1.1 Overview and History

Step 1: Define information content of a single event

The "surprise" or information gained from observing an event with probability p should be: (a) a decreasing function of p (rare events are more informative), and (b) additive for independent events. The unique function satisfying both is:

$$I(p) = -\log_2 p \quad \text{(measured in bits)}$$

Step 2: Average information content -- Shannon entropy

For a random variable X with outcomes $x_1, x_2, \ldots, x_n$ occurring with probabilities $p_1, p_2, \ldots, p_n$, the expected information per observation is:

$$\boxed{H(X) = -\sum_{i=1}^{n} p_i \log_2 p_i}$$

Step 3: Apply to DNA with a 4-letter alphabet

If each base (A, T, G, C) occurs with equal probability $p = 1/4$:

$$H = -4 \times \frac{1}{4}\log_2\frac{1}{4} = -\log_2\frac{1}{4} = \log_2 4 = 2 \text{ bits per position}$$

Step 4: Total information for a sequence of length L

For L independent positions, the total information capacity is:

$$I_{total} = L \times H = L \times 2 = 2L \text{ bits}$$

Step 5: Information loss at each step of the Central Dogma

Transcription (DNA to RNA) preserves information (1:1 mapping). Translation (RNA to protein) uses a degenerate code: 64 codons map to 20 amino acids. The protein alphabet has $H_{protein} = \log_2 20 \approx 4.32$ bits per residue, but it takes 3 codons ($3 \times 2 = 6$ bits of DNA) to encode one residue. The coding efficiency is:

$$\eta = \frac{\log_2 20}{3 \times \log_2 4} = \frac{4.32}{6} \approx 72\%$$

Step 6: Effect of biased base composition

When bases are not equally frequent (e.g., AT-rich organisms like P. falciparum with 80.6% AT), information content per position decreases. For an organism with $p_{AT} = 0.3$ and $p_{GC} = 0.2$ (per base):

$$H = -2(0.3)\log_2(0.3) - 2(0.2)\log_2(0.2) \approx 1.97 \text{ bits}$$

This is less than the maximum 2 bits, reflecting the reduced information density in biased genomes.

Step 1: Counting possible sequences

For a polymer of length L built from an alphabet of size A, each position has A independent choices. By the multiplication principle:

$$N = A^L$$

Step 2: DNA sequence space

For DNA (A = 4), a gene of 1000 bp has $N = 4^{1000} \approx 10^{602}$ possible sequences. Even a short 20-mer primer has $4^{20} \approx 10^{12}$ possibilities.

Step 3: Protein sequence space

For proteins (A = 20 amino acids), a modest 100-residue protein has:

$$N = 20^{100} \approx 10^{130}$$

For comparison, the number of atoms in the observable universe is only ~$10^{80}$.

Step 4: The Levinthal paradox for search

If evolution sampled one sequence per second since the origin of life ($\sim 4 \times 10^9$ years $\approx 10^{17}$ seconds), and all $\sim 10^{30}$ organisms that ever lived each tested one sequence:

$$N_{sampled} \approx 10^{17} \times 10^{30} = 10^{47} \ll 10^{130}$$

Step 5: Fraction of sequence space explored

$$f = \frac{10^{47}}{10^{130}} = 10^{-83}$$

An infinitesimally small fraction. This means evolution does not search sequence space randomly -- it navigates fitness landscapes guided by natural selection, exploring neighboring sequences through point mutations, recombination, and gene duplication.

Step 6: Practical constraint -- mutational distance

The number of sequences reachable by k point mutations from a given protein of length L is:

$$N_k = \binom{L}{k} \times 19^k$$

For L = 100 and k = 1: $N_1 = 100 \times 19 = 1900$ neighbors. For k = 2: $N_2 = \binom{100}{2} \times 19^2 = 1,789,050$. Evolution efficiently explores this local neighborhood around functional sequences.