Lipids: Practice Questions (Biology 1.5)

Answer: (A). Olive oil is rich in unsaturated fatty acids. Each double bond kinks the tail, so the tails cannot line up closely. With fewer attractions between tails, the oil stays liquid at a lower temperature. Butter is rich in straight saturated tails that pack tightly.

• (B) is backwards: unsaturated tails have fewer hydrogen atoms (each double bond means two fewer). Mass is not the reason anyway.
• (C) is false: both are fats, and every fat (triglyceride) contains glycerol.
• (D) confuses melting with breaking covalent bonds. Melting separates whole molecules from each other; the tails stay intact.

Answer: (B). The heads are polar and hydrophilic, so they are stable next to water. The tails are nonpolar and hydrophobic, so they gather away from water, next to each other. A bilayer satisfies both.

• (A) puts the hydrophobic tails in the water, which is the least stable position. (At an air–water surface, phospholipids do form a single layer, but with their tails pointing up into the air.)
• (C) reverses the orientation: the hydrophobic tails would be in contact with water.
• (D) ignores the tails. The molecule is only partly polar, so it does not dissolve as single molecules; it gathers into structures that hide the tails.

Answer: (A). mass = energy ÷ energy per gram = 555 kJ ÷ 37 kJ g⁻¹ = 15.0 g.

• (B) uses the carbohydrate value: 555 ÷ 17 = 32.6 g. That is the mass of carbohydrate, more than twice as much.
• (C) divides by 9, which is the energy of fat in kilocalories per gram, not kilojoules. Mixing units gives 61.7 g.
• (D) multiplies instead of dividing: 555 × 37 = 20 535 g. The units would be kJ² g⁻¹, not grams, which shows the error.

(a) As the mean number of double bonds per tail increases, the melting point decreases. It falls from 52 °C for P (0.2) to −14 °C for S (2.0): a drop of 66 °C as the double bonds rise by 1.8 per tail.

(b) An oil is liquid if 25 °C is above its melting point. Q, R and S are liquid (melting points 21, 3 and −14 °C). P (52 °C) is solid.

(c) Gradient from Q to R = (3 − 21) ÷ (1.3 − 0.8) = −18 ÷ 0.5 = −36 °C per double bond. From Q: 21 + (−36 × 0.2) = 21 − 7.2 = 13.8, so about 14 °C. (A value read from a line drawn on the graph, between 12 and 16 °C, is acceptable.)

(d) More double bonds mean more kinks in the tails. Kinked tails cannot pack closely, so there are fewer attractions between neighbouring tails. Less energy is needed to let the molecules move past each other, so the oil melts at a lower temperature.

Do not award the last point for “double bonds are weaker”. The covalent bonds inside the tails do not break when a fat melts.

(a) The phosphate head is hydrophilic: it is charged or polar, so it attracts the δ⁺ and δ⁻ ends of water molecules. The two fatty acid tails are hydrophobic: they are made of nonpolar C–H bonds and cannot form attractions with water.

(b) Water molecules are most stable when they can hydrogen-bond with each other or with polar groups. If the heads face the water and the tails face each other, no tail touches water. Water keeps its hydrogen bonds and the heads interact with water. This is the most stable arrangement, so it forms on its own.

(c) A stable bilayer would not form. A triglyceride has three fatty acids and no polar head, so it is hydrophobic all over. It has no part that is stable facing the water. Triglycerides would gather into oily droplets instead. Without a continuous bilayer, the cell could not keep a barrier between its contents and the outside, so ions and other substances would leak, and the cell would probably die.

Accept “the fats would form droplets” for the prediction point if the reason (no hydrophilic part) is given.

(a) Control: (6 + 8 + 7) ÷ 3 = 21 ÷ 3 = 7%. Cholesterol removed: (21 + 25 + 23) ÷ 3 = 69 ÷ 3 = 23%. That is about 3.3 times as many damaged cells (23 ÷ 7 = 3.29), or 16 percentage points more. The ranges (6–8 and 21–25) do not overlap.

(b) The independent variable is whether cholesterol was removed from the membrane. The control group shows how many cells are damaged by the stress and the handling alone, so any extra damage in the treated group can be linked to the loss of cholesterol.

(c) Cholesterol sits among the phospholipids of animal cell membranes and gives the membrane structural stability. Without it, the membrane is less able to hold together under stress, so more membranes tear or become leaky and let the dye in.

(d) Damage would probably increase further. At a warmer temperature, phospholipid tails move more and the membrane becomes more fluid. Cholesterol normally limits this movement. Without it, the membrane would become even less stable, so more cells would be damaged. (A good answer also says the prediction should be tested by comparing treated and control cells at both temperatures.)

Do not award the third point for “cholesterol stores energy” or “cholesterol is a hormone”: cholesterol’s role here is structural.

(a) Energy = 12 g × 37 kJ g⁻¹ = 444 kJ.

(b) Time = 444 kJ ÷ 45 kJ h⁻¹ = 9.87 h, so about 9.9 hours.

(c) Mass of carbohydrate = 444 kJ ÷ 17 kJ g⁻¹ = 26.1 g. Extra mass = 26.1 − 12 = 14.1 g, more than doubling the store. (Equivalently, 12 g of carbohydrate would hold only 12 × 17 = 204 kJ, enough for about 4.5 hours.)

(d) Fat tails are long chains of C–H bonds with very little oxygen, so a gram of fat releases more energy when respired. Fat is also hydrophobic, so it is stored in compact droplets without drawing in water. For a flying animal, every extra gram costs energy to carry, so the denser store is a clear advantage.

Suggested mark points (4): 1 for 444 kJ; 1 for about 9.9 h with units; 1 for 26.1 g and about 14 g extra; 1 for linking energy density to C–H-rich, oxygen-poor tails and to the cost of carrying mass in flight.

Common errors: multiplying 12 × 45 = 540 mixes up the store with the rate; dividing 45 ÷ 12 = 3.75 has no meaning here. Check that the units work out (kJ ÷ kJ h⁻¹ = h).