DNA
DISTANT NUCLEOTIDE APOGEE
OBJECTIVE: This exercise provides students with a startling insight pertaining to the
linear quantity of DNA contained within a cell, and within their entire body. Students are
challenged to utilize their algebraic skills to calculate the length of DNA, as well as to
determine a 'compaction ratio' which should enable them to more fully appreciate the
complexity and organization of cells.
PURPOSE: To allow students to appreciate the concept and application of molecular
packing (supercoiling) in living cells. In addition, students are challenged to apply the
skills of dimensional analysis and exponential notation to a novel engineering problem.
MATERIALS: Each student team needs only one clear gelatin capsule (1" capsules,
rounded to 2 cm. for this exercise, will suffice), one spool of thread, and a pair of
scissors. A rod, thin enough to fit into the holes of the spool, may also facilitate the
exercise. In addition, a tape measure may be used to determine a 20 meter distance, if the
teacher has not already done so in advance.
PROCEDURE:
1. Each group measures and cuts a piece of thread 20 meters long. (Two pieces may be
used in the exercises to better simulate the double helix. Two different colors are
suggested for dramatic effect.)
2. The thread should then be coiled, wrapped, or condensed in some way to prepare for
insertion into the capsule.
3. After insertion, the students are to calculate or determine the following:
a) the length of the total DNA, uncoiled and laid end-to-end, within one human somatic
cell.
b) The length of the entire amount of DNA in their body (again, assume uncoiled and laid
out).
c) The closest astronomical object to that distance.
d) OPTIONAL: The total number of trips to the sun and back that this length of DNA
would make (total # of astronomical units).
e) compare the 'compaction ratio' of DNA in a cell nucleus to their compaction ratio of
thread in a 2 cm. capsule.
KEY DATA:
-one set of human chromosomes represents about 3 billion (3 x 109) base pairs.
-one base pair is approximately 0.34 nanometers (3.4 x 10-10 meters)
-the human body contains about 10 trillion (1 x 1013) nucleated cells. (actually, probably
between 1-100 trillion, so we will use 10 trillion as a conservative estimate)
-the distance from earth to the sun (1 astronomical unit) is about 1.5 x 108 kilometer (or
1.5 x 1011 meter)
-the volume of a nucleus is about 1 micrometer (um) cubed or 1 x 10-18 cubic meters.
QUESTIONS/CALCULATIONS (Show all work)
________1. What is the length (meters) of the DNA of one cell uncoiled and laid end to
end?
________2. What is the length (meters) of all the DNA in your body uncoiled and laid
end to end?
3. How does this length compare to the distance to the sun?
________4. How many times could this DNA reach to the sun?
5. What astronomical object is closest to this distance?
6. What is the 'compaction ratio' of your thread/capsule system? In other words, divide
your length of thread by the length of capsule. Also determine the length/volume
compaction ratio by dividing the length of string by the volume of the capsule (assume
2cm x 1cm x 1cm). Remember to express in meters and cubic meters.
7. What are the 'compaction ratios' of your DNA in cell nuclei? [Assume the volume of
the nuclei to approximate a cube of dimensions 1um/side. Thus, volume = (1 x 10-6
meters)3]
8. How do the above values compare mathematically?
9. At this point, it is obvious that compacting a thread-like molecule such as DNA to this
degree seems almost impossible. Therefore, it seems that our model or simulation must
have a flaw. What difference in the dimensions of DNA and your thread must exist? Do
you think this accounts for the wide variation in 'compaction ratio'?
10. When do you think DNA is most tightly compacted in a cell? Why might this be?
Explain.
11. List any mechanisms or structures which contribute to DNA packing (compaction).
Drawings may be helpful.
D.N.A. Solution
(numbers rounded for ease of explanation/calculation)
A) 3 x 109 base pairs x 0.34 x 10-9 meters/base pair = 1 x 100 meter = 1 meter
Diploidy, thus 1 meter x 2 = 2 meters total DNA length/cell
B) 2 meters/cell x (10 x 1012 cells) = 20 x 1012 meters = 2 x 1013 meters
Notice that 10 trillion cells is probably an underestimate of cell population of an adult
C) distance to sun (1 astronomical unit) = 1.5 x 108 kilometers = 1.5 x 1011 meters.
Dividing B by C--- 2 x 1013 meters/1.5 x 1011 meters = 1.3 x 102 A.U.’s (130 trips to
sun) or about 65 round trips.
D) Comparing DNA distance to other astronomical objects:
Average distance to Pluto = 63 A.U. (DNA could make a round trip to Pluto)
Average distance to Neptune = 29 A.U. = 2 round trips to this planet
Compaction Ratio Comparisons:
Length:length
20 meters thread into .02 meter capsule length = about a 1,000/1 length:length
compaction ratio. 2 meters DNA into a nucleus (assume large 1 micrometer long
nucleus) = 2,000,000/1 length: length ratio. Notice that cells accomplish over a 1,000
greater length compaction ratio.
Length:volume
If considering length/volume compaction (a better analysis): 20 meters thread into a
capsule volume of .02 x .01 x .01 meter or (2 x 10-6 m3). Cell manages to fit 2 meters
of DNA into approximately (1 x 10-6m)3 or 1 x 10-18 m3. The difference in
compaction ratios is on the order of 1013, or thirteen orders of magnitude. This
incredible difference appears impossible, considering how difficult it was to get 20
meters of string into the capsule. There must be a flaw in the simulation. Students
should be challenged to discover the weakness (flaw) in the simulation exercise. It
might help top remind them of the Watson/Crick/Franklin dimensions for DNA (a
labeled drawing or overhead should suffice). Notice that the width of DNA is about 2
nm (2 x 10-9 meter). The string is on the order of 0.1 mm or 10-4 meter (teachers can
direct students to examine this under a microscope to obtain the best estimate for the
string that they used). That great difference partially explains the discrepancy.