Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 3

Let $T$ be the time since the two sampled mitochondrial genomes shared their ancestor. Under the stipulated one-change-per-difference approximation, both descendant lineages contribute, giving a total evolutionary duration $2T$. Therefore the per-site, per-lineage mutation rate estimate isThis is not a whole-genome event rate; multiplying by the number of usable sites would give that rate. Without equal lineage rates, the pairwise data identify their sum rather than either rate separately.

The calculation assumes an accurate sequence alignment of corresponding ancestral sites, negligible errors in DNA sequencing, and an appropriate divergence-time calibration for the sampled mitochondrial genomes. It assumes a constant molecular clock on each lineage, comparable rates across sites for this simple average, and no extra differences introduced by incorrectly aligned insertions or deletions. It treats observed nucleotide substitutions as a record of mutations; interpreting a substitution-based rate as the underlying mutation rate additionally requires a suitable neutral model rather than strong filtering by natural selection. Independence between sites is useful for a simple sampling model and uncertainty calculation. The supplied time must refer to the common ancestor of these sequences: a species split need not coincide exactly with a sampled gene's ancestral split.

The printed “otherwise” branch requires correction. Its factor $3/4$ is the probability of ending at any of the three bases different from the initial base, not of ending at one specified different base. Assigning that probability to each of three alternatives would make a row of the transition matrix sum to $5/2-(3/2)e^{-\mu t}$, which exceeds one for $t>0$. This distinction is present in the source PDF, not introduced by the transcription.

For a consistent Jukes-Cantor model, let each ordered change from one nucleotide to a specified different nucleotide have instantaneous rate $q$. Its continuous-time Markov chain has diagonal generator entries $-3q$ and off-diagonal entries $q$. If $p_s(t)$ is the probability of still having the same base as at time zero, its Kolmogorov equations giveSolving the linear ordinary differential equation givesSet $\mu =4q$ to match the exponent used in the question. Symmetry then gives the properly normalized transition probabilitiesSumming the three different-base outcomes yieldsThe assumptions are a time-homogeneous Markov process, the same change rate between every distinct ordered base pair, identical rates across independent sites, and no insertions or deletions. The stationary distribution is uniform over the four bases; starting the process in that distribution is not required for this conditional calculation.

There is also a parameter-convention issue. The actual total mutation rate out of a base is $r=3q=3\mu /4$. Thus the exponent parameter $\mu$ in (b)–(d) is $4/3$ times the event rate used in the elementary counting interpretation of (a). Keeping this conversion explicit avoids changing the meaning of “mutation rate” halfway through the calculation.

Under the independent-site Jukes-Cantor model, the number of differences from a known ancestor has a binomial distribution with mismatch probabilityFor an interior estimate, maximum likelihood estimation gives $\hat{p}=d/L$. Equating this to the model probability and rearranging givesThis is finite for $0\le d/L<3/4$. At $d/L=3/4$, the estimate tends to infinity: the endpoint distribution has reached its limiting saturation and no finite rate can reproduce it exactly. For $d/L>3/4$, the displayed logarithm is not defined as a real number; within this model the constrained likelihood is maximized only as $\mu \to \infty$. For $d=0$ the boundary estimate is zero. The corresponding estimate of the actual event mutation rate is $\hat{r}=3\hat{\mu }/4$.

The two descendant branches of an equal-rate Jukes-Cantor model compose to a single branch of duration $2T$. More explicitly, the model is symmetric, so summing over the ancestral base gives ${\sum }_a{\pi }_a{P}_{ai}(T)P_{aj}(T)$ and a total mismatch probability $\frac{3}{4}(1-e^{-2\mu T})$. Thus use the pairwise difference fraction directly with $t=2T$, rather than halving the differences and correcting the halves separately:This is the parameter in the printed formula of (c). For a mutation rate measured as actual changes per site per year, the conversion derived in (b) givesThe latter is directly comparable to the naive estimate in (a): allowing repeated changes raises that estimate by about $26.4\%$. The exponent parameter and actual event rate are two conventions for the same fitted process, not two different biological estimates.