Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S12 / 2

Natural selection requires variation among individuals, heritable contributions to that variation, and differences in reproductive success associated with it. Given those conditions, variant frequencies can change across generations. Competition and finite resources help make reproductive success unequal, but selection does not require every interaction to be direct combat. Mutation and recombination supply variation; they do not generally arise because an organism needs a particular adaptation. Drift also changes frequencies, and common ancestry and branching descent connect population changes to the evolution of species. Selection is a mechanism within evolution, not a synonym for every evolutionary change.^[ref]

The chemical mismatch types are A–C and G–T, with either strand acting as template. In the fixed ring alignment shown, A–C has no correct hydrogen bonds while G–T retains one. Relative probabilities must specify the template: template A loses two bonds versus A–T, giving $e^{-10}$; template G loses two versus G–C, giving $e^{-10}$; template C loses three versus C–G, giving $e^{-15}$; template T loses one versus T–A, giving $e^{-5}$. Equal incoming activities and $5k_{B}T$ per bond give approximate relative weights $4.5\times 10^{-5},4.5\times 10^{-5},3.1\times 10^{-7},6.7\times 10^{-3}$ respectively.

For roughly equal template frequencies the mean error probability is order $1.7\times 10^{-3}$, or several million errors across $3\times 10^9$ copied base pairs. Restricting the estimate to purine templates instead gives order $10^5$ errors; neither is an accurate replication model. Both greatly exceed the source's order-one surviving mutation input. Polymerase kinetic discrimination, proofreading and mismatch repair suppress errors, and a transient mismatch need not become an inherited mutation. Ring relaxation, stacking, solvent and rare tautomers are missing from this simple hydrogen-bond calculation. Real G–T wobble geometry can form additional bonds. The source's mutation rate is illustrative, not a universal measured human rate.^[ref]

A base pair has four possible sequence identities, hence two bits, not four bits from treating complementary strands as independent. B-DNA has diameter approximately 2 nm, axial rise 0.34 nm per base pair and about ten base pairs per helical turn. The tightly packed-cylinder upper density is thereforeA haploid human genome has about $6\times 10^9$ raw bits; in a nucleus of several hundred cubic micrometres its bulk density is order $10^{25}$ bits m${}^{-3}$, far lower. Cellular DNA must allow hydration, packaging, regulatory access and copying/repair machinery, and cannot fill a cell as parallel bare cylinders. Diploidy and repeated or constrained sequence also reduce independent functional information compared with the raw four-letter capacity; those concepts should not be confused with geometric storage density.

The original molecular figure in question 3 shows the approximate helical dimensions.

For independent equiprobable bases there are $4^3=64$ codons, three of them stops, so $p_{\mathrm{stop}}=3/64$. In a fixed reading frame, and conditional on starting translation, surviving at least $L$ successive non-stop codons has probability $\boxed{(61/64{)}^L}$. For 300 it is about $5.6\times 10^{-7}$. Strictly “exceeding 300” requires 301 non-stop codons, giving $(61/64{)}^{301}$. A specified start codon, codon bias, correlations or looking for a long sequence anywhere in a large genome changes the sampling problem and cannot be silently folded into this single-frame probability.