The whole Genetics, Cell and Molecular Biology formula sheet, to print

For two independent events, e.g. the genotype at one unlinked locus combined with the genotype at another unlinked locus.

For mutually exclusive outcomes, e.g. two different genotypes that give the same phenotype.

n = number of heterozygous gene pairs. When all n loci assort independently, each gamete type has frequency $(1/2{)}^n$.

The probability that exactly s of n offspring are of one kind (probability p each) and t are of the other kind (probability q each), with s + t = n and p + q = 1. Each offspring is treated as an independent event.

O = observed count and E = expected count in each class under the hypothesised ratio. Use counts, not percentages. Degrees of freedom = number of classes − 1. If the calculated ${\chi }^2$ is below the tabulated critical value at the chosen significance level (commonly 0.05), the deviation is attributed to chance and the hypothesised ratio is not rejected.

Made up of n homozygotes and $\frac{n(n-1)}{2}$ heterozygotes. For n = 3 (ABO): 3 + 3 = 6 genotypes.

Measured in a testcross (heterozygote × fully recessive homozygote), where each offspring's phenotype shows the gamete it received from the heterozygote. The observed RF never exceeds 50%, the value for independent assortment. Over long intervals it underestimates the true map distance, because two (or any even number of) crossovers between the markers restore the parental combination and go undetected. Map distances are therefore built by adding short intervals.

r = observed recombination frequency as a fraction (r < 0.5); d = map distance in morgans (multiply by 100 for cM). Assumes crossovers occur independently along the chromosome (no interference). For small r, d ≈ r.

Expected double crossovers = RF(interval 1) × RF(interval 2) × total offspring. Both RFs are taken as fractions and each already includes the double-crossover classes. This expectation assumes that crossovers in the two intervals are independent.

I = 1: complete interference (no double crossovers observed). I = 0: crossovers in the two intervals occur independently (observed = expected). I < 0 (c.o.c. > 1): negative interference, more double crossovers than expected.

t₁ and t₂ are the times of entry (minutes after mixing) of two markers transferred by the same Hfr strain in the same direction. Distances from different Hfr strains are compared only after aligning each strain's order on the circular map.

C is a fraction (multiply by 100 for a percentage). Select for one donor marker, then score the unselected marker. Cotransformation frequency is defined the same way among transformants.

C is the cotransduction frequency (a fraction), d the distance between the two markers, and L the length of the packaged fragment. d and L must be in the same unit, either kb or map minutes; for P1, L is about 2 min of the E. coli map. The formula assumes the fragment ends fall at random and holds for d < L. C falls to 0 as d approaches L and is taken as 0 for d ≥ L.

Use this to find d from a measured cotransduction frequency C (as a fraction, not a percentage). The same conditions as the Wu formula apply.

RF is expressed as a percentage, and 1% RF = 1 map unit (cM). It applies to crosses scored for recombinant and parental classes (including phage crosses). This scale cannot be converted directly into the minutes of an interrupted-mating map.

$p$ = frequency of allele $A$, $q$ = frequency of allele $a$ in the gene pool.

Valid whether or not the population is in Hardy-Weinberg equilibrium: each heterozygote carries one copy of each allele.

At equilibrium: $p^2$ = frequency of $AA$, $2pq$ = frequency of $Aa$, $q^2$ = frequency of $aa$.

Homozygote frequencies are the squares; each heterozygote frequency is twice the product of its two allele frequencies.

The fraction of males showing an X-linked recessive trait estimates $q$ directly. The fraction of affected females is $q^2$, which is why such traits are far more frequent in males.

Expected counts $E$ come from $p^{2}N$, $2pqN$, $q^{2}N$, using $p$ and $q$ estimated from the same data, which costs one degree of freedom. With two alleles, df = 1 (critical value 3.84 at the 5% level).

$F$ = inbreeding coefficient (the probability that the two alleles in an individual are identical by descent). Inbreeding raises homozygosity without changing allele frequencies. $F=1-\frac{H_{\text{obs}}}{2pq}$.

Fitnesses $AA=1$, $Aa=1$, $aa=1-s$. Selection is slow when $q$ is small, because almost all copies of $a$ are hidden in heterozygotes.

From $q^{\prime }=q/(1+q)$, each generation adds 1 to $1/q$. $n$ is the number of generations to go from $q_0$ to $q_n$.

Fitnesses $AA=1-s$, $Aa=1$, $aa=1-t$; $q$ = frequency of $a$. The equilibrium is stable. The allele whose homozygote pays the larger cost stays rarer.

$\mu$ = mutation rate per generation towards the deleterious allele; $s$ = selection coefficient against the affected genotype. These approximations hold when $\mu \ll s$.

$m$ = fraction of the recipient population's genes that come from migrants each generation; $q_m$ = allele frequency among migrants.

$N_e$ = effective population size; $t$ = generations. The smaller $N_e$, the faster variation is lost.

$N_m$, $N_f$ = numbers of breeding males and females. $N_e$ is always ≤ the census number of breeders, and is dominated by the rarer sex.

$S$ = selection differential (mean of selected parents − population mean); $R$ = response (mean of offspring − population mean). $h^2$ = narrow-sense heritability (additive variance / phenotypic variance); $H^2$ = broad-sense heritability (total genetic variance / phenotypic variance). Only $h^2$ predicts the response to selection.

In a diploid population of $N$ individuals, $2N\mu$ new neutral mutations arise per generation, and each is eventually fixed with probability $1/(2N)$. The neutral substitution rate $k$ therefore equals the neutral mutation rate $\mu$, independent of population size. This is the basis of the molecular clock.

$d_N$ = non-synonymous substitutions per non-synonymous site; $d_S$ = synonymous substitutions per synonymous site. $\omega <1$: purifying (negative) selection; $\omega \approx 1$: neutral evolution; $\omega >1$: positive (adaptive) selection.

Here $N_0$ is the initial cell count, $N$ the count after time $t$, and $t_d$ the doubling time. It holds only while every cell is cycling with the same doubling time, with no cell death and outside any lag or stationary phase.

This follows directly from the exponential-growth formula under the same conditions.

This is the same exponential law written with base $e$. $\mu$ has units of reciprocal time.

$T$ is the total cell-cycle time and $n_{\text{phase}}/n_{\text{total}}$ is the fraction of cells observed in the phase; for M phase this fraction is the mitotic index. This is an approximation for an asynchronous culture that treats cells as evenly spread across cycle ages. In an exponentially growing culture young cells slightly outnumber old ones, so the estimate is not exact.

Here $n$ is the haploid chromosome number. The count ignores crossing over, which adds further variation. For humans ($n=23$), $2^{23}=8,388,608$.

For a simple 1:1 interaction. $K_d$ has units of concentration ($k_{off}$ in ${\text{s}}^{-1}$, $k_{on}$ in ${\text{M}}^{-1}{\text{s}}^{-1}$). A smaller $K_d$ means tighter (higher-affinity) binding. The association constant is $K_a=1/K_d$.

At $[L]=K_d$, $\theta =0.5$. Rearranged, $[L]=K_d\frac{\theta }{1-\theta }$, so moving from 10% to 90% occupancy needs an 81-fold rise in $[L]$ (from $K_d/9$ to $9K_d$). Valid only for one independent binding site per receptor.

$B$ = bound ligand ($[RL]$), $F$ = free ligand, $B_{max}$ = total receptor concentration. For a single class of independent sites, plotting $B/F$ against $B$ gives a straight line with slope $-1/K_d$ and x-intercept $B_{max}$. A curved Scatchard plot indicates cooperativity or more than one class of binding site.

Here $K_d$ has units of concentration${}^n$ and equals $(K_{0.5}{)}^n$, where $K_{0.5}$ is the ligand concentration that gives half saturation. $n$ is the Hill coefficient: $n>1$ means positive cooperativity, $n<1$ means negative cooperativity (or several classes of site), and $n=1$ reduces to the single-site equation. $n$ is an empirical measure: it cannot exceed the number of binding sites and is usually smaller.

A straight line whose slope is the Hill coefficient $n$.

$G$ = genome (or region) length in bp and $\ell$ = average Okazaki fragment length in nt. Every base pair is copied by exactly one fork, and each fork makes one strand continuously and the other discontinuously. So the total lagging-strand DNA equals $G$ nt, whatever the number of origins. Primer lengths and end effects at origins and termini are ignored.

4 bases in 3 positions give 64 codons: 61 sense codons (including AUG) and 3 stop codons (UAA, UAG, UGA).

$L_{\mathrm{ORF}}$ = nucleotides from the first base of the start codon to the last base of the stop codon, inclusive. $N_{\mathrm{aa}}$ = number of amino acids in the primary translation product, including the initiator Met. The +1 is the stop codon, which is not translated. If the initiator Met is removed after translation, add it back: a mature chain of $n$ residues gives $N_{\mathrm{aa}}=n+1$. Any signal peptide or propeptide that is cleaved off must be added back in the same way.