When you’re first exposed to nanotechnology, it can be difficult to grasp the immense difference in scale that nano-sized objects have. It’s helpful to think through the actual mechanics of how objects that size interact. For example, we use DNA linkers to bind DNP cubes together into useful structures. Just how strong are the bonds between DNA linkers? Imagine you’re playing a nano-sized version of the claw game, except instead of a mechanical claw, you’re going to lift objects with using DNA. It would look something like this:
How large of a DNP cube could you lift?
While there are many ways to bind nanoparticles together (e.g. covalent bonds, electrostatic attraction, hydrophilic/hydrophobic interactions, etc.), we typically use the stickiness of double stranded DNA. We do this by first painting single-stranded DNA on the cube's faces:
When two cubes get close, they will bind together to form double stranded DNA, if and only if they have complementary sequences:
At least a couple sources list the binding force of DNA as being in the piconewton (pN) range.1Let’s assume it takes a force F= 4 pN to break apart one of the short DNA strands holding the cubes together. If you have two DNA strands binding the cubes together, it should take 10 pN of force. If you have 3 DNA strands, it should take 15 pN. If 10 DNA strands, it should take 50 pN. The more DNA strands you have, the stronger the cubes will bind together.
How many DNA strands can you fit on the face of a cube? Clearly it depends on the size of the cube. Suppose you have cube of length L= 10 nanometer (nm). The area Acubeof a cube face with that length would be
Acube= L × L = 10 nm × 10 nm = 100 nm2
DNA is 2 nm wide, giving it an area of roughly2
ADNA= 2 nm × 2 nm = 4 nm2
Assuming you’re covering the surface of the cube completely, we should be able to fit roughly N = 25 DNA strands with a 4 nm2area onto a cube face with a 100 nm2area,
N= Acube/ ADNA= 100 nm2/ 4 nm2= 25 strands
Since DNA has a force of 4 pN per strand, the total force binding the cubes together will be
FDNA= F× N= (4 pN/strand) × (25 strands) = 100 pN
What if we make the cube larger? We can rearrange the equations above to get an equation for the DNA force that depends on the length of the cube,
FDNA= FL2/ ADNA
FDNAtells us the strength of the force that binds cubes together. In order to lift a cube, it needs to overcome it’s weight, i.e. the force of gravity. The force of gravity is given by the equations
Fgrav= m g,
where m is the mass and g= 9.8 m/s2is the acceleration of gravity. We can calculate the mass using the density of cube and its length.
m= ρ L3
We’ll assume the silver nanocube has the same density of silver, ρ= 10.5 g/cm3. Combining the two previous equations, we get
Fgrav= ρ g L3= (10.5 g/cm3) × (9.8 m/s2) × L3,
Notice that Fgrav~ L3, whereas FDNA~ L2. This means that as the cube grows in size, the force of gravity pulling the cube down grows faster than the force of the DNA binding pulling it up. Eventually, the weight of the cube will be too much, and the DNA strands will not be strong enough to lift it. We can see that on a plot of force versus cube length:
We see that the gravitational force acting on the cube (green) grows faster than the DNA binding force (orange). The forces intersect at a cube length L= 2.4 m. That’s an 8 foot long cube weighing 165 tons! Evidently, DNA molecules are very strong if you have enough of them.
I should point out that for a real silver cube, the largest size the DNA can lift is almost certainly smaller than the result computed here. Nanocube faces are very flat, whereas bulk silver is rough and bumpy. When you stack bumpy cubes on top of each other, the bumps that stick out will be in contact with the other cube, but the dimples won’t be in contact with the other cube. Since the bumps and dimples will be larger than the short DNA strands, the strands located in the dimples will never reach far enough to bind with the DNA on the other cube.
 The cross-sectional area of DNA is more accurately described as a circle, but for an order of magnitude estimation such as this, the difference between the area of a square and the area of a circle will not be significant.