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A-Level Biology – Biological Molecules and Cells

Carbohydrates, lipids, proteins, enzymes, nucleic acids, cell structure, transport.

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12 questions  ·  42 marks  ·  answers included  ·  A-Level Biology (Edexcel/AQA/OCR)
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This worksheet covers:

  • This worksheet covers carbohydrates
  • Lipids
  • Proteins
  • Nucleic acids
  • Enzymes
  • Cell structure and transport

A calculator is allowed for the numerical questions.

Questions

Section A – Biological Molecules

1

Two $\alpha$-glucose molecules join to form maltose.(a) Name the type of reaction that joins them and the bond formed.(b) Name the small molecule released during this reaction.(c) Name the reaction, and the type of enzyme, needed to break this bond during digestion.

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(a) Condensation reaction; a glycosidic bond is formed.(b) Water ($\text{H}_2\text{O}$).(c) Hydrolysis, catalysed by a carbohydrase (e.g. maltase).
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2

Starch, glycogen and cellulose are all polymers of glucose but have different roles.(a) State the role of starch in plants and of glycogen in animals.(b) Explain how the structure of cellulose (unbranched chains of $\beta$-glucose linked by hydrogen-bonded microfibrils) relates to its function in plant cell walls.

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(a) Starch is the main storage carbohydrate in plants; glycogen is the main storage carbohydrate in animals (and fungi).(b) The long, straight, unbranched chains of $\beta$-glucose allow many hydrogen bonds to form between adjacent parallel chains, bundling them into strong microfibrils; this gives cellulose high tensile strength, allowing it to provide structural support and rigidity to the plant cell wall.
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3

(a) Draw (describe in words) the general structure of a triglyceride, naming the two types of molecule that combine and the bond formed.(b) Explain why a phospholipid, unlike a triglyceride, is described as amphipathic, and explain the significance of this property for cell membranes.

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(a) One glycerol molecule joins with three fatty acid molecules by condensation reactions, each forming an ester bond (with loss of a water molecule).(b) A phospholipid has a hydrophilic phosphate ‘head’ and two hydrophobic fatty acid ‘tails’, so part of the molecule interacts with water and part does not (amphipathic); this causes phospholipids to spontaneously arrange into a bilayer in water, with hydrophilic heads facing the aqueous environments on each side and hydrophobic tails facing inwards, forming the basic structure of the cell membrane.
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4

(a) Describe what is meant by the primary, secondary and tertiary structure of a protein.(b) Name one type of bond that maintains tertiary structure and explain how it forms.(c) State what is meant by the quaternary structure of a protein, using haemoglobin as an example.

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(a) Primary structure is the specific sequence of amino acids in the polypeptide chain. Secondary structure is the folding of the chain into $\alpha$-helices or $\beta$-pleated sheets, held by hydrogen bonds between nearby amino acids. Tertiary structure is the further folding of the whole polypeptide into a specific 3D shape, held by interactions (ionic bonds, hydrogen bonds, disulfide bridges, hydrophobic interactions) between R groups.(b) A disulfide bridge (covalent bond) forms between the sulfur atoms of two cysteine R groups that are brought close together as the chain folds.(c) Quaternary structure is the arrangement of two or more polypeptide chains (and any non-protein groups) together to form a functional protein; haemoglobin’s quaternary structure consists of four polypeptide subunits (two $\alpha$ and two $\beta$ chains), each with an associated haem group.
[5]

Section B – Enzymes

5

Amylase was mixed with a starch solution and iodine solution, and the time taken for the blue-black colour to disappear (indicating starch had been fully hydrolysed) was measured as $40\,\text{s}$.(a) Calculate the initial rate of reaction, in $\text{s}^{-1}$, using rate $= 1/\text{time}$.(b) Suggest one variable that should be controlled to make this a fair test.

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(a) $\text{rate} = 1/40 = 0.025\,\text{s}^{-1}$(b) Any one of: temperature, pH, volume/concentration of starch solution, volume/concentration of amylase solution (any variable other than the one being investigated).
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6

Explain, using the induced-fit model, why an enzyme is specific to one substrate (or a small group of substrates) rather than catalysing many different reactions.

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The enzyme’s active site has a specific tertiary structure (a specific arrangement of R groups) that is complementary in shape and chemical properties to only one substrate (or very similar substrates); only a molecule with this complementary shape can bind and induce the slight change in active-site shape needed for catalysis, so other substrates cannot bind effectively.
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7

A competitive inhibitor and a non-competitive inhibitor are both added, in separate experiments, to an enzyme-catalysed reaction.(a) Explain how a competitive inhibitor reduces the rate of reaction, and how this effect can be overcome.(b) Explain how a non-competitive inhibitor reduces the rate of reaction, and why increasing substrate concentration does not overcome this effect.

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(a) A competitive inhibitor has a similar shape to the substrate and binds to the active site, blocking the substrate from binding; increasing the substrate concentration increases the chance of substrate (rather than inhibitor) binding, so the effect can be overcome (largely restoring $V_{max}$) at high substrate concentration.(b) A non-competitive inhibitor binds to a site other than the active site, changing the tertiary structure of the enzyme and distorting the active site so the substrate can no longer bind (or catalysis cannot occur); since the inhibitor does not compete with the substrate for the active site, increasing substrate concentration does not reduce the degree of inhibition.
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8

Some industrial processes use enzymes immobilised on an inert support (e.g. in beads of alginate gel) rather than free enzymes in solution.State two advantages of using immobilised enzymes rather than free enzymes in an industrial process.

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Any two of: the enzyme can easily be separated from the product and reused, reducing cost; the product is not contaminated with enzyme; immobilised enzymes are often more stable (more resistant to changes in temperature/pH) than free enzymes; the process can be run continuously (e.g. substrate flowing over immobilised enzyme in a column).
[2]

Section C – Cell Structure and Transport

Exam-style, includes calculations
9

(a) State two structural features that distinguish a eukaryotic cell from a prokaryotic cell.(b) Name the organelle responsible for synthesising and folding proteins destined for secretion, and name the organelle that then modifies, packages and sorts these proteins.

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(a) Any two of: eukaryotic cells have a nucleus surrounded by a nuclear envelope (prokaryotic DNA is free in the cytoplasm, not membrane-bound); eukaryotic cells have membrane-bound organelles (e.g. mitochondria, Golgi apparatus) which prokaryotic cells lack; eukaryotic cells are generally much larger than prokaryotic cells; eukaryotic ribosomes are larger (80S) than prokaryotic ribosomes (70S).(b) The rough endoplasmic reticulum (with bound ribosomes) synthesises and folds the proteins; the Golgi apparatus then modifies, packages (into vesicles) and sorts them for secretion.
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10

An electron micrograph shows a mitochondrion with an image length of $50\,\text{mm}$. The actual length of the mitochondrion is $5\,\mu\text{m}$.Calculate the magnification of the image, showing your working and converting units where necessary.

Show answer
Convert image length to the same unit as actual length: $50\,\text{mm} = 50\,000\,\mu\text{m}$.$\text{magnification} = \dfrac{\text{image size}}{\text{actual size}} = \dfrac{50\,000}{5} = 10\,000$ (i.e. $\times10\,000$)
[3]
11

Cylinders of potato tissue, each with an initial mass of $4.20\,\text{g}$, were placed in distilled water. After 30 minutes, a cylinder was removed and had a mass of $4.62\,\text{g}$.(a) Calculate the percentage change in mass of the potato cylinder.(b) Explain, in terms of water potential, why the potato cylinder gained mass in distilled water.

Show answer
(a) $\text{percentage change} = \dfrac{4.62-4.20}{4.20}\times100 = +10.0\%$(b) Distilled water has a higher (less negative) water potential than the cell sap inside the potato cells, so there is a water potential gradient from the water into the cells; water therefore moves into the cells by osmosis, across the partially permeable cell surface membrane, down the water potential gradient, increasing the mass of the tissue.
[4]
12

Explain the difference between facilitated diffusion and active transport of substances across a cell surface membrane, including the role of membrane proteins and the energy source (or lack of one) used in each.

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Facilitated diffusion is the passive movement of a substance (e.g. a polar molecule or ion) down its concentration gradient through a specific channel or carrier protein; it requires no metabolic energy (ATP), since movement is driven by the concentration gradient. Active transport moves a substance against its concentration gradient using a carrier protein, and requires metabolic energy in the form of ATP (produced by respiration) to change the shape of the carrier protein and move the substance across the membrane.
[3]

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