IB Biology Quiz: Apply Carbohydrates And Lipids
19 questions · exam conditions
0:00
Apply Carbohydrates And LipidsQuestion 1 of 19

Cellulose and chitin are both structural polysaccharides. Cellulose is found in plant cell walls, and chitin is found in fungal cell walls and the exoskeletons of arthropods. Both are polymers of glucose or a glucose derivative. What key structural feature do they share that contributes to their strength?

A highly branched structure that allows for extensive cross-linking and insolubility.
They are both composed of α-glucose monomers, which allows them to form compact helical structures.
Their monomers are joined by β-1,4 glycosidic bonds, which results in straight, unbranched chains that can align in parallel.
They both contain peptide cross-links between chains, forming a rigid peptidoglycan-like matrix.
← Back to quizzes

IB Biology Quiz

IB Biology Quiz: Apply Carbohydrates And Lipids

Practice Apply Carbohydrates And Lipids in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Carbohydrates And Lipids, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Cellulose and chitin are both structural polysaccharides. Cellulose is found in plant cell walls, and chitin is found in fungal cell walls and the exoskeletons of arthropods. Both are polymers of glucose or a glucose derivative. What key structural feature do they share that contributes to their strength?

  1. A highly branched structure that allows for extensive cross-linking and insolubility.
  2. They are both composed of α-glucose monomers, which allows them to form compact helical structures.
  3. Their monomers are joined by β-1,4 glycosidic bonds, which results in straight, unbranched chains that can align in parallel. (correct answer)
  4. They both contain peptide cross-links between chains, forming a rigid peptidoglycan-like matrix.
Explanation: Both cellulose (a polymer of β-glucose) and chitin (a polymer of N-acetylglucosamine, a β-glucose derivative) are linked by β-1,4 glycosidic bonds. This type of linkage causes each successive monomer to be flipped 180 degrees relative to the previous one, resulting in long, straight, unbranched chains. These chains can lie parallel to each other and form extensive hydrogen bonds between them, creating strong microfibrils. Distractor A is incorrect; they are unbranched. Distractor B is incorrect; they are polymers of β-glucose or its derivative, which forms straight chains, not helices (which are characteristic of α-glucose polymers like starch). Distractor D is incorrect; peptide cross-links are characteristic of bacterial peptidoglycan, not cellulose or chitin.

Question 2

A food scientist hydrolyzes three different unknown disaccharides. Hydrolysis of Sample 1 yields only glucose. Hydrolysis of Sample 2 yields glucose and fructose. Hydrolysis of Sample 3 yields glucose and galactose. What are the identities of the three original disaccharides?

  1. 1: Sucrose, 2: Maltose, 3: Lactose
  2. 1: Maltose, 2: Lactose, 3: Sucrose
  3. 1: Lactose, 2: Sucrose, 3: Maltose
  4. 1: Maltose, 2: Sucrose, 3: Lactose (correct answer)
Explanation: This question requires knowledge of the composition of common disaccharides. Maltose is composed of two glucose units. Sucrose (table sugar) is composed of one glucose and one fructose unit. Lactose (milk sugar) is composed of one glucose and one galactose unit. Therefore, Sample 1 (glucose + glucose) is Maltose, Sample 2 (glucose + fructose) is Sucrose, and Sample 3 (glucose + galactose) is Lactose.

Question 3

A scientist is studying a newly discovered polysaccharide from an exotic plant. It is found to be a polymer of β-glucose linked by 1,4-glycosidic bonds, forming straight, unbranched chains that cross-link with adjacent chains. Which function is this polysaccharide most likely to have?

  1. A short-term energy storage molecule in the plant's seeds, easily hydrolysed for germination.
  2. A structural component of the plant's cell wall, providing high tensile strength. (correct answer)
  3. The primary solute for regulating water potential within the phloem sap.
  4. A long-term energy reserve in the plant's roots, with a highly branched structure for rapid mobilization.
Explanation: The description of β-glucose, 1,4-glycosidic bonds, and straight, cross-linked chains perfectly matches the structure of cellulose. This structure provides high tensile strength, making it the ideal structural component for plant cell walls. Distractor A describes the function of starch, which is a polymer of α-glucose. Distractor C is incorrect; polysaccharides are generally insoluble and not used to regulate water potential in sap (sucrose, a disaccharide, is used). Distractor D describes the function of starch in roots, but incorrectly states it is highly branched (glycogen is highly branched, amylopectin in starch is moderately branched, and the molecule described in the stem is unbranched).

Question 4

Glycogen and amylopectin are both branched polymers of α-glucose used for energy storage. However, glycogen has a more highly branched structure than amylopectin. Which statement correctly deduces a functional consequence of this structural difference?

  1. The higher branching in glycogen makes it less soluble in water, facilitating more compact storage in animal cells.
  2. The greater number of branches in glycogen provides more terminal ends for enzymes to act on, allowing for more rapid glucose mobilization. (correct answer)
  3. The reduced branching in amylopectin allows plant cells to pack the molecules into denser, crystalline granules in amyloplasts.
  4. The branching in glycogen involves β-1,6 glycosidic bonds, which are stronger and store more energy than the α-1,6 bonds in amylopectin.
Explanation: Enzymes that hydrolyze these polymers, such as phosphorylase, act on the non-reducing terminal ends of the chains. A more highly branched structure, as seen in glycogen, means there are many more terminal ends available simultaneously. This allows for a much faster rate of glucose release, which is critical for animals with higher metabolic rates and the need for rapid energy mobilization (e.g., for muscle contraction). Distractor A is incorrect; branching tends to increase solubility. Distractor C is plausible but the primary advantage of glycogen's branching is rapid mobilization, not packing. Distractor D is incorrect; both molecules use α-1,6 glycosidic bonds for branching.

Question 5

Cholesterol is often discussed in the context of diet and health, but it is also an essential molecule. Which of the following describes a primary function of cholesterol that is distinct from the function of triglycerides?

  1. Acting as a precursor molecule for the synthesis of steroid hormones like testosterone. (correct answer)
  2. Serving as the primary long-term energy storage molecule in adipose tissue.
  3. Forming the hydrophilic head of phospholipid molecules in the plasma membrane.
  4. Providing thermal insulation and cushioning for vital organs.
Explanation: Cholesterol is a steroid, a type of lipid characterized by a four-ring structure. This structure serves as the backbone for synthesizing all other steroid hormones, including testosterone, estrogen, and cortisol. This role as a precursor is unique to steroids and not a function of triglycerides. Distractor A is the primary function of triglycerides, not cholesterol. Distractor C is incorrect; the hydrophilic head of a phospholipid is a phosphate group. Distractor D describes functions of triglyceride deposits (adipose tissue), not cholesterol itself.

Question 6

A scientist is designing a synthetic cell membrane for a liposome that must remain fluid at very low temperatures. Which combination of phospholipid fatty acid tails would be most effective for this purpose?

  1. Long and saturated, to maximize hydrophobic interactions and stabilize the membrane.
  2. Short and saturated, to decrease the melting point by reducing intermolecular forces.
  3. Long and polyunsaturated, to create significant kinks that disrupt close packing. (correct answer)
  4. A mixture of long saturated and short trans-unsaturated tails for optimal packing density.
Explanation: To maintain fluidity at low temperatures, the phospholipids must be prevented from packing tightly together and solidifying. Two features contribute to this: short chain length (fewer van der Waals forces) and unsaturation (kinks disrupt packing). Long, polyunsaturated tails provide the most significant disruption to packing due to multiple kinks, thereby lowering the freezing point of the membrane most effectively. Short and saturated (B) would also increase fluidity compared to long and saturated (A), but the effect of polyunsaturation is more pronounced. Trans-unsaturated tails (D) are relatively straight and would not increase fluidity as much as cis-unsaturated tails and are not the optimal choice.

Question 7

A research study investigates the effects of two different diets on blood lipid profiles. Diet 1 is rich in saturated fats from butter and red meat. Diet 2 is rich in cis-unsaturated fats from olive oil and nuts. Which outcome is most likely to be observed in the group consuming Diet 1 compared to the group consuming Diet 2?

  1. A decrease in LDL (low-density lipoprotein) cholesterol and a decrease in the risk of atherosclerosis.
  2. An increase in HDL (high-density lipoprotein) cholesterol and a decrease in the risk of atherosclerosis.
  3. An increase in LDL (low-density lipoprotein) cholesterol and an increased risk of atherosclerosis. (correct answer)
  4. A decrease in total blood triglycerides because saturated fats are more readily used for energy.
Explanation: Diets high in saturated fats are strongly correlated with an increase in levels of LDL cholesterol in the blood. LDL is often referred to as 'bad cholesterol' because high levels can lead to the formation of atherosclerotic plaques in arteries, increasing the risk of cardiovascular disease. Conversely, diets rich in cis-unsaturated fats are generally associated with lower LDL and higher HDL ('good cholesterol'). Therefore, Diet 1 is likely to lead to increased LDL and an increased risk of atherosclerosis compared to Diet 2. A and B describe outcomes associated with healthier diets. D is incorrect; saturated fats are not more readily used for energy and are in fact linked to higher triglyceride levels.

Question 8

A student suggests that since both lipids and carbohydrates are used for energy, an animal could survive on a diet completely lacking one of them, as long as the other is provided in sufficient caloric quantity. Which statement provides the strongest scientific counterclaim to this suggestion?

  1. Carbohydrates are required to form the sugar-phosphate backbone of DNA and RNA, and cannot be synthesized from lipids.
  2. Lipids are required for the absorption of all vitamins, so a lipid-free diet would lead to immediate deficiencies in vitamins like Vitamin C.
  3. The brain and central nervous system can only use glucose for energy and cannot metabolize fatty acids under any circumstances.
  4. Certain fatty acids, known as essential fatty acids, cannot be synthesized by the body and must be obtained from dietary lipids. (correct answer)
Explanation: While the body can interconvert many carbohydrates and lipids for energy, there are certain molecules that are 'essential,' meaning they cannot be synthesized and must be consumed in the diet. This includes essential amino acids and essential fatty acids (e.g., omega-3 and omega-6). A diet completely lacking lipids would lead to deficiencies in these fatty acids, which are crucial for forming cell membranes, brain development, and regulating inflammation. Distractor A is incorrect because the pentose sugars for nucleic acids can be synthesized from other precursors in the body. Distractor C is a common misconception; while the brain prefers glucose, it can adapt to use ketone bodies (derived from fatty acids) during periods of starvation or very low carbohydrate intake. Distractor D is partially correct in that lipids are needed for fat-soluble vitamins (A, D, E, K), but incorrect in that they are not needed for water-soluble vitamins like Vitamin C.

Question 9

A food product is advertised as being made with 'healthy, natural plant-based fats'. Analysis reveals it contains high levels of partially hydrogenated vegetable oil. A student evaluates this claim. Which is the most scientifically accurate conclusion?

  1. The claim is accurate because hydrogenation makes unsaturated plant fats more stable and less prone to oxidation, which is a health benefit.
  2. The claim is misleading because partial hydrogenation converts cis-unsaturated fatty acids into saturated and trans-unsaturated fatty acids, which are linked to negative health outcomes. (correct answer)
  3. The claim is accurate because the fats originate from plants, and all plant-derived lipids are inherently healthier than animal-derived lipids like butter or lard.
  4. The claim is misleading because hydrogenation removes all double bonds, converting the fat into a pure saturated fat that is identical to unhealthy animal fats.
Explanation: Partial hydrogenation is an industrial process that adds hydrogen to unsaturated vegetable oils. This process saturates some double bonds but also converts many of the remaining natural cis-double bonds into a trans configuration. Trans fats have been strongly linked to increased risk of cardiovascular disease. Therefore, the claim is misleading. Distractor A states a reason for hydrogenation (stability) but incorrectly frames it as a health benefit. Distractor C makes an overly broad and incorrect generalization. Distractor D is incorrect because partial hydrogenation, by definition, does not remove all double bonds; complete hydrogenation would.

Question 10

A marathon runner's liver cells store energy primarily as glycogen instead of an osmotically equivalent concentration of free glucose. Which of the following provides the most significant advantage for this storage strategy?

  1. Glycogen can be mobilized and transported through the bloodstream more rapidly than individual glucose molecules.
  2. The synthesis of glycogen from glucose via condensation reactions generates water molecules, which helps to hydrate the cell.
  3. Storing glucose as a large, insoluble polymer prevents a dangerous influx of water into the cell due to osmosis. (correct answer)
  4. Each glycosidic bond in glycogen contains more chemical energy than a single molecule of glucose, making it more energy-dense.
Explanation: A high concentration of free glucose molecules would drastically lower the water potential inside the liver cell, causing a massive and potentially damaging influx of water by osmosis. By polymerizing glucose into glycogen, a large, less soluble molecule, the cell stores the same amount of energy with a minimal effect on the overall solute concentration, thus maintaining osmotic stability. Distractor A is incorrect; glycogen is stored in the liver and muscle and is not transported through the blood. Glucose is the transported form. Distractor B is incorrect; while condensation does produce water, the amount is negligible and not a primary advantage. Distractor D is incorrect; the energy is stored within the chemical bonds of the glucose monomers, not created in the glycosidic bonds themselves.

Question 11

A migrating bird preparing for a long-distance flight builds up significant energy reserves. From a biochemical perspective, why is storing this energy as lipids (fat) more advantageous than storing an equivalent amount of energy as carbohydrates (glycogen)?

  1. Lipids can be metabolized anaerobically during high-altitude flight, whereas carbohydrates require constant oxygen.
  2. Lipids are less dense and store more energy per unit mass, and their storage does not require associated water, reducing the overall mass the bird must carry. (correct answer)
  3. The C-H bonds in lipids are more easily broken than the C-O and O-H bonds in carbohydrates, allowing for faster energy release.
  4. Carbohydrates must be converted to lipids before they can be used for energy, so direct lipid storage is more efficient.
Explanation: Lipids have two key advantages for mobile energy storage. First, they are more reduced (have more C-H bonds) and thus yield more energy upon oxidation per gram (~38 kJ/g) compared to carbohydrates (~17 kJ/g). Second, glycogen is hydrophilic and stored with a significant amount of water (about 2g of water per gram of glycogen), whereas lipids are hydrophobic and stored in an almost anhydrous state. Both factors lead to a dramatic reduction in the mass required to store the same amount of energy, which is a critical advantage for flight. A is incorrect; lipid metabolism is strictly aerobic. C is incorrect; the high energy yield is due to the state of reduction, not the ease of breaking bonds. D is incorrect; carbohydrates can be used directly for energy via glycolysis.

Question 12

When forming a disaccharide such as sucrose from the monosaccharides glucose and fructose, a specific chemical reaction occurs. Which statement accurately describes this reaction?

  1. A hydrolysis reaction occurs, where a water molecule is consumed to break the bond between the two monosaccharides.
  2. A phosphorylation reaction occurs, where a phosphate group is added to activate the monosaccharides before they can join.
  3. An oxidation reaction occurs, where electrons are removed from the monosaccharides to form a stable covalent bond.
  4. A condensation (dehydration) reaction occurs, forming a glycosidic bond and releasing a molecule of water. (correct answer)
Explanation: The formation of a disaccharide from two monosaccharides is a synthesis reaction. Specifically, it is a condensation reaction (also known as a dehydration synthesis) because a molecule of water is removed to form a new covalent bond, called a glycosidic bond, between the two sugar units. Hydrolysis (A) is the reverse reaction, used for digestion. Oxidation (C) and phosphorylation (D) are important reactions in cellular respiration but do not describe the fundamental process of joining two monosaccharides.

Question 13

The fat reserves of an arctic mammal are composed primarily of saturated triglycerides, while the lipids in the cell membranes of an Antarctic fish contain a high proportion of unsaturated fatty acids. What is the most likely functional reason for this difference?

  1. Saturated fats are more energy-dense, which is critical for the high metabolic rate of a warm-blooded mammal.
  2. Unsaturated fatty acids maintain membrane fluidity at low temperatures, which is essential for the cold-water fish. (correct answer)
  3. The fish consumes a diet rich in unsaturated fats, while the mammal consumes a diet rich in saturated fats, directly determining their lipid composition.
  4. Saturated triglycerides are hydrophobic, providing better insulation for the mammal than the more hydrophilic unsaturated fats.
Explanation: Unsaturated fatty acids have 'kinks' in their tails due to double bonds, which prevent them from packing closely together. This increases the fluidity of the cell membrane, which is a crucial adaptation for organisms living in cold environments to prevent their membranes from becoming too rigid. Saturated fats, which lack these kinks, pack tightly and are solid at room temperature, making membranes too viscous in the cold. Distractor A is incorrect because saturated and unsaturated fats have very similar energy densities. Distractor C is an oversimplification; while diet plays a role, the composition is also a result of adaptation and regulated synthesis. Distractor D is incorrect as both saturated and unsaturated triglycerides are strongly hydrophobic.

Question 14

In the human digestive system, the breakdown of a triglyceride molecule must occur before its components can be absorbed. Which statement correctly identifies the process, enzyme, and products?

  1. Process: Hydrolysis; Enzyme: Pancreatic lipase; Products: One glycerol and three fatty acids. (correct answer)
  2. Process: Condensation; Enzyme: Pancreatic lipase; Products: Two glycerol and three fatty acids.
  3. Process: Hydrolysis; Enzyme: Pepsin; Products: One glycerol and three amino acids.
  4. Process: Anabolism; Enzyme: Amylase; Products: Glucose and fatty acids.
Explanation: The breakdown of a triglyceride is a hydrolysis reaction, as it involves the chemical addition of water to break the ester bonds linking the fatty acids to the glycerol. The primary enzyme responsible for this in the small intestine is pancreatic lipase. The breakdown of one triglyceride molecule yields one glycerol molecule and three fatty acid molecules. Distractor B incorrectly identifies the process as condensation. Distractor C incorrectly identifies the enzyme as pepsin (a protease) and the products as amino acids. Distractor D incorrectly identifies the process and enzyme (amylase digests starch) and products.

Question 15

Lactose, the sugar in milk, is a disaccharide. An individual with lactase persistence can digest it. Which statement accurately describes the chemical process and products of lactose digestion?

  1. A hydrolysis reaction, catalyzed by lactase, breaks lactose into glucose and galactose. (correct answer)
  2. A condensation reaction breaks lactose into two molecules of glucose.
  3. A hydrolysis reaction, catalyzed by sucrase, breaks lactose into glucose and fructose.
  4. A condensation reaction joins lactose molecules together to form a polysaccharide for storage.
Explanation: Digestion involves the breakdown of large molecules into smaller ones by the addition of water, a process called hydrolysis. The enzyme specific for lactose is lactase. Lactose is a disaccharide composed of the monosaccharides glucose and galactose. Therefore, hydrolysis of lactose yields glucose and galactose. Distractor A is incorrect because digestion is hydrolysis (not condensation) and the products are wrong. Distractor C is incorrect because the enzyme is wrong (sucrase digests sucrose) and one of the products is wrong (fructose comes from sucrose). Distractor D describes synthesis, not digestion.

Question 16

What is the key structural difference between α-glucose and β-glucose, and how does this difference lead to the functional distinction between starch and cellulose?

  1. α-glucose has a six-membered ring while β-glucose has a five-membered ring; this makes starch digestible and cellulose indigestible by humans.
  2. α-glucose is the L-isomer and β-glucose is the D-isomer; only D-isomers can be polymerized into polysaccharides like starch, while L-isomers form storage lipids.
  3. β-glucose contains an additional carboxyl group that allows for peptide bonding, which is why cellulose is a structural component unlike starch.
  4. They are stereoisomers differing in the orientation of the hydroxyl (-OH) group on carbon-1; this leads to helical polymers in starch and straight-chain polymers in cellulose. (correct answer)
Explanation: α-glucose and β-glucose are isomers that differ only in the spatial orientation of the hydroxyl group on the first carbon (C1). In α-glucose, the -OH group is below the plane of the ring, while in β-glucose it is above. This seemingly small difference has massive functional consequences. Polymerization of α-glucose (as in starch) leads to curved chains that form helices, suitable for compact energy storage. Polymerization of β-glucose (as in cellulose) requires each monomer to be flipped 180°, leading to straight, rigid chains suitable for forming strong structural fibers. Distractor A is incorrect; both are six-membered rings. Distractor C is incorrect; glucose does not have a carboxyl group. Distractor D incorrectly describes the isomerism.

Question 17

When phospholipids are dispersed in water, they spontaneously form structures like micelles or bilayers. What is the primary driving force behind this self-assembly?

  1. The formation of strong covalent bonds between the phosphate heads of adjacent phospholipid molecules.
  2. The attraction between the hydrophobic fatty acid tails and the polar water molecules, which draws the tails outwards.
  3. The hydrophobic exclusion of the non-polar fatty acid tails by water, and the hydrophilic attraction of the polar phosphate heads to water. (correct answer)
  4. The enzymatic catalysis of lipid assembly, which requires ATP to position the molecules correctly in an aqueous environment.
Explanation: This phenomenon is driven by the amphipathic nature of phospholipids. The hydrophobic fatty acid tails are repelled by water (more accurately, they disrupt the hydrogen bonding network of water, so it is energetically favorable to minimize their contact with water). The hydrophilic phosphate heads are attracted to water. This results in the spontaneous arrangement where tails are shielded from water and heads are in contact with it, forming bilayers or micelles. A is incorrect; phospholipids in a membrane are not covalently bonded to each other. B describes the opposite of what happens. D is incorrect; this self-assembly is a spontaneous thermodynamic process and does not require enzymes or ATP.

Question 18

The majority of lipids in the human diet are triglycerides, but they cannot be directly absorbed from the intestine into the blood. Which statement best explains why?

  1. Triglycerides are too large and non-polar to pass through the plasma membranes of the intestinal epithelial cells and are insoluble in the aqueous blood plasma. (correct answer)
  2. The blood is maintained at a pH that would cause the immediate denaturation of any absorbed triglyceride molecules.
  3. Triglycerides are electrically charged molecules that are repelled by the phospholipid bilayer of the intestinal cells.
  4. Specific protein carriers for triglycerides are absent in the blood, so they must be converted to carbohydrates first.
Explanation: Triglycerides are large molecules. For absorption, they must first be digested by lipase into smaller components (monoglycerides and fatty acids). These smaller, non-polar molecules can diffuse into the intestinal cells. Once inside, they are re-formed into triglycerides and then packaged into chylomicrons (a type of lipoprotein) to be transported in the lymph and blood, because their non-polar nature makes them insoluble in the aqueous environment of blood plasma. Distractor B is incorrect; pH does not denature lipids. Distractor C is incorrect; triglycerides are not charged. Distractor D is incorrect; they are transported in lipoproteins, not converted to carbohydrates.

Question 19

The waterproof coating on the leaves of many plants is a waxy cuticle primarily composed of lipids. Which property of lipids is most directly responsible for this function?

  1. The high number of C-H bonds, which allows for dense energy storage.
  2. The amphipathic nature of lipid molecules, allowing them to interface with both water and air.
  3. The non-polar and therefore hydrophobic nature of the long hydrocarbon chains. (correct answer)
  4. The ability of lipids to form rigid, crystalline structures at ambient temperatures.
Explanation: Water is a polar molecule. Lipids, such as the cutin that makes up the plant cuticle, are composed of long hydrocarbon chains that are non-polar. Due to the principle of 'like dissolves like', non-polar molecules do not mix with polar molecules like water. This hydrophobic property creates a barrier that prevents water from evaporating from the leaf surface. A relates to energy storage, not waterproofing. B describes phospholipids, not typically the lipids in waxy cuticles, which are mainly hydrophobic. D is incorrect; a waxy cuticle needs to be flexible, not rigid and crystalline.