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Organisation of the Organism


5 questions, 38 marks, every one with its mark scheme, 1 of them Higher Tier or Extended only.

Or open them one at a time, as you finish each question.

Question 1

9 marks

A student uses a light microscope to look at cells from a leaf of a pondweed and cells from the lining of a human cheek.

1(a)[4 marks]

Complete the table to give the function of each structure.

StructureFunction
Nucleus
Cytoplasmwhere chemical reactions take place
Cell membrane
Mitochondria
Ribosomes
Mark scheme
  • contains DNA1
  • controls what goes into and out of the cell1
  • aerobic respiration1
  • protein synthesis1
1(b)[2 marks]

The pondweed cells have a cell wall but the cheek cells do not.

State two functions of the cell wall.

Mark scheme
  • provides support1
  • prevents the cell bursting1
1(c)[1 mark]

Name the substance that makes up the cell wall of a pondweed cell.

Mark scheme
  • cellulose1
1(d)[2 marks]

State two functions of the large vacuole in a pondweed cell.

Mark scheme
  • contains cell sap1
  • stores water1

Question 2

7 marks

Lactobacillus is a bacterium used to make yoghurt.

2(a)[1 mark]

Some other bacteria have a flagellum.

What does a bacterium use its flagellum for?

Mark scheme
  • movement1
2(b)[1 mark]

Lactobacillus cells contain plasmids.

State one use of plasmids by scientists.

Mark scheme
  • used in genetic modification1
2(c)[5 marks]

Compare the structure of a Lactobacillus cell with the structure of a palisade mesophyll cell.

Mark scheme

Any 5 from:

  • both have a cell wall1
  • both have a cell membrane1
  • both have cytoplasm1
  • both have ribosomes1
  • bacterium has circular DNA free in the cytoplasm1
  • bacterium has no nucleus1
  • bacterium has plasmids1
  • bacterium has no mitochondria1
  • palisade mesophyll cell has chloroplasts1
  • palisade mesophyll cell has a large permanent vacuole1
  • palisade mesophyll cell has a cellulose cell wall1

Question 3

11 marks

Animals and flowering plants contain many types of specialised cell.

3(a)[3 marks]

Complete the table to give the function of each specialised cell.

Specialised cellFunction
Ciliated cell in the trachea
Red blood cell
Neurone
Root hair cellabsorption
Mark scheme
  • move mucus1
  • transports oxygen1
  • conduction of electrical impulses1
3(b)[2 marks]

Define the term tissue.

Mark scheme
  • a group of cells with similar structures1
  • working together to perform a shared function1
3(c)[4 marks]

Complete the table to give the level of organisation of each structure.

StructureLevel of organisation
a layer of palisade mesophyll cells
a leaf
a sperm cell
the circulatory system
Mark scheme
  • tissue1
  • organ1
  • cell1
  • organ system1
3(d)[2 marks]

Define the term organ.

Mark scheme
  • a structure made up of a group of tissues1
  • working together to perform a specific function1

Question 4

5 marks

A student makes a drawing of a leaf from a moss plant.

4(a)[1 mark]

Write the equation that links magnification, image size and actual size.

Mark scheme
  • magnification = image size ÷ actual size1
4(b)[2 marks]

On the drawing the leaf measures 68 mm. The real leaf measures 4.0 mm.

Calculate the magnification of the drawing.

Mark scheme
  • 68 ÷ 4.01
  • ×171

Do not accept: ×17 mm

4(c)[2 marks]

The student also has a photograph of a dust mite. The length of the mite on the photograph is 45 mm. The magnification of the photograph is ×150.

Calculate the actual length of the dust mite.

Give the unit of your answer.

Mark scheme
  • 45 ÷ 1501
  • 0.3 mm1

Question 5

Extended6 marks

A student measures cells on photographs taken with a microscope.

5(a)[1 mark]

A cheek cell has an actual width of 60 µm.

Convert 60 µm into millimetres.

Mark scheme
  • 0.06 mm1
5(b)[2 marks]

The width of the same cheek cell on the photograph is 48 mm.

Calculate the magnification of the photograph.

Mark scheme
  • 48 ÷ 0.061
  • ×8001
5(c)[3 marks]

On another photograph, the image of a bacterial cell is 36 mm long. The magnification of this photograph is ×12 000.

Calculate the actual length of the bacterial cell in µm.

Mark scheme
  • 36 ÷ 12 000 = 0.003 mm1
  • 0.003 × 10001
  • 3 µm1