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Biology · 85

Every formula in this course and the easiest ways to remember them, on one page.

1. The living world2. The plant kingdom3. Animal systematics4. The human body and health5. Cell division and genetics6. Evolution and ecology7. Biology for upper grades8. University biology

1The living worldBeginner

What is alive? Characteristics, research methods and classification

Open lesson
M = M₁ × M₂
where:
  • Mtotal magnification of the microscope (how many times)
  • M₁magnification of the eyepiece
  • M₂magnification of the objective

Magnifications are multiplied, not added: a 10× eyepiece with a 40× objective gives 10 × 40 = 400×.

Real size = image size ÷ M
where:
  • Mtotal magnification

The other way round: image size = real size × M. Watch the units: 1 mm = 1000 µm.

The cell: the building block of life

Open lesson
Total magnification = eyepiece magnification × objective magnification

The eyepiece is the lens you look into; the objective is the lens close to the specimen.

Plants and photosynthesis

Open lesson
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
where:
  • CO₂carbon dioxide — enters the leaf from the air through the stomata
  • H₂Owater — absorbed from the soil by the roots
  • C₆H₁₂O₆glucose — the food the plant makes
  • O₂oxygen — released into the air

The process needs light and chlorophyll.

2The plant kingdomBeginner

Plant cell, tissues and the root

Open lesson

Shoot, buds, stem and leaf

Open lesson
Age = n
where:
  • nthe number of annual rings in the cross-section

One light (spring) band plus one dark (autumn) band make one year. The age counted on a stump is roughly the age of the whole tree; a cut higher up shows fewer rings.

First year = year of cutting − n + 1
where:
  • nthe number of rings (including the ring of the year of cutting)

The +1 appears because both the first year’s ring and the ring of the cutting year are counted (the tree was cut in autumn, after the ring was complete).

Modified organs and vegetative propagation

Open lesson

The flower, flower formulas and inflorescences

Open lesson
n = K + L + E + D
where:
  • nthe sum of the structural elements in the flower formula
  • K, Lnumbers of sepals and petals (with a simple perianth, Ç instead)
  • E, Dnumbers of stamens and pistils

Brackets and “+” do not change the count: add up all the numbers of each letter. 0 counts as zero; if the formula has ∞, the sum cannot be found.

Double fertilisation, fruits and seeds

Open lesson
H = 4 · t, S = 2 · t, T = t
where:
  • tnumber of seeds formed (= fertilised embryo sacs, ovules)
  • Hcells taking part directly in double fertilisation
  • Ssperm cells needed
  • Tminimum number of pollen grains that must germinate

In each embryo sac 4 cells take part directly: 2 sperm cells, the egg cell and the central cell. Synergids and antipodal cells are not counted.

3Animal systematicsBeginner

The animal kingdom: common features and invertebrates

Open lesson
Kingdom → Subkingdom → Phylum → Subphylum → Class → Order → Family → Genus → Species
where:
  • Phylumunites related classes; in plants this rank is called a division
  • Orderunites related families; in plants it is also called an order
  • Speciesindividuals that resemble each other and interbreed to give fertile offspring — the smallest rank

Taxonomic ranks: from left to right the group gets smaller and its members more alike. Subkingdom and subphylum are helper ranks.

Unicellular animals

Open lesson

Cnidarians

Open lesson

Flatworms and roundworms

Open lesson

Chordates: the lancelet and fish

Open lesson
Chordate = notochord + dorsal nerve tube + gill slits in the pharynx
where:
  • notochordthe axial skeleton; replaced by the backbone in vertebrates
  • nerve tubea hollow tube lying on the back above the notochord; in vertebrates its front part becomes the brain
  • gill slitsopenings in the wall of the pharynx; aquatic chordates keep them for life, land animals have them only as embryos

All three features are present in every chordate at least as an embryo. In addition, the heart lies on the belly side, circulation is closed, symmetry is bilateral and the body cavity is secondary.

Paired fins: 2 pectoral + 2 pelvic = 4 | Unpaired fins: 2 dorsal + 1 anal + 1 tail = 4
where:
  • pectoral, pelvicturning, stopping, moving up and down
  • tailthe main organ of movement — pushes the fish forward
  • dorsal, analkeep the body upright and stop it rolling over

In the perch the number of paired and unpaired fins is equal: 4 = 4, eight fins in all.

ventricle → ventral aorta (venous) → gill capillaries (+O₂) → dorsal aorta (arterial) → organs (−O₂) → veins → atrium

The fish’s single circuit: blood passes through the heart once per round, so the heart holds only venous blood.

Amphibians and reptiles

Open lesson
right atrium (venous) + left atrium (arterial) → ventricle (mixed) ⇒ 3 kinds of blood in the heart
where:
  • pulmonary circuitventricle → lungs and skin → left atrium
  • systemic circuitventricle → organs of the body → right atrium

The frog heart has three chambers: 2 atria + 1 ventricle. Two circuits first appear in amphibians, but blood mixes in the ventricle.

Birds

Open lesson
Forelimb girdle = shoulder blades + coracoids + collarbones (each paired)

In a bird the breastbone belongs, with the ribs, to the chest — the trunk skeleton; in the frog it is counted as part of the forelimb girdle. The upper-arm bone, ulna and radius are not the girdle but the skeleton of the free limb, the wing.

breathing in: air → lungs (gas exchange) → air sacs; breathing out: air sacs → lungs (gas exchange) → outside
where:
  • air sacsstore air and make it pass through the lungs twice; they also cool and lighten the body; no gas exchange takes place in them

Double breathing: in one breathing cycle gas exchange happens in the lungs twice — on breathing in and on breathing out. In flight the wingbeats drive the breathing movements.

Mammals

Open lesson
I 3/3 · C 1/1 · P 4/4 · M 2/3 → 2 · (3 + 1 + 4 + 2) + 2 · (3 + 1 + 4 + 3) = 20 + 22 = 42I 3/3 · C 1/1 · P 4/4 · M 2/3 → 2 · (3 + 1 + 4 + 2) + 2 · (3 + 1 + 4 + 3) = 20 + 22 = 42
where:
  • I, Cincisors and canines
  • P, Mpremolars and molars
  • 3/33/3above the line — teeth in one half of the upper jaw, below — in one half of the lower jaw

The dental formula of the domestic dog. It shows only one half of each jaw, so each jaw’s sum is multiplied by 2: a dog has 42 teeth.

4The human body and healthIntermediate

Digestion, circulation and breathing

Open lesson
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy

Cellular respiration — the reverse of photosynthesis.

5Cell division and geneticsAdvanced

DNA and heredity

Open lesson
A = T, G = C, A + G = T + C

Chargaff's rule: in double-stranded DNA the amount of adenine equals thymine, and guanine equals cytosine. All the nucleotides together make 100%.

Mendel's laws of inheritance

Open lesson
Aa × Aa → 1 AA : 2 Aa : 1 aa

Genotype ratio 1 : 2 : 1, phenotype ratio 3 : 1 (3 yellow : 1 green).

6Evolution and ecologyAdvanced

Ecosystems and food chains

Open lesson
Eₙ₊₁ ≈ 0.1 · Eₙ
where:
  • Eₙthe energy of one link
  • Eₙ₊₁the energy passed to the next link

This is an approximate rule: in real ecosystems the share passed on can be a little lower or higher.

7Biology for upper gradesAdvanced

The human skeleton and muscles

Open lesson
F₁ · l₁ = F₂ · l₂
where:
  • F₁force of the muscle, N
  • l₁distance from the joint to where the muscle is attached (effort arm)
  • F₂weight of the load, N
  • l₂distance from the joint to the load (load arm)

The lever rule: the joint is the pivot and the bone is the lever.

The endocrine system and hormones

Open lesson
c (mmol/L) = c (mg/dL) ÷ 18c (mmol/L) = c (mg/dL) ÷ 18
where:
  • cblood glucose concentration
  • 18the molar mass of glucose (180 g/mol) ÷ 10

Some countries (for example the USA and Türkiye) measure glucose in mg/dL, while others (for example Azerbaijan and Russia) use mmol/L.

Excretion and homeostasis

Open lesson
Fraction reabsorbed = (V filtrate − V urine) ÷ V filtrate × 100%
where:
  • V filtratevolume of filtrate formed per day
  • V urinevolume of urine passed per day

Reproduction and development

Open lesson
N = 2ⁿ
where:
  • Nnumber of cells
  • nnumber of successive divisions starting from the zygote

Each division doubles the number of cells (if all cells divide at the same time).

Transport, transpiration and hormones in plants

Open lesson
V = π · r² · l, v = V ÷ t
where:
  • Vvolume of water taken up, mm³
  • rradius of the potometer’s capillary tube, mm
  • ldistance moved by the air bubble, mm
  • v, trate of water uptake and time of the measurement

A potometer measures how fast a shoot takes up water, which is roughly equal to the rate of transpiration.

Protein synthesis: transcription and translation

Open lesson
DNA → (transcription) → mRNA → (translation) → protein

The central idea of molecular biology: information flows from DNA to RNA and from RNA to protein.

Number of nucleotides = 3 × number of amino acids + 3 (stop codon)

For the coding part of the mRNA; introns and non-coding ends are not counted.

Biotechnology and genetic engineering

Open lesson
N = N₀ · 2ⁿ
where:
  • Nnumber of DNA copies after n cycles
  • N₀starting number of DNA molecules
  • nnumber of cycles

Ideally, the amount of DNA doubles in every cycle.

8University biologyUniversity

Biomolecules and enzyme kinetics

Open lesson
n C₆H₁₂O₆ → H–(C₆H₁₀O₅)ₙ–OH + (n − 1) H₂O
where:
  • nnumber of glucose units (degree of polymerisation)
  • (n − 1) H₂Oeach glycosidic bond releases one water molecule

Condensation (polymerisation); the reverse process, hydrolysis, happens during digestion.

v = Vmax · [S] / (Km + [S])v = Vmax · [S] / (Km + [S])
where:
  • vinitial reaction rate, e.g. µmol/min
  • Vmax = kcat · [E]₀maximum rate, when all the enzyme is saturated with substrate
  • [S]substrate concentration, mM
  • Km = (k₋₁ + kcat) / k₁Km = (k₋₁ + kcat) / k₁Michaelis constant: the [S] at which v = Vmax/2, mM
  • kcatturnover number: substrate molecules converted per enzyme molecule per second, s⁻¹

When [S] ≪ Km, v ≈ (Vmax/Km)[S] — first order; when [S] ≫ Km, v ≈ Vmax — zero order. The ratio kcat/Km measures catalytic efficiency.

1/v = (Km/Vmax) · (1/[S]) + 1/Vmax1/v = (Km/Vmax) · (1/[S]) + 1/Vmax
where:
  • Km/VmaxKm/Vmaxslope of the straight line
  • 1/Vmax1/Vmaxy-intercept
  • −1/Km−1/Kmx-intercept

Lineweaver–Burk (double-reciprocal) plot: taking the reciprocal of both sides of the Michaelis–Menten equation turns the hyperbola into a straight line.

v = Vmax · [S] / (α · Km + [S]), α = 1 + [I]/Kiv = Vmax · [S] / (α · Km + [S]), α = 1 + [I]/Ki
where:
  • [I]inhibitor concentration
  • Kidissociation constant of the EI complex — the inhibitor's “strength” (small Ki = strong inhibitor)

Competitive inhibition: apparent Km = α · Km, Vmax unchanged.

Cellular respiration and metabolism

Open lesson
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O, ΔG°′ ≈ −2870 kJ/molC₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O, ΔG°′ ≈ −2870 kJ/mol
where:
  • ΔG°′change in Gibbs energy under standard conditions at pH 7; the minus sign means energy is released

Glucose is oxidised (loses electrons) and oxygen is reduced (gains electrons). The electrons are carried by the coenzymes NAD⁺ and FAD: NAD⁺ + 2e⁻ + H⁺ → NADH, FAD + 2e⁻ + 2H⁺ → FADH₂.

Glucose + 2 NAD⁺ + 2 ADP + 2 Pᵢ → 2 pyruvate + 2 NADH + 2 H⁺ + 2 ATP + 2 H₂O

The net equation of glycolysis (Pᵢ is inorganic phosphate).

NADH → ≈ 2.5 ATP; FADH₂ → ≈ 1.5 ATP
where:
  • P/O ratioP/O ratioATP made per two electrons passed to oxygen. Electrons from NADH pump ≈ 10 H⁺, electrons from FADH₂ (which bypass complex I) pump ≈ 6 H⁺; one ATP (including transport) needs ≈ 4 H⁺

10 ÷ 4 = 2.5 and 6 ÷ 4 = 1.5. The values 3 and 2 in older textbooks are rounded early estimates.

RQ = volume of CO₂ released ÷ volume of O₂ consumed
where:
  • RQrespiratory quotient: carbohydrates ≈ 1.0, proteins ≈ 0.8, fats ≈ 0.7

At the same temperature and pressure, gas volumes are proportional to numbers of moles, so RQ can be found from the coefficients in the equation.

Photosynthesis in depth

Open lesson
6 CO₂ + 12 H₂O → C₆H₁₂O₆ + 6 O₂ + 6 H₂O, ΔG°′ ≈ +2870 kJ/mol6 CO₂ + 12 H₂O → C₆H₁₂O₆ + 6 O₂ + 6 H₂O, ΔG°′ ≈ +2870 kJ/mol
where:
  • 12 H₂O12 water molecules provide 12 oxygen atoms, i.e. 6 O₂ — all the oxygen released comes from water
  • ΔG°′ > 0the process is not spontaneous; light supplies the energy

Writing water on both sides shows where the oxygen comes from; the short form is 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂.

E = h · c / λ, Eₘ = NA · h · c / λE = h · c / λ, Eₘ = NA · h · c / λ
where:
  • hPlanck's constant, 6.626 · 10⁻³⁴ J·s
  • cspeed of light, 3.00 · 10⁸ m/s
  • λwavelength, m
  • NAAvogadro's number, 6.022 · 10²³ mol⁻¹ (Eₘ is the energy of one mole of photons)
2 H₂O → O₂ + 4 H⁺ + 4 e⁻

Water splitting happens at the oxygen-evolving complex of photosystem II, on the lumen side. For one O₂, 4 electrons must pass through both photosystems: at least 8 photons.

3 CO₂ + 9 ATP + 6 NADPH → G3P + 9 ADP + 8 Pᵢ + 6 NADP⁺

Simplified balance of the Calvin cycle (water and H⁺ not shown). Per CO₂: 3 ATP and 2 NADPH; one of the nine phosphates stays in G3P.

Molecular genetics and gene regulation

Open lesson
Fraction of hybrid molecules = 2 ÷ 2ⁿ (n ≥ 1)
where:
  • nnumber of generations in ¹⁴N medium
  • 2the two original heavy strands are never lost and always sit in two hybrid molecules

One heavy molecule gives 2ⁿ molecules after n generations; all the rest are light.

t = G / (2 · v · n)t = G / (2 · v · n)
where:
  • treplication time, s
  • Glength of DNA to copy, base pairs
  • vspeed of one fork, nucleotides/s
  • nnumber of origins firing together (2 forks per origin)

Valid if the origins are evenly spaced and start at the same time.

Population genetics and evolution

Open lesson
p = (2·NAA + NAa) / (2N), q = 1 − pp = (2·NAA + NAa) / (2N), q = 1 − p
where:
  • p, qfrequencies of alleles A and a (their shares of the gene pool)
  • NAA, NAanumbers of individuals with each genotype
  • 2Ntotal number of alleles in N diploid individuals
p² + 2pq + q² = 1
where:
  • p²frequency of AA homozygotes
  • 2pqfrequency of Aa heterozygotes (carriers)
  • q²frequency of aa homozygotes

Hardy–Weinberg equilibrium is the state of a population when “nothing happens”; deviations from it are a sign of evolutionary forces at work.

χ² = Σ (O − E)² / Eχ² = Σ (O − E)² / E
where:
  • O, Eobserved and expected count for each genotype class
  • df = 1degrees of freedom: 3 classes − 1 − 1 (p was estimated from the data); the critical value at the 0.05 level is 3.84
Hₜ = H₀ · (1 − 1/(2N))ᵗ; qₜ = q₀ / (1 + t·q₀)Hₜ = H₀ · (1 − 1/(2N))ᵗ; qₜ = q₀ / (1 + t·q₀)
where:
  • Hₜ, H₀heterozygosity after t generations and at the start (drift in an ideal population of N)
  • 1/(2N)1/(2N)also the probability that a new neutral mutation eventually becomes fixed
  • qₜfrequency of a lethal recessive allele (s = 1) after t generations
t = K / (2r)t = K / (2r)
where:
  • ttime since the two species split, years
  • Kdifferences between the two sequences (substitutions per site)
  • rsubstitution rate per site per year in each lineage

For example, with K = 0.02 and r = 10⁻⁹: t = 0.02 ÷ (2 · 10⁻⁹) = 10⁷ years = 10 million years. The clock must be calibrated with fossils.

Immunology

Open lesson
D = (VH · DH · JH) · (VL · JL)
where:
  • VH, DH, JHnumbers of functional heavy-chain V, D and J segments
  • VL, JLnumbers of light-chain V and J segments (κ and λ are counted separately and added)

Combinatorial diversity only; junctional diversity and hypermutation multiply this number many times over.

H = 1 − 1/R₀, Vc = (1 − 1/R₀) / E, R = R₀ · sH = 1 − 1/R₀, Vc = (1 − 1/R₀) / E, R = R₀ · s
where:
  • R₀basic reproduction number: the average number of people one case infects in a fully susceptible population
  • Hherd-immunity threshold — the share that must be immune
  • Vc, Erequired vaccine coverage and vaccine effectiveness (0–1)
  • R, seffective reproduction number and the susceptible fraction; the epidemic dies out when R < 1

H comes from the condition R = R₀ · (1 − H) = 1.

Neuroscience basics

Open lesson
E = (RT / zF) · ln([X]out / [X]in) ≈ (61,5 mV / z) · log₁₀([X]out / [X]in)E = (RT / zF) · ln([X]out / [X]in) ≈ (61,5 mV / z) · log₁₀([X]out / [X]in)
where:
  • Eequilibrium (Nernst) potential of the ion
  • R, T, Fgas constant 8.314 J/(mol·K), absolute temperature (37 °C = 310 K), Faraday constant 96,485 C/mol
  • zcharge of the ion (+1 for K⁺ and Na⁺, −1 for Cl⁻, +2 for Ca²⁺)

At 37 °C RT/F ≈ 26.7 mV; switching from the natural to the base-10 logarithm multiplies by 2.303 and gives 61.5 mV.

Vm = 61,5 mV · log₁₀((PK[K⁺]o + PNa[Na⁺]o + PCl[Cl⁻]i) / (PK[K⁺]i + PNa[Na⁺]i + PCl[Cl⁻]o))Vm = 61,5 mV · log₁₀((PK[K⁺]o + PNa[Na⁺]o + PCl[Cl⁻]i) / (PK[K⁺]i + PNa[Na⁺]i + PCl[Cl⁻]o))
where:
  • Vmmembrane potential
  • PK, PNa, PClrelative permeabilities of the ions; at rest roughly PK : PNa : PCl = 1 : 0.04 : 0.45
  • o, ioutside and inside concentrations (Cl⁻ is negative, so its “i” and “o” are swapped)

The Goldman–Hodgkin–Katz equation: the membrane potential is like an average of the Nernst potentials, weighted by permeability.