Edexcel International GCSE Biology 4BI1: Level of organisation and biological molecules – Study Guide

This study guide teaches sub-topics 2(a) Level of organisation (point 2.1) and 2(c) Biological molecules (points 2.7–2.14B) of the Pearson Edexcel International GCSE Biology (4BI1) specification, Issue 3. The qualification is untiered. Point 2.14B is printed in bold with a B reference, so it is examined on Paper 2 only; everything else can appear on either paper. It applies to the June and November series examined under Issue 3.

Use it with the revision notes and the practice questions. Sub-topic 2(b), cell structure, is taught in the cell structure and biological molecules overview, with its own revision notes and practice set. The full course is on the Edexcel IGCSE Biology hub, and the printable checklist lists every point.

What this unit covers

Spec point	What you must be able to do	Paper
2.1	Describe the levels of organisation: organelles, cells, tissues, organs and systems	1 and 2
2.7	Identify the elements in carbohydrates, proteins and lipids	1 and 2
2.8	Describe starch, glycogen, protein and lipid as large molecules built from smaller units	1 and 2
2.9	Practical: test food samples for glucose, starch, protein and fat	1 and 2
2.10	Understand enzymes as biological catalysts in metabolic reactions	1 and 2
2.11	Understand how temperature affects enzyme function, including the active site	1 and 2
2.12	Practical: investigate the effect of temperature on enzyme activity	1 and 2
2.13	Understand how pH affects enzyme function by altering the active site	1 and 2
2.14B	Practical: investigate the effect of pH on enzyme activity	Paper 2 only

The practical points (2.9, 2.12, 2.14B) are assessed through written questions, so you must be able to describe each method, name the variables and explain the results.

2(a) Level of organisation (2.1)

A multicellular organism is built up in levels. Each level is made of units from the level below.

Level	Meaning	Animal example	Plant example
Organelle	A structure inside a cell with a particular job	Mitochondrion	Chloroplast
Cell	The basic unit of a living organism	Muscle cell	Palisade cell
Tissue	A group of similar cells working together to carry out a function	Muscle tissue	Palisade mesophyll
Organ	A structure made of several different tissues working together	Stomach, heart	Leaf, root
System	A group of organs working together to carry out a function	Digestive system	Shoot system

The organism itself sits above the system level.

A tissue is made of similar cells; an organ is made of different tissues. The stomach is an organ because it contains muscle tissue (which churns food), glandular tissue (which secretes enzymes) and epithelial tissue (which lines it).

Worked example: placing structures in order

Put these in order, starting with the smallest: digestive system, epithelial cell, small intestine, ribosome, epithelium.

1. A ribosome is inside a cell, so it is an organelle.
2. An epithelial cell is a single cell.
3. Epithelium is a layer of similar epithelial cells, so it is a tissue.
4. The small intestine contains epithelium, muscle and other tissues, so it is an organ.
5. The digestive system contains the small intestine, stomach, pancreas and other organs, so it is a system.

Answer: ribosome → epithelial cell → epithelium → small intestine → digestive system.

2(c) Biological molecules

Elements (2.7)

Molecule	Elements
Carbohydrates	Carbon, hydrogen, oxygen
Proteins	Carbon, hydrogen, oxygen, nitrogen (often sulfur as well)
Lipids (fats and oils)	Carbon, hydrogen, oxygen

Carbohydrates and lipids contain the same three elements, but lipids have much less oxygen in proportion to their carbon and hydrogen. Glucose, a carbohydrate, is C₆H₁₂O₆. If a molecule contains nitrogen, it is the protein.

Large molecules from small units (2.8)

Large molecule	Built from	Where it is found
Starch	Many glucose (simple sugar) units	Storage carbohydrate in plants
Glycogen	Many glucose (simple sugar) units	Storage carbohydrate in animals and fungi
Protein	Many amino acids	Enzymes, muscle, haemoglobin
Lipid	Fatty acids and glycerol	Fats in animals, oils in plants

Points examiners look for:

• Starch and glycogen are both made of glucose; they differ in branching and in which organisms store them.
• Proteins are long chains of amino acids. There are about 20 different amino acids, and the order of amino acids decides the protein’s shape. This is why enzymes can have a precise active site.
• A lipid molecule is made of one glycerol joined to fatty acids (usually three). A lipid is not a chain of repeating identical units, so do not call it a chain of fatty acids alone.

Digestion (point 2.29, sub-topic 2(e)) reverses 2.8: amylase and maltase break starch to glucose, proteases break proteins to amino acids, and lipases break lipids to fatty acids and glycerol.

Food tests (2.9)

Substance	Reagent and method	Negative result	Positive result
Glucose	Add Benedict’s solution and heat in a water bath (about 80 °C) for a few minutes	Stays blue	Green → yellow → orange → brick-red as the amount of glucose rises
Starch	Add a few drops of iodine solution	Stays orange-brown	Blue-black
Protein	Biuret test: add sodium hydroxide solution, then a few drops of dilute copper sulfate solution (or biuret reagent)	Stays blue	Purple (lilac)
Fat	Emulsion test: shake the sample with ethanol, then pour the ethanol into water	Stays clear	Cloudy white emulsion

Practical points:

• Grind solid food with a little water first, so the reagent can mix with it.
• Heat Benedict’s tests in a water bath, not a direct flame. Keep ethanol away from flames.
• The Benedict’s colour gives a rough idea of how much glucose is present. It does not give an exact value.
• Sucrose (table sugar) gives a negative Benedict’s result. A sucrose solution stays blue.
• Run a control with distilled water in place of the food, so you can see the colour with no food present.

Worked example: interpreting food test results

A student tests three drinks.

Drink	Benedict’s	Iodine	Biuret	Emulsion
P	Orange	Orange-brown	Purple	Cloudy
Q	Blue	Orange-brown	Blue	Clear
R	Green	Orange-brown	Blue	Clear

1. P: positive Benedict’s, biuret and emulsion tests, so it contains glucose, protein and fat, and no starch. This fits milk with added glucose.
2. Q: all tests negative. It might be water, or it might contain a sugar such as sucrose that Benedict’s does not detect. You cannot conclude it has no sugar at all.
3. R: green Benedict’s means a small amount of glucose, less than in P.

Enzymes (2.10–2.14B)

Enzymes as biological catalysts (2.10)

A catalyst speeds up a chemical reaction without being used up. Enzymes are biological catalysts: proteins made by living cells. Metabolic reactions are the chemical reactions in a cell; without enzymes they would be far too slow at body temperature.

Enzymes catalyse both kinds of metabolic reaction:

• breaking down large molecules, for example amylase breaking starch into sugars
• building up large molecules, for example joining glucose units into starch or glycogen, or amino acids into proteins.

Each enzyme has an active site with a specific shape. Only a substrate with a complementary shape fits it. The substrate binds to form an enzyme-substrate complex, the reaction happens, and the products leave. The enzyme is unchanged and can be used again. This is why each enzyme is specific: amylase acts on starch but not on protein.

Temperature (2.11)

• Below the optimum: molecules have less kinetic energy, so enzyme and substrate collide less often. Fewer enzyme-substrate complexes form and the rate is low. The enzyme is not damaged; warming it up restores activity.
• Rising to the optimum: more frequent successful collisions, faster rate. Many human enzymes have an optimum near 37 °C.
• Above the optimum: the bonds holding the protein in shape break. The active site changes shape, so the substrate no longer fits. The enzyme is denatured. The rate falls steeply. Denaturation is permanent.

Practical: temperature and amylase (2.12)

1. Put starch solution and amylase solution in separate tubes in a water bath at the chosen temperature. Leave them for a few minutes to reach that temperature.
2. Put one drop of iodine solution in each well of a spotting tile.
3. Mix the amylase and starch and start a stopwatch.
4. Every 30 seconds, transfer one drop of the mixture to a fresh well of iodine.
5. Record the time when the iodine stays orange-brown: all the starch has been digested.
6. Repeat at other temperatures, and repeat each temperature three times to calculate a mean.

Independent variable: temperature. Dependent variable: time for the starch to disappear. Control variables: volume and concentration of amylase and starch, pH (use a buffer), and the interval between samples. A shorter time means a faster rate, so rate = 1/time.

Worked example: processing temperature results

Temperature (°C)	20	30	40	50	60
Time for starch to disappear (s)	240	120	60	150	No end point

Calculate the rate at 40 °C and explain the pattern.

rate = 1 / time
     = 1 / 60
     = 0.0167 s⁻¹ (3 s.f.)

At 20 °C the rate is 1/240 = 0.00417 s⁻¹, so the rate at 40 °C is four times faster. From 20 °C to 40 °C, more kinetic energy means more frequent successful collisions. At 50 °C the rate has fallen because some of the amylase has denatured: its active site has changed shape. At 60 °C the amylase is fully denatured, so starch is never broken down. The optimum lies between 30 °C and 50 °C; to find it more precisely, repeat with smaller temperature steps in that range.

pH (2.13)

Each enzyme works fastest at its optimum pH. A stomach protease works best at about pH 2. Amylase in saliva works best at about pH 7. Away from the optimum, extra hydrogen ions (acid) or hydroxide ions (alkali) disturb the bonds that hold the enzyme’s shape. The active site changes shape, the substrate fits less well, and fewer enzyme-substrate complexes form. Far from the optimum the enzyme is denatured.

Practical: pH and amylase (2.14B) – Paper 2 only

The method is the same as the temperature practical, with two changes:

• Independent variable: pH, set by adding a buffer solution of known pH (for example pH 4, 5, 6, 7 and 8) to each tube.
• Temperature is now a control variable: keep every tube in the same water bath, for example at 35 °C.

Worked example: pH results (Paper 2 only)

pH	4	5	6	7	8
Time for starch to disappear (s)	360	180	90	45	120

Identify the optimum and compare pH 5 with pH 7.

1. Shortest time is 45 s at pH 7, so the optimum in this range is about pH 7.
2. Rate at pH 7 = 1/45 = 0.0222 s⁻¹. Rate at pH 5 = 1/180 = 0.00556 s⁻¹.
3. 180 ÷ 45 = 4, so the reaction at pH 7 is four times faster than at pH 5.
4. At pH 5 the active site of amylase is altered, so fewer starch molecules fit and fewer enzyme-substrate complexes form per second.

Common errors

• Calling an organ “a group of cells”. It is a group of different tissues.
• Naming a single cell, such as a red blood cell, as a tissue.
• Listing nitrogen as an element in carbohydrates or lipids.
• Saying glycogen is made of amino acids, or that lipids are made only of glycerol.
• Heating the Benedict’s test without stating a water bath, or writing “turns red” for iodine.
• Heating the biuret test. It needs no heat.
• Saying enzymes are “killed”. They are molecules; they are denatured.
• Saying low temperature denatures an enzyme. It only slows it.
• Reading the longest time as the fastest rate.

Next steps

Recap with the revision notes, then try the practice questions and the free diagnostics.

Official syllabus

Pearson Edexcel International GCSE in Biology (4BI1) specification, Issue 3, first examination June 2019, Pearson Education Limited – Topic 2, Structures and functions in living organisms: sub-topics (a) Level of organisation and (c) Biological molecules.