The smaller the cell the greater its surface-to-volume ratio?

The smaller the cell the greater its surface-to-volume ratio?

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I read in a book that the smaller the cell the greater its surface-to-volume ratio (the surface area of a cell compared to its volume).

And larger cells have limited surface area compared to its volume.

How is this possible? I don't understand.

This is a basic property of geometry: for any given shape, area will be generally proportional to the square of the length, and volume will be proportional to its cube. (this isn't true for all shapes by any means; the "generally" refers to the principle that length, area and volume are 1, 2 and 3-dimensional properties respectively, not that all or even most shapes behave this way… although a lot of the shapes that happen to be common in the Universe do. See end of the answer).

For example, if a sphere has a radius of R, its surface will have an area of 4*Pi*R², and its interior will have a volume of 4/3*Pi*R³. This means its surface-to-volume ratio will be 3/R, which is inversely proportional to R, meaning when R increases it will decrease and vice-versa.

Like so:

egin{array} {|l|l|} hline R~(sim length) & R^2~(sim surface~area) & R^3~(sim volume) & 1/R~(sim surface~area / volume) hline 1 & 1 & 1 & 1~(= 1/1) hline 2 & 4 & 8 & 1/2~(= 4/8) hline 3 & 9 & 27 & 1/3~(=9/27) hline 4 & 16 & 64 & 1/4~(=16/64) hline 5 & 25 & 125 & 1/5~(=25/125) hline end{array}

Put another way, if your shape is growing in 3D then whenever you add a bit of length you'll tend to add much more area (because area is increasing along two directions of length) but you'll be adding even more volume (because volume is increasing along three directions of length), and more volume than area means the area-to-volume ratio becomes smaller.

It helps that cells are convex and can often be approximated as spheres, so the rule absolutely applies to them. It's different with squiggly shapes, or cases where you aren't growing along every axis (for example this principle is true if you increase a cylinder's radius and length, but not if you just increase its length; at that point length, surface area and volume are all increasing along a single dimension and thus all at the same rate; check the equations to see this). You often do find such squiggly, long or thin shapes in nature, and this is often an adaptation to get around the surface-to-volume decrease issue. Examples include your intestines, or the interface between the placenta and the uterus.

The answer of Rozenn Keribin and Gerardo Furtado already explains the basic principle very well. I'd like to add another aspect, though. The effect of the increased surface to volume ratio also comes into play in the case off nanotechnology where small particles are compared to bulk material.

Consider the following example, where a cube with edge length 1 m is devided into smaller cubes with edge length s:

egin{array} {c c c} hline edge length ~s & number~of~particles & surface~area ~A (m²) & total~volume hline 1~m & 10^0 & 6*10^0 & 1~m^3 1~cm & 10^6 & 6*10^3 & 1~m^3 1~mm & 10^9 & 6*10^5 & 1~m^3 1~µm & 10^{18} & 6*10^{11} & 1~m^3 1~nm & 10^{27} & 6*10^{17} & 1~m^3 hline end{array}

As you can see, the total volume of all particles is always 1 m³ since you only cut the original cube of 1x1x1 m into smaller pieces. However, this example highlights that you have a much larger total surface area for all particles if they are smaller when compared to larger particles.
This means that the effective surface area of the obtained particles increases with the inverse of the square of the size while the total mass (and volume) stays constant. The surface of the particles is proportional to their (bio-)activity.

Let's approximate a cell by the simplest possible shape, a cube. Let's say the sides of the cube are 1mm (that'd be a large cell… ). It's volume is obviously 1 mm3, and since it has 6 sides, a surface area of 6 mm2. That gives us a surface to volume ratio of 6. Now let's cut this cube in half. Now every part has a volume of 0.5 mm3. But what happens to the surface area? Four of the six sides are cut in half, and we keep one of the original side. But we also get a new side from where the cut took place. So our new area is 1+1+4*0.5 = 4 mm2. This gives us a surface to volume ratio of 4/0.5 = 8, which is greater than the original 6.

Others have mentioned cubes, and the principle is the same, but I prefer to think in terms of spheres.

The volume of a sphere is $frac43 pi r^3$

The surface area of a sphere is $4 pi r^2$

The surface-volume ratio is thus $frac3{r}$.

You can plot the function $frac3{r}$ (see here), or just know that this is an inverse relationship -- as $r$ goes up, the ratio goes down; as $r$ goes down, the ratio goes up.

This is a different way of saying -- a smaller object (which has smaller $r$) has a greater surface-volume ratio.

The smaller the cell the greater its surface-to-volume ratio? - Biology

1.) Cell growth depends on nutrient uptake and waste disposal. You might imagine, therefore, that the rate of movement of nutrients and waste production across the cell membrane would be an important determinant of the rate of cell growth. Is there a correlation between a cell’s growth rate and its surface-to-volume ratio? Assuming that the cells are spheres, compare a bacterium (radius of 1 μm), which divides every 20 minutes, with a human cell (radius 10 μm), which divides every 24 hours. Is there a match between the surface-to-volume ratios and the doubling times for these cells? Explain why your results make sense.

There are theoretical upper and lower limits on cell size. This is primarily due to the metabolic requirements needed to sustain the cell and maintain the greatest efficiency. When an object increases in size, its volume grows proportionately more than its surface area. Thus, the smaller the object, the greater its ratio of surface area to volume. The cells with radius of 1 micrometer have a surface area to volume ratio of 3 to 1 (surface area 3 pi while volume is (3pi)/4) while the cells with the radius of 10 micrometers have a surface area to volume ratio of 3 to 10 (surface area 400pi while volume is (4000pi)/3). As can be seen here, there is an inverse relationship because of the aforementioned rule: the smaller the object, the greater its ratio of surface area to volume.

2.) An adult human is composed of about 10 13 cells, all of which are derived by cell divisions from a single fertilized egg.

a. Assuming that all cells continue to divide (like bacteria in rich media), how many generations of cell divisions would be required to produce 10 13 cells?

Assuming that all cells continue to divide, it would require more than log2(10 13 ) divisions, which becomes 43.19, which, rounding up would give 44 cell divisions

b. Human cells in culture divide about once per day. Assume all cells continue to divide at this rate during development, how long would it take to generate and adult organism?

It would take 44 days, rounded up, but to be more specific, would give one 43 days, 4 hours, 33 minutes, and 36 seconds.

c. Why is it, do you think, that adult humans take longer to develop than these calculations might suggest?

A variety of reasons exist for adult humans taking longer to develop than these calculations suggest. For one, there are many different types of cells in the human body, each with its own size and function. These different types of cells need different requirements to function. Also, unlike in a culture with bacteria in rich media, humans do not have an infinite supply of food while developing. One more significant reason exists and that is cells also die as the body needs to recycle cells that have gotten old or have already performed the necessary functions. So as cells are being made, some are also dying, taking the adult human much longer to develop than one might assume.

3.) Why do eukaryotic cells require a nucleus as a separate compartment when prokaryotic cells manage perfectly well without?

Eukaryotic cells are vastly complicated, much more so than their prokaryotic counterparts. In a prokaryotic cell, the DNA is located in a dense region in the cytoplasm known as the nucleoid. The corresponding location of DNA in a eukaryotic cell is in the nucleus. Because the eukaryotic cell is much more complex, it increases its efficiency by compartmentalizing its organelles. This allows the cell to perform its various functions simultaneously. Therefore, the existence of a nucleus and its cellular function devoted to just the diverse work of DNA (DNA replication, transcription etc.), increases cellular efficiency. Then, the eukaryotic cell can transcript DNA into pre-mRNA, a primary transcript, which can then undergo RNA processing and then translation. Prokaryotic cells do not need a nucleus because not only do they have the ability to perform transcription and translation concurrently, but they also lack introns (something that eukaryotic mRNA does have and hence, another reason why the nucleus is useful).

4.) What is the fate of a protein with no sorting signal?

A protein with no sorting signal does not leave the cell with much choice. Basically, what a protein with no sorting signal means to the cell, is that the cell does not know where the protein should go and cannot guide it to where it DOES need to go. Hence, they may accumulate in one place (cytoplasm where free ribosomes produce them or rough ER), in which case the cell might elect to destroy these proteins since they are not particularly useful while their amino acids themselves could be.

5.) ) Is it really true that all human cells contain the same basic set of membrane enclosed organelles? Do you know of any examples of human cells that do not have a complete set of organelles?

Not all human cells contain the same basic set of membrane enclosed organelles. While it is true that most do, there are some that do not have the same set of organelles. A prime example of this can be seen in erythrocytes, or red blood cells. They do not have a nucleus (and hence do not have DNA) or mitochondria to facilitate the production of ATP. The reason some cells do not have these common organelles is that it would impede their function instead of helping. It is also possible for human cells to have more than one nucleus (polynucleated) and a chief example of this is skeletal muscle cells, called myocytes, which become polynucleated during development. This shows one that there are different variations to every rule.

Random Fact: China surpasses Germany to become the world's third-largest economy, based on revised 2007 GDP figures.

Many reactions occur within the cell. Substances need to be taken into the cell to fuel these reactions and the wast products of the reactions need to be removed. When the cell increases in size so does its chemical activity. This means that more substances need to be taken in and more need to be removed. The surface area of the cell is vital for this. Surface area affects the rate at which particles can enter and exit the cell (the amount of substances that it takes up from the environment and excretes into the environment), whereas the volume affects the rate at which material are made or used within the cell, hence the chemical activity per unit of time.

As the volume of the cell increases so does the surface area however not to the same extent. When the cell gets bigger its surface area to volume ratio gets smaller. To illustrate this we can use three different cubes. The first cube has a side of 1 cm, the second 3 cm and the third 4 cm. If we calculate the surface area to volume ratio we get:

Cube 1
Surface area: 6 sides x 12 = 6 cm2
Volume: 13 = 1 cm3
Ratio = 6:1

Cube 2
Surface area: 6 sides x 32 = 54 cm2
Volume: 33 = 27 cm3
Ratio = 2:1

Cube 3
Surface area: 6 sides x 42 = 96 cm2
Volume : 43 = 64 cm3
Ratio = 1.5:1

As we can see the cube with the largest surface area and volume has the smallest surface area to volume ratio. If the surface area to volume ratio gets too small then substances won’t be able to enter the cell fast enough to fuel the reactions and wast products will start to accumulate within the cell as they will be produced faster than they can be excreted. In addition, cells will not be able to lose heat fast enough and so may overheat. Therefor the surface area to volume ratio is very important for a cell.


- Substances need to be taken in to the cell to fuel reactions & waste products need to be removed
- Increase in cell size leads to increase in chemical reactions --> more substances needed in and more substances needing to be removed
- Surface area affects the rate at which particles enter and exit the cell
- Volume affects the rate of the chemical activities
- When the volume increases so does the surface area but not to the same extent
- As the cell gets larger, its surface area to volume ratio gets smaller
- If the ratio gets too small, particles will not be able to enter and exit the cell fast enough
- Results in accumulation of waste products and overheating of the cell

AP Biology Question 458: Answer and Explanation

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Question: 458

3. Most animal cells, regardless of species, are relatively small and about the same size. Relative to larger cells, why is this?

  • A. Smaller cells avoid excessive osmosis and subsequent lysis.
  • B. Smaller cells have a smaller surface-to-volume ratio.
  • C. Smaller cells have a larger surface-to-volume ratio.
  • D. Smaller cells fit together more tightly.

Correct Answer: C


C. It is important that cells have a large surface area relative to their volume in order to maximize the ability of cells to import necessary nutrients and export wastes. The greater the surface area is, the greater the area for carrying out these processes. A large cell may not have enough surface area to accommodate the transport needs of the cell. Another limitation to cell size is the genome-to-volume ratio. The genome of a cell (the cell’s genetic material or chromosomes) remains fixed in size and fixed in its ability to control the activity of the cell (by producing RNA, which in turn produces proteins). The genome may not be able to accommodate the protein (and enzyme) needs of a large cell.

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Why Are Cells so Small?

Brooklyn College explains that cells are small because they must have a large surface area relative to the amount of volume they contain to function properly. As a sphere grows larger, its volume increases much more rapidly than its surface area does. This presents logistical problems for the cell, as it tries to transport resources and products through a large volume without the resources available via a large surface.

As an example, a typical animal cell requires oxygen to survive. The size of the cell partially dictates the amount of oxygen it needs, while the surface area of the cell limits the amount of oxygen that can be absorbed at a time. Accordingly, when the size of a cell grows, its demand for oxygen and other resources rises at a rapid rate, while its capacity for absorbing oxygen increases more slowly. At some point, the size of the cell will cause the cell to divide or die, according to Brooklyn College.

Despite the limitations on cell size that are imposed by the surface-to-volume ratio, a 2013 study published in the journal Nature Cell Biology demonstrates that gravity also limits cell size. While gravity is a negligible force at very small scales, cells that attain about 1 millimeter in diameter must include structural elements to keep some organelles stable under the influence of gravity. Without such elements, cell components can lose their structural integrity.

The smaller the cell the greater its surface-to-volume ratio? - Biology

Surface are to volume ratio is an important concept that you need to understand.
Essentially, it is area of an object that is exposed to the external environment (surface area), compared to the amount within an object (volume).

Therefore an elephant has a lower surface area to volume ratio than a mouse.

The smaller an object is the greater its surface area to volume ratio.

Surface area = Length x height x number of sides

Volume = Length x height x width

SA:V = Surface area ÷ Volume

This cube has a surface area to volume ratio of 6

This cube has a surface area to volume ratio of 1.5

A high surface area to volume ratio, allows objects to diffuse nutrients and heat at a high rate.

You will often see small mammals shirving constantly, because they are quickly loosing body heat to the enviroment and need to generate more heat to survive.

This is also why we have over 1,000,000 small cells, instead of 5 large cells.

As you can see, the large an object becomes, the smaller its surface area to volume ratio is.

The video to the left uses sugar cubes and sulphuric acid.

Observe the amount of time for the reaction to take place.

Now observe the video to the right.

In this experiment, we are using sugar powder instead of cubes.

Notice how much faster the reaction occurs. This is due to the increase in surface area.

Why is surface area to volume ratio bigger in smaller animals and vice versa?

In surface area to volume ratio (SA:Vol) we are comparing the outside surface area of an object to its internal volume. So imagine 3 rubber balls, one's small, one's medium, one's large. The smallest ball has little internal content (volume) compared to it's outside surface area. Therefore it has a big surface area compared to it's volume, so a high SA:Vol. In the medium ball, compared to the smaller ball, the Surface area hasn't increased hugely, but the internal content (volume) will be much much greater, this means that surface area to volume ratio is reduced. In the biggest ball again, the surface area covering the ball hasn't increased significantly but the internal contents have increased greatly bringing the surface area to volume ratio down (making it smaller) because the internal volume is far greater than the outside surface area. If we apply that to animals, things like polar bears which have a great deal of central mass and very few appendages (eg. they don't have large ears)- these animals will have small surface area to volume ratio because their internal volume is much greater than their outside surface area. Animals such as the fennec fox have very large ears, and are very lean (not fat), the big ears increase the surface area and the leanness reduces the volume of the animal, increasing the surface area to volume ratio. In general, remember: Small things have big SA:Vol. Big things have small SA:Vol.

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The smaller the cell the greater its surface-to-volume ratio? - Biology


These exercises are designed to introduce you to the concept of surface-to-volume ratios (S/V) and their importance in biology. S/V ratio refers to the amount of surface a structure has relative to its size. Or stated in a more gruesome manner, the amount of "skin" compared to the amount of "guts". To calculate the S/V ratio, simply divide the surface area by the volume.

EXERCISE 1. INFLUENCE OF SIZE ON S/V RATIOS. We will use a cube to serve as a model cell (or organism). Cubes are especially nice because surface area (length x width x number of sides) and volume (length x width x height) calculations are easy to perform. To calculate the surface-to-volume ratio divide the surface area by the volume. Complete the table below for a series of cubes of varying size:

EXERCISE 2. SHAPE AND S/V RATIOS: In this exercise we will explore the impact of shape on surface to volume ratios. The three shapes given below have approximately the same volume. For each, calculate the volume, surface area and S/V ratio and complete the table.

Shape Dimensions (mm) Volume (mm 3 ) Surface Area (mm 2 ) S/V ratio Volume of environment within 1.0 mm
Sphere 1.2 diameter
Cube 1 x 1 x1
Filament 0.1 x 0.1 x 100

note: volume of a sphere = 4/3 ( p )r 3 = 4.189r 3 surface area of sphere = 4( p )r 2 = 12.57 r 2

EXERCISE 3. S/V RATIOS IN FLATTENED OBJECTS: In this exercise we will explore how flattening an object impacts the surface to volume ratio. Consider a box that is 8 x 8 x 8 mm on a side. Then, imagine that we can flatten the box making it thinner and thinner while maintaining the original volume. What will happen to the surface area, and s/v ratio as the box is flattened? Complete the table below.

EXERCISE 4. S/V RATIOS IN ELONGATED OBJECTS: In this exercise we will explore how elongating an object impacts the surface to volume ratio. Consider a box that is 8 x 8 x 8 mm on a side. Then, imagine that we pull on the ends to make it longer and longer while maintaining the original volume. What will happen to the surface area, and s/v ratio as the box is flattened? Complete the table below.

EXERCISE 5. WHY ARE CELLS SMALL? The typical eukaryotic cell is rather small - approximately 100 m m in diameter. This exercise is designed to help provide an explanation why cells are not normally larger.

Obtain 2 cell models, one small and one large. Measure the length and diameter of each and then record your data in the table below. Place each cell in a bowl containing clear vinegar (BE CAREFUL!). Allow to sit for a few minutes, or until most of the blue color is gone from the smallest cell. Remove the models with a plastic spoon (CAUTION: don't get the vinegar on your hands. ) and place it on a piece of paper towel. Then, measure the size of the colored area remaining and record these data in the table below. Complete the calculations.

Theory: The cell models are made of a gelatin-like material called agar. The agar has an acid-sensitive dye incorporated into it. The dye turns from blue to yellow (clear in the presence of acid). The uptake of acid, and hence colorless areas of the cell models represents the uptake of food/nutrients by the cell. From this, we can calculate the percent of each cell that was fed during the incubation period.

volume of a cylinder = p r 2 l where p = 3.14
surface area of cylinder = 2 p r 2 + 2 p rh
percent cell fed = (initial volume - final volume)/initial volume x 100

EXERCISE 6. WHY DO MICE HAVE GREATER METABOLIC RATES THAN ELEPHANTS? - It is well known that there is an inverse relationship between body size and metabolic rate. The purpose of this exercise is to determine the reason for this relationship.

Heat content (joules) = temp. (C) x vol. (cm 3 ) x specific heat of water (4.2 joules/cm 3 C)

Calculate the total heat loss during the two minute period (row 11) by subtracting the final heat content (row 10) from the initial heat content (row 8).

How does surface area to volume ratio affect gas exchange?

The surface area to volume ratio of a cell must be such that the cell membrane has enough surface area to adequately serve the internal contents (volume) of the cell, including the adequate exchange of gases. As the cell grows, its surface area to volume ratio decreases, reducing the rate of gas exchange. This is because as the surface to volume ratio decreases, there is not enough surface area (cell membrane) for adequate gas exchange to occur in order to serve the needs of the internal contents (volume) of the cell. If the surface area to volume ratio of a cell gets too small, the cell must either undergo mitotic cell division or it will no longer be able to function.

In terms of why volume increases faster than surface area as the cell grows, you can think of it in multiple ways. One is to simply realize that volume is increasing in three dimensions whereas surface area is increasing in only two. You can also just look at calculations of the two. Here I'll pretend the cell is a cube it's not, but it makes the numbers simpler to grasp and the same principles apply with other shapes.

So you can see, as the cell size increases, the surface area/volume ratio decreases. The cell needs this ratio to be high to ensure it can perform gas exchange at the necessary rate.