n the fatty acid and glycerol is an **ester bond**. Naming it precisely earns the second mark.","highlights":[{"text":"Name the covalent bond formed between a fatty acid and glycerol.","location":"specification"}]},{"type":"text","order":2,"title":"Name the reaction","content":"Because water is removed as the ester bond forms, this is a **condensation reaction**. This earns the third mark.","highlights":[{"text":"Name the type of reaction involved in forming this bond.","location":"specification"}]}]}],"generatedAt":"2026-07-21T04:02:25.137Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"11eb0c93-ea59-4ddc-9596-2124c638289c","label":"2","content":"Table 1 shows information about three different 18-carbon fatty acids.\n\nTable 1\n\n| Name of fatty acid | Number of carbon-carbon double bonds (\$\\text{C}=\\text{C}\$) | Melting point (\$^\\circ\\text{C}\$) |\n| :--- | :---: | :---: |\n| **Stearic acid** | 0 | 70 |\n| **Oleic acid** | 1 | 13 |\n| **Linolenic acid** | 3 | -11 |\n\n1. Name the fatty acid from Table 1 that is a saturated fatty acid.\n2. Use Table 1 to describe the relationship between the structure of a fatty acid and its melting point.","markscheme":"- Identifies **Stearic acid** B1\n- Describes the relationship: as the number of carbon-carbon double bonds ($\\text{C}=\\text{C}$) increases, the melting point decreases B1","marks":2,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"read_table","label":"Read the table","steps":[{"type":"text","order":0,"title":"Identify the saturated acid","content":"A **saturated** fatty acid has no carbon-carbon double bonds. In the table, stearic acid has $0$ $\text{C}=\text{C}$ bonds, so it is saturated.\n\n**Answer: stearic acid.** This earns the first mark.","highlights":[{"text":"Name the fatty acid from Table 1 that is a saturated fatty acid.","location":"specification"}]},{"type":"text","order":1,"title":"Compare all three rows","content":"All three fatty acids have 18 carbon atoms, so the key structural difference in Table 1 is the number of carbon-carbon double bonds. Across the rows, this increases from $0$ in stearic acid to $1$ in oleic acid and $3$ in linolenic acid, while the melting point falls from $70\\,^\\circ\\text{C}$ to $13\\,^\\circ\\text{C}$ and then $-11\\,^\\circ\\text{C}$.\n\nTherefore, **as the number of carbon-carbon double bonds increases, the melting point decreases**. This earns the second mark.","highlights":[{"text":"Use Table 1 to describe the relationship between the structure of a fatty acid and its melting point.","location":"specification"}]}]}],"generatedAt":"2026-07-21T04:03:37.626Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"f71afbef-1fa9-4bcd-82a7-5eef08e23f6a","label":"3","content":"The ratio of saturated to unsaturated fatty acids in a cell-surface membrane determines the extent of the membrane's fluidity.\n\nScientists cultured the single-celled eukaryote *Amoeba proteus* in a liquid medium enriched with unsaturated fatty acids. They observed:\n* An increase in the proportion of membrane phospholipids containing unsaturated fatty acids.\n* A significantly higher rate of endocytosis (the process of engulfing extracellular nutrients).\n\nSuggest and explain why there was an increase in the rate of endocytosis.","markscheme":"- Explains that incorporating more unsaturated fatty acids increases membrane fluidity (due to double bonds introducing kinks and preventing tight packing) B1\n- Explains that increased fluidity makes the cell-surface membrane more flexible / easier to deform B1\n- Explains that a more flexible membrane allows easier folding / pinching off to form endocytic vesicles, thereby increasing the rate of endocytosis B1","marks":3,"order":2,"markingRubric":null,"explanation":{"methods":[{"id":"membrane_fluidity","label":"Link membrane fluidity to endocytosis","steps":[{"type":"text","order":0,"title":"Identify the membrane change","content":"The cultured amoebae

Lipids

Question 2

Lipids Questions

Lipids

Question 2

Lipids Questions