I85PREP
⌘ K
Assam HS Class 11-12 Biology study materials
Assam HS Class 11-12 Biology study resources

Assam HS Class 11-12 Biology practice questions

One question at a time. Build your Assam HS Class 11-12 Biology understanding.

First 50 of 199 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Class XI Biology (first-year HS course content)

In the five-kingdom classification proposed by R. H. Whittaker, the kingdom that includes unicellular eukaryotes such as Amoeba, Paramecium and Euglena is:

  1. Monera
  2. Animalia
  3. Fungi
  4. Protista Correct answer.

Explanation

Whittaker's five kingdoms are Monera, Protista, Fungi, Plantae and Animalia. Protista is defined as the kingdom of eukaryotic organisms that are mainly unicellular, and it collects forms such as Amoeba, Paramecium, Euglena and the diatoms.

Question 2Class XI Biology (first-year HS course content)

A student observes a slide of onion root tip stained with acetocarmine and identifies a cell in which the sister chromatids have just separated and are moving towards opposite poles. This cell is in which stage of mitosis?

  1. Telophase
  2. Metaphase
  3. Anaphase Correct answer.
  4. Prophase

Explanation

Anaphase begins when the centromeres split and the two chromatids of each chromosome are pulled to opposite poles by the spindle fibres. Prophase shows condensing chromosomes, metaphase shows chromosomes aligned at the equator, and telophase shows nuclear envelopes reforming.

Question 3Class XI Biology (first-year HS course content)

Which cell organelle is described as the 'suicidal bag' of the cell because it contains hydrolytic enzymes that can digest the cell's own components?

  1. Golgi apparatus
  2. Lysosome Correct answer.
  3. Peroxisome
  4. Ribosome

Explanation

Lysosomes are membrane-bound vesicles formed from the Golgi apparatus and are rich in acid hydrolases. Because these enzymes can digest the cell's own organelles during autophagy, lysosomes are commonly called suicidal bags.

Question 4Class XI Biology (first-year HS course content)

In a flowering plant, the tissue that conducts water and dissolved minerals upward from the roots to the shoot is:

  1. Phloem
  2. Xylem Correct answer.
  3. Cambium
  4. Cortex

Explanation

Xylem is the water-conducting tissue; its tracheary elements are dead at maturity and form continuous tubes for the ascent of sap. Phloem, by contrast, conducts sugars and other organic solutes in both directions.

Question 5Class XI Biology (first-year HS course content)

Which of the following is a nitrogenous base found in RNA but not in DNA?

  1. Thymine
  2. Adenine
  3. Uracil Correct answer.
  4. Guanine

Explanation

RNA contains the pyrimidine uracil in place of the thymine found in DNA. Adenine, guanine and cytosine occur in both nucleic acids, so only uracil distinguishes the base composition of RNA.

Question 6Class XI Biology (first-year HS course content)

During inspiration in a healthy human, contraction of the diaphragm and external intercostal muscles produces which set of changes in the thoracic cavity?

  1. Decrease in thoracic volume and decrease in intrapulmonary pressure
  2. Decrease in thoracic volume and increase in intrapulmonary pressure
  3. Increase in thoracic volume and decrease in intrapulmonary pressure Correct answer.
  4. Increase in thoracic volume and increase in intrapulmonary pressure

Explanation

Contraction of the diaphragm lowers the floor of the thorax while the external intercostals lift the ribs, so thoracic volume increases. According to Boyle's law this expansion lowers intrapulmonary pressure below atmospheric pressure, and air flows in.

Question 7Class XI Biology (first-year HS course content)

A plant physiologist supplies a leaf with radioactive carbon dioxide and finds the label first incorporated into a four-carbon organic acid. This plant most probably belongs to which photosynthetic type?

  1. C3 plants
  2. Non-photosynthetic plants
  3. CAM plants only
  4. C4 plants Correct answer.

Explanation

In C4 plants the primary carbon dioxide acceptor is phosphoenolpyruvate, and the first stable labelled product is the four-carbon acid oxaloacetate (or malate). C3 plants instead fix carbon dioxide directly onto RuBP, giving the three-carbon compound 3-phosphoglycerate.

Question 8Class XI Biology (first-year HS course content)

Which of the following statements about the fluid mosaic model of the plasma membrane is correct?

  1. The lipid bilayer is rigid and the proteins are fixed in place.
  2. The bilayer is made of cellulose microfibrils embedded in protein.
  3. Proteins are embedded in or associated with a fluid lipid bilayer. Correct answer.
  4. The membrane consists only of proteins with no lipid component.

Explanation

The fluid mosaic model describes the membrane as a fluid phospholipid bilayer in which proteins are dispersed like tiles in a mosaic. Some proteins are integral, spanning the bilayer, while others are peripheral and attached to its surface.

Question 9Class XI Biology (first-year HS course content)

A cell is placed in a hypotonic solution. Which observation is expected for a plant cell with an intact cell wall?

  1. The cell wall dissolves and the cell bursts immediately.
  2. Water enters the cell, the protoplast swells and presses against the wall. Correct answer.
  3. The cell shrinks and the protoplast pulls away from the wall.
  4. There is no net movement of water across the membrane.

Explanation

In a hypotonic solution the external solute concentration is lower, so water enters the cell by osmosis. In a plant cell the rigid wall resists expansion, the protoplast becomes turgid and exerts turgor pressure, which is why plant cells do not usually burst.

Question 10Class XI Biology (first-year HS course content)

In the light reactions of photosynthesis, the splitting of water at photosystem II is significant mainly because it:

  1. Directly phosphorylates ADP to ATP without any electron transport.
  2. Produces carbon dioxide for the Calvin cycle.
  3. Provides electrons to replace those lost by the reaction centre and releases oxygen. Correct answer.
  4. Reduces NADP+ to NADPH without involving photosystem I.

Explanation

Photolysis of water at photosystem II supplies electrons that fill the hole left in the reaction centre after excitation, and molecular oxygen is released as a by-product. The electrons then travel down the electron transport chain, and NADPH is generated at the end via photosystem I.

Question 11Class XI Biology (first-year HS course content)

A patient's blood report shows a total count of leucocytes far above the normal range with a large proportion of abnormal immature white cells. This picture is most consistent with:

  1. Anaemia
  2. Thrombocytopenia
  3. Leukaemia Correct answer.
  4. Polycythaemia

Explanation

Leukaemia is a cancer of blood-forming tissues marked by uncontrolled proliferation of white blood cell precursors, so the leucocyte count rises and immature forms appear in blood. Anaemia and thrombocytopenia involve red cells and platelets respectively, and polycythaemia is an excess of red cells.

Question 12Class XI Biology (first-year HS course content)

Which of the following correctly matches a phylum with its characteristic feature?

  1. Arthropoda - water vascular system
  2. Annelida - cnidoblasts for stinging
  3. Mollusca - jointed appendages
  4. Echinodermata - water vascular system Correct answer.

Explanation

The water vascular system, with tube feet used in locomotion and feeding, is a defining feature of the phylum Echinodermata. The other pairings assign a feature to a phylum that does not possess it.

Question 13Class XI Biology (first-year HS course content)

The enzyme that unwinds the DNA double helix ahead of the replication fork during DNA replication is:

  1. Primase
  2. Helicase Correct answer.
  3. DNA ligase
  4. DNA polymerase

Explanation

Helicase breaks the hydrogen bonds between complementary base pairs and separates the two strands, creating the replication fork. DNA polymerase then adds nucleotides, primase lays down the RNA primer, and ligase joins the Okazaki fragments.

Question 14Class XI Biology (first-year HS course content)

In a dihybrid cross between two heterozygous pea plants (RrYy × RrYy), assuming independent assortment, what fraction of the progeny is expected to show the recessive phenotype for both traits?

  1. 1/4
  2. 3/16
  3. 9/16
  4. 1/16 Correct answer.

Explanation

Each trait behaves as a monohybrid cross in which the recessive phenotype appears in one quarter of the progeny. Independent assortment allows multiplication of the probabilities, so the double recessive class is 1/4 × 1/4 = 1/16.

Question 15Class XI Biology (first-year HS course content)

Which statement about the fate of the products of light-dependent and light-independent reactions is correct?

  1. The Calvin cycle produces ATP and NADPH for the light reactions.
  2. Oxygen released in photosynthesis is produced during the reduction of carbon dioxide.
  3. ATP and NADPH formed in the light reactions are used to reduce carbon dioxide in the Calvin cycle. Correct answer.
  4. ATP and NADPH are used to split water into hydrogen and oxygen.

Explanation

The light reactions convert light energy into chemical energy as ATP and NADPH, and these two products drive the reduction of 1,3-bisphosphoglycerate in the Calvin cycle. The Calvin cycle consumes ATP and NADPH rather than producing them.

Question 16Class XI Biology (first-year HS course content)

A moss plant is described as amphibious in habit mainly because:

  1. It grows on land but needs water for the transfer of male gametes to the egg. Correct answer.
  2. It can live equally well in water and on dry desert sand.
  3. It possesses a well-developed vascular system with true roots.
  4. Its sporophyte is completely independent of the gametophyte.

Explanation

Bryophytes are terrestrial plants, but their flagellated antherozoids must swim through a film of water to reach the archegonium. This dependence on water for fertilisation is why they are called amphibians of the plant kingdom.

Question 17Class XI Biology (first-year HS course content)

In the human heart, the bundle of His and the Purkinje fibres are important because they:

  1. Conduct the impulse from the atrioventricular node to the ventricular muscle. Correct answer.
  2. Prevent backflow of blood from the ventricles into the atria.
  3. Initiate the heartbeat by generating the first electrical impulse.
  4. Secrete hormones that regulate the rate of heart contraction.

Explanation

The impulse initiated at the sinoatrial node passes to the atrioventricular node and then travels through the bundle of His and Purkinje fibres to the ventricular myocardium. This specialised conducting system ensures coordinated ventricular contraction.

Question 18Class XI Biology (first-year HS course content)

A bacterial culture is shifted from 30 °C to 45 °C. Which change in the enzyme-catalysed reactions of the cells is most likely to be observed first?

  1. An immediate fall in rate because enzymes become completely inactive at any temperature above 30 °C.
  2. A gradual increase in reaction rate with no upper limit.
  3. An initial rise in rate followed by a fall as enzymes begin to denature. Correct answer.
  4. No change at all, because enzyme activity is independent of temperature.

Explanation

Raising temperature increases kinetic energy and hence the frequency of enzyme-substrate collisions, so the rate rises initially. Beyond the optimum, however, heat disrupts hydrogen bonds and other weak interactions, denaturing the enzyme and lowering the rate.

Question 19Class XI Biology (first-year HS course content)

Which of the following best explains why the oxygen dissociation curve of haemoglobin is sigmoid in shape?

  1. Carbon dioxide increases the affinity of haemoglobin for oxygen.
  2. Each haemoglobin molecule binds only one oxygen molecule.
  3. Binding of the first oxygen molecule increases the affinity of the remaining subunits for oxygen. Correct answer.
  4. Oxygen binds irreversibly to haemoglobin in the lungs.

Explanation

Haemoglobin is a tetramer, and the four subunits interact cooperatively. After the first oxygen binds, the conformation changes so that the remaining sites bind oxygen more readily, producing the characteristic sigmoid curve.

Question 20Class XI Biology (first-year HS course content)

Which of the following is a correct difference between the cell wall of a typical plant cell and the cell wall of a bacterium?

  1. Plant walls contain chitin, while bacterial walls contain cellulose.
  2. Plant walls are made mainly of cellulose, while bacterial walls contain peptidoglycan. Correct answer.
  3. Plant walls are absent in young cells, while bacterial walls are always present.
  4. Plant walls are selectively permeable, while bacterial walls are fully permeable.

Explanation

The plant cell wall is composed chiefly of cellulose microfibrils embedded in a matrix of hemicellulose and pectin. Bacterial cell walls are built of peptidoglycan, a polymer of sugars cross-linked by short peptides.

Question 21Class XI Biology (first-year HS course content)

A mutation changes a codon from GAA to GAG but the amino acid incorporated remains glutamic acid. This mutation is best described as:

  1. A silent mutation Correct answer.
  2. A frameshift mutation
  3. A nonsense mutation
  4. A chromosomal deletion

Explanation

GAA and GAG both code for glutamic acid, so the amino acid sequence of the protein is unchanged. Such a change is called a silent mutation because the degeneracy of the genetic code absorbs it.

Question 22Class XI Biology (first-year HS course content)

In a food chain, the amount of energy transferred from one trophic level to the next is generally about 10 per cent. Which statement best explains the fate of the remaining energy?

  1. It is lost as heat during respiration and used in metabolism, excretion and undigested material. Correct answer.
  2. It is converted completely into biomass of the next trophic level.
  3. It returns to the sun as reflected light.
  4. It is stored indefinitely in the body tissues of the next trophic level.

Explanation

Assimilation is inefficient: much of the ingested energy is used in respiration and released as heat, some is excreted, and some remains undigested. Only the small fraction retained as new biomass can be passed on, which is why trophic levels form a pyramid.

Question 23Class XI Biology (first-year HS course content)

Which of the following observations would most strongly support the conclusion that a particular plant species is pollinated by wind rather than by insects?

  1. Sticky pollen grains with spiny surfaces and a strong scent.
  2. Small, inconspicuous flowers with large feathery stigmas and abundant light pollen. Correct answer.
  3. Flowers that open only at night and are white in colour.
  4. Large, brightly coloured petals with nectar guides.

Explanation

Wind-pollinated flowers are typically small and inconspicuous, without scent or nectar, and produce large amounts of light, dry pollen. Feathery, exposed stigmas increase the chance of catching airborne pollen grains.

Question 24Class XI Biology (first-year HS course content)

A cell with 2n = 24 undergoes meiosis. What is the chromosome number in each daughter cell at the end of meiosis II, and what is the number of chromatids in a cell at metaphase I?

  1. 24 chromosomes and 24 chromatids
  2. 12 chromosomes and 48 chromatids Correct answer.
  3. 12 chromosomes and 24 chromatids
  4. 24 chromosomes and 48 chromatids

Explanation

Meiosis halves the chromosome number, so each haploid product of a 2n = 24 cell contains 12 chromosomes. At metaphase I the 24 chromosomes are each duplicated into two chromatids, giving 48 chromatids in the cell.

Question 25Class XI Biology (first-year HS course content)

Two pure-breeding lines of snapdragon, one with red flowers and one with white flowers, are crossed. The F1 plants all have pink flowers, and selfing of F1 gives red, pink and white in the ratio 1:2:1. Which interpretation is correct?

  1. The pink F1 plants are heterozygous for a lethal allele.
  2. The trait is controlled by two genes showing epistasis.
  3. The red allele is completely dominant over the white allele.
  4. The alleles show incomplete dominance, so the heterozygote has an intermediate phenotype. Correct answer.

Explanation

The F1 heterozygote is phenotypically intermediate between the two homozygotes, and the F2 shows a 1:2:1 phenotypic ratio matching the genotypic ratio. This is the classic pattern of incomplete dominance, not of complete dominance.

Question 26Class XI Biology (first-year HS course content)

Which taxonomic category is placed immediately above the class in the standard hierarchy used for classifying plants and animals?

  1. Phylum Correct answer.
  2. Genus
  3. Family
  4. Order

Explanation

The standard Linnaean hierarchy runs kingdom, phylum (division in plants), class, order, family, genus and species, so the category directly above class is phylum. Order lies below class, and family and genus lie further down the hierarchy.

Question 27Class XI Biology (first-year HS course content)

In the five-kingdom classification, the kingdom Monera is distinguished from all other kingdoms mainly because its members:

  1. have cell walls made only of cellulose
  2. reproduce exclusively by sexual means
  3. possess a true nucleus bounded by a nuclear envelope
  4. are prokaryotic, lacking a membrane-bound nucleus Correct answer.

Explanation

Monera contains the bacteria, which are prokaryotes: their genetic material is not enclosed by a nuclear membrane and they lack membrane-bound organelles. All remaining four kingdoms are eukaryotic, which is the fundamental dividing line in this system.

Question 28Class XI Biology (first-year HS course content)

A student examines a slide of onion peel cells and records the position of the nucleus. In a typical mature onion epidermal cell the nucleus is located:

  1. dispersed as many small nuclei throughout the cytoplasm
  2. pushed to the periphery near the cell wall by a large central vacuole Correct answer.
  3. at the centre of the cell, surrounded by a large vacuole
  4. inside the chloroplast, since onion peel is green tissue

Explanation

Mature plant cells contain a large central vacuole that occupies most of the cell volume and presses the cytoplasm and nucleus towards the periphery. The nucleus is therefore seen near the cell wall rather than in the middle.

Question 29Class XI Biology (first-year HS course content)

Which biomolecule class is built from amino acids joined by peptide bonds, and includes enzymes and antibodies?

  1. Nucleic acids
  2. Lipids
  3. Proteins Correct answer.
  4. Carbohydrates

Explanation

Proteins are polymers of amino acids linked by peptide bonds, and this class includes enzymes, antibodies, hormones and structural fibres. Carbohydrates are built from sugars, lipids from fatty acids and glycerol, and nucleic acids from nucleotides.

Question 30Class XI Biology (first-year HS course content)

During which phase of mitosis do the sister chromatids separate and move towards opposite poles of the cell?

  1. Anaphase Correct answer.
  2. Telophase
  3. Prophase
  4. Metaphase

Explanation

In anaphase the centromeres split and the two sister chromatids of each chromosome are pulled to opposite poles. Prophase involves chromosome condensation, metaphase their alignment at the equator, and telophase nuclear reorganisation.

Question 31Class XI Biology (first-year HS course content)

Which tissue in a flowering plant conducts water and dissolved minerals upward from the roots to the shoot?

  1. Xylem Correct answer.
  2. Epidermis
  3. Collenchyma
  4. Phloem

Explanation

Xylem is the water-conducting tissue; its tracheary elements are dead, hollow cells that form continuous tubes from root to leaf. Phloem conducts sugars, while collenchyma and epidermis serve support and protection.

Question 32Class XI Biology (first-year HS course content)

Which of the following is a distinguishing feature of the phylum Arthropoda?

  1. Presence of a water vascular system with tube feet
  2. Presence of a notochord at some stage of life
  3. Presence of a mantle and radula
  4. Presence of a jointed exoskeleton and paired jointed appendages Correct answer.

Explanation

Arthropods are named for their jointed legs and possess a chitinous exoskeleton that is shed during moulting. The water vascular system belongs to echinoderms, the notochord to chordates, and the mantle and radula to molluscs.

Question 33Class XI Biology (first-year HS course content)

In the human alimentary canal, the enzyme pepsin that digests proteins is secreted in an active or activatable form into the:

  1. mouth
  2. large intestine
  3. duodenum
  4. stomach Correct answer.

Explanation

The stomach's gastric glands release pepsinogen, which hydrochloric acid converts to active pepsin in the acidic lumen. Protein digestion begins in the stomach, and pepsin works optimally at a low pH.

Question 34Class XI Biology (first-year HS course content)

A red blood cell is placed in distilled water. Which change is expected and why?

  1. It swells and may burst because water enters by osmosis Correct answer.
  2. It swells because solutes move into the cell by active transport
  3. It shrinks because water leaves the cell by osmosis
  4. It remains unchanged because the cell wall resists pressure

Explanation

Distilled water is hypotonic to the cytoplasm, so water moves in by osmosis across the plasma membrane. An animal cell lacks a rigid wall, so it swells and eventually haemolyses.

Question 35Class XI Biology (first-year HS course content)

An enzyme-catalysed reaction is studied at increasing substrate concentration with enzyme concentration fixed. The rate rises and then levels off. The plateau is best explained by:

  1. saturation of all active sites of the enzyme molecules Correct answer.
  2. removal of the products from the reaction mixture
  3. denaturation of the enzyme at high substrate concentration
  4. a fall in the activation energy of the reaction

Explanation

At the plateau every enzyme molecule is engaged with substrate, so the active sites are saturated and further substrate cannot increase the rate. The maximum velocity is set by enzyme concentration and turnover number.

Question 36Class XI Biology (first-year HS course content)

A plant is kept in darkness for 48 hours and then a leaf is partly covered with black paper and exposed to sunlight for six hours. After decolourisation and iodine testing, starch is detected:

  1. only in the covered part of the leaf
  2. equally in both covered and exposed parts
  3. only in the exposed part of the leaf Correct answer.
  4. in neither part of the leaf

Explanation

Only the illuminated part can photosynthesise and accumulate starch, so iodine gives a blue-black colour there. The covered part receives no light and cannot make starch, so it stains brown with iodine.

Question 37Class XI Biology (first-year HS course content)

In the electron transport chain of aerobic respiration, the final acceptor of electrons is:

  1. oxygen Correct answer.
  2. cytochrome c
  3. NAD+
  4. pyruvate

Explanation

Oxygen accepts electrons at the end of the chain and combines with protons to form water, allowing the chain to keep running. NAD+ and cytochrome c are carriers within the chain, not the terminal acceptor.

Question 38Class XI Biology (first-year HS course content)

A person hyperventilates voluntarily for one minute. Which immediate change in blood chemistry is expected?

  1. A fall in oxygen with no change in carbon dioxide
  2. A rise in carbon dioxide with a fall in pH
  3. A fall in carbon dioxide with a rise in pH Correct answer.
  4. A rise in both oxygen and carbon dioxide

Explanation

Hyperventilation removes carbon dioxide faster than it is produced, so the partial pressure of carbon dioxide falls. Less carbon dioxide means less carbonic acid, so the blood pH rises, producing respiratory alkalosis.

Question 39Class XI Biology (first-year HS course content)

Which vessel carries deoxygenated blood from the heart to the lungs in a human?

  1. Systemic vein
  2. Aorta
  3. Pulmonary vein
  4. Pulmonary artery Correct answer.

Explanation

The pulmonary artery leaves the right ventricle and carries deoxygenated blood to the lungs, making it the exception to the rule that arteries carry oxygenated blood. The pulmonary vein returns oxygenated blood to the left atrium.

Question 40Class XI Biology (first-year HS course content)

In a cross between a heterozygous tall pea plant (Tt) and a dwarf plant (tt), what proportion of the offspring is expected to be dwarf?

  1. 100 per cent
  2. 25 per cent
  3. 50 per cent Correct answer.
  4. 0 per cent

Explanation

The cross Tt × tt gives gametes T and t from the heterozygote and only t from the dwarf parent. Half the offspring are Tt (tall) and half tt (dwarf), a 1:1 ratio.

Question 41Class XI Biology (first-year HS course content)

Which statement correctly describes the role of the centromere during cell division?

  1. It forms the nuclear envelope after telophase
  2. It is the site where DNA replication begins
  3. It is the region where the two sister chromatids are held together and attach to spindle fibres Correct answer.
  4. It synthesises the proteins of the spindle fibres

Explanation

The centromere is the constricted region joining sister chromatids, and it bears the kinetochore to which spindle microtubules attach. This attachment is what allows chromatids to be pulled apart in anaphase.

Question 42Class XI Biology (first-year HS course content)

In a flowering plant, the megaspore mother cell undergoes meiosis and the functional megaspore then divides mitotically to form the embryo sac. The typical mature embryo sac contains:

  1. sixteen nuclei in eight cells
  2. eight nuclei in seven cells, including two polar nuclei in the central cell Correct answer.
  3. four nuclei in four cells, each haploid
  4. one egg cell and one synergid only

Explanation

The Polygonum type embryo sac is eight-nucleate and seven-celled: three antipodals, two synergids, one egg and a central cell containing two polar nuclei. This is the structure that receives the pollen tube during double fertilisation.

Question 43Class XI Biology (first-year HS course content)

A patient's diet lacks sufficient protein over a long period while energy intake is adequate. Which change in the blood is most directly expected?

  1. A fall in plasma albumin concentration Correct answer.
  2. A rise in plasma glucose concentration
  3. A fall in plasma cholesterol concentration
  4. A rise in plasma albumin concentration

Explanation

Plasma albumin is synthesised by the liver from dietary amino acids, so prolonged protein deficiency lowers its concentration. This reduces the colloid osmotic pressure of plasma and contributes to oedema.

Question 44Class XI Biology (first-year HS course content)

Which feature is characteristic of the phylum Mollusca but absent in Annelida?

  1. A mantle that secretes the shell and a muscular foot for locomotion Correct answer.
  2. A digestive tract with mouth and anus
  3. A true coelom lined by mesoderm
  4. Bilateral symmetry of the body

Explanation

Molluscs possess a mantle, a fleshy fold that secretes the calcareous shell and encloses the mantle cavity bearing the gills, together with a muscular foot. Annelids are coelomate, bilaterally symmetrical animals with a complete gut but no mantle.

Question 45Class XI Biology (first-year HS course content)

A freshwater protozoan is transferred to sea water. Which response would best maintain its internal water balance?

  1. Increasing the permeability of its membrane to water
  2. Decreasing the activity of its contractile vacuole Correct answer.
  3. Increasing the activity of its contractile vacuole
  4. Increasing the rate of active salt uptake from the medium

Explanation

Sea water is hypertonic to the fresh-water organism, so water tends to leave rather than enter. Reducing contractile vacuole activity limits further water loss, since the vacuole normally expels excess water in hypotonic surroundings.

Question 46Class XI Biology (first-year HS course content)

A segment of DNA has the template strand sequence 3'-TAC GGA-5'. Assuming standard genetic code reading, which amino acid sequence is translated?

  1. Tyrosine–Glycine
  2. Methionine–Proline Correct answer.
  3. Tyrosine–Proline
  4. Methionine–Glycine

Explanation

The mRNA transcribed from the template 3'-TAC GGA-5' is 5'-AUG CCU-3'. AUG codes for methionine and CCU for proline, so the dipeptide is methionine–proline.

Question 47Class XI Biology (first-year HS course content)

In the C4 pathway, the first stable product of carbon fixation is oxaloacetate, formed in mesophyll cells. The advantage of this pathway to a tropical grass is that it:

  1. eliminates the need for light reactions
  2. fixes carbon dioxide directly into glucose without any intermediate
  3. concentrates carbon dioxide around rubisco and reduces photorespiration Correct answer.
  4. allows stomata to remain wide open in hot dry conditions

Explanation

PEP carboxylase in mesophyll cells fixes carbon dioxide into a four-carbon acid that is decarboxylated in bundle-sheath cells, raising the carbon dioxide concentration around rubisco. This suppresses the oxygenase activity of rubisco and cuts photorespiration losses in hot climates.

Question 48Class XI Biology (first-year HS course content)

Two genes are located close together on the same chromosome. In a test cross, the proportion of recombinant offspring observed is much lower than the 50 per cent expected for independent assortment. The best interpretation is that:

  1. the two genes are linked and crossing over between them is infrequent Correct answer.
  2. the two genes are on different chromosomes but one shows incomplete dominance
  3. the genes assort independently but the sample size was too small
  4. a mutation has occurred in one of the genes

Explanation

Genes on the same chromosome tend to be inherited together, and recombination between them arises only through crossing over. The closer the genes, the rarer the crossover, so recombinant frequency falls well below 50 per cent, which is the signature of linkage.

Question 49Class XI Biology (first-year HS course content)

A cell with 2n = 12 completes meiosis. Which statement about the products is correct?

  1. Four cells, each with 6 chromosomes, genetically varied due to crossing over and independent assortment Correct answer.
  2. Two cells, each with 6 chromosomes, genetically identical
  3. Two cells, each with 12 chromosomes, genetically varied
  4. Four cells, each with 12 chromosomes, all genetically identical

Explanation

Meiosis halves the chromosome number, so each of the four haploid products contains 6 chromosomes. Crossing over in prophase I and independent assortment in metaphase I make the four products genetically different from one another.

Question 50Class XI Biology (first-year HS course content)

In an ecosystem, net primary productivity is calculated as gross primary productivity minus respiration of the producers. If gross primary productivity is 2000 kcal m⁻² yr⁻¹ and plant respiration is 800 kcal m⁻² yr⁻¹, the net primary productivity available to herbivores is:

  1. 1200 kcal m⁻² yr⁻¹ Correct answer.
  2. 2800 kcal m⁻² yr⁻¹
  3. 800 kcal m⁻² yr⁻¹
  4. 200 kcal m⁻² yr⁻¹

Explanation

Net primary productivity equals gross primary productivity minus the respiration losses of autotrophs, so 2000 minus 800 gives 1200 kcal m⁻² yr⁻¹. This stored energy is the maximum that primary consumers can draw upon.