Offspring inherit one allele for each gene from each parent. Mendel’s crosses with pea plants revealed the rules that predict the results.
🟨 Monohybrid crosses
A dominant allele (A) shows in the phenotype of the heterozygote (Aa), while a recessive allele (a) shows only in aa. Cross two heterozygotes and the offspring come out in the ratio 3 dominant : 1 recessive.
Aa × Aa → 1 AA : 2 Aa : 1 aa
P(aa) = 1/4 = 0.25
80 offspring → 60 dominant, 20 recessive
Test cross Aa × aa → 1 Aa : 1 aa
half are recessive
Incomplete dominance: red × white → all pink
pink × pink → 1 red : 2 pink : 1 white
🖼️ See it
Phenotype ratio 3 : 1 when T is dominant over t.
🎲 Two genes and statistics
Genes on different chromosomes assort independently (Mendel’s second law). A dihybrid cross AaBb × AaBb gives 9 : 3 : 3 : 1. To test whether results fit a ratio use χ² = Σ (O − E)² / E.
The product rule: the chance of two independent events together is the product of their separate chances.
🧪 Worked examples: χ² tests and gene interactions
χ² test. Null hypothesis: the observed results fit the expected ratio and any difference is due to chance. Compare χ² with the critical value at p = 0.05, with degrees of freedom = classes − 1.
cross Aa × Aa, 80 offspring: expected 60 dominant : 20 recessive; observed 63 : 17
χ² = Σ (O − E)² ÷ E = (3² ÷ 60) + (3² ÷ 20) = 0.15 + 0.45 = 0.60
df = 2 − 1 = 1; critical value (p = 0.05) = 3.84
0.60 < 3.84 → accept the null hypothesis: no significant difference from 3 : 1
Epistasis. One gene masks the expression of another, so the 9 : 3 : 3 : 1 ratio is modified.
recessive epistasis (e.g. coat colour in mice): 9 : 3 : 4
complementary genes (both dominant alleles needed): 9 : 7
in 160 offspring of AaBb × AaBb with a 9 : 3 : 4 ratio:
the “4” class = 4 ÷ 16 × 160 = 40 offspring
Dihybrid test cross. AaBb × aabb gives 1 : 1 : 1 : 1 if the genes assort independently.
Choosing the critical value. For 1, 2 and 3 degrees of freedom at p = 0.05, the critical values are 3.84, 5.99 and 7.81.
🎯 Exam focus: writing genetic crosses properly
Show every step: parental phenotypes → parental genotypes → gametes → offspring genotypes → offspring phenotypes and ratio. Define the symbols at the start.
Let A = tall (dominant), a = dwarf (recessive)
Aa × Aa → gametes A, a × A, a → AA, Aa, Aa, aa
phenotypes: 3 tall : 1 dwarf
Common mistakes. Using different letters for the same gene (such as T and D). Writing gametes as pairs of alleles. Using the percentage in place of the number of observed individuals in χ². Concluding that “the results prove the theory” (you can only fail to reject the null hypothesis). Using the total df as the number of offspring.
Evaluating. A χ² test is valid only with large samples (expected values over 5) and counts, not percentages.
✏️ Quick check · 1 of 11
Genotype
An individual with the genotype Aa is:
The two alleles differ.
✏️ Quick check · 2 of 11
Test cross
A test cross is used to find out if a dominant-looking individual is:
The partner is aa.
✏️ Quick check · 3 of 11
Codominance
In codominance, a heterozygote:
Both appear.
✍️ Type your answer · 4 of 11
Two heterozygotes (Aa × Aa) are crossed.
What is the probability of an aa offspring, as a decimal?
Enter an exact answer, without rounding.
Draw the Punnett square.
✍️ Type your answer · 5 of 11
Aa × Aa produces 80 offspring.
How many are expected to show the dominant phenotype?
Enter an exact answer, without rounding.
The 3 : 1 ratio.
✍️ Type your answer · 6 of 11
A dihybrid cross AaBb × AaBb produces 160 offspring.
How many are expected to show both dominant traits (9/16)?
Enter an exact answer, without rounding.
Use the 9 : 3 : 3 : 1 ratio.
✍️ Type your answer · 7 of 11
A cross should give 60 dominant : 20 recessive, but you observe 63 : 17.
What is χ² (Σ(O − E)²/E)?
Enter an exact answer, without rounding.
Work out each term separately.
✍️ Type your answer · 8 of 11
A cross Aa × Aa gives 80 offspring. The expected numbers are 60 and 20; the observed are 63 and 17.
What is the value of χ²?
Enter an exact answer, without rounding.
Sum (O − E)² ÷ E for each class.
✍️ Type your answer · 9 of 11
A χ² test compares the observed numbers in four phenotype classes with the expected numbers.
How many degrees of freedom does the test have?
Enter an exact answer, without rounding.
Classes minus 1.
✍️ Type your answer · 10 of 11
A dihybrid cross AaBb × AaBb shows recessive epistasis with a ratio of 9 : 3 : 4, and 160 offspring are produced.
How many offspring are expected in the class of 4?
Enter an exact answer, without rounding.
The ratio adds up to 16 parts.
✏️ Quick check · 11 of 11
Interpreting χ²
A calculated χ² value is 2.1 with 1 degree of freedom (critical value 3.84 at p = 0.05). What is the conclusion?
Compare with the critical value.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
Alleles are:
Question 2 of 5
Mendel’s law of segregation states that:
Question 3 of 5
The phenotype ratio of a cross between two heterozygotes with complete dominance is:
Question 4 of 5
The number of different gametes made by an AaBb individual is:
Question 5 of 5
A genotype shows what the organism’s:
Lesson 2 · Genetics
🧬 Sex-Linkage, Linkage & Pedigrees
Genes on the sex chromosomes, and genes that sit close together on the same chromosome, do not follow Mendel’s simple ratios.
♂️♀️ Sex-linked inheritance
Females are XX and males XY. A gene on the X chromosome with no copy on the Y shows in a male even if only one allele is recessive. That is why X-linked recessive conditions, such as red–green colour blindness and haemophilia, are more common in males.
carrier mother XᴴXʰ × unaffected father XᴴY
daughters: XᴴXᴴ, XᴴXʰ (none affected)
sons: XᴴY, XʰY (half affected)
P(any child affected) = 1/4 = 0.25
Barr body = one inactivated X
XX cell: 1 XXY cell: 1 XXXX cell: 3
Turner (XO): 45 chromosomes Klinefelter (XXY): 47
🖼️ See it
XhY sons have no second X to mask the recessive allele.
🔗 Linked genes and crossing over
Genes on the same chromosome tend to be inherited together. Crossing over in meiosis separates some, producing recombinants. The recombination frequency shows how far apart genes are: 1% = 1 map unit (centimorgan).
recombination frequency = recombinants / total offspring · 100
12 recombinants in 200 → 6% → 6 map units
20 of 400 → 5 map units
Autosomal recessive pedigree: two carriers (Aa × Aa)
P(affected) = 0.25 P(unaffected) = 0.75
Pedigrees: if two unaffected parents have an affected child, the allele is recessive and both parents are carriers.
🧪 Worked examples: crosses, map distances and pedigrees
X-linked recessive. Let Xᴮ be normal vision and Xᵇ be colour blindness.
carrier mother XᴮXᵇ × colour-blind father XᵇY
daughters: XᴮXᵇ, XᵇXᵇ sons: XᴮY, XᵇY
P(a daughter is colour-blind) = 1/2 P(a son is colour-blind) = 1/2
Recombination frequency. A test cross of a double heterozygote gives parental and recombinant classes.
Ordering three genes. The largest distance is between the outer genes.
A–B = 10, B–C = 6, A–C = 16 → order A – B – C (10 + 6 = 16)
Pedigree reasoning. Two unaffected parents with an affected child means the trait is recessive. If the trait appears in both sexes equally, it is likely to be autosomal.
Carrier risk. The unaffected sibling of someone with an autosomal recessive condition (both parents Aa) has a 2/3 chance of being a carrier, because aa is excluded: 2 of the remaining 3 outcomes are Aa.
🎯 Exam focus: telling the types of inheritance apart
X-linked recessive more males affected; no father-to-son transmission; skips generations through carrier mothers
autosomal recessive both sexes equally; usually unaffected parents (carriers) of an affected child
autosomal dominant every affected person has an affected parent; vertical pattern
X-linked dominant an affected father passes it to ALL daughters and NO sons
In a pedigree answer, give the genotypes of the key individuals and explain how you eliminated the other modes of inheritance.
Common mistakes. Dividing the number of recombinants by the number of parental offspring. Saying that a map unit is a physical distance in nucleotides. Writing an X-linked allele as a plain letter without X (such as “H” instead of Xᴴ). Ignoring that a male can only be hemizygous. Concluding “recessive” just because the trait skips a generation.
Barr bodies. One X chromosome in each female cell is randomly inactivated, so females who are carriers can show patches of both phenotypes (tortoiseshell cats).
✏️ Quick check · 1 of 11
X-linked recessive
Why are X-linked recessive disorders more common in males?
Males have one copy.
✏️ Quick check · 2 of 11
Recombinants
Recombinant offspring are produced by:
Think prophase I.
✏️ Quick check · 3 of 11
Pedigree
Two unaffected parents have an affected child. The trait is:
It skipped a generation.
✍️ Type your answer · 4 of 11
A carrier mother (XᴴXʰ) and an unaffected father (XᴴY) have a child.
What is the probability that the child is affected, as a decimal?
Enter an exact answer, without rounding.
Combine P(son) and P(son affected).
✍️ Type your answer · 5 of 11
In a test cross, 12 of 200 offspring are recombinant.
What is the recombination frequency, in %?
Enter an exact answer, without rounding.
%
recombinants ÷ total × 100.
✍️ Type your answer · 6 of 11
Two carriers of an autosomal recessive disorder have a child.
What is the probability that the child is unaffected, as a decimal?
Enter an exact answer, without rounding.
P(aa) = 0.25.
✍️ Type your answer · 7 of 11
A person has Turner syndrome (XO).
How many chromosomes are in each of their body cells?
Enter an exact answer, without rounding.
One sex chromosome is missing.
✍️ Type your answer · 8 of 11
A test cross gives 440 parental and 60 recombinant offspring.
What is the recombination frequency, as a percentage?
Enter an exact answer, without rounding.
%
Recombinants ÷ total.
✍️ Type your answer · 9 of 11
A carrier mother (XᴮXᵇ) has children with a colour-blind father (XᵇY).
What is the probability that a daughter is colour-blind, as a decimal?
Enter an exact answer, without rounding.
List the daughter’s genotypes.
✍️ Type your answer · 10 of 11
Genes A, B and C are on one chromosome. A–B = 10, B–C = 6 and B lies between A and C.
What is the map distance from A to C, in map units?
Enter an exact answer, without rounding.
map units
Add the two distances.
✏️ Quick check · 11 of 11
Pedigree
Two unaffected parents have an affected daughter. What does this show about the condition?
Check whether the father could pass on an X-linked allele.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A Barr body is:
Question 2 of 5
Genes that are close together on the same chromosome are:
Question 3 of 5
A man with an X-linked recessive disorder will pass the allele to:
Question 4 of 5
Non-disjunction leads to:
Question 5 of 5
A recombination frequency of 50% means the genes:
Lesson 3 · Evolution
🦎 Natural Selection & Evolution
Populations change over generations because some individuals leave more offspring than others. Natural selection explains how adaptations arise and species diverge.
🐢 How selection works
Darwin’s argument has four parts: variation exists, much of it is inherited, more offspring are born than can survive (competition), and individuals with advantageous traits survive and reproduce more. Over time the advantageous alleles become more common.
relative fitness w = offspring of genotype / offspring of the best genotype
40 vs 50 → w = 0.8, selection coefficient s = 1 − w = 0.2
bacteria doubling every 30 min, 10 resistant at the start
2 h = 4 doublings → 10 · 2⁴ = 160
Selection differential S = mean of parents − mean of population
9.8 − 9.0 = 0.8 response R = h²·S = 0.5·0.8 = 0.4
🖼️ See it
Selection acts on individuals but evolution is the change in a population’s allele frequencies over time.
🌳 Evidence and speciation
Evidence for evolution includes fossils, homologous structures (the same bones in a bat wing and a human arm), and molecular similarities in DNA and proteins. A new species forms when populations become reproductively isolated, often after being geographically separated (allopatric speciation).
Molecular clock
two species differ by 20 substitutions
each lineage gains 1 per million years → 2 per million years in total
divergence = 20 / 2 = 10 million years
Selection acts on phenotypes, but evolution is a change in allele frequency in a population over generations.
🧪 Worked examples: selection and sequence divergence
Allele frequency after selection. If the aa genotype has a relative fitness of 0.5, the new frequency q′ = (pq + 0.5 q²) ÷ w̄.
Sequence divergence. Count the differences and divide by the sequence length.
36 amino-acid differences in a protein of 120 residues → 30% divergence
the species split 60 million years ago → 30 ÷ (2 × 60) = 0.25% per million years per lineage
(the factor of 2 is because both lineages change)
Types of selection acting on a continuous trait:
stabilising the extremes are removed (birth mass in humans)
directional one extreme is favoured (antibiotic resistance, peppered moth in industrial areas)
disruptive both extremes are favoured (beak size with two seed types)
Gene flow and drift. Gene flow (migration) makes populations more similar; genetic drift and selection make them differ. Barriers prevent gene flow and can lead to speciation.
🎯 Exam focus: explaining a change in a population
Use the pattern: (1) variation already exists (random mutation) → (2) a selective pressure → (3) individuals with the advantageous allele survive and reproduce more → (4) the allele frequency rises in the next generation.
antibiotic resistance
1 a few bacteria have a resistance mutation
2 antibiotic kills the sensitive ones
3 resistant bacteria survive and reproduce, passing on the allele
4 the population becomes mostly resistant (not “the bacteria adapt”)
Common mistakes. Writing that organisms “need” a trait, or “try” to evolve it. Saying that individuals evolve (populations evolve). Confusing acquired characteristics with inherited ones. Treating mutation as directed by the environment. Mixing up “fitness” (reproductive success) with strength or health.
Speciation. Allopatric speciation needs geographic isolation; sympatric speciation happens without it, for example by polyploidy in plants.
✏️ Quick check · 1 of 11
Fitness
In biology, an individual’s fitness is a measure of its:
Not strength.
✏️ Quick check · 2 of 11
Homologous
The same bone pattern in a human arm and a bat wing is evidence of:
Homologous means shared ancestry.
✏️ Quick check · 3 of 11
Antibiotic resistance
Antibiotic resistance spreads in bacteria mainly because:
Think natural selection.
✍️ Type your answer · 4 of 11
Genotype A leaves 40 offspring while the best genotype leaves 50.
What is the relative fitness w of A, as a decimal?
Enter an exact answer, without rounding.
Divide by the best genotype.
✍️ Type your answer · 5 of 11
A bacterial colony has 10 resistant cells that double every 30 minutes.
How many are there after 2 hours?
Enter an exact answer, without rounding.
Count the doublings first.
✍️ Type your answer · 6 of 11
The breeder’s equation is R = h²·S. Heritability h² = 0.5 and the selection differential S = 0.8.
What is the response R?
Enter an exact answer, without rounding.
Multiply the two numbers.
✍️ Type your answer · 7 of 11
Two species differ by 20 substitutions. Each lineage gains 1 per million years.
How long ago did they diverge, in millions of years?
Enter an exact answer, without rounding.
Myr
Count both lineages.
✍️ Type your answer · 8 of 11
A protein has 120 amino-acid residues. Two species differ at 36 positions.
What is the percentage divergence?
Enter an exact answer, without rounding.
%
Differences divided by length.
✍️ Type your answer · 9 of 11
The two species above separated 60 million years ago, with 30% divergence.
What is the rate of change per lineage, in % per million years?
Enter an exact answer, without rounding.
% per Myr
Both lineages change independently.
✍️ Type your answer · 10 of 11
In a population, q = 0.20 and p = 0.80. The genotype aa has a relative fitness of 0.5, and the others have a fitness of 1.
What is q after one generation of selection, to 2 decimal places?
Round your answer as the question asks.
Divide the contribution of a to the next generation by the mean fitness.
✏️ Quick check · 11 of 11
Types of selection
Peppered moths became mostly dark in a polluted industrial area. This is an example of:
Which way did the population mean move?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Which statement about mutations is correct?
Question 2 of 5
A population bottleneck:
Question 3 of 5
Convergent evolution gives rise to:
Question 4 of 5
Which is NOT a requirement for natural selection?
Question 5 of 5
Reproductive isolation means two populations can no longer:
Lesson 4 · Evolution
📐 Population Genetics: Hardy–Weinberg
How common is an allele in a population, and is it changing? The Hardy–Weinberg model gives a baseline to compare with real data.
🧮 The Hardy–Weinberg equations
For a gene with two alleles, the dominant allele frequency is p and the recessive is q. Then p + q = 1, and the genotype frequencies in the next generation are p² + 2pq + q² = 1 (AA, Aa, aa).
4% of a population is aa → q² = 0.04, q = 0.2
p = 1 − 0.2 = 0.8
carriers 2pq = 2 · 0.8 · 0.2 = 0.32
Population of 1000, q² = 0.09:
q = 0.3, 2pq = 0.42 → 420 carriers
36 AA, 48 Aa, 16 aa (100 people)
p = (2·36 + 48) / 200 = 0.6
🖼️ See it
Heterozygotes are most common when the two alleles are equally frequent (p = q = 0.5).
⚖️ When does the model hold?
Allele frequencies stay constant only if there is no mutation, no migration, no selection, random mating and a very large population. Real populations break these conditions, which is why they evolve. A small population can change by chance alone: genetic drift.
p = 0.6 → AA = p² = 0.36
aa = q² = 0.4² = 0.16 → 16 of 100
q² = 0.16 → q = 0.4, p = 0.6
2pq = 2 · 0.6 · 0.4 = 0.48
Always start from the only frequency you can see: the recessive homozygotes, q², because their genotype can be read from their phenotype.
🧪 Worked examples: X-linked traits, three alleles and equilibrium
X-linked recessive. The frequency of affected males equals q, because males have one X. Affected females have the frequency q².
Testing for equilibrium. Work out p and q from the genotype counts, calculate expected counts, and use a χ² test with 1 degree of freedom (3 genotypes − 1 − 1 for the estimated allele frequency).
🎯 Exam focus: stating the assumptions and the working
Always start by identifying the one number you can read directly. For a recessive phenotype, that is q², so q = √(q²). Then p = 1 − q.
conditions for Hardy–Weinberg equilibrium
• very large population (no genetic drift)
• random mating
• no mutation
• no migration (no gene flow)
• no natural selection
Common mistakes. Using the frequency of the recessive phenotype as q (it is q²). Calculating carriers as q × p (it is 2pq). Applying p² + 2pq + q² = 1 to a population that is not in equilibrium. Giving frequencies greater than 1. Forgetting that the HW calculation gives the genotype frequencies for the next generation, from the allele frequencies of this one.
Founder effect and bottlenecks. A small group that starts a new population, or a population reduced to a few survivors, can have very different allele frequencies by chance.
✏️ Quick check · 1 of 11
Conditions
Which condition is NOT required for Hardy–Weinberg equilibrium?
Which one changes the allele frequencies?
✏️ Quick check · 2 of 11
Genetic drift
Genetic drift has the biggest effect in:
Think of chance.
✍️ Type your answer · 3 of 11
4% of a population is homozygous recessive (aa), so q² = 0.04.
What is q?
Enter an exact answer, without rounding.
Take the square root.
✍️ Type your answer · 4 of 11
q = 0.2 in a population at Hardy–Weinberg equilibrium.
What is the frequency of carriers (2pq)?
Enter an exact answer, without rounding.
First find p.
✍️ Type your answer · 5 of 11
A population of 1000 has q² = 0.09.
How many individuals are expected to be carriers?
Enter an exact answer, without rounding.
Find q, then p, then 2pq.
✍️ Type your answer · 6 of 11
A population has 36 AA, 48 Aa and 16 aa individuals (100 in total).
What is the frequency p of allele A?
Enter an exact answer, without rounding.
Count the A alleles.
✍️ Type your answer · 7 of 11
p = 0.6 and q = 0.4 in a population of 100.
How many aa individuals are expected?
Enter an exact answer, without rounding.
q² · 100.
✍️ Type your answer · 8 of 11
8% of males (0.08) have a recessive X-linked condition.
What is the expected frequency of affected females? Give it as a decimal.
Enter an exact answer, without rounding.
Females need two copies.
✍️ Type your answer · 9 of 11
Blood group alleles have the frequencies f(Iᴬ) = 0.3, f(Iᴮ) = 0.1 and f(i) = 0.6.
What is the frequency of blood group AB?
Enter an exact answer, without rounding.
Use 2pq for the two alleles.
✍️ Type your answer · 10 of 11
Selection against the AA genotype is 0.2 and against aa is 0.8. Use q = s_AA ÷ (s_AA + s_aa).
What is the equilibrium frequency q of the a allele?
Enter an exact answer, without rounding.
Substitute into the equilibrium formula.
✏️ Quick check · 11 of 11
Genetic drift
A few individuals colonise an island, and their allele frequencies differ from the original population by chance. This is called:
Think about a small sample.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
In the Hardy–Weinberg equation, 2pq represents the frequency of:
Question 2 of 5
If p = 0.7, then q equals:
Question 3 of 5
Migration of individuals into a population:
Question 4 of 5
The founder effect occurs when:
Question 5 of 5
If a population is NOT in Hardy–Weinberg equilibrium, this suggests that:
Lesson 5 · Technology
🧪 Biotechnology & Genetic Engineering
We can copy, cut, read and edit DNA. These tools make medicines, crops and forensic evidence, and they raise new ethical questions.
🧫 Copying and sorting DNA
PCR copies a short DNA sequence millions of times by repeating three steps: denature (about 95 °C), anneal primers (about 55 °C) and extend with heat-stable Taq polymerase (72 °C). Each cycle doubles the number of copies.
copies = 2ⁿ n = 5 → 32 n = 10 → 1024
Gel electrophoresis: DNA is negative,
so it moves toward the + electrode;
smaller fragments travel farther.
Restriction enzymes cut at specific sequences
linear DNA, 3 cuts → 4 fragments
plasmid (circular), 3 cuts → 3 fragments
5000 bp = 1200 + 2300 + 1500
🖼️ See it
The same enzyme leaves matching “sticky ends”, so the gene fits into the plasmid like a key into a lock.
✂️ Engineering genomes
To make a recombinant plasmid, cut the gene and the plasmid with the same restriction enzyme and join them with DNA ligase. The plasmid is put into bacteria (transformation). CRISPR–Cas9 uses a guide RNA to bring the Cas9 enzyme to a chosen sequence, where it cuts the DNA so it can be changed.
transformation efficiency = colonies / μg of DNA
40 colonies from 0.5 μg → 80 per μg
guide RNA is about 20 nucleotides long
Marker genes such as antibiotic resistance let you pick out the bacteria that actually took up the plasmid.
🧪 Worked examples: PCR, digests and DNA concentration
PCR yield. Copies = starting copies × 2ⁿ for n cycles (in theory).
Restriction digest of a circular plasmid. Two cuts on a circle give two fragments.
plasmid 5000 bp, EcoRI cuts at positions 800 and 2000
fragments: 2000 − 800 = 1200 bp and 5000 − 1200 = 3800 bp
in a gel: the 1200 bp band runs further from the wells
DNA concentration from absorbance. For double-stranded DNA, A₂₆₀ = 1.0 corresponds to 50 μg/mL.
Blue–white screening. The plasmid carries an antibiotic resistance gene and lacZ. Inserting the gene inside lacZ breaks it, so colonies with the insert are white, and those without are blue on X-gal plates.
Controls. Include a no-template PCR tube (to detect contamination) and a known-size DNA ladder in the gel.
🎯 Exam focus: explaining each step of a genetic engineering procedure
1 isolate the gene (restriction enzyme, or reverse transcriptase from mRNA)
2 cut the vector with the same enzyme → complementary sticky ends
3 join with DNA ligase (phosphodiester bonds)
4 transform host cells (heat shock, electroporation)
5 select with antibiotic resistance marker; confirm with a reporter gene or PCR
Explain why Taq polymerase is used: it is stable at 95 °C, so it is not denatured in the denaturing steps and doesn’t need to be added again each cycle.
Common mistakes. Saying that the ligase “glues” the DNA without naming the bonds. Writing that PCR requires cells (it is in vitro). Confusing a restriction enzyme with a ligase. Saying that all the bacteria take up the plasmid (transformation is inefficient, so selection is needed). Forgetting that smaller fragments move further in a gel.
Ethics and safety. Genetically modified crops: benefits include yield and nutrition (golden rice); concerns include gene flow to wild relatives, effects on biodiversity, and patents.
✏️ Quick check · 1 of 11
Ligase
Which enzyme joins DNA fragments together?
To ligate means to tie.
✏️ Quick check · 2 of 11
Electrophoresis
In gel electrophoresis the smallest DNA fragments:
Think of an obstacle course.
✏️ Quick check · 3 of 11
CRISPR
In CRISPR–Cas9, the guide RNA:
It is the address label.
✍️ Type your answer · 4 of 11
A PCR reaction starts with one DNA molecule.
How many copies are there after 5 cycles?
Enter an exact answer, without rounding.
It doubles each cycle.
✍️ Type your answer · 5 of 11
A PCR reaction starts with one molecule.
How many copies are there after 10 cycles?
Enter an exact answer, without rounding.
2 to the power of the cycles.
✍️ Type your answer · 6 of 11
A linear piece of DNA is cut by a restriction enzyme at 3 sites.
How many fragments result?
Enter an exact answer, without rounding.
One more fragment than cuts.
✍️ Type your answer · 7 of 11
A 5000 bp plasmid is cut into 1200 bp and 2300 bp pieces and one other piece.
How long is the other piece, in bp?
Enter an exact answer, without rounding.
bp
The pieces add up to 5000.
✍️ Type your answer · 8 of 11
A PCR reaction starts with 3 template molecules and runs for 8 cycles with perfect doubling.
How many copies are produced?
Enter an exact answer, without rounding.
Multiply by 2 for each cycle.
✍️ Type your answer · 9 of 11
A circular plasmid of 5000 bp is cut by a restriction enzyme at positions 800 and 2000.
What is the length of the smaller fragment, in bp?
Enter an exact answer, without rounding.
bp
Subtract the two positions.
✍️ Type your answer · 10 of 11
A DNA solution diluted 1 : 50 has A₂₆₀ = 0.40. A₂₆₀ = 1.0 corresponds to 50 μg/mL.
What is the concentration of the original solution, in μg/mL?
Enter an exact answer, without rounding.
μg/mL
Convert absorbance, then multiply by the dilution factor.
✏️ Quick check · 11 of 11
Blue–white screening
A gene is inserted into the lacZ gene of a plasmid. On X-gal plates, bacteria that have taken up the recombinant plasmid form:
What happens to lacZ?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Taq polymerase is used in PCR because it:
Question 2 of 5
A plasmid is:
Question 3 of 5
Restriction enzymes cut DNA at:
Question 4 of 5
DNA profiling compares:
Question 5 of 5
A concern about genetically modified crops is:
Lesson 6 · Ecology
🌍 Ecosystems: Energy Flow & Cycles
Energy enters ecosystems as sunlight and leaves as heat, while matter such as carbon and nitrogen is recycled again and again.
☀️ Energy flow
Producers capture light energy. At each step along a food chain only about 10% of the energy passes to the next trophic level, since the rest is lost as heat from respiration, in waste, or in uneaten parts. That is why food chains are short.
producers 10000 kJ → primary 1000 → secondary 100 → tertiary 10
trophic efficiency = energy at next level / energy at this level · 100
200 / 2000 · 100 = 10%
NPP = GPP − R
8000 − 3000 = 5000 kJ/m²/yr
Toxins magnify: a persistent poison like DDT is not broken down, so its concentration rises at each level (biomagnification).
🖼️ See it
Respiration by every organism returns carbon to the air as CO₂.
♻️ Nutrient cycles
In the carbon cycle photosynthesis removes CO₂ from the air, and respiration, decomposition and combustion return it. In the nitrogen cycle bacteria turn nitrogen gas into usable ammonium (fixation), then nitrate (nitrification), and others return it to the air (denitrification).
Where does the energy go? Heat from respiration, undigested material in faeces, parts that are not eaten (bone, roots) and excretion. That is why only 5–20% is transferred.
Eutrophication. Fertiliser runs into a lake → algae grow rapidly (bloom) → they block light → plants below die → decomposers use up oxygen (high BOD) → fish suffocate.
Pyramids. Pyramids of energy are never inverted. Pyramids of numbers and biomass can be (one tree supports many insects; phytoplankton have a small biomass but reproduce quickly).
🎯 Exam focus: the nitrogen cycle and managing ecosystems
Crop rotation with legumes puts nitrogen back into the soil, so fewer fertilisers are needed. Drainage and ploughing add oxygen, which promotes nitrification and prevents denitrification.
Common mistakes. Calling the transfer efficiency a “loss of energy” (energy is transferred to the surroundings as heat, not destroyed). Using NPP when the question asks for GPP. Saying that plants take in nitrogen gas. Mixing up nitrification and nitrogen fixation. Leaving out the units (kJ m⁻² year⁻¹).
Biomagnification. A persistent, fat-soluble pollutant (such as DDT) is not excreted, so its concentration increases at each trophic level.
✏️ Quick check · 1 of 11
Decomposers
The role of decomposers in an ecosystem is to:
Recycling.
✏️ Quick check · 2 of 11
Nitrogen fixation
Nitrogen fixation is the conversion of:
From the air to a usable form.
✏️ Quick check · 3 of 11
Pyramid
Why are pyramids of energy always pyramid-shaped?
Think about the 10% rule.
✍️ Type your answer · 4 of 11
Producers capture 10000 kJ. Each transfer is 10% efficient.
How much energy reaches the secondary consumers, in kJ?
Enter an exact answer, without rounding.
kJ
Apply 10% twice.
✍️ Type your answer · 5 of 11
Producers capture 10000 kJ. Each transfer is 10% efficient.
How much energy reaches the tertiary consumers, in kJ?
Enter an exact answer, without rounding.
kJ
Apply 10% three times.
✍️ Type your answer · 6 of 11
A primary consumer level holds 2000 kJ and the secondary level holds 200 kJ.
What is the trophic efficiency, in %?
Enter an exact answer, without rounding.
%
next ÷ this level × 100.
✍️ Type your answer · 7 of 11
A field has GPP = 8000 kJ/m²/yr and plant respiration R = 3000 kJ/m²/yr.
What is the net primary productivity, in kJ/m²/yr?
Enter an exact answer, without rounding.
Subtract respiration.
✍️ Type your answer · 8 of 11
The gross primary productivity of a field is 15 000 kJ m⁻² year⁻¹ and the plants use 40% of it in respiration.
What is the net primary productivity, in kJ m⁻² year⁻¹?
What percentage of the producers’ energy reaches the secondary consumers?
Enter an exact answer, without rounding.
%
Divide the last level by the first.
✏️ Quick check · 10 of 11
Eutrophication
Why do fish die in a lake after fertiliser runs into it?
Follow the chain of events.
✏️ Quick check · 11 of 11
Nitrogen cycle
Which process converts nitrate in waterlogged soil back into nitrogen gas?
It occurs without oxygen.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Which process removes CO₂ from the atmosphere?
Question 2 of 5
Denitrification returns nitrogen to the atmosphere by:
Question 3 of 5
A food web differs from a food chain because it:
Question 4 of 5
An ecosystem includes:
Question 5 of 5
Which organisms are the first trophic level?
Lesson 7 · Ecology
📈 Population Ecology & Sampling
How many individuals live in an area, and how does the number change? Ecologists answer with sampling methods and growth models.
📊 Growth and carrying capacity
Population size changes by births − deaths + immigration − emigration. With unlimited resources growth is exponential. In real habitats growth is logistic and levels off at the carrying capacity K, with the fastest growth near K/2.
births 60, deaths 40, immigration 10, emigration 0
change = 60 − 40 + 10 − 0 = 30
100 bacteria doubling every hour for 5 h
N = 100 · 2⁵ = 3200
K = 1000 → fastest growth at 500
density = N / area: 120 / 40 m² = 3 per m²
🖼️ See it
Growth levels off where resources limit the population, at the carrying capacity.
🔎 Estimating population size
It is usually impossible to count everything. With quadrats you count a sample area and scale up. With mark–recapture you use the Lincoln index: N = (M × C) / R, where M animals are marked, C are caught later and R of them carry marks.
quadrats: mean 4 plants per 1 m², field 500 m² → 4 · 500 = 2000
Lincoln: M = 50, C = 40, R = 10
N = 50 · 40 / 10 = 200
Simpson’s diversity D = 1 − Σ (n/N)²
two species, 5 and 5: D = 1 − (0.25 + 0.25) = 0.5
Sample randomly and take enough samples so that the estimate is not biased by where you happen to look.
🧪 Worked examples: Lincoln index, logistic growth and diversity
Mark–release–recapture. N = (M × C) ÷ R.
100 marked and released; 60 caught later, 12 of them marked
N = 100 × 60 ÷ 12 = 500
Logistic growth. dN/dt = r N (1 − N ÷ K). The growth rate is largest at K/2.
r = 0.40, N = 250, K = 1000
dN/dt = 0.40 × 250 × (1 − 0.25) = 0.40 × 250 × 0.75 = 75 per year
Simpson’s index of diversity. D = 1 − Σ (n ÷ N)². A value near 1 means high diversity.
counts: species A 6, B 3, C 1 (N = 10)
Σ(n/N)² = 0.36 + 0.09 + 0.01 = 0.46
D = 1 − 0.46 = 0.54
Assumptions of the Lincoln index: marking does not affect survival or behaviour, marks are not lost, the animals mix randomly, and there is no birth, death or migration between the samples.
r- and K-strategists. r-strategists (insects, weeds) reproduce early, with many small offspring; K-strategists (elephants) have few offspring with a large parental investment and live near the carrying capacity.
🎯 Exam focus: limiting factors and sampling methods
Density-dependent factors (competition, predation, disease) have a stronger effect when the population is dense. Density-independent factors (fire, frost, drought) act regardless of the density.
quadrats random sampling of plants / slow animals; use random coordinates
transects systematic sampling along an environmental gradient
pitfall traps ground beetles; kick sampling: stream invertebrates
mean per quadrat × (total area ÷ quadrat area) = population estimate
Common mistakes. Placing quadrats where the plants look thickest. Using too few samples. Treating the Lincoln index as exact. Using a transect when random sampling is required (or the reverse). Saying that the carrying capacity is the maximum population ever reached (it is the maximum a habitat can support in the long term). Using D = Σ(n/N)² (without the “1 −”).
Statistics. To test whether two species are associated, use a χ² test or Spearman’s rank correlation; to compare two means, use a t-test.
✏️ Quick check · 1 of 11
Carrying capacity
The carrying capacity of an environment is the:
K.
✏️ Quick check · 2 of 11
Density-dependent
Which of these is a density-dependent factor?
Related to how crowded it is.
✏️ Quick check · 3 of 11
r-strategy
Organisms with many offspring and little parental care are called:
Think of mice versus elephants.
✍️ Type your answer · 4 of 11
A population has 60 births, 40 deaths, 10 immigrants and 0 emigrants.
What is the net change in population size?
Enter an exact answer, without rounding.
Add the gains and subtract the losses.
✍️ Type your answer · 5 of 11
A colony of 100 bacteria doubles every hour.
How many are there after 5 hours?
Enter an exact answer, without rounding.
Five doublings.
✍️ Type your answer · 6 of 11
In a mark–recapture study M = 50, C = 40 and R = 10.
What is the estimated population size (Lincoln index)?
Enter an exact answer, without rounding.
N = (M × C) / R.
✍️ Type your answer · 7 of 11
The mean is 4 plants per 1 m² quadrat and the field is 500 m².
How many plants are estimated in the field?
Enter an exact answer, without rounding.
Scale the mean up to the whole area.
✍️ Type your answer · 8 of 11
100 animals are marked and released. A later sample of 60 contains 12 marked animals.
What is the estimated population size?
Enter an exact answer, without rounding.
Use the Lincoln index.
✍️ Type your answer · 9 of 11
A population has r = 0.40 per year, N = 250 and K = 1000.
What is the growth rate dN/dt, in individuals per year?
Enter an exact answer, without rounding.
per year
Use the logistic equation.
✍️ Type your answer · 10 of 11
A sample contains 6 individuals of species A, 3 of species B and 1 of species C.
What is Simpson’s index of diversity D?
Enter an exact answer, without rounding.
Remember to subtract from 1.
✏️ Quick check · 11 of 11
Limiting factors
Which of these is a density-dependent factor?
Does the effect depend on how crowded the population is?
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
Logistic growth is characterised by:
Question 2 of 5
For mark–recapture to be valid, the marked animals must:
Question 3 of 5
A high Simpson’s diversity index means:
Question 4 of 5
Immigration:
Question 5 of 5
The best way to place quadrats for an unbiased sample is:
Lesson 8 · Health
🛡️ Immunity & Disease
The immune system recognises invaders and remembers them. Vaccines train it in advance, so a later infection is stopped quickly.
🧱 Innate and adaptive immunity
Innate defences act fast and are the same for every invader: skin, mucus, stomach acid and phagocytes that engulf pathogens. Adaptive immunity is specific and has memory: B cells make antibodies (Y-shaped proteins with 4 chains, two heavy and two light), and T cells kill infected cells or help other immune cells.
first infection: slow response, a few days
second infection: fast and large response (memory cells)
an antibody is made of 4 polypeptide chains
HIV infects helper T cells → AIDS
🖼️ See it
First exposure at day 0. Memory cells make the second response faster and much stronger. This is why vaccines work.
💉 Vaccines and herd immunity
A vaccine contains a weakened, killed or partial pathogen, so the body makes memory cells without the disease. Herd immunity occurs when enough people are immune that the disease cannot spread easily. The basic reproduction number R₀ is the number of people one case infects in a fully susceptible population.
immune fraction needed = 1 − 1/R₀
R₀ = 5 → 1 − 0.2 = 0.8 → 80%
2000 people, 75% needed → 1500 immune
R₀ = 3 with no immunity, from 1 case:
new cases per generation: 3, 9, 27 → the third is 27
Antibiotics kill bacteria, not viruses. Overusing them selects for resistant bacteria.
🧪 Worked examples: vaccines, herd immunity and diagnostic tests
Herd immunity threshold = 1 − 1 ÷ R₀.
R₀ = 4 → 1 − 0.25 = 0.75 → 75% of the population must be immune
in a town of 20 000, that is 15 000 people
Vaccine efficacy = (attack rate in unvaccinated − attack rate in vaccinated) ÷ attack rate in unvaccinated × 100.
100 people truly infected: the test is positive in 85 and negative in 15
sensitivity = true positives ÷ all infected = 85 ÷ 100 = 85%
1000 truly uninfected: negative in 950
specificity = 950 ÷ 1000 = 95%
Clonal selection. A pathogen’s antigen binds to the one B cell with the matching receptor; this cell divides (clonal expansion) into plasma cells, which make antibodies, and memory cells, which allow a faster response next time.
Active against passive. Active immunity: the body makes its own antibodies and memory cells (infection, vaccination). Passive: ready-made antibodies are given (breast milk, injection), so protection is immediate but short-lived.
🎯 Exam focus: primary and secondary responses, and HIV
For the secondary response: faster, larger and longer-lasting, because memory B cells are already present and are quickly activated. A graph of antibody concentration against time shows a steeper, higher peak.
HIV infects helper T cells (CD4+) → fewer T cells → B cells receive less stimulation
→ fewer antibodies → AIDS: death from opportunistic infections
antibiotics do not work on viruses (they have no cell wall or ribosomes of their own)
Common mistakes. Saying that a vaccine “contains antibodies” (it contains antigens). Confusing antibiotics with vaccines. Saying that antigens are made by the body to fight pathogens (antibodies are). Writing that herd immunity protects the vaccinated (it protects the unvaccinated). Calculating sensitivity as true positives ÷ all positive results.
Monoclonal antibodies. They are made by fusing a B cell with a tumour cell (a hybridoma). They are used in pregnancy tests, diagnosis and targeted cancer therapy.
✏️ Quick check · 1 of 11
B cells
B cells are responsible for producing:
Humoral immunity.
✏️ Quick check · 2 of 11
Memory cells
Why is the second response to an infection faster?
The immune system remembers.
✏️ Quick check · 3 of 11
Vaccine
A vaccine works by:
Training, not treatment.
✍️ Type your answer · 4 of 11
An antibody molecule has 2 heavy and 2 light polypeptide chains.
How many polypeptide chains does it have in total?
Enter an exact answer, without rounding.
Add the two types.
✍️ Type your answer · 5 of 11
A disease has R₀ = 5.
What percentage of the population must be immune for herd immunity (1 − 1/R₀)?
Enter an exact answer, without rounding.
%
Use the formula.
✍️ Type your answer · 6 of 11
A school has 2000 people and 75% need to be immune for herd immunity.
How many must be immune?
Enter an exact answer, without rounding.
Multiply the fraction by the total.
✍️ Type your answer · 7 of 11
In a fully susceptible population one case infects 3 others (R₀ = 3), and so on.
How many new cases are there in the third generation of spread?
Enter an exact answer, without rounding.
3 to the power of the generation.
✍️ Type your answer · 8 of 11
A disease has a basic reproduction number R₀ = 4.
What percentage of the population must be immune for herd immunity?
Enter an exact answer, without rounding.
%
Use 1 − 1/R₀.
✍️ Type your answer · 9 of 11
During a trial, 20% of unvaccinated people and 2% of vaccinated people catch the disease.
What is the vaccine efficacy, as a percentage?
Enter an exact answer, without rounding.
%
Compare the attack rates.
✍️ Type your answer · 10 of 11
A test is positive for 85 of 100 truly infected people and negative for the other 15.
What is the sensitivity of the test, as a percentage?
Enter an exact answer, without rounding.
%
True positives ÷ all infected.
✏️ Quick check · 11 of 11
Secondary response
Why is the secondary immune response faster and larger than the primary response?
What remains in the body after the first infection?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Which of these is part of innate immunity?
Question 2 of 5
HIV damages the immune system by infecting:
Question 3 of 5
An antigen is:
Question 4 of 5
Autoimmune diseases occur when the immune system:
Question 5 of 5
Antibiotic resistance becomes more common because:
Final Challenge
🏆 16-Question Challenge Quiz
Questions from every topic in this course. Pick an answer for each one, then hit Check My Answers.
0 / 16
💪 Keep practicing — go back through the lessons and retake the quiz!
Q1 · Mendelian Genetics
A cross between an Aa and an aa individual gives:
Q2 · Mendelian Genetics
A genotype shows what the organism’s:
Q3 · Sex-Linkage, Linkage & Pedigrees
A son inherits his X chromosome from his:
Q4 · Sex-Linkage, Linkage & Pedigrees
A recombination frequency of 50% means the genes:
Q5 · Natural Selection & Evolution
Evolution can be defined as a change in:
Q6 · Natural Selection & Evolution
Reproductive isolation means two populations can no longer:
Q7 · Population Genetics: Hardy–Weinberg
If q² = 0.16, then p is:
Q8 · Population Genetics: Hardy–Weinberg
If a population is NOT in Hardy–Weinberg equilibrium, this suggests that:
Q9 · Biotechnology & Genetic Engineering
During the annealing step of PCR:
Q10 · Biotechnology & Genetic Engineering
A concern about genetically modified crops is:
Q11 · Ecosystems: Energy Flow & Cycles
The main source of energy for nearly all ecosystems is the:
Q12 · Ecosystems: Energy Flow & Cycles
Which organisms are the first trophic level?
Q13 · Population Ecology & Sampling
A limiting factor that gets stronger as population density rises is called:
Q14 · Population Ecology & Sampling
The best way to place quadrats for an unbiased sample is:
Q15 · Immunity & Disease
Herd immunity protects people who:
Q16 · Immunity & Disease
Antibiotic resistance becomes more common because: