2026 and 2025 paper questions

Class 11 Biology, FBISE HSSC-I 2026 and 2025 (1st Annual). 210 questions across 15 units, in paper order.

Set B

105 questions

2026 Annual I, Set B, MCQ 11 markUnit 1
Which organelle is responsible for modifying and packaging proteins?
  1. ARibosome
  2. BGolgi apparatusAnswer
  3. CNucleus
  4. DLysosome
Why

The Golgi apparatus (B) collects the secretory proteins that the rough ER makes, modifies them so that they can do their specific work (it also forms glycoproteins and lipoproteins), and packs them into vesicles for export. The ribosome (A) only makes the protein, the nucleus (C) holds the DNA, and the lysosome (D) digests material; none of these modifies and packages proteins.

2025 Annual I, Set B, MCQ 11 markUnit 15
When energy flows from one trophic level to the next, the amount of energy lost is about:
  1. A10%
  2. B50%
  3. C90%Answer
  4. D100%
Why

Only about 10% of the energy of one trophic level reaches the next (Figure 15.3), so about 90% is lost, as heat, in respiration, growth, reproduction, non-predatory death and excretory products or faeces. 10% is the amount passed on, not lost, which is the tempting wrong choice; 50% and 100% do not fit the 10% rule.

2026 Annual I, Set B, MCQ 21 markUnit 1
Cellulose is mainly found in:
  1. AAnimal cells
  2. BFungal cell walls
  3. CPlant cell wallsAnswer
  4. DBacterial capsules
Why

Cellulose forms the microfibrils of the plant cell wall (C). The tempting option is B, because fungi also have a cell wall, but the fungal wall is made of chitin. Bacterial walls are made of peptidoglycan, and animal cells have no cell wall at all.

2025 Annual I, Set B, MCQ 21 markUnit 1
The part involved in detoxification of drugs is:
Figure for MCQ 2.
  1. AP
  2. BQAnswer
  3. CR
  4. DS
Why

Detoxification of drugs and poisons is done by enzymes of the smooth ER, which has no ribosomes and forms branched tubules. In the figure that is Q, the tubular structures at the end of the ER. P is the nucleus, and S and R are flattened sheets studded with ribosomes, the rough ER, which makes secretory proteins. The tempting wrong choices are S and R, which lie next to the nucleus and look like membrane, but they are rough ER.

Answer figure: P nucleus, Q smooth ER, R and S rough ER.

Note. Q is the ribosome-free tubular smooth ER; S and R are the ribosome-studded sheets of rough ER.

2026 Annual I, Set B, MCQ 31 markUnit 4
During anaerobic respiration in human muscles, pyruvate is converted into:
  1. AEthanol
  2. BCarbon dioxide
  3. CLactic acidAnswer
  4. DGlucose
Why

In muscle during strenuous exercise, oxygen runs short and the cell uses lactic acid fermentation: pyruvate is reduced by NADH to lactic acid, and NADH returns as so that glycolysis can continue.

Ethanol (A) and (B) are the products of alcoholic fermentation in yeast, where pyruvate is first decarboxylated. Glucose (D) is the starting material of glycolysis, not a product.

2025 Annual I, Set B, MCQ 31 markUnit 11
The main function of progesterone in human female reproductive system is:
  1. AOvulation
  2. BFollicle development
  3. CEndometrium developmentAnswer
  4. DImplantation of embryo
Why

Progesterone from the corpus luteum acts on the estrogen-primed endometrium and makes it glandular and secretory, so the main function is endometrium development (preparing it for implantation). Ovulation is triggered by the LH surge, follicle development is stimulated by FSH, and implantation is the embryo's attachment to the prepared endometrium, not a function of the hormone itself.

2026 Annual I, Set B, MCQ 41 markUnit 3
The minimum energy required to start a chemical reaction is called:
  1. APotential energy
  2. BActivation energyAnswer
  3. CKinetic energy
  4. DHeat energy
Why

The energy that must be added to make molecules react is the energy of activation. In the book it is the energy needed to break the existing bonds and begin the reaction; an enzyme lowers it. Potential energy is energy stored in a system and kinetic energy is energy of motion, so neither names the starting energy of a reaction. Heat is only the way the energy of activation is supplied in a non-living system, not the name of the energy itself.

2025 Annual I, Set B, MCQ 41 markUnit 13
When RNA polymerase binds to promoter for transcription, the unit which detaches form it is:
  1. A (Alpha)
  2. B (Beta)
  3. C (Gamma)Answer
  4. D (Theta)
Why

The unit that leaves the RNA polymerase once it is attached to the promoter is the **sigma () factor**. It is needed only to bind the promoter; then it is removed and the core enzyme (two and the subunits) carries on. The printed options have no sigma, so the intended letter is probably the one that looks closest in print, option C. The core subunits and (A and B) stay on the DNA, and there is no subunit.

Note. None of the printed options is sigma. Option C is the likely misprint for , so C is the answer.

2025 Annual I, Set B, MCQ 51 markUnit 4
Which one of the following sequence of reactions is correct regarding Calvin cycle?
  1. A3PGA 1-3BPG G3PAnswer
  2. B3PGA G3P RuP
  3. CG3P RuBP 3PGA
  4. DG3P RuP 3PGA
Why

In the reduction phase of the Calvin cycle, 3-PGA takes a phosphate from ATP to become 1,3-bisphosphoglycerate, which NADPH then reduces (losing a phosphate) to G3P. So the sequence is 3PGA, 1,3-BPG, G3P (option A).

B leaves out 1,3-BPG; G3P to RuP is also only part of the regeneration, and 3PGA does not become RuP directly. C and D start from G3P and reverse the cycle: RuBP is the acceptor that combines with to make 3PGA, and G3P is not changed into RuBP in one step.

2026 Annual I, Set B, MCQ 51 markUnit 5
Viruses are classified structurally based on:
  1. ASize only
  2. BShape and nucleic acid typeAnswer
  3. CColour
  4. DHabitat
Why

The book classifies viruses by the structure and composition of the virion: the capsid (shape: helical, polyhedral, enveloped or complex) and the genome (DNA or RNA, single or double stranded). Size alone (A) and colour (C) are not used. Habitat (D) is close to the classification by host (bacteriophage, plant and animal viruses), which is a separate scheme, not a structural one.

2025 Annual I, Set B, MCQ 61 markUnit 1
The stem cells which can differentiate into few cell types are:
  1. ATotipotent
  2. BUnipotent
  3. CPluripotent
  4. DOligopotentAnswer
Why

The book classification by potency puts oligopotent stem cells (D) at "a few different cell types", for example lymphoid or myeloid stem cells. The tempting option is B: unipotent cells make only one kind of cell, their own type. Totipotent cells (A) form all cell types and pluripotent cells (C) almost any cell.

MCQ 62026 Annual I, Set B, MCQ 61 markUnit 6
Which vitamin is commonly produced by intestinal bacteria?
  1. AVitamin A
  2. BVitamin C
  3. CVitamin KAnswer
  4. DVitamin D
Why

The bacterial flora of the colon make vitamin K and B vitamins, which the body absorbs. Vitamins A, C and D are not made by gut bacteria: A and C come from the diet, and D is made in the skin in sunlight.

2025 Annual I, Set B, MCQ 71 markUnit 2
The protein present in tendons and ligaments is:
  1. ACasein
  2. BKeratin
  3. CElastinAnswer
  4. DFibrinogen
Why

Elastin provides support for connective tissues such as tendons and ligaments (collagen is also a component of them). Keratin (B), the tempting option because it is also a fibrous structural protein, strengthens hair, nails, feathers, horns and beaks. Casein (A) is the milk protein and fibrinogen (D) is a blood plasma protein used in clotting.

2026 Annual I, Set B, MCQ 71 markUnit 9
Which plant traps insects using sticky leaves?
  1. ANepenthesAnswer
  2. BSaltwort
  3. CSunflower
  4. DPine
Why

Of the four plants, only Nepenthes (the pitcher plant) is a carnivorous plant that traps insects in modified leaves. Saltwort is a halophyte of saline soils, and sunflower and pine are ordinary photosynthetic plants that do not trap insects, so B, C and D are wrong.

Note. The plant that really traps insects on sticky leaves is the sundew (Drosera), which is not among the options. The pitcher plant is the only carnivorous plant listed, so A is the best key.

2026 Annual I, Set B, MCQ 81 markUnit 2
A triglyceride molecule is formed by:
  1. AOne fatty acid and one glycerol
  2. BTwo fatty acids and one phosphate
  3. CThree fatty acids and one glycerolAnswer
  4. DAmino acids and glycerol
Why

A triacylglycerol (triglyceride) is the ester of one glycerol and three fatty acids: each of the three –OH groups of glycerol joins a fatty acid and three water molecules are released. Option A describes a monoacylglycerol (only one fatty acid). Option B is tempting but a phosphate group belongs to a phospholipid, which has glycerol, two fatty acids and a phosphate. Amino acids (D) are the units of proteins, not of lipids.

2025 Annual I, Set B, MCQ 81 markUnit 3
The compound malonate is competitive inhibitor of:
  1. AFumarate
  2. BSuccinate
  3. CSuccinic dehydrogenaseAnswer
  4. DHistidine decarboxylase
Why

Malonate has a shape similar to succinate, the substrate of the enzyme succinate dehydrogenase (the book writes succinic dehydrogenase). It enters the active site, forms an enzyme-inhibitor complex and cannot be converted into product, so it competitively inhibits the enzyme. Succinate (B) is the substrate that malonate imitates, not the thing it inhibits, and fumarate (A) is the product of the reaction. Histidine decarboxylase (D) is a lyase and has nothing to do with malonate.

2026 Annual I, Set B, MCQ 91 markUnit 7
Mycorrhiza is an association between:
  1. ABacteria and roots
  2. BAlgae and fungi
  3. CPlants and insects
  4. DFungi and plant rootsAnswer
Why

A mycorrhiza is a mutually beneficial partnership between soil fungi and the roots of plants. The fungal hyphae increase the absorbing surface of the root and the plant gives the fungus carbohydrates from photosynthesis.

Algae and fungi (B) is the tempting option, but that association is a lichen, not a mycorrhiza. Bacteria and roots (A) describes a different association, and plants and insects (C) is not a fungal association at all.

2025 Annual I, Set B, MCQ 91 markUnit 13
The condition in which centromere of a chromosome is located slightly away from the center is termed as:
  1. AAcrocentric
  2. BSub metacentricAnswer
  3. CTelocentric
  4. DMetacentric
Why

A centromere located slightly away from the centre gives two arms of unequal length (one slightly shorter): a submetacentric chromosome. In a metacentric chromosome (D) it is exactly in the centre; in an acrocentric (A) it is near the end; in a telocentric (C) it is at the end.

2025 Annual I, Set B, MCQ 101 markUnit 7
Which fungus is poisonous?
  1. AAgaricus
  2. BMorchella
  3. CAspergillus
  4. DAmanitaAnswer
Why

Amanita (for example Amanita virosa, the destroying angel, and Amanita phalloides, the death cap) is the poisonous genus; one mushroom can be fatal.

Agaricus (A) is the main edible mushroom genus and Morchella (B), the morel, is edible. Aspergillus (C) is a mold used, for example, in soy sauce, though some strains make aflatoxin. The poisonous mushroom is Amanita.

2026 Annual I, Set B, MCQ 101 markUnit 10
Holozoic nutrition is found in:
  1. APlants
  2. BFungi
  3. CBacteria
  4. DAnimalsAnswer
Why

In holozoic nutrition an organism takes in (ingests) solid food and digests it inside its body. The book defines an animal as an eukaryotic, multicellular, ingestive heterotroph, so this is the mode of nutrition of animals. Plants make their own food (autotrophs), so they are not holozoic. Fungi and bacteria do not swallow food: they absorb dissolved nutrients from their surroundings.

Note. The word holozoic is not used in the book; the question is answered from the book's description of animals as ingestive heterotrophs.

2025 Annual I, Set B, MCQ 111 markUnit 5
Hepatitis-D is caused by:
  1. ABacteria
  2. BVirus
  3. CPrions
  4. DViroidsAnswer
Why

The book names the only human disease known to be caused by a viroid as hepatitis D: a short, circular, single stranded RNA with no protein coat. The tempting answer is the virus (B): hepatitis A, B and C are caused by viruses, and hepatitis D needs the hepatitis B virus to supply its capsid, but the cause of hepatitis D is the viroid. Prions (C) cause kuru and mad cow disease.

2026 Annual I, Set B, MCQ 111 markUnit 8
Example of vascular plant:
  1. AFernAnswer
  2. BAlgae
  3. CMoss
  4. DFungi
Why

A fern is a vascular plant: it has xylem and phloem, and its sporophyte has true roots, stems and leaves. Algae and fungi have no vascular tissue (they are not even in the plant groups with xylem and phloem), and a moss is a bryophyte, the non-vascular group, so options B, C and D are wrong.

2025 Annual I, Set B, MCQ 121 markUnit 9
Bark is produced by:
  1. ACork cambiumAnswer
  2. BLateral meristem
  3. CApical meristem
  4. DIntercalary meristem
Why

The cork cambium (phellogen) is the meristem that adds cork to the outside and secondary cortex to the inside. The cork cells die and, with the epidermis, form the rough outer bark of a mature tree, so the bark is produced by the cork cambium.

Lateral meristem (B) is tempting because the cork cambium is a lateral meristem, but the vascular cambium is also a lateral meristem and it makes wood, so the general name does not identify the bark-producing tissue. Apical meristem (C) gives primary growth in length and intercalary meristem (D) gives growth at the nodes or bases of leaves, not bark.

Note. The book defines bark as everything outside the vascular cambium, which includes secondary phloem made by the vascular cambium. The question asks for the layer that gives the outer bark (cork), so the key is A.

2026 Annual I, Set B, MCQ 121 markUnit 10
Which phylum includes animals with jointed appendages?
  1. AArthropodaAnswer
  2. BAnnelida
  3. CMollusca
  4. DEchinodermata
Why

Arthropoda means "animals with jointed legs" (joint-footed). The body is covered by a chitinous exoskeleton that is flexible at many places to allow articulation, and there are several pairs of appendages. Annelida are also segmented, but they move with circular and longitudinal muscles and setae, not jointed appendages.

2026 Annual I, Set B, MCQ 131 markUnit 12
Polygenic inheritance results in:
  1. ASingle trait expression
  2. BContinuous variationAnswer
  3. CDominant traits only
  4. DMutation only
Why

Polygenic traits are controlled by two or more gene pairs at different loci, all acting on the same trait in an additive way. The intensity of the phenotype depends on the number of effect-causing alleles, so many small steps appear between the extremes and the trait shows continuous variation (height, skin colour, wheat grain colour). Single phenotype, dominant-only and mutation-only are not what polygenic inheritance means.

2025 Annual I, Set B, MCQ 131 markUnit 12
The sex linked traits which are more common in males than female are:
  1. AAutosomal recessive
  2. BY- linked dominant
  3. CX- linked dominant
  4. DX- linked recessiveAnswer
Why

An X-linked recessive allele shows in a man with a single copy, because he has only one X, but a woman is affected only when she has two copies (). The trait is therefore more common in males (haemophilia, colour blindness). Autosomal traits are equally common in both sexes. A Y-linked trait occurs only in males, so it is 'only in males', not 'more common', and X-linked dominant traits are, if anything, more common in females.

MCQ 142025 Annual I, Set B, MCQ 141 markUnit 6
The organism having Chlorophyll - a, Phycocyanin and Heterocyst may be a/an:
  1. ABacterium
  2. BCyanobacteriumAnswer
  3. CFungus
  4. DAlga
Why

Chlorophyll-a with the blue pigment phycocyanin (a phycobilin) and nitrogen-fixing heterocysts are the features of cyanobacteria. A plain bacterium (A) does not have these combined; fungi (C) are non-photosynthetic; algae (D) have chlorophyll but no phycocyanin-bearing heterocysts and are eukaryotes. Cyanobacteria are called blue-green algae but are bacteria.

2026 Annual I, Set B, MCQ 141 markUnit 12
Epistasis modifies:
  1. ADNA
  2. BGenetrix ratio
  3. CPhenotypic ratioAnswer
  4. DProtein
Why

Epistasis is an interaction between two genes at different loci in which one gene hides the effect of the other. The genotypes are formed exactly as in Mendel's dihybrid cross, but the classes that look alike merge, so the usual 9 : 3 : 3 : 1 phenotypic ratio is modified (to 9 : 3 : 4 in the Labrador retriever). The DNA itself and the proteins are not changed by the interaction, and the genotypic ratio is unchanged.

Note. Option B is printed as 'Genetrix ratio' in the paper (probably 'genotypic ratio'). Even read as genotypic ratio, it is not the answer: epistasis changes the phenotypic ratio only.

2025 Annual I, Set B, MCQ 151 markUnit 10
Which of the following structure is common to all chordate embryo?
  1. APlacenta
  2. BNotochordAnswer
  3. CVertebral column
  4. DLungs
Why

The notochord is one of the four basic features of every chordate, present at some stage, so it is found in every chordate embryo. The vertebral column is found only in vertebrates (it replaces the notochord), the placenta only in eutherian mammals, and lungs are not a chordate feature (gill slits are).

2026 Annual I, Set B, MCQ 151 markUnit 11
Menstrual cycle is controlled by:
  1. AEnzymes
  2. BVitamins
  3. CMinerals
  4. DHormonesAnswer
Why

The menstrual cycle is controlled by hormones: the gonadotropins FSH and LH from the anterior pituitary, and estrogen and progesterone from the ovary. Enzymes, vitamins and minerals are not the chemical messengers that coordinate the ovarian and uterine cycles.

2025 Annual I, Set B, MCQ 161 markUnit 12
Identify the trait which is NOT controlled by polygenes?
  1. AIntelligence
  2. BHeight
  3. CSkin color
  4. DTongue rollingAnswer
Why

Intelligence, height and skin colour are polygenic traits with continuous variation. Tongue rolling is the book's example of a simple Mendelian trait of one gene pair (roller dominant, non-roller ) with two clear classes.

2026 Annual I, Set B, MCQ 161 markUnit 14
Molecular evidence of evolution is obtained from:
  1. ADNA analysisAnswer
  2. BEmbryology
  3. CAnatomy
  4. DFossils
Why

Molecular evidence comes from the common biomolecules in living organisms: DNA with a universal genetic code, ATP, the 20 amino acids and cytochrome c. Embryology and anatomy are different fields (comparative anatomy studies structures, not molecules), and fossils are evidence from palaeontology.

2026 Annual I, Set B, MCQ 171 markUnit 13
Griffith and Hershey-Chase experiments proved that:
  1. AProteins are hereditary material
  2. BDNA is hereditary materialAnswer
  3. CRNA is genetic material
  4. DLipids carry genes
Why

Griffith showed that a heritable factor passes from dead S-type to living R-type bacteria (transformation), and Hershey and Chase showed that only the DNA of a bacteriophage enters the bacterium to direct new viruses, while its protein coat stays outside. Together they point to DNA as the hereditary material. Protein (option A) was the older guess, which these experiments disproved; neither experiment involved RNA or lipids.

2025 Annual I, Set B, MCQ 171 markUnit 14
Which of the following is an example of homologous organs?
  1. AWing of an insect & a bird
  2. BForelimb of human & wing of a batAnswer
  3. CWing of an insect & a bat
  4. DGills of fish & lungs of mammals
Why

Homologous organs have the same basic internal pattern because of a common ancestor. The forelimb of a human and the wing of a bat are both mammalian forelimbs with the same bone pattern. The other pairs are analogous (insect wing and bird wing, insect wing and bat wing: same function, different origin) or unrelated (gills and lungs).

2026 Annual I, Set B, Q.2 (i)1x3 = 3 marksUnit 1
Differentiate between smooth and rough endoplasmic reticulum (any three points).
Solution
PointSmooth ERRough ER
RibosomesNone attached, so it looks smoothAttached to the side facing the cytoplasm, so it looks rough under the electron microscope
ShapeBranched tubules, continuous with the rough ERFlattened sacs (cisternae)
Main functionSynthesises lipids (oil, cholesterol, phospholipids, steroids and sex hormones) and detoxifies drugs and poisonsMakes secretory proteins on its ribosomes; also a membrane factory that adds proteins and phospholipids to its own membrane
Where abundantLiver cells, and skeletal muscle cells (as sarcoplasmic reticulum)Cells that secrete proteins
2025 Annual I, Set B, Q.2 (i)3 marksUnit 1
How does Golgi complex play role in Cell secretion and Cell division?
Solution
Cell secretion
  • The cell secretions are mostly proteins. The Golgi complex collects them from the rough ER (through the smooth ER) in transport vesicles at the cis face.
  • It modifies the proteins so that they can do their specific work, and forms glycoproteins and lipoproteins.
  • At the trans face the cisternae break up into secretory vesicles that carry the finished products out of the cell. That is why the Golgi complex is abundant in secretory cells.
Cell division (plant cells)
  • During cytokinesis the Golgi complex gives rise to vesicles that contain cell wall synthesising material.
  • These vesicles line up on the cell equator and fuse to form the phragmoplast; the new cell wall between the daughter cells is made from it.
2025 Annual I, Set B, Q.2 (i) OR3 marksUnit 3
Compare the characteristics of Cofactors of enzymes with one example of each.
Solution

A cofactor is a non-protein part that some enzymes need to become active. There are three types (1 mark each):

PointInorganic ions (activators)CoenzymesProsthetic group
NatureMetallic ions such as Fe, Mg, Cu, ZnOrganic molecules, derivatives of vitaminsOrganic part covalently bonded to the enzyme
AttachmentDetachable; bound when the substrate bindsDetachable; bound when the substrate bindsPermanently attached; does not detach after the reaction
One exampleMg with hexokinaseNAD with alcohol dehydrogenaseIron-containing porphyrin ring of cytochromes
2026 Annual I, Set B, Q.2 (i) OR1x3 = 3 marksUnit 8
State any three general characteristics of Whisk Ferns.
Solution

Whisk ferns (Psilopsida) are the simplest vascular plants. Three general characteristics:

  • No true roots: they have underground stems called rhizomes that bear rhizoids for anchorage.
  • No true leaves: the aerial stems carry only very small green scales, and the stems themselves carry out photosynthesis.
  • Sporangia are borne at the apical tips or along the sides of the aerial branches, and the gametophyte is a flat, colourless structure that lives beneath the soil with a fungus inside its cells.
2026 Annual I, Set B, Q.2 (ii)1+3 = 3 marksUnit 3
Classify enzymes on the basis of the substrates they act upon, giving any three examples.
Solution

The book classifies enzymes by the substrate they act upon into four groups. Any three, each with examples, earn the marks (1 mark per group with its examples).

Class (1 mark each)Acts on (substrate)ProductsExamples
ProteasesProteins and peptidesSmall polypeptides, peptones, dipeptides or amino acidsPepsin, trypsin, erypsin
LipasesLipidsFatty acids and glycerolPancreatic lipases
CarbohydrasesCarbohydratesDisaccharides or monosaccharidesAmylase (starch to maltose), maltase, sucrase, lactase
Nucleases (fourth group)DNA, RNA (and ATP)Ribonucleotides or deoxyribonucleotidesRNAases, DNAases, ATPases
2025 Annual I, Set B, Q.2 (ii)3 marksUnit 8
Write any three main features of Bryophytes.
Solution

Bryophytes (liverworts, hornworts and mosses) are the only non-vascular plants. Three main features:

  • They lack xylem and phloem, so they have no true roots, stems or leaves (only rhizoids and root-like, stem-like and leaf-like parts).
  • The gametophyte is the main plant body, green and independent. The sporophyte is smaller and depends on the gametophyte for food.
  • They have multicellular sex organs that produce the embryo, and the sporophyte is homosporous (one kind of spore). Water is needed for fertilization because the sperms swim.
2026 Annual I, Set B, Q.2 (ii) OR3 marksUnit 4
Outline steps of preparatory phase of glycolysis.
Solution
What the phase is (1 mark)

The preparatory (investment) phase is steps 1 to 5 of glycolysis in the cytoplasm. It uses 2 ATP and changes one glucose (6C) into two molecules of G3P (3C). No ATP is made and no NADH is formed in this phase.

The five steps (2 marks)
StepNameEnzymeWhat happens
1PhosphorylationHexokinaseATP gives a phosphate to glucose: glucose 6-phosphate + ADP
2IsomerizationPhosphohexose isomeraseGlucose 6-phosphate becomes its isomer fructose 6-phosphate
3PhosphorylationPhosphofructokinaseA second ATP gives a phosphate: fructose 1,6-bisphosphate + ADP
4SplittingAldolaseFructose 1,6-bisphosphate (6C) splits into two 3C sugars: G3P and dihydroxyacetone 3-phosphate
5IsomerizationPhosphotriose isomeraseDihydroxyacetone 3-phosphate becomes G3P, so there are two G3P

Because step 5 gives a second G3P, everything in the pay-off phase happens twice for each glucose.

Q.2 (ii) OR2025 Annual I, Set B, Q.2 (ii) OR1x3 = 3 marksUnit 6
Enlist any three physical methods used to control bacteria with their mechanisms.
Solution
  1. Moist heat (steam, boiling, pasteurisation): the heat causes coagulation of protein in the microbe, so it dies. Milk is pasteurised at for 15 seconds.
  2. Dry heat (hot air): heat oxidises the parts of the microbe and kills it. Both dry and moist heat are used in microbiology laboratories.
  3. Radiation: electromagnetic radiation below 300 nm, for example gamma rays, kills micro-organisms and is used for sterilisation. (A membrane filter removes bacteria from heat-sensitive liquids such as antibiotics and sera.)
2025 Annual I, Set B, Q.2 (iii)3 marksUnit 2
Justify the statement "Water is a universal solvent".
Solution
Polarity of water (1 mark)

In a water molecule the shared electrons are pulled towards oxygen, so oxygen is slightly negative () and each hydrogen slightly positive (). Water is a polar molecule.

Dissolving ionic and polar substances (1 mark)

The charged ends of water molecules attract the ions of ionic compounds (electrolytes) and the polar groups of other molecules, surround them and separate them. Non-polar molecules that carry charged groups also dissolve because their charged regions attract water.

Importance in living things (1 mark)

On dissolving, substances dissociate into positive and negative ions, which are in a more favourable state to react. This is why all the chemical reactions of living organisms occur in an aqueous medium. Water is therefore called the universal solvent (for polar substances).

2026 Annual I, Set B, Q.2 (iii)1x3 = 3 marksUnit 10
Describe any three general characteristics of sponges.
Solution

Phylum Porifera ("pore bearer") has these characteristics (any three, 1 mark each):

  • Body organization: sponges are sessile, aquatic (mostly marine) animals. The body is multicellular but not organized into tissues or organs and has no symmetry.
  • Pores and cavity: the body is perforated by many pores, the ostia, and has a single central cavity, the spongocoel. Water enters by the ostia and leaves by the osculum, bringing food with it.
  • Skeleton: it is made of needle-like spicules (calcareous or siliceous) or of spongin fibre (bath sponge).
  • Body wall and reproduction: three layers (pinacoderm, mesenchyme, choanoderm); asexual reproduction by budding or gemmules, sexual by egg and sperm; they regenerate from a small fragment.

Example: Spongilla, Sycon.

Q.2 (iii) OR2025 Annual I, Set B, Q.2 (iii) OR3 marksUnit 6
Endospore formation is a mechanism in bacteria to survive under unfavorable conditions. Justify.
Solution
Why the bacterium forms a spore (1 mark)

When nutrients run short or temperature and dryness become extreme, bacteria such as Bacillus and Clostridium undergo sporulation instead of dying: the cell forms one endospore inside itself.

How the spore resists (1 mark)
  1. The core holds the DNA, a little cytoplasm and ribosomes, with very little water.
  2. Several membranes, a peptidoglycan cortex and a thick keratin-like protein coat surround it.
  3. So it withstands high temperature, drying, chemicals and lack of nutrients.
Structure of an endospore inside the mother cell: the layers protect the core.
Return to active life (1 mark)

The spore stays dormant until the mother cell ruptures and releases it. When conditions are favourable again it takes in water, germinates and grows into a vegetative cell. Since one cell makes one spore, this is survival, not reproduction.

2026 Annual I, Set B, Q.2 (iii) OR1x3 = 3 marksUnit 13
State three conclusions of Griffith control experiments.
Solution

Griffith injected mice with different bacterial samples. The control experiments (live R-type alone, heat-killed S-type alone, and the mixture) lead to three conclusions.

Griffith control experiments
Control experimentResultConclusion
Live R-type (rough, no capsule) injected aloneMice stay healthyR-type is non-virulent: it cannot cause disease by itself
Heat-killed S-type injected aloneMice stay healthyKilling by heat destroys virulence; the dead capsule material alone does not make the mice ill
Live R-type mixed with heat-killed S-typeMice develop disease and die; living S-type bacteria are found in their bloodSome hereditary information (the transforming principle) passed from the dead S-type to the living R-type and changed it into virulent S-type: transformation
2026 Annual I, Set B, Q.2 (iv)3 marksUnit 2
Describe three effects of amino acid sequence change in haemoglobin.
Solution

The sequence of amino acids is the primary structure of a protein and it decides how the protein works. In haemoglobin the change of one amino acid out of 146 in the β-chain (HbA to HbS) has three effects.

Effect 1: the haemoglobin molecule itself changes (1 mark)

Glutamic acid at position 6 of the β-chain is replaced by valine, so normal haemoglobin (HbA) becomes abnormal sickle haemoglobin (HbS).

Effect 2: the red blood cells change shape (1 mark)

Normal red blood cells are discs, like doughnuts without holes, and move easily through vessels. Cells with HbS become sickle or crescent shaped, stiff and sticky.

Effect 3: blood flow is blocked and the disease sickle cell anaemia results (1 mark)

Sickle cells block blood flow in the vessels of the limbs and organs, which causes pain and organ damage.

How one nucleotide leads to the disease
  1. A point mutation in the β-globin gene replaces one nucleotide by another, so the codon changes.
  2. One amino acid of the β-chain changes: glutamic acid is replaced by valine at position 6 (Thr-Pro-Glu-Glu becomes Thr-Pro-Val-Glu).
  3. The abnormal sickle haemoglobin (HbS) makes the red blood cells sickle or crescent shaped, stiff and sticky.
  4. The sickle cells block blood flow in the vessels of the limbs and organs, causing pain and organ damage.
2025 Annual I, Set B, Q.2 (iv)3 marksUnit 11
List the Accessory glands associated with male reproductive system with their respective locations.
Solution
Accessory gland (1 mark each)Location
Seminal vesicles (a pair)At the junction of the sperm duct (vas deferens) and the ejaculatory duct
Prostate glandEncircles the urethra just below the bladder
Bulbourethral (Cowper's) glands (a pair)At the junction of the ejaculatory duct and the urethra

Their secretions make up the liquid part of the semen.

2026 Annual I, Set B, Q.2 (iv) OR3 marksUnit 5
Enlist any six stages of HIV life cycle.
Solution

HIV has twelve numbered stages in the book. Any six, in order (half a mark each):

No.StageWhat happens
1Attachment (adsorption)gp 120 binds the CD4 receptor of the T cell
2Penetration and uncoatingEnvelope fuses with the membrane using gp 41; the virion enters by endocytosis and sheds its coat
3Reverse transcriptionReverse transcriptase makes cDNA from the viral RNA
4Breakdown of viral RNARibonuclease destroys the viral genomic RNA
5ReplicationSingle stranded cDNA is copied into double stranded cDNA
6IntegrationIntegrase inserts the DNA into the host DNA; it is then the provirus
2025 Annual I, Set B, Q.2 (iv) OR1+2 = 3 marksUnit 12
What is incomplete dominance? Explain with a suitable example.
Solution
Definition (1 mark)

Incomplete dominance is a dominance relation in which neither allele of a pair is completely dominant, so the heterozygote shows a phenotype intermediate between those of its two homozygous parents.

Example: flower colour of the 4 O'clock plant (2 marks)
  1. Red () x white () gives all pink () F1 plants, as Carl Correns found in 1899.
  2. F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1, which is also the genotypic ratio ().
Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.
2025 Annual I, Set B, Q.2 (v)1+2 = 3 marksUnit 4
Why and how cyclic Photophosphorylation occurs in light reactions?
Solution
Why it occurs (1 mark)

Non-cyclic flow makes NADPH and ATP, but the Calvin cycle needs more ATP than NADPH. When NADPH rises and ATP is short, the chloroplast shifts for a while to cyclic flow, which makes only ATP, until the supply of ATP catches up with the demand.

How it occurs (2 marks)
  1. Only PS-I is used. P700 absorbs light and its electrons are boosted to a higher energy level.
  2. The excited electrons are captured by the primary electron acceptor of PS-I.
  3. They pass through a chain of **ferredoxin, cytochrome bf complex and plastocyanin**.
  4. As electrons pass through the cytochrome complex, protons are pumped and ATP is made by chemiosmosis.
  5. The electrons, having lost energy, return to P700.
Figure 4.9: Cyclic electron flow during photophosphorylation. The electrons (accent) return to P700, so only ATP is produced.
2026 Annual I, Set B, Q.2 (v)1x3 = 3 marksUnit 9
Briefly explain three types of plant movements.
Solution

Plants cannot move from place to place, but their organs move by differential growth in response to an external stimulus. These growth movements are called tropisms (tropic movements). Three of them are:

1. Phototropism (1 mark)

Movement of a plant part towards (positive) or away from (negative) light. Shoots grow towards light (positive); roots grow away from light (negative).

2. Geotropism (1 mark)

Movement of a plant part in response to gravity. Shoots grow upward, away from gravity (negative); roots grow downward, towards gravity (positive).

3. Thigmotropism (1 mark)

Movement in response to touch. Climbing plants coil round a wall, rod or rope that supports them.

(Chemotropism, the growth of fungal hyphae towards nutrients, may be written instead of any one of these.)

2025 Annual I, Set B, Q.2 (v) OR3 marksUnit 5
Write steps of lytic cycle in bacteriophage.
Solution

The lytic cycle of a bacteriophage (virulent phage) has these steps, in order:

  1. Adsorption. The tail fibres and pins attach to receptors on the bacterial cell.
  2. Penetration and genome injection. The sheath contracts, lysozyme digests the cell wall and the DNA is injected through the core tube.
  3. Disintegration of host DNA. A viral DNAase breaks down the host DNA; the viral DNA takes control.
  4. Synthesis of phage components. Phage mRNAs and proteins (head, tail and lysis protein) are made.
  5. Assembly and maturation. DNA is packed into heads and tails are added; about 200 phages in 20 to 25 minutes.
  6. Lysis and release. Lysozyme accumulates, the cell bursts and the phages are released.
Q.2 (v) OR2026 Annual I, Set B, Q.2 (v) OR1x3 = 3 marksUnit 6
Compare photosynthetic bacteria and chemosynthetic bacteria. (Three points)
Solution
PointPhotosynthetic bacteriaChemosynthetic bacteria
Source of energySunlight (light energy)Energy released by oxidising inorganic substances (for example sulphur compounds, ammonia)
PigmentContain bacteriochlorophyll, built into the membrane or scattered in the cytoplasmColourless; no light-trapping pigment
Hydrogen source and by-productUse hydrogen sulphide () instead of water, so sulphur is releasedEnergy comes from oxidising the inorganic compound; no light needed
ExamplesGreen sulphur, purple sulphur and purple non-sulphur bacteriaNitrifying bacteria and sulphur bacteria

Both groups are autotrophic: they build carbohydrate from carbon dioxide. They differ only in the source of energy used to do it.

Note. Three points are asked; the table gives four, so write any three.

2025 Annual I, Set B, Q.2 (vi)1+2 = 3 marksUnit 7
What is mycorrhiza? Differentiate between its types.
Solution
Definition (1 mark)

A mycorrhiza is a mutually beneficial partnership between soil fungi and the roots of plants. The hyphae increase soil contact and surface area and help the root absorb phosphorus, zinc and copper; the plant gives the fungus carbohydrates made by photosynthesis.

Types (2 marks)
Endomycorrhiza and ectomycorrhiza
FeatureEndomycorrhizaEctomycorrhiza
Position of the fungusPenetrates the outer (cortex) cells of the rootForms a protective sheath outside the root
Growth in the rootCoils, swellings and tiny branches inside the cellsHyphae grow between the cell walls and do not enter the cells
Plants involvedNot named in the bookTypically trees such as pines and firs
2026 Annual I, Set B, Q.2 (vi)1x3 = 3 marksUnit 8
Give three characteristics of bryophytes.
Solution

Bryophytes (liverworts, hornworts and mosses) are the non-vascular plants. Three characteristics:

  • They lack specialized vascular tissue (no xylem or phloem), so they have no true roots, stems or leaves.
  • The gametophyte is the dominant, green, independent plant body, while the sporophyte is smaller and obtains its food from the gametophyte.
  • They form multicellular sex organs (antheridia and archegonia) and a multicellular embryo, and the sporophyte is homosporous; their swimming sperms need water for fertilization.
Q.2 (vi) OR2026 Annual I, Set B, Q.2 (vi) OR1x3 = 3 marksUnit 6
Briefly explain three physical methods to control bacteria.
Solution
  1. Moist heat (steam, boiling, pasteurisation): the heat causes coagulation of protein in the microbe, so it dies. Milk is pasteurised at for 15 seconds.
  2. Dry heat (hot air): heat oxidises the parts of the microbe and kills it. Both dry and moist heat are used in microbiology laboratories.
  3. Radiation: electromagnetic radiation below 300 nm, for example gamma rays, kills micro-organisms and is used for sterilisation. (A membrane filter removes bacteria from heat-sensitive liquids such as antibiotics and sera.)
2025 Annual I, Set B, Q.2 (vi) OR1+2 = 3 marksUnit 13
What is point mutation? How it causes sickle cell anaemia?
Solution
Point mutation (1 mark)

A point (gene) mutation is a change in a single gene produced by a change of one or a few bases of the DNA. It may be a base substitution (one base replaced by another), an insertion or a deletion of bases.

How it causes sickle cell anaemia (2 marks)
  1. A single base substitution occurs in the gene for the beta chain of haemoglobin: the DNA triplet CTT for glutamic acid becomes CAT, so the mRNA codon GAA becomes GUA.
  2. The sixth amino acid of the beta chain is therefore valine in place of glutamic acid. This abnormal haemoglobin is Hb.
  3. Hb has lost its normal oxygen-binding capacity and forms strands, so the red blood cells become sickle shaped, block the flow of blood and deliver less oxygen to the tissues. The disease is autosomal recessive: it appears when the gene is inherited from both parents.

Note. The question carries a unit note for units 2 and 12; the base change and the disease are taught in unit 13 (gene or point mutations).

2026 Annual I, Set B, Q.2 (vii)1+2 = 3 marksUnit 3
What is meant by minimum temperature? How it affects enzymatic action?
Solution
Meaning of minimum temperature (1 mark)

The minimum temperature is the temperature at which an enzyme that has been inactivated by cold becomes active again when it is warmed.

Effect on enzymatic action (2 marks)
  • Near or below freezing point enzymes are inactivated, not denatured: molecular motion is too slow for enzyme and substrate to react, but the enzyme structure is not damaged.
  • When the temperature is raised the activity returns, and the rate then rises (about double for every 10 degrees Celsius) up to the optimum, 37 to 38 degrees Celsius for human enzymes.
TermMeaningEffect on the enzyme
Minimum temperatureTemperature at which an inactive (cold) enzyme becomes active againInactivated at freezing, not damaged; recovers on warming
Optimum temperatureTemperature of maximum activityHighest rate
Maximum temperatureTemperature that causes denaturationStructure and active site destroyed
2025 Annual I, Set B, Q.2 (vii)3 marksUnit 9
Compare apoplast and symplast pathways involved in uptake of water by roots in plants.
Solution

Water entering the root hairs reaches the xylem by the apoplast, symplast or vacuolar pathway. The first two compare as follows:

PointApoplast pathwaySymplast pathway
Made ofThe system of adjacent cell walls (and the spaces in them), continuous through the plantThe system of interconnected protoplasts (cytoplasm) of the cells
Linked bySpaces in the cell wallsPlasmodesmata between neighbouring cells, helped by cytoplasmic streaming
Membranes crossedNone until the endodermisNone after the first cell (water moves from cytoplasm to cytoplasm)
At the endodermisWater is stopped by the Casparian strips (suberin and lignin) and must enter the cytoplasmWater passes on through the cytoplasm to the xylem
Figure 9.4: Pathways of water in the root. The three pathways are in the accent colour, each with its own line style.
2026 Annual I, Set B, Q.2 (vii) OR1x3 = 3 marksUnit 1
Draw and briefly explain three events of Endosymbiotic theory.
Solution
Endosymbiotic theory: the three events, shown for the mitochondrion (row A) and the chloroplast (row B).
Event 1: engulfment (1 mark)

Mitochondria and chloroplasts show many similarities with bacteria, so evolutionists believe that they were once independent organisms. An early ancestor of the eukaryotic cell (the host) engulfed an aerobic bacterium, which became the mitochondrion. In the same way a host cell engulfed a photosynthetic bacterium (a cyanobacterium), which became the chloroplast.

Event 2: the engulfed cell stays alive and forms a relationship (1 mark)

The engulfed cell was not digested. It formed a relationship with the host cell in which it was enclosed and became an endosymbiont, a cell living within another cell. The host gave it protection and raw materials, and the endosymbiont gave the host energy (ATP from the aerobic bacterium; food from the photosynthetic one).

Event 3: the endosymbiont becomes an organelle (1 mark)

Over time the endosymbiont became a permanent part of the host, an organism within an organism. It still divides by itself (fission) and keeps its own small circular DNA, all kinds of RNA and 70S ribosomes, so its number doubles before cell division and it makes some of its own proteins.

2025 Annual I, Set B, Q.2 (vii) OR3 marksUnit 14
Molecular biology provides evidence of evolution. Justify the statement with reference to any three examples.
Solution

Molecular biology shows biochemical homology: the same kinds of molecules are present in a complete range of living organisms, just as organs and tissues show anatomical homology. Even taxonomically very remote organisms share them, so they reflect an evolutionary relationship and a common ancestor. Three examples:

  1. DNA: the hereditary material of all living organisms is DNA made of the same kind of nucleotides, and the genetic code is universal.
  2. ATP: the energy currency molecule is the same from prokaryotes to the most advanced plants and animals.
  3. Amino acids: all living organisms make their proteins from the same 20 types of amino acid.

Cytochrome c, a respiratory protein of the respiratory chain of all aerobic organisms, is a fourth example. Because such molecules are shared even by remote species, they point to the origin of living organisms from a common ancestor.

2026 Annual I, Set B, Q.2 (viii)1x3 = 3 marksUnit 15
Explain three ways of nitrogen fixation.
Solution

Nitrogen gas () has a very strong triple bond between its two atoms, so it is chemically unreactive and a large amount of energy is needed to break the bond. It must be fixed (combined into compounds) before organisms can use it. It is fixed in three ways.

1. Atmospheric fixation (1 mark)

Lightning supplies the energy for nitrogen and oxygen of the air to combine and form various oxides of nitrogen. Rain carries these oxides into the soil, where plants can use them. This is spontaneous and fixes only a small amount (5 to 8%).

2. Industrial fixation (1 mark)

Man makes nitrogen-containing fertilizers in industry, for example by combining nitrogen and hydrogen to form ammonia.

3. Biological fixation (1 mark)

Nitrogen-fixing bacteria convert atmospheric nitrogen to ammonia. This is the largest share (about 60%). Free-living bacteria such as Azotobacter (aerobic) and Clostridium (anaerobic) make fixed nitrogen available to plants when they die and lyse. Rhizobium is symbiotic and lives in the root nodules of legume plants.

2025 Annual I, Set B, Q.2 (viii)1.5+1.5 = 3 marksUnit 15
Write about de-nitrification and assimilation processes in nitrogen cycle.
Solution
Denitrification (1.5 marks)

In the absence of oxygen, certain soil bacteria (denitrifying bacteria, e.g. Pseudomonas) break down nitrates and use the oxygen for their own respiration. Nitrogen is released back into the atmosphere as a gas, so this is a loss of nitrogen from the soil. It completes the cycle.

Assimilation (1.5 marks)

Assimilation is the utilisation of nitrogenous compounds in living bodies. Plants take up nitrogen from the soil as inorganic compounds such as ammonia and nitrates and build amino acids, proteins and nucleic acids from it. Animals get their nitrogen by eating plants or other animals. (Many micro-organisms can use free nitrogen directly from the atmosphere.)

Q.2 (viii) OR2026 Annual I, Set B, Q.2 (viii) OR1x3 = 3 marksUnit 6
Enlist three advantages of endospore formation.
Solution
  1. Survival of extreme conditions: the spore resists high temperature, drying (desiccation), and exposure to chemicals and most poisons, so the bacterium outlives shortage of water and nutrients.
  2. Protection of the genetic material: the DNA, a little cytoplasm and ribosomes lie in the core, shielded by several membranes, a peptidoglycan cortex and a thick protein (keratin-like) coat.
  3. Dormancy and recovery: the dehydrated spore stays dormant for a long time and, when conditions improve, germinates into an active vegetative cell that can grow and reproduce.
2025 Annual I, Set B, Q.2 (viii) OR2+1 = 3 marksUnit 11
Write about the structure and functions of ovary and oviduct in female reproductive system.
Solution
Structure (2 marks)
OrganStructure
OvaryA pair of solid, ovoid gonads, 3 to 5 cm long and 2 to 3 cm wide, lying on the sides of the uterus and held in the peritoneal cavity by ligaments. Inside are many tiny sac-like ovarian follicles, each with an immature egg (oocyte); after ovulation the ruptured follicle becomes the glandular corpus luteum.
Oviduct (Fallopian tube)The initial part of the female duct system, about 10 cm long, running from the region of the ovary to the uterus; it receives the ovulated oocyte.
Functions (1 mark)
  • Ovary: produces ova (eggs) by oogenesis and releases hormones (estrogen and progesterone).
  • Oviduct: receives the oocyte, is the usual site of fertilization, and transfers the developing ovum towards the uterus.
2026 Annual I, Set B, Q.2 (ix)1x3 = 3 marksUnit 11
Name three Hormones of male reproductive system and write role of each Hormone.
Solution
HormoneMade byRole (1 mark each)
FSHAnterior pituitaryStimulates the Sertoli cells of the testes to facilitate spermatogenesis
LHAnterior pituitaryStimulates the interstitial (Leydig) cells of the testes to make and release testosterone
TestosteroneLeydig cellsStimulates spermatogenesis and is responsible for the secondary sexual characteristics (deepening of voice, facial, axillary and pubic hair); inhibits GnRH, FSH and LH by negative feedback

Any three of these are accepted. GnRH (from the hypothalamus, makes the anterior pituitary release FSH and LH) and inhibin (from the Sertoli cells, inhibits GnRH and FSH when the sperm count is too high) are the other two.

2025 Annual I, Set B, Q.2 (ix)1+1+1 = 3 marksUnit 12
What is Y- linked inheritance? How these traits are transmitted? Also give two examples.
Solution
Y-linked inheritance (1 mark)

Y-linked (holandric) traits are controlled by genes on the non-homologous region of the Y chromosome, which have no allele on the X chromosome.

Transmission (1 mark)
  • They are expressed only in males, as only males have a Y chromosome.
  • A son always receives the Y from his father, so the trait passes from father to all his sons and never to daughters.
Examples (1 mark)

Hypertrichosis (hair on the rim of the pinna), porcupine man (straight hair on the body) and webbing of the toes.

Q.2 (ix) OR2025 Annual I, Set B, Q.2 (ix) OR1x3 = 3 marksUnit 6
Name and briefly explain Growth Phases of bacteria. (Any three)
Solution
PhaseWhat happens
Lag phaseBacteria adapt to the new medium and may make new enzymes to digest its nutrients; growth has not yet reached its maximum rate.
Log (exponential) phaseConditions are favourable; cells divide repeatedly and growth is at its maximum rate, a straight line when is plotted against time.
Stationary phaseNutrients are used up and toxic wastes build up, so the death rate equals the reproduction rate; the number of living cells stays constant.
Death (decline) phaseThe death rate exceeds reproduction, cells stop multiplying and the number of living bacteria falls.
Growth curve of a bacterial population
O
Typical growth curve ( is the logarithm of the number of living bacteria): lag, log (exponential), stationary and death (decline) phases.
2026 Annual I, Set B, Q.2 (ix) OR1x3 = 3 marksUnit 11
State three events of menstrual phase of menstrual cycle.
Solution
Events of the menstrual phase (days 1 to 5), 1 mark each
  1. The uterus sheds all but the deepest part of its endometrium: the thick, hormone-dependent functional layer detaches from the uterine wall.
  2. The detached tissue and blood pass out through the vagina as the menstrual flow, with bleeding for 3 to 5 days.
  3. The ovarian hormones are at their lowest normal levels, and the FSH level begins to rise (a new follicle starts to grow).
2025 Annual I, Set B, Q.2 (x)3 marksUnit 3
Complete the table with reference to enzymes: [Table]
EnzymeMechanism of actionExample
a______Break large molecules with water______
bLyases____________
c____________Cytochrome oxidase
Solution

Use the six groups of enzymes classified by reaction type. Each row is worth 1 mark.

EnzymeMechanism of actionExample
aHydrolasesBreak large molecules into smaller ones by adding water (hydrolysis)Pepsin (also renin, trypsin)
bLyasesBreak specific covalent bonds and remove groups without hydrolysisHistidine decarboxylase
cOxidoreductasesCatalyse oxidation and reduction by removing or adding electrons or H ionsCytochrome oxidase
2026 Annual I, Set B, Q.2 (x)3 marksUnit 12
Explain through a cross that incomplete dominance is a deviation from Mendel's law of segregation.
Solution

Mendel's monohybrid cross gives a 3 : 1 phenotypic ratio in F2 because one allele is completely dominant and hides the other in the heterozygote. In incomplete dominance neither allele is completely dominant, so the ratio changes. The 4 O'clock plant (Mirabilis jalapa) shows this.

Cross (1 mark for parents and F1)
  • Parents: pure red () x pure white (). Gametes: and .
  • F1: all with pink flowers. The heterozygote is intermediate and resembles neither parent, whereas Mendel's F1 resembled one parent.
F2 (1 mark)
F1 gametes
red pink
pink white

Genotypic ratio 1 : 2 : 1 and phenotypic ratio 1 : 2 : 1 (red : pink : white).

Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.
Why it is a deviation (1 mark)

The alleles still separate (segregate) into different gametes, but because dominance is incomplete the heterozygote can be told apart from both homozygotes. The F2 phenotypic ratio is therefore 1 : 2 : 1 instead of Mendel's 3 : 1, and the genotypic and phenotypic ratios become the same.

2025 Annual I, Set B, Q.2 (x) OR2+1 = 3 marksUnit 12
Coat color in Labrador retriever shows epistasis. If a black(BbEe) and a chocolate (bbEe) Labradors retrievers are crossed: (a) Show genotypes and phenotypes of offspring(s). (b) Calculate the probability of black and yellow coat color.
Solution
(a) Genotypes and phenotypes (2 marks)

Parents: black x chocolate . Gametes of : , , , . Gametes of : , .

Eight equally likely combinations.
Gametes
black black
black yellow
chocolate chocolate
chocolate yellow
PhenotypeGenotypesOut of 8
Black (1), (2)3
Chocolate (1), (2)3
Yellow (1), (1)2

Phenotypic ratio black : chocolate : yellow . Yellow appears because is epistatic: it hides both and .

(b) Probabilities (1 mark)

Black (37.5%). Yellow (25%).

2026 Annual I, Set B, Q.2 (x) OR1x3 = 3 marksUnit 13
Describe three steps of Polypeptide elongation during protein synthesis.
Solution

Polypeptide elongation is repeated in three steps while the ribosome moves along the mRNA from the 5′ end towards the 3′ end.

  1. Codon recognition: the codon exposed at the A site is read by the aminoacyl-tRNA whose anticodon is complementary to it. The tRNA binds there with the help of an elongation factor.
  2. Peptide bond formation: the enzyme peptidyl transferase takes the amino acid (or the growing chain) from the tRNA at the P site and joins it by a peptide bond to the new amino acid on the tRNA at the A site.
  3. Translocation: the ribosome moves one codon along the mRNA (5′ to 3′). The empty tRNA reaches the E site and leaves, the tRNA carrying the chain moves from the A site to the P site, and a new codon is exposed at the A site.
2025 Annual I, Set B, Q.2 (xi)3 marksUnit 4
Illustrate the conversion of pyruvic acid to acetyl CoA as link reaction between glycolysis and Krebs cycle.
Solution

Pyruvate made in the cytosol by glycolysis enters the mitochondrion by active transport through pyruvate translocase. There the pyruvate dehydrogenase complex (three enzymes) converts it to acetyl CoA, which enters the Krebs cycle. This is why it is the link reaction.

Figure 4.13: Pathway of oxidation of pyruvate. One and one NADH are released for each pyruvate.
  1. Decarboxylation: a is removed from pyruvate (3C), forming acetaldehyde (2C).
  2. Oxidation (dehydrogenation): removes hydrogen from acetaldehyde, giving a two-carbon acetyl group and NADH.
  3. Addition of coenzyme A: the acetyl group joins CoA to form acetyl CoA.

For one glucose (2 pyruvate) this gives 2 acetyl CoA, 2 NADH and 2 . No ATP is made directly.

2026 Annual I, Set B, Q.2 (xi)1x3 = 3 marksUnit 10
State three characteristics which are common in all chordates.
Solution

All chordates possess four basic characteristics (any three, 1 mark each):

  • Notochord: a solid, unjointed rod in the mid-dorsal line between the gut and the nerve cord; it is the axial endoskeleton.
  • Dorsal hollow nerve cord: a single, tubular, fluid-filled central nervous system lying above the notochord. Non-chordates have a ventral, solid nerve cord.
  • Gill slits (pharyngeal pouches): paired openings from the pharynx to the outside, present at some stage of life.
  • Post-anal tail (the fourth characteristic): a tail extending beyond the anus, present at least in the embryo.
2026 Annual I, Set B, Q.2 (xi) OR1x3 = 3 marksUnit 7
Name and explain any three diseases caused by fungi in humans.
Solution
Fungal diseases of humans (any three)
Disease (1 mark each)Causative fungusExplanation
RingwormMicrosporum audouiniA superficial infection: only the skin, hair or nails are infected, giving a ring-shaped scaly patch.
Athlete's footTinea pedisA superficial infection of the skin, usually between the toes.
CandidiasisCandida yeastProduces oral thrush in humans (white patches in the mouth).
HistoplasmosisHistoplasma (spores are inhaled)A systemic, serious infection of the lungs.

Any three rows earn full marks: one mark for each disease named with its cause and a short explanation.

2025 Annual I, Set B, Q.2 (xi) OR3 marksUnit 13
Interpret the results of Meselson and Stahl experiment when they grew bacteria in N_{15} and then in N_{14} medium for tube 2, 3 and 4. [Figure]
The figure as set: bands of DNA in four tubes (1 control, 2 at 0 min, 3 at 20 min, 4 at 40 min).
Solution

Bacteria were grown for many generations in ^{15}\mathrm{N} (heavy) medium, then moved to ^{14}\mathrm{N} (light) medium. DNA was extracted at intervals and spun in a CsCl density gradient; heavier DNA settles lower in the tube.

TubeBand positionInterpretation
1 (control)Top (lightest)DNA with ^{14}\mathrm{N} in both strands, the reference for light DNA
2 (0 min)LowestAll DNA is heavy: both strands contain ^{15}\mathrm{N} (no replication has happened in ^{14}\mathrm{N} yet)
3 (20 min, one generation)One band at the middleAll DNA is a hybrid: each daughter duplex has one old ^{15}\mathrm{N} strand and one new ^{14}\mathrm{N} strand
4 (40 min, two generations)Two bands: top and middleHalf the molecules are hybrid (one ^{15}\mathrm{N} and one ^{14}\mathrm{N} strand) and half are light (both strands ^{14}\mathrm{N})

One middle band after one generation rules out the conservative model (it predicts a heavy and a light band), and the light band at 40 minutes rules out the dispersive model (it predicts a single band that only moves up). The results fit semi-conservative replication.

2025 Annual I, Set B, Q.2 (xii)3 marksUnit 8
Write difference in flower, stem and seed of dicots and monocots.
Solution

Differences between dicots and monocots in flower, stem and seed:

PartDicotsMonocots
FlowerTypically tetra- or penta-merous; calyx and corolla differ; usually insect pollinatedParts usually in threes (trimerous); no distinction between calyx and corolla; often wind pollinated
StemVascular bundles in a ring; vascular cambium present, so secondary growth occursVascular bundles scattered; cambium usually absent, so no secondary growth
SeedEmbryo has two cotyledonsEmbryo has one cotyledon
2026 Annual I, Set B, Q.2 (xii)1x3 = 3 marksUnit 14
Briefly elaborate three drawbacks of Larmarckism.
Solution

Lamarck was right that the environment can change an individual, but his mechanism fails for three main reasons:

  1. No experimental proof: there is no experimental evidence that acquired characters are inherited.
  2. Use and disuse cannot alter the genotype: the changes are non-genetic, so the organism's genes stay the same and nothing is passed on.
  3. Somatic changes are not heritable: changes of the body made during life (such as a blacksmith's biceps) are not carried to the offspring. Organs are also not modified by the wish or requirements of an organism.
2026 Annual I, Set B, Q.2 (xii) OR3 marksUnit 8
Describe three characteristics of angiosperms, that make them successful land plants.
Solution

Angiosperms (flowering plants) are the dominant land plants today. Three characteristics that make them successful:

  • Flowers and fruits: the ovules are enclosed in the ovary, which becomes the fruit after fertilization. The seed is protected, and the fruit helps to protect and spread it, which enhances reproductive success.
  • Well-developed vascular system: vessels and fibres in the xylem and companion cells in the phloem conduct water and food efficiently, so the plant can grow large on land.
  • Dominant sporophyte with heteromorphic alternation of generations, and a pollen tube that carries the male gametes to the ovule, so fertilization does not depend on water.
2025 Annual I, Set B, Q.2 (xii) OR3 marksUnit 13
Name any three enzymes with their functions involved in DNA replication.
Solution

Three enzymes of DNA replication and their functions (any three):

EnzymeFunction
DNA helicaseBreaks the hydrogen bonds between the base pairs and opens the duplex at the origin to form the replication bubble and forks
PrimaseMakes short RNA primers on the template strands, which DNA polymerase needs to start
DNA polymerase IIIMain enzyme: adds nucleotides to the 3′ end of the primer (5′ to 3′) to make the leading and lagging strands, and proofreads
DNA polymerase IReplaces the RNA primers with DNA nucleotides
DNA ligaseJoins the Okazaki fragments by forming phosphodiester bonds
DNA gyraseWorks ahead of helicase and reduces the tension created by unwinding
2025 Annual I, Set B, Q.2 (xiii)3 marksUnit 1
Elaborate the endosymbiont theory regarding evolution of chloroplast and mitochondria.
Solution

The theory says that chloroplasts and mitochondria were once free-living bacteria that came to live inside an early host cell.

Endosymbiotic origin of mitochondria (row A) and chloroplasts (row B).
  1. An early ancestor of the eukaryotic cell engulfed an aerobic bacterium (later the mitochondrion) and a photosynthetic bacterium (later the chloroplast).
  2. The engulfed cells were not digested. They formed a relationship with the host and became endosymbionts, cells living within a cell.
  3. They became permanent organelles, an organism within an organism, still dividing by fission and keeping their own circular DNA, RNA and 70S ribosomes.
Evidence from the book
  • Both organelles have a small circular DNA, all kinds of RNA and 70S ribosomes, like a bacterium.
  • Both are self-replicating organelles and divide by fission.
  • Both have their own metabolic machinery and can survive outside the cell in an artificial medium if carefully fractionated.
  • Both are enclosed by a double membrane.
2026 Annual I, Set B, Q.2 (xiii)1x3 = 3 marksUnit 14
Differentiate between convergent and divergent evolution. (Three points)
Solution

Convergent and divergent evolution are both ways in which species change, but they differ in ancestry, habitat and the structures produced:

Convergent and divergent evolution.
FeatureConvergent evolutionDivergent evolution
AncestryDifferent, unrelated ancestorsSpecies evolved from a common ancestor
HabitatSpecies share a common habitatSpecies live in different habitats
Structures producedAnalogous structures (wings of bat, bird and insect)Homologous structures (forelimbs of man, horse, whale and bat)
2025 Annual I, Set B, Q.2 (xiii) OR1x3 = 3 marksUnit 2
How is maltose different from sucrose in glycosidic linkage, monosaccharide units and specific role?
Solution

Both are disaccharides (\ce{C12H22O11}) formed by a condensation reaction that releases one water molecule, but they differ as follows.

PointMaltoseSucrose
Glycosidic linkageα-1,4: the –OH on C1 of the first glucose joins the H of the –OH on C4 of the second glucoseα-1,2 (α-1,β-2): the –OH on C1 of α-glucose joins the H of the –OH on C2 of β-fructose
Monosaccharide unitsTwo α-glucose moleculesOne α-glucose and one β-fructose
Specific roleMalt sugar: an intermediate in the breakdown of starch and glycogen and an energy source in germinating seeds and sprouting potatoesCane sugar: the transport disaccharide of plants (very soluble and relatively unreactive), and the common household sweetener
2026 Annual I, Set B, Q.2 (xiii) OR1x3 = 3 marksUnit 4
State three roles of ATP-Synthetase during chemiosmosis.
Solution

ATP synthetase (ATP synthase) sits in the inner mitochondrial membrane (and in the thylakoid membrane in chloroplasts). In chemiosmosis it does three jobs.

  1. Proton channel: its F_0 part, embedded in the membrane, lets protons flow back down the gradient from the intermembrane space (high \ce{H+}) to the matrix (low \ce{H+}).
  2. Uses the proton gradient: it is activated by the gradient built by the electron transport chain, and changes the energy of the flowing protons into chemical energy.
  3. Makes ATP: its F_1 head, projecting into the matrix, joins ADP and inorganic phosphate (Pi) by a high-energy bond to form ATP.
Chemiosmosis: protons pumped into the intermembrane space flow back through the F0 channel, and F1 uses the flow to make ATP from ADP and Pi.
2025 Annual I, Set B, Q.2 (xiv)3 marksUnit 14
What are the limitations of Lamarckism?
Solution

The limitations (drawbacks) of Lamarckism are:

  1. No experimental proof: there is no experimental evidence that acquired characters are inherited.
  2. Use and disuse cannot alter the genotype: the changes are non-genetic, so the organism's genes stay the same and nothing is passed on.
  3. Somatic changes are not heritable: changes of the body made during life (such as a blacksmith's biceps) are not carried to the offspring. Organs are also not modified by the wish or requirements of an organism.

In short, the environment can change an individual, but such acquired traits are non-genetic and cannot be inherited.

2025 Annual I, Set B, Q.2 (xiv) OR3 marksUnit 8
Write characteristics of angiosperms with reference to vascular tissues and alternation of generation.
Solution

Angiosperms are flowering plants whose seeds are enclosed in fruits.

Vascular tissues (1.5 marks)
  • The vascular system is well developed: the xylem has vessels and fibres (gymnosperms have only tracheids) and the phloem has companion cells, which gymnosperms lack.
  • This allows efficient conduction of water, minerals and food, so angiosperms can grow large on land.
Alternation of generations (1.5 marks)
  • They show heteromorphic alternation of generations: the sporophyte and gametophyte look different.
  • The sporophyte is dominant and independent (the plant we see, with the flowers), while the gametophyte is small, less prominent and dependent on the sporophyte.
2026 Annual I, Set B, Q.31+3+2 = 6 marksUnit 5
Identify the cycle of Bacteriophage and explain its infection and replication in host cell.
Six drawings of a bacteriophage cycle joined by clockwise arrows.
Solution

Identification (1 mark)

The figure shows the lytic cycle of a bacteriophage. The infected bacterium is filled with new phages and ends as a burst cell with many free phages. The phage is a lytic (virulent) phage and has a master-slave relationship with its host.

The lytic cycle of a bacteriophage, the paper figure with its five stages numbered and named.

Infection process (3 marks)

  1. Adsorption. The phage attaches to the bacterial cell. The tail fibres and tail pins (spikes) bind to specific receptors on the bacterial surface.
  2. Penetration. Binding makes the sheath contract and releases lysozyme from the base of the tail. Lysozyme digests part of the bacterial cell wall and the hollow core tube is pushed through the envelope.
  3. Genome injection. The phage DNA is injected through the core tube into the bacterial cytoplasm. Only the DNA enters; the protein coat stays outside.

Replication in the host cell (2 marks)

  1. Host DNA breaks down and phage parts are made. A viral DNAase digests the host DNA (the phage DNA is chemically modified and is not affected). The viral DNA takes over the cell, and phage mRNAs direct the making of head and tail proteins and the lysis protein.
  2. Assembly, maturation and lysis. DNA is packed into heads and tails are added; about 200 phages are made in 20 to 25 minutes. Lysozyme builds up, the cell lyses and the new phages are released.
2025 Annual I, Set B, Q.35+2 = 7 marksUnit 5
Explain the steps of life cycle of Human Immunodeficiency Virus (HIV). Also draw its life cycle.
Solution

HIV is specific to CD4 cells: its gp 120 spikes bind the CD4 receptors on helper T cells, macrophages and certain brain cells (section 5.4.2). Its life cycle has twelve numbered stages, shown in the figure in the order of the book.

Steps of the life cycle (5 marks)

  1. Attachment. The gp 120 envelope protein binds the CD4 receptor on the T cell.
  2. Penetration and uncoating. The viral envelope fuses with the cell membrane, using gp 41, and the virion enters the cell by endocytosis. Uncoating releases the single-stranded viral RNA and the viral enzymes into the cytoplasm.
  3. Reverse transcription. The enzyme reverse transcriptase synthesizes a single-stranded DNA complementary to the viral RNA, called cDNA.
  4. RNA breakdown. Ribonuclease (RNAase) breaks down the viral genomic RNA after reverse transcription.
  5. Replication of cDNA. The single-stranded cDNA is copied to form double-stranded cDNA.
  6. Integration. Integrase, an enzyme encoded by the virus, inserts the double-stranded DNA into the host cell DNA.
  7. Provirus. The integrated DNA is now called the provirus.
  8. Transcription. Host RNA polymerase transcribes viral mRNA from the provirus. Viral genomic RNA is also produced, not only mRNA.
  9. Translation. Host ribosomes translate the viral mRNA into several large proteins, which the virus-encoded protease cuts into the virion's structural proteins.
  10. Assembly and maturation. The viral components are assembled and mature virions are produced.
  11. Release. Mature virions bud off from the host cell, each enclosing a portion of host membrane. The host cell becomes smaller and non-functional.

Life cycle diagram (2 marks)

Figure 5.5: Life cycle of HIV. 1 Attachment (adsorption), 2 Penetration, 3 Reverse transcription, 4 Breakdown of viral genomic RNA, 5 Replication, 6 Integration, 7 Provirus, 8 and 9 Transcription, 10 Biosynthesis of protein, 11 Maturation, 12 Release.
2026 Annual I, Set B, Q.3 OR3+3 = 6 marksUnit 13
Identify parts A - F in DNA replication process and briefly explain each part.
The figure as set: a replication fork with parts A to F.
Solution

The drawing shows a replication fork moving to the right: the unopened parental DNA is at the right, and the two daughter duplexes trail away at the left. Each part is named below, with its work (1 mark for naming and 1 mark for the explanation in each of the two halves, 3+3).

The answer figure: parts A to F of the replication fork named.
Identification and explanation (3 + 3 marks)
PartNameWhat it is and what it does
ADNA polymerase IIIThe main replicating enzyme. It adds nucleotides to the 3′-OH end of a primer, always in the 5′ to 3′ direction, and proofreads. Here it is working on the upper (lagging) arm.
BLagging strandThe new strand made discontinuously, as short Okazaki fragments, away from the fork. Each fragment needs its own primer; the pieces are later joined by DNA ligase.
CTemplate strandThe old (parental) strand that is copied. Its sequence decides the order of bases in the new strand, by base pairing.
DLeading strandThe new strand made continuously in one piece towards the fork by DNA polymerase III, after one primer.
EDNA helicaseWorks at the fork, breaks the hydrogen bonds between base pairs and opens the duplex so that the two strands separate (gyrase works ahead of it to relieve tension).
FParental DNAThe original double helix that has not yet been unwound. Each strand of it becomes the template for a daughter duplex.
2025 Annual I, Set B, Q.3 OR7 marksUnit 15
Explain different stages of xerarch succession on bare rock starting form lichens to flowering plants.
Solution

Xerarch succession is primary succession that begins on bare rock where water is deficient. It passes through seral stages and ends in a climax community.

Xerarch succession on bare rock, in order of time, ending in the climax community.
Stages of xerarch succession in order.
StageMembers and what happens
1. Crustose lichen (pioneer)Licanora, Rhinodina. Slow growers that withstand extreme desiccation; absorb water like sponges in rain and decompose the rock by secreting acids
2. Foliose lichenPermelia, Dermatocarpon. Grow on the little soil that has collected; their activities add a thin layer of soil
3. MossXerophytic mosses appear when soil has gathered in crevices. They add soil; their dead mat holds water, making the habitat suitable for herbs
4. HerbAnnual weeds first; roots penetrate and secrete acids (more weathering); humus is added; shading reduces evaporation; then biennial and perennial herbs and xeric grasses
5. ShrubXeric shrubs (from seeds or rhizomes) overshadow the herbs, which are replaced; birds help disperse seeds, giving dense growth
6. TreeTree saplings establish among the shrubs; a canopy shades the area; leaf litter and decaying roots add humus and weather the soil
7. ClimaxXeric trees give place to mesophytic trees as soil deepens; a forest develops, a relatively stable community
2026 Annual I, Set B, Q.44+2 = 6 marksUnit 13
Explain the double helical structure of DNA as proposed by Watson and Crick. Also draw diagram.
Solution

In 1953 James Watson and Francis Crick built a model of DNA from the X-ray diffraction pictures of Wilkins and Franklin (a duplex 2 nm wide, with one turn every 3.4 nm) and Chargaff's rule (A = T and G = C). Their model says:

Structure (4 marks)
  • DNA is made of two polynucleotide chains wound round each other as a right-handed double helix and held together by base pairs.
  • The two chains are antiparallel: one runs 5′ to 3′ downward and the other 5′ to 3′ upward. The sugar-phosphate backbones are on the outside and the bases are inside.
  • The helix has a constant width of 2 nm; each turn is 3.4 nm long and contains 10 base pairs.
  • Pairing is specific: adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds), so a purine always pairs with a pyrimidine and the two chains are complementary.
  • There is no restriction on the order of nucleotides along one strand; the sequence is specific for each species and individual and carries the genetic information.
Diagram (2 marks)
Watson and Crick model of DNA: (a) key features of the double helix; (b) partial chemical structure with antiparallel backbones and hydrogen-bonded base pairs.

Note. The double helix is taught in unit 2 (section "Watson and Crick model of DNA") but the paper places this question with unit 13 (chromosomes and DNA).

2025 Annual I, Set B, Q.42+2+1+2 = 7 marksUnit 13
Explain translation process starting from initiation complex till termination. Also draw diagram of initiation complex and elongation steps.
Solution

Translation (the second step of gene expression) is the decoding of the mRNA by the ribosome into a polypeptide. The amino acids have already been activated, that is joined to their own tRNAs by aminoacyl-tRNA synthetase (aminoacyl-tRNA complexes). The three phases that follow are given below.

1. Initiation complex (2 marks)
  1. The tRNA carrying (N-formyl) methionine binds the smaller ribosomal subunit with an initiation factor.
  2. The 5′ end of the mRNA binds the small subunit with another initiation factor, the AUG start codon pairing with the initiator anticodon.
  3. The larger subunit sits on the smaller one. The initiator tRNA is at the P site; the A site is free for the next tRNA and the E site is the exit.
2. Elongation (2 marks)
  1. Codon recognition: the aminoacyl-tRNA with the complementary anticodon enters the A site, helped by an elongation factor.
  2. Peptide bond: peptidyl transferase moves the amino acid or chain from the P-site tRNA onto the amino acid at the A site.
  3. Translocation: the ribosome moves one codon along the mRNA (5′ to 3′); the empty tRNA leaves from the E site, the chain-bearing tRNA shifts from A to P, and a new codon is exposed at the A site. The three steps repeat for every amino acid.
3. Termination (1 mark)

When a stop codon (UAA, UAG or UGA) is exposed at the A site, no tRNA binds. Release factors recognise it, the polypeptide is released from the tRNA, the tRNA leaves, and the ribosomal subunits separate from the mRNA.

Diagrams (2 marks)
Initiation complex: the initiator tRNA (methionine) and the 5′ end of the mRNA bind the small subunit, then the large subunit joins and gives the E, P and A sites.
Chain elongation in three repeated steps: codon recognition at the A site, peptide bond formation, and translocation.
2025 Annual I, Set B, Q.4 OR5+2 = 7 marksUnit 4
How electrons move through electron transport chain till they reach final acceptor? Explain. Draw diagram of electron transport chain.
Solution
Passage of electrons (5 marks)

After glycolysis, the link reaction and the Krebs cycle, most of the energy of glucose is held in **NADH and \ce{FADH2}. They are oxidised in the electron transport chain (ETC) in the inner mitochondrial membrane, and their electrons fall down a series of carriers to the final acceptor, oxygen. The passage is a downhill journey**: energy is released in small steps instead of all at once.

No.ComponentFurther composed of
1NADH dehydrogenase complex (I)FMN and Fe-S
2FADH dehydrogenase complex (II)
3Coenzyme Q (ubiquinone)mobile carrier
4Cytochrome reductase complex (III)Cyt b, Fe-S, Cyt c_1
5Cytochrome cmobile carrier
6Cytochrome oxidase complex (IV)Cyt a, Cyt a_3, copper
  1. NADH is oxidised at complex I. Electrons and protons pass to FMN, then to the Fe-S complex, which pumps protons into the intermembrane space.
  2. **\ce{FADH2}** is oxidised at complex II and hands its electrons to coenzyme Q (NADH electrons also reach Q).
  3. Electrons go from Q to complex III, which passes them to cytochrome c and pumps protons.
  4. Cytochrome c carries the electrons to complex IV (Cyt a, a_3, copper), which is an electron carrier and a proton pump.
  5. Finally the electrons reach oxygen, the final acceptor. Oxygen combines with them and with protons to form water.
\ce{1/2 O2 + 2H+ + 2e- -> H2O}

The energy released pumps protons from the matrix into the intermembrane space at three sites: between FMN and CoQ (complex I), between Cyt b and Cyt c_1 (complex III) and between Cyt a and Cyt a_3 (complex IV). The resulting gradient is used by ATP synthetase to make ATP, so the chain is also called oxidative phosphorylation.

Diagram (2 marks)
Figure 4.16: Sequence of electron carriers in the respiratory ETC. Electrons (accent) lose energy at each step; protons are pumped at I, III and IV.
2026 Annual I, Set B, Q.4 OR3+3 = 6 marksUnit 15
Explain kinds and general stages of succession.
Solution
Kinds of succession (3 marks)

Ecological succession is the process by which species are replaced over time. There are two kinds.

PointPrimary successionSecondary succession
Starting habitatNever inhabited before: bare rock, new volcanic lava, rock scraped clean by glaciersAlready substantially modified by a pre-existing community
SoilNone at the start; it has to be madeAlready present
ExamplesSuccession on bare rockAbandoned farm fields reverting to forest; a forest devastated by fire, flood or cyclone

Succession is also named by the habitat. Hydrarch succession begins in water (ponds, lakes, marshes) and its stages are hydroseres. Xerarch succession begins on bare rock, sand dunes and rocky slopes where water is deficient, and its stages are xeroseres.

General stages of succession (3 marks)
  1. Plants and animals migrate from surrounding areas into the bare area and aggregate there.
  2. Pioneers. The first migrants take hold of the soil and increase in number.
  3. By their death and decay the pioneers add organic matter, moisture and nitrogen to the soil, which becomes suitable for the next group of invaders.
  4. Seral communities. The next groups of invaders form one seral stage after another. Their activities change the environment so that it becomes unsuitable for themselves, and a new group invades.
  5. Climax community. After several seral communities, the end product is a relatively stable community.
General stages of succession: pioneers, seral communities, climax.
2026 Annual I, Set B, Q.52+2+2 = 6 marksUnit 1
Describe the structure and functions of mitochondria. Also draw suitable figure of mitochondria.
Solution
Structure (2 marks)

Mitochondria are cylindrical or rod-shaped organelles present in all eukaryotic cells. Each is enclosed by a double membrane.

  • Outer membrane: smooth and sieve-like because of the porins, special proteins that let molecules pass freely into the intermembrane space.
  • Inner membrane: selectively permeable and folded inwards into cristae, which increase the surface area. Its inner surface carries the granular F0-F1 particles (ATP synthase) and the electron carrier complexes of the electron transport chain.
  • Intermembrane space: the narrow region between the two membranes.
  • Matrix: the jelly-like material enclosed by the inner membrane. It holds a small circular DNA, all kinds of RNA, 70S ribosomes and enzymes.
Functions (2 marks)

Mitochondria are the sites of cellular respiration: they use oxygen to make ATP by extracting energy from sugars, fats and other organic compounds.

StructureFunction
Outer membrane and porinsFree passage of small molecules into the intermembrane space
Inner membrane and cristaeLarge surface carrying the electron transport chain
F0-F1 particles (ATP synthase)Make ATP
Matrix enzymesCatalyse some steps of cellular respiration, such as the Krebs cycle
Matrix DNA and 70S ribosomesMake some of the proteins the mitochondrion needs (own genetic system)
Labelled figure (2 marks)
Structure of a mitochondrion: outer and inner membranes, intermembrane space, cristae, matrix with DNA and 70S ribosomes, and F0-F1 particles (after Fig. 1.25).
2025 Annual I, Set B, Q.54+2 = 6 marksUnit 1
Describe the pathways involved in moving signals from outside to inside the cell with their diagrams.
Solution
Pathways (4 marks)

The path depends on the ligand. A signal is passed through three stages: reception, transduction and response. The book describes two pathways.

1. Protein (peptide) signal
  1. A protein or peptide hormone (the first messenger) is water soluble, so it cannot cross the lipid bilayer of the plasma membrane.
  2. It binds to its specific receptor on the outer surface of the plasma membrane (reception).
  3. The receptor starts a chain of relay molecules in the cytoplasm; each step activates more molecules (transduction).
  4. A second messenger such as cAMP is made.
  5. The second messenger triggers the response: activation of enzymes or of genes.
2. Steroid signal
  1. A steroid hormone is lipophilic, so it diffuses freely across the plasma membrane.
  2. It binds to a receptor in the cytoplasm or nucleus, forming an active hormone-receptor complex.
  3. The complex moves into the nucleus and binds to sites on the chromatin (DNA).
  4. It acts as a transcription factor and activates mRNA transcription: a change in gene expression is the response.
PointProtein signalSteroid signal
LigandWater soluble protein or peptideLipophilic steroid
Crosses membrane?NoYes, by diffusion
ReceptorOn the plasma membraneIn the cytoplasm or nucleus
Second messengerYes (cAMP)Not needed
Diagrams (2 marks)
Protein signal (top row) and steroid signal (bottom row) pathways from outside the cell to inside (after Fig. 1.11 and 1.12).
2025 Annual I, Set B, Q.5 OR3+3 = 6 marksUnit 2
Elaborate the structures and roles of steroids and prostaglandins as important group of lipids.
Solution
Steroids (3 marks)

Structure. Steroids are lipids of high molecular weight which can be crystalline. The steroid nucleus has 17 carbon atoms arranged in four attached rings: three rings (A, B and C) of six carbons and a fourth ring (D) of five carbons. The side chains that extend from these rings distinguish one steroid from another. Steroids are synthesised from isoprene units.

Steroid nucleus: rings A, B and C are six-membered and ring D is five-membered. Side chains attached to these rings distinguish one steroid from another.

Roles. Cholesterol is a structural component of the cell membrane and the precursor of other steroids: bile acids and bile salts (they emulsify fats), the male sex hormone testosterone and the female sex hormones progesterone and oestrogen. Vitamin D, which helps to regulate calcium metabolism, is also a steroid.

Prostaglandins (3 marks)

Structure. Prostaglandins are lipids derived from arachidonic acid, a fatty acid. They occur in virtually every mammalian tissue and act as local hormones.

Prostaglandins are made from arachidonic acid. Aspirin reduces fever and pain by inhibiting their synthesis.

Roles. Their functions depend on the tissue: some lower blood pressure while others raise it; they help to induce fever and inflammation and intensify pain; and they regulate the aggregation of platelets, an early step in blood clotting. Aspirin reduces fever and pain because it inhibits prostaglandin synthesis.

GroupStructureRoles
SteroidsNucleus of 17 carbons in four rings (three of six carbons, one of five)Cholesterol in membranes and as precursor; bile salts; testosterone, progesterone, oestrogen; vitamin D
ProstaglandinsDerived from arachidonic acidLocal hormones: blood pressure, fever, inflammation, pain, platelet aggregation
2026 Annual I, Set B, Q.5 OR3+3 = 6 marksUnit 9
Describe the mechanism involved in opening and closing of stomata by \mathrm{K^+} ion influx / efflux theory.
Solution

Stomata open when their guard cells are turgid and close when the guard cells are flaccid. According to the \ce{K+} theory the movement of potassium ions in and out of the guard cells changes their water potential.

Opening of stomata in the morning (3 marks)
  1. Light starts photosynthesis, so the \ce{CO2} level in the guard cells falls.
  2. Low \ce{CO2} stimulates the influx of \ce{K+} ions into the guard cells from the surrounding epidermal cells by active transport.
  3. Blue light ionises malic acid into malate and \ce{H+} ions; the \ce{H+} ions are pumped out to the epidermal cells to keep the pH steady.
  4. \ce{K+} combines with malate to form soluble potassium malate, which lowers the solute and water potential of the guard cells.
  5. Water enters from the epidermal cells by osmosis, the guard cells become turgid and the stoma opens.
Closing of stomata in the evening (3 marks)
  1. Photosynthesis stops and the \ce{CO2} level in the guard cells rises.
  2. High \ce{CO2} stimulates the efflux of \ce{K+} ions from the guard cells to the epidermal cells by active transport.
  3. Without blue light, malate and \ce{H+} recombine into malic acid, so no potassium malate remains.
  4. The solute and water potential of the guard cells rise, water moves out to the epidermal cells, the guard cells become flaccid and the stoma closes.
Figure 9.8: (i) Stoma opening, (ii) stoma closing. The \ce{K+} ions are shown as accent dots. When \ce{K+} ions are inside the guard cells, water enters and the stoma opens; when they leave, water leaves and the stoma closes.
2026 Annual I, Set B, Q.62+2+2 = 6 marksUnit 4
Draw events of non-cyclic photophosphorylation and explain absorption of light by PS-II and chemiosmotic synthesis of ATP during light reaction.
Solution
Events of non-cyclic photophosphorylation (2 marks)

Non-cyclic photophosphorylation is the main pathway of the light reaction. Both photosystems and two electron transport chains are used, and the electrons do not return: they leave PS-II, pass to PS-I and end up in NADPH. The zigzag path of the electrons is called the Z-scheme.

Figure 4.8: Non-cyclic photophosphorylation (Z-scheme). Each photosystem raises the electrons to a higher energy level; they lose energy on the way down, which is used to make ATP.

Products: **ATP, NADPH and \ce{O2}**. Order of carriers: PS-II (P680), primary acceptor, PQ, cytochrome b_6f, PC, PS-I (P700), primary acceptor, Fd, NADP reductase, \ce{NADP+}.

Absorption of light by PS-II (2 marks)
  1. Light strikes the antenna complex of PS-II. Two photons are enough for one cycle of the reaction centre.
  2. The pigments pass the energy from one to the next until it reaches the reaction centre, the special chlorophyll a called P680.
  3. P680 absorbs the energy and two electrons are excited to a higher energy level.
  4. The primary electron acceptor of PS-II captures the two electrons. This leaves two "electron holes" in P680, which becomes a strong oxidising agent.
  5. Photolysis of water fills the holes: \ce{H2O} splits into 2\ce{H+}, 2\ce{e-} and \tfrac12\ce{O2}. The electrons refill P680; the protons stay in the thylakoid space and oxygen is released.
Chemiosmotic synthesis of ATP (2 marks)
  1. Excited electrons pass from the primary acceptor to plastoquinone (PQ) and then through the **cytochrome b_6f complex**.
  2. The cytochrome complex is a proton pump: the energy of the falling electrons pumps \ce{H+} from the stroma into the thylakoid space (lumen), so a proton gradient builds up (high \ce{H+} inside the lumen).
  3. The gradient activates ATP synthase in the thylakoid membrane. Protons flow back to the stroma through its F_0 channel.
  4. The flow drives F_1 to join ADP and Pi into ATP (photophosphorylation). This ATP supplies the Calvin cycle.
Figure 4.7: Chemiosmotic synthesis of ATP during the light reaction. Electron flow is shown in accent, proton movement in grey.
2025 Annual I, Set B, Q.61+1+2+2 = 6 marksUnit 9
Explain the mechanism of photoperiodism.
Solution
Photoperiodism (1 mark)

The effect of the photoperiod (the total duration of light in the day) on the development of flowers is photoperiodism. Short-day plants flower when the day is shorter than a critical length, long-day plants when it is longer, and day-neutral plants flower in any day length.

Dark period is the real stimulus (1 mark)

Night-interruption experiments show that the uninterrupted dark period is critical. A short-day plant (cocklebur) will not flower if its long night is interrupted by light, so short-day plants are really long-night plants. A long-day plant (henbane) behaves in the opposite way: it is a short-night plant. Red light (660 nm) is the effective light and far-red light (730 nm) reverses it; the last treatment decides the response.

Phytochrome and its two forms (2 marks)

The mechanism depends on the phytochromes, blue-green protein pigments that exist in two interconvertible forms: Pr (\mathrm{P_{660}}), which absorbs red light and changes into Pfr (\mathrm{P_{730}}), and Pfr, which absorbs far-red light and reverts to Pr. Pr changes into Pfr in the day; Pfr changes slowly back into Pr during the night, and this rate of conversion works as a biological clock that measures the length of the night.

Figure 9.19: Interconversion of the two forms of phytochrome. Pfr is the active form (accent outline); it leads to a physiological response such as seed germination, stomatal opening or flowering.
GroupRatio of Pfr to Pr at end of nightReason
Short-day plantLowA long enough dark period (cocklebur: 8.5 hours or more) converts all Pfr to Pr
Long-day plantHighA short dark period (henbane: 13 hours or less) cannot convert all Pfr to Pr
Florigen (2 marks)
  1. The night length is measured by the conversion of Pfr to Pr.
  2. The right ratio of Pfr to Pr at the end of the night is reached.
  3. This ratio triggers the production of the hormone florigen.
  4. Florigen changes the vegetative buds into floral buds, so flowers are produced.
2025 Annual I, Set B, Q.6 OR2x3 = 6 marksUnit 10
Write three distinguishing characteristics and one example of each of the phylum Cnidaria Arthropoda and Mollusca.
Solution
Phylum Cnidaria (2 marks)
  • Diploblastic with mesoglea between ectoderm and endoderm; radial or biradial symmetry.
  • Sac-like gastrovascular cavity with one opening (mouth) surrounded by tentacles.
  • Stinging cells (cnidocytes with nematocysts) on tentacles and body.

Example: Hydra, Obelia, jellyfish.

Phylum Arthropoda (2 marks)
  • Segmented body with several pairs of jointed appendages, bilateral symmetry.
  • Body covered by a chitinous exoskeleton; the coelom is reduced to a haemocoel.
  • Sexes separate; fertilization mostly internal; development through metamorphosis.

Example: prawn, butterfly, spider, cockroach.

Phylum Mollusca (2 marks)
  • Soft, unsegmented, bilateral, coelomate animals; body of head, ventral muscular foot and dorsal visceral mass.
  • A mantle covers the body and secretes the shell (external, internal or absent); a radula may be present.
  • Development through trochophore larvae.

Example: snail, oyster, freshwater mussel, octopus.

2026 Annual I, Set B, Q.6 OR (a)3 marksUnit 12
Explain genetics of hemophilia with reference to pattern of inheritance.
Solution
What the disease is

Haemophilia is a hereditary disease in which blood does not clot properly after injury, because a clotting factor is reduced, faulty or absent. Haemophilia A (factor VIII, 80%) and B (factor IX, 20%) are X-linked recessive; haemophilia C is autosomal recessive.

Genotypes
GenotypePhenotype
X^HX^HNormal female
X^HX^hCarrier female (normal)
X^hX^hHaemophiliac female
X^HYNormal male
X^hYHaemophiliac male
Pattern of inheritance
  • It is carried on the X chromosome and is recessive, so a man with one X^h is affected (he has no second X to mask it), while a woman must be X^hX^h. It is therefore much more common in men.
  • A carrier mother passes X^h to half of her eggs, so half of her sons are affected and half of her daughters are carriers.
  • An affected father gives his X only to daughters (all carriers) and his Y to sons, so it never passes directly from father to son. It 'zigzags' from the maternal grandfather through a carrier daughter to a grandson.
Carrier mother x normal father.
Mother / fatherX^HY
X^HX^HX^H normal daughterX^HY normal son
X^hX^HX^h carrier daughterX^hY haemophiliac son
2026 Annual I, Set B, Q.6 OR (b)3 marksUnit 12
Can two normal parents have a hemophilic child? Work out the probability.
Solution

Yes, but only when the mother is a carrier. The haemophilic son gets his single X from his mother, and the allele is recessive, so a normal-looking mother can be X^HX^h while the father is X^HY (normal).

Mother / fatherX^H (1/2)Y (1/2)
X^H (1/2)X^HX^H normal daughterX^HY normal son
X^h (1/2)X^HX^h carrier daughterX^hY haemophiliac son
  1. Probability that the egg carries X^h is 1/2.
  2. Probability that the sperm carries Y (to make a son) is 1/2.
  3. Probability of an affected son = 1/2\times 1/2 = 1/4.

So the probability of a haemophilic child is 1/4 (25%); all the affected children are sons, and half of the sons (50%) are affected. No daughter can be haemophilic, because the normal father gives every daughter X^H. If the mother were X^HX^H, no child could be affected.

Note. Matches Genetic Problem 12.13 of the book.

Set D

105 questions

2026 Annual I, Set D, MCQ 11 markUnit 1
Smooth endoplasmic reticulum is mainly involved in:
  1. AProtein synthesis
  2. BDNA replication
  3. CLipid synthesisAnswer
  4. DRespiration
Why

The smooth ER makes lipids (oil, cholesterol, phospholipids and steroids), so the answer is C. Protein synthesis (A) is the job of the ribosomes on the rough ER, DNA replication (B) happens in the nucleus, and respiration (D) takes place in the mitochondria.

2025 Annual I, Set D, MCQ 11 markUnit 13
In tRNA molecule, the main function of middle loop is to:
  1. ARecognize enzyme activity
  2. BBind codon on mRNAAnswer
  3. CRecognize binding site on ribosome
  4. DAttach amino acid to CCA
Why

The middle loop of a tRNA is the anticodon loop. It is made of 7 bases, the middle three of which are the anticodon, which pairs with the complementary codon on the mRNA. The amino acid is attached to the CCA at the 3′ end of the molecule (option D), not by the middle loop.

2026 Annual I, Set D, MCQ 21 markUnit 2
Which polysaccharide is storage carbohydrate in animals?
  1. AStarch
  2. BGlycogenAnswer
  3. CCellulose
  4. DChitin
Why

Glycogen is the storage carbohydrate of animals, kept mainly in the liver and muscles, so it is called animal's starch. Starch (A) is the tempting wrong answer: it is also a storage polysaccharide of α-glucose, but it is stored in plants. Cellulose (C) and chitin (D) are structural, not storage, polysaccharides.

2025 Annual I, Set D, MCQ 21 markUnit 2
Prostaglandins are derivatives of:
  1. AAmino acids
  2. BSteroids
  3. CArchidonic acidAnswer
  4. DNucleic acids
Why

Prostaglandins are derived from arachidonic acid, a fatty acid (spelled "Archidonic" in the paper). Steroids (B) are another group of lipids, but they are synthesised from isoprene units, not the source of prostaglandins. Amino acids and nucleic acids are the units of proteins and the nucleic acids, not lipid precursors.

2025 Annual I, Set D, MCQ 31 markUnit 1
What is correct for protein signaling? I Water soluble; II Binds receptor on plasma membrane; III Lipophilic; IV Do not require cAMP
  1. AI & IIAnswer
  2. BII & III
  3. CIII & IV
  4. DI & IV
Why

Protein and peptide signals are water soluble (I) and cannot cross the membrane, so they bind a receptor on the plasma membrane (II). The answer is A. Statements III and IV describe the steroid signal: steroids are lipophilic and act without a second messenger, while the protein signal does use a second messenger such as cAMP.

2026 Annual I, Set D, MCQ 31 markUnit 4
End product of anaerobic respiration in yeast is:
  1. ALactic acid
  2. BWater
  3. COxygen
  4. DEthanolAnswer
Why

Yeast carries out alcoholic fermentation: pyruvate is decarboxylated to acetaldehyde (releasing \ce{CO2}), and acetaldehyde is reduced by NADH to ethanol. Ethanol (with \ce{CO2}) is the end product.

Lactic acid (A) is the end product in muscles and some bacteria. Water (B) and oxygen (C) are not products of fermentation: water is formed only when oxygen accepts electrons in aerobic respiration, and oxygen is not made at all.

2025 Annual I, Set D, MCQ 41 markUnit 1
Which labelled part in nucleus is responsible for ribosome synthesis?
Figure for MCQ 4.
  1. AP
  2. BQ
  3. CR
  4. DSAnswer
Why

Ribosomes are assembled in the nucleolus, the dense, non-membrane-bound body in the nucleoplasm, so the answer is S (D). Its granular area holds ribosomal subunits and its fibrillar area holds rRNA and rDNA. P is the nuclear envelope, and Q and R mark the chromatin, which holds the DNA but does not build ribosomes.

Answer figure: P nuclear envelope, Q and R chromatin, S nucleolus.

Note. S points to the dense body inside the nucleus, the nucleolus.

2026 Annual I, Set D, MCQ 41 markUnit 3
Enzymes speed up reactions by:
  1. AIncreasing activation energy
  2. BDecreasing activation energyAnswer
  3. CConsuming substrates
  4. DChanging products
Why

Enzymes are biological catalysts. They speed up a reaction by lowering the energy of activation (Figure 3.6 of the book): the reactants and products stay at the same energy levels, but the hill between them is lower. A is the opposite of what happens. An enzyme is not consumed (it comes out unchanged, so C is wrong) and it does not change what the products are (D is wrong).

2025 Annual I, Set D, MCQ 51 markUnit 3
Which enzymes break specific covalent bonds without hydrolysis?
  1. AOxidoreductases
  2. BHydrolases
  3. CTransferases
  4. DLyasesAnswer
Why

Lyases break specific covalent bonds and remove groups without hydrolysis (for example histidine decarboxylase). Hydrolases (B) break bonds too, but they do it by adding water. Oxidoreductases (A) add or remove electrons or H^+, and transferases (C) move a functional group from one substrate to another.

2026 Annual I, Set D, MCQ 51 markUnit 5
Viruses that infect bacteria are called:
  1. APlant viruses
  2. BAnimal viruses
  3. CBacteriophagesAnswer
  4. DRetroviruses
Why

A virus that infects bacteria is a bacteriophage (phage), the tadpole-shaped virus of the book. Plant viruses (A) infect plants (for example tobacco mosaic virus) and animal viruses (B) infect animals. Retroviruses (D), such as HIV, have an RNA genome and infect animal cells (HIV attacks CD4 cells).

2025 Annual I, Set D, MCQ 61 markUnit 5
Viroids cause:
  1. AHepatitis A
  2. BHepatitis B
  3. CHepatitis C
  4. DHepatitis DAnswer
Why

The book names the only human disease caused by a viroid as hepatitis D (a short, circular RNA with no coat). Hepatitis A is caused by the hepatitis A virus (A), hepatitis B by the hepatitis B virus (B) and hepatitis C by the hepatitis C virus (C). Hepatitis D needs the hepatitis B virus to enter a liver cell, which is why B can look right.

MCQ 62026 Annual I, Set D, MCQ 61 markUnit 6
Gut bacteria helps in the synthesis of:
  1. AVitamin KAnswer
  2. BProteins
  3. CLipids
  4. DDNA
Why

The bacterial flora of the gut make vitamin K (and B vitamins). Proteins, lipids and DNA are made by the body's own cells from the digested food; gut bacteria help digestion and vitamin supply but are not the source of these.

MCQ 72025 Annual I, Set D, MCQ 71 markUnit 6
Type of flagella arrangement shown in bacteria is:
  1. ALophotrichous
  2. BPeritrichousAnswer
  3. CAmphitrichous
  4. DMonotrichous
Why

The drawing shows flagella spread all over the surface of the cell. That is the peritrichous arrangement (peri = around). Lophotrichous has a tuft of flagella at a pole, amphitrichous has one flagellum at each end, and monotrichous has a single flagellum.

Arrangements of flagella in bacteria (the figure in the question is the last one).
2026 Annual I, Set D, MCQ 71 markUnit 9
Venus flytrap responds to:
  1. ATouch stimulusAnswer
  2. BHeat stimulus
  3. CLight stimulus
  4. DGravity stimulus
Why

The Venus flytrap (Dionaea muscipula) has modified leaves that act as traps. When an insect touches the sensitive hairs on the leaf, the trap closes, so it responds to a touch stimulus.

Heat (B), light (C) and gravity (D) are not the stimulus that closes the trap. Light and gravity are the stimuli for phototropism and geotropism of shoots and roots.

2026 Annual I, Set D, MCQ 81 markUnit 2
A phospholipid molecule contains:
  1. AThree fatty acids only
  2. BGlycerol, two fatty acids a phosphate groupAnswer
  3. COnly glucose units
  4. DOnly amino acids
Why

A phospholipid is built from glycerol, two fatty acids and a phosphate group (with a compound such as choline attached to the phosphate). Option A, three fatty acids only, describes a triacylglycerol, which has no phosphate. Glucose units only (C) describes a polysaccharide and amino acids only (D) a protein.

MCQ 82025 Annual I, Set D, MCQ 81 markUnit 6
Genetic material moves from donor to recipient bacterium in:
  1. ABinary fission
  2. BConjugationAnswer
  3. CTransduction
  4. DTransformation
Why

In conjugation a plasmid (sometimes part of the chromosome) passes directly from a donor to a recipient cell through a bridge-like connection. Binary fission (A) only copies DNA into daughter cells and moves nothing between cells. Transduction (C) and transformation (D) also transfer DNA, but by a bacteriophage and by free DNA taken up from the surroundings; the question describes direct donor-to-recipient transfer, which is conjugation.

Note. Transduction and transformation also involve a donor and a recipient cell in the book, so the question relies on conjugation being the direct cell-to-cell transfer.

2026 Annual I, Set D, MCQ 91 markUnit 7
In lichens, the fungus mainly provides:
  1. AFood
  2. BProtection and moistureAnswer
  3. COxygen
  4. DCarbon dioxide
Why

In a lichen the fungal hyphae form the body, wrap round the algal cells and give them protection from drying out (moisture). The alga or cyanobacterium makes the food by photosynthesis and the fungus gains nutrients from it.

Food (A) is the tempting option, but it is supplied by the photosynthetic partner, not by the fungus. Oxygen (C) and carbon dioxide (D) are exchanged gases, not what the fungus is said to provide.

2025 Annual I, Set D, MCQ 91 markUnit 7
Mutualistic fungi which grows between root cells of plants and does not penetrate cells is:
  1. ALichens
  2. BEndomycorrhiza
  3. CEctomycorrhizaAnswer
  4. DEpiphytes
Why

Ectomycorrhiza forms a sheath outside the root and its hyphae grow between the cell walls without entering the cells.

Endomycorrhiza (B) is the tempting option, but it penetrates the cortex cells. Lichens (A) are fungus and alga associations and have nothing to do with roots. Epiphytes (D) are plants that grow on other plants.

2026 Annual I, Set D, MCQ 101 markUnit 7
Organisms that obtain food from dead organic matter are:
  1. AAutotrophs
  2. BParasites
  3. CSaprophytesAnswer
  4. DPredators
Why

Organisms that feed on dead organic matter are saprophytes (saprotrophs). Fungi are the standard example: they secrete enzymes onto dead plants and animals and absorb the digested food.

Parasites (B) feed on a living host and harm it, so they are the tempting wrong choice. Autotrophs (A) make their own food and predators (D) kill and eat living prey.

2025 Annual I, Set D, MCQ 101 markUnit 9
Growth movements shown by climbing plants with a supporting structure is:
  1. APhototropism
  2. BGeotropism
  3. CThigmotropismAnswer
  4. DChemotropism
Why

Thigmotropism is the growth movement in response to touch. Climbing plants coil round a wall, rod or rope that supports them because they touch it.

Phototropism (A) is a response to light, geotropism (B) to gravity and chemotropism (D) to a chemical such as nutrients, so none of them is the response to a supporting structure.

2026 Annual I, Set D, MCQ 111 markUnit 8
Which group includes enclosed seed plants?
  1. AGymnosperms
  2. BFern
  3. CBryophytes
  4. DAngiospermAnswer
Why

Angiosperms are the plants whose seeds are enclosed in fruits (the name means 'enclosed seed'). Gymnosperms have naked seeds on cone scales, and ferns and bryophytes produce no seeds at all, so A, B and C are wrong.

2025 Annual I, Set D, MCQ 111 markUnit 13
During unwinding of DNA double helix, the hydrogen bonds between base pairs are broken by enzyme:
  1. AHelicaseAnswer
  2. BPrimase
  3. CPolymerase
  4. DGyrase
Why

DNA helicase opens the double helix by breaking the hydrogen bonds between base pairs. Gyrase (D) only works ahead of it to reduce the tension made by unwinding; primase (B) makes RNA primers and polymerase (C) builds new strands.

2026 Annual I, Set D, MCQ 121 markUnit 10
Which animal group has a water vascular system?
  1. AArthropods
  2. BMolluscs
  3. CEchinodermsAnswer
  4. DAnnelids
Why

The water vascular system is a system of water-filled canals in the coelom of echinoderms (starfish, sea urchin). It works the tube feet and is used for locomotion, attachment, gas exchange and feeding. Annelids and molluscs have a coelom but no such canal system, and arthropods have a haemocoel.

2025 Annual I, Set D, MCQ 121 markUnit 10
Radial symmetry and water vascular system are found in phylum:
  1. AHemichordata
  2. BEchinodermataAnswer
  3. CChordata
  4. DVertebrata
Why

Adult echinoderms (starfish, sea urchin) show radial symmetry and have a water vascular system of canals and tube feet. Hemichordata (acorn worm) and Chordata have neither; Vertebrata is a subphylum, not a phylum.

2025 Annual I, Set D, MCQ 131 markUnit 11
Uterine lining in female starts breaking due to less concentration of:
  1. AProgesteroneAnswer
  2. BEstrogen
  3. CLuteinizing hormone
  4. DFSH
Why

Menstruation begins when the corpus luteum degenerates and the progesterone level falls, depriving the endometrium of hormonal support so that its cells die. Estrogen is also low then, but progesterone is the hormone the book names as the one whose fall sets off the breakdown; LH falls too, but it acts on the corpus luteum, not on the lining directly, and FSH rises at this time.

2026 Annual I, Set D, MCQ 131 markUnit 12
Polygenic inheritance results in:
  1. ADiscrete variation
  2. BMutation only
  3. CNo variation
  4. DContinuous variationAnswer
Why

Polygenic traits are controlled by several gene pairs whose effects add up, giving many small steps between the extremes: continuous variation. Discrete variation (clear separate classes) is what single-gene Mendelian traits and epistasis give, which is why it is the tempting wrong option.

2026 Annual I, Set D, MCQ 141 markUnit 12
In Labrador retrievers, coat colour inheritance is an example of:
  1. AIncomplete dominance
  2. BCodominance
  3. CMutation
  4. DEpistasisAnswer
Why

Two genes at different loci decide the coat: B/b (black or chocolate) and E/e. A dog that is ee is yellow whatever its B gene, so one gene hides the effect of another, which is epistasis and gives a 9 : 3 : 4 ratio. In incomplete dominance and codominance only one gene (two alleles at one locus) is involved.

2025 Annual I, Set D, MCQ 141 markUnit 15
A forest was completely destroyed by fire. After about hundred year, a new forest was evolved. The type of succession shown is:
  1. APrimary and xerarch
  2. BSecondary and xerarchAnswer
  3. CPrimary and hydrarch
  4. DSecondary and hydrarch
Why

The fire destroyed the vegetation but the soil was left, so the habitat was already modified by a previous community. That is secondary succession. It is xerarch because it is a land (dry-habitat) succession; hydrarch succession begins in water. Primary succession needs a habitat never inhabited before (bare rock, lava), so A and C are wrong.

2026 Annual I, Set D, MCQ 151 markUnit 11
Ovulation occurs around:
  1. ADay 1
  2. BDay 7
  3. CDay 14Answer
  4. DDay 28
Why

In a 28-day cycle the sudden release of LH at the end of the proliferative phase causes ovulation on about day 14. Day 1 is the first day of menstrual bleeding, day 7 falls in the middle of the proliferative phase while the follicle is still maturing, and day 28 is the end of the secretory phase when menstruation is about to begin.

2025 Annual I, Set D, MCQ 151 markUnit 12
Holandric traits show all of the following characteristics EXCEPT:
  1. ASons of affected father are affected
  2. BShow Y-linked inheritance
  3. CPorcupine man is an example
  4. DDaughters of affected father are carriersAnswer
Why

Holandric traits are Y-linked: the Y goes from father to every son, so sons of an affected father are affected, and porcupine man is a book example. A daughter receives her father's X, not his Y, so she gets no holandric gene and cannot be a carrier. Carrier daughters of an affected father occur in X-linked recessive traits, not Y-linked ones.

2025 Annual I, Set D, MCQ 161 markUnit 9
In TS of a dicot stem, which substances are expected to pass through labelled parts P and Q respectively?
Transverse section of a dicot stem: vascular bundles in a ring round the pith; in each bundle the phloem (P) is outside and the xylem (Q, solid) is inside.
  1. AWater and sucrose
  2. BSucrose and amino acids
  3. CSucrose and waterAnswer
  4. DWater and amino acids
Why

In the vascular bundle of a dicot stem the phloem lies on the outer side and the xylem on the inner side (towards the pith). P is the outer part, the phloem, which carries organic food such as sucrose and amino acids. Q is the inner part, the xylem, which carries water and minerals.

So the order is sucrose then water (C). Option A reverses the two tissues. Options B and D are wrong because B names no water-carrying tissue and D names no sucrose-carrying tissue.

2026 Annual I, Set D, MCQ 161 markUnit 14
Which molecule provides strong evidence for evolution due to similarity among species?
  1. ADNAAnswer
  2. BLipids
  3. CCarbohydrates
  4. DMinerals
Why

DNA is found in all organisms, is built of the same kind of nucleotides and uses a universal genetic code, so its similarity among species shows common ancestry. Lipids, carbohydrates and minerals are not named in the book as molecular evidence of evolution (the book names DNA, ATP, the 20 amino acids and cytochrome c).

MCQ 172025 Annual I, Set D, MCQ 171 markUnit 6
The structures which help bacteria to survive under unfavorable conditions include:
  1. AFlagella and pili
  2. BPili and akinetes
  3. CEndospores and plasmidAnswer
  4. DFlagella and heterocyst
Why

Endospores are the survival structures of bacteria such as Bacillus and Clostridium. Of the pairs offered, C is the best match: it names endospores and plasmids (which carry extra genes). A is wrong because flagella and pili are for movement and attachment or gene transfer; B pairs pili (not a survival structure) with akinetes (resting cells of cyanobacteria); D pairs flagella with heterocysts, which fix nitrogen.

Note. Endospores are certainly right. The book does not say plasmids help survival, but B is excluded by pili and D by flagella and heterocyst, so C is the answer by elimination.

2026 Annual I, Set D, MCQ 171 markUnit 13
According to Griffith and Hershy - Chase genetic material proved to be:
  1. ADNAAnswer
  2. BProtein
  3. CRNA
  4. DLipid
Why

Griffith found that something from heat-killed S-type bacteria transformed living R-type bacteria, and Hershey and Chase showed that the DNA of the bacteriophage (labelled with ^{32}\mathrm{P}) entered the bacteria while its protein coat (^{35}\mathrm{S}) stayed outside. Protein (option B) was the tempting older idea, but it was shown not to enter the host cell and not to carry the information.

2026 Annual I, Set D, Q.2 (i)1x3 = 3 marksUnit 8
State any three general characteristics of club mosses.
Solution

Club mosses (Lycopsida, for example Lycopodium). Three general characteristics:

  • The plant body has a branching rhizome that produces aerial stems, usually not more than 30 cm tall.
  • The stems are densely covered with scale-like leaves called microphylls, each with only a single strand of vascular tissue. Club mosses are the only living plants with microphylls.
  • The sporangia are borne on terminal clusters of leaves called strobili, which look club-shaped, and the gametophyte is mainly underground.
2025 Annual I, Set D, Q.2 (i)1x3 = 3 marksUnit 11
The diagram shows the section through testis: (a) Identify the parts A & B. (b) Write functions of C & D. (c) What is the function of inhibin hormone in males?
Solution
Section through a testis with the four labelled parts. A, outer coat; B, lobules with coiled seminiferous tubules; C, rete testis; D, epididymis.
(a) Identify A and B (1 mark)
  • A: the outer coat (covering) of the testis (the tunica), which encloses the lobules.
  • B: the testis lobules, each containing one to four tightly coiled seminiferous tubules (250 to 300 lobules in each testis).
(b) Functions of C and D (1 mark)
  • C: rete testis, the tubular network that receives the sperms from the seminiferous tubules and in which they undergo further maturation; the efferent ductules then carry them out of the testis.
  • D: epididymis, coiled on the outer surface of the testis; it is the place for transport and storage of the sperms, and it opens into the vas deferens.
(c) Inhibin (1 mark)

Inhibin is made by the Sertoli cells and released into the blood when the sperm count is too high. It acts on the hypothalamus and anterior pituitary and inhibits the release of GnRH and FSH, so spermatogenesis slows down (negative feedback). When the sperm count falls to 20 million per ml, inhibin is no longer released and the sperm count rises again.

Note. C is the rete testis (where the tubules converge) and D is the epididymis (the coiled tube on the outer edge).

2026 Annual I, Set D, Q.2 (i) OR1x3 = 3 marksUnit 1
State three structural features of mitochondria.
Solution
Double membrane (1 mark)

The outer membrane is smooth and sieve-like because of porins. The inner membrane is selectively permeable and folded into cristae that increase its surface area.

F0-F1 particles and two compartments (1 mark)

The inner surface of the cristae has granular F0-F1 particles (ATP synthase). The inner membrane divides the organelle into the narrow intermembrane space and the matrix.

Matrix with its own genetic system (1 mark)

The jelly-like matrix contains a small circular DNA, all kinds of RNA, 70S ribosomes and enzymes.

Structure of a mitochondrion: outer and inner membranes, intermembrane space, cristae, matrix with DNA and 70S ribosomes, and F0-F1 particles (after Fig. 1.25).
2025 Annual I, Set D, Q.2 (i) OR3 marksUnit 14
Write the major points of theory of natural selection. (Any three)
Solution

In The Origin of Species Darwin explained natural selection through four observations (any three are enough for 3 marks):

  1. Over production: each species produces more offspring than will survive to maturity, so populations can increase exponentially.
  2. Variations: the individuals of a population differ in their traits, and some traits improve the chances of survival and reproduction.
  3. Struggle for existence: there are more individuals than the limited food, water, light and growing space can support, so they compete (interspecific, intraspecific and environmental struggle).
  4. Survival of the fittest: individuals with the most favourable combination of characteristics survive and reproduce, passing their heritable traits on. This is natural selection.

Result: favourable alleles increase, unfavourable alleles decrease, and over generations the population becomes better adapted and new species can arise.

2026 Annual I, Set D, Q.2 (ii)1+3 = 3 marksUnit 3
Write any three examples of enzymes and their specific substrates.
Solution

Each enzyme has a specific substrate, because its active site is shaped for that substrate. Any three of these (1 mark each):

EnzymeSpecific substrateProduct
PepsinProteinsSmall polypeptides (peptones)
AmylaseStarch or glycogenMaltose
LactaseLactoseGalactose and glucose
SucraseSucroseGlucose and fructose
Pancreatic lipaseLipidsFatty acids and glycerol
UreaseUrea (absolute specificity)Ammonia and carbon dioxide
2025 Annual I, Set D, Q.2 (ii)3 marksUnit 9
Compare any three adaptations for osmotic adjustments in hydrophytes and mesophytes.
Solution

Hydrophytes live in water (hypotonic conditions, risk of flooding); mesophytes live where water is moderate (nearly isotonic). Their adaptations differ as follows:

AdaptationHydrophytesMesophytes
Rate of transpirationVery high (broad leaves, large surface area)Moderate (medium-sized leaves)
Position of stomataOn the upper epidermisOn the lower epidermis
Stomatal openingOpen all the time, day and nightOpen in the day and close at night
CuticleVery thin or absentLess thick
Storage of waterNoneVery little

Any three rows give the three marks.

2026 Annual I, Set D, Q.2 (ii) OR3 marksUnit 4
Outline steps of payoff phase of Glycolysis.
Solution
What the phase is (1 mark)

The pay-off (oxidative) phase is steps 6 to 10. It starts with 2 G3P and ends with 2 pyruvate. ATP is made by substrate level phosphorylation and NADH is made. Each step happens twice for one glucose.

The five steps (2 marks)
StepNameEnzymeWhat happens
6Dehydrogenation and phosphorylationG3P dehydrogenaseG3P gives hydrogen to \ce{NAD+} (NADH forms) and takes inorganic phosphate: 1,3-bisphosphoglycerate
7Formation of ATPPhosphoglycerate kinase1,3-Bisphosphoglycerate gives its phosphate to ADP: 3-phosphoglycerate + ATP
8IsomerizationPhosphoglycerate mutaseThe phosphate moves from C3 to C2: 2-phosphoglycerate
9DehydrationEnolase2-Phosphoglycerate loses water: phosphoenolpyruvate (PEP)
10Formation of ATPPyruvate kinasePEP gives its phosphate to ADP: pyruvate + ATP
Pay-off phase (steps 6 to 10)
ATP4 ATP made (steps 7 and 10, two molecules each)
NADH2 NADH made (step 6)
End product2 pyruvate
2025 Annual I, Set D, Q.2 (ii) OR1+2 = 3 marksUnit 13
Why post-transcriptional modifications occur in eukaryotic mRNA? Summarize the changes which occur during mRNA processing.
Solution
Why (1 mark)

In eukaryotes transcription occurs in the nucleus but translation in the cytoplasm. The newly made pre-mRNA must be changed into mature, functional mRNA and protected from enzymes while it travels to the cytoplasm. (In prokaryotes there is no nucleus and no such delay.)

Changes during mRNA processing (2 marks)
  1. Capping: a cap of 7-methyl GTP is added to the 5′ end.
  2. Poly-A tail: a chain of 30 to 500 adenine nucleotides is added to the 3′ end. Cap and tail protect the mRNA from degradation.
  3. Splicing: spliceosomes cut out the non-coding introns, and the coding exons are joined by RNA ligase to give mature mRNA, which leaves through a nuclear pore.
2025 Annual I, Set D, Q.2 (iii)2+1 = 3 marksUnit 5
Write about the structure of prions. Enlist two diseases caused by them.
Solution
Structure of prions (2 marks)
  1. A prion is an infectious protein particle with no nucleic acid: it contains neither DNA nor RNA.
  2. It consists of a single protein encoded by a single cellular gene. It is smaller than a virus and much more resistant than a virus to inactivation by UV light and heat.
Two diseases (1 mark)

Kuru in humans and mad cow disease in cattle: fatal neurodegenerative diseases with clumping of proteins in the brain, memory loss and permanent damage. There is no treatment or cure.

2026 Annual I, Set D, Q.2 (iii)3 marksUnit 10
State three characteristics of cnidarians.
Solution

Phylum Cnidaria ("stinging celled organisms") (any three, 1 mark each):

  • They are diploblastic: the body wall has ectoderm and endoderm with a jelly-like mesoglea between them, and the symmetry is radial or biradial.
  • They have a sac-like gastrovascular cavity with a single opening, the mouth, usually surrounded by tentacles.
  • The tentacles and body carry stinging cells (cnidocytes) containing nematocysts, which give the phylum its name.
  • They occur as polyps (sessile, feeding) and medusae (free swimming, reproductive); the nervous system is a net; reproduction is by budding and by gametes.

Examples: Hydra, Obelia, jellyfish.

2025 Annual I, Set D, Q.2 (iii) OR3 marksUnit 8
Write any three distinguishing characteristics of vascular plants.
Solution

Vascular plants (tracheophytes) are distinguished by:

  • Xylem and phloem present for conducting water, minerals and food, with the sporophyte as the main plant body bearing true roots, stems and leaves.
  • Protected sporangia, which ultimately led to the evolution of seeds, and a pollen tube in seed plants that allows water-independent transfer of male gametes.
  • Heteromorphic alternation of generations, and the evolution of flowers and fruits, which gives flexibility and reproductive success on land.
2026 Annual I, Set D, Q.2 (iii) OR1x3 = 3 marksUnit 13
What were the findings of Hershey-Chase experiment? Write briefly.
Solution

Hershey and Chase (1952) labelled the DNA of bacteriophages with ^{32}\mathrm{P} and their protein coats with ^{35}\mathrm{S}, let them infect bacteria, then blended and centrifuged the mixture. Their findings were:

Findings of the Hershey-Chase experiment
FindingMeaning
^{32}\mathrm{P} (DNA label) was found inside the bacterial cells (pellet)The DNA of the virus enters the bacterium
^{35}\mathrm{S} (protein label) was found in the medium, outside the cellsThe protein coat stays outside and does not enter
The infected bacteria went on to make new bacteriophagesThe injected DNA carries the hereditary information for making new viruses, so DNA is the genetic material
2026 Annual I, Set D, Q.2 (iv)1x3 = 3 marksUnit 2
Compare sickle cell hemoglobin and normal hemoglobin. (Any three points)
Solution

Both are haemoglobin, the iron-rich protein of red blood cells that carries oxygen; they differ in one amino acid of the β-chain.

PointNormal haemoglobin (HbA)Sickle haemoglobin (HbS)
Amino acid at position 6 of the β-chainGlutamic acid (sequence Thr-Pro-Glu-Glu)Valine (sequence Thr-Pro-Val-Glu)
CauseNormal β-globin genePoint mutation in the β-globin gene: one nucleotide replaced by another
Shape of the red blood cellsDisc-shaped, like doughnuts without holesSickle or crescent shaped
Effect on blood flowCells move easily through blood vesselsCells are stiff and sticky and block blood flow, causing pain and organ damage
2025 Annual I, Set D, Q.2 (iv)1x3 = 3 marksUnit 14
What are the homologous organs? Write an example. Which type of evolution is caused by them?
Solution

Homologous organs (structures) look externally different and may do different jobs, but are internally alike because they were inherited from a common ancestor.

  1. Example: the forelimbs of man, horse, whale and bat have the same basic pattern of bones, although they serve as arm, leg, flipper and wing.
  2. Type of evolution: they are caused by divergent evolution, in which different species have evolved from a common ancestor at different habitats.
  3. Other homologous structures: sepals, petals, stamens and carpels (all modified leaves) and vestigial organs such as the human appendix.
2026 Annual I, Set D, Q.2 (iv) OR3 marksUnit 5
Draw a labeled HIV life cycle.
Solution

Draw the twelve numbered stages of Figure 5.5 in order and label each with its stage name (the key enzymes are shown on the stages they act in). 1 mark for the diagram, 1 for correct order and 1 for labels.

Figure 5.5: Life cycle of HIV. 1 Attachment (adsorption), 2 Penetration, 3 Reverse transcription, 4 Breakdown of viral genomic RNA, 5 Replication, 6 Integration, 7 Provirus, 8 and 9 Transcription, 10 Biosynthesis of protein, 11 Maturation, 12 Release.
  • Stages 1 to 2: gp 120 binds the CD4 receptor; fusion by gp 41; entry by endocytosis; uncoating.
  • Stages 3 to 7: reverse transcriptase (RNA to cDNA), RNA broken down by RNAase, double stranded DNA, integrase inserts it into the host DNA (provirus).
  • Stages 8 to 12: transcription by host RNA polymerase, translation on host ribosomes, protease cuts the large proteins, maturation, release by budding.
Q.2 (iv) OR2025 Annual I, Set D, Q.2 (iv) OR1x3 = 3 marksUnit 6
Write any three chemical methods with their mechanisms used to control bacteria.
Solution
  1. Disinfectants: kill or inhibit bacteria on surfaces and non-living objects in hospitals, laboratories and homes. Examples: bleach (sodium hypochlorite), hydrogen peroxide, quaternary ammonium compounds.
  2. Antiseptics: chemicals used on living tissues such as skin and mucous membranes to prevent or treat infection. Examples: alcohol-based hand sanitisers, iodine-based antiseptics.
  3. Preservatives: added to food and cosmetics to inhibit the growth of bacteria so that the product lasts longer. Examples: sodium benzoate, potassium sorbate, parabens. (Chlorine and chloramine disinfect drinking water; sulphonamides and nitrofurans are used in chemotherapy.)
2025 Annual I, Set D, Q.2 (v)3 marksUnit 8
Complete the table (fill the blank cells for rows a, b and c). Given: row a Group is Bryophytes; row b Character is Seedless plants; row c Example is Rose.
GroupCharacterExample
a) Bryophytes
b)Seedless plants
c)Rose
Solution

Each row gives one clue, and the other two cells follow from it.

GroupCharacterExample
a) BryophytesNon-vascular plants (no xylem or phloem); gametophyte is dominant; need water for fertilizationPolytrichum (moss), Marchantia (liverwort) or Funaria
b) Pteridophytes (seedless vascular plants, ferns)Seedless plantsAdiantum (fern); also Lycopodium or Equisetum
c) Angiosperms (flowering plants)Seeds enclosed in fruits; flowers are the reproductive structuresRose

Row a: bryophytes are non-vascular. Row b: seedless plants that have vascular tissue are the pteridophytes. Row c: the rose is a dicot flowering plant, so it belongs to the angiosperms.

2026 Annual I, Set D, Q.2 (v)1x3 = 3 marksUnit 9
State three stimuli affecting plant movement.
Solution

Plant organs move by differential growth in response to external stimuli. Three important stimuli are:

StimulusMovement it causesExample (as in the book)
LightPhototropismShoots grow towards light (positive); roots grow away from it (negative)
GravityGeotropismShoots grow upward (negative); roots grow downward (positive)
TouchThigmotropismClimbing plants grow over a wall, rod or rope

A fourth stimulus is a chemical (nutrients), which causes chemotropism in fungal hyphae.

2026 Annual I, Set D, Q.2 (v) OR1x3 = 3 marksUnit 7
Describe three differences between saprotrophic and parasitic nutrition.
Solution
Saprotrophic and parasitic nutrition
PointSaprotrophic nutritionParasitic nutrition
Source of foodDead plants, animals and other organic matterThe body of a living host
Effect on the organism fed onNone, because it is already dead; the saprotroph acts as a decomposerThe host is harmed and may be weakened or killed
MethodEnzymes are secreted onto the dead matter (external digestion) and the simple products are absorbedFood is absorbed from the living host, often with disease as a result
ExamplesFungi such as Rhizopus on stale breadFungi that cause ringworm or rusts and smuts of plants

Any three differences are enough. In both modes the organism is heterotrophic and cannot make its own food.

2025 Annual I, Set D, Q.2 (v) OR1+2 = 3 marksUnit 12
Name the type of pattern of sex determination in humans. Also write mechanism of sex determination in man.
Solution
Type of pattern (1 mark)

Humans follow the XX-XY type of sex determination: the female is XX (homogametic) and the male is XY (heterogametic).

Mechanism (2 marks)
  1. Humans have 23 pairs of chromosomes: 22 pairs of autosomes and one pair of sex chromosomes.
  2. All eggs of a woman carry one X chromosome. Half of the sperms of a man carry X and half carry Y, with equal chance.
  3. An X sperm fertilising the egg gives XX (a girl); a Y sperm gives XY (a boy). The sperm therefore decides the sex of the child, and the sex ratio is 1 : 1.
GametesX (egg)X (egg)
X (sperm)XX femaleXX female
Y (sperm)XY maleXY male
XX-XY sex determination in man (after Figure 12.14): the sperm that fertilises the egg decides the sex of the child.
2026 Annual I, Set D, Q.2 (vi)1x3 = 3 marksUnit 8
Why are bryophytes called amphibious plants. Give three reasons.
Solution

Bryophytes are called amphibious plants because they live on land but depend on water for reproduction, like amphibians. Three reasons:

  • They are classified as land plants, with land adaptations such as a cuticle, rhizoids and multicellular embryos.
  • Their male gametes (sperms) are motile and swim through a film of water to reach the egg in the archegonium, so fertilization cannot occur without water.
  • They lack vascular tissue and so mostly live in wet or moist places, though some can survive in dry environments.
2025 Annual I, Set D, Q.2 (vi)3 marksUnit 8
Write the difference between leaf, seed and root of monocots and dicots.
Solution

Differences between monocots and dicots:

PartMonocotsDicots
LeafLong, narrow, monofacial, with parallel venationBroad, generally bifacial, with reticulate venation
SeedOne cotyledonTwo cotyledons
RootAdventitious roots from the base of the stem form a fibrous root system; more than 8 patches of xylem; no secondary growthPrimary root is a tap root with lateral roots; 2 to 8 patches of xylem; secondary growth occurs
2025 Annual I, Set D, Q.2 (vi) OR3 marksUnit 1
Write down any three roles of smooth endoplasmic reticulum in a cell.
Solution
1. Lipid synthesis (1 mark)

Enzymes of the smooth ER make lipids: oil, cholesterol, phospholipids and steroids.

2. Steroid hormones (1 mark)

It makes the sex hormones of vertebrates and the steroid hormones of the adrenal glands.

3. Detoxification (1 mark)

Its enzymes detoxify drugs and poisons, especially in liver cells. In skeletal muscle cells, where it is well developed, it is called the sarcoplasmic reticulum.

Q.2 (vi) OR2026 Annual I, Set D, Q.2 (vi) OR1x3 = 3 marksUnit 6
Briefly explain three chemical methods to control bacteria.
Solution
  1. Disinfectants: kill or inhibit bacteria on surfaces and non-living objects in hospitals, laboratories and homes. Examples: bleach (sodium hypochlorite), hydrogen peroxide, quaternary ammonium compounds.
  2. Antiseptics: chemicals used on living tissues such as skin and mucous membranes to prevent or treat infection. Examples: alcohol-based hand sanitisers, iodine-based antiseptics.
  3. Preservatives: added to food and cosmetics to inhibit the growth of bacteria so that the product lasts longer. Examples: sodium benzoate, potassium sorbate, parabens. (Chlorine and chloramine disinfect drinking water; sulphonamides and nitrofurans are used in chemotherapy.)
2025 Annual I, Set D, Q.2 (vii)3 marksUnit 1
Peroxisomes are more abundant in liver cells. Justify.
Solution
Liver detoxifies alcohol (1 mark)

Peroxisomes are oxidative organelles whose main task is the detoxification of alcohol, and the liver is the organ where this is done.

Two-enzyme pathway (1 mark)
  1. Peroxidase oxidises alcohol and produces hydrogen peroxide.
  2. Hydrogen peroxide is itself toxic, so catalase immediately breaks it down to water and oxygen.
Number adapts to need (1 mark)

Peroxisomes increase in number and size when there is a lot of alcohol to break down. Liver cells handle such a heavy load of toxins, so they contain many peroxisomes.

2026 Annual I, Set D, Q.2 (vii)1+2 = 3 marksUnit 3
What is meant by maximum temperature. How it affects enzymatic action?
Solution
Meaning of maximum temperature (1 mark)

The maximum temperature is the temperature that causes denaturation of the enzyme.

Effect on enzymatic action (2 marks)
  • Above the optimum (37 to 38 degrees Celsius for human enzymes) the hydrogen bonds break and the secondary and tertiary structures are disrupted. The enzyme unfolds and the precise shape of the active site is lost.
  • The substrate can no longer bind, so the rate falls steeply even though collisions are more frequent. At the maximum temperature the enzyme is fully denatured and the reaction stops. This cannot be reversed.
Effect of temperature on the rate of an enzyme-controlled reaction
The rate doubles for every 10^\circ\mathrm{C} rise up to the optimum, then falls as the enzyme denatures.
2026 Annual I, Set D, Q.2 (vii) OR1.5+1.5 = 3 marksUnit 1
Briefly explain origin of mitochondria and chloroplasts as explained by Endosymbiotic theory.
Solution
Origin of mitochondria (1.5 marks)

An early ancestor of the eukaryotic cell engulfed a free-living aerobic bacterium. It was not digested but lived on inside the host as an endosymbiont and, over time, became the mitochondrion, the organelle that makes ATP by respiration.

Origin of chloroplasts (1.5 marks)

A host cell engulfed a photosynthetic bacterium (a cyanobacterium) in the same way. It became a permanent endosymbiont and developed into the chloroplast, the organelle of photosynthesis.

Endosymbiotic origin of mitochondria (row A) and chloroplasts (row B).
Evidence from the book
  • Both organelles have a small circular DNA, all kinds of RNA and 70S ribosomes, like a bacterium.
  • Both are self-replicating organelles and divide by fission.
  • Both have their own metabolic machinery and can survive outside the cell in an artificial medium if carefully fractionated.
  • Both are enclosed by a double membrane.
2025 Annual I, Set D, Q.2 (vii) OR3 marksUnit 4
Outline the process of non-cyclic photophosphorylation with the help of a labelled flow sheet.
Solution

Non-cyclic photophosphorylation uses both photosystems and two electron transport chains. Electrons flow from water to \ce{NADP+} and do not return.

The Z scheme of non-cyclic photophosphorylation: electrons rise in PS II, fall through the chain, rise in PS I and are used to make NADPH.
  1. PS-II absorbs light. P680 loses two excited electrons to the primary acceptor.
  2. Photolysis of water fills the holes of PS-II and releases \ce{O2} and \ce{H+}.
  3. Electrons pass through **PQ, the cytochrome b_6f complex and PC; protons are pumped into the thylakoid space and ATP is made by chemiosmosis**.
  4. PS-I absorbs light. P700 gives its electrons to the primary acceptor; its holes are filled by electrons from PC.
  5. Electrons go through ferredoxin and NADP reductase to \ce{NADP+}, forming NADPH.

Products: ATP, NADPH and oxygen.

2025 Annual I, Set D, Q.2 (viii)2+1 = 3 marksUnit 2
The diagram shows molecule of Lecithin: (a) Identify parts A, B, C and D. (b) Mention the polarity of E and F.
Lecithin as given in the paper: three stacked parts (A, B, C) form the head E, and three tails form F; D is the bent tail.
Solution

Lecithin (phosphatidylcholine) is the commonest phospholipid. It is built from glycerol, two fatty acids (one saturated, one unsaturated) and a phosphate group with choline attached.

(a) Identify parts A, B, C and D. (2 marks)
PartName
ACholine (the nitrogenous base)
BPhosphate group
CGlycerol (a three-carbon molecule)
DUnsaturated fatty acid: the tail with a cis double bond that bends the chain
Lecithin (phosphatidylcholine) labelled: choline, phosphate and glycerol form the polar head; the two fatty acids are the non-polar tails.
(b) Mention the polarity of E and F. (1 mark)

E (choline, phosphate and glycerol, the head) is polar or hydrophilic: it is readily soluble in water. F (the fatty acid tails) is non-polar or hydrophobic: it is insoluble in water. A molecule with both a polar and a non-polar end is amphipathic.

Note. In the paper's drawing the right-hand tail has a bend, which stands for the cis double bond of an unsaturated fatty acid.

Q.2 (viii)2026 Annual I, Set D, Q.2 (viii)3 marksUnit 6
Enlist six stages of endospore formation.
Solution
  1. Initiation: under unfavourable conditions (lack of nutrients, extreme temperature) sporulation begins and the DNA replicates, giving two identical copies.
  2. Asymmetric division: the cell divides unequally into a larger mother cell and a smaller forespore.
  3. Membrane development: several protective membranes form around the forespore.
  4. Peptidoglycan layer: the mother cell lays down layers of peptidoglycan around the forespore (the cortex).
  5. Spore coat formation: the mother cell makes a tough protein coat around the spore.
  6. Maturation: the endospore dehydrates, so its water content falls greatly.

The book lists a seventh stage, release: the mother cell ruptures and frees the mature endospore, which stays dormant until conditions improve. Write this too if the question allows seven.

Note. The book gives seven stages; the question asks for six, so the first six are listed and release is added as a remark.

2025 Annual I, Set D, Q.2 (viii) OR3 marksUnit 5
Following is the diagram of HIV: (a) Identify the parts A, B & C. (b) Write the functions of protein gp40, gp120 and enzyme integrase in HIV. [Figure]
HIV with the parts A, B and C marked.
Solution
(a) Identify A, B and C (1.5 marks)
HIV with the parts of the question named.
PartNameNote
AEnvelopeLipoprotein bilayer borrowed from the host cell; carries the glycoprotein spikes
BMatrix proteinProtein shell between the envelope and the capsid
CCapsidConical, made of capsomeres; encloses the two RNA strands and the enzymes
(b) Functions (1.5 marks)
Protein or enzymeFunction
gp 120 (outer glycoprotein)Attaches the virus to the CD4 receptor of the host cell
gp 41 (transmembrane; printed gp40 in the paper)Critical for cell fusion: the envelope fuses with the host membrane
Integrase (enzyme)Incorporates the viral DNA into the host cell DNA to form the provirus

Note. A, B and C mark the envelope (A), matrix protein (B) and capsid (C), the three layers in order from outside. The paper prints gp40; it should be gp 41.

2026 Annual I, Set D, Q.2 (viii) OR3 marksUnit 15
State the events of nitrification in nitrogen cycle.
Solution

Nitrification is the conversion of ammonia and ammonium ions (formed by ammonification) into nitrates by nitrifying bacteria. It takes place in two steps, carried out by two groups of bacteria.

  1. Ammonia and ammonium ions in the soil are converted into nitrites by the first group of bacteria (e.g. Nitrosomonas).
  2. Nitrites are converted into nitrates by the second group of bacteria (e.g. Nitrobacter).
  3. The bacteria are aerobic, so nitrification occurs only in well-aerated soils. The nitrate formed is the form of nitrogen that plants take up from the soil.
\ce{NH3 ->[\textit{Nitrosomonas}] NO2^- ->[\textit{Nitrobacter}] NO3^-}
2025 Annual I, Set D, Q.2 (ix)3 marksUnit 4
Differentiate between absorption spectrum of Chlorophyll- a, b and Carotenoids.
Solution

An absorption spectrum is a graph of the relative absorption of different wavelengths of light by a particular pigment. The three pigments absorb differently.

FeatureChlorophyll aChlorophyll bCarotenoids
Regions of main absorptionViolet-blue (400 to 470 nm) and orange-red (630 to 660 nm)Violet-blue and orange-redBlue-green, 430 to 500 nm only
Violet-blue peakAt the shorter wavelength end (about 420 nm)At the longer wavelength end (about 460 nm)Broad peak around 430 to 500 nm
Red peakAt the longer wavelength end (about 660 nm)At the shorter wavelength end (about 640 nm)None; absorbs little or no red
Reason for the differenceMethyl group on the 2nd pyrrole ringAldehyde group on the 2nd pyrrole ringDifferent structure; accessory pigment
Absorption spectra of different pigments
Figure 4.4: Absorption spectra of chlorophyll a, chlorophyll b and the carotenoids. Chlorophylls absorb most in violet-blue and orange-red; carotenoids absorb most between about 430 and 500 nm.
2026 Annual I, Set D, Q.2 (ix)1x3 = 3 marksUnit 11
Enlist names and location of three accessory glands of male reproductive system.
Solution
Accessory gland (1 mark each)Location
Seminal vesicles (a pair)At the junction of the sperm duct (vas deferens) and the ejaculatory duct
Prostate glandEncircles the urethra just below the bladder (the ejaculatory duct enters it)
Bulbourethral (Cowper's) glands (a pair)At the junction of the ejaculatory duct and the urethra

Their secretions form the liquid part of the semen, which nourishes and protects the sperms and carries them.

2025 Annual I, Set D, Q.2 (ix) OR3 marksUnit 5
How can a virus survive under unfavorable conditions outside the host?
Solution

Outside a host a virus survives by going into a dormant state (3 points, 1 mark each):

  1. It is non-living outside the host. It needs no food and has no metabolism, so it remains dormant or inactive.
  2. Its genome stays viable for a long time in inactive form. The only threats are unfavourable pH and temperature, which denature its proteins.
  3. Non-enveloped viruses survive for long periods (up to several days), while enveloped viruses survive for shorter periods, because their envelope is degraded by sunlight and cleaning.
2026 Annual I, Set D, Q.2 (ix) OR1x3 = 3 marksUnit 11
State three events of secretory phase of menstrual cycle.
Solution
Events of the secretory phase (days 15 to 28), 1 mark each
  1. The corpus luteum secretes rising levels of progesterone, which acts on the estrogen-primed endometrium and converts its functional layer into a glandular secretory layer.
  2. The uterine glands enlarge, coil and secrete nutritious glycogen into the uterine cavity to sustain the embryo until it implants.
  3. If fertilization does not occur, the corpus luteum degenerates as LH declines; progesterone falls, the endometrial cells die and menstruation begins on day 28 (day 1 of the next cycle).
2026 Annual I, Set D, Q.2 (x)3 marksUnit 12
When Red 4'O clock flower is crossed with white flower, what is the probability of red, white and pink flowers?
Solution

Flower colour in the 4 O'clock plant shows incomplete dominance, so red is R_1R_1, white is R_2R_2 and the heterozygote is pink (R_1R_2).

Red x white (the cross asked)
GametesR_2R_2
R_1R_1R_2 pinkR_1R_2 pink
R_1R_1R_2 pinkR_1R_2 pink

All the F1 plants are pink: probability of pink = 1 (100%), red = 0 and white = 0.

If the pink F1 plants are crossed with each other (F2)
F1 gametesR_1R_2
R_1R_1R_1 redR_1R_2 pink
R_2R_1R_2 pinkR_2R_2 white

Red = 1/4 (25%), pink = 2/4 = 1/2 (50%), white = 1/4 (25%), a 1 : 2 : 1 ratio.

Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.

Note. The question can be read as the first cross (all pink) or as the F2 of the book's Figure 12.3; both answers are given.

2025 Annual I, Set D, Q.2 (x)3 marksUnit 12
Justify that O^{-ve} persons are universal donors and AB^{+ve} are universal recipients of blood.
Solution

A transfusion works on the donor's antigens and the recipient's antibodies: if the donor's antigen meets the matching antibody, the red cells clump (agglutination).

O negative (universal donor)AB positive (universal recipient)
Antigens on red cellsNone: no A, no B and no Rh (D) antigenA, B and Rh (D)
Antibodies in plasmaAnti-A and anti-BNone: no anti-A, no anti-B and no anti-Rh
ResultIts red cells carry nothing that a recipient's antibodies can attack, so it can be given to anyoneIts plasma has no antibody to attack any donor's cells, so it can receive blood of any group

A donor must be free of antigens, and a recipient must be free of the matching antibodies. A person of group O negative can only receive from O negative (he has both antibodies), and AB positive can only give to AB positive (his cells carry all the antigens).

Q.2 (x) OR2025 Annual I, Set D, Q.2 (x) OR1x3 = 3 marksUnit 6
How is photosynthesis in Cyanobacteria different from Bacteria? (any three differences)
Solution
PointCyanobacteriaOther photosynthetic bacteria (purple and green)
PigmentsChlorophyll a, chlorophyll b and phycobiliproteins (phycocyanin, phycoerythrin)Bacteriochlorophylls (purple bacteria) or chlorosomes with bacteriochlorophyll c (green sulphur bacteria)
SiteThylakoids stacked as grana, as in chloroplastsPigments in the cell membrane, in infoldings called chromatophores; no thylakoids
OxygenWater is split and oxygen is released, as in plantsMost do not release oxygen; they use sulphur compounds or other molecules as electron donors
Light absorbedA broad spectrum including blue and redMainly infrared and near-infrared
2026 Annual I, Set D, Q.2 (x) OR1x3 = 3 marksUnit 13
Describe three stages of formation of initiation complex during translation process.
Solution

The initiation complex is formed by the ribosomal subunits, the mRNA and the first aminoacyl-tRNA. It forms in three stages.

  1. A tRNA carrying (N-formyl) methionine, the initiator, binds to the smaller ribosomal subunit with the help of an initiation factor.
  2. The 5′ end of the mRNA binds to the same small subunit with the help of another initiation factor, with the start codon AUG paired to the initiator tRNA.
  3. The larger ribosomal subunit is placed on the smaller one. This completes the complex and sets up the E, P and A sites, with the initiator tRNA at the P site.
2025 Annual I, Set D, Q.2 (xi)3 marksUnit 2
How are Starch and Glycogen different from each other although both are polysaccharides?
Solution

Both are homopolysaccharides made of many α-glucose units, and both store energy. They differ in the following points.

PointStarchGlycogen
Where storedStorage carbohydrate of plants: roots, stems and seeds (cereals, potato tubers)Storage carbohydrate of animals: liver and muscles (animal's starch)
BranchingTwo types: amylose (unbranched, α-1,4) and amylopectin (branched with α-1,6 links)Resembles amylopectin but is much more highly branched
Colour with iodineBlue (blue-black)Red
2026 Annual I, Set D, Q.2 (xi)1+2 = 3 marksUnit 10
What is notochord? What is fate of notochord in all vertebrates?
Solution
Notochord (1 mark)

The notochord is a solid, unjointed rod located in the mid-dorsal line between the gut and the central nervous system, outside the coelom. It is the axial endoskeleton of chordates: it supports the body and gives space for muscle attachment.

Fate in vertebrates (2 marks)
  • In vertebrates (Craniata) the notochord is present in the embryo but is replaced in the adult by the jointed backbone (vertebral column).
  • It is partly or wholly replaced; it persists throughout life only in some lower chordates (protochordates).
2025 Annual I, Set D, Q.2 (xi) OR3 marksUnit 1
Why Lysosomes are also called suicidal bags?
Solution

Lysosomes are single-membrane vesicles filled with about 40 acidic digestive (hydrolytic) enzymes that can break down every major macromolecule of the cell.

  • During development a cell may have to disintegrate. In this programmed cell death, autolysis, the lysosomes burst.
  • Their enzymes spill into the cytoplasm and digest the cell from inside, so the cell destroys itself.
  • The fragments of the dead cell are then taken up and digested by other cells (phagocytosis).

Because their enzymes can kill the cell that carries them, lysosomes are called suicidal bags.

2026 Annual I, Set D, Q.2 (xi) OR1x3 = 3 marksUnit 7
Name and explain three diseases caused by fungi in plants.
Solution
Fungal diseases of plants (any three)
Disease (1 mark each)Group and causeExplanation
Powdery mildewAscomycetesA white powdery growth on leaves; the plant grows poorly.
Rusts and smutsBasidiomycetesAttack many plants, such as the cereal crops wheat and corn, and cause serious crop loss.
Ergot of ryeClaviceps purpurea (a sac fungus)Infects the rye grain; eating the infected grain causes ergotism in humans.
Brown rotMonilinia fruticola (Ascomycetes)Attacks fruit such as cherries, peaches, plums and apricots and makes them rot.

Other diseases the book names are Dutch elm disease, chestnut blight and apple scab. Fungal diseases show as leaf spots, wilting and rotting, and can destroy whole crops.

2025 Annual I, Set D, Q.2 (xii)1+1+1 = 3 marksUnit 7
Enlist any three fungal diseases in humans with one symptom each.
Solution
Fungal diseases of humans (any three)
DiseaseCausative fungusOne symptom
RingwormMicrosporum audouiniScaly, ring-shaped patches on the skin (skin, hair or nails infected)
Athlete's footTinea pedisItching, cracked skin between the toes
CandidiasisCandida yeastOral thrush: white patches in the mouth
HistoplasmosisHistoplasmaSerious lung infection after inhaling spores
ErgotismClaviceps purpureaNervous spasm, convulsions and even gangrene

One mark for each disease with its symptom.

2026 Annual I, Set D, Q.2 (xii)1.5+1.5 = 3 marksUnit 14
Give one example each supporting use and disuse of organs according to Lamarck theory.
Solution

Lamarck's theory rests on use and disuse of organs. One example of each:

  • Use (development by use): a blacksmith develops a bigger biceps in the arm that works the hammer. Another example is the webbed feet of aquatic birds, which result from constant spreading of the toes and skin in swimming.
  • Disuse (degeneration by disuse): snakes creep into small holes and crevices and have lost their legs. Another example is the loss of the ability of humans to position the head to collect sound waves (the ear muscles).

Lamarck believed that such changes, gained during life, were inherited by the offspring.

2026 Annual I, Set D, Q.2 (xii) OR1x3 = 3 marksUnit 8
Give three differences between monocots and dicots.
Solution

Monocots and dicots are the two subclasses of angiosperms.

FeatureMonocotsDicots
CotyledonsOne cotyledon in the seedTwo cotyledons in the seed
LeafLong, narrow, monofacial, with parallel venationBroad, generally bifacial, with reticulate venation
RootFibrous root system from adventitious roots; no secondary growthTap root with lateral roots; secondary growth occurs
StemVascular bundles scattered; no cambiumVascular bundles in a ring; cambium present

Any three rows earn full marks.

2025 Annual I, Set D, Q.2 (xii) OR3 marksUnit 13
Name the types of chromosomes on the basis of position of centromere mentioning its location in each type. (Any three)
Solution

Chromosomes are classified by the position of the centromere:

TypePosition of the centromereArms
MetacentricIn the centreTwo arms of about equal length
SubmetacentricSlightly away from the centreTwo unequal arms
AcrocentricNear the endOne very short and one long arm
TelocentricAt the endOnly one arm
2025 Annual I, Set D, Q.2 (xiii)3 marksUnit 11
Write about the structure and function of each of the three layers of uterus.
Solution

The wall of the uterus (a hollow, muscular, inverted-pear-shaped organ) has three layers, listed from outside to inside.

Layer (1 mark each)StructureFunction
PerimetriumOutermost, thin covering layer of the uterusCovers and protects the uterus
MyometriumMiddle, thick layer made of bundles of smooth muscleContracts rhythmically during childbirth to expel the baby from the mother's body
EndometriumInnermost, spongy lining of the uterine cavityReceives the young embryo, which implants in it and stays there for the rest of its development; its functional layer is shed in menstruation
2026 Annual I, Set D, Q.2 (xiii)1x3 = 3 marksUnit 14
Compare homologous and analogous structures. (three points)
Solution

Homologous and analogous structures are both found by comparing the anatomy of different species, but they differ in origin, appearance and internal structure:

Homologous and analogous structures compared.
PointHomologous structuresAnalogous structures
OriginCommon ancestorDifferent ancestors (unrelated organisms)
Appearance and functionLook externally different; may have different functionsLook externally alike; perform similar functions
Internal structureAlike (same basic pattern)Different (different internal organization)
ExampleArm of man, forelimb of horse, flipper of whale, wing of batWings of bats, birds and insects

Homologous structures arise by divergent evolution; analogous structures arise by convergent evolution.

2026 Annual I, Set D, Q.2 (xiii) OR1+2 = 3 marksUnit 4
Write down composition and roles of parts of ATP synthetase.
Solution
Composition

ATP synthetase (ATP synthase) is a membrane enzyme complex found in the inner mitochondrial membrane (thylakoid membrane in chloroplasts). It has two parts: F_0 and F_1.

PartComposition and positionRole
F_0Embedded in the inner membrane; forms a channel across itChannel through which protons flow from the intermembrane space (high \ce{H+}) to the matrix (low \ce{H+})
F_1Head-like particle (elementary particle) that projects from the membrane into the matrixCatalyses the joining of ADP and inorganic phosphate (Pi) into ATP, using the energy of the proton flow
Chemiosmosis: protons pumped into the intermembrane space flow back through the F0 channel, and F1 uses the flow to make ATP from ADP and Pi.
2025 Annual I, Set D, Q.2 (xiii) OR3 marksUnit 14
How evolution of Giraffe neck can be explained on the basis of Lamarckism?
Solution

According to Lamarck the long neck of the modern giraffe was the result of generations of short-necked ancestors feeding on leaves at progressively higher levels of trees. It is explained in three steps:

  1. Use of the neck. Each short-necked giraffe constantly stretched its neck during its lifetime to reach the leaves at the top of the trees. By use, the neck became a little longer (an acquired character).
  2. Inheritance of the acquired character. The longer neck gained by stretching was passed to the offspring, so each offspring was born with a slightly longer neck than its parents.
  3. Accumulation over generations. The same stretching and inheritance went on in every generation, so the neck grew longer step by step until the modern long-necked giraffe appeared.
2025 Annual I, Set D, Q.2 (xiv)3 marksUnit 13
Write down the sources of mutagens with their examples. (Any three)
Solution

A mutagen is an agent that causes mutations. Its sources are physical, chemical and biological agents.

Source of mutagensExamples
PhysicalRadiation of any kind: ultraviolet rays, gamma rays, X-rays, radioactive rays; high temperature alterations
ChemicalColchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes, free oxygen particles
BiologicalCertain viruses, transposons, and errors that occur during meiosis or DNA replication

Note. Mutagen sources are taught under mutations; the unit note also lists unit 12.

2025 Annual I, Set D, Q.2 (xiv) OR3 marksUnit 15
Name any three main greenhouse gases and their sources of emission.
Solution
Greenhouse gasSources of emission
Carbon dioxide (\ce{CO2})Burning of coal, oil and natural gas for energy; deforestation reduces the trees that absorb it
Methane (\ce{CH4})Production and transport of coal, oil and gas; livestock and other agricultural practices
Nitrous oxide (\ce{N2O})Agricultural and industrial activities; combustion of fossil fuels and solid waste

(Fluorinated gases, synthetic gases used in industry and released during their production and use, are the fourth group.)

2026 Annual I, Set D, Q.31+3+2 = 6 marksUnit 5
Identify the cycle of bacteriophage and explain its infection and replication in host cell.
Six numbered stages of a bacteriophage infection.
Solution

Identification (1 mark)

The figure shows the lysogenic cycle of a bacteriophage. The phage DNA enters the bacterium, but the cell is not lysed: it divides (stage 5) into two daughter cells that still carry the phage DNA. The phage is a lysogenic (temperate) phage and has a host-guest relationship with its host.

The lysogenic cycle of a bacteriophage, the numbers of the paper figure with the stages named.

Infection process (3 marks)

  1. Adsorption. The phage attaches to the bacterial cell. The tail fibres and tail pins (spikes) bind to specific receptors on the bacterial surface.
  2. Penetration. Binding makes the sheath contract and releases lysozyme from the base of the tail. Lysozyme digests part of the bacterial cell wall and the hollow core tube is pushed through the envelope.
  3. Genome injection. The phage DNA is injected through the core tube into the bacterial cytoplasm. Only the DNA enters; the protein coat stays outside.

Replication in the host cell (2 marks)

  1. Integration. The phage DNA inserts itself into the bacterial chromosome and becomes a prophage. It is copied with the chromosome before binary fission.
  2. Division of the host. Binary fission gives two daughter cells, each with the prophage. They survive and the prophage can change their properties (lysogenic conversion). Adverse conditions such as UV radiation cause induction: the phage DNA leaves the chromosome and starts the lytic cycle.
2025 Annual I, Set D, Q.33+2+2 = 7 marksUnit 13
Explain Meselson-Stahl experiment and results to prove that DNA replication is semi conservative. Also draw the diagram.
Solution
The experiment (3 marks)

Matthew Meselson and Franklin Stahl (1958) tested the three models (semi-conservative, conservative, dispersive) using nitrogen isotopes.

  1. Bacteria were grown for several generations in a medium containing heavy nitrogen, ^{15}\mathrm{N}, which became part of the bases of their DNA.
  2. The bacteria were shifted to a medium containing light nitrogen, ^{14}\mathrm{N}.
  3. DNA was sampled at 0, 20 and 40 minutes (and a control grown only in ^{14}\mathrm{N}).
  4. Each sample was dissolved in cesium chloride (CsCl) and spun in an ultracentrifuge; DNA settles at the level of its own density, so heavy, hybrid and light DNA form separate bands.
Results and explanation (2 marks)
SampleBandMeaning
Control (^{14}\mathrm{N})TopLight DNA
0 minBottomHeavy DNA, both strands ^{15}\mathrm{N}
20 minMiddleHybrid: each molecule has one ^{15}\mathrm{N} and one ^{14}\mathrm{N} strand
40 minTop and middleHalf hybrid, half light

One hybrid band after one generation rejects the conservative model (which gives heavy and light bands), and the appearance of a light band after two generations rejects the dispersive model (which gives only intermediate bands). Hence DNA replication is semi-conservative: every daughter molecule keeps one old strand and gets one new strand.

Diagram (2 marks)
Meselson-Stahl experiment: bacteria grown in ^{15}\mathrm{N} are moved to ^{14}\mathrm{N}; the tubes show a heavy band at 0 minutes, one hybrid band at 20 minutes and hybrid plus light bands at 40 minutes.
2025 Annual I, Set D, Q.3 OR5+2 = 7 marksUnit 4
Explain how the movement of protons across mitochondrial membrane is involved in ATP synthesis. Also draw diagram showing movement of electrons.
Solution
Protons and ATP synthesis (5 marks)

ATP is made in the electron transport chain by chemiosmosis (oxidative phosphorylation). The electron transport chain itself makes no ATP; it builds a proton gradient that ATP synthetase then uses.

  1. NADH and \ce{FADH2} give their electrons to the chain. As the electrons fall from carrier to carrier, energy is released.
  2. Complexes I, III and IV use this energy to **pump protons (\ce{H+}) from the matrix into the intermembrane space**. Protons cannot cross the inner membrane by themselves.
  3. The intermembrane space now has a higher \ce{H+} concentration than the matrix: a proton gradient across the inner membrane.
  4. The gradient makes ATP synthetase active. Protons flow back down the gradient into the matrix through the **F_0 channel**.
  5. This flow makes the **F_1 head work: it joins ADP and Pi with a high-energy bond to form ATP**.

One NADH gives 3 ATP and one \ce{FADH2} gives 2 ATP by this mechanism.

Figure 4.17: Mechanism of chemiosmosis in the respiratory electron transport chain. (1) The ETC pumps H^+ into the intermembrane space; (2) H^+ flows back through ATP synthase, which makes ATP.
Diagram showing movement of electrons (2 marks)
Figure 4.16: Sequence of electron carriers in the respiratory ETC. Electrons (accent) lose energy at each step; protons are pumped at I, III and IV.
2026 Annual I, Set D, Q.3 OR1x6 = 6 marksUnit 13
Identify parts A - F in DNA replication process and briefly explain each part.
The figure as set: a replication fork with parts A to F.
Solution

The figure shows a replication fork (parental DNA at the left). One new strand is made in one piece and the other in short pieces. Each of the six parts earns 1 mark (name plus its role).

The answer figure: parts A to F of the replication fork named.
Identification and explanation (6 x 1 mark)
PartNameExplanation
ALeading strandThe new strand made continuously, in one piece, towards the replication fork, by DNA polymerase III from a single primer.
BRNA primerA short stretch of RNA nucleotides made by primase. DNA polymerase III cannot start a chain, so it needs this primer with a free 3′-OH to add the first DNA nucleotide. Later DNA polymerase I replaces it with DNA.
COkazaki fragmentsThe short pieces of the lagging strand, made away from the fork, each from its own primer.
DDNA polymerase IIIThe main enzyme that adds nucleotides at the 3′ end of the primer in the 5′ to 3′ direction and proofreads.
EDNA helicaseBreaks the hydrogen bonds between the base pairs at the fork and opens the double helix (with gyrase reducing the tension, and SSB proteins keeping the strands apart).
FDNA ligaseJoins the Okazaki fragments, after the primers are replaced by DNA, by forming phosphodiester bonds so that the lagging strand becomes continuous.
2026 Annual I, Set D, Q.44+2 = 6 marksUnit 13
Explain the double helical structure of DNA as proposed by Watson and Crick. Draw diagram as well.
Solution

In 1953 James Watson and Francis Crick built a model of DNA from the X-ray diffraction pictures of Wilkins and Franklin (a duplex 2 nm wide, with one turn every 3.4 nm) and Chargaff's rule (A = T and G = C). Their model says:

Structure (4 marks)
  • DNA is made of two polynucleotide chains wound round each other as a right-handed double helix and held together by base pairs.
  • The two chains are antiparallel: one runs 5′ to 3′ downward and the other 5′ to 3′ upward. The sugar-phosphate backbones are on the outside and the bases are inside.
  • The helix has a constant width of 2 nm; each turn is 3.4 nm long and contains 10 base pairs.
  • Pairing is specific: adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds), so a purine always pairs with a pyrimidine and the two chains are complementary.
  • There is no restriction on the order of nucleotides along one strand; the sequence is specific for each species and individual and carries the genetic information.
Diagram (2 marks)
Watson and Crick model of DNA: (a) key features of the double helix; (b) partial chemical structure with antiparallel backbones and hydrogen-bonded base pairs.

Note. The double helix is taught in unit 2 (section "Watson and Crick model of DNA") but the paper places this question with unit 13 (chromosomes and DNA).

2025 Annual I, Set D, Q.45+2 = 7 marksUnit 15
Describe Nitrogen Fixation, Ammonification and Nitrification stages involved in Nitrogen cycle. Also draw labelled Nitrogen cycle flow sheet.
Solution

The nitrogen cycle is a series of processes that convert nitrogen gas to organic substances and back to nitrogen in nature. It is maintained by decomposers and other bacteria.

1. Nitrogen fixation (about 2 marks)

Nitrogen gas has a strong triple bond and is unreactive, so it must be fixed. It is fixed in three ways: atmospheric (lightning makes oxides of nitrogen that rain carries into the soil), industrial (nitrogen fertilizers are manufactured, e.g. \ce{N2 + 3H2 -> 2NH3}) and biological (nitrogen-fixing bacteria convert nitrogen to ammonia: free-living Azotobacter and Clostridium, and symbiotic Rhizobium in the root nodules of legumes).

2. Ammonification (about 1.5 marks)

Nitrogenous wastes of animals and compounds of dead organisms are decomposed by saprophytic soil bacteria and fungi into water, carbon dioxide, amino acids and energy. The amino acids are converted into ammonia or ammonium ions. This production of ammonia in the decomposition of organic matter is ammonification (aerobic soil).

3. Nitrification (about 1.5 marks)

Ammonia and ammonium ions are converted into nitrates by two groups of aerobic bacteria: Nitrosomonas converts ammonia to nitrites and Nitrobacter converts nitrites to nitrates. It occurs only in well-aerated soils.

\ce{NH3 ->[\textit{Nitrosomonas}] NO2^- ->[\textit{Nitrobacter}] NO3^-}
Labelled nitrogen cycle flow sheet (2 marks)
Nitrogen cycle flow sheet: fixation, ammonification, nitrification, assimilation and denitrification.

Plants take up nitrates and ammonia (assimilation) and animals eat plants; denitrifying bacteria return nitrogen to the air to complete the cycle.

2025 Annual I, Set D, Q.4 OR4+3 = 7 marksUnit 3
Explain mechanism of enzyme action through 'induced fit' and 'lock and key' models.
Solution

The mechanism of enzyme action is the same in both models: the enzyme (E) and substrate (S) form an enzyme-substrate complex (ES), the substrate is changed into product while attached (EP), and the product is released, leaving the enzyme unchanged. The models differ in how the active site fits the substrate.

Lock and key model

  • Proposed by Emil Fischer in 1894.
  • The enzyme is the lock and the substrate is the key. The active site has a shape complementary to the substrate, so only that substrate fits, like a key in its lock.
  • The enzyme is rigid: the shape of the active site does not change on binding.
  • It explains the high specificity of enzymes. Only a few enzymes follow it (non-regulatory enzymes), for example sucrase and maltase; urease shows absolute specificity.
Lock and key model: the substrate fits the rigid active site exactly.

Induced fit model

  • Proposed by Koshland in 1959.
  • The active site is flexible. At first it does not fit the substrate completely; the amino acids of the active site are moulded into the precise shape as the substrate binds. This change of shape makes the catalysis effective.
  • When the reaction is over the product leaves and the active site regains its original shape.
  • Because it is flexible, related substrates can bind, so one enzyme can carry out more than one related reaction. Enzymes that follow it are regulatory or allosteric enzymes, for example hexokinase; carbonic anhydrase is the book's example of one enzyme acting on related substrates.
Induced fit model: the flexible active site changes shape as the substrate binds, then returns to its original shape.

Comparison

PointLock and key modelInduced fit model
Proposed byEmil Fischer (1894)Koshland (1959)
Active siteRigid; shape does not changeFlexible; moulded when the substrate binds
FitComplementary to the substrate from the startImproves by a conformational change
Enzyme typeNon-regulatory (sucrase, maltase)Regulatory or allosteric (hexokinase)
2026 Annual I, Set D, Q.4 OR1x6 = 6 marksUnit 15
Describe xerarch succession on bare rock from lichens to flowering plants.
Solution

Xerarch succession is primary succession that begins on bare rock where water is deficient. It passes through seral stages and ends in a climax community.

Xerarch succession on bare rock, in order of time, ending in the climax community.
Stages of xerarch succession in order.
StageMembers and what happens
1. Crustose lichen (pioneer)Licanora, Rhinodina. Slow growers that withstand extreme desiccation; absorb water like sponges in rain and decompose the rock by secreting acids
2. Foliose lichenPermelia, Dermatocarpon. Grow on the little soil that has collected; their activities add a thin layer of soil
3. MossXerophytic mosses appear when soil has gathered in crevices. They add soil; their dead mat holds water, making the habitat suitable for herbs
4. HerbAnnual weeds first; roots penetrate and secrete acids (more weathering); humus is added; shading reduces evaporation; then biennial and perennial herbs and xeric grasses
5. ShrubXeric shrubs (from seeds or rhizomes) overshadow the herbs, which are replaced; birds help disperse seeds, giving dense growth
6. TreeTree saplings establish among the shrubs; a canopy shades the area; leaf litter and decaying roots add humus and weather the soil
7. ClimaxXeric trees give place to mesophytic trees as soil deepens; a forest develops, a relatively stable community
2026 Annual I, Set D, Q.53+3 = 6 marksUnit 9
Describe the mechanism involved in opening and closing of stomata by starch sugar inter conversion theory.
Solution

Stomata open when the guard cells are turgid and close when they are flaccid. The theory says that the guard cells are the only photosynthesising cells of the leaf epidermis (they have more chlorophyll than the neighbouring epidermal cells), and that the change of sugar and starch alters their water potential.

Opening of stomata in the morning (3 marks)
  1. In the light, photosynthesis starts in the guard cells.
  2. The concentration of glucose rises, so the solute potential and water potential of the guard cells become lower (more negative).
  3. The surrounding epidermal cells have a higher water potential, so water moves into the guard cells by osmosis.
  4. The guard cells become turgid and the stomata open.
Closing of stomata in the evening (3 marks)
  1. Photosynthesis stops, so the glucose concentration falls. Some glucose is used in respiration, some is changed to sucrose for transport, and the rest is converted into insoluble starch.
  2. The solute potential and water potential of the guard cells rise.
  3. Water moves out of the guard cells into the surrounding epidermal cells.
  4. The guard cells become flaccid and the stomata close.
Starch sugar interconversion theory: the sequence in the morning (opening) and in the evening (closing).
2025 Annual I, Set D, Q.56 marksUnit 12
Epistasis is different from dominance. Explain with the example of coat color in Labrador retriever.
Solution

Epistasis and dominance

DominanceEpistasis
Involves one pair of alleles at one locus.Involves two pairs of alleles at different loci.
A gene suppresses the expression of its own allele.A gene suppresses the expression of an allele of another gene.
Only the recessive allele is suppressed.Both dominant and recessive alleles of the other gene can be suppressed.
The effect is due only to the dominant allele.Either a dominant or a recessive allele can be epistatic.

Coat colour in the Labrador retriever

  • Gene 1 (B/b): black (B) is dominant to chocolate (b).
  • Gene 2 (E/e) at a different locus: it decides whether pigment is deposited. A dog that is ee is yellow, whatever its B gene, because ee is epistatic and hides B and b.
How the two genes decide the coat colour of a Labrador retriever: ee is epistatic and makes the dog yellow whatever its B gene.

Cross of BBEE (black) x bbee (yellow)

  1. F1: all BbEe, black.
  2. F1 x F1: each parent makes gametes BE, Be, bE, be; the 16 combinations are in the table.
GametesBEBebEbe
BEBBEE blackBBEe blackBbEE blackBbEe black
BeBBEe blackBBee yellowBbEe blackBbee yellow
bEBbEE blackBbEe blackbbEE chocolatebbEe chocolate
beBbEe blackBbee yellowbbEe chocolatebbee yellow
PhenotypeGenotypesOut of 16
BlackB\_E\_9
ChocolatebbE\_3
YellowB\_ee and bbee3 + 1 = 4

Why this is not dominance

Mendel's dihybrid ratio would be 9 : 3 : 3 : 1, but the F2 ratio is 9 : 3 : 4: the B\_ee and bbee classes both look yellow, because ee masks the other gene. This interaction between two different loci, not between two alleles of one locus, is epistasis.

Note. Combines the book's long question 9 with the dominance and epistasis comparison (short question 15).

2026 Annual I, Set D, Q.5 OR2+2+2 = 6 marksUnit 1
Describe the structure and functions of Golgi apparatus. Also draw a suitable figure of Golgi apparatus.
Solution
Structure (2 marks)

The Golgi apparatus (Golgi complex or Golgi bodies, discovered by Camillo Golgi in 1898) is found in all eukaryotic cells. It is a stack of flattened, membrane-bound sacs called cisternae, with associated Golgi vesicles.

  • Forming face (cis face): convex, close to the ER; new cisternae form here by the fusion of vesicles from the smooth ER.
  • Maturing face (trans face): concave; the cisternae break up into vesicles again here.
Functions (2 marks)
  1. Secretory proteins arrive from the rough ER in transport vesicles at the cis face.
  2. The Golgi complex modifies them to do their specific work and makes conjugated molecules such as glycoproteins and lipoproteins.
  3. The products are packed into secretory vesicles at the trans face and exported from the cell.
  4. It gives rise to lysosomes, peroxisomes and glyoxysomes.
  5. In plant cells at cytokinesis its vesicles gather at the cell equator to form the phragmoplast, from which the new cell wall is made.
Labelled figure (2 marks)
Role of the Golgi complex: protein from the rough ER enters at the cis face, passes through the cisternae, and leaves in secretory vesicles from the trans face (after Fig. 1.19).
2025 Annual I, Set D, Q.5 OR2x3 = 6 marksUnit 2
Identify and explain the properties of water which enable it to: (a) Act as temperature stabilizer (b) Help organisms to live in warm environment (c) Help organisms to live under ice
Solution
PartProperty of waterReason in one line
(a) Temperature stabiliserHigh specific heat capacityHydrogen bonds absorb heat first, so the temperature rises slowly
(b) Warm environmentHigh heat of vaporisation (574 cal per g)Sweat evaporating carries away much heat from the body
(c) Under iceLower density of iceIce floats and insulates the liquid water beneath
(a) Act as temperature stabilizer (2 marks)

The property is the high specific heat capacity of water: 1 calorie (4.18 joules) is needed to raise the temperature of 1 g of water by 1 °C, far more than for most substances.

Explanation. Much of the heat that water absorbs is used first to break hydrogen bonds between its molecules, so it does not appear as a rise in temperature. A very large amount of heat therefore raises the temperature of water very little. Water protects living material against sudden thermal change and acts as a temperature stabiliser or regulator, for example for organisms in a hot environment.

(b) Help organisms to live in warm environment (2 marks)

The property is the high heat of vaporisation: 574 calories are needed to convert 1 g of liquid water into gas.

Explanation. A large amount of heat can be lost with a very small loss of water. When an animal sweats, body heat is used to vaporise the sweat and the animal is cooled. The evaporation of only 2 ml out of one litre of water lowers the temperature of the remaining 998 ml by 1 °C. This is an efficient way to release excess heat in a hot environment.

(c) Help organisms to live under ice (2 marks)

The property is the lower density of ice than liquid water.

Explanation. Ice has a giant lattice-like structure with the maximum number of hydrogen bonds between water molecules, so it is less dense and floats. In freezing weather ice forms on the surface of ponds and lakes and acts as an insulating layer above the water below. The water under it stays liquid, so organisms can live there until the ice melts.

A frozen pond: floating ice insulates the liquid water below, where organisms survive.
2025 Annual I, Set D, Q.64+2 = 6 marksUnit 9
Describe the components of theory explaining the ascent of sap in plants. Also draw the diagram.
Solution

The upward movement of water and minerals from the roots to the leaves through the xylem, against gravity, is the ascent of sap. It is explained by the TACT theory: Transpiration pull, Adhesion, Cohesion and Tension together make one collective force.

Components of the TACT theory (4 marks)
ComponentMeaningRole
Transpiration pullPulling force produced by evaporation of water from the leafOpen stomata let water vapour out; the dry air spaces pull water from the cells, and these pull it from the xylem, so the sap rises
AdhesionAttraction between water molecules and the polar cellulose walls of the xylemWater sticks to the walls of the narrow xylem tubes, helping it rise and holding it when transpiration stops
CohesionAttraction among water molecules by hydrogen bondsKeeps the water as a continuous, unbroken column, as strong as a steel wire of the same diameter
TensionThe hydrogen bonds between water molecules, with the column kept taut by the pull from aboveKeeps the string of water tense so it is pulled up as one unit

Mechanism: transpiration at the leaf pulls on the column, cohesion and adhesion stop it breaking, and fresh water enters the root hairs by osmosis. The energy comes from the sun, which drives evaporation.

Diagram (2 marks)
Figure 9.7: Movement of water in the xylem through the TACT mechanism. Leaf: transpiration pull. Stem: cohesion between water molecules and adhesion to the cell wall (enlarged circle). Root: water uptake from the soil.
2026 Annual I, Set D, Q.6 OR2+2+2 = 6 marksUnit 4
Draw events of non-cyclic photophosphorylation and explain absorption of light by PS-I and electron flow from PS-I to \mathrm{NADP^+}.
Solution
Events of non-cyclic photophosphorylation (2 marks)

Both PS-II and PS-I work, with two electron transport chains. Electrons excited in PS-II (replaced from water) pass down a chain to PS-I and are finally accepted by \ce{NADP+}. The path is a zigzag, the Z-scheme. Products: ATP, NADPH and \ce{O2}.

Figure 4.8: Non-cyclic photophosphorylation (Z-scheme). Each photosystem raises the electrons to a higher energy level; they lose energy on the way down, which is used to make ATP.
Figure 4.7: Chemiosmotic synthesis of ATP during the light reaction. Electron flow is shown in accent, proton movement in grey.
Absorption of light by PS-I (2 marks)
  1. Electrons arriving from PS-II reach plastocyanin (PC), which carries them into PS-I and fills the electron holes of P700.
  2. The antenna pigments of PS-I absorb light and pass the energy to the reaction centre, the chlorophyll a called P700.
  3. P700 absorbs two photons and its electrons are boosted to a higher energy level.
  4. The primary electron acceptor of PS-I captures the electrons and leaves electron holes in P700. These holes are filled by the electrons that came from P680 (PS-II) through the transport chain.
Electron flow from PS-I to NADP^+ (2 marks)
  1. The primary acceptor of PS-I passes the electrons to a second electron transport chain.
  2. They are accepted by ferredoxin (Fd), an iron-containing protein.
  3. The enzyme NADP reductase (a flavoprotein) transfers the electrons from Fd to \ce{NADP+}.
  4. \ce{NADP+} combines with the electrons and hydrogen ions to form NADPH.
\ce{NADP+ + 2e- + 2H+ -> NADPH + H+}

NADPH carries the reducing power to the Calvin cycle in the stroma.

2025 Annual I, Set D, Q.6 OR2x3 = 6 marksUnit 10
Write three distinguishing characteristics and one example of each of the phylum Porifera, Platyhelminthes and Annelida.
Solution
Phylum Porifera (2 marks)
  • Sessile aquatic animals without tissues or organs and without symmetry.
  • Body perforated by pores (ostia) leading to a spongocoel; water leaves by the osculum.
  • Skeleton of calcareous or siliceous spicules, or spongin fibre.

Example: Spongilla, Sycon.

Phylum Platyhelminthes (2 marks)
  • Flattened dorsiventrally, triploblastic, bilaterally symmetrical.
  • Acoelomate: no coelom, the spaces are filled with mesenchyme.
  • Free living (Planaria) or parasitic; mostly monoecious with internal fertilization.

Example: Planaria (Dugesia), liver fluke (Fasciola), tapeworm (Taenia).

Phylum Annelida (2 marks)
  • Body metamerically segmented, bilateral.
  • True coelom divided into compartments; circular and longitudinal muscles with chitinous setae.
  • Closed circulatory system and ventral nerve cord with ganglia.

Example: Nereis, earthworm (Pheritima), leech (Hirudo).

2026 Annual I, Set D, Q.6 (a)3 marksUnit 12
Explain genetics of colour-blindness with reference to pattern of inheritance.
Solution
Colour vision

Normal (trichromatic) vision depends on three kinds of cone cells in the retina, sensitive to red, green and blue, each with its own opsin protein. The red and green opsin genes are on the X chromosome and the blue opsin gene is on autosome 7.

Types
  • Dichromacy: one opsin is missing, so two primary colours are seen (protanopia red blind, deuteranopia green blind, tritanopia blue blind).
  • Monochromacy: only one colour is seen (true colour blindness); blue cone monochromacy is X-linked recessive.
Pattern of inheritance of red-green colour blindness
GenotypePhenotype
X^CX^CNormal female
X^CX^cCarrier female (normal vision)
X^cX^cColour-blind female
X^CYNormal male
X^cYColour-blind male
  • It is X-linked recessive, so it is more common in men: one X^c is enough in a man, but a woman needs two.
  • A carrier mother passes X^c to half of her eggs, so half of her sons are colour blind.
  • It never passes directly from father to son; it zigzags from the maternal grandfather through a carrier daughter to a grandson.
Carrier mother x normal father.
Mother / fatherX^CY
X^CX^CX^C normal daughterX^CY normal son
X^cX^CX^c carrier daughterX^cY colour-blind son
2026 Annual I, Set D, Q.6 (b)3 marksUnit 12
Can two normal parents have a colourblind child? Work out the probability.
Solution

Yes, if the mother is a carrier. The colour-blind son gets his only X from his mother, so a normal-sighted mother X^CX^c and a normal father X^CY can have a colour-blind son.

Mother / fatherX^C (1/2)Y (1/2)
X^C (1/2)X^CX^C normal daughterX^CY normal son
X^c (1/2)X^CX^c carrier daughterX^cY colour-blind son
  1. Probability that the egg carries X^c is 1/2.
  2. Probability that the sperm carries Y (to make a son) is 1/2.
  3. Probability of an affected son = 1/2\times 1/2 = 1/4.

The probability of a colour-blind child is 1/4 (25%), always a son; half of the sons are colour blind, and no daughter is affected because the normal father gives her X^C. If the mother is X^CX^C, no child is colour blind.

Note. Matches Genetic Problem 12.14 (carrier woman x normal man).