Cell Shape and Surface Area (Katie Brennan)
Investigating and Measuring Surface Area : Volume Ratios using Moldable Models of Cells.
Principle(s) Investigated
1. Cell size is limited by the need to maintain a sufficient surface area:volume ratio to ensure adequate supply of raw materials for the metabolic activity within (all materials needed by the cell enter across the plasma membrane).
2. Surface area (SA):volume ratios decrease as objects get larger.
3. A sphere is the most compact shape, and thus minimizes the SA:volume ratio. Therefore, SA:volume ratios increase as shapes deviate from the sphere.
4. Although a sphere is a good representation of a generic cell, many cells are non-spherical. Sometimes this is an adaptation to maximize SA:volume ratios. Cells that need to maximize their surface areas include cells with absorptive functions (eg intestinal epithelial cells) or cells that need lots of contact points with other cells (eg nerve cells and immune dendritic cells). Since the sphere is the most compact shape, any non-spherical cell has a larger SA:volume ratio than a sphere. Particular cell shape adaptations that maximixe SA:volume ratios include flattening (red blood cells), microvilli (intestinal epithelial cells), branched structures (neurons), and ruffles (immune dendritic cells).
California Content Standards
Focus on Life Sciences (7th grade)
5a. Students know plants and animals have levels of organization for structure and function, including cells, tissues, organs, organ systems, and the whole organism.
Investigation and Experimentation (7th grade)
7. Scientific progress is made by asking meaningful questions and conducting careful investigations. As a basis for understanding concepts and addressing content in the other three strands, students should develop their own questions and perform investigations. Students will:
a. Select and use appropriate tools and technology (including calculators, computers, balances, spring scales, microscopes, and binoculars) to perform tests, collect data, and display data.
c. Communicate the logical connection among hypotheses, science concepts, tests conducted, data collected, and conclusions drawn from the scientific evidence.
Biology (9th – 12th grades)
1a. Students know cells are enclosed within semipermeable membranes that regulate their interaction with their surroundings.
Investigation and Experimentation (high school)
1. Students will be able to:
a. Select and use appropriate tools and technology (such as spreadsheets and graphing calculators) to perform tests, collect data, analyze relationships, and display data.
l. Analyze situations and solve problems that require combining and applying concepts from more than one area of science.
Next Generation National Science Standards
Dimension 1: Scientific and Engineering Practices
1. Asking questions (for science) and defining problems (for engineering)
2. Developing and using models
3. Planning and carrying out investigations
4. Analyzing and interpreting data
5. Using mathematics and computational thinking
Dimension 2: Cross Cutting Concepts that have Common Applications Across Fields
1. Patterns
3. Scale, proportion, and quantity
4. Systems and system models
6. Structure and function
Dimension 3: Core Ideas in Four Disciplinary Areas
Life Sciences 1: From molecules to organisms: structures and processes
AP Biology Curriculum Framework: Big Ideas and Science Practices
Big Idea 2: Biological systems use free energy and molecular building blocks to grow, to reproduce, and to maintain dynamic homeostasis
Essential Knowledge 2.A.3b: Surface area-to-volume ratios affect a biological systems’s ability to obtain necessary resources or eliminate waste products.
Learning Objective 2.7 Students will be able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination
Science Practices
1.1 The student can create representations of models of natural or manmade phenomena and systems.
1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively.
2.1 The student can apply mathematical routines to quantities that describe natural phenomena.
5.1 The student can analyse data to identify patterns or relationships.
Materials:
play-doh or other similar moldable material
(need 2 oz (57g) per student/group)
a scale or balance
a large beaker of water
sharpie markers
weigh boats or other disposable container
(8” x 11” paper cut in quarters can work)
1-2 lbs of cheap granular material such as sand or salt, in a Ziploc bag.
recipe for homemade play-doh from www.fun.familyeducation.com:
1 cup water
1 tablespoon vegetable oil
1/2 cup salt
1 tablespoon cream of tartar
Saucepan
1 cup flour
a. Combine water, oil, salt, and cream of tartar, and food coloring in a saucepan and heat until warm.
b. Remove from heat and add flour.
c. Stir, then knead until smooth.
d. Will last several months if stored in airtight container.
Procedure:
- Start with a brief discussion of cells, and their typical size (about 50 um for a generic mammalian cell). Although some cells are larger, one limiting factor in cell size is the surface area:volume ratio (see if students know this or can figure this out). All materials needed by the cell enter by crossing the plasma membrane. A larger cell needs more materials, but as the SA:volume ratio is inversely proportional to size, the larger a cell is, the less surface area is available per unit volume.
We can consider the sphere as the prototypical cell shape. Review the formulae for
sphere surface area and volume.
SA = 4 p r2
V = 4/3 p r3
- Then derive the formula for surface area:volume ratio for a particular sphere size.
SA : V
4 p r2 : 4/3 p r3
3 p r2 : p r3
3 : r
(Another way to think about this, in terms of cells, is that for a given radius, a surface area of 9.4 r2 is supplying a volume of r3. When r is small, this relationship might be manageable, but when r is large, the volume really gets out of hand.)
(Yet another way to think about this is that uptake is proportional to the radius, but the amount of nutrients needed is proportional to the volume. (ref Charles Brokaw Caltech website))
3. Then bring out one 2 oz piece of play-doh, and roll it into a sphere for the students to see. It will have a radius of about 1 cm. So, it has a surface area of about:
SA = 4 p r2
SA = 9.4 cm2
It has a SA:vol ratio of:
3 : R or, 3cm-1
4. Then run through some calculations for SA:volume ratio for spheres of different sizes (radius lengths). Just looking at the numbers, have students sketch what they predict a graph of these relationships will look like. Then quickly graph the numbers using the teacher laptop (eg using Excel or Google Spreadsheet). Project the graph, and discuss how quickly volume outstrips surface area as the sphere size increases. Again, discuss the implications for a cell: if a cell is too big, the membrane will not be big enough for transport to supply all the metabolic functions within.
5. Discuss how some cells need more surface area than others, especially if their function requires them to contact a lot of other cells, or transport a lot of material across their cell membrane. Have students brainstorm about what kinds of cells might have such functions.
6. Brainstorm the options for cells to maximize their surface area:volume ratio. The 2 main options are (a) shrinking and (b) changing shape (since the sphere is the most compact shape, any deviation from spherical will increase the SA:volume ratio).
7. Discuss why shrinking is not really an option (there is a minimum size for the nucleus, and so even the smallest cell will have to be slightly larger than this). But lots of cells are non-spherical, and there are specialized structures for maximizing surface area.
8. Show a picture of a red blood celll (flattened disc)(see powerpoint link). Have students estimate how much this might increase the SA:volume ratio.
9. Take one 58g (2 oz) piece of play-doh and roll it into a ball for the students to see. Explain the activity. The radius of this sphere is about 1 cm. So the SA:volume ratio is 3/cm. Each group of 3 students will get 3 similar sized (2 oz) pieces of playdoh, and is asked to mould it into the following 3 shapes:
- the flattened (biconcave) disc similar to red blood cells
- a cube
- a shape of their design that maximizes surface area
10. Once the students have molded their cell shapes, discuss calculating the new SA:volume ratios. How do we calculate the volume? We don’t have to: it is the same as the starting volume. How do we calculate the SA? Very tricky calculation! Instead, we will employ an indirect method to measure surface area.
11. Instead of measuring the surface area, we will weigh the amount of material (sand, salt, etc) that will adhere to the surface. This will be proportional to the surface area.
12. Demonstrate using the sphere of play-doh. Briefly dip it in water (to enhance stickiness), and then put it in the Ziploc of sand, and coat it on all sides. Remove the sand-coated play-doh ball to the weigh boat, and weigh. In our hands, the weight of the playdoh ball increases from 57g to 63g, once coated. This means the surface area of this sphere can hold 6 g of sand. We already calculated the surface area of the sphere to be 9.4 cm2.
We can use the ratio of 9.4 cm2 (or 1.56667 cm2)
6g g
to calculate the surface areas of the new cell shapes.
13. Have the students repeat the dipping-in-water then coating-in-sand procedures to determine the weight of the sand that adheres to their shape. They must remember to subtract the weight of the playdoh in order to determine the weight of the adhered sand.
14. Then have the students calculate their surface area using the ratio above. Or, the students could just calculate the fold increase in surface area they achieved with their different shapes. To do this, students could use a data table like the following:
15. Then have the students calculate the fold-increase of surface area they achieved with their cell shape.
For example, a student whose coated shape weighs 70g would perform the following calculations:
1. weight of sand = total weight – weight of playdoh = 70g – 58g = 12g
(right away, the student may see that the weight of the adhered sand is
double, and infer that the surface area must have doubled)
2. ratio of surface area : weight of sand = 1.56667 cm2
1 g
so … 12 g represents a surface area of 18.8 cm2
3. The surface area of the sphere was 9.4 cm2. 18.8/9.4 = 2.
The surface area has doubled.
16. If time and technology permit, have the students use sharpies to write the fold increase in surface area they achieved (eg 1.6x, 2x, 3.1x) on the weigh boat with the coated playdoh shape, and take a photo and email it to the teacher’s picasa website. Then a quick comparison of shapes and surface area effects will be possible (with all of them projected on the screen).
17. Finish with a recap of some of the effects that were seen. Reiterate how cell size is limited by the ability of the plasma membrane area to support the volume of the cell with needed metabolites.
Introduce another consideration of cell size (and shape), which is the need for materials that cross the plasma membrane to diffuse rapidly to all parts of the cell. Since diffusion time increases with the square of the diffusion distance (radius), doubling the cell size will make cause the time required for diffusion from the outside to the center increase fourfold. (etc) (Is this a better explanation for flattened shape of RBCs?)
Show some more photos of cells with shapes that maximize surface area (in addition to RBC: neuron (axonal projections and branching arborization to maximize contact with other cells)(see again attached powerpoint), intestinal epithelial cells (microvilli to maximixe surface for absorption and contact with beneficial microbes), immune dendritic cell (membrane folds to increase contact with T cells), and even photoreceptor neurons (rods and cones with extensions and internal membrane stacks to maximize pigments and light absorption).
Student Prior Knowledge:
That cells are the unit of life.
A level of comfort with geometry of spheres.
Simple algebra.
Questions:
Q1. When you think of a cell that has a greatly enlarged surface area, what modifications to its metabolism would you expect to find?
Q2. The strategies a dendritic cell and a red blood cell use to increase SA:vol ratio are very different. Why do you think this might be?
Q3. Cells could also increase their SA:volume ratio by shrinking. Why do we not see many examples of this?
... and Answers:
A1. Since the plasma membrane is largely composed of lipids, one would expect the lipid biosynthetic capacities of such cells to be greatly increased.
A2. A red blood cell must be able to flow smoothly through small blood vessels, such as capillaries. Therefore, it must maintain a smooth exterior, and the flattened biconcave disc is a compromise that increases surface area without increasing "spikiness". The dendritic cell has no such limitations, and thus uses lots of ruffles to increase SA.
A3. The lower limit to the size of a eukaryotic cell is probably the need to fit all the components (nucleus, other organelles) inside. Quiescent leukocytes (white blood cells) are not much bigger that the nucleus itself.
Applications to Everyday Life:
1. See the powerpoint for images of cells that have different reasons for needing large surface areas, and the different ways in which they have expanded their surface area.
2. Pharmacologists who design timed release capsules experiment with different shapes of the particles to achieve the desired rate of release.
Photographs:
1. Supplies needed: scale, weigh boats, bag of sand, sharpie markers, play-doh predivided into equal-sized servings (here: 2 oz). Not shown: beakers of water for dipping playdoh shapes in (to help sand stick).
2. Weighing the sand that sticks to the sphere: The playdoh alone weighs 57g. With the sand, it weighs 63g. Therefore 6g is the amount of sand that sticks to the sphere.
3. Some different shapes that can be made by the students.
4. The different shapes coated with sand.
5. Weighing the sand that sticks to the different shapes. Don't forget to subtract the weight of the playdoh itself.
6. Showing the fold increase in surface area (compared to the sphere) for the different shapes. You can see that it is possible to more than double the surface area (and therefore the SA:vol ratio). What is the biggest increase the students can achieve?!
Acknowledgements:
I developed the concept for this demonstration myself. It was modified on the recommendations of my classmates in SED525f12, including adding a table to scaffold the calculations needed to determine the fold increase in surface area.