AS 3 2017-style practical skills question · 6 marks
Explain how you would carry out a serial dilution and use it to estimate the water potential of potato tissue.
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CCEA AS Biology 2017 past paper and mark scheme
Summer 2017 is part of the current CCEA GCE Biology (2016) specification, so every question style in it is still examinable. First series of the 2016 specification. The papers and mark schemes themselves are published by CCEA; BioCCEA is an independent study tool and is not endorsed by CCEA.
What you sit in AS 2017
AS 1: Molecules and Cells
AS 1 written paper (1 hour 30 minutes)
Biological molecules, enzymes, cell structure, transport across membranes, cell division, nucleic acids and the immune response.
AS 1 topic list and exam focusAS 2: Organisms and Biodiversity
AS 2 written paper (1 hour 30 minutes)
Gaseous exchange, transport in plants and mammals, adaptation, biodiversity, classification, sampling and human impact.
AS 2 topic list and exam focusAS 3: Practical Skills in AS Biology
AS 3 practical booklet plus a written practical-skills paper (1 hour)
Planning, manipulating apparatus, recording, processing and evaluating experimental data at AS level.
AS 3 topic list and exam focus
How to sit the 2017 paper properly
- 1Print the paper and sit it in one block, to time, with no notes. A paper done in pieces tells you nothing about your exam performance.
- 2Mark it with the official CCEA mark scheme in front of you and award a mark only where the scheme's point is actually made.
- 3Log every dropped mark by unit topic, not by question number, so patterns appear across papers.
- 4Rewrite the two worst answers in full, using mark-scheme wording, then re-mark them.
- 5Come back to the same paper three weeks later and redo only the questions you lost marks on.
Sit the AS 2017 style questions
Answer each one under exam conditions. Your wording is marked against every mark scheme point before the full-mark answer appears.
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These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Worked AS 2017 questions with mark schemes
Question styles that carried the most marks in the Summer 2017 papers, each with the full list of creditworthy mark scheme points, a model answer at full marks, and the marking point candidates most often missed.
AS 3 2017-style practical skills question · 6 marks
Explain how you would carry out a serial dilution and use it to estimate the water potential of potato tissue.
Command word: Explain — how to answer itMark scheme — 6 creditworthy points
- Prepare a range of sucrose concentrations by serial dilution from a 1.0 mol dm⁻³ stock, stating the method
- Cut potato cylinders of equal length and diameter with a cork borer, blot dry and weigh each
- Place one cylinder in each concentration for a fixed time at constant temperature
- Blot dry and reweigh; calculate percentage change in mass for each concentration
- Plot percentage change in mass against concentration and find where the line crosses zero
- At that concentration there is no net movement of water, so the water potential of the solution equals that of the potato tissue
Full-mark answer
I would prepare a serial dilution from a 1.0 mol dm⁻³ sucrose stock, mixing measured volumes of stock and distilled water to give, for example, 0.8, 0.6, 0.4, 0.2 and 0 mol dm⁻³. Using a cork borer I would cut potato cylinders of equal length and diameter, blot each dry to remove surface liquid, and weigh it. One cylinder would be placed in each concentration for the same fixed time at a constant temperature. Each cylinder would then be blotted dry, reweighed, and the percentage change in mass calculated. I would plot percentage change in mass against sucrose concentration and read off the concentration at which the line crosses zero. At that point there is no net movement of water into or out of the tissue, so the water potential of the sucrose solution is equal to the water potential of the potato tissue.
Examiner insight: Percentage change in mass — not change in mass — is required, because the cylinders will not have identical starting masses. Blotting before both weighings is also a discrete mark.
AS 1 2017-style digestion and absorption question · 5 marks
Explain how the ileum is adapted for the absorption of the products of digestion.
Command word: Explain — how to answer itMark scheme — 5 creditworthy points
- Villi and microvilli give a very large surface area for absorption
- The epithelium is one cell thick, giving a short diffusion pathway
- A dense capillary network maintains a steep concentration gradient by removing absorbed products
- Co-transport: sodium ions are actively pumped out of the epithelial cell, and glucose enters with sodium down its gradient
- Numerous mitochondria supply ATP for active transport; lacteals absorb fatty acids and glycerol as chylomicrons
Full-mark answer
The inner surface of the ileum is folded into villi, and each epithelial cell bears microvilli, so the surface area available for absorption is enormous. The epithelium is only one cell thick, so the diffusion pathway into the blood is very short. Each villus contains a dense network of capillaries that carries absorbed products away, maintaining a steep concentration gradient between the gut contents and the blood. Glucose and amino acids are absorbed by co-transport: sodium ions are actively pumped out of the epithelial cell into the blood, and sodium then re-enters from the lumen through a co-transporter protein, bringing glucose in with it against its own concentration gradient. Epithelial cells contain many mitochondria to supply the ATP needed for this active transport, and each villus also contains a lacteal, which absorbs fatty acids and glycerol.
Examiner insight: The co-transport mark is only given if the sodium gradient is created by active transport first. 'Glucose is actively transported into the blood' is a common and costly simplification.
AS 2 2017-style haemoglobin question · 6 marks
Describe how the structure of haemoglobin allows it to load and unload oxygen, and explain the effect of a high carbon dioxide concentration on the oxygen dissociation curve.
Command word: Describe — how to answer itMark scheme — 6 creditworthy points
- Haemoglobin has four polypeptide chains, each with a haem prosthetic group containing iron
- Each molecule can carry four oxygen molecules; oxygen binds to form oxyhaemoglobin
- Binding of the first oxygen changes the shape of the molecule so further oxygen binds more easily - cooperative binding, giving the S-shaped curve
- In the lungs the high partial pressure of oxygen means haemoglobin is almost fully saturated (loading)
- In respiring tissues the low partial pressure of oxygen causes unloading / dissociation
- High carbon dioxide lowers the pH, changing the shape of haemoglobin so its affinity for oxygen falls - the curve shifts right (Bohr effect), so more oxygen is released to the tissues
Full-mark answer
A haemoglobin molecule is made of four polypeptide chains, each associated with a haem prosthetic group containing an iron ion, so each molecule can bind four oxygen molecules to form oxyhaemoglobin. When the first oxygen binds, the shape of the molecule changes so that it becomes easier for the next oxygen molecules to bind - cooperative binding - which is why the dissociation curve is S-shaped. In the lungs the partial pressure of oxygen is high, so haemoglobin is almost fully saturated and loads oxygen, while in a respiring tissue the partial pressure of oxygen is low and oxygen dissociates and is unloaded. Where the respiring tissue is producing a lot of carbon dioxide, the carbon dioxide dissolves and lowers the pH, which slightly changes the shape of the haemoglobin and reduces its affinity for oxygen. The dissociation curve shifts to the right - the Bohr effect - so at any given partial pressure of oxygen haemoglobin is less saturated and even more oxygen is released to the actively respiring tissue.
Examiner insight: The S-shape must be explained by cooperative binding, and the Bohr mark needs the words 'shifts right' or 'lower affinity' plus the pH change.
These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Download the official CCEA paper and mark schemePractise the AS 2017 paper in the app
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Get your 2017 answers marked
Photograph a written answer from this paper and see it marked line by line in CCEA mark-scheme language, with the exact phrase you needed to earn each missing mark.