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GCSE · Biology

B1 Cell Biology

AQA 8461 · Calculator allowed · about 68 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about the structure of cells.

(a)Which structure controls what enters and leaves an animal cell?(1)

A) Nucleus
B) Cell membrane
C) Cytoplasm
D) Mitochondria

(b)Name the part of a plant cell in which photosynthesis takes place.(1)

(c)State one structure found in a plant cell but not in an animal cell, other than a chloroplast.(1)

(Total for Question 1 is 3 marks)

This question is about units used to measure cells.

(a)Put these units of length in order of size, from smallest to largest: mm, nm, um.(1)

(b)How many micrometres (um) are there in 1 millimetre (mm)?(1)

(Total for Question 2 is 2 marks)

Bacterial cells are prokaryotic. Animal cells are eukaryotic.

(a)State two ways the structure of a bacterial cell is different from the structure of an animal cell.(2)

(b)A bacterial cell has a diameter of about 2 um. An animal cell has a diameter of about 20 um. How many times larger is the diameter of the animal cell?(1)

(Total for Question 3 is 3 marks)

Cells can be specialised for their function.

(a)A sperm cell is adapted to reach and fertilise an egg. Describe one adaptation of a sperm cell and explain how it helps the cell carry out its function.(2)

(b)State one way a root hair cell is adapted for absorbing water from the soil.(1)

(Total for Question 4 is 3 marks)

Differentiation produces specialised cells.

(a)What is meant by differentiation?(1)

(b)In most animal cells, at what stage of life is the ability to differentiate mainly lost, so that later it is used only for repair and replacement?(1)

(c)Name the plant tissue in which cells keep the ability to differentiate throughout the plant's life.(1)

(Total for Question 5 is 3 marks)

Scientists use light microscopes and electron microscopes to study cells.

(a)Give one advantage of using an electron microscope instead of a light microscope to study cells.(1)

(b)A student uses a light microscope with a x10 eyepiece lens and a x40 objective lens. Calculate the total magnification.(1)

(Total for Question 6 is 2 marks)

A student measures a cell using a microscope. The image of the cell is 48 mm wide. The actual width of the cell is 0.06 mm.

(a)Calculate the magnification of the image. Use the equation: magnification = size of image / size of real object.(2)

(b)The student then views a chloroplast at a magnification of x2000. The image of the chloroplast is 10 mm long. Calculate the real length of the chloroplast. Give your answer in micrometres (um).(3)

(Total for Question 7 is 5 marks)

A micrograph of a cell has a scale bar. The scale bar is 20 mm long and represents an actual length of 10 um.

(a)On the same micrograph the image of a cell measures 60 mm across. Calculate the actual width of the cell. Give your answer in micrometres (um).(3)

(Total for Question 8 is 3 marks)

Substances move in and out of cells by diffusion.

(a)Define diffusion.(2)

(b)A cube-shaped model cell has sides of length 2 cm. Calculate the surface area to volume ratio of the cube. Give your answer in the form n : 1.(2)

(c)Explain why large multicellular organisms need specialised exchange surfaces, but a single-celled organism does not.(2)

(Total for Question 9 is 6 marks)

Osmosis is a special case of diffusion.

(a)Define osmosis.(2)

(b)A cylinder of potato is placed in pure water. Predict what will happen to the mass of the cylinder and explain your answer.(2)

(Total for Question 10 is 4 marks)

Some substances are moved by active transport.

(a)State one way active transport is different from diffusion.(1)

(b)Root hair cells absorb mineral ions from the soil. The concentration of mineral ions is lower in the soil than inside the root hair cell. Explain why the root hair cell must use active transport to absorb these ions.(2)

(Total for Question 11 is 3 marks)

A student investigates osmosis. Cylinders of potato of equal size are placed in sugar solutions of different concentrations. The mass of each cylinder is measured before and after being left for 30 minutes.

(a)Name the independent variable in this investigation.(1)

(b)State two variables that should be controlled to make this a fair test.(2)

(c)One cylinder had a mass of 3.20 g before and 2.72 g after. Calculate the percentage change in mass.(2)

(d)Explain, in terms of water movement and concentration, why this cylinder lost mass.(2)

(Total for Question 12 is 7 marks)

A student prepares a slide of onion epidermal cells to view under a light microscope.

(a)Describe how the student should prepare the slide.(3)

(b)Explain why iodine solution is added to the slide.(1)

(c)The student should start with the lowest-power objective lens. Give one reason why.(1)

(Total for Question 13 is 5 marks)

Cells divide by mitosis as part of the cell cycle.

(a)During the first stage of the cell cycle (before mitosis), the cell grows. State two other things that happen to the contents of the cell during this stage.(2)

(b)Mitosis produces two daughter cells. State how the daughter cells compare genetically with the parent cell.(1)

(c)Give one reason why the body needs mitosis.(1)

(Total for Question 14 is 4 marks)

In ideal conditions, bacteria can divide very quickly by binary fission.

(a)One type of bacterium divides every 20 minutes. A culture starts with 800 bacteria. Calculate the number of bacteria after 2 hours, assuming ideal conditions.(2)

(b)Write your answer to part (a) in standard form.(1)

(Total for Question 15 is 3 marks)

Stem cells can be used to treat some medical conditions. Stem cells can be taken from human embryos or, for some treatments, from the patient's own adult tissue (for example bone marrow).

Evaluate the use of stem cells from human embryos to treat human diseases. Your answer should consider the benefits and the objections.

(Total for Question 16 is 6 marks)

Mark scheme · B1 Cell Biology

Question 1

(a) B1 B (cell membrane)
(b) B1 chloroplast(s)
(b) Answer: Chloroplast
(c) B1 cell wall / permanent vacuole (allow: made of cellulose)
(c) Answer: Cell wall (or permanent vacuole)

Question 2

(a) B1 nm, um, mm (in this order)
(a) Answer: nm, um, mm
(b) B1 1000
(b) Answer: 1000

Question 3

(a) B1 bacterial cell has no nucleus / genetic material is a single loop of DNA (free in cytoplasm)
(a) B1 any one of: has a cell wall / has plasmid(s) / is (much) smaller / may have flagella
(a) Answer: No nucleus (single loop of DNA); plus one of: cell wall / plasmids / much smaller / flagella
(b) B1 10 (times)
(b) Answer: 10

Question 4

(a) B1 one adaptation, e.g. has a tail / flagellum (allow: many mitochondria; acrosome/enzymes in head; streamlined)
(a) B1 linked function, e.g. tail allows it to swim to the egg (mitochondria release energy for swimming; enzymes digest the egg membrane)
(a) Answer: e.g. tail/flagellum -> allows the cell to swim to the egg
(b) B1 large surface area (long projection) / thin cell wall / many mitochondria (for active transport of ions)
(b) Answer: Large surface area (long, thin projection into the soil)

Question 5

(a) B1 the process by which a cell becomes specialised (for a particular function/job)
(a) Answer: A cell becoming specialised for a particular function
(b) B1 early (in the development of the embryo) / at an early stage
(b) Answer: At an early stage (early development)
(c) B1 meristem (allow: meristem tissue)
(c) Answer: Meristem

Question 6

(a) B1 higher magnification OR higher resolution / resolving power (oe: can see smaller sub-cellular structures)
(a) Answer: Higher resolution (and magnification), so smaller structures can be seen
(b) B1 400 (from 10 x 40)
(b) Answer: 400

Question 7

(a) M1 48 / 0.06
(a) A1 800 (no units) cao
(a) Answer: 800
(b) M1 real size = image / magnification = 10 / 2000
(b) M1 0.005 mm converted to um (x 1000)
(b) A1 5 um cao
(b) Answer: 5 um

Question 8

(a) M1 uses scale: 20 mm represents 10 um, so 1 mm represents 0.5 um (oe, e.g. magnification = 20000/10 = 2000)
(a) M1 60 x 0.5 (oe: 60 mm = 60000 um, /2000)
(a) A1 30 um cao
(a) Answer: 30 um

Question 9

(a) B1 (net) movement of particles/molecules from a region of higher concentration to a region of lower concentration
(a) B1 down a concentration gradient (oe)
(a) Answer: Net movement of particles from higher to lower concentration, down a concentration gradient
(b) M1 surface area = 6 x (2 x 2) = 24 cm2 AND volume = 2 x 2 x 2 = 8 cm3
(b) A1 3 : 1 (from 24 : 8)
(b) Answer: 3 : 1
(c) B1 large organisms have a small(er) surface area to volume ratio
(c) B1 so diffusion (across the outer surface) is too slow / distances too great to supply all cells, so an exchange surface (with a large surface area) is needed
(c) Answer: Large organisms have a small SA:V ratio, so diffusion alone is too slow to meet their needs

Question 10

(a) B1 movement of water (molecules) from a dilute solution / higher water concentration to a more concentrated solution / lower water concentration
(a) B1 through a partially permeable membrane
(a) Answer: Movement of water from a dilute to a concentrated solution through a partially permeable membrane
(b) B1 mass increases
(b) B1 water moves into the cells by osmosis because the cell contents are more concentrated than pure water (oe)
(b) Answer: Mass increases; water enters the cells by osmosis (cell contents more concentrated than pure water)

Question 11

(a) B1 active transport moves substances against a concentration gradient (low to high) OR requires energy (from respiration); diffusion does not
(a) Answer: Active transport moves substances against the concentration gradient and requires energy
(b) B1 ions must move from a lower concentration (soil) to a higher concentration (cell), i.e. against the concentration gradient
(b) B1 this needs energy (from respiration), which diffusion cannot provide
(b) Answer: Ions move against the concentration gradient (low soil to high cell), which requires energy from respiration

Question 12

(a) B1 concentration of the sugar solution
(a) Answer: Concentration of the sugar solution
(b) B1 any one of: volume of solution / temperature / time left in solution / size (surface area) of potato cylinder / same type of potato / same drying method before reweighing
(b) B1 a second, different, correct control variable from the list
(b) Answer: Any two of: volume of solution, temperature, time, size of cylinder, type of potato, drying method
(c) M1 (2.72 - 3.20) / 3.20 x 100 (oe: change in mass = -0.48 g)
(c) A1 -15 (%) (allow 15% decrease)
(c) Answer: -15%
(d) B1 the sugar solution was more concentrated (lower water concentration) than the cell contents
(d) B1 so water moved out of the cells by osmosis (down the water concentration gradient), reducing the mass
(d) Answer: The solution was more concentrated than the cells, so water left the cells by osmosis

Question 13

(a) B1 add a drop of water (or iodine solution) to the middle of a clean glass slide
(a) B1 peel a thin layer of onion epidermis and place it flat on the drop (using forceps)
(a) B1 lower a cover slip onto the sample (using a mounting needle) to avoid trapping air bubbles
(a) Answer: Drop of water/iodine on slide; thin onion epidermis layer placed flat; cover slip lowered gently
(b) B1 it stains the cells / increases contrast so the structures can be seen more clearly
(b) Answer: It stains the cells and increases contrast so structures are easier to see
(c) B1 easier to find the cells / larger field of view / reduces risk of the lens hitting/breaking the slide
(c) Answer: Easier to locate the cells (wider field of view; less risk of hitting the slide)

Question 14

(a) B1 the DNA / chromosomes are replicated (copied) so there are two copies of each
(a) B1 the number of sub-cellular structures (e.g. ribosomes and mitochondria) increases
(a) Answer: DNA/chromosomes are copied; number of sub-cellular structures (ribosomes, mitochondria) increases
(b) B1 they are genetically identical (to the parent and to each other)
(b) Answer: They are genetically identical to the parent cell
(c) B1 growth / repair / replacement of damaged cells (allow: asexual reproduction)
(c) Answer: Growth, repair, or replacement of cells

Question 15

(a) M1 number of divisions = 120 / 20 = 6, so number = 800 x 2⁶ (oe)
(a) A1 51200 cao
(a) Answer: 51200
(b) B1 5.12 x 10⁴ (ft from part a)
(b) Answer: 5.12 x 10⁴

Question 16

Level 3 (5-6): A detailed and coherent evaluation that considers benefits AND objections, leading to a justified conclusion. The response uses relevant scientific ideas throughout and is logically sequenced.
Level 2 (3-4): Some relevant benefits and objections are given with some linking. There may be an attempt at a judgement. Scientific ideas are mostly correct.
Level 1 (1-2): One or more isolated relevant points (a benefit or an objection) with no development or judgement.
Level 0 (0): No relevant content.
Indicative content:
- Benefit: embryonic stem cells can differentiate into any cell type, so could replace faulty or damaged cells.
- Benefit: could treat conditions such as paralysis (nerve cells), type 1 diabetes (insulin-producing cells) or blindness.
- Benefit: using stem cells from the patient (therapeutic cloning) reduces the risk of rejection.
- Objection (ethical): the embryo is destroyed; some people believe an embryo has a right to life.
- Objection (religious/personal): some hold that using embryos in this way is wrong.
- Risk: transplanted stem cells could be infected with a virus and pass it on.
- Risk: uncontrolled cell division could lead to tumours/cancer.
- Practical: research is expensive and treatments are still being developed.
- Regulation: use of embryonic stem cells is controlled by law.
- A judgement: e.g. the potential to cure serious diseases can be weighed against the ethical cost and the risks.

Cell Biology - Worksheets, Questions and Revision