Question

The Xenopus laevis oocyte is roughly 1mm in diameter. After fertilization, it undergoes mitosis, a process requiring the coordination of proteins at various locations within the large cell.

a) If this event were coordinated by diffusion alone, estimate how long it would take for a message to reach from one side of the cell to the other if it was passed through a Ca2+ ion?
b) How long would it take for a message to be transmitted through the diffusion of a small protein?
c) How long would it take if the message to reach the other side of the cell if it were carried on a transmembrane protein with diffusion coefficient of 0.2

Answer 1

a) the diffusion length = 1000 micrometers;

so if the diffusion co-efficient is also known or given; we can therefore deduce the time taken( i.e how long it would take for a message to reach from one side of the cell to the other).

b) We need to be given the data value concerning the diffusion coefficient of the protein.

c) 1.234 × 10⁷ seconds

Step-by-step explanation:

a)

In a controlled transportation that occurs in diffusion; the time taken can be expressed as:

`time (t) = (diffusion length^2)/(diffusion co-efficient )`

Now; if a message is passed through a
`Ca^(2+)` ion, From studies; we know that these
`Ca^(2+)` ions are infinitesimally smaller in size in the
`Ca^(2+)` channels.

Also the information being carried through the
`Ca^(2+)` ions are bound to be transported diagonally across the cell.

However; the diffusion length = 1000 micrometers;

so if the diffusion co-efficient is also known or given; we can therefore deduce the time taken( i.e how long it would take for a message to reach from one side of the cell to the other).

b) Since; we don't know the information concerning the diffusion coefficient of the protein, then it is quite not possible to determine how long it would take for a message to be transmitted through the diffusion of a small protein.

c) Here; we are given the value of the diffusion coefficient to be = 0.2 micrometer²/seconds

In a transmembrane protein; the diffusion length is half the circumference of the cell which is =
`\pi*r`

=
`\pi * (1000)/(2)`

= 1570.8 micrometers

Now;the time taken can now be calculated as:

`time (t) = (diffusion length^2)/(diffusion co-efficient )`

`time (t) = ((1570.8)^2)/(0.2 )`

`time (t) =1.234*10^7 seconds`