1 Based on your results for flagella regeneration calculate

1) Based on your results for flagella regeneration; calculate how many amino acids are being polymerized per minute into the tubulin component of the regenerating flagella. 2) Assume that all of the mRNA needed for this process is being synthesized de novo, from scratch. How many mRNA bases (assume no introns) must be transcribed per minute if every mRNA is translated only once? What if each mRNA is translated 100 times? 3) If RNA polymerase can travel no faster than 2500 bases per minute, it possible for all of the RNA to be transcribed de novo? Explain your answer.

Solution

A (1).

Mass of a heterodimer of Tubulin = 100kD

1kb of DNA encodes 333 amino acids = 3.7 x 10⁴ Da

or, 333 amino acids = 37000 Da = 37 kD

So, in a 100 kD protein (such as Tubulin heterodimer), there are 333/37 x 100 = 900 amino acids.

Therefore, 1 Tubulin molecule contains 900 amino acids.

Now, each Tubulin heterodimer has a diameter of 8 nm (as per the figure supplied)

Rate of synthesis = 2.5 micrometers/hr = 2500 nm/hr

So, In 60 min (1 hr), increase in length = 2500 nm

Therefore, in 1 min, increase in length = 2500/60 = 42 nm

Since the length of 1 Tubulin heterodimer is 8 nm (comprising of 900 amino acids)

Total number of Tubulin molecules synthesized / min = 42/8 = 5.25 molecules. This equilibrates to 900 x 5.25 = 4,725 amino acids / min.

Therefore, 4,725 amino acids are added per minute (Answer).

A (2).

From the above, we see that 4,725 amino acids are incorporated per minute.

We know that 3 bases (triplet) of mRNA encode 1 amino acid.

So, 4,725 amino acids, are encoded by 4,725 x 3 = 14,175 bases of mRNA.

Therefore, 14,175 bases of mRNA need to be transcribed per minute (Answer).