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JAMB Biology 2011 Objective Past Questions

All 60 questions from the Joint Admissions and Matriculation Board (JAMB) Biology 2011 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Biology 2011 Objective — Question 1

The branch of Biology that deals with the principles of classification of organisms is known as

  • A. A. biological index
  • B. B. nomenclature
  • C. C. taxonomyCorrect
  • D. D. ecology

Explanation

The branch of biology that deals with the principles of classification of organisms is known as taxonomy.

Biology 2011 Objective — Question 2

Which of the following structures is a tissue?

  • A. A. Vessel element
  • B. B. BloodCorrect
  • C. C. Sieve tube element
  • D. D. Erythrocyte

Explanation

Blood contains many cells such as red blood cells (erythrocytes), white blood cells (leucocytes) and blood platelets (thrombocytes), so it is classed as a tissue.

Biology 2011 Objective — Question 3

Which of the following cells are not regarded as specialized cells?

  • A. A. Sperm cells
  • B. B. Root tip cells
  • C. C. Muscle cells
  • D. D. Somatic cellsCorrect

Explanation

Sperm cells, root tip cells and muscle cells are all specialized cells. The term 'somatic cell' is too ambiguous to be considered specialized.

Biology 2011 Objective — Question 4

Which of the following pairs of cells carry out the same function?

  • A. A. Check cell and red blood cell
  • B. B. Spermatozoon and ovumCorrect
  • C. C. Palisade cell and epidermal cell
  • D. D. Root tip cell and guard cell

Explanation

Spermatozoon and ovum perform the same function i.e. they are both used for reproduction.

Biology 2011 Objective — Question 5

If Amoeba is placed in a salt solution, the contractile vacuoles would

  • A. A. be bursting more frequently
  • B. B. be more numerous
  • C. C. be formed less frequentlyCorrect
  • D. D. grow bigger before they burst

Explanation

The contractile vacuole removes excess water from the protozoan's body. In salt solution the organism loses water instead of gaining it, so the vacuole forms less frequently.

Biology 2011 Objective — Question 6

In which of the following habitats is Paramecium not found?

  • A. A. Pond
  • B. B. AquariumCorrect
  • C. C. Lake
  • D. D. Puddle

Explanation

Paramecium is found in pond, lake, puddle etc., not typically in an aquarium.

Biology 2011 Objective — Question 7

The following processes are involved in water movement in the endodermis except

  • A. A. osmosis
  • B. B. vacuolar pathwayCorrect
  • C. C. diffusion
  • D. D. active transport

Explanation

Osmosis, diffusion and active transport all play a part in water movement in the endodermis; 'vacuolar pathway' is not a recognised term here.

Biology 2011 Objective — Question 8

Cells that utilize a lot of energy are characterized by the presence of a large number of

  • A. A. vacuoles
  • B. B. mitochondriaCorrect
  • C. C. endoplasmic reticulum
  • D. D. ribosomes

Explanation

Mitochondria are the powerhouse of the cell, so cells needing lots of energy have many mitochondria.

Biology 2011 Objective — Question 9

The part labeled II in the diagram is the

Diagram for question 9
  • A. A. centriole
  • B. B. chloroplast
  • C. C. chromatidCorrect
  • D. D. tonoplast

Explanation

The part labeled 'II' is the chromatid; a chromosome comprises two chromatids.

Biology 2011 Objective — Question 11

Secondary growth is brought about by the activities of the

  • A. A. phellogen and phelloderm
  • B. B. phellogen and procambium
  • C. C. vascular cambium and phelloderm
  • D. D. vascular cambium and phellogenCorrect

Explanation

In gymnosperms and most dicots, secondary growth (increase in diameter) results from activity of the vascular cambium and cork cambium (phellogen).

Biology 2011 Objective — Question 12

A monocot root is different from a dicot root by having

  • A. A. endodermis
  • B. B. cambium
  • C. C. wide pithCorrect
  • D. D. root hair

Explanation

A monocot root has a wide pith while a dicot root does not.

Biology 2011 Objective — Question 13

Which of the following statements best describes haemoglobin? It is

  • A. A. yellowish in colour
  • B. B. a red blood cell
  • C. C. an oxygen carrying pigmentCorrect
  • D. D. needed for blood clotting

Explanation

Haemoglobin is the pigment in red blood cells responsible for transporting oxygen.

Biology 2011 Objective — Question 14

Inhaled air is made warm and moist in the

  • A. A. epiglottis
  • B. B. nasal cavityCorrect
  • C. C. trachea
  • D. D. mouth

Explanation

Inhaled air is warmed and moistened as it passes through the paranasal sinuses surrounding the nasal cavity.

Biology 2011 Objective — Question 15

Which of the following structures is not involved in respiration?

  • A. A. Lung books
  • B. B. Mouth
  • C. C. StomachCorrect
  • D. D. Trachea

Explanation

The stomach is not adapted for respiration; lung books (arachnids), mouth (amphibian buccal respiration) and trachea (insects) are respiratory structures.

Biology 2011 Objective — Question 16

Filtrate in the Bowman's capsule contains vitamins because

  • A. A. only a little amount is required by the body
  • B. B. they can be reabsorbed into the blood
  • C. C. they have low molecular mass
  • D. D. most of them are fat solubleCorrect

Explanation

Vitamins appear in the filtrate mainly because most of them are fat soluble.

Biology 2011 Objective — Question 17

More sweat is produced during muscular exercise because

  • A. A. the contracting muscles produce water
  • B. B. fermentation occurs in muscles
  • C. C. the temperature of the body risesCorrect
  • D. D. the muscle fatigues

Explanation

Sweating is an adaptation to lower body temperature, which rises during exercise; sweat cools the body as it evaporates.

Biology 2011 Objective — Question 18

Which of the following neurons may not have myelin sheath?

  • A. A. Effector cells
  • B. B. Intermediate neuronsCorrect
  • C. C. Motor neurons
  • D. D. Sensory neurons

Explanation

The myelin sheath insulates axons and speeds impulse conduction. Intermediate (association) neurons are typically unmyelinated.

Biology 2011 Objective — Question 19

Which of the following is an effector organ?

  • A. A. Tongue
  • B. B. SkinCorrect
  • C. C. Nose bridge
  • D. D. Ear glands

Explanation

Skin is regarded as an effector organ since it can react to impulses from the brain.

Biology 2011 Objective — Question 20

During sexual reproduction in Paramecium, how many times does the zygote divide to produce eight nuclei?

  • A. A. 1
  • B. B. 2
  • C. C. 3Correct
  • D. D. 4

Explanation

Reproduction in Paramecium is by conjugation; the zygote in each conjugant divides three times to form eight nuclei.

Biology 2011 Objective — Question 21

The testes in male mammals descend into the scrotal sac because

  • A. A. there is congestion in the lower abdomen
  • B. B. they run the risk of being destroyed
  • C. C. they need special support
  • D. D. they require a relatively low temperatureCorrect

Explanation

The testes must be at a temperature slightly lower than body temperature for optimum sperm production.

Biology 2011 Objective — Question 22

Some animals return to water bodies to reproduce because

  • A. A. natural enemies destroy their eggs on land
  • B. B. water nourishes their embryos and young ones
  • C. C. they are close to their ancestors and imitate them
  • D. D. the temperature on land is not suitable for the development of their embryosCorrect

Explanation

The land temperature may not be suitable for the development of their embryos, so they return to water to reproduce.

Biology 2011 Objective — Question 23

Which of the following methods is appropriate for the cultivation of cassava?

  • A. A. Budding
  • B. B. Fragmentation
  • C. C. Root cutting
  • D. D. Stem cuttingCorrect

Explanation

Cassava is propagated by stem cutting.

Biology 2011 Objective — Question 24

What will be the chromosome number of the hybrid of two varieties of a plant with 36 chromosomes in the endosperm cell?

  • A. A. 12
  • B. B. 24Correct
  • C. C. 36
  • D. D. 48

Explanation

Endosperm cells are triploid (3n). If 3n = 36, then 2n = 36/3 x 2 = 24.

Biology 2011 Objective — Question 25

Kreb's cycle occurs in the

  • A. A. mitochondriaCorrect
  • B. B. cytoplasm
  • C. C. nucleus
  • D. D. ribosomes

Explanation

Glycolysis occurs in the cytoplasm, while the Krebs (citric acid) cycle occurs in the mitochondria.

Biology 2011 Objective — Question 26

The process whereby electrons are emitted from chlorophyll and returned to it unchanged is known as

  • A. A. non-cyclic phosphorylation
  • B. B. photochemical reaction
  • C. C. phosphorylation
  • D. D. cyclic photophosphorylationCorrect

Explanation

In cyclic photophosphorylation, electrons emitted from P700 cycle through the electron transport chain and return to the same photosystem; NADPH and oxygen are not produced.

Biology 2011 Objective — Question 27

Which of the following statements about photosynthesis is not true?

  • A. A. Plants can photosynthesize without an increase in dry weightCorrect
  • B. B. Carbon dioxide is absorbed in bright light
  • C. C. Oxygen is produced during photosynthesis from the breakdown of water
  • D. D. Photosynthesis occurs in green plants

Explanation

Photosynthesis always results in an increase in dry weight of the plant, so statement A is false.

Biology 2011 Objective — Question 28

The equation below represents the process of glycolysis: Glucose+ATP → X+ADP. X in the equation represents

  • A. A. glucose diphosphate
  • B. B. glucose phosphateCorrect
  • C. C. glucose triphosphate
  • D. D. fructose phosphate

Explanation

Glucose is phosphorylated by addition of a phosphate group to form glucose phosphate.

Biology 2011 Objective — Question 29

The removal of all phloem tissues of the stem of a plant close to the root system for a long period of time is likely to

  • A. A. provide more energy to the roots
  • B. B. accumulate more starch in the roots
  • C. C. cause the underground roots to develop buds
  • D. D. cause the plant to wither and dieCorrect

Explanation

Removing all phloem tissue for a long time will kill the plant because conduction of food to the roots is stopped (ring-barking).

Biology 2011 Objective — Question 30

The conversion of glucose to starch in the leaf during the day principally

  • A. A. enables photosynthesis
  • B. B. prevents osmotic problems
  • C. C. enables the leaf to store the starch to be used upCorrect
  • D. D. enables glucose to be used up

Explanation

Plants store carbohydrate as starch (animals and fungi store it as glycogen); converting glucose to starch allows storage for later use.

Biology 2011 Objective — Question 31

Which of the following groups of substances are not six-carbon compounds?

  • A. A. Glucose and Lactose
  • B. B. Lactose and Cellulose
  • C. C. Glucose, Lactose and PyruvateCorrect
  • D. D. Pyruvate and Cellulose

Explanation

Pyruvate is not a carbohydrate; it is formed when a six-carbon glucose molecule is broken down into three-carbon molecules.

Biology 2011 Objective — Question 32

A purple colour was obtained when sodium hydroxide solution and a drop of copper sulphate solution was added to a food substance. The food substance is likely to be a

  • A. A. carbohydrate
  • B. B. fat
  • C. C. proteinCorrect
  • D. D. sugar

Explanation

This is the Biuret test, which gives a purple/violet colour in the presence of protein.

Biology 2011 Objective — Question 33

Which of the following enzymes is active in the duodenum?

  • A. A. Pepsin
  • B. B. Renin
  • C. C. TrypsinCorrect
  • D. D. Amylase

Explanation

Trypsin is produced in the pancreas but transported (with other enzymes in pancreatic juice) to the duodenum where it converts peptone to polypeptides.

Biology 2011 Objective — Question 34

The following organisms are producers except

  • A. A. Hibiscus
  • B. B. MushroomCorrect
  • C. C. Cactus
  • D. D. Spirogyra

Explanation

Hibiscus, cactus and spirogyra all contain chlorophyll and are producers; mushrooms are achlorophyllous (fungi) and are not producers.

Biology 2011 Objective — Question 35

Which of the following structures produces the greatest variety of digestive enzymes?

  • A. A. Salivary glands
  • B. B. PancreasCorrect
  • C. C. Stomach
  • D. D. Colon

Explanation

The pancreas produces the greatest variety of digestive enzymes: amylase (amylopsin), lipase, trypsin, chymotrypsin, carboxypeptidase, ribonuclease and deoxyribonuclease.

Biology 2011 Objective — Question 36

The following are abiotic components of an ecosystem except

  • A. A. oxygen
  • B. B. bacteriaCorrect
  • C. C. water
  • D. D. soil

Explanation

Abiotic components are non-living; bacteria are living organisms and so are biotic, not abiotic.

Biology 2011 Objective — Question 37

[Capture-recapture data: First capture = 200, second capture = 120; Number of organisms with mark in second capture = 40] The total number of organisms therefore is

  • A. A. 200
  • B. B. 360
  • C. C. 600Correct
  • D. D. 800

Explanation

N = (first capture x second capture) / marked recaptures = (200 x 120) / 40 = 600.

Biology 2011 Objective — Question 38

Ecological investigation in a habitat includes the following procedures except

  • A. A. Choosing a habitat
  • B. B. identification of specimens
  • C. C. determining the genetic make-up of specimensCorrect
  • D. D. measuring abiotic and biotic factors

Explanation

Investigating the genetic make-up of specimens is not part of ecological investigation, which studies interactions of organisms with each other and their environment.

Biology 2011 Objective — Question 39

Which of the following substances when lost from the body of a mammal will not be returned to the ecosystem?

  • A. A. Sweat
  • B. B. Urea
  • C. C. Heat energyCorrect
  • D. D. Carbon dioxide

Explanation

Sweat, urea and carbon dioxide are returned to the ecosystem; heat energy dissipates and is not returned/recycled in the ecosystem.

Biology 2011 Objective — Question 40

The rate of decomposition of organisms is faster in the tropical rain forest than in other biomes because

  • A. A. of the relatively constant daylight
  • B. B. there are more plants per square feet
  • C. C. of abundance of waterCorrect
  • D. D. of constant cool temperature

Explanation

Tropical rainforests receive abundant rainfall (2000-4000mm annually) and have high temperatures/humidity, which speed up decomposition.

Biology 2011 Objective — Question 41

A Xerophyte conserves water by possession of the following features except

  • A. A. thick cuticle
  • B. B. sunken stomata
  • C. C. broad leavesCorrect
  • D. D. fleshy stem

Explanation

Xerophytes have thick cuticle, sunken stomata and fleshy stems to conserve water; broad leaves increase water loss through transpiration, so this is the exception.

Biology 2011 Objective — Question 42

The falling off of leaves of deciduous trees is helpful to the plant because it

  • A. A. reduces the rate of transpirationCorrect
  • B. B. enables the plant to conveniently eliminate its excretory products
  • C. C. enables the plant to bear more fruits
  • D. D. ensures that the limited mineral salts get to only growing regions

Explanation

Shedding leaves reduces the surface area for transpiration, helping the plant conserve water.

Biology 2011 Objective — Question 43

A sample of wet garden soil of known weight was heated to constant weight. The loss of weight is due to loss of

  • A. A. water
  • B. B. organic matter
  • C. C. water and organic matterCorrect
  • D. D. water and inorganic matter

Explanation

Continual heating causes loss of water first; if heating persists to a constant weight, the organic matter (humus) is also burnt off.

Biology 2011 Objective — Question 44

Assuming that a period of stable population size is followed by a period when natality increases by ten (10) percent and immigration by twenty (20) percent of population size, the population size will

  • A. A. be at equilibrium
  • B. B. increase by forty (40) percent
  • C. C. begin to increaseCorrect
  • D. D. begin to decrease

Explanation

An increase in natality (birth rate) and immigration rate causes the population to begin to increase.

Biology 2011 Objective — Question 45

Which of the following methods does not make water fit for drinking?

  • A. A. Addition of chlorine
  • B. B. Boiling
  • C. C. Distillation
  • D. D. Addition of alumCorrect

Explanation

Alum coagulates solid impurities in water (the water must still be boiled to kill bacteria); it does not by itself make water fit for drinking.

Biology 2011 Objective — Question 46

Vaccination results in

  • A. A. aiding red blood cells to carry more oxygen
  • B. B. production of antibodies which destroy toxins of germsCorrect
  • C. C. arresting excessive bleeding
  • D. D. production of white blood cells which engulf and digest bacteria

Explanation

Vaccination works by stimulating the body to produce antibodies against the antigen introduced.

Biology 2011 Objective — Question 47

Leguminous crops are incorporated into crop rotation in order to

  • A. A. improve aeration of soil
  • B. B. promote nitrogen fixationCorrect
  • C. C. increase the rate of soil formation
  • D. D. improve upon the water holding capacity of the soil

Explanation

Leguminous crops host nitrogen-fixing bacteria (e.g. Rhizobium leguminosarium) in their root nodules, enriching the soil with nitrogen.

Biology 2011 Objective — Question 48

Which of the following practices improves crop yield in a clayey soil? Addition of

  • A. A. more water and humus
  • B. B. lime and humusCorrect
  • C. C. fertilizers
  • D. D. weedicides and fertilizers

Explanation

Adding lime and humus improves crop yield in clayey soil.

Biology 2011 Objective — Question 49

Which of the following activities promotes forest conservation?

  • A. A. Lumbering
  • B. B. Use of firewood for cooking
  • C. C. AfforestationCorrect
  • D. D. Production of paper

Explanation

Afforestation (planting trees) promotes forest conservation, while the other options tend to encourage deforestation.

Biology 2011 Objective — Question 50

Which of the following statements is not true about continuous variation? It

  • A. A. is usually controlled by several genes
  • B. B. can be influenced by environmental factors
  • C. C. follows a normal distribution curve
  • D. D. is usually controlled by one or two pair(s) of genesCorrect

Explanation

Continuous variation is usually controlled by many genes (polygenic), not just one or two gene pairs.

Biology 2011 Objective — Question 51

Acquired characters are

  • A. A. received from parents
  • B. B. passed to offspring
  • C. C. caused by the environmentCorrect
  • D. D. caused by mutation

Explanation

Acquired characters are caused by the environment; they cannot be passed from one generation to the next.

Biology 2011 Objective — Question 52

Differences in the characteristics observed between individuals of the same species is known as

  • A. A. trait
  • B. B. phenotype
  • C. C. mutation
  • D. D. variationCorrect

Explanation

Variation refers to the difference in characteristics observed between individuals of the same species.

Biology 2011 Objective — Question 53

Which of the following statements best describes protein synthesis?

  • A. A. DNA is directly involved in translation
  • B. B. Translocation of the ribosome exposes a new codon for base pairing with an amino acidCorrect
  • C. C. Each tRNA with a particular anticodon always carries a different protein in the process
  • D. D. One amino acid is always carried by more than one tRNA

Explanation

During translation, the ribosome moves (translocates) along the mRNA, exposing a new codon each time for base pairing.

Biology 2011 Objective — Question 54

If a person has two alleles of the sickle cell anaemia gene, the person is

  • A. A. a heterozygous carrier of the disease
  • B. B. immune to the disease and cannot pass it on to an offspring
  • C. C. a sufferer of the diseaseCorrect
  • D. D. probably of Asian ancestry

Explanation

A person with two alleles of the sickle cell gene (homozygous) is a sufferer of the disease; a person with just one allele is a carrier.

Biology 2011 Objective — Question 55

What is the probability of a man of blood group AB married to a woman of blood group O producing a child of blood group O?

  • A. A. 0%Correct
  • B. B. 25%
  • C. C. 50%
  • D. D. 75%

Explanation

It is not possible for a man of blood group AB married to a woman of blood group O to produce a child of blood group O; the child will be either A or B.

Biology 2011 Objective — Question 56

A man with blood group IA IA is married to a woman with blood group IA IO. The blood group of their son is likely to be

  • A. A. ACorrect
  • B. B. AB
  • C. C. O
  • D. D. B

Explanation

Since allele A is dominant over O, the child of this cross can only be blood group A.

Biology 2011 Objective — Question 57

Natural selection is a consequence of

  • A. A. distribution of organisms
  • B. B. adverse conditionsCorrect
  • C. C. variation in organisms
  • D. D. inbreeding

Explanation

Natural selection is a consequence of adverse conditions, which cause organisms to strive and compete for survival.

Biology 2011 Objective — Question 58

Which of the following components of Lamarck's theory of evolution is considered faulty?

  • A. A. Individuals of the same species growing under different environmental conditions differ from each other
  • B. B. Use of certain organs results in development of those parts
  • C. C. Unused organs degenerate
  • D. D. Changes that result in individuals of the same species are transmitted to offspringCorrect

Explanation

The loophole in Lamarck's theory of use and disuse is the claim that acquired traits are transmittable to offspring, which is false.

Biology 2011 Objective — Question 59

A vestigial structure in humans is

  • A. A. earlobe
  • B. B. toe bone
  • C. C. tail boneCorrect
  • D. D. spleen

Explanation

Vestigial organs have no popular function in humans; the tail bone (caudal vertebrae) is a classic example, alongside the appendix.

Biology 2011 Objective — Question 60

Adenine pairs with thymine because

  • A. A. the two occur in the same nucleic acid
  • B. B. one is a strong base and the other a weak base
  • C. C. two purine bases easily pair up
  • D. D. one is a purine and the other a pyrimidineCorrect

Explanation

Guanine and adenine are purines; cytosine and thymine are pyrimidines. A purine always pairs with a pyrimidine (adenine-thymine, guanine-cytosine).

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