Chapter 1
Biomolecules & Cell Biology (Botany): biomolecules, cell structure, cell division
NEB Class 11 past exam questions
Past questions and answers
19 questions set from this chapter. Most repeated first.
- CDC Class 11 Biology Model Question · 1 mark
Identify the correct bonding between nitrogen bases in DNA molecule from given pair.
(a) A = C
(b) A = G
(c) A = U
(d) A = T
Answer
(d) A = T
In DNA, adenine (A) always pairs with thymine (T) by two hydrogen bonds, while guanine (G) pairs with cytosine (C). A = U occurs only in RNA, and A = C or A = G are incorrect pairings.
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- CDC Class 11 Biology Model Question · 2+2+3 marks
Why is sexual reproduction important in lower grade plant like fungi? Explain it with respect to sexual reproduction of Yeast along with diagram.
Figure: Diagram of sexual reproduction of Yeast.
Answer
Why sexual reproduction is important in lower plants like fungi (2 marks)
- Genetic variation: Sexual reproduction involves fusion of two different nuclei (karyogamy) followed by meiosis, producing spores with new gene combinations. This variation helps fungi adapt to changing environments.
- Survival in adverse conditions: Sexual spores such as ascospores and zygospores have thick walls and are highly resistant to heat, drought and unfavourable conditions, so the species survives when the parent thallus dies.
- Recombination and evolution: It mixes hereditary material from two parents, giving rise to new strains and thus driving evolution and speciation in fungi.
Sexual reproduction in Yeast (Saccharomyces cerevisiae) (2 marks)
Yeast is a unicellular ascomycetous fungus. It reproduces sexually by conjugation (mating) between two haploid cells of opposite mating types, usually denoted a and α. It does not reproduce sexually by gametangial fusion; gametangial fusion is seen in some other fungi such as Rhizopus.
Steps:
- Two haploid yeast cells of opposite mating types come close and produce mating factors. They form projections/conjugation tubes toward each other.
- Plasmogamy: the cytoplasm of the two cells fuses.
- Karyogamy: the two haploid nuclei fuse to form a diploid zygote ().
- The diploid cell may multiply by budding for some time. Under unfavourable conditions, such as nitrogen starvation, it acts as an ascus mother cell.
- Meiosis occurs in the ascus mother cell and produces four haploid ascospores ().
- The ascus wall ruptures, ascospores are liberated, and each germinates into a new haploid yeast cell.
Thus yeast shows a haplo-diplontic life cycle; both haploid and diploid cells can multiply by budding, and meiosis occurs in the diploid ascus mother cell. It is not correctly described as simply haplontic or diplontic.
Diagram of sexual reproduction in Yeast (3 marks)
Haploid yeast cells (n)
a-type α-type
● ●
\ /
\ /
\ /
\ /
\ /
conjugation tube
\ /
\/
┌──────────────┐
│ Diploid zygote│ (2n)
│ ● │
└──────────────┘
│
│ meiosis (sporulation)
▼
┌──────────────┐
│ Ascus │
│ ● ● ● ● │ four haploid ascospores (n)
└──────────────┘
│
│ ascus wall ruptures
▼
● ● ● ●
liberated ascospores
│
│ germination
▼
New haploid yeast cells (n)
Labelled points to write under the diagram:
- Fusion of two haploid cells of opposite mating types (a and α) by conjugation.
- Plasmogamy: cytoplasmic fusion.
- Karyogamy: nuclear fusion forming diploid zygote ().
- Meiosis in ascus mother cell producing four haploid ascospores ().
- Liberation and germination of ascospores into new haploid yeast cells.
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- CDC Class 11 Biology Model Question · 8 marks
Meiosis cell division is important in life of organism since it maintains the chromosomal stability in offspring which is happen due to reduction and disjunction of homologous chromosome. It is a multi-step processes and two successive steps within this process is given.
Fig- 1 Fig-2
a. In reference to disjuction,what changes do you find between fig-1 & fig-2 ? Describe it.
b. What had happened in chromosomal stability if changes not occurs from figure-1 to figure-2 ?
c. Draw the diagram of succesive next one step after figure-2 with two distinguishing features.
Figure: Fig-1 and Fig-2 showing stages of meiosis.
Answer
(a) Change between Fig-1 and Fig-2 with respect to disjunction
Fig-1 represents Metaphase I and Fig-2 represents Anaphase I of meiosis I.
| Feature | Fig-1 (Metaphase I) | Fig-2 (Anaphase I) |
|---|---|---|
| Position of chromosomes | Homologous chromosomes lie paired (bivalents/tetrads) at the equatorial plate | Homologous chromosomes have separated and are moving to opposite poles |
| Spindle fibres | Fibres from opposite poles attach to the centromeres of the two homologues of each pair | Fibres shorten and pull the two homologues apart |
| Chromatids | Each chromosome has 2 sister chromatids (tetrad = 4 chromatids) | Each moving chromosome still has its 2 sister chromatids joined at the centromere (dyad) |
Change found: In Fig-1 the homologous chromosomes are held together as bivalents at the equator, whereas in Fig-2 the homologous chromosomes of each bivalent have disjoined (separated) and migrated to opposite poles. Thus the change is the disjunction of homologous chromosomes, which is the actual event of reduction division. Note that here only the homologous chromosomes separate; the sister chromatids do not separate (unlike anaphase of mitosis).
(b) Effect on chromosomal stability if this change does not occur
If disjunction fails to occur (non-disjunction), both homologous chromosomes of a pair pass to the same pole. As a result:
- One daughter cell receives an extra chromosome (n + 1) and the other becomes deficient (n − 1).
- Such aneuploid gametes, on fertilisation, produce aneuploid zygotes, so the offspring do not get the normal, constant diploid number (2n) of chromosomes.
- Hence chromosomal stability is lost, causing chromosomal (genetic) disorders such as Down's syndrome (trisomy 21), Klinefelter's syndrome (XXY) and Turner's syndrome (XO), and may also cause miscarriage or sterility.
(c) Diagram of the next successive step after Fig-2
The next step after Anaphase I is Telophase I (followed by cytokinesis).
Telophase I
,-----------------. ,-----------------.
/ nuclear envelope \ / nuclear envelope \
| reforming (x x x) | | reforming (x x x) |
\ (dyads) / \ (dyads) /
`--------+--------' `--------+--------'
| |
| <--- cleavage furrow ---> |
\____________________________/
Two distinguishing features:
- Two haploid daughter nuclei are formed — the chromosome number is reduced from 2n to n, i.e. reduction division is completed (each pole gets only one chromosome of every homologous pair).
- Each chromosome is still a dyad — the two sister chromatids remain joined at the centromere; they separate only in Anaphase II of meiosis II.
(Other supporting features: nuclear envelope and nucleolus reappear, spindle fibres disappear, and the cytoplasm divides by cleavage furrow/cytokinesis.)
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- Class 11 Biology Model Question Set D · 1 mark
Which of the following biomolecules acts as a primary energy currency in cellular metabolism?
(a) DNA
(b) ATP
(c) Cellulose
(d) Phospholipid
Answer
(b) ATP
ATP (adenosine triphosphate) stores energy in its high-energy phosphate bonds and releases it when the terminal phosphate is removed, so it is the cell's immediate energy currency. DNA stores information, cellulose is structural and phospholipids form membranes.
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- Class 11 Biology Model Question Set A · 1 mark
Which of the following cell organelles is known as the "powerhouse of the cell"?
(a) Ribosome
(b) Mitochondria
(c) Lysosome
(d) Golgi apparatus
Answer
(b) Mitochondria
Mitochondria carry out aerobic respiration (Krebs cycle and electron transport chain) and produce most of the cell's ATP, hence the name powerhouse.
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- Class 11 Biology Model Question Set A · 1 mark
Which molecule carries genetic information in most organisms?
(a) Protein
(b) DNA
(c) Lipid
(d) Glucose
Answer
(b) DNA
DNA stores hereditary information as a sequence of nucleotides and passes it on during cell division. Only some viruses use RNA instead.
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- Class 11 Biology Model Question Set C · 1 mark
Which of the following is a purine nitrogenous base?
(a) Cytosine
(b) Thymine
(c) Adenine
(d) Uracil
Answer
(c) Adenine
Purines have a double-ring structure and include adenine and guanine. Cytosine, thymine and uracil are single-ring pyrimidines.
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- Class 11 Biology Model Question Set E · 1 mark
Which of the following biomolecules acts as a primary biological catalyst (enzyme)?
(a) Carbohydrate
(b) Protein
(c) Lipid
(d) Nucleic Acid
Answer
(b) Protein
Almost all enzymes are globular proteins that lower the activation energy of biochemical reactions. (A few RNA molecules, ribozymes, are catalytic, but proteins are the main biological catalysts.)
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- Class 11 Biology Model Question Set D · 1 mark
During karyokinesis in plant cells, the spindle fibers attach to which specialized structure on chromosomes?
(a) Telomere
(b) Centromere (Kinetochore)
(c) Satellite
(d) Nucleolus
Answer
(b) Centromere (Kinetochore)
Each chromosome has a kinetochore, a protein disc built on its centromere, to which the spindle fibres attach so the chromatids can be pulled to opposite poles. Telomeres are chromosome ends and satellites are small terminal segments.
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- Class 11 Biology Model Question Set B · 1 mark
Which of the following organelles is directly involved in photorespiration?
(a) Mitochondria only
(b) Chloroplast only
(c) Peroxisome, chloroplast, and mitochondria
(d) Golgi body and endoplasmic reticulum
Answer
(c) Peroxisome, chloroplast, and mitochondria
In photorespiration RuBisCO acts as an oxygenase in the chloroplast, forming phosphoglycolate; this is processed in the peroxisome and mitochondrion, where glycine is converted to serine with release of CO.
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- Class 11 Biology Model Question Set B · 1 mark
Mesosomes in bacteria function analogously to which eukaryotic organelle?
(a) Ribosome
(b) Lysosome
(c) Mitochondria
(d) Nucleolus
Answer
(c) Mitochondria
Mesosomes are infoldings of the bacterial plasma membrane that carry respiratory enzymes and help in respiration, like the cristae of mitochondria. They also assist in DNA attachment during cell division.
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- Class 11 Biology Model Question Set B · 1 mark
Crossing over takes place during which sub-stage of meiotic prophase I?
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diakinesis
Answer
(c) Pachytene
In pachytene the homologous chromosomes are fully paired as bivalents and non-sister chromatids exchange segments at chiasma sites, with the help of recombination nodules.
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- Class 11 Biology Model Question Set D · 5 marks
- Class 11 Biology Model Question Set E · 5 marks
- Class 11 Biology Model Question Set C · 5 marks
Differentiate between prokaryotic and eukaryotic cells with any five points.
Answer
Differences between prokaryotic and eukaryotic cells (any five)
| Feature | Prokaryotic cell | Eukaryotic cell |
|---|---|---|
| Size | Small (about 1-10 μm) | Larger (about 10-100 μm) |
| Nucleus | No true nucleus; no nuclear membrane, genetic material lies in the nucleoid | True nucleus enclosed by a double nuclear membrane |
| Genetic material | Single circular DNA, usually without histones | Linear DNA in several chromosomes, associated with histone proteins |
| Membrane-bound organelles | Absent (no mitochondria, ER, Golgi, etc.) | Present (mitochondria, ER, Golgi bodies, lysosomes, plastids) |
| Ribosomes | 70S type | 80S type in cytoplasm (70S in mitochondria and chloroplasts) |
| Cell division | Binary fission; no mitosis or meiosis | Mitosis and meiosis |
| Examples | Bacteria, cyanobacteria | Protists, fungi, plants, animals |
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- Class 11 Biology Model Question Set B · 5 marks
Describe the ultrastructure of a typical plant cell with a labeled diagram.
Answer
Ultrastructure of a typical plant cell
A plant cell seen under the electron microscope is a eukaryotic cell with a rigid wall outside the plasma membrane.
Labelled diagram (to be drawn): an outer cell wall (with middle lamella), inside it the plasma membrane; within the cytoplasm label the nucleus (nuclear envelope, nucleolus, chromatin), chloroplast, mitochondrion, endoplasmic reticulum, Golgi apparatus, ribosomes, large central vacuole with tonoplast, and peroxisome.
Structure and function of the parts
- Cell wall: Non-living, made mainly of cellulose, hemicellulose and pectin, with the middle lamella between neighbouring cells. It gives shape and protection and prevents bursting.
- Plasma membrane: A fluid mosaic of a phospholipid bilayer and proteins that is selectively permeable and controls entry and exit of substances.
- Cytoplasm: The jelly-like matrix holding the organelles, where many metabolic reactions occur.
- Nucleus: Double-membrane-bound, with nucleolus and chromatin. It controls cell activities and stores DNA.
- Mitochondria: Double-membrane organelles with cristae, the site of aerobic respiration and ATP production.
- Chloroplasts: Double-membrane plastids with grana and stroma, the site of photosynthesis.
- Endoplasmic reticulum and ribosomes: Rough ER with 80S ribosomes makes proteins; smooth ER makes lipids.
- Golgi apparatus: Stacks of flattened sacs that modify, package and secrete materials and help form the cell wall.
- Vacuole: Large central sac bounded by the tonoplast, storing water, salts and pigments and maintaining turgidity.
- Absent in plant cells: Centrioles and lysosomes are generally absent (or rare).
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- Class 11 Biology Model Question Set B · 8 marks
What are enzymes? Classify enzymes based on their mode of action and chemical nature.
Answer
Enzymes
Enzymes are biological catalysts, mostly globular proteins, that speed up biochemical reactions by lowering the activation energy without being used up. Features: they are highly specific, needed in small amounts, work best at an optimum temperature and pH, and are reusable. The substance acted on is the substrate, which binds the enzyme's active site to form an enzyme-substrate complex that releases the product(s).
Mode of action
- Lock-and-key model (Fischer): the active site is rigid and exactly fits the substrate.
- Induced-fit model (Koshland): the substrate induces a change in the shape of the active site so that it fits closely and the reaction proceeds. This model better explains enzyme flexibility.
Classification by chemical nature
- Simple enzymes: Made of protein only (e.g. pepsin, trypsin, amylase).
- Conjugated enzymes (holoenzymes): Protein part (apoenzyme) plus a non-protein part (cofactor). The cofactor may be:
- a prosthetic group, firmly bound (e.g. haem in catalase);
- a coenzyme, an organic molecule loosely bound, often a vitamin derivative (e.g. NAD, FAD);
- a metal ion activator (e.g. Zn²⁺ in carbonic anhydrase, Mg²⁺).
Classification by type of reaction catalysed (IUBMB, six classes)
| Class | Reaction | Example |
|---|---|---|
| 1. Oxidoreductases | Oxidation-reduction | Dehydrogenase, oxidase |
| 2. Transferases | Transfer of a group | Transaminase, kinase |
| 3. Hydrolases | Hydrolysis using water | Amylase, lipase, protease |
| 4. Lyases | Removal of a group without water | Decarboxylase, aldolase |
| 5. Isomerases | Rearrangement within a molecule | Glucose phosphate isomerase |
| 6. Ligases | Joining of two molecules using ATP | DNA ligase, synthetase |
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- Class 11 Biology Model Question Set A · 8 marks
Explain the process of mitosis cell division. Why is mitosis often referred to as equational division? Describe its significance in living organisms.
Answer
Mitosis
Mitosis is the division of a somatic cell into two daughter cells, each with the same chromosome number and genetic content as the parent. It has karyokinesis (nuclear division) followed by cytokinesis.
Stages (after interphase, in which DNA replicates in S phase)
- Prophase: Chromatin condenses into visible chromosomes, each of two sister chromatids joined at the centromere. The nucleolus and nuclear envelope disappear, the spindle begins to form, and centrioles move to the poles in animal cells.
- Metaphase: Chromosomes line up at the equatorial plate and spindle fibres attach to their centromeres at the kinetochores. This is the best stage to count chromosomes.
- Anaphase: Centromeres split and the sister chromatids (now daughter chromosomes) move to opposite poles, pulled by the shortening spindle fibres.
- Telophase: Chromosomes reach the poles and uncoil, the nuclear envelope and nucleolus reappear, and the spindle disappears.
- Cytokinesis: In animal cells a cleavage furrow pinches the cell; in plant cells a cell plate forms from the centre outward and becomes the new wall.
Why equational division?
The chromosome number stays the same () in the daughter cells. Each daughter gets an equal and identical set of chromosomes because sister chromatids separate, so the division is called equational.
Significance
- Growth: increases cell number in a multicellular body.
- Repair and regeneration: replaces damaged or dead cells and heals wounds.
- Genetic stability: daughter cells are genetically identical to the parent, so the chromosome number is kept constant.
- Asexual reproduction: the basis of vegetative reproduction and budding.
- Maintains the nucleo-cytoplasmic ratio by dividing a cell when it grows too large.
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- Class 11 Biology Model Question Set D · 8 marks
- Class 11 Biology Model Question Set B · 5 marks
Compare mitosis and meiosis cell division with respect to their occurrence, stages, genetic variation, and chromosome number changes. (Draw appropriate diagrams.)
Answer
Comparison between mitosis and meiosis
| Basis | Mitosis | Meiosis |
|---|---|---|
| Occurrence | Somatic (body) cells; also in cells of growing regions | Reproductive cells (germ cells) to form gametes or spores |
| Number of divisions | One nuclear division | Two successive divisions (meiosis I and II) |
| Daughter cells | Two | Four |
| Stages | Prophase, metaphase, anaphase, telophase (single set) | Same four stages twice; prophase I has five substages (leptotene, zygotene, pachytene, diplotene, diakinesis) |
| Pairing of homologues (synapsis) | Absent | Occurs in zygotene |
| Crossing over | Absent | Occurs in pachytene |
| Genetic variation | None; daughter cells are identical to the parent | Present due to crossing over and independent assortment |
| Chromosome number | Remains the same () | Halved (), so it is called reductional division |
| Role | Growth, repair, asexual reproduction | Gamete formation, keeps chromosome number constant over generations, variation |
Diagrams (to be drawn): (i) a mitosis sequence of a cell with showing metaphase with 4 chromosomes on the plate and anaphase with sister chromatids moving apart; (ii) meiosis I with bivalents at metaphase I and homologues separating in anaphase I, then meiosis II giving four haploid cells.
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- Class 11 Biology Model Question Set C · 8 marks
Describe Prophase-I of meiosis in detail with well-labeled descriptions. How does meiosis differ from mitosis?
Answer
Prophase-I of meiosis
Prophase-I is the longest and most complex stage of meiosis I. It has five substages.
- Leptotene (thin threads): Chromosomes begin to condense and appear as long thin threads, each already replicated but with sister chromatids not yet distinct. Chromosome ends are attached to the nuclear envelope.
- Zygotene (paired threads): Homologous chromosomes pair side by side along their length (synapsis) by the synaptonemal complex, forming bivalents.
- Pachytene (thick threads): Chromosomes shorten and thicken, and each bivalent shows four chromatids (a tetrad). Crossing over takes place between non-sister chromatids of homologues at points called recombination nodules, exchanging segments of genetic material.
- Diplotene (double threads): The synaptonemal complex dissolves and homologues start to separate but remain joined at the points of crossing over, the chiasmata.
- Diakinesis: Chromosomes condense maximally, the chiasmata move toward the ends (terminalisation), the nucleolus and nuclear envelope disappear and the spindle begins to form.
Significance: crossing over produces new combinations of genes (genetic recombination).
How meiosis differs from mitosis
| Mitosis | Meiosis |
|---|---|
| One division, two daughter cells | Two divisions, four daughter cells |
| No synapsis or crossing over | Synapsis and crossing over occur |
| Chromosome number remains | Chromosome number halved to |
| Daughter cells genetically identical | Daughter cells genetically different |
| Occurs in somatic cells | Occurs in germ cells |
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- Class 11 Biology Model Question Set E · 8 marks
Sketch a well labelled diagram of a typical plant cell and describe the structure and function of mitochondria and nucleus.
Answer
Labelled diagram of a plant cell (to be drawn)
Show an outer cell wall, plasma membrane, cytoplasm, nucleus with nucleolus, mitochondria, chloroplasts, endoplasmic reticulum, Golgi apparatus, ribosomes and a large central vacuole with tonoplast.
Mitochondria
Structure: Rod-shaped or oval organelles, about 0.5-1 μm wide, bounded by two unit membranes. The outer membrane is smooth; the inner membrane is folded inward into cristae, which bear stalked particles (F1 particles, ATP synthase). The inner space is the matrix, which contains enzymes, circular DNA, 70S ribosomes and RNA. Mitochondria are semi-autonomous since they can make some of their own proteins and divide independently.
Function: Site of aerobic respiration. The Krebs cycle takes place in the matrix and the electron transport chain on the cristae, producing ATP, so mitochondria are called the "powerhouse of the cell".
Nucleus
Structure: A large, usually spherical organelle (the control centre), with these parts:
- Nuclear envelope: double membrane with nuclear pores for exchange of materials.
- Nucleoplasm: the ground substance inside the nucleus.
- Chromatin: thread-like DNA bound with histone proteins; it condenses into chromosomes during division.
- Nucleolus: dense, membraneless body rich in rRNA and protein; it makes ribosomal subunits.
Function: It controls all cellular activities, stores hereditary information (DNA), directs protein synthesis through RNA, and passes genetic information to the next generation during cell division.
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Questions from CDC Class 11 Biology Model Question (New Course) (CDC model paper, Sub.Code 2011, new course), Practice questions written for this site (Written from the official syllabus and specification grid where no past papers cover a chapter), Class 11 Biology Model Question Set A (Sub. code 301, 75 marks) (Model question paper Set A published by nebexam.com (College Info Nepal), following the CDC Class 11 Biology grid; not an actual board exam, year not stated), Class 11 Biology Model Question Set B (Sub. code 301, 75 marks) (Model question paper Set B published by nebexam.com (College Info Nepal), following the CDC Class 11 Biology grid; not an actual board exam, year not stated), Class 11 Biology Model Question Set C (Sub. code 301, 75 marks) (Model question paper Set C published by nebexam.com (College Info Nepal), following the CDC Class 11 Biology grid; not an actual board exam, year not stated), Class 11 Biology Model Question Set D (Sub. code 301, 75 marks) (Model question paper Set D published by nebexam.com (College Info Nepal), following the CDC Class 11 Biology grid; not an actual board exam, year not stated) and Class 11 Biology Model Question Set E (Sub. code 301, 75 marks) (Model question paper Set E published by nebexam.com (College Info Nepal), following the CDC Class 11 Biology grid; not an actual board exam, year not stated). Answers are written for this site; check them against your class notes.
Chapter titles and hours from the official syllabus ↗