Animal and plant cells

AQA·Biology (Combined Science)·4.1.1.2·Trilogy — Foundation

Plant and animal cell comparison

Plant and animal cells share several structures that carry out the same basic activities. A typical plant cell also has structures that support it and enable photosynthesis.

Structures shared by plant and animal cells

  • The cell membrane controls the movement of substances into and out of the cell.
  • The cytoplasm is where many chemical reactions occur.
  • The nucleus contains genetic material and controls the activities of the cell.
  • Ribosomes make proteins.
  • Mitochondria are the sites of aerobic respiration, which releases energy for the cell.

Structures of a typical plant cell

A typical plant cell has three additional structures:

  • The cell wall lies outside the cell membrane. It supports the cell and helps maintain its shape.
  • Chloroplasts absorb light for photosynthesis.
  • A large vacuole contains cell sap and helps keep the cell firm.

Animal cells do not have a cell wall or chloroplasts. Not every plant cell contains chloroplasts, but in diagrams and images, the combination of a cell wall, chloroplasts and a large vacuole identifies a typical plant cell. The shared structures can be identified in both plant and animal cells.

Bacterial cell structure

A bacterial cell has an outer cell wall surrounding a cell membrane. Inside the membrane are the cytoplasm, ribosomes and genetic material. Unlike a cell with a nucleus, a bacterial cell has its main DNA in the cytoplasm, not enclosed inside a nucleus.

Bacterial cell structures and their functions

  • Cell wall: supports and protects the cell. In a diagram, it forms the outer layer around the cell membrane.
  • Cell membrane: controls the movement of substances into and out of the cell. It lies just inside the cell wall.
  • Cytoplasm: the material filling the inside of the cell, where many chemical reactions occur.
  • Ribosomes: small structures in the cytoplasm that make proteins. They are often shown as dots in diagrams.

Bacterial DNA and plasmids

The bacterium’s main genetic material is circular DNA in the cytoplasm. In a diagram, it is often drawn as a large loop, which may appear tangled or folded.

Some bacterial cells also contain plasmids: smaller circles of DNA separate from the main circular DNA. In a diagram, these can be recognised as small loops in the cytoplasm. Plasmids are not present in every bacterial cell.

Bacterial plasmids

A plasmid is a small, circular molecule of DNA found in some bacteria. It is separate from the bacterium’s main chromosome and carries additional genes.

A plasmid can replicate independently of the main chromosome: it can make copies of its DNA without the chromosome having to be copied at the same time.

Plasmids can also be transferred between bacteria. When a bacterium receives a plasmid, it receives the genes carried by that plasmid. These genes may give the bacterium an advantage in particular conditions, so transferring a plasmid can pass that advantage between cells.

Cellulose cell walls

Plant and algal cells have a cell wall outside the cell membrane. The wall forms a strong outer layer that supports the cell.

The wall is made of cellulose, which forms strong fibres. These fibres reinforce the wall and help it resist stretching. This strengthens the cell and helps it maintain its shape.

Estimating sub-cellular size and area

An image can be used to estimate the size or area of a sub-cellular structure when an exact measurement is unnecessary or its boundary is difficult to identify. The estimate starts with a measurement on the image, then uses a scale bar or known magnification to relate it to the actual structure.

Estimating length from an image

A scale bar represents a stated actual length. Measure both the structure and the scale bar on the image using the same units. Then use their ratio to find the structure’s actual length:

Actual length=measured image length of structuremeasured image length of scale bar×actual length represented by scale bar\text{Actual length} = \frac{\text{measured image length of structure}}{\text{measured image length of scale bar}} \times \text{actual length represented by scale bar}

For example, suppose a scale bar labelled $$2\ \mu\text{m}$$ measures $$10\ \text{mm}$$ on the image. A structure measuring $$15\ \text{mm}$$ is $$15/10$$ times the length of the bar, so its estimated actual length is $$3\ \mu\text{m}$$.

If the magnification is known, the actual length can instead be found by dividing the image length by the magnification:

Actual length=image lengthmagnification\text{Actual length} = \frac{\text{image length}}{\text{magnification}}

Use consistent units for image and actual lengths when working with magnification. For example, $$1\ \text{mm} = 1000\ \mu\text{m}$$. The stated magnification must apply to the image as measured; resizing an image changes its magnification.

Comparing relative size

Relative linear size compares lengths. If one structure measures $$15\ \text{mm}$$ and another measures $$10\ \text{mm}$$ on images at the same scale, their length ratio is $$15:10$$, or $$3:2$$. The first is therefore estimated to be $$1.5$$ times as long as the second.

Estimating and comparing area

For a structure with an irregular outline, its area can be approximated in either of two ways:

  • Simple shape: approximate the outline with a shape such as a rectangle, measure its dimensions and calculate its area.
  • Grid squares: count the whole squares within the outline and combine partial squares approximately. Multiply the estimated number of squares by the area of one square.

To compare relative areas, use areas measured on the same scale. For example, if two outlines cover approximately $$20$$ and $$5$$ equal grid squares, their estimated area ratio is $$20:5$$, or $$4:1$$.

When converting an image area to an actual area, the linear scale factor must be squared. With a known magnification:

Actual area=image area(magnification)2\text{Actual area} = \frac{\text{image area}}{(\text{magnification})^2}

The same principle applies when using a scale bar. In the earlier example, $$10\ \text{mm}$$ on the image represents $$2\ \mu\text{m}$$. If an outline has an estimated image area of $$50\ \text{mm}^2$$, its actual area is:

50 mm2×(2 μm10 mm)2=2 μm250\ \text{mm}^2 \times \left(\frac{2\ \mu\text{m}}{10\ \text{mm}}\right)^2 = 2\ \mu\text{m}^2

All these results are estimates. Their accuracy depends on image quality, how closely the chosen shape or grid follows the outline, and which section of the structure is visible. A visible section may not show the structure’s full size.