WAEC Biology 2013 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 2013 WAEC Biology paper, with answers and explanations. We hold 61 questions from this paper, and the Skolra app has all of them.

WAEC Biology 2013 questions and answers

1. 1.The producers in a food chain in an aquatic environment are
   1. A
      decomposers that break down dead organic material
   2. B
      phytoplankton and aquatic green plants
      Correct
   3. C
      fish that occupy the first trophic level
   4. D
      zooplankton that feed on dissolved organic matter
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   Answer: B

   Producers are autotrophs — organisms that manufacture their own food through photosynthesis. In aquatic ecosystems, the producers are phytoplankton (microscopic photosynthetic organisms such as diatoms and green algae) and larger aquatic plants (macrophytes). They form the base of the food chain, converting solar energy into organic matter. Zooplankton are primary consumers, decomposers recycle dead matter, and fish are generally secondary or tertiary consumers.

2. 2.Which of the following associations is an example of epiphytism?
   1. A
      Lichen formed by a fungus and a green alga living together
   2. B
      Oxpecker birds removing ticks from the backs of cattle
   3. C
      Orchid plants growing on the branches of a mango tree for support
      Correct
   4. D
      Tapeworm living and feeding inside the intestine of a pig
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   Answer: C

   An epiphyte is a plant that grows on another plant, using it only for physical support and not deriving nutrients from it. The host plant is neither harmed nor benefited — making it a form of commensalism. Orchids growing on mango tree branches are a classic example. Tapeworms represent parasitism; lichens represent mutualism; oxpeckers and cattle represent commensalism (or mutualism), but not epiphytism, which specifically refers to plants growing on plants.

3. 3.The maintenance of a stable internal environment within the body of a mammal is known as
   1. A
      excretion
   2. B
      thermoregulation
   3. C
      homeostasis
      Correct
   4. D
      osmoregulation
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   Answer: C

   Homeostasis is the maintenance of a stable internal environment in the body, including constancy of temperature, blood sugar, water balance and pH. Thermoregulation and osmoregulation are specific aspects of homeostasis, while excretion is the removal of metabolic waste products.

4. 4.

   (a)(i) What is ecological succession?
   (ii) Describe the process of ecological succession on a bare plot of land.

   (b) Explain the following terms:
   (i) biosphere
   (ii)population
   (iii)community
   (iv)ecosystem.

   (c) Name one instrument each used for measuring:
   (i) relative humidity
   (ii) wind speed
   

   1. A
      The movement of animal populations from one habitat to another in search of food
   2. B
      The gradual and orderly sequential change in the structure of a community in an area over time
      Correct
   3. C
      The sudden extinction of species caused by catastrophic environmental events
   4. D
      The rapid invasion and spread of exotic species into a new environment
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   Answer: B

   Ecological succession is the process by which a community of organisms in a given area changes in a predictable, sequential manner over time, beginning with pioneer species and progressing through a series of stages until a stable climax community is reached. On a bare plot of land, mosses and lichens arrive first, followed by grasses, then shrubs, and finally trees. This is primary succession. The process is directional and relatively orderly, not random or rapid.

5. 5.Spirogyra is regarded as a multicellular plant because
   1. A
      it has a nucleus and cytoplasm like all true plants
   2. B
      its body consists of many cells joined end to end in a filament
      Correct
   3. C
      it photosynthesises using chlorophyll in its chloroplasts
   4. D
      it reproduces asexually by fragmentation of its filaments
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   Answer: B

   An organism is described as multicellular when its body is composed of more than one cell. In Spirogyra, individual cells are joined end to end to form long, unbranched filaments, making it a multicellular alga. Possessing a nucleus, photosynthesising and reproducing by fragmentation are features it shares with both unicellular and multicellular organisms and therefore cannot alone justify the multicellular classification.

6. 6.If a boy has blood type O and his mother has blood type A,what is the genotype of his father?
   1. A
      Homozygous for blood type B (two copies of the dominant B allele)
   2. B
      Heterozygous for blood type A (one dominant A allele and one recessive allele)
   3. C
      Homozygous dominant for blood type A (two copies of the dominant A allele)
   4. D
      Type O (carrying two recessive alleles with no dominant A or B)
      Correct
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   Answer: D

   A boy with blood type O has genotype ii (two recessive alleles). He received one i allele from his mother and one from his father. His mother has blood type A, so she must be genotype I^Ai (heterozygous) since she contributed an i allele to the boy. For the father to have contributed an i allele without having the A or B phenotype dominate, the simplest conclusion is that his father is genotype ii — blood type O — carrying only recessive alleles. This is the most directly determinable answer.

7. 7.In humans, excretion of hypertonic urine is associated with the
   1. A
      loop of Henle creating a high solute concentration gradient in the medulla
      Correct
   2. B
      proximal convoluted tubule secreting excess solutes
   3. C
      Bowman's capsule increasing the filtration rate
   4. D
      collecting duct preventing any reabsorption of water
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   Answer: A

   Hypertonic urine (urine more concentrated than blood plasma) is produced when the loop of Henle establishes a high osmotic gradient in the renal medulla by the countercurrent multiplier mechanism. This gradient allows water to be drawn out of the collecting duct by osmosis — an effect enhanced by antidiuretic hormone (ADH). The result is concentrated, hypertonic urine. The loop of Henle is therefore the structure most directly responsible for the kidney's ability to concentrate urine.

8. 8.Which of the following substances is not complex carbohydrate?
   1. A
      Cellulose
   2. B
      Glucose
      Correct
   3. C
      Glycogen
   4. D
      Starch
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   Answer: B

   Complex carbohydrates are polysaccharides — large molecules made of many monosaccharide units joined by glycosidic bonds. Starch (plant storage), cellulose (plant cell wall) and glycogen (animal storage in the liver and muscles) are all polysaccharides. Glucose is a monosaccharide (a simple sugar) with the formula C₆H₁₂O₆ and is the building block from which all three polysaccharides are made. It is therefore not a complex carbohydrate.

9. 9.The gene that is only located on X-chromosome is said to be
   1. A
      sex-linked (X-linked)
      Correct
   2. B
      autosomal dominant
   3. C
      codominant
   4. D
      pleiotopic
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   Answer: A

   Genes located exclusively on the X-chromosome (with no corresponding locus on the Y-chromosome) are described as sex-linked or more specifically X-linked. Because males have only one X chromosome (XY), they express X-linked recessive traits whenever the allele is present, with no second allele to mask it. Examples include haemophilia and colour blindness. Autosomal genes are on non-sex chromosomes, codominance involves two alleles equally expressed, and pleiotrophy is when one gene affects multiple traits.

10. 10.A man's who suffers from haemophilia marries a woman who is a carrier, what percentage of their children are likely to be haemophilie?
    1. A
       50%
       Correct
    2. B
       25%
    3. C
       75%
    4. D
       100%
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    Answer: A

    Haemophilia is an X-linked recessive disorder. The affected father's genotype is X^hY and the carrier mother's genotype is X^HX^h. A cross gives offspring: X^HX^h (carrier female), X^hX^h (affected female), X^HY (normal male) and X^hY (affected male). Two out of four offspring (one affected female and one affected male) will have haemophilia, giving a probability of 50%. The remaining 50% will be either carriers or normal.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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