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A-Level Biology Calculators

Free interactive calculators for every maths skill in A-level Biology — each one shows the full step-by-step working an examiner wants to see, with worked biology examples and the formula symbols explained. Covers AQA, WJEC, Eduqas, OCR and Edexcel.

AQA WJEC / Eduqas OCR Edexcel

Basic Maths Skills

Percentage Change

% Change = [(New − Original) ÷ Original] × 100

where Original = the starting value · New = the value after the change. A positive answer = increase, negative = decrease. Always divide by the original, never the new value.

⚡ Quick Test Examples

A potato chip starts at 5.2 g and, after an hour in salt solution, has a mass of 4.7 g.

% change = (4.7 − 5.2) ÷ 5.2 × 100 = −9.6% (a 9.6% decrease in mass — water left the cells by osmosis).

Do I divide by the old value or the new value?
Always the original (old) value. Dividing by the new value is the single most common mistake and loses the mark.
Why use percentage change instead of just the difference?
It standardises results so you can compare samples of different starting sizes — e.g. potato chips of different initial masses, or populations of different sizes.

Percentage Error

% Error = |Experimental − Theoretical| ÷ Theoretical × 100

where Experimental = your measured/result value · Theoretical = the true or accepted value · | | = take the positive (absolute) difference.

⚡ Quick Test Examples

You measure a solution’s concentration as 0.48 mol dm⁻³; the true value is 0.50 mol dm⁻³.

% error = |0.48 − 0.50| ÷ 0.50 × 100 = 4.0%.

What’s the difference between percentage error and percentage uncertainty?
Percentage error compares your result to a known true value (accuracy). Percentage uncertainty comes from the resolution of the apparatus (see the Measurement Uncertainty calculator). Don’t confuse the two.

Standard Form Converter

a × 10ⁿ (where 1 ≤ a < 10)

where a = the mantissa, a number between 1 and 10 · n = the power of 10 (positive for large numbers, negative for small).

⚡ Quick Test Examples

A ribosome is about 0.000000025 m across.

In standard form that’s 2.5 × 10⁻⁸ m — far easier to handle than a string of zeros.

Why do cell biologists use standard form so much?
Biological sizes span a huge range — from nanometre-scale organelles (10⁻⁹ m) to metre-scale organisms — so standard form keeps the numbers readable and makes unit conversions easier.

Rounding (DP & SF)

Round numbers to specified precision

⚡ Quick Test Examples

A calculated reaction rate comes out as 0.008267 s⁻¹.

To 2 s.f. that’s 0.0083 s⁻¹. Quote your answer to the same number of significant figures as your least precise measurement.

Decimal places or significant figures — which should I use?
Significant figures are usually safer in Biology, and you should match the precision of your raw data. Don’t quote more figures than your apparatus justifies — it implies false precision.

Ratio Simplifier

Simplify ratios to lowest terms

where the simplified ratio is found by dividing every value by their highest common factor (HCF). Useful for genetic crosses and surface-area:volume.

⚡ Quick Test Examples

A monohybrid cross gives 152 dominant and 48 recessive offspring.

152 : 48 simplifies to roughly 3 : 1 — the expected Mendelian ratio for a heterozygous cross.

How do I turn counts into a 3:1 (or 9:3:3:1) ratio?
Divide every count by the smallest count, then round to whole numbers. Real data won’t be exact — use a chi-squared test (Genetics tab) to check whether it fits the expected ratio.

Measurement Uncertainty & % Error

Analogue: ± ½ × resolution  |  Digital: ± resolution  |  % error = (uncertainty ÷ reading) × 100

📚 Exam rules

  • Analogue: uncertainty = ± half the smallest division.
  • Digital: uncertainty = ± the full resolution (last digit).
  • Two readings (e.g. start & end of a burette) → add the absolute uncertainties.
  • OCR: never call a result “reliable”; higher resolution ≠ more precise.

Percentage Yield

% Yield = (actual amount ÷ theoretical amount) × 100

where actual = the amount you really obtained · theoretical = the maximum amount possible in theory. The answer can never exceed 100%.

⚡ Quick Test Examples

Genetically modified bacteria could in theory produce 120 g of insulin from a batch; the fermenter actually yields 90 g.

% yield = (90 ÷ 120) × 100 = 75%. The “lost” 25% reflects cells that didn’t express the gene, protein lost in purification, etc.

How is percentage yield different from percentage change?
Percentage yield compares what you got to the maximum possible (it can’t go above 100%). Percentage change compares a new value to a starting value and can be any size, positive or negative.

Logarithms & Antilogs

log₁₀(x)  |  ln(x) = logₑ(x)  |  antilog: 10ˣ or eˣ

where log₁₀ = “what power of 10 gives this number” · ln = natural log (base e ≈ 2.718) · antilog reverses a log. Used for pH, log-scale graphs and microbial growth over many orders of magnitude (A-level).

⚡ Quick Test Examples

A bacterial culture grows from 1×10³ to 1×10⁹ cells. Plotting raw numbers is impossible on normal axes.

Taking log₁₀ gives 3 → 9, a straight line. The number of divisions = log₁₀(10⁹ ÷ 10³) ÷ log₁₀(2) = 6 ÷ 0.301 ≈ 19.9 divisions.

Why do biologists plot microbial growth on a log scale?
Bacterial numbers rise exponentially, spanning many orders of magnitude. A log₁₀ scale turns that exponential curve into a straight line, making the growth rate easy to read and compare.
What’s the difference between log and ln?
log (log₁₀) is base 10 — used for pH and log-scale graphs. ln is base e (≈2.718) — used where things change continuously, like some growth and decay equations. Your calculator’s “log” button is base 10; “ln” is natural log.

Circle: Circumference & Area

Circumference = 2πr = πd  |  Area = πr²

where r = radius · d = diameter = 2r · π ≈ 3.14159. OCR & WJEC require you to recall these.

⚡ Quick Test Examples

A spherical cell seen under the microscope has a radius of 5 µm. Find the cross-sectional area.

Area = πr² = π × 5² = 78.5 µm². Circumference = 2πr = 2 × π × 5 = 31.4 µm.

I only have the diameter — what do I do?
Halve it to get the radius (r = d ÷ 2), then use πr². Or use circumference = πd directly. A common error is using the diameter in πr² instead of the radius.

Straight Line (y = mx + c)

y = mx + c

where m = gradient (rate of change) · c = y-intercept (value of y when x = 0) · enter any value of x to predict y. Used to predict from calibration lines and to read off intercepts (e.g. the compensation point).

A calibration line for a reaction has gradient 0.8 cm³ s⁻¹ and passes through the origin (c = 0). Predict the gas volume after 30 s.

y = mx + c = (0.8 × 30) + 0 = 24 cm³. The y-intercept (c = 0) here means no product at t = 0.

How do I find the intercept from a graph?
The y-intercept (c) is where the line crosses the y-axis, i.e. the value of y when x = 0. The x-intercept is where it crosses the x-axis (y = 0), found from x = −c ÷ m. In biology the intercept can be meaningful — e.g. the light compensation point where photosynthesis equals respiration.

Percentage Difference / % of a Total

% difference = |A − B| ÷ (chosen base) × 100  |  % of total = part ÷ total × 100

where for % difference you must choose which value is the base (denominator) — exams test spotting the right one. For % of a total always divide by the whole, never by another part.

An allele appears 0.72 of the time and a total of 0.90 of measurements were usable. What percentage of the usable total is that allele?

% of total = 0.72 ÷ 0.90 × 100 = 80%. The common mistake is dividing by another part (e.g. the 0.18 non-allele) instead of the 0.90 total.

Which value is the “base” for a percentage difference?
It depends what you’re comparing to. If you ask “how much bigger is A than B”, divide by B. Exam questions test whether you pick the denominator the question implies — always state which value you used as the base.

Power Law (y = a × xᵏ)

y = a × xᵏ  (k can be negative or fractional)

where a = constant · x = the variable (e.g. body mass) · k = the power (often negative for metabolic-rate scaling, e.g. MR = 63 × BM⁻⁰·²⁷). A negative power means xᵏ = 1 ÷ x^|k|.

⚡ Quick Test Examples

Metabolic rate scales with body mass as MR = 63 × BM⁻⁰·²⁷. Find MR for a 30 kg mammal.

MR = 63 × 30⁻⁰·²⁷ = 63 × 0.399 = 25.1. The negative power means larger animals have a lower metabolic rate per kg — the common error is forgetting the reciprocal (30⁻⁰·²⁷ = 1 ÷ 30⁰·²⁷) and getting a huge number.

How do I handle a negative power on a calculator?
x⁻ᵏ means 1 ÷ xᵏ. Use the xʸ (or ^) button with a negative exponent, e.g. 30 ^ (−0.27). Putting the minus in the wrong place, or forgetting the reciprocal, gives an answer that’s far too big — sense-check it.

Linear Interpolation (Read Between Two Points)

y = y₁ + (x − x₁) × (y₂ − y₁) ÷ (x₂ − x₁)

where you have two known points (x₁,y₁) and (x₂,y₂) on a line/calibration curve, and you want the y at an in-between x (or rearrange to find x from y). Used to read values off calibration graphs, e.g. water-potential curves.

On a water-potential calibration line, 0.2 mol dm⁻³ gives −0.5 MPa and 0.4 mol dm⁻³ gives −1.1 MPa. What is the value at 0.3 mol dm⁻³?

y = −0.5 + (0.3 − 0.2) × (−1.1 − −0.5) ÷ (0.4 − 0.2) = −0.5 + 0.1 × (−0.6 ÷ 0.2) = −0.5 + (−0.3) = −0.8 MPa.

When do I use interpolation?
When a value falls between two measured points on a graph or table — you assume the line is straight between them and read off the value in proportion. For a value beyond your data range you’d extrapolate instead (less reliable).

Statistics

Standard Deviation & Standard Error

s = √[Σ(x−x̄)² ÷ (n−1)]  |  SE = s ÷ √n

where s = standard deviation · x = each individual value · x̄ = sample mean · Σ = sum of · n = sample size (number of values) · SE = standard error of the mean

⚡ Quick Test Examples

You measure the height (cm) of 7 wheat plants grown with a fertiliser: 31, 34, 29, 36, 33, 30, 35. To compare this treatment with a control you need a measure of spread and the standard error so you can plot error bars.

Mean = 32.6 cm, n = 7, so s ≈ 2.64 cm and SE = s ÷ √7 ≈ 1.00 cm. Plot the mean ± SE (or ± 2×SE ≈ 95% confidence) as error bars: if the bars of two treatments don’t overlap, the difference is likely to be significant and a t-test is justified.

Why divide by (n − 1) and not n?
At A-level you work with a sample, not the whole population. Dividing by (n − 1) — Bessel’s correction — gives an unbiased estimate of the population standard deviation. Dividing by n alone underestimates the true spread.
What’s the difference between standard deviation and standard error?
Standard deviation (s) measures the spread of the individual data values. Standard error (SE = s ÷ √n) measures how precisely you have estimated the mean; it gets smaller as sample size increases. Use SE for error bars when comparing means.
What do overlapping error bars tell me?
If 95% confidence error bars (≈ mean ± 2 SE) overlap, any difference between the means is probably not significant. If they clearly do not overlap, there is likely a significant difference — confirm with a t-test.

Chi-Squared Test (χ²)

χ² = Σ[(O − E)² ÷ E]

where χ² = chi-squared statistic · O = observed frequency · E = expected frequency · Σ = sum of (over all categories)

📚 Critical Values (p = 0.05)

  • df = 1: 3.84 | df = 2: 5.99 | df = 3: 7.82 | df = 4: 9.49

⚡ Quick Test Examples

A monohybrid cross predicts a 3:1 ratio of purple to white flowers. From 1064 offspring you observe 787 purple and 277 white. Is the deviation from the expected 3:1 ratio significant?

Null hypothesis: there is no significant difference between observed and expected ratios. Expected = 798 purple : 266 white. χ² = (787−798)²/798 + (277−266)²/266 ≈ 0.15 + 0.45 = 0.61. df = categories − 1 = 1, critical value = 3.84 at p = 0.05. Since 0.61 < 3.84, accept the null hypothesis: the difference is not significant and the data fit a 3:1 ratio.

When do I use chi-squared instead of a t-test?
Use chi-squared for frequency / count data placed in categories (e.g. numbers of each phenotype), to test goodness-of-fit to an expected ratio or association between categories. Use a t-test to compare the means of measured continuous data.
How do I work out the degrees of freedom?
For a goodness-of-fit test, df = number of categories − 1. For a contingency table, df = (rows − 1) × (columns − 1). A monohybrid 3:1 cross has 2 categories, so df = 1.
What does it mean if χ² is greater than the critical value?
If χ² ≥ the critical value at p = 0.05, you reject the null hypothesis: the difference between observed and expected is significant (less than 5% probability of arising by chance) — the data do not fit the predicted ratio.

Student’s t-Test (Unpaired)

t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂)

where t = t statistic · x̄₁, x̄₂ = means of group 1 and group 2 · s₁, s₂ = standard deviations of each group · s² = variance (standard deviation squared) · n₁, n₂ = sample size of each group

⚡ Quick Test Examples

You compare the width (mm) of limpet shells on a sheltered shore (12, 14, 13, 15, 11) and an exposed shore (18, 20, 19, 22, 17) to test whether wave exposure affects shell size.

Null hypothesis: there is no significant difference between the mean shell widths of the two shores. Means = 13.0 and 19.2 mm; the calculated t ≈ 8.6. df = n₁ + n₂ − 2 = 8, critical value = 2.31 at p = 0.05. Since 8.6 ≥ 2.31, reject the null hypothesis: the difference in mean shell width between the two shores is significant.

When can I use a t-test?
Use an unpaired t-test to compare the means of two independent groups of continuous data that are approximately normally distributed. If the data are skewed or ordinal, use Mann-Whitney U instead.
What are the degrees of freedom for an unpaired t-test?
df = n₁ + n₂ − 2 (the two sample sizes added together, minus two). Use this df to read the critical value from the t-table at p = 0.05.
How do I decide if the difference is significant?
If the calculated t value ≥ the critical value at p = 0.05, the difference between the two means is significant and you reject the null hypothesis. If t is less than the critical value, the difference is not significant.

Spearman’s Rank Correlation (rs)

rs = 1 − (6Σd²) ÷ n(n² − 1)

where rs = Spearman’s rank correlation coefficient · d = difference in ranks for each pair · Σd² = sum of the squared rank differences · n = number of pairs of values

⚡ Quick Test Examples

Along a transect from a river you record soil moisture (%) and the number of rushes per quadrat at 7 stations to test whether there is a correlation between moisture and rush abundance.

Null hypothesis: there is no significant correlation between soil moisture and rush abundance. After ranking both variables and finding Σd², suppose rs = +0.89 with n = 7 pairs. The critical value at p = 0.05 (n = 7) is 0.714. Since 0.89 ≥ 0.714, reject the null hypothesis: there is a significant positive correlation. (Remember: correlation does not prove causation.)

How do I read the significance for Spearman’s?
Compare the calculated rs (ignore the sign) with the critical value for your n pairs at p = 0.05. If |rs| ≥ the critical value, the correlation is significant and you reject the null hypothesis.
What does the sign of rs mean?
rs ranges from −1 to +1. A positive value means as one variable increases the other tends to increase; a negative value means as one increases the other decreases. A value near 0 means little or no correlation.
Does a significant correlation prove cause and effect?
No. A significant rs shows the two variables are associated, but a third factor could be responsible. Always state that correlation does not, by itself, demonstrate causation.

Mean, Median, Mode & Range

Mean = Σx ÷ n  |  Median = middle value  |  Range = max − min  |  IQR = Q₃ − Q₁

where Σx = sum of all values · n = number of values · Q₁ = lower quartile · Q₃ = upper quartile · IQR = interquartile range (the middle 50% of the data)

📚 Which average?

  • Mean — interval/continuous data with no extreme outliers.
  • Median — skewed data or when an outlier would distort the mean.
  • Mode — categorical/discrete data (most frequent value).

⚡ Quick Test Examples

You count the number of stomata in 7 fields of view on a leaf epidermis peel: 12, 15, 11, 14, 13, 16, 13. You need an average and a measure of spread for your results table.

Mean = 94 ÷ 7 ≈ 13.4 stomata; ordered data 11,11,13,13,14,15,16 gives median = 13 and mode = 13; range = 16 − 11 = 5. Because the values are fairly symmetrical with no extreme outlier, the mean is the most appropriate average to quote.

When should I use the median instead of the mean?
Use the median when the data are skewed or contain an outlier that would drag the mean away from the bulk of the values (e.g. one unusually large count). The median is more representative of the typical value in those cases.
What’s the difference between range and interquartile range?
Range (max − min) uses only the two extreme values, so a single outlier can inflate it. The interquartile range (Q₃ − Q₁) describes the spread of the middle 50% and is far less affected by outliers.

Mean from a Frequency Table

Mean = Σ(value × frequency) ÷ Σfrequency

where each value occurs a number of times given by its frequency. Multiply each value by its frequency, add those up, and divide by the total frequency (the number of items).

⚡ Quick Test Examples

You record the number of offspring per nest for 46 nests as a frequency table: 0 (×5), 1 (×12), 2 (×18), 3 (×9), 4 (×2).

Σ(value × frequency) = 0 + 12 + 36 + 27 + 8 = 83. Σfrequency = 46. Mean = 83 ÷ 46 ≈ 1.80 offspring per nest.

Why can’t I just average the values 0,1,2,3,4?
Because each value occurs a different number of times. Averaging the values alone ignores how common each one is. You must weight each value by its frequency, then divide by the total number of items.

Paired t-Test

t = d̄ ÷ (s_d ÷ √n)  |  df = n − 1

where t = t statistic · d̄ = mean of the differences between paired readings · s_d = standard deviation of those differences · n = number of pairs · df = degrees of freedom

For “before vs after” data from the same individuals (e.g. heart rate before/after caffeine). Enter the two readings for each pair, in the same order.

⚡ Quick Test Examples

You measure each volunteer’s resting heart rate (bpm) before and after a caffeinated drink: before = 64, 70, 61, 72, 68; after = 78, 85, 74, 88, 80. Because each pair comes from the same person, a paired t-test is appropriate.

Null hypothesis: there is no significant difference between heart rate before and after caffeine. The differences are 14, 15, 13, 16, 12 (mean d̄ ≈ 14), giving a large t value. df = n − 1 = 4, critical value = 2.78 at p = 0.05. As t ≥ 2.78, reject the null hypothesis: caffeine produced a significant increase in heart rate.

When do I use a paired rather than an unpaired t-test?
Use a paired t-test when each value in one set is naturally linked to a value in the other — typically the same individual measured twice (before/after) or matched pairs. Pairing removes variation between individuals, making the test more sensitive.
What are the degrees of freedom for a paired t-test?
df = n − 1, where n is the number of pairs (not the total number of readings). With 5 pairs, df = 4.

Mann-Whitney U Test

U = n₁n₂ + [n₁(n₁+1) ÷ 2] − ΣR₁  (use the smaller U)

where U = Mann-Whitney U statistic · n₁, n₂ = sample sizes of the two groups · ΣR₁ = sum of the ranks of group 1 · the smaller of the two U values is compared with the critical value

Non-parametric test for a difference between two medians — use for skewed/ordinal data, or small samples that aren’t normally distributed (AQA/Edexcel).

⚡ Quick Test Examples

You compare the abundance of mayfly nymphs (an ordinal pollution indicator) at a clean site (12, 15, 11, 9, 14) and a polluted site (18, 22, 17, 25, 20). The counts are small and not normally distributed, so a Mann-Whitney U test is used.

Null hypothesis: there is no significant difference between the median abundances of the two sites. After ranking all values together and summing the ranks, the smaller U = 0. For n₁ = n₂ = 5, the critical value at p = 0.05 is 2. Since U (0) ≤ 2, reject the null hypothesis: the difference in median abundance between the two sites is significant.

How is significance judged for Mann-Whitney U?
Unlike chi-squared or t, the result is significant when the calculated U is small. If the smaller U ≤ the critical value at p = 0.05, the difference between the two medians is significant and you reject the null hypothesis.
When should I choose Mann-Whitney over a t-test?
Use Mann-Whitney U when the data are ordinal, skewed, or come from small samples that are not normally distributed — situations where a t-test’s assumptions are not met. It compares medians rather than means.
Why do I rank all the data together?
You pool both samples and rank every value from smallest to largest, giving tied values the mean of the ranks they share. The test then asks whether the ranks of one group are systematically higher than the other.

Unpaired t-Test from Given Means & Variances

t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂)  |  df = (n₁ − 1) + (n₂ − 1)

where x̄ = sample mean · s² = variance (= SD²) · n = sample size. Use this when the question gives you the means and variances (common in WJEC/Eduqas) rather than raw data.

Only have standard deviations? Tick to enter SD instead of variance (the calculator squares it).

⚡ Quick Test Examples

Topshell lengths: exposed shore mean 23.2 mm, variance 30.6, n=10; sheltered shore mean 28.0 mm, variance 14.0, n=10.

t = |23.2 − 28.0| ÷ √(30.6/10 + 14.0/10) = 4.8 ÷ √4.46 = 4.8 ÷ 2.11 = 2.27, df = 18. Compare with the critical value (2.10 at p=0.05): 2.27 > 2.10, so the difference is significant.

Variance or standard deviation — which do I enter?
The formula uses variance (s²). If the question gives you the standard deviation (s), tick the box and the calculator will square it for you. Variance = SD², so a SD of 5 means a variance of 25.
What degrees of freedom do I use?
For the unpaired t-test, df = (n₁ − 1) + (n₂ − 1) = n₁ + n₂ − 2. Use this row of the critical-value table at p = 0.05.

P-value Interpreter (Significance Decision)

If P ≤ 0.05 → significant, reject H₀  |  If P > 0.05 → not significant, accept H₀

where P = probability the result is due to chance · 0.05 = the usual 5% significance threshold (95% confidence). A smaller P = stronger evidence. Tells you the decision and the exam wording.

⚡ Quick Test Examples

A t-test comparing two means gives P = 0.03.

0.03 < 0.05, so the result is significant: reject the null hypothesis. Say “there is a significant difference between the means” — not “the results are significant”.

What exactly does P < 0.05 mean?
There is less than a 5% probability that the difference/association/correlation you found is due to chance alone. That’s strong enough evidence to reject the null hypothesis at the 95% confidence level.
What’s the wording examiners want?
Name what is significant: a “significant difference” (t/Mann-Whitney), a “significant association” (chi-squared), or a “significant correlation” (Spearman’s). Saying “the results are significant” is too vague and loses the mark.

Which Statistical Test Should I Use?

Answer two questions → the right test

how it works Pick what you’re investigating and the type of data, and this tells you the standard A-level test to choose and why.

You want to know if mean shell length differs between two shores, and the data are normally distributed and unpaired.

→ Use an unpaired Student’s t-test. For categories/frequencies you’d use chi-squared; for a relationship between two ranked variables, Spearman’s rank.

How do I know if my data are “normally distributed”?
Plot a frequency histogram — normally distributed data form a symmetrical bell shape. If it’s skewed or bimodal, or the data are ordinal (ranked categories), use a non-parametric test (Mann-Whitney or Spearman’s) instead.

Ecology & Sampling

Simpson’s Diversity Index

D = 1 − [Σn(n−1) ÷ N(N−1)]  or  D = N(N−1) ÷ Σn(n−1)

where D = Simpson’s diversity index · n = number of individuals of one species · N = total number of individuals of all species · Σ = sum across every species

⚡ Quick Test Examples

A student samples ground beetles in woodland using pitfall traps and records four species with counts of 10, 12, 8 and 6 (N = 36). Calculate Simpson’s Diversity Index.

Σn(n−1) = (10×9)+(12×11)+(8×7)+(6×5) = 90+132+56+30 = 308. N(N−1) = 36×35 = 1260. D = 1 − (308 ÷ 1260) = 1 − 0.244 = 0.76. A value close to 1 indicates high diversity, so this community is relatively diverse and likely stable.

What does a high Simpson’s index value mean?
Using this form (D = 1 − Σ…), values run from 0 to almost 1. A value near 1 means high diversity — many species with evenly spread numbers — which usually indicates a more stable ecosystem better able to resist environmental change. A value near 0 means low diversity, often where one species dominates.
What’s the difference between species richness and diversity?
Species richness is simply the number of different species present. Diversity (Simpson’s index) considers both richness and evenness — how the individuals are spread between species. Two habitats can have the same richness but very different diversity if one is dominated by a single abundant species.
Why use N(N−1) rather than N²?
Sampling is done without replacement — once an individual is counted it isn’t returned. N(N−1) gives the number of ways to draw two different individuals, which correctly models the probability that two randomly chosen individuals belong to the same species.

Lincoln Index (Mark-Release-Recapture)

N = (n₁ × n₂) ÷ m₂

where N = estimated total population size · n₁ = number caught, marked and released in the first sample · n₂ = total number caught in the second sample · m₂ = number of marked individuals recaptured in the second sample

⚡ Quick Test Examples

An ecologist catches 60 minnows from a pond, marks them with a harmless fin clip and releases them. A few days later 80 minnows are caught, of which 20 carry the mark. Estimate the population.

N = (n₁ × n₂) ÷ m₂ = (60 × 80) ÷ 20 = 4800 ÷ 20 = 240 minnows. This estimate assumes the marked fish mixed randomly back into the population before the second sample.

What assumptions does mark-release-recapture make?
The method assumes: no migration in or out and no significant births or deaths between samples (closed population); marks do not wear off, harm the animal or make it more visible to predators; and marked individuals mix randomly and have time to redistribute before the second sample. Breaking any assumption biases the estimate.
Why is the estimate sometimes inaccurate?
If marked animals avoid recapture (trap-shyness), or if the marks affect survival, m₂ falls and N is over-estimated. Small samples also give large random error. It’s an estimate, not an exact count, so results are usually reported to appropriate significant figures.

Percentage Cover / Density

% Cover = (hits ÷ total) × 100

where hits = number of point-frame pins (or quadrat squares) touching the species · total = total number of pins or squares sampled · % cover = proportion of ground occupied by that species

⚡ Quick Test Examples

Using a point quadrat with 50 pins lowered onto a patch of grassland, a student records that clover touches the pin on 18 occasions. Calculate the percentage cover of clover.

% cover = (hits ÷ total) × 100 = (18 ÷ 50) × 100 = 36%. Percentage cover is an estimate of abundance that works well for plants that are hard to count as individuals, such as creeping or matted species.

When is percentage cover used instead of counting individuals?
Cover is used for plants where individuals are hard to define or count — grasses, mosses, lichens and creeping species. It estimates abundance by the proportion of ground occupied rather than by number of organisms.
Can percentage cover add up to more than 100%?
Yes. Because plants grow in layers, a single point can touch more than one species, so the total cover across all species in a quadrat can exceed 100%. This is expected and not an error.

Quadrat Sampling – Population Estimate

Population = Density × Total area

where Density = mean number of individuals per quadrat ÷ quadrat area (individuals per m²) · Quadrat area = area of one quadrat (m²) · Total area = total size of the habitat being estimated (m²)

⚡ Quick Test Examples

A student counts dandelions in six 0.25 m² quadrats placed at random co-ordinates: 5, 8, 3, 6, 7 and 4. The whole field is 5000 m². Estimate the dandelion population.

Mean per quadrat = (5+8+3+6+7+4) ÷ 6 = 33 ÷ 6 = 5.5. Density = 5.5 ÷ 0.25 = 22 per m². Population = 22 × 5000 = 110 000 dandelions. Quadrats must be placed using random co-ordinates so the sample is unbiased and representative.

Why must quadrats be placed randomly?
Random placement (using random number co-ordinates on a grid) removes sampling bias, so the mean density fairly represents the whole area. Taking more quadrats reduces the effect of chance and gives a more reliable estimate.
When would you use a belt or line transect instead?
Use a transect when there is an environmental gradient — for example moving up a rocky shore or away from a path. Random quadrats are used when distribution is fairly even and you want overall abundance, not change along a gradient.

Population Growth

Nₜ = N₀ × eʳᵗ

where Nₜ = population size after time t · N₀ = starting (initial) population size · r = growth rate (per unit time) · t = time elapsed · e = the natural exponential constant (≈ 2.718)

⚡ Quick Test Examples

A bacterial culture starts with 200 cells and grows exponentially with a growth rate r = 0.35 per hour. Estimate the population after 8 hours.

Nₜ = N₀ × e^(rt) = 200 × e^(0.35 × 8) = 200 × e^2.8 = 200 × 16.44 ≈ 3289 cells. Exponential growth like this only continues while resources are unlimited (the log phase); it slows once a limiting factor such as nutrients or space takes effect.

Why is e used in the growth equation?
e (≈ 2.718) is the natural base for continuous growth. When a population reproduces continuously rather than in fixed steps, N₀eʳᵗ describes how the size changes smoothly over time. A positive r means growth, a negative r means decline.
When does exponential growth stop?
In the real world, growth slows when limiting factors (food, space, oxygen, build-up of waste) reduce the per-capita growth rate. The population then levels off near the carrying capacity, giving the S-shaped (sigmoid) curve rather than continued exponential rise.

Microscopy & Cells

Magnification Calculator (MAI)

M = I ÷ A  |  I = M × A  |  A = I ÷ M
  • M = magnification — how many times bigger the image is (no units, written as ×)
  • I = image size — the measured size of the drawing/photograph
  • A = actual size — the true size of the real specimen
  • Golden rule: I and A must be in the same unit before dividing. Convert mm → µm by ×1000 (and µm → nm by ×1000).

⚡ Quick Test Examples

A photomicrograph of a cell measures 45 mm across and the magnification is ×900. What is the actual diameter of the cell?

A = I ÷ M = 45 mm ÷ 900 = 0.05 mm. Convert to µm: 0.05 × 1000 = 50 µm. (A typical animal cell — sensible answer.)

Why must I and A be in the same unit?
Magnification is a ratio, so the units must cancel. If the image is in mm and the actual size in µm, convert one of them first (mm → µm by ×1000). A common exam error is dividing 45 mm by 50 µm and getting a nonsense answer.
How do I use a scale bar?
Measure the printed length of the scale bar with a ruler (this is the image size, I) and read the real length it represents (this is the actual size, A). Magnification = measured bar length ÷ real length it represents — converting to the same unit first.
Does magnification have units?
No. It is a pure ratio of two lengths, so it is written as a number with a × symbol, e.g. ×400. Don’t write mm or µm after it.

Graticule Calibration

1 EPU = (Stage divs × µm/div) ÷ EPU aligned
  • EPU = eyepiece (graticule) unit — the arbitrary divisions on the scale in the eyepiece
  • Stage divisions = divisions on the stage micrometer of known size
  • µm per stage division = the true size of one stage division (usually 10 µm)
  • Result (1 EPU) = the real size in µm represented by one eyepiece unit at that magnification

⚡ Quick Test Examples

Looking down the microscope, 40 eyepiece units line up exactly with 10 stage-micrometer divisions. Each stage division is 10 µm. What is one eyepiece unit worth?

1 EPU = (10 × 10 µm) ÷ 40 = 100 ÷ 40 = 2.5 µm per eyepiece unit. A cell measuring 16 EPU is therefore 16 × 2.5 = 40 µm across.

Why do I have to recalibrate for each objective lens?
The eyepiece graticule is fixed, but the stage micrometer image is magnified differently by each objective. So the µm value of one eyepiece unit changes every time you switch magnification — you must re-align the two scales and recalculate.
What is the stage micrometer for?
It is a slide with a scale of known, true dimensions (commonly 100 divisions of 10 µm each). It is used once to calibrate the unknown eyepiece graticule; after that you measure real specimens in eyepiece units and convert.

Mitotic Index

MI = (Dividing ÷ Total) × 100
  • Mitotic index = the proportion of cells in a sample that are visibly in mitosis = cells in mitosis ÷ total cells (often ×100 for a %)
  • Cells in mitosis = those showing condensed chromosomes (prophase, metaphase, anaphase or telophase)
  • Total cells = every cell counted in the field, including those in interphase
  • A higher mitotic index means a faster-dividing tissue (e.g. a root tip or a tumour).

⚡ Quick Test Examples

In a stained root-tip squash, a student counts 200 cells and finds 15 of them showing condensed chromosomes. What is the mitotic index?

MI = (15 ÷ 200) × 100 = 7.5%. Expressed as a proportion this is 0.075. A high value is expected in a root tip, where cells divide rapidly to grow the root.

Is the mitotic index a percentage or a fraction?
Either — check the question. The raw mitotic index is dividing ÷ total (a decimal such as 0.075). Multiply by 100 to give a percentage (7.5%). Read the unit asked for carefully.
Which cells count as “in mitosis”?
Only cells where chromosomes are visible and condensed — prophase, metaphase, anaphase and telophase. Cells in interphase (no visible chromosomes) are not counted as dividing, even though most cells in a sample are usually in interphase.
Why is a high mitotic index linked to cancer?
Cancer cells lose control of the cell cycle and divide far more often than normal, so a tumour sample shows an unusually high mitotic index. Measuring it helps assess how aggressive a tumour is.

Water Potential

ψ = ψs + ψp
  • ψ (psi) = water potential in kPa — for any solution this is always ≤ 0; pure water is exactly 0 (the maximum possible)
  • ψs = solute (osmotic) potential — always negative; the more solute dissolved, the more negative it becomes
  • ψp = pressure potential — usually positive in a turgid plant cell (wall pushing in)
  • Direction of flow: water moves by osmosis from HIGH (less negative) ψ to LOW (more negative) ψ.

⚡ Quick Test Examples

A turgid plant cell has a solute potential of −500 kPa and a pressure potential of +100 kPa. What is its water potential, and which way will water move if it is placed in pure water?

ψ = ψs + ψp = (−500) + (+100) = −400 kPa. Pure water has ψ = 0, which is higher (less negative) than −400 kPa, so water moves from the pure water INTO the cell.

Why is water potential always negative for a solution?
Pure water has the highest possible water potential, defined as 0 kPa. Adding any solute lowers it, so any real solution has a negative water potential. You can’t get a value above 0.
Which way does water actually move?
By osmosis, from a region of higher (less negative) water potential to a region of lower (more negative) water potential, down the water-potential gradient — until the two are equal.
What does pressure potential represent?
In a plant cell it is the inward push of the cellulose cell wall on the contents as the cell takes up water and becomes turgid. It is normally positive, which raises (makes less negative) the overall water potential.

Surface Area : Volume Ratio

Cube: SA=6s², V=s³ | Sphere: SA=4πr², V=⁴⁄₃πr³ | Cylinder: SA=2πr(r+h), V=πr²h
  • SA:V = the ratio of total surface area to volume (e.g. 6:1, often written as a single number, 6)
  • s = side length of a cube; r = radius; h = height of a cylinder
  • Key idea: smaller objects have a LARGER surface area : volume ratio, so they exchange substances faster relative to their size
  • This is why cells are small, and why large organisms need specialised exchange surfaces (lungs, gills) and transport systems (blood).

⚡ Compare Different Sizes (cube)

Compare a cube-shaped “cell” of side 1 unit with one of side 5 units. Which exchanges materials more efficiently with its surroundings?

Side 1: SA = 6×1² = 6, V = 1³ = 1, ratio = 6:1. Side 5: SA = 6×5² = 150, V = 5³ = 125, ratio = 1.2:1. The smaller cube has the larger SA:V, so substances diffuse in and out faster relative to its volume — which is why real cells stay small.

Why do small cells have a bigger SA:V ratio?
As an object gets bigger, its volume (length³) grows faster than its surface area (length²). So the ratio of surface to volume falls as size increases. Small cells therefore have proportionally more membrane for exchange per unit of contents.
Why does SA:V matter in biology?
A high SA:V means substances (oxygen, glucose, waste, heat) can be exchanged across the surface quickly enough to supply the whole volume by diffusion. Large organisms have a low SA:V, so they need specialised exchange surfaces (alveoli, gills, villi) and a transport system (circulation) to overcome this.
Does shape affect the ratio too?
Yes. For the same volume, a flat or elongated shape has more surface area than a sphere, giving a higher SA:V. This is why exchange surfaces such as root hair cells and the small intestine are folded, flattened or elongated to maximise surface area.

Rates & Enzymes

Rate of Reaction

Rate = 1 ÷ time (s⁻¹)  or  Rate = amount ÷ time
  • time = time taken for the reaction (or for a measurable end-point), usually in seconds (s)
  • Rate (1/time) = how fast the reaction goes; units s⁻¹
  • amount = quantity of product made or reactant used (e.g. cm³ of gas, mass)

A faster reaction takes less time, so 1/time is larger. Using 1/time lets you compare reactions when all you measured was how long they took.

In a disappearing-cross (sodium thiosulfate + acid) experiment, the cross took 45 s to vanish. What is the rate?

Rate = 1 ÷ time = 1 ÷ 45 = 0.022 s⁻¹. A reaction that finished in 20 s would have a higher rate (1 ÷ 20 = 0.050 s⁻¹).

Why use 1 ÷ time as a measure of rate?
When you can only measure how long a reaction takes (e.g. time for a cross to disappear), 1/time is proportional to rate. A shorter time means a faster reaction, so 1/time goes up — making it easy to plot against temperature or concentration.
What units does 1 ÷ time give?
If time is in seconds, 1/time is in s⁻¹. If you measured an amount of product instead, divide that amount by time to get units like cm³ s⁻¹ or mg min⁻¹.

Q₁₀ Temperature Coefficient

Q₁₀ = (R₂ ÷ R₁)^(10 ÷ ΔT)
  • R₁ = rate of reaction at the lower temperature T₁
  • R₂ = rate of reaction at the higher temperature T₂
  • T₁, T₂ = the two temperatures in °C
  • ΔT = T₂ − T₁ (the temperature rise)
  • Q₁₀ = the factor by which rate changes per 10 °C rise

Q₁₀ measures how sensitive a reaction is to temperature. For many enzyme-controlled reactions Q₁₀ ≈ 2 (rate roughly doubles per 10 °C). This logic only holds up to about 40 °C — above that enzymes denature and the rate falls, so Q₁₀ breaks down.

An enzyme reaction has a rate of 5 (arb. units) at 20 °C and 10 at 30 °C. Find Q₁₀.

Q₁₀ = (R₂ ÷ R₁)^(10 ÷ ΔT) = (10 ÷ 5)^(10 ÷ 10) = 2^1 = 2.0. The rate doubles for a 10 °C rise — typical of an enzyme reaction below its optimum.

What does a Q₁₀ of 2 mean?
The reaction rate doubles for every 10 °C rise in temperature. A Q₁₀ near 1 means temperature has little effect; a Q₁₀ much greater than 2 means the reaction is very temperature-sensitive.
Why doesn’t Q₁₀ work at high temperatures?
Above roughly 40 °C the enzyme starts to denature: its active site changes shape and rate falls instead of rising. Q₁₀ assumes rate keeps increasing with temperature, so it is only valid below the optimum.
Does ΔT have to be exactly 10 °C?
No. The formula corrects for any temperature gap using the 10 ÷ ΔT exponent, so you can use any two temperatures and still get the rate change per 10 °C.

Respirometer: O₂ Uptake (Distance → Volume → Rate)

Volume = π r² × distance  |  Rate = volume ÷ time  (÷ mass for per-gram rate)
  • distance = how far the manometer fluid / bubble moved along the capillary, in mm
  • r = radius of the capillary tube bore, in mm (= diameter ÷ 2)
  • π r² = cross-sectional area of the tube — multiply by distance to get the volume of O₂ absorbed (mm³)
  • time = time over which the fluid moved (min)
  • mass = mass of organisms (g), if you want a fair per-gram rate

O₂ is used up and CO₂ is absorbed by soda lime/KOH, so pressure falls and fluid moves toward the organisms. The volume of O₂ used = distance moved × cross-sectional area of the tube. Dividing by mass lets you fairly compare organisms of different sizes (mm³ g⁻¹ min⁻¹).

Woodlice in a respirometer move the fluid 15 mm in 5 minutes; the capillary tube has a bore radius of 0.5 mm.

Volume of O₂ = πr² × distance = π × 0.5² × 15 = 11.78 mm³. Rate = 11.78 ÷ 5 = 2.36 mm³ min⁻¹ of oxygen taken up.

Why convert distance into a volume?
Distance alone (mm min⁻¹) depends on how wide your tube is, so it can’t be compared between set-ups. Volume of O₂ (mm³ min⁻¹), found by multiplying by the tube’s cross-sectional area πr², is the true measure of respiration rate.
Why divide by mass?
A bigger organism uses more oxygen simply because it’s bigger. Dividing by mass gives the rate per gram (mm³ g⁻¹ min⁻¹), so you can fairly compare the metabolic rate of different organisms or sizes.
I only know the tube diameter, not the radius.
Halve it: radius = diameter ÷ 2. A common mistake is to put the diameter straight into πr², which makes the volume four times too big.

Photosynthesis Rate (Bubbles or O₂ Volume)

Bubble method: rate = bubbles ÷ time  |  Volume method: volume = π r² × distance, rate = volume ÷ time
  • bubbles = number of O₂ bubbles from the pondweed (e.g. Elodea/Cabomba)
  • distance = length of the O₂ bubble/gas column drawn along the capillary (mm) — the more accurate method
  • r = radius of the capillary tube bore (mm); π r² converts the bubble length into a volume of O₂ (mm³)
  • time = time of collection (min)

O₂ is a product of photosynthesis, so a faster rate gives more bubbles, or a longer/larger gas volume, per minute. Measuring the volume (capillary bubble length × πr², or gas collected in a syringe) is more accurate than counting bubbles, because bubble size varies.

Pondweed produced a gas bubble that filled 22 mm of a capillary tube (bore radius 0.5 mm) in 3 minutes.

Volume of O₂ = πr² × length = π × 0.5² × 22 = 17.3 mm³. Rate = 17.3 ÷ 3 = 5.76 mm³ min⁻¹ — a true volume rate, unlike a bubble count.

Why is the volume method better than counting bubbles?
Bubbles vary in size, so the same number of bubbles can be very different volumes of gas. Drawing the gas into a capillary and measuring its length × πr² (or collecting it in a syringe) gives an accurate, quantitative volume of oxygen per minute.
What factors change the rate of photosynthesis?
Light intensity, carbon dioxide concentration and temperature are the main limiting factors. Increasing one only raises the rate until another becomes limiting. Light intensity follows the inverse-square law — see the Light Intensity calculator.

Potometer: Rate of Water Uptake (Transpiration)

Volume = π r² × distance  |  Rate = volume ÷ time  (÷ leaf area for a fair rate)
  • distance = how far the air bubble moved along the capillary scale (mm)
  • r = radius of the capillary tube bore (mm); π r² = cross-sectional area → volume of water taken up (mm³)
  • time = time over which the bubble moved (min)
  • leaf area = total surface area of leaves (cm² or mm²), if you want rate per unit area — the fairest comparison

A potometer measures water uptake, which is assumed to equal water lost in transpiration. Distance moved × πr² gives the volume of water taken up. Dividing by leaf surface area gives a rate per unit area, so plants with different leaf sizes can be compared fairly.

In a potometer the air bubble moved 60 mm in 10 minutes; the capillary bore radius is 0.5 mm and the shoot has 25 cm² of leaf.

Volume = πr² × distance = π × 0.5² × 60 = 47.1 mm³. Rate = 47.1 ÷ 10 = 4.71 mm³ min⁻¹. Per leaf area = 4.71 ÷ 25 = 0.19 mm³ cm⁻² min⁻¹.

Why divide by leaf surface area?
A plant with more leaves transpires more simply because it has more surface area. Dividing the rate by total leaf area (giving mm³ cm⁻² min⁻¹) lets you fairly compare transpiration between shoots of different sizes or species.
Does the potometer measure uptake or transpiration?
It measures water uptake. We assume this equals the water lost by transpiration, but a small amount is used in photosynthesis and to keep cells turgid, so uptake slightly overestimates transpiration.
How do I turn mm³ into cm³?
Divide by 1000 (1 cm³ = 1000 mm³). So 47.1 mm³ = 0.0471 cm³. Use the Unit Converter for any volume conversion.

Light Intensity (Inverse-Square Law)

light intensity ∝ 1 ÷ distance²  |  I₂ = I₁ × (d₁ ÷ d₂)²

where light intensity falls with the square of the distance from the lamp. I = intensity (relative units, or = 1/d²) · d = distance from the lamp. Doubling the distance gives a quarter of the intensity.

⚡ Quick Test Examples

In a pondweed photosynthesis experiment, the lamp is moved from 10 cm to 20 cm from the plant. How does the light intensity change?

I₂ = I₁ × (d₁ ÷ d₂)² = 100 × (10 ÷ 20)² = 100 × 0.25 = 25 units. Doubling the distance quarters the light intensity, so the rate of photosynthesis should fall sharply.

Why use 1 ÷ d² as the light intensity?
Light spreads out over a larger area as it travels, so its intensity falls with the square of the distance. In photosynthesis practicals we plot rate against 1/d² (a relative light intensity) because we can’t easily measure the actual intensity in lux.
Why must distance be controlled carefully?
Because intensity changes with the square of distance, even a small error in measuring the lamp distance causes a large error in the light intensity — so it must be measured precisely and kept constant when it’s a control variable.

Rate from a Graph (Gradient / Tangent)

Rate = gradient = Δy ÷ Δx = (y₂ − y₁) ÷ (x₂ − x₁)
  • x₁, y₁ = coordinates of the first point on the line/tangent
  • x₂, y₂ = coordinates of the second point
  • Δy = y₂ − y₁ (the rise, the change up the y-axis)
  • Δx = x₂ − x₁ (the run, the change along the x-axis)
  • gradient = Δy ÷ Δx = rise over run = the rate

For a straight line, read any two points. For a curve, draw a tangent that just touches the curve at the point of interest and take two points off the tangent — this gives the rate at that instant. The initial rate is the gradient of the tangent drawn at t = 0.

Read two points off a straight line, or off a tangent drawn to a curve (for the rate at an instant). Enter both points.

A tangent drawn at t = 0 on a graph of gas volume against time passes through (0, 0) and (30, 24), with volume in cm³ and time in s. Find the initial rate.

Gradient = Δy ÷ Δx = (24 − 0) ÷ (30 − 0) = 24 ÷ 30 = 0.8 cm³ s⁻¹. Because the tangent was drawn at t = 0, this is the initial rate of reaction.

How do I find the rate at a single instant on a curve?
Draw a tangent — a straight line that just touches the curve at that point — then pick two well-spaced points on the tangent and work out its gradient (Δy ÷ Δx). That gradient is the rate at that moment.
What is the “initial rate” and why does it matter?
The initial rate is the gradient of the tangent drawn at t = 0, when the reaction is fastest and conditions are clearest. It is the fairest value to compare between experiments because product/substrate levels haven’t yet changed.
Why must I label units on the gradient?
The gradient’s units come from the axes: y-axis units ÷ x-axis units. For example cm³ on the y-axis and s on the x-axis give cm³ s⁻¹. Marks are often lost for a missing or wrong unit.

Genetics

Hardy-Weinberg Equilibrium

p + q = 1  |  p² + 2pq + q² = 1
  • p = frequency of the dominant allele
  • q = frequency of the recessive allele
  • p² = frequency of homozygous dominant genotype
  • 2pq = frequency of heterozygous genotype
  • q² = frequency of homozygous recessive genotype

Allele frequencies (p, q) describe single alleles in the gene pool; genotype frequencies (p², 2pq, q²) describe the resulting individuals. A recessive phenotype percentage equals q², so take its square root (√q²) to find q. Assumptions: large population, random mating, no selection, no mutation, no migration.

1 in 2500 people in a population has a recessive condition. What proportion are carriers (heterozygous)?

q² = 1/2500 = 0.0004, so q = √0.0004 = 0.02. Then p = 1 − q = 0.98. Carrier frequency 2pq = 2 × 0.98 × 0.02 = 0.0392, i.e. about 3.9% (roughly 1 in 26) are carriers.

Why do I square-root the frequency of affected individuals?
Affected individuals show the recessive phenotype, so they are homozygous recessive, with genotype frequency q². To recover the recessive allele frequency q you take the square root: q = √q². You cannot square-root carrier or dominant frequencies this way, because those genotype groups (2pq, p²) are not simple squares of a single allele frequency.
What’s the difference between allele frequency and genotype frequency?
Allele frequencies (p and q) measure how common each allele is in the whole gene pool and always sum to 1 (p + q = 1). Genotype frequencies (p², 2pq, q²) measure the proportion of individuals of each genotype and also sum to 1. A common exam error is reporting q² when the question asks for q, or vice versa.
What assumptions must hold for Hardy-Weinberg equilibrium?
A large population (no genetic drift), random mating, no natural selection, no mutation, and no migration (no gene flow in or out). If these hold, allele and genotype frequencies stay constant between generations.

Genetics Chi-Squared

Test observed vs expected Mendelian ratios
  • O = observed count (the numbers you actually counted)
  • E = expected count (from the predicted Mendelian ratio)
  • Σ = sum of the terms across all phenotype classes

χ² = Σ (O − E)² / E. Degrees of freedom = (number of phenotype classes − 1). Compare χ² to the critical value at p = 0.05: if χ² ≥ the critical value, the observed ratio differs significantly from the expected ratio (reject the null hypothesis).

A monohybrid cross predicts a 3:1 ratio. From 423 offspring you observe 315 tall and 108 short. Does this differ significantly from 3:1?

Expected: 423 × 3/4 = 317.25 tall, 423 × 1/4 = 105.75 short. χ² = (315−317.25)²/317.25 + (108−105.75)²/105.75 = 0.016 + 0.048 = 0.064. df = 2 − 1 = 1; critical value at p = 0.05 is 3.84. As 0.064 < 3.84, the difference is not significant — the data fit a 3:1 ratio.

How many degrees of freedom should I use?
df = (number of phenotype classes − 1). A 3:1 ratio has 2 classes, so df = 1; a 9:3:3:1 ratio has 4 classes, so df = 3. Use df to read the correct critical value from the χ² table.
What does it mean if χ² is greater than the critical value?
If χ² ≥ the critical value at p = 0.05, the difference between observed and expected is statistically significant: the observed ratio differs significantly from the expected ratio and the difference is unlikely to be due to chance alone. You reject the null hypothesis (which states there is no significant difference).
Why do I need expected counts, not the ratio itself?
Chi-squared compares actual numbers, so you must convert the ratio into expected counts by sharing the total offspring in the ratio’s proportions (e.g. for 3:1 with 423 offspring, 3/4 and 1/4 of 423). Using raw ratio numbers instead of expected counts gives a wrong χ².

Punnett Square Generator

Monohybrid (e.g. Aa × Aa → 3:1) or Dihybrid (e.g. AaBb × AaBb → 9:3:3:1)
2 letters = monohybrid · 4 letters = dihybrid. Use the same letter in upper/lower case for each gene (e.g. AaBb).

Cross two pea plants heterozygous for two genes, AaBb × AaBb. What phenotypic ratio is expected in the offspring?

Each parent produces four gamete types (AB, Ab, aB, ab) by independent assortment. The 16-box Punnett square gives a 9 : 3 : 3 : 1 phenotypic ratio — 9 showing both dominant traits, 3 dominant-A/recessive-b, 3 recessive-a/dominant-B, 1 showing both recessive traits.

Why does AaBb × AaBb give a 9:3:3:1 ratio?
Because the two genes assort independently (independent assortment, Mendel’s second law), each behaves like a separate 3:1 monohybrid cross. Combining them (3:1) × (3:1) gives 9:3:3:1 across the 16 equally likely combinations.
How do I enter a monohybrid versus a dihybrid cross?
Use 2 letters (e.g. Aa) for a monohybrid cross and 4 letters (e.g. AaBb) for a dihybrid cross. Use the same letter in upper and lower case for each gene, where the capital represents the dominant allele.
Does a 9:3:3:1 ratio always appear in real data?
No — it is the expected ratio. Real offspring numbers vary by chance, and the ratio can be altered by linkage, epistasis, or small sample sizes. A chi-squared test is used to decide whether observed numbers fit the expected 9:3:3:1.

DNA/RNA Base Calculator

Chargaff’s Rules: A=T, G=C
  • %A = %T (adenine pairs with thymine)
  • %G = %C (guanine pairs with cytosine)

In double-stranded DNA the four bases total 100%, so %A + %T + %G + %C = 100. Because complementary bases are equal, %A + %G = 50% and %T + %C = 50%. (In RNA, uracil (U) replaces thymine, so A pairs with U.)

A sample of double-stranded DNA contains 30% adenine. Work out the percentage of the other three bases.

By Chargaff’s rules %T = %A = 30%. That leaves 100 − 30 − 30 = 40% shared equally between G and C, so %G = %C = 20%. Final: A 30%, T 30%, G 20%, C 20%.

What are Chargaff’s rules?
In double-stranded DNA the amount of adenine equals thymine (%A = %T) and the amount of guanine equals cytosine (%G = %C). This follows from complementary base pairing: A always pairs with T, and G always pairs with C.
If I know %A, how do I find the rest?
Set %T equal to %A. Then subtract (%A + %T) from 100 to find what is left for G and C, and split it equally because %G = %C. For example, if A = 30%, then T = 30%, and G = C = (100 − 60)/2 = 20%.
Do Chargaff’s rules apply to RNA or single-stranded DNA?
Not reliably. The rules depend on complementary base pairing between two strands, so they hold for double-stranded DNA. Single-stranded DNA or RNA has no paired complementary strand, so %A need not equal %U and %G need not equal %C.

Genetic Biodiversity (Polymorphic Loci)

Proportion of polymorphic gene loci = polymorphic loci ÷ total loci

where a polymorphic locus has two or more alleles in the population (it varies) · total loci = all gene loci studied. A higher proportion = greater genetic biodiversity. (OCR-named formula.)

⚡ Quick Test Examples

A study of a population examines 25 gene loci and finds that 4 of them are polymorphic (have more than one allele).

Proportion = 4 ÷ 25 = 0.16 (16%). The higher this proportion, the greater the genetic biodiversity, which helps a population adapt to environmental change.

What does “polymorphic” mean?
A gene locus is polymorphic if it has two or more alleles present in the population at a meaningful frequency (more than one variant). A locus with only one allele is monomorphic. More polymorphic loci means more genetic variation.
Why does genetic biodiversity matter?
Greater genetic biodiversity means a wider range of alleles, so a population is more likely to contain individuals that can survive a new selection pressure (disease, climate change). Low genetic biodiversity makes a population vulnerable.

Molecular Clock (Divergence Time)

divergence time = number of differences ÷ rate of change  (rate in differences per unit time)

where differences = number of base/amino-acid differences between two species’ DNA or protein · rate = how many changes accumulate per unit time. More differences = longer since the two species shared a common ancestor.

⚡ Quick Test Examples

Two species’ cytochrome c differs at a rate of 2×10⁻⁶ changes per year and they show 24 differences.

Divergence time = 24 ÷ (2×10⁻⁶) = 1.2×10⁷ = 12 million years since they shared a common ancestor.

What is a molecular clock?
DNA and protein sequences accumulate mutations at a roughly steady rate over time. By counting the differences between two species and dividing by that rate, you can estimate how long ago they diverged from a common ancestor.

Energy & Productivity

Rf Value (Chromatography)

Rf = Pigment distance ÷ Solvent distance
  • Rf = retention factor (a ratio between 0 and 1, no units)
  • Pigment distance = distance moved by the centre of the spot from the origin/baseline (mm)
  • Solvent distance = distance moved by the solvent front from the origin/baseline (mm)

Each pigment has a characteristic Rf in a given solvent, so an unknown spot can be identified by comparing its Rf to known reference values. The pigment always travels less far than the solvent, so Rf is never greater than 1.

A chlorophyll spot moved 45 mm up the chromatography paper while the solvent front moved 80 mm. Calculate its Rf value.

Rf = 45 ÷ 80 = 0.56 (no units). Comparing to a reference table, an Rf of ~0.56 matches chlorophyll a.

Why does Rf never have units?
It is a ratio of two distances (pigment ÷ solvent), so the units of length cancel out. Rf is always a number between 0 and 1.
Why is Rf always less than 1?
The pigment is carried up the paper by the solvent, so it can never travel further than the solvent front itself. The solvent distance is therefore always the larger value.
How is Rf used to identify a pigment?
Rf is constant for a given pigment in a given solvent and temperature. You compare your calculated Rf with published reference values to identify unknown pigments (e.g. chlorophyll a, chlorophyll b, carotene, xanthophyll).

Energy Transfer Efficiency

Efficiency = (Output ÷ Input) × 100
  • Efficiency = percentage of energy (or biomass) passed on to the next trophic level (%)
  • Output = energy/biomass transferred to (and stored in) the next trophic level (kJ)
  • Input = energy/biomass available at the previous trophic level (kJ)

Energy transfer between trophic levels is typically only about 10%. Energy is lost as heat from respiration, through movement, and in indigestible/uneaten material (egestion) and excretion. These large losses are why food chains rarely have more than 4–5 trophic levels.

Producers in a field contain 50 000 kJ m⁻² yr⁻¹ of energy. Primary consumers store 5 000 kJ m⁻² yr⁻¹. Calculate the efficiency of energy transfer.

Efficiency = (5 000 ÷ 50 000) × 100 = 10%. The remaining 90% is lost mainly through respiration (heat), movement, and indigestible/uneaten material.

Why is energy transfer only about 10%?
Most energy is lost before it can be passed on: as heat from respiration, in movement, and as material that is not eaten or not digested (egestion/excretion). Only the energy stored as new biomass is available to the next level.
Why are food chains usually short?
Because only ~10% of energy passes between levels, there is not enough energy left to support many trophic levels — typically only 4 or 5 before too little energy remains.
What is the difference between input and output here?
Input is the energy available at the lower trophic level; output is the energy actually transferred to and stored in the next level. Efficiency compares the two as a percentage.

Net Primary Productivity

NPP = GPP − R
  • GPP = gross primary productivity — the total energy/biomass fixed by producers in photosynthesis
  • R = respiratory losses — energy used by the producers in their own respiration
  • NPP = net primary productivity — energy/biomass left after respiration (NPP = GPP − R)

NPP is the energy stored as new plant biomass, so it is the amount actually available for plant growth and to the next trophic level (the herbivores). Units are typically kJ m⁻² yr⁻¹ (energy) or g m⁻² yr⁻¹ (biomass).

A grassland fixes a gross primary productivity (GPP) of 20 000 kJ m⁻² yr⁻¹. The plants use 8 000 kJ m⁻² yr⁻¹ in respiration. Calculate the net primary productivity (NPP).

NPP = GPP − R = 20 000 − 8 000 = 12 000 kJ m⁻² yr⁻¹. This is the energy stored as new biomass and available to herbivores.

What is the difference between GPP and NPP?
GPP is the total energy fixed by producers in photosynthesis. NPP is what remains after the producers’ own respiration (R) is subtracted: NPP = GPP − R. NPP is the energy available for growth and to the next trophic level.
Why does NPP matter for the rest of the food chain?
NPP is the amount of energy stored as new plant biomass, so it sets the limit on how much energy is available to herbivores and the trophic levels above them.
What units are used for NPP?
Usually energy per unit area per unit time, e.g. kJ m⁻² yr⁻¹, or biomass per area per time, e.g. g m⁻² yr⁻¹.

Respiratory Quotient (RQ)

RQ = CO₂ produced ÷ O₂ consumed
  • RQ = respiratory quotient (a ratio, no units)
  • CO₂ produced = volume (or moles) of carbon dioxide released by respiration
  • O₂ consumed = volume (or moles) of oxygen taken up by respiration

The RQ reveals the respiratory substrate being used: ~1.0 for carbohydrate, ~0.7 for lipid, and ~0.8–0.9 for protein. An RQ greater than 1.0 suggests some anaerobic respiration (fermentation) is also occurring.

A respiring organism produces 16 cm³ of CO₂ while consuming 23 cm³ of O₂. Calculate the RQ and identify the likely substrate.

RQ = 16 ÷ 23 = 0.70. An RQ of about 0.7 indicates that lipid is being used as the respiratory substrate.

What does the RQ tell you about the substrate?
RQ ≈ 1.0 indicates carbohydrate, ≈ 0.7 indicates lipid, and ≈ 0.8–0.9 indicates protein. The value reflects the ratio of CO₂ released to O₂ used during respiration of that substrate.
What does an RQ greater than 1.0 mean?
It suggests that anaerobic respiration (fermentation) is occurring alongside aerobic respiration, because extra CO₂ is released without a matching uptake of O₂.
Why does RQ have no units?
It is a ratio of two volumes (or moles) of gas, CO₂ ÷ O₂, so the units cancel and RQ is just a number.

ATP Yield Calculator

Glucose → ATP via respiration
  • Glucose molecules = number of glucose molecules respired
  • Aerobic = full oxidation using oxygen (glycolysis + link reaction + Krebs cycle + oxidative phosphorylation) — a net yield of about 38 ATP per glucose (often quoted as ~30–32 when transport/proton-leak losses are included)
  • Anaerobic = no oxygen; only glycolysis yields a net 2 ATP per glucose

Aerobic respiration releases far more ATP because the electrons carried by reduced NAD and FAD pass along the electron transport chain to drive oxidative phosphorylation; without oxygen this cannot happen.

How much ATP is produced when 2 molecules of glucose are fully respired aerobically?

Aerobic yield ≈ 38 ATP per glucose, so 2 × 38 = 76 ATP (≈ 60–64 ATP if the lower 30–32 per glucose estimate is used).

Why does aerobic respiration yield so much more ATP than anaerobic?
Aerobic respiration completes the link reaction, Krebs cycle and oxidative phosphorylation, regenerating NAD and producing most of the ATP via the electron transport chain. Anaerobic respiration only runs glycolysis, giving a net 2 ATP per glucose.
Why is the aerobic figure sometimes 38 and sometimes ~30–32?
38 is the theoretical maximum. The lower figures account for ATP used to transport molecules across the mitochondrial membrane and protons leaking back across the membrane, so less ATP is made in practice.

Calorimetry: Energy in Food / Biomass (Q = mcΔT)

Q = m × c × ΔT  |  energy per gram = Q ÷ mass of sample burned

where Q = heat energy transferred (J) · m = mass of water heated (g; 1 cm³ water = 1 g) · c = specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹ · ΔT = temperature rise (°C). Divide Q by the mass of food burned for energy per gram.

⚡ Quick Test Examples

Burning 0.5 g of a food sample raises the temperature of 20 g of water by 15°C. Find the energy released per gram.

Q = mcΔT = 20 × 4.18 × 15 = 1254 J. Energy per gram = 1254 ÷ 0.5 = 2508 J g⁻¹ (≈ 2.5 kJ g⁻¹). Real values are higher — a lot of heat is lost to the surroundings in a simple calorimeter.

Why is the measured energy lower than the true value?
Much of the heat is lost to the surroundings and the apparatus rather than going into the water, and the food may not burn completely. A bomb calorimeter reduces these losses for a more accurate value.
What’s the value of c I should use?
For water, c = 4.18 J g⁻¹ °C⁻¹ (sometimes rounded to 4.2). Since 1 cm³ of water has a mass of 1 g, the volume of water in cm³ equals its mass in g.

Physiology

Cardiac Output

CO = HR × SV
Symbol key:
  • CO = cardiac output (cm³/min) — total volume of blood pumped by a ventricle each minute
  • HR = heart rate (beats/min)
  • SV = stroke volume (cm³/beat) — volume ejected per beat

At rest a student has a heart rate of 72 beats/min and a stroke volume of 70 cm³/beat. What is their cardiac output?

CO = HR × SV = 72 × 70 = 5040 cm³/min (≈ 5.04 dm³/min). During exercise both HR and SV rise, so CO can climb to 20–30 dm³/min to deliver more oxygen and glucose to working muscles.

Why does cardiac output increase during exercise?
Because CO = HR × SV, raising either factor raises CO. During exercise sympathetic stimulation increases heart rate, and increased venous return stretches the ventricle so it contracts more forcefully (Starling’s law), raising stroke volume. Both effects together greatly increase the blood — and therefore oxygen and glucose — delivered to muscles.
What units should I use?
Keep stroke volume in cm³ (or dm³) per beat and heart rate in beats per minute, giving CO in cm³ per minute (or dm³ per minute). 1000 cm³ = 1 dm³, so divide a cm³/min answer by 1000 to quote it in dm³/min.

Buffer pH (Henderson-Hasselbalch)

pH = pKa + log([A⁻] ÷ [HA])
Symbol key:
  • pH = measure of acidity (−log[H⁺])
  • pKa = the pH at which the acid is half dissociated; characteristic of each weak acid
  • [A⁻] = concentration of the conjugate base (salt form) in mol/dm³
  • [HA] = concentration of the undissociated weak acid in mol/dm³

When [A⁻] = [HA] the log term is 0, so pH = pKa — buffers work best near their pKa.

An ethanoic acid / ethanoate buffer (pKa = 4.76) contains 0.1 M conjugate base and 0.1 M acid. What is its pH?

pH = pKa + log([A⁻] ÷ [HA]) = 4.76 + log(0.1 ÷ 0.1) = 4.76 + log(1) = 4.76 + 0 = 4.76. With equal amounts of acid and base the buffer sits exactly at its pKa, where it resists pH change most effectively.

How does a buffer resist changes in pH?
A buffer contains both a weak acid and its conjugate base. Added H⁺ is mopped up by the conjugate base (A⁻), and added OH⁻ is neutralised by the acid (HA). Because both reservoirs are present, the H⁺ concentration — and so the pH — changes only slightly. This is vital for keeping blood pH near 7.4 and maintaining enzyme activity.
When is a buffer most effective?
A buffer works best within about ±1 pH unit of its pKa, and is most effective when [A⁻] = [HA] (pH = pKa). The blood hydrogencarbonate buffer is supported by respiratory and renal control to hold pH within its narrow physiological range.

Dilution Calculator (C₁V₁ = C₂V₂)

C₁V₁ = C₂V₂
Symbol key:
  • C₁ = starting (stock) concentration
  • V₁ = starting (stock) volume needed
  • C₂ = final (diluted) concentration
  • V₂ = final total volume after adding solvent

C₁ and V₁ must use the same units as C₂ and V₂. This relationship holds only for dilutions — making a solution weaker by adding solvent.

You have a 1.0 M glucose stock and want to make a 0.1 M solution using 10 cm³ of stock. What final volume do you need?

C₁V₁ = C₂V₂ → V₂ = C₁V₁ ÷ C₂ = (1.0 × 10) ÷ 0.1 = 100 cm³. So add solvent to the 10 cm³ of stock until the total volume reaches 100 cm³ (i.e. add 90 cm³ of water).

When can I use C₁V₁ = C₂V₂?
Only for dilutions — making a solution less concentrated by adding solvent. The number of moles of solute stays the same (concentration × volume on each side), so the equation balances. It does not apply when you add more solute or mix two different solutions.
Do the units have to match?
Yes. C₁ and C₂ must be in the same concentration units, and V₁ and V₂ in the same volume units. The volume unit cancels, so as long as each pair is consistent (e.g. both in cm³) the answer is correct.

Spirometer: Volume from a Trace (Tidal Volume / Vital Capacity)

volume = trace height × (calibration volume ÷ calibration height)

where trace height = vertical height of the breath on the trace (cm) · calibration = a known volume marked over a known height (e.g. 1 dm³ over 2.5 cm). Use the average breath height for tidal volume, or the big forced breath for vital capacity.

⚡ Quick Test Examples

On a spirometer trace, 2.5 cm of height was calibrated as 1 dm³. At rest the average breath has a height of 1.3 cm.

Tidal volume = 1.3 × (1 ÷ 2.5) = 0.52 dm³. (After exercise a 2.4 cm breath = 0.96 dm³ — it nearly doubles.)

Tidal volume or vital capacity — which height do I use?
Use the average height of the small, normal breaths for tidal volume. Use the height of the single big forced breath (maximum in to maximum out) for vital capacity. Same calibration, different part of the trace.
Why average several breaths?
No two breaths are identical, so averaging several normal breaths gives a more reliable tidal volume than reading just one.

Breathing Rate from a Trace

breathing rate = (number of breaths ÷ time) × 60

where number of breaths = peaks (or troughs) you count on the trace · time = the time those breaths took (seconds) · result is breaths per minute. Read time from the chart speed (e.g. 2 small squares = 1 s).

⚡ Quick Test Examples

At rest, a subject takes 4 breaths in 26 seconds on the trace.

Breathing rate = (4 ÷ 26) × 60 = 9 breaths min⁻¹. After exercise, 11 breaths in 34 s = 19 breaths min⁻¹ — it more than doubles.

How do I get the time from the trace?
Use the chart-recorder speed. If the drum runs at 0.5 cm s⁻¹, then each cm of horizontal distance = 2 s; if “2 small squares = 1 second”, count squares and halve. Multiply or divide accordingly to get the number of seconds your breaths cover.

Oxygen Consumption from a Spirometer Trace

O₂ used per min = volume drop (over a measured time) scaled to 60 s

where the soda lime absorbs CO₂, so the trace baseline falls as O₂ is used. Measure the fall in volume over a known time (the gradient of the line through the low points), then scale to one minute. Enter the drop as a volume, or as a trace height + calibration.

⚡ Quick Test Examples

The baseline of a spirometer trace fell by 160 cm³ over 30 seconds at rest.

In one minute (60 s) the drop would be 160 × (60 ÷ 30) = 320 cm³ min⁻¹ of oxygen consumed. After exercise an 880 cm³ min⁻¹ value shows ~2.75× more O₂ used.

Why does the spirometer trace slowly fall?
The subject removes oxygen from the chamber as they respire, and the carbon dioxide they breathe out is absorbed by the soda lime. Because gas is removed but not replaced, the total volume — and so the baseline of the trace — steadily falls. The steeper the fall, the faster the oxygen consumption.
Why draw a line through the low points?
The trace wobbles up and down with each breath. Drawing a line of best fit through the lowest points (the troughs) isolates the steady downward drift caused by oxygen being used, ignoring the breathing wobble.

Basal Metabolic Rate (BMR) from O₂ Consumption

energy = O₂ used × 20.1 kJ per dm³  |  BMR (kJ/day) = O₂ (dm³ min⁻¹) × 20.1 × 1440

where O₂ used = oxygen consumption · 20.1 kJ dm⁻³ = the energy released per dm³ of oxygen used in respiration (the “energy equivalent of oxygen”, ~20.1–20.2 kJ) · 1440 = minutes in a day. BMR is the energy used at complete rest.

⚡ Quick Test Examples

A person at complete rest consumes 0.25 dm³ of oxygen per minute. Estimate their BMR.

Energy per minute = 0.25 × 20.1 = 5.03 kJ min⁻¹. BMR per day = 5.03 × 1440 ≈ 7240 kJ day⁻¹ (about 1730 kcal — a typical adult BMR).

Why is it hard to measure your true BMR in a lesson?
BMR must be measured lying down, completely at rest, awake but relaxed, in a warm room, and at least 12 hours after eating. A noisy classroom after walking to the lab can’t meet those conditions, so any school measurement is really an over-estimate of resting metabolic rate, not true BMR.
Where does the 20.1 kJ per dm³ come from?
It’s the average energy released when the body respires using one cubic decimetre (litre) of oxygen — the “energy equivalent of oxygen”. It varies slightly with substrate (≈20.1 kJ for carbohydrate, a little less for fat), so exam questions usually give you the value to use.

Minute Ventilation (Pulmonary)

Minute ventilation = tidal volume × ventilation (breathing) rate

where tidal volume = volume of air per breath (dm³) · ventilation rate = breaths per minute · minute ventilation = total air moved per minute (dm³ min⁻¹). The lung equivalent of cardiac output. (Edexcel-named.)

⚡ Quick Test Examples

At rest a person has a tidal volume of 0.45 dm³ and breathes 15 times a minute.

Minute ventilation = 0.45 × 15 = 6.75 dm³ min⁻¹. During exercise both tidal volume and rate rise, so minute ventilation increases sharply to deliver more oxygen.

How is this read from a spirometer trace?
Tidal volume is the height of each normal breath on the trace; ventilation rate is the number of breaths in one minute (or scale up from a shorter time). Multiply them for minute ventilation.

Fick’s Law: Diffusion Time / Distance

diffusion ∝ (surface area × conc. difference) ÷ distance  |  time ≈ distance² ÷ D

where for the timed form, distance = diffusion distance · D = diffusion coefficient · time ≈ distance² ÷ D, so distance = √(time × D). Diffusion time rises with the square of distance — why exchange surfaces are thin.

⚡ Quick Test Examples

An ion diffuses with D = 5×10⁴ µm² s⁻¹. How far can it diffuse in 5 seconds?

distance = √(time × D) = √(5 × 50000) = √250000 = 500 µm. Because time ∝ distance², doubling the distance quadruples the time — so exchange surfaces (alveoli, capillary walls) are extremely thin.

Why does diffusion distance matter so much?
Diffusion time rises with the square of distance, so even a small increase in distance hugely slows diffusion. That’s why gas-exchange surfaces are only one or two cells thick and why large organisms need specialised exchange and transport systems.
What is Fick’s law in full?
Rate of diffusion ∝ (surface area × concentration difference) ÷ diffusion distance. A bigger surface area and steeper concentration gradient speed diffusion up; a longer diffusion distance slows it down.

Levers & Moments (Muscle Force)

F_effort × d_effort = F_load × d_load  |  weight (N) = mass (kg) × 9.81

where at balance, the moment (force × distance from pivot) on each side is equal. F = force (N) · d = distance from the pivot. To turn a mass into a force (weight), multiply by g = 9.81 N kg⁻¹.

⚡ Quick Test Examples

A weight produces a load force of 50 N at the hand, 30 cm from the elbow pivot. The biceps attaches 3 cm from the pivot. What muscle force is needed to hold it?

F_effort = (F_load × d_load) ÷ d_effort = (50 × 30) ÷ 3 = 500 N. The biceps must pull with 500 N — far more than the load — because its lever arm is short (a third-class lever trades force for speed/range).

Why does the muscle need such a big force?
In the arm the muscle attaches very close to the joint (a short effort arm), while the load is far away (a long load arm). To balance the moments, the muscle force must be much larger than the load — the trade-off is that a small muscle movement makes the hand move a long way, quickly.
How do I turn a mass into a force?
Weight (in newtons) = mass (in kg) × g, where g = 9.81 N kg⁻¹ (often rounded to 9.8 or 10). Tick the box and the calculator multiplies your mass by 9.81 for you.

Microbiology & Bacterial Growth

Population Growth: N = N₀ × 2ⁿ (Bacteria / Cells / PCR)

N = N₀ × 2ⁿ  |  n = time ÷ generation time
  • N₀ = starting (initial) number of cells
  • n = number of divisions (generations) that have occurred
  • N = final number of cells after n divisions
  • generation time = time for one division; n = total time ÷ generation time

Each division doubles the population, so growth is exponential. Plotting log(N) against time turns the curve into a straight line, which makes exponential growth easier to read and compare.

For organisms dividing by binary fission. Find the final number, or work out how many divisions / how long it took.

50 bacteria are grown for 180 minutes with a generation time of 30 minutes. How many bacteria are there at the end?

n = time ÷ generation time = 180 ÷ 30 = 6 divisions. N = N₀ × 2ⁿ = 50 × 2⁶ = 50 × 64 = 3200 bacteria.

How do I find the number of divisions (n)?
Divide the total time by the generation time: n = time ÷ generation time. For example, 180 min ÷ 30 min per division = 6 divisions.
Why plot bacterial growth on a log scale?
Exponential growth produces enormous numbers very quickly, which are hard to read on a normal axis. Taking the logarithm (e.g. log₁₀N) straightens the exponential curve into a straight line, making the growth rate easy to measure.
Does N = N₀ × 2ⁿ assume anything?
Yes — it assumes ideal exponential (log) growth where every cell divides and none die. In reality nutrients run out and waste builds up, so real cultures eventually reach a stationary then death phase.

Exponential Growth: N = N₀eʳᵗ

Nₜ = N₀ × eʳᵗ  |  t = ln(Nₜ ÷ N₀) ÷ r

where N₀ = starting number · Nₜ = number after time t · r = growth rate constant (per unit time) · t = time · e ≈ 2.718. The continuous form of exponential growth (use this when given a rate constant r, rather than a doubling time).

⚡ Quick Test Examples

A yeast culture starts at 2000 cells and grows with rate constant r = 0.5 hr⁻¹. How many cells after 10 hours?

Nₜ = N₀eʳᵗ = 2000 × e^(0.5×10) = 2000 × e⁵ = 2000 × 148.4 ≈ 2.97 × 10⁵ cells.

When do I use this instead of N = N₀ × 2ⁿ?
Use N = N₀ × 2ⁿ when you’re told a doubling/generation time (the population doubles each generation). Use N = N₀eʳᵗ when you’re given a continuous growth rate constant r. They describe the same exponential growth — r and the doubling time are linked by r = ln2 ÷ doubling time.

Relative Growth Rate (R = (ln W₂ − ln W₁) ÷ t)

R = (ln W₂ − ln W₁) ÷ t  |  t = (ln W₂ − ln W₁) ÷ R

where W₁ = starting size/mass · W₂ = final size/mass · t = time · R = relative (specific) growth rate. Used for plant biomass or height growth. Watch the hidden W₁: “grew BY 268 mm to 387 mm” means W₁ = 119.

A plant shoot grows from 119 mm to 387 mm over 12 days. Find its relative growth rate.

R = (ln 387 − ln 119) ÷ 12 = (5.958 − 4.779) ÷ 12 = 1.179 ÷ 12 ≈ 0.098 day⁻¹.

Why use ln (natural log) and not log₁₀?
The relative growth rate is defined using natural logs because growth is continuous and exponential — the maths of continuous compounding uses base e. Using log₁₀ would give a different (wrong) value of R for this formula.

Cell Count from a Log Graph (Antilog)

number of cells = 10^(log₁₀ value)  |  log₁₀(count) places a count on the axis

where microbial growth is often plotted as log₁₀(number of cells) on the y-axis. To get the real number from a graph reading, take the antilog: count = 10^(reading). To go the other way, take log₁₀ of the count. (Classic AQA log-scale question.)

⚡ Quick Test Examples

In an AQA question, a log₁₀ graph of bacteria in the bladder reads 2.85 for one group and 1.25 for another.

Group R: 10^2.85 = 708 bacteria. Group A: 10^1.25 = 18 bacteria. Convert from the log scale first, then compare — here R has ~40× more bacteria than A.

Why can’t I just subtract the two log values?
Because log values are powers of 10, not the actual numbers. log 2.85 − log 1.25 = 1.60 does not mean “160%”. You must take the antilog (10^x) of each reading to get the real cell counts first, then compare or take percentages.
Each gridline on a log scale — how big a jump is it?
One unit up the log₁₀ axis (e.g. 1 → 2) is a 10-fold increase, not double. Going from 1 to 5 looks small but is a factor of 10 000. That’s exactly why a log scale is used — to fit huge ranges onto one graph.

Viable Count (CFU Plate Count)

Viable count = number of living cells only (each living cell grows into one visible colony). Use a plate/CFU count when you need living cells — e.g. testing a disinfectant.

CFU/cm³ = Colonies × Dilution ÷ Volume
Symbol key:
  • CFU/cm³ = colony-forming units per cm³ — the estimated number of viable cells in the original sample
  • Colonies = number of colonies counted on the plate (use a countable plate, 30–300)
  • Dilution = total dilution factor applied before plating (e.g. 10⁴)
  • Volume = volume of diluted sample spread on the plate (cm³)

Each colony is assumed to have grown from a single viable cell, so the count estimates the live cell number.

You plate 0.1 cm³ of a 10⁻⁴ dilution and count 150 colonies. How many CFU per cm³ were in the original culture?

CFU/cm³ = colonies × dilution factor ÷ volume plated = 150 × 10⁴ ÷ 0.1 = 1.5 × 10⁷ CFU/cm³. This assumes each colony arose from one cell and that the plate count fell in the reliable 30–300 range.

Why only count plates with 30–300 colonies?
Below 30 the count is too small to be statistically reliable; above 300 colonies merge and overlap, so cells are undercounted. Counting a plate in the 30–300 range, then scaling up by the dilution factor, gives the most accurate estimate of the original cell number.
What does “colony-forming unit” mean and why not just “cells”?
We can only see colonies that grew, so we count colony-forming units. The method assumes each colony came from a single viable cell, but if cells clumped together one colony may represent several cells, so CFU is an estimate of viable cells rather than an exact total cell count.

Total Count (Haemocytometer)

Total count = all cells counted under the microscope, living and dead (a haemocytometer can’t tell them apart). Use it for a quick total cell density.

Cells per cm³ = (mean count ÷ volume of one square in cm³) × dilution factor

where a haemocytometer square has a known area and depth (0.1 mm), giving a tiny known volume · dilution factor accounts for any dilution before counting. Counting a known volume lets you scale up to cells per cm³.

⚡ Quick Test Examples

A standard haemocytometer square is 1 mm × 1 mm × 0.1 mm deep = 0.1 mm³ = 1×10⁻⁴ cm³. (A small “triple-line” square is 1×10⁻⁷ cm³.) You count a mean of 80 cells in a 1×10⁻⁷ cm³ square after a 10× dilution.

Cells per cm³ = (80 ÷ 1×10⁻⁷) × 10 = 8×10⁹ cells cm⁻³. Remember to multiply by the dilution factor to get back to the original suspension.

Why multiply by the dilution factor?
You usually dilute a dense suspension so the cells are countable. The count reflects the diluted sample, so you multiply by the dilution factor (e.g. ×10) to scale back up to the concentration of the original, undiluted suspension.
How do I get the volume of a square?
Multiply the square’s area by the chamber depth (always 0.1 mm on a haemocytometer), then convert mm³ to cm³ by dividing by 1000. A 1 mm² square gives 0.1 mm³ = 1×10⁻⁴ cm³; smaller ruled squares give proportionally smaller volumes.

Serial Dilution

Each dilution = Previous ÷ Factor
Symbol key:
  • Initial concentration = concentration of the first (undiluted) tube
  • Dilution factor = how many times each step is diluted (e.g. 10 = a tenfold dilution at every step)
  • Number of dilutions = how many sequential dilution steps are made

Each step divides the previous concentration by the factor. Two 10× steps give a 100× total dilution, three give 1000×, and so on.

A dense bacterial culture is too concentrated to count. You make a tenfold serial dilution: 1 cm³ of culture into 9 cm³ of sterile water, repeated five times. What is the final dilution?

Each step multiplies the dilution by 10, so after 5 steps the total dilution is 10 × 10 × 10 × 10 × 10 = 10⁵ (a 1 in 100 000 dilution). This spreads cells out enough to give a countable plate of 30–300 colonies.

Why use a serial dilution instead of one big dilution?
Pipetting a single 1 in 100 000 dilution accurately is impractical. A serial dilution builds up the same factor through several easy-to-measure steps (e.g. five 1-in-10 steps), reducing pipetting error and giving a range of plates so at least one falls in the countable 30–300 colony range.
How do I work out the total dilution factor?
Multiply the step factors together. A 1-in-10 dilution repeated n times gives a total dilution of 10ⁿ. To find the original concentration from a plate count, multiply the count by the total dilution factor (and divide by the volume plated).

Unit Converter

Convert Between Units

Pick a quantity, then convert from one unit to another

how it works Every unit is converted via its SI base unit, so you can go between any two units in a category — e.g. nanometres → micrometres, dm³ → cm³, mol dm⁻³ → mmol dm⁻³.

Exam tip: A-level Biology mixes units constantly — cell sizes in µm, organelle sizes in nm, volumes in cm³/dm³, concentrations in mol dm⁻³. Convert before you put numbers into a formula (e.g. magnification needs image and actual size in the same unit).

📚 The metric ladder (most-used)

  • Length: 1 m = 10³ mm = 10⁶ µm = 10⁹ nm. So ×1000 each step down (mm→µm→nm).
  • Volume: 1 dm³ = 1 litre = 1000 cm³ = 1000 ml; 1 cm³ = 1 ml = 1000 µl.
  • Mass: 1 g = 1000 mg = 10⁶ µg; 1 kg = 1000 g.
  • Concentration: 1 mol dm⁻³ = 1000 mmol dm⁻³ = 10⁶ µmol dm⁻³.
  • Area scales by the square: 1 cm² = 100 mm² (not 10).
Why is 1 cm² = 100 mm² and not 10 mm²?
Because area is two-dimensional. 1 cm = 10 mm, so 1 cm² = (10 mm)² = 100 mm². The same logic makes 1 cm³ = (10 mm)³ = 1000 mm³ for volume. Always square the linear factor for area and cube it for volume.
How do I convert °C to a temperature coefficient or to kelvin?
For kelvin, K = °C + 273.15 (this converter does it for you). You don’t usually need kelvin in A-level Biology, but it appears in some respiration/energy questions. Temperature differences (e.g. for Q₁₀) are the same number in °C and K.
What does mol dm⁻³ mean?
It’s the same as “moles per litre” (mol/L) — dm³ is just the SI-correct way to write a litre. 1 dm³ = 1000 cm³, so a 0.5 mol dm⁻³ solution has 0.5 mol in every 1000 cm³.