than can survive; (2) struggle for existence — limited resources cause competition (intraspecific, interspecific, environmental); (3) variation — individuals show heritable variations; (4) survival of the fittest / natural selection — individuals with favourable variations are better adapted, survive and reproduce, passing on those traits; (5) inheritance and origin of new species — over many generations, accumulation of favourable variations leads to new, better-adapted forms and ultimately new species. Thus nature 'selects' the fittest, driving evolution.
Group 'C'
Give long answer to the following questions.
Why is sexual reproduction important in lower grade plant like fungi? Explain it with respect to sexual reproduction of Yeast along with diagram.
OR
Explain the diagnostic floral characters of family Solanaceae with floral formula and floral diagram. Write the botanical name of any two plants of this family regarding medicinal and edible values.
Importance of sexual reproduction in fungi: it introduces genetic variation (recombination) which aids adaptation and evolution; it restores the diploid phase and produces resistant resting structures (spores) for survival in unfavourable conditions and dispersal.
Sexual reproduction in Yeast: Under unfavourable conditions, two haploid yeast cells (or a cell with a bud) act as gametes; they fuse — plasmogamy (fusion of cytoplasm) followed by karyogamy (fusion of nuclei) — forming a diploid zygote. The diploid nucleus then undergoes meiosis to form four (or eight) haploid ascospores enclosed within the parent cell wall, which now acts as an ascus. On release in favourable conditions, the ascospores germinate into new haploid yeast cells.
OR — Solanaceae: Diagnostic floral characters: flowers bisexual, actinomorphic, hypogynous; calyx 5 sepals (gamosepalous, persistent); corolla 5 petals (gamopetalous); androecium 5 stamens epipetalous; gynoecium bicarpellary, syncarpous, superior ovary with axile placentation, obliquely placed. Floral formula: . Examples: Solanum tuberosum (potato — edible), Atropa belladonna / Withania somnifera (medicinal); Solanum melongena (brinjal — edible).
How is excretion performed in earthworm? Explain in analytical approach of the structure of septal nephridia of earthworm with labeled diagram.
Excretion in earthworm: carried out by segmentally arranged coiled tubular structures called nephridia (three types: septal, integumentary, pharyngeal). They remove nitrogenous wastes (urea, ammonia) and excess water.
Septal nephridia (structure): located on both sides of the inter-segmental septa from segment 15 backward; each consists of (1) the nephrostome (a ciliated funnel opening into the coelom that draws in coelomic fluid), (2) a short neck, (3) the body of the nephridium — a long twisted tube differentiated into a straight lobe and a twisted loop with ciliated and glandular regions, (4) the terminal duct leading to a common septal excretory canal, which opens through supra-intestinal excretory ducts into the gut (enteronephric type).
Function: coelomic fluid and wastes enter via the nephrostome; as the fluid passes along the tubule, water and useful materials are reabsorbed and nitrogenous wastes are concentrated; the waste is finally discharged into the alimentary canal and expelled with faeces (enteronephric excretion conserves water — an adaptation to terrestrial life).
(Labelled diagram of septal nephridium: nephrostome, neck, body, terminal duct — to be drawn.)
Meiosis cell division maintains chromosomal stability in offspring through reduction and disjunction of homologous chromosomes. Two successive steps within this process are given.

a. In reference to disjunction, what changes do you find between Fig-1 & Fig-2? Describe it.
b. What had happened in chromosomal stability if changes did not occur from Fig-1 to Fig-2?
c. Draw the diagram of the successive next one step after Fig-2 with two distinguishing features.
(a) Fig-1 shows Metaphase I — homologous chromosomes (bivalents/tetrads) are arranged in pairs at the equatorial plate, spindle fibres attached. Fig-2 shows Anaphase I — the homologous chromosomes of each bivalent disjoin (separate) and move to opposite poles (whole chromosomes, each still with two chromatids, move apart). The change is the disjunction/segregation of homologous chromosomes, reducing the chromosome number to half (reduction division) and distributing one of each pair to each pole, giving genetic variation.
(b) If disjunction did not occur (non-disjunction), the homologous chromosomes would not separate properly, the gametes would have an abnormal chromosome number (aneuploidy) — chromosomal instability — leading to disorders (e.g. Down's syndrome) and loss of chromosomal stability in offspring.
(c) The next step is Telophase I (followed by cytokinesis): two distinguishing features — (1) chromosomes reach the two opposite poles and a nuclear membrane reforms around each haploid set; (2) the cell divides (cytokinesis) into two haploid daughter cells. (Diagram to be drawn.)
Frequently asked questions
- Where can I find the NEB Class 11 Biology question paper 2078?
- The full NEB Class 11 Biology 2078 (Model questions) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
- Does the Biology 2078 paper come with solutions?
- Yes. Every question on this Biology past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
- How many marks is the NEB Class 11 Biology 2078 paper?
- The NEB Class 11 Biology 2078 paper carries 75 full marks and is meant to be completed in 180 minutes, across 22 questions.
- Is practising this Biology past paper free?
- Yes — reading and attempting this Biology past paper on Kekkei is completely free.