Question

BIOCHEMISTRY: Independent mutations in 4 different gene types are required for the cell to become cancerous. Assume that a hypothetical intestinal stem cell divides every 4 days. The general DNA replication error is 10^-8 per nucleotide per replication, average gene length is 8000 nucleotides, and the person average lifespan is 70 years. If we assume that mutations in both alleles of 4 specific genes are required for the cell to become cancerous, what is an average random chance for that intestinal stem cell to acquire these specific mutations during its lifetime? After calculating this number, think how it compares to the lifetime chances of a person getting cancer, and why these numbers do not seem to match (remember that human body has 10^13 cells).

Answer 1

Replication error is 1 mutation for 10⁸ nucleotides.

Each gene has average length of 8000 nucleotides, and a total of 32,000 nucleotides combined together. If mutations of both the alleles are required, the total number of nucleotides is 64,000 or 6.4 x 10⁴

Now, the required number of nucleotides is 10⁸ for one mutation to occur. To get that many nucleotides, the number of times the cell has to replicate is given by the equation

N_t = N₀ 2ⁿ

10⁸ = 6.4 x 10⁴ 2ⁿ

10⁸ = 6.4 x 10⁴ x n log 2

10⁸ = 6.4 x 10⁴ x n 0.301

n = 0.504 x 10⁴ times a cell has to replicate to accumulate one mutation in all the 4 genes.

Though human body has far more cells than the one needed to accumulate mutations, in reality, those many number of mutations actually does not accumulate due to the following reasons:

* DNA repair and proof reading mechanisms remove the mutated base

* Immune cells eliminate the cancerous cells from circulation.

If any of these mechanisms fail, then only it leads to cancers.