AP Biology Practice Test 11 – 60 Questions with Answers (College Board)

Work through all 60 AP Biology questions, checking each answer as you go to build confidence for your AP Biology exam.

Chemistry of Life

Q1. A lake stays liquid under winter ice while air temperature falls well below freezing. Which property of water best explains why aquatic organisms survive beneath the ice?

  • (A) Ice is less dense than liquid water, so it floats and insulates the water below
  • (B) Water has a low specific heat, so it cools rapidly
  • (C) Water molecules are nonpolar and repel dissolved salts
  • (D) Ice conducts heat away from the lake very efficiently
Show Answer

Answer: A – Because hydrogen bonds hold water molecules farther apart in ice, ice floats and insulates the liquid water beneath it.

Q2. Two solutions differ by 3 pH units. How do their hydrogen ion concentrations compare?

  • (A) The more acidic solution has 3 times the H+ concentration
  • (B) The more acidic solution has 1,000 times the H+ concentration
  • (C) The more acidic solution has 30 times the H+ concentration
  • (D) The more basic solution has 1,000 times the H+ concentration
Show Answer

Answer: B – pH is logarithmic, so each unit is a tenfold change; 103 = 1,000.

Q3. A researcher replaces the hydroxyl group on a sugar with a hydrogen atom. Which functional-group change most likely reduces the molecule’s solubility in water?

  • (A) Gain of a methyl group that donates electrons to water
  • (B) Loss of a carbonyl group that repels water molecules
  • (C) Gain of a phosphate group that neutralises the molecule
  • (D) Loss of a polar hydroxyl group that could hydrogen-bond with water
Show Answer

Answer: D – Hydroxyl groups are polar and form hydrogen bonds with water; removing one lowers solubility.

Q4. Which feature of carbon most directly allows the enormous diversity of biological molecules?

  • (A) Carbon forms four stable covalent bonds and can bond to other carbons in chains and rings
  • (B) Carbon has a full outer shell and rarely bonds
  • (C) Carbon forms only ionic bonds with hydrogen
  • (D) Carbon atoms cannot bond to nitrogen or oxygen
Show Answer

Answer: A – Tetravalent carbon builds chains, branches and rings, the backbone of organic molecules.

Q5. Starch, glycogen and cellulose are all polymers of glucose. What mainly accounts for their very different functions?

  • (A) Differences in the monomer, which is fructose in cellulose
  • (B) Differences in glycosidic linkage and branching change their shape and digestibility
  • (C) Differences in the number of carbon atoms per glucose
  • (D) Differences in whether they contain nitrogen
Show Answer

Answer: B – Alpha 1-4 and 1-6 linkages give storage polymers, while beta 1-4 linkages give strong structural fibres.

Q6. A phospholipid bilayer is exposed to a nonpolar solvent. Which part of the membrane is most likely to interact with the solvent?

  • (A) The phosphate heads on the outer surfaces
  • (B) The fatty-acid tails in the interior of the bilayer
  • (C) The integral proteins spanning the membrane
  • (D) The carbohydrate chains on the cell surface
Show Answer

Answer: B – Hydrophobic tails face inward and interact with nonpolar substances, while polar heads face water.

Q7. A protein is heated until it loses its three-dimensional shape but its amino-acid sequence is unchanged. Which level of structure is lost?

  • (A) Secondary and tertiary structure, while primary structure remains
  • (B) Primary structure only
  • (C) Quaternary structure only, with tertiary structure intact
  • (D) All levels including the amino-acid sequence
Show Answer

Answer: A – Denaturation disrupts hydrogen bonds and other weak interactions, not the peptide bonds of the primary sequence.

Cell Structure and Function

Q8. A spherical cell doubles its radius. How does its surface-area-to-volume ratio change?

  • (A) It is halved, so exchange across the membrane becomes less efficient per unit volume
  • (B) It doubles, improving exchange
  • (C) It stays the same because both increase equally
  • (D) It falls to one quarter of its original value
Show Answer

Answer: A – Surface area grows with r2 and volume with r3, so SA/V falls as 1/r; doubling r halves the ratio.

Q9. A white blood cell engulfs a bacterium into a vesicle that then fuses with a lysosome. Which sequence best describes this?

  • (A) Pinocytosis, followed by exocytosis of the bacterium
  • (B) Osmosis, followed by active transport out of the cell
  • (C) Facilitated diffusion, followed by hydrolysis in the cytosol
  • (D) Phagocytosis, followed by enzymatic digestion inside the phagolysosome
Show Answer

Answer: D – Phagocytosis brings the particle in, and lysosomal enzymes digest it.

Q10. Cells are placed in a solution and swell until some burst. What can be concluded about the solution?

  • (A) It is hypertonic, so water entered by osmosis
  • (B) It is isotonic, so solutes left the cells
  • (C) It is hypertonic, so solutes entered the cells
  • (D) It is hypotonic to the cells, so water entered by osmosis
Show Answer

Answer: D – In a hypotonic solution water moves into the cell, causing swelling and possible lysis.

Q11. Which evidence best supports the endosymbiotic origin of mitochondria?

  • (A) Mitochondria are the largest organelle in every cell
  • (B) Mitochondria contain ribosomes identical to cytosolic ribosomes
  • (C) Mitochondria have their own circular DNA and double membrane and divide independently
  • (D) Mitochondria are enclosed by a single membrane only
Show Answer

Answer: C – Circular DNA, 70S-like ribosomes and a double membrane point to a free-living bacterial ancestor.

Q12. A plant cell is placed in a concentrated salt solution and its cytoplasm shrinks away from the cell wall. This is best explained by

  • (A) water leaving the central vacuole and cytoplasm by osmosis
  • (B) active transport of salt into the vacuole
  • (C) facilitated diffusion of water into the cell
  • (D) rupture of the cell wall by turgor pressure
Show Answer

Answer: A – Water leaves a cell in a hypertonic solution, causing plasmolysis.

Q13. Which transport process moves glucose into an intestinal cell against its concentration gradient by coupling it to sodium moving down its gradient?

  • (A) Secondary active transport (cotransport)
  • (B) Simple diffusion through the lipid bilayer
  • (C) Facilitated diffusion through a channel
  • (D) Osmosis through aquaporins
Show Answer

Answer: A – Cotransport uses the sodium gradient, established by the sodium-potassium pump, to drive glucose uptake.

Q14. A mutation prevents ribosomes from binding to the rough endoplasmic reticulum. Which process is most directly impaired?

  • (A) Synthesis of lipids for the plasma membrane
  • (B) Production of ATP by oxidative phosphorylation
  • (C) Secretion of proteins such as insulin and antibodies
  • (D) Replication of nuclear DNA
Show Answer

Answer: C – Proteins destined for secretion are translated into the rough ER for processing and export.

Cellular Energetics

Q15. An enzyme-catalysed reaction is run at increasing substrate concentrations. The rate levels off at high substrate concentration. What best explains the plateau?

  • (A) The enzyme is denatured by excess substrate
  • (B) The reaction has reached equilibrium with no product formed
  • (C) All active sites are saturated with substrate
  • (D) Competitive inhibitors accumulate as substrate rises
Show Answer

Answer: C – At Vmax every enzyme molecule is working at full capacity, so more substrate cannot increase the rate.

Q16. A noncompetitive inhibitor binds to an enzyme. Which change is expected?

  • (A) Vmax increases and Km decreases
  • (B) Km increases with no change in Vmax
  • (C) Both Vmax and Km increase equally
  • (D) Vmax decreases, while Km is largely unchanged
Show Answer

Answer: D – Noncompetitive inhibition reduces the number of functional enzymes, lowering Vmax without changing substrate affinity.

Q17. For a reaction with a negative change in Gibbs free energy, which statement is true?

  • (A) The reaction requires a constant input of ATP
  • (B) The products have more free energy than the reactants
  • (C) The reaction is exergonic and can proceed spontaneously
  • (D) The reaction is endergonic and never occurs in cells
Show Answer

Answer: C – A negative delta G means free energy is released and the reaction is spontaneous.

Q18. In the light reactions of photosynthesis, water is split and O2 is released. Where do the electrons from water ultimately go?

  • (A) To NADP+, forming NADPH for the Calvin cycle
  • (B) Directly to carbon dioxide to form glucose
  • (C) To ADP, forming ATP by substrate-level phosphorylation
  • (D) Back to photosystem II to restart the cycle
Show Answer

Answer: A – Electrons from water travel through the electron transport chain to reduce NADP+ to NADPH.

Q19. A plant is kept in light with carbon dioxide but no water. Which Calvin cycle outcome is expected?

  • (A) The cycle runs faster using stored oxygen
  • (B) Glucose is made directly from carbon dioxide alone
  • (C) Only the light reactions continue normally
  • (D) The cycle stalls because no NADPH or ATP is supplied by the light reactions
Show Answer

Answer: D – Without water as an electron donor, the light reactions cannot supply ATP and NADPH to the Calvin cycle.

Q20. During cellular respiration, most ATP is produced by which process?

  • (A) Oxidative phosphorylation using a proton gradient across the inner mitochondrial membrane
  • (B) Substrate-level phosphorylation in glycolysis only
  • (C) Fermentation in the cytoplasm
  • (D) Photophosphorylation in the thylakoid membrane
Show Answer

Answer: A – The electron transport chain and ATP synthase generate the bulk of ATP in aerobic respiration.

Q21. Yeast fermenting glucose in the absence of oxygen produces ethanol and CO2. What is the main purpose of fermentation for the yeast?

  • (A) To produce more ATP than aerobic respiration
  • (B) To oxidise glucose completely to water
  • (C) To regenerate NAD+ so glycolysis can continue
  • (D) To store oxygen for later use
Show Answer

Answer: C – Fermentation oxidises NADH back to NAD+, allowing glycolysis to keep making a small ATP yield.

Q22. A muscle cell is poisoned with cyanide, which blocks cytochrome c oxidase. Which immediate effect is expected?

  • (A) Glycolysis stops because no glucose enters the cell
  • (B) The Krebs cycle produces extra NADH to compensate
  • (C) Oxygen consumption rises as the chain speeds up
  • (D) Electrons back up in the transport chain and ATP synthesis by chemiosmosis falls sharply
Show Answer

Answer: D – Blocking the final electron acceptor step halts electron flow, the proton gradient and most ATP production.

Cell Communication and Cell Cycle

Q23. Epinephrine binds a receptor on a liver cell and triggers glycogen breakdown, but it does not enter the cell. What does this imply about the receptor?

  • (A) It is a nuclear receptor that binds DNA directly
  • (B) It is an enzyme that digests glycogen itself
  • (C) It is a gap junction allowing epinephrine to pass between cells
  • (D) It is a membrane receptor that activates an intracellular signalling cascade
Show Answer

Answer: D – Water-soluble hormones bind cell-surface receptors and act through second messengers.

Q24. In a signal transduction pathway, cyclic AMP is produced after a hormone binds. What is the role of cyclic AMP?

  • (A) It acts as a second messenger that activates protein kinase A
  • (B) It is the hormone itself, entering the nucleus
  • (C) It blocks the receptor so the signal ends
  • (D) It transports the hormone across the membrane
Show Answer

Answer: A – Second messengers relay and amplify the signal inside the cell.

Q25. A cell receives a signal that activates a kinase cascade. What is the main advantage of such a cascade?

  • (A) It prevents any response until the signal is removed
  • (B) Each step can amplify the signal, so a few molecules trigger a large response
  • (C) It ensures only one protein is ever activated
  • (D) It makes the signal independent of the receptor
Show Answer

Answer: B – Sequential activation multiplies the signal at each step.

Q26. Insulin is released when blood glucose rises, and it lowers blood glucose. This is an example of

  • (A) negative feedback, which returns a variable toward its set point
  • (B) positive feedback, which amplifies the change
  • (C) feedforward control, which ignores the variable
  • (D) competitive inhibition of glucose transport
Show Answer

Answer: A – The response opposes the original change, restoring homeostasis.

Q27. During labour, oxytocin release increases uterine contractions, which stimulate more oxytocin release. This is an example of

  • (A) negative feedback, which stabilises contractions
  • (B) positive feedback, which amplifies the change until the process ends
  • (C) a signal transduction cascade with no amplification
  • (D) apoptosis triggered by a hormone
Show Answer

Answer: B – Positive feedback reinforces the change, driving labour to completion.

Q28. A cell in G1 has damaged DNA that cannot be repaired. At which checkpoint is the cycle most likely to arrest?

  • (A) The G2 checkpoint, after mitosis is complete
  • (B) The metaphase checkpoint, during chromosome separation
  • (C) The G1 checkpoint, before DNA replication begins
  • (D) The cytokinesis checkpoint, after the nucleus divides
Show Answer

Answer: C – The G1 checkpoint assesses DNA integrity before committing to S phase.

Q29. Which event is controlled by the spindle assembly checkpoint?

  • (A) Completion of DNA replication before mitosis
  • (B) Growth of the cell to sufficient size in G1
  • (C) Attachment of all chromosomes to spindle microtubules before anaphase begins
  • (D) Separation of the cytoplasm during cytokinesis
Show Answer

Answer: C – The metaphase checkpoint ensures proper kinetochore attachment, preventing missegregation.

Q30. A drug blocks cyclin-dependent kinases. Which process is most directly disrupted?

  • (A) Translation of mRNA into protein
  • (B) Replication of mitochondrial DNA only
  • (C) Progression through the cell cycle, because Cdks drive transitions between phases
  • (D) Active transport across the plasma membrane
Show Answer

Answer: C – Cdks, activated by cyclins, phosphorylate targets that advance the cycle.


Heredity

Q31. In meiosis, homologous chromosomes pair and exchange segments. What is the main genetic consequence?

  • (A) Identical daughter cells with no variation
  • (B) Doubling of the chromosome number in gametes
  • (C) Loss of all recessive alleles from the population
  • (D) New combinations of alleles on chromosomes, increasing variation
Show Answer

Answer: D – Crossing over creates recombinant chromosomes with new allele combinations.

Q32. A diploid cell with 2n = 12 undergoes meiosis. How many chromosomes are in each gamete, and how many chromatids are in a secondary oocyte in metaphase II?

  • (A) 12 chromosomes; 24 chromatids
  • (B) 6 chromosomes; 6 chromatids
  • (C) 6 chromosomes; 12 chromatids
  • (D) 3 chromosomes; 6 chromatids
Show Answer

Answer: C – Meiosis halves the number to 6 chromosomes, each still consisting of two sister chromatids in metaphase II.

Q33. A cross AaBb x aabb produces offspring in a 1:1:1:1 ratio. What does this indicate?

  • (A) The two genes are tightly linked on the same chromosome
  • (B) One gene is sex-linked and the other is not
  • (C) The cross is a test of codominance only
  • (D) The two genes assort independently
Show Answer

Answer: D – Equal proportions of all four phenotypes are expected only with independent assortment.

Q34. In a pedigree, two unaffected parents have an affected daughter. What is the most likely mode of inheritance?

  • (A) Autosomal dominant
  • (B) X-linked dominant
  • (C) Autosomal recessive
  • (D) Y-linked
Show Answer

Answer: C – Unaffected parents producing an affected child indicates recessive inheritance, and an affected daughter rules out Y-linkage.

Q35. A plant heterozygous for flower colour (Rr) is self-crossed. What is the probability of a white-flowered (rr) offspring?

  • (A) 1/2
  • (B) 3/4
  • (C) 1/4
  • (D) 1/16
Show Answer

Answer: C – A monohybrid cross gives a 1:2:1 genotype ratio, so rr occurs in one quarter of offspring.

Q36. Genes A and B are 20 map units apart. What percentage of gametes from an AB/ab heterozygote are expected to be recombinant?

  • (A) 10 percent
  • (B) 20 percent
  • (C) 40 percent
  • (D) 50 percent
Show Answer

Answer: B – One map unit equals 1 percent recombination, so 20 map units gives 20 percent recombinant gametes.

Q37. Nondisjunction in meiosis I in a human produces gametes with which chromosome numbers?

  • (A) Two gametes with n + 1 and two with n – 1
  • (B) Four gametes with n + 1
  • (C) Two gametes with n and two with n + 1
  • (D) Four gametes with the normal number n
Show Answer

Answer: A – Failure of homologues to separate in meiosis I gives two disomic and two nullisomic gametes.

Q38. A chi-square test on a dihybrid cross gives a value of 9.2 with 3 degrees of freedom. The critical value at p = 0.05 is 7.81. What is the correct conclusion?

  • (A) Reject the null hypothesis; the observed ratio differs significantly from the expected ratio
  • (B) Accept the null hypothesis; the fit is good
  • (C) The data prove the genes are linked
  • (D) The degrees of freedom were calculated incorrectly
Show Answer

Answer: A – A chi-square value above the critical value means the deviation is unlikely by chance alone.

Gene Expression and Regulation

Q39. During translation, a tRNA molecule arrives at the ribosome carrying the amino acid methionine. Which feature of the tRNA allows it to pair with the correct mRNA codon?

  • (A) Its amino acid attachment site, which reads the codon directly
  • (B) Its anticodon, a three-base sequence complementary to the mRNA codon
  • (C) Its 5′ cap, which binds the ribosome to the mRNA
  • (D) Its poly-A tail, which stabilises the messenger RNA
Show Answer

Answer: B – The anticodon base-pairs with the mRNA codon, ensuring the correct amino acid is added to the growing polypeptide.

Q40. A gene’s coding strand reads 5′-ATG CCA TGG-3′. What is the sequence of the mRNA transcribed from this gene?

  • (A) 5′-TAC GGT ACC-3′
  • (B) 3′-UAC GGU ACC-5′
  • (C) 5′-ATG CCA TGG-3′
  • (D) 5′-AUG CCA UGG-3′
Show Answer

Answer: D – The mRNA sequence matches the coding strand (5′ to 3′) with uracil in place of thymine.

Q41. A mutation changes a codon from GAA to GAG. Both code for glutamate. What type of mutation is this?

  • (A) A frameshift mutation
  • (B) A nonsense mutation
  • (C) A silent mutation
  • (D) A missense mutation
Show Answer

Answer: C – The amino acid is unchanged, so the protein sequence is unaffected.

Q42. In the lac operon, lactose is present and glucose is absent. What is the expected state of the operon?

  • (A) Transcription is off because the repressor is bound
  • (B) Transcription is high because the repressor is inactive and CAP is bound
  • (C) Transcription is low because CAP cannot bind
  • (D) Transcription is high because glucose activates the repressor
Show Answer

Answer: B – Allolactose inactivates the repressor and, without glucose, cAMP-CAP activates transcription.

Q43. A eukaryotic gene has introns removed from its primary transcript. What is this process called?

  • (A) Translation
  • (B) RNA splicing
  • (C) Replication
  • (D) Reverse transcription
Show Answer

Answer: B – Splicing removes introns and joins exons to form mature mRNA.

Q44. Alternative splicing of a single gene can produce several different proteins. What does this demonstrate?

  • (A) One protein is made from several genes
  • (B) Introns code for the same protein in all tissues
  • (C) Splicing changes the DNA sequence permanently
  • (D) One gene can give rise to multiple polypeptides
Show Answer

Answer: D – Different exon combinations from one transcript yield different proteins.

Q45. A researcher uses PCR to amplify a DNA fragment. Which component is essential for the reaction to start synthesis at the target site?

  • (A) Restriction enzymes that cut the plasmid
  • (B) Primers complementary to the sequences flanking the target
  • (C) RNA polymerase to transcribe the fragment
  • (D) Ligase to join the amplified fragments
Show Answer

Answer: B – Primers define the region to be amplified and provide a 3′ hydroxyl group for DNA polymerase.

Q46. A single base substitution creates a premature stop codon in a gene for a metabolic enzyme. What is the most likely effect on the protein?

  • (A) A truncated, usually nonfunctional protein
  • (B) A longer protein with extra amino acids
  • (C) No change, because stop codons are ignored
  • (D) A protein with one amino acid substituted
Show Answer

Answer: A – A nonsense mutation terminates translation early, producing a shortened polypeptide.

Natural Selection

Q47. In a population of beetles, green beetles are eaten more often by birds after the foliage darkens, while brown beetles survive better. This is best described as

  • (A) stabilising selection favouring the average phenotype
  • (B) disruptive selection favouring both extremes
  • (C) genetic drift with no effect on fitness
  • (D) directional selection favouring the brown phenotype
Show Answer

Answer: D – The environment shifts the favoured phenotype in one direction.

Q48. A population is in Hardy-Weinberg equilibrium and 9 percent of individuals show a recessive phenotype. What is the frequency of heterozygotes?

  • (A) 0.09
  • (B) 0.30
  • (C) 0.49
  • (D) 0.42
Show Answer

Answer: D – q2 = 0.09, so q = 0.3, p = 0.7, and 2pq = 0.42.

Q49. Which condition would most directly violate Hardy-Weinberg equilibrium by changing allele frequencies through chance alone?

  • (A) Random mating in a very large population
  • (B) No mutation occurring over many generations
  • (C) A small population in which a flood kills most individuals at random
  • (D) No migration into or out of the population
Show Answer

Answer: C – A bottleneck in a small population causes genetic drift, a random change in allele frequencies.

Q50. Two populations of salamanders are separated by a river and gradually accumulate different mating calls. If they later meet and cannot interbreed, what has occurred?

  • (A) Sympatric speciation with no barrier
  • (B) Artificial selection by humans
  • (C) Hybridisation producing a single species
  • (D) Allopatric speciation by geographic isolation
Show Answer

Answer: D – Geographic separation followed by reproductive isolation is allopatric speciation.

Q51. A phylogenetic tree shows that species A and B share a more recent common ancestor than either shares with species C. Which statement is correct?

  • (A) A and C are sister taxa
  • (B) B and C diverged most recently
  • (C) A and B are more closely related to each other than to C
  • (D) All three species are equally related
Show Answer

Answer: C – Sharing a more recent node indicates a closer evolutionary relationship.

Q52. Fossils of marine organisms are found in rock layers on a mountaintop. What does this best support?

  • (A) Marine organisms evolved to live on mountains
  • (B) The fossils were placed there by recent floods only
  • (C) Rock layers form in a single rapid event
  • (D) The area was once under water and the rock layers were uplifted
Show Answer

Answer: D – Marine fossils in uplifted strata record past environments and geological change.

Q53. A gene sequence in humans and chimpanzees differs by 1 percent, while the same gene in humans and mice differs by 12 percent. What is the best inference?

  • (A) The gene evolves at the same rate in all lineages
  • (B) Humans share a more recent common ancestor with chimpanzees than with mice
  • (C) Mice and chimpanzees are more closely related
  • (D) The gene is nonfunctional in humans
Show Answer

Answer: B – Smaller sequence divergence generally reflects a more recent common ancestor.

Ecology

Q54. A plant opens its stomata at night and fixes CO2 into organic acids, then closes them during the day. This adaptation is most associated with

  • (A) C4 photosynthesis in cool, wet habitats
  • (B) CAM photosynthesis in arid environments, reducing water loss
  • (C) Photorespiration in shaded forests
  • (D) Fermentation in waterlogged soils
Show Answer

Answer: B – CAM plants fix carbon at night when stomata can open with less water loss.

Q55. In a food web, removing a top predator causes herbivore numbers to rise and producer biomass to fall. This is best described as

  • (A) competitive exclusion
  • (B) primary succession
  • (C) a trophic cascade
  • (D) density-independent regulation
Show Answer

Answer: C – Effects propagate down trophic levels when a top predator is removed.

Q56. A population grows logistically and approaches its carrying capacity. What happens to the per-capita growth rate as population size nears K?

  • (A) It approaches zero
  • (B) It increases exponentially
  • (C) It becomes negative and the population goes extinct
  • (D) It stays constant at the intrinsic rate
Show Answer

Answer: A – In logistic growth, growth slows as resources become limiting and stops at K.

Q57. Two species of warblers feed in different parts of the same tree. This is best explained by

  • (A) mutualism, in which both species benefit directly
  • (B) resource partitioning, which reduces competition
  • (C) predation, in which one species eats the other
  • (D) commensalism with no effect on either species
Show Answer

Answer: B – Using different parts of a shared resource lowers niche overlap.

Q58. A mark-recapture study marks 40 fish, later catches 50 fish and finds 10 marked. What is the estimated population size?

  • (A) 200 fish
  • (B) 90 fish
  • (C) 100 fish
  • (D) 400 fish
Show Answer

Answer: A – N = (marked x total caught) / recaptured = (40 x 50) / 10 = 200.

Q59. An island has many endemic bird species found nowhere else. Which factor best explains this pattern?

  • (A) High immigration from the mainland
  • (B) Geographic isolation allowing independent evolution
  • (C) A uniform climate across all islands
  • (D) Frequent gene flow between islands
Show Answer

Answer: B – Isolation restricts gene flow, so island populations diverge into unique species.

Q60. Fertiliser runoff causes an algal bloom in a lake, followed by a sharp fall in dissolved oxygen. What is the most likely cause of the oxygen fall?

  • (A) Algae produce oxygen only at night
  • (B) Decomposition of dead algae by bacteria consumes oxygen
  • (C) Fertiliser directly removes oxygen from water
  • (D) Fish consume oxygen faster in warm water only
Show Answer

Answer: B – Bacterial decomposition of the bloom creates a high biological oxygen demand.

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