Microbiomechanics 442 Professor James McGrath
spring/2015
Prepared by Jon Strauss
Problem 1. Howard, Problem 2.3. During mitosis, the chromosomes move several micrometers over the
course of about 30 minutes. Calculate the average speed. If the viscosity of the cytoplasm is 1000 times that
of water, estimate the required force.
I am, in this preparation going to take Professor Howards exercise a step farther by
assessing work done by the microtubules during mitosis.
As we know,
work = force x distance.
.Force = ma
But Professor Howard points out that “The net force is a sum of
individual forces.” Given the crowded nature of the cytoplasm it would
be easy to conclude that more than just viscosity is at work. Certainly
one can envision a collision force or two involved as our chromosome
bumps a myriad of cytoplasmic proteins heading for polar ends of the
cell. For the sake of simplicity though, now we will work with just
viscosity forces and consider the significance of other forces on the
return trip.
According to Newton’s second law chromosomes acted upon by the
force of motor proteins kinesin and dynein along microtubule tracks
should accelerate, according to the above F = ma, unless opposing
forces act to interfere. Therefore we may conclude that the movement of
mitosis at a speed of 14 𝜇𝑚/ℎ𝑜𝑢𝑟 derived below, is not accelerating due to the
net quantity of opposing forces, which leaves the cell with a restrained velocity. Still,
if we are to say that F=ma is only part of the equation, then we must add the
counteracting force of viscous drag as follows:
The Drag coefficient (𝛾) 𝑜𝑟 𝑔𝑎𝑚𝑚𝑎, 𝑖𝑠 𝑠𝑖𝑚𝑝𝑙𝑦 = 6πμr, according to Dr. Stokes
so that the total Force related to drag is the negative product of the drag coefficient
and the velocity.
𝐹𝑑 = −(𝛾)(𝑉)
But this is the negative product as drag force acts against the primary force enacted
by the microtubules attempting to accelerate the chromosomes leaving us with a
remaining force of
𝐹𝑚𝑎 + 𝐹𝑑 = 𝐹𝑛𝑒𝑡
The net force, 𝐹𝑛𝑒𝑡 , is not the force we see but rather the force used to create net
movement. Interestingly it has been suggested that the drag force is equivalent to
the work force (“Regulation of Chromosome Speeds in Mitosis”, by MD Betterton,
Cel. Mol. Bioeng. [2013]6:418-430). The idea has as its basis that, since there is no
acceleration there is no net force and so drag force and microtubule force cancel
each other out. My objection here is emanates from the fact that, the movement of
the chromosomes does work, and so there must be a net force, above that of the
drag force, which I am suggesting as the above equation.
The question becomes, how does one calculate 𝐹𝑚𝑎 , from the acceleration, which
did not occur?
𝑚
𝑚
Acceleration in 𝑠2 = velocity in 𝑠 divided by time in seconds. If we were to look at the
movement of the chromosomes as a function of time it would be linear. Looking at
this movement as a function on time squared would be exponential.
SO, I am going to assume that if the velocity is ≅ 3.89 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠/ second then
acceleration will continue along those lines or ≅ 3.89 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠/ 𝑠𝑒𝑐𝑜𝑛𝑑 2 .
The mass of a given chromosome will differ. Generally yeast chromosome mass is
recognized in the 108 𝑑𝑎𝑙𝑡𝑜𝑛 𝑜𝑟 𝑎𝑡𝑜𝑚𝑖𝑐 𝑢𝑛𝑖𝑡 𝑟𝑎𝑛𝑔𝑒/𝑐ℎ𝑟𝑜𝑚𝑜𝑠𝑜𝑚𝑒, 𝑝𝑒𝑟
𝑃𝑟𝑜𝑐 𝑁𝑎𝑡𝑙 𝐴𝑐𝑎𝑑 𝑜𝑓 𝑆𝑐𝑖, 1973 𝑣𝑜𝑙 70 𝑛𝑜. 11, 𝑇ℎ𝑜𝑚𝑎𝑠 𝑃𝑒𝑡𝑒𝑠
(A gram is ~6.022 𝑥 1023 𝑑𝑎𝑙𝑡𝑜𝑛𝑠, 𝑎𝑘𝑎 𝑎𝑣𝑎𝑔𝑎𝑑𝑟𝑜𝑠 𝑛𝑢𝑚𝑏𝑒𝑟. ) Apparently the human
chromosomal mass is not substantially different. (“Mass characteristics of DNA…”,
Shawn Strickel, “Cromosoma 1985 92:234-41)
This therefore would be ~ (1.661 𝑥 10−27 𝑘𝑔/𝑑𝑎𝑙𝑡𝑜𝑛) x (108 𝑑𝑎𝑙𝑡𝑜𝑛) =
1.661 x 10−19 𝑘𝑔/ chromosome = mass.
To reiterate:
Force = ma, then
𝐹𝑜𝑟𝑐𝑒𝑚𝑎 = (1.661 x 10−19 𝑘𝑔) (3.89 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠/ 𝑠𝑒𝑐𝑜𝑛𝑑 2 )=
=6.46 x x 10−28kg – meter/ 𝑠𝑒𝑐 2 .
Where 𝜇 = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦, and if we are saying water then, the dynamic
viscosity of water is 6.92 × 10−4 Pa s = kg/m-sec at 37°C but
Howard suggests cytoplasmic viscosity is 1000x that of water therefore equaling
0.692 Pa s = kg/m-sec at 37°C.
[Katherine Luby-Phelps explains this in her article, “Cytoarchitecture and
Physical Properties of Cytoplasm…”, (International Review of Cytology. Vol 192)
by saying, in effect that the assumption of a dilute fluid as describing the
cytoplasm would be tantamount to calling times square on new years eve a small
gathering of people. The cytoplasm is so crowded with proteins as to make it’s
consistency more comparable to a plastic solid rather than a liquid with plastic
characteristics.]
Howard is also suggesting here that a chromosome has the same drag
coefficient as a bacterium. He further states that according to Stokes law said E
coli drag coefficient (in water) ≅ 20 nN s/m or 20 x 10−9N-sec/meter (a N=kgm/𝑠𝑒𝑐 2 ) so we are talking 20 x 10−9 kg/sec (also a unit measure of resistance).
Unfortunately that not what I get when I multiply 6πμr;
As E coli radius is ~250nm or 250 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠 and
Viscosity,𝜇 = 6.92 × 10−4 Pa s = kg/m-sec at 37°C
so
6π(6.92 × 10−4 )( 250 x 10−9 ) = so I get 3 x 10−9 kg/sec as a drag coefficient for
E coli in water.(Off by as factor of 6.)
And since the viscosity of cytoplasm is 1000x that of water, then the drag will be
in my calculation 3.26 x 10−6 kg/sec.
Human ES (embryonic stem) cells measure approximately 14 μm while mouse ES cells are
closer to 8 μm.[5]
VELOCITY:
Assuming no more than half the distance of the human ES (7 𝜇𝑚) in 30 minutes
𝜇𝑚
𝜇𝑀
gives us an average speed of 14 ℎ𝑜𝑢𝑟 , 0.23 𝑀𝐼𝑁𝑈𝑇𝐸 or 14 x 10−6 𝑚𝑒𝑡𝑒𝑟𝑠/3600 sec =
3.89 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠/ second = 3.89 nm/ sec which is comparable to the rates
𝜇𝑀
generated by other researchers. (Prof. Betterton above gives 1𝑀𝐼𝑁𝑈𝑇𝐸 as the typically
accepted speed for chromosome movement during anaphase and actually concludes
with speeds ≅ 16 nm/ sec .)
In multiplying Drag of chromosomes by the above velocity during
anaphase we get 3.26 x 10−6 kg/sec x 3.89 x 10−9 𝑚𝑒𝑡𝑒𝑟𝑠/ second =
1.27 x 10−14 𝑘𝑔 − 𝑚/𝑠𝑒𝑐 2 or .0127 pN drag force.
To reiterate:
𝐹𝑚𝑎 + 𝐹𝑑 = 𝐹𝑛𝑒𝑡
𝑚
6.46 x x 10−28kg – meter/ 𝑠𝑒𝑐 2 +1.27 x 10−14 𝑘𝑔 − 𝑠𝑒𝑐 2 =
and we can see that for all intents and purposes drag force is = net force but that the
work is still done by this seemly small relatively insignificant
Work as potential energy = Force x the distance or
6.46 x x 10−28kg – meter x 7x 10−5 𝑚𝑒𝑡𝑒𝑟𝑠 = 4.5 x 10−32 kg- 𝑚2 /𝑠𝑒𝑐 2 .
I have some questions here. Professors Betterton, Howard and the majority of the
Biomicromechanical community suggests that drag force must be used to determine
the relationship between force and speed. While this seems reasonable given how
insignificant 𝐹𝑚𝑎 𝑖𝑠 𝑖𝑛 𝑐𝑜𝑚𝑝𝑎𝑟𝑖𝑠𝑜𝑛 𝑡𝑜 𝑑𝑟𝑎𝑔 I am having difficulty getting past how
small the actual work is in relationship to the force and am wondering if my
calculations and logic are in err.
Future exercises which I would like calculate include reconfiguring the chromosome
as it is truly in terms of size and shape taking into account the fact that it represents
the motion of a combination of mechanical forces and then the extent to which that
combination number is equivalent to a “slinky”.
Another question would be in the evaluation of the kinesin and dynein energy,
which brings together the separate parental results of meiosis and consequently
leads to their being fused into one nuclear unit. If separating the chromosomes
twice leads to their reunification once why would that not leave the cell with half of
the potential energy?
Finally I would be interested in tying structure to function with proteins more
succinctly. The pictures presented by professor Howard figures 12.7 and 12.9 do not
appear similar enough to gain insight into how the protein motors work on a
molecular level. Linus Pauling developed a beautifully symmetric secondary
structure. How we go from this sort of symmetry to a globule, which works to create
life, is unclear to me.