📚 Honors Biology Final — Study Hub
Everything you need for Monday's exam, all in one place. Click any card or use the tabs above to navigate.
🗂️ Your Resources
3-Day Study Plan
Sat / Sun / Mon checklist with interactive checkboxes and progress tracking.
★ Start herePractice Test Grader
Punch in answers, get an instant grade + personalized study plan.
Interactive graderCheat Sheet
All 10 topics condensed with reveal-answer self-tests.
Topic summaryFlashcards
199 cards. Hide/reveal answers, star cards, filter by topic, shuffle.
Active recallDeep Dives — All 11 Topics
Focused walkthroughs with worked examples and practice problems for every topic. Pick the one you need.
Topic-specific🎯 Top 10 Highest-Yield Facts
If you only have 5 minutes before the exam:
- AUG = start codon; UAA / UAG / UGA = stop
- Transcription → nucleus; Translation → ribosome
- Crossing over happens in Prophase I of meiosis
- Producers = most energy; 10% transfers up each trophic level
- DNA similarity = best evidence for common ancestry
- Mutation = source of new genetic variation
- q² = % showing recessive trait (Hardy-Weinberg starting move)
- Sex-linked recessive = more common in males (only 1 X)
- Carrier = heterozygous (Bb), doesn't show recessive trait
- Bb × Bb = 3:1 dominant:recessive ratio
⚡ Quick Strategy Recap:
- 70% of your study time = testing yourself, not rereading
- Spaced over 3 days beats one big cramming session
- Sleep is part of the plan — memory consolidates overnight
- Top 3 weak topics from the practice test = where you spend most of Sunday
- Say flashcard answers OUT LOUD before clicking reveal
📅 3-Day Study Plan
Saturday → Sunday → Monday. Check off tasks as you go.
The whole plan is built around ACTIVE RECALL. You'll spend ~70% of your time testing yourself rather than rereading. That feels harder than passive reading — that's the point. The struggle IS the learning.
🎯 Guiding Principles
- Test yourself FIRST, then check.
- Write down what you missed. Build a "weak list."
- Sleep is non-negotiable. Cramming all night → worse on Monday.
- Eat breakfast Monday. Glucose = brain fuel.
- Spaced repetition beats massed practice.
- Take real breaks. 50 min work → 10 min off.
📅 Saturday — Diagnostic + Weak-Spot Mapping
0 / 0
Goal: Find what you know vs. what you think you know.
🟢 Block 1: Take the practice test BLIND 60 min
☕ 10-min break — water, snack, stretch
🟢 Block 2: Grade & build your weak list 45 min
☕ 15-min break
🟢 Block 3: First pass through the cheat sheet 45 min
☕ 10-min break
🟢 Block 4: Flashcards Round 1 (first half) 40 min
🌙 End of Saturday
📅 Sunday — Deep Work on Weak Spots
0 / 0
Goal: Drill your weak spots. Heaviest study day.
🟢 Block 1: Flashcards Round 2 (second half) 60 min
☕ 15-min break
🟢 Block 2: Deep-dive your TOP 3 weak topics 60 min
☕ 30-min break — eat lunch, recharge
🟢 Block 3: Re-attempt missed practice test questions 45 min
☕ 10-min break
🟢 Block 4: Practice the math problems 45 min
🟢 Block 5: "Teach it" exercise 30 min
🌙 End of Sunday
📅 Monday — Light Review + Exam
0 / 0
Goal: Stay confident, refresh memory, don't burn out.
🟢 Morning routine
🟢 Pre-exam review 60-90 min
DON'T: Try to learn anything new. Re-do the whole practice test. Panic.
🟢 During the exam
✅ Practice Test Grader
Interactive grader with personalized study recommendations.
How to use: Take the practice test on paper first. Then click your A/B/C/D answers below and hit "Grade My Test." You'll see your score, which questions you got wrong (red), the correct answers (green), and a personalized study plan ranked by priority.
📝 Click your answers
Note: Questions 11, 21, 53, and 58 only have A/B/C.
0 / 65
0% F
🎯 Your Personalized Study Plan
Based on what you missed/skipped, ranked by priority:
📖 Full Answer Key with Explanations
★ = high-frequency concept
Q 1–10
1. C — fossil record ★
Best evidence of physical changes over geologic time. Homologous structures suggest common ancestry, but fossils literally show change over time.
Best evidence of physical changes over geologic time. Homologous structures suggest common ancestry, but fossils literally show change over time.
2. D — DNA "unzips" along hydrogen bonds
Replication starts when the double helix unzips. Free nucleotides then pair up.
Replication starts when the double helix unzips. Free nucleotides then pair up.
3. D — phenylalanine ★
tRNA anticodon AAA pairs with mRNA codon UUU → phenylalanine. Flip A↔U, C↔G.
tRNA anticodon AAA pairs with mRNA codon UUU → phenylalanine. Flip A↔U, C↔G.
4. D — mutations and nonrandom mating ★
Both violate Hardy-Weinberg conditions → gene frequencies change.
Both violate Hardy-Weinberg conditions → gene frequencies change.
5. D — crossing-over ★
Homologous chromosomes during prophase I exchanging segments = crossing over.
Homologous chromosomes during prophase I exchanging segments = crossing over.
6. B — more similar sequences = more closely related ★
Molecular similarity = recent common ancestor. High-frequency concept.
Molecular similarity = recent common ancestor. High-frequency concept.
7. D — level D (bottom = producers) ★
Most stored energy is always at the producer level. 10% rule.
Most stored energy is always at the producer level. 10% rule.
8. D — GAU to GAC
Silent mutation — both code for aspartic acid. Others change the amino acid.
Silent mutation — both code for aspartic acid. Others change the amino acid.
9. D — different jawbone structure
Different structural anatomy = NOT closely related.
Different structural anatomy = NOT closely related.
10. B — energy flows bottom to top, decreases at each level ★
10% rule. Energy decreases as it flows up.
10% rule. Energy decreases as it flows up.
Q 11–20
11. C — both DNA and RNA
Adenine is in both.
Adenine is in both.
12. B — 75%
Rr × Rr → 3 round : 1 oval = 75% round.
Rr × Rr → 3 round : 1 oval = 75% round.
13. D — ii (type O) ★
AB father has no i allele to pass → no ii child possible.
AB father has no i allele to pass → no ii child possible.
14. C — best adapted to their environment ★
Darwin's survival of the fittest.
Darwin's survival of the fittest.
15. D — grasshopper population
Wrens eat grasshoppers — direct food source decreases first.
Wrens eat grasshoppers — direct food source decreases first.
16. B — energy decreases from consumer 2 to consumer 3 ★
Energy always decreases moving up trophic levels.
Energy always decreases moving up trophic levels.
17. C — predator, carnivore, consumer ★
Scorpion hunts (predator), eats meat (carnivore), is heterotroph (consumer).
Scorpion hunts (predator), eats meat (carnivore), is heterotroph (consumer).
18. B — ribosome
Translation happens at the ribosome.
Translation happens at the ribosome.
19. C — DNA sequences are 90% identical ★
DNA similarity = best evidence of common ancestry.
DNA similarity = best evidence of common ancestry.
20. B — natural selection ★
Predators eat visible bright termites → dark survive → classic natural selection.
Predators eat visible bright termites → dark survive → classic natural selection.
Q 21–30
21. B — RNA only
mRNA, tRNA, rRNA interact with ribosomes.
mRNA, tRNA, rRNA interact with ribosomes.
22. D — Iᴬi × Iᴮi ★
Only cross that produces all 4 blood types.
Only cross that produces all 4 blood types.
23. B — half of male offspring affected ★
Carrier mom × unaffected dad: 50% of sons get the affected X.
Carrier mom × unaffected dad: 50% of sons get the affected X.
24. A
Only crossing over exchanges segments between homologous chromosomes.
Only crossing over exchanges segments between homologous chromosomes.
25. C — haploid cells produced by meiosis ★
Supports Mendel's idea: one factor per gamete.
Supports Mendel's idea: one factor per gamete.
26. C — rabbit food supply increases
Fewer mice → more grass → rabbits' food supply increases.
Fewer mice → more grass → rabbits' food supply increases.
27. B — common ancestors ★
Bat wing + human arm + dolphin flipper = homologous structures.
Bat wing + human arm + dolphin flipper = homologous structures.
28. D — offspring of more resistant termites ★
Natural selection — survivors pass on resistance genes.
Natural selection — survivors pass on resistance genes.
29. C — diverges into two species ★
Disruptive selection — small and large beaks survive; medium loses out.
Disruptive selection — small and large beaks survive; medium loses out.
30. D — gene frequencies remain constant ★
Hardy-Weinberg equilibrium = no evolution.
Hardy-Weinberg equilibrium = no evolution.
Q 31–40
31. C — meiosis, produces gametes ★
One cell → 4 cells with half the chromosomes = meiosis.
One cell → 4 cells with half the chromosomes = meiosis.
32. A — predator populations ★
Biotic (living) factor.
Biotic (living) factor.
33. B — high survival value ★
Traits spread if they help survival AND reproduction.
Traits spread if they help survival AND reproduction.
34. B — it is evolving ★
Allele frequencies changed = evolution by definition.
Allele frequencies changed = evolution by definition.
35. A — rainfall in northern Africa ★
Abiotic = non-living.
Abiotic = non-living.
36. B
Large fish removed → small fish reproduce → distribution shifts smaller.
Large fish removed → small fish reproduce → distribution shifts smaller.
37. A — Ty, ty
Ttyy gametes: only Ty and ty possible.
Ttyy gametes: only Ty and ty possible.
38. C — A-A-A
Phe codon UUU → DNA template AAA.
Phe codon UUU → DNA template AAA.
39. B — grasses and shrubs ★
Producers always have the most energy.
Producers always have the most energy.
40. C — DNA replication ★
Must happen before any cell division.
Must happen before any cell division.
Q 41–50
41. C — taller around 3,000m
Reading the graph: peak height is around 3,000m.
Reading the graph: peak height is around 3,000m.
42. A — increased
Birth rate > death rate the whole time.
Birth rate > death rate the whole time.
43. B — average body size decreased
Big fish removed before reproducing → smaller fish dominate.
Big fish removed before reproducing → smaller fish dominate.
44. A — 0 children
TT × TT = all TT = all can roll.
TT × TT = all TT = all can roll.
45. A — nucleus
mRNA is made in the nucleus (transcription).
mRNA is made in the nucleus (transcription).
46. B — nucleus to ribosome ★
mRNA's job: carry code FROM nucleus TO ribosome.
mRNA's job: carry code FROM nucleus TO ribosome.
47. A — mRNA U-A-C → U-A-G
DNA ATG → mRNA UAC. DNA ATC → mRNA UAG (stop codon — nonsense mutation).
DNA ATG → mRNA UAC. DNA ATC → mRNA UAG (stop codon — nonsense mutation).
48. D — fewer foxes and wolves
Remove rabbits → less food up the chain → fewer of everyone above.
Remove rabbits → less food up the chain → fewer of everyone above.
49. A — heterozygous dominant
Green offspring of green × yellow cross must be Gg.
Green offspring of green × yellow cross must be Gg.
50. D — 22 autosomes + XY ★
Normal male: 46 chromosomes = 22 pairs autosomes + XY.
Normal male: 46 chromosomes = 22 pairs autosomes + XY.
Q 51–65
51. C — tRNA transports amino acids ★
tRNA's textbook function.
tRNA's textbook function.
52. C — sex-linked recessive ★
Mostly males affected, females are carriers — X-linked recessive pattern.
Mostly males affected, females are carriers — X-linked recessive pattern.
53. B — RNA only
UCG contains U → only RNA.
UCG contains U → only RNA.
54. C — Normal × Feather shedder
FF × ff → all Ff = all frizzle.
FF × ff → all Ff = all frizzle.
55. A — secondary consumers
Foxes, snakes, sparrows eat primary consumers.
Foxes, snakes, sparrows eat primary consumers.
56. B — mutation ★
Original source of new genetic variation.
Original source of new genetic variation.
57. D — comparative anatomy
Studying homologous structures = comparative anatomy.
Studying homologous structures = comparative anatomy.
58. B — RNA only
tRNA transports amino acids.
tRNA transports amino acids.
59. A — gene mutation ★
Change in DNA base sequence = gene mutation by definition.
Change in DNA base sequence = gene mutation by definition.
60. C — independent assortment ★
Height and color sort independently → Mendel's law.
Height and color sort independently → Mendel's law.
61. D — not highly beneficial ★
Vestigial structures lose function with no selection pressure.
Vestigial structures lose function with no selection pressure.
62. A
Producers → rodents → snakes → hawks/owls (correct order).
Producers → rodents → snakes → hawks/owls (correct order).
63. A — equal births and deaths ★
Stable population = births equal deaths.
Stable population = births equal deaths.
64. (No answer in key — best guess: B: 4)
Count only FULLY SHADED squares (affected males). Half-shaded circles are carriers and don't count. X-linked recessive disorders typically affect ~10-25% of a family pedigree, so 3 or 4 is realistic.
Count only FULLY SHADED squares (affected males). Half-shaded circles are carriers and don't count. X-linked recessive disorders typically affect ~10-25% of a family pedigree, so 3 or 4 is realistic.
65. B — adapted to specific food source ★
Adaptive radiation — Darwin's finches concept.
Adaptive radiation — Darwin's finches concept.
📘 Cheat Sheet
Condensed notes on all 10 topics. Use the reveal buttons for active recall.
Topics:
1. Meiosis
Big idea: Meiosis makes haploid gametes with half the chromosomes. Two divisions, one DNA replication → 4 genetically unique haploid cells.
- Purpose: produce gametes for sexual reproduction
- Haploid (n): one set of chromosomes; diploid (2n) = two sets
- Includes crossing over (Prophase I) — alleles swap → variation
- Independent assortment (Metaphase I) → more variation
- PMAT happens twice; DNA replicates once
| Mitosis | Meiosis | |
|---|---|---|
| Purpose | Growth/repair | Gamete production |
| # divisions | 1 | 2 |
| Daughter cells | 2 | 4 |
| Chromosome # | Same (2n→2n) | Half (2n→n) |
| Variation | None | Lots |
🎯 Self-Test
1. Why is crossing over important?
Creates genetic variation by mixing alleles between homologous chromosomes.
2. If a parent has 46 chromosomes, how many do gametes have?
23 (half — gametes are haploid).
3. What two events create variation in meiosis?
Crossing over (Prophase I) and independent assortment (Metaphase I).
2. Protein Synthesis (DNA → RNA → Protein)
Central Dogma: DNA → mRNA → Protein. Transcription = nucleus. Translation = ribosome.
| DNA | RNA | |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Strands | Double | Single |
| Bases | A, T, C, G | A, U, C, G |
- mRNA — carries code to ribosome
- tRNA — brings amino acids (has anticodon)
- rRNA — makes up ribosomes
- Codon = 3 bases on mRNA = 1 amino acid
- Anticodon = 3 bases on tRNA, complementary to codon
- AUG = start codon (Methionine); UAA, UAG, UGA = stop
Worked example: DNA T-A-C G-G-A C-C-T → mRNA A-U-G C-C-U G-G-A → Met-Pro-Gly
🎯 Self-Test
1. Where do transcription and translation occur?
Transcription = nucleus. Translation = ribosome.
2. Codon vs. anticodon?
Codon = 3 bases on mRNA. Anticodon = 3 complementary bases on tRNA.
3. Transcribe TAC-GGC-ATA
AUG-CCG-UAU. (T→A, A→U, G→C, C→G)
4. Translate AUG-GCU-UAA
Methionine — Alanine — STOP.
3. Inheritance & Punnett Squares
- Genotype = genetic makeup (BB, Bb, bb)
- Phenotype = physical trait
- Homozygous = BB or bb; Heterozygous = Bb
- Carrier = heterozygous; doesn't show recessive trait but can pass it
Bb × Bb Punnett square:
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
Result: 1 BB : 2 Bb : 1 bb → 75% dominant phenotype
Blood types (codominance): See .
- Iᴬ Iᴬ or Iᴬi = A; Iᴮ Iᴮ or Iᴮi = B; Iᴬ Iᴮ = AB; ii = O
- All 4 types from Iᴬi × Iᴮi
Sex-linked (X-linked): Males (XY) only need 1 recessive allele to show trait. Examples: hemophilia, colorblindness.
Pedigree: Square = male, Circle = female. Shaded = affected. Half-shaded = carrier.
🎯 Self-Test
1. Aa × aa: % heterozygous offspring?
50%.
2. Tt × tt: probability of non-roller child?
50%.
3. Both parents homozygous dominant — recessive child possible?
No (0%). TT × TT → all TT.
4. Why are X-linked recessive disorders more common in males?
Males have only one X — a single recessive allele is enough to show. Females need two copies.
4. Gene & Chromosomal Mutations
| Gene mutation | What happens |
|---|---|
| Substitution | One base swapped |
| Insertion | Extra base(s) added |
| Deletion | Base(s) removed |
| Frameshift | Reading frame shifts (caused by insertion/deletion, NOT substitution) |
- Silent — no amino acid change
- Missense — one amino acid changes
- Nonsense — creates a stop codon, shortens protein
| Chromosomal mutation | What happens |
|---|---|
| Deletion | Section lost |
| Duplication | Section repeated |
| Inversion | Segment flips |
| Translocation | Piece moves to different chromosome |
| Nondisjunction | Chromosomes fail to separate → Down syndrome (trisomy 21) |
Disorders: Sickle cell (substitution); Down syndrome (nondisjunction).
🎯 Self-Test
1. What causes a frameshift mutation?
Insertion or deletion of bases (NOT substitution).
2. Silent vs. missense vs. nonsense?
Silent = no change. Missense = one amino acid changes. Nonsense = creates stop codon.
3. What chromosomal error causes Down syndrome?
Nondisjunction — chromosome 21 fails to separate, leading to trisomy 21.
5. Natural Selection & Evolution
Darwin's 4 conditions:
- Variation in populations
- Overproduction — more offspring than env can support
- Struggle for existence
- Survival of the fittest — survive AND reproduce
Key: Populations evolve, not individuals. Reproduction is the key — not just surviving.
| Type | Effect | Example |
|---|---|---|
| Directional | Favors one extreme | Pesticide-resistant termites |
| Stabilizing | Favors the average | Human birth weight |
| Disruptive | Favors both extremes | Beak sizes with only large/small seeds |
- Mutation = source of new variation
- Adaptive radiation = one ancestor → many species (Darwin's finches)
🎯 Self-Test
1. 4 conditions for natural selection?
Variation, overproduction, struggle, differential survival/reproduction.
2. Directional vs. stabilizing vs. disruptive?
Directional = one extreme. Stabilizing = average. Disruptive = both extremes.
3. Why doesn't an individual "evolve"?
Evolution = allele frequency change in a POPULATION over generations.
4. Source of new variation?
Mutation.
6. Evidence for Evolution
| Evidence | What it shows |
|---|---|
| Fossil record | Physical changes over geologic time |
| Homologous structures | Common ancestor (bat wing, human arm, dolphin flipper) |
| Analogous structures | NOT common ancestor (bird vs. insect wings) |
| Vestigial structures | Reduced function (appendix, whale hip bones) |
| Embryology | Similar early embryos = common ancestor |
| Molecular / DNA | Strongest evidence — similar DNA = closer relationship |
💡 Shortcut: "Best evidence for common ancestry?" → usually DNA / molecular evidence.
🎯 Self-Test
1. Homologous vs. analogous?
Homologous = same structure, different function (common ancestry). Analogous = same function, different structure (NOT common ancestry).
2. What does the appendix tell us?
Vestigial structure — useful in ancestors, lost function over time.
3. Why is DNA evidence strongest?
Quantifiable; changes accumulate predictably; you can count base differences directly.
7. Biodiversity & Food Webs
- Biodiversity = variety of life; increases ecosystem stability
- Producers (bottom, autotrophs) → most energy
- Primary consumers (herbivores)
- Secondary consumers
- Tertiary consumers (top) → least energy
- Decomposers recycle nutrients
10% Rule: Only ~10% of energy passes up. Rest lost as heat. Producers have MOST energy; top consumers LEAST.
Biotic = living. Abiotic = non-living (rainfall, temp, sunlight).
🎯 Self-Test
1. What % of energy transfers between trophic levels?
About 10%. The rest is lost as heat.
2. Which level has the most stored energy?
Producers (bottom).
3. Are predators biotic or abiotic?
Biotic (living).
8. Hardy-Weinberg
Full walkthrough in the .
- p + q = 1
- p² + 2pq + q² = 1
- p² = AA; 2pq = Aa; q² = aa (start here when given recessive phenotype %)
5 conditions for equilibrium (no evolution): large pop, no migration, no mutation, random mating, no selection.
9. Mark & Recapture
N = (M × C) / R
N = population estimate, M = marked first capture, C = total in second capture, R = recaptured marked.
Examples: 40 marked, 60 caught, 12 marked → N = 200. 25 marked, 50 caught, 5 marked → N = 250. 80 marked, 100 caught, 20 marked → N = 400.
10. Population Growth
| Exponential | Logistic | |
|---|---|---|
| Shape | J-curve | S-curve |
| Resources | Unlimited | Limited |
Carrying capacity (K): max population the environment can sustain.
Stable population: births ≈ deaths.
⚡ Quick Cheat-Sheet Recap
- AUG = start; UAA/UAG/UGA = stop
- Transcription = nucleus; Translation = ribosome
- Crossing over = Prophase I
- 10% rule; producers = most energy
- DNA evidence = strongest for common ancestry
- Mutation = source of new variation
- q² = % showing recessive trait
- N = MC/R for mark-recapture
- Sex-linked recessive = more common in males
- Carrier = heterozygous
🃏 Flashcards
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🔬 Topic Deep Dives
Pick any topic for a focused walkthrough with worked examples + practice problems.
🔄 Meiosis
Why meiosis trips people up: Two divisions instead of one, with different things happening at each stage. Just remember: 4 unique haploid cells at the end, and the unique part comes from crossing over + independent assortment.
1. The Big Picture
Meiosis takes 1 diploid cell (2n) → 4 haploid cells (n). It does this with:
- 1 DNA replication at the start (S phase before Meiosis I)
- 2 divisions (Meiosis I and Meiosis II)
- Each division goes through PMAT: Prophase → Metaphase → Anaphase → Telophase
2. What Happens in Each Phase
| Phase | What happens |
|---|---|
| Prophase I | Homologous chromosomes pair up. Crossing over happens — alleles exchange. ★ |
| Metaphase I | Homologous pairs line up randomly at the equator. Independent assortment. ★ |
| Anaphase I | Homologous pairs SEPARATE (not sister chromatids yet). |
| Telophase I | Two haploid cells form. Each has chromosomes made of 2 sister chromatids. |
| Prophase II → Telophase II | Like mitosis — sister chromatids separate. Result: 4 haploid cells. |
3. Meiosis vs. Mitosis (KNOW THIS COLD)
| Mitosis | Meiosis | |
|---|---|---|
| Purpose | Growth, repair | Make gametes |
| Divisions | 1 | 2 |
| Daughter cells | 2 (diploid) | 4 (haploid) |
| Identical? | Yes (clones) | No (unique) |
| Crossing over? | No | Yes |
4. Worked Example
Q: A human cell starts meiosis with 46 chromosomes. After Meiosis I, how many chromosomes are in each cell? After Meiosis II?
After Meiosis I: 23 chromosomes per cell (homologs separated), but each is still 2 sister chromatids.
After Meiosis II: 23 chromosomes per cell (sister chromatids separated). Now there are 4 total cells.
After Meiosis II: 23 chromosomes per cell (sister chromatids separated). Now there are 4 total cells.
5. Practice Problems
Practice 1
Why must gametes be haploid?
So that fertilization (n + n) restores the diploid chromosome number in the offspring. If gametes were diploid, each generation would double the chromosomes.
Practice 2
Name the two events in meiosis that create genetic variation.
Crossing over (Prophase I) and independent assortment (Metaphase I).
Practice 3
During which phase of meiosis do homologous chromosomes exchange alleles?
Prophase I — this is crossing over.
Practice 4
If a parent cell has 8 chromosomes, how many chromosomes are in each gamete it produces?
4 (half — gametes are haploid).
Practice 5
Both mitosis and meiosis require what to occur first?
DNA replication (during S phase, before division starts).
Practice 6
If meiosis didn't happen, what would go wrong in sexual reproduction?
Gametes would be diploid. Fertilization would create tetraploid offspring, and chromosome number would double each generation — nonviable.
🧬 Protein Synthesis (DNA → mRNA → Protein)
The Central Dogma: DNA → mRNA → Protein. Two steps: transcription (DNA → mRNA, in the nucleus) and translation (mRNA → protein, at the ribosome).
1. Transcription (DNA → mRNA)
- DNA "unzips" along weak hydrogen bonds
- RNA polymerase reads the DNA template strand
- Free RNA nucleotides pair with DNA bases (A→U, T→A, C→G, G→C)
- An mRNA strand is built
- mRNA leaves the nucleus and travels to a ribosome
2. Translation (mRNA → Protein)
- Ribosome reads mRNA 3 bases at a time (each 3-base group = a codon)
- tRNA brings the matching amino acid; its anticodon pairs with the codon
- Amino acids link together via peptide bonds → protein
- Stop codon (UAA, UAG, UGA) → ribosome releases protein
3. The Critical Rules
- AUG = start codon (Methionine)
- UAA, UAG, UGA = stop codons
- DNA → mRNA flips: A→U, T→A, C→G, G→C (T becomes U!)
- mRNA codon ↔ tRNA anticodon flips: A↔U, C↔G (no T in either)
4. Worked Example (Full Pipeline)
DNA template: T-A-C G-G-A C-C-T A-T-T
Transcribe to mRNA (T→A, A→U, G→C, C→G):
mRNA:
Translate (read codons): Met — Pro — Gly — STOP
tRNA anticodons (complement of codons):
UAC — GGA — CCU (no anticodon needed for stop)
mRNA:
A-U-G C-C-U G-G-A U-A-ATranslate (read codons): Met — Pro — Gly — STOP
tRNA anticodons (complement of codons):
UAC — GGA — CCU (no anticodon needed for stop)
5. Practice Problems
Practice 1
Where does transcription occur? Translation?
Transcription: nucleus. Translation: ribosome (in cytoplasm).
Practice 2
Transcribe DNA: T-A-C C-G-A T-T-A
mRNA: A-U-G G-C-U A-A-U. (Remember: T→A, A→U, C→G, G→C.)
Practice 3
Translate mRNA: AUG-UUU-GGA-UAA
Methionine — Phenylalanine — Glycine — STOP. (UUU = Phe; GGA = Gly; UAA = stop.)
Practice 4
A tRNA has anticodon C-G-A. What mRNA codon does it pair with, and what amino acid does it carry?
mRNA codon = G-C-U (complement: C↔G, A↔U). Amino acid = Alanine.
Practice 5
If DNA changes from A-T-G to A-T-C, what does the mRNA change to? Effect?
Original mRNA: U-A-C (codes for Tyrosine in middle of protein). New mRNA: U-A-G — a STOP codon! Protein terminates early. This is a nonsense mutation.
Practice 6
What three molecules are made up of RNA?
mRNA (carries code), tRNA (brings amino acids), rRNA (makes up ribosomes).
Practice 7
What's the difference between a codon and an anticodon?
Codon = 3 bases on mRNA, codes for an amino acid. Anticodon = 3 complementary bases on tRNA, pairs with the codon.
👨👩👧 Punnett Squares & Inheritance
The basics: A Punnett square predicts offspring genotype probabilities. Each parent contributes one allele per gene. List parent gametes on the sides, fill in the boxes, count the ratios.
1. Vocabulary You MUST Know
| Term | Means |
|---|---|
| Allele | Version of a gene (B or b) |
| Genotype | Genetic makeup (BB, Bb, bb) |
| Phenotype | Physical trait expressed (brown vs. blue eyes) |
| Homozygous | Two same alleles (BB or bb) |
| Heterozygous | Two different alleles (Bb) |
| Dominant | Expressed with one copy (capital) |
| Recessive | Only expressed when homozygous (lowercase) |
| Carrier | Heterozygous; has recessive allele but doesn't show trait |
2. Standard Monohybrid Crosses
Bb × Bb (the classic):
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
Genotype ratio 1:2:1 | Phenotype ratio 3:1 dominant:recessive → 75% dominant.
Bb × bb (test cross):
| B | b | |
|---|---|---|
| b | Bb | bb |
| b | Bb | bb |
50% dominant : 50% recessive.
3. Sex-Linked (X-linked) Crosses
Carrier mom (Xᴴ Xʰ) × unaffected dad (Xᴴ Y) — for hemophilia or colorblindness:
| Xᴴ | Xʰ | |
|---|---|---|
| Xᴴ | Xᴴ Xᴴ (normal F) | Xᴴ Xʰ (carrier F) |
| Y | Xᴴ Y (normal M) | Xʰ Y (AFFECTED M) |
Result: 50% of sons affected, 50% of daughters are carriers, no daughters affected.
4. Reading Pedigrees
- Square = male, Circle = female
- Shaded = affected, Half-shaded = carrier
- Autosomal recessive: unaffected parents → affected child; skips generations; both sexes
- Autosomal dominant: appears every generation; affected parent → usually affected child
- Sex-linked recessive: mostly males affected; passes from carrier mother to son
5. Practice Problems
Practice 1
Tt × Tt cross. What % of offspring will be heterozygous?
50% (Tt). The cross gives 1 TT : 2 Tt : 1 tt.
Practice 2
A heterozygous tongue roller (Tt) × non-roller (tt). Probability of a non-roller child?
50%. Cross Tt × tt → ½ Tt (roller), ½ tt (non-roller).
Practice 3
Two homozygous dominant parents (TT × TT). Can they have a recessive child?
No. All offspring will be TT. There's no recessive allele anywhere to pass on.
Practice 4
Why are sex-linked recessive disorders much more common in males than females?
Males are XY — they only have one X chromosome. A single recessive allele on that X is enough to show the trait. Females (XX) need two copies of the recessive allele to show the trait, which is much rarer.
Practice 5
A pedigree shows two unaffected parents who have an affected daughter. What inheritance pattern?
Autosomal recessive. Both parents must be carriers (heterozygous). It's autosomal (not sex-linked) because a daughter is affected, AND the unaffected parents giving an affected child is the classic recessive pattern.
Practice 6
Genotype Ttyy — what gametes can this organism produce?
Ty and ty. The T/t gene contributes T or t; the yy gene only contributes y (homozygous).
Why blood types feel hard: Almost every other trait uses TWO alleles. Blood types use THREE: Iᴬ, Iᴮ, i. Once you get past that, it's the same logic as everything else.
1. The Three Alleles
- Iᴬ — codes for A antigen
- Iᴮ — codes for B antigen
- i — codes for NO antigen (recessive)
2. How They Interact
- Iᴬ and Iᴮ are CODOMINANT with each other → both expressed = AB
- Both are DOMINANT over i
i is fully recessive — only shows up if you're ii.
3. Genotype → Phenotype (MEMORIZE)
| Genotype | Blood Type |
|---|---|
| IᴬIᴬ or Iᴬi | A |
| IᴮIᴮ or Iᴮi | B |
| IᴬIᴮ | AB (codominance) |
| ii | O |
💡 Type AB is ALWAYS heterozygous. Type O is ALWAYS homozygous recessive.
4. Key Punnett Squares
Iᴬi × Iᴮi (the famous "all 4 types" cross)
| Iᴮ | i | |
|---|---|---|
| Iᴬ | IᴬIᴮ AB | Iᴬi A |
| i | Iᴮi B | ii O |
25% each: A, B, AB, O.
AB × O
| i | i | |
|---|---|---|
| Iᴬ | Iᴬi A | Iᴬi A |
| Iᴮ | Iᴮi B | Iᴮi B |
50% A, 50% B. NO AB or O possible!
5. Critical Rules
- AB parent cannot pass an i → no O children possible.
- O parent cannot pass Iᴬ or Iᴮ → no AB children possible.
6. Practice Problems
Practice 1
Type A man + Type B woman have a Type O child. Parent genotypes?
Both parents must be heterozygous: Iᴬi × Iᴮi. The O child is ii, meaning each parent contributed an i.
Practice 2
Two type A parents have a type O child. Possible?
Yes — but both parents must be heterozygous (Iᴬi × Iᴬi). Cross gives ¼ ii (type O).
Practice 3
AB × AB: possible offspring blood types and ratio?
25% A, 50% AB, 25% B. NO type O (no i alleles to pass).
Practice 4
Type O mom + Type AB dad. Possible child blood types?
50% A, 50% B. NEVER AB or O.
Practice 5
Child has type O. Mother is type A. What blood types could father be?
A (het), B (het), or O. The father must carry at least one i allele. He CAN'T be AB or homozygous A/B.
Practice 6 (Like Q22)
Which cross gives all 4 blood types? (a) IᴬIᴬ × IᴬIᴮ (b) ii × Iᴬi (c) IᴬIᴮ × IᴬIᴮ (d) Iᴬi × Iᴮi
(d) Iᴬi × Iᴮi. Only cross with both i alleles AND both Iᴬ and Iᴮ.
Practice 7 (Like Q13)
AB man × B woman. Which genotype is impossible for their child? (a) Iᴬi (b) IᴮIᴮ (c) IᴬIᴮ (d) ii
(d) ii. AB father has no i allele to give.
Practice 8 — Tough
A child has AB blood. Could both parents be type O?
No. Both parents would be ii → can only pass i → child would be ii (type O). Impossible to get AB.
7. Summary
- 3 alleles: Iᴬ, Iᴮ, i
- Iᴬ, Iᴮ codominant. i recessive.
- AB = always heterozygous; O = always homozygous recessive
- AB parent → no O children
- O parent → no AB children
- All 4 types possible: Iᴬi × Iᴮi only
⚠️ Mutations
The big distinction: Gene mutations affect ONE gene's DNA sequence. Chromosomal mutations affect LARGE pieces or entire chromosomes.
1. Gene Mutations (Point Mutations)
| Type | What happens | Effect |
|---|---|---|
| Substitution | One base swapped | Changes 1 codon at most |
| Insertion | Base(s) added | Frameshift! Changes every codon after |
| Deletion | Base(s) removed | Frameshift! Changes every codon after |
Key rule: Substitutions never cause frameshifts. Only insertions and deletions do.
2. Three Outcomes of Substitution Mutations
| Type | What happens |
|---|---|
| Silent | Base change doesn't change the amino acid (genetic code is redundant) |
| Missense | Changes one amino acid in the protein |
| Nonsense | Creates a stop codon → protein cut short |
3. Frameshift Example
Original mRNA: AUG-CCU-GGA-UAA → Met-Pro-Gly-STOP
Insert one base (G) after AUG:
New mRNA: AUG-GCC-UGG-AUA-A...
New protein: Met-Ala-Trp-Ile-... (completely different!)
That's why frameshifts are usually devastating — everything downstream is scrambled.
New mRNA: AUG-GCC-UGG-AUA-A...
New protein: Met-Ala-Trp-Ile-... (completely different!)
That's why frameshifts are usually devastating — everything downstream is scrambled.
4. Chromosomal Mutations
| Type | What happens |
|---|---|
| Deletion | Section of chromosome lost |
| Duplication | Section repeated |
| Inversion | Segment flips direction |
| Translocation | Piece moves to a different chromosome |
| Nondisjunction | Chromosomes fail to separate in meiosis → wrong # of chromosomes |
5. Disorder Examples (KNOW THESE)
- Sickle cell anemia — substitution mutation (single base change in hemoglobin gene)
- Down syndrome — nondisjunction → trisomy 21 (extra copy of chromosome 21)
6. Practice Problems
Practice 1
What causes a frameshift mutation?
Insertion or deletion of bases (NOT substitution). They shift the reading frame, scrambling every codon after the mutation point.
Practice 2
Original DNA: A-T-G. New DNA: A-T-A. What kind of mutation, and what's the effect?
Substitution. Original mRNA = UAC (Tyrosine). New mRNA = UAU (also Tyrosine — same amino acid). So this is a silent mutation — no phenotypic change.
Practice 3
Which type of mutation typically has a bigger effect — gene or chromosomal? Why?
Chromosomal. They affect MANY genes at once (whole sections or chromosomes). Gene mutations only affect one gene.
Practice 4
What error in meiosis causes Down syndrome?
Nondisjunction — chromosome 21 fails to separate, resulting in a gamete with two copies. After fertilization the offspring has trisomy 21 (3 copies).
Practice 5
DNA changes from ATG to ATC. What does the mRNA become and what kind of mutation is this?
mRNA changes from UAC (Tyrosine) to UAG (STOP codon). This is a nonsense mutation — protein is cut short.
Practice 6
Are all mutations harmful?
No. Some are harmful, some neutral, some beneficial. Beneficial mutations are the source of new genetic variation that natural selection acts on.
🦎 Natural Selection
The mantra: Populations evolve, not individuals. And it's not just about surviving — it's about surviving AND reproducing.
1. Darwin's Four Conditions
- Variation — individuals differ in traits
- Overproduction — more offspring than environment can support
- Struggle for existence — competition for limited resources
- Differential survival/reproduction — better-suited individuals leave more offspring
2. The Three Types of Selection
| Type | Effect | Example |
|---|---|---|
| Directional | One extreme favored | Pesticide-resistant termites; dark mice on dark rocks |
| Stabilizing | Average favored | Human birth weight (too small or too big = problems) |
| Disruptive | Both extremes favored, average disadvantaged | Drought leaves only small & large seeds — medium beaks lose out |
3. Key Vocab
- Mutation = original source of new genetic variation
- Adaptation = trait that improves survival/reproduction
- Fitness = ability to survive AND reproduce
- Artificial selection = humans breed for traits (dog breeds, crops)
- Speciation = formation of new species
- Adaptive radiation = one ancestor → many species in different niches (Darwin's finches)
- Genetic drift = random changes, especially in small populations
4. Worked Example
Scenario: Termites are sprayed with insecticide. Initially 5% survive. After 6 generations, 80% survive. Why?
Not because individual termites "got used to it." Individuals don't adapt.
The few resistant termites (random pre-existing variation) survived AND reproduced. Their offspring inherited the resistance gene. Over generations, the frequency of resistance alleles increased. This is natural selection — directional selection.
The few resistant termites (random pre-existing variation) survived AND reproduced. Their offspring inherited the resistance gene. Over generations, the frequency of resistance alleles increased. This is natural selection — directional selection.
5. Practice Problems
Practice 1
What are the 4 conditions for natural selection?
Variation, overproduction, struggle for existence, differential survival/reproduction.
Practice 2
Do individual organisms evolve?
No — populations evolve over generations. Individuals are born with their genes and can't change them. What changes is the allele frequency in the population.
Practice 3
A drought kills medium-seed plants. Birds with medium beaks decline; small- and large-beaked birds thrive. What type of selection?
Disruptive selection. Both extremes favored, average against. Could lead to speciation over time.
Practice 4
What's the original source of new genetic variation?
Mutation. (Cloning produces no variation. Selective breeding and natural selection act on existing variation but don't create new variation.)
Practice 5
Fishermen only keep fish > 60cm. After 10 generations, what happens to the fish population?
Population shifts toward smaller sizes. Large fish are removed before reproducing; smaller fish escape nets and reproduce. Directional selection.
Practice 6
Why is "fitness" not the same as just being strong or fast?
Fitness = surviving AND reproducing. A super-strong individual that never reproduces has zero fitness — its genes don't get passed on. Reproduction is the whole point.
🦕 Evidence for Evolution
Six lines of evidence support evolution. The strongest? DNA / molecular evidence — more similar DNA = more closely related.
1. The Six Lines of Evidence
| Evidence | What it shows |
|---|---|
| Fossil record | Physical changes in species over geologic time |
| Homologous structures | Same underlying structure, different function → common ancestor |
| Analogous structures | Same function, different structure → NOT common ancestor |
| Vestigial structures | Reduced/non-functional remnants → ancestors had different lifestyles |
| Embryology | Similar early embryos → shared developmental genes |
| DNA / molecular | ★ Similar DNA = closer relationship (strongest evidence) |
2. The Big Confusion: Homologous vs. Analogous
Homologous: Look at the BONE STRUCTURE.
- Bat wing, human arm, dolphin flipper — all have the same bones (humerus, radius, ulna, carpals).
- Different FUNCTIONS (flying, grasping, swimming) but same underlying structure.
- → Common ancestor.
Analogous: Look at the FUNCTION.
- Bird wings vs. insect wings — both fly, but the structures are completely different.
- Same FUNCTION but different underlying structure.
- → NOT common ancestor. Evolved independently (convergent evolution).
3. Vestigial Structures
- Whale hip bones — whales' ancestors walked on land
- Human appendix — ancestors processed cellulose-heavy diets
- Snake leg remnants — ancestors had legs
Why do they lose function? No selection pressure to keep them once they're no longer useful.
4. Practice Problems
Practice 1
Best evidence for common ancestry between two species?
DNA / molecular evidence — more similar DNA = more recently shared ancestor. It's quantifiable and direct.
Practice 2
Bat wings and bird wings — homologous or analogous?
Tricky! Both have similar bones (homologous TO their common ancestor's forelimb), BUT in the context of "flying things," bat and bird wings are often called analogous because their flight structure (skin membrane vs. feathers) evolved independently. The most common textbook answer: bird and insect wings are ANALOGOUS (truly independent origin).
Practice 3
What does the human appendix tell us?
It's a vestigial structure — useful in our ancestors (likely for digesting plants) but not now. Shows that structures lose function over evolutionary time when no longer beneficial.
Practice 4
Why is DNA evidence considered the strongest?
It's quantifiable (you can count base differences), changes accumulate at predictable rates, and it's the direct genetic record. Other evidence is comparative; DNA evidence is direct.
Practice 5
Five finch species evolved from one ancestor on different islands. Each has a different beak shape. What is this an example of?
Adaptive radiation. One ancestor diversifies into many species, each adapted to a different niche/food source.
Practice 6
Two species have very similar embryos in early development. What does this suggest?
Common ancestry — they share developmental genes inherited from a recent common ancestor.
🌳 Ecology & Food Webs
The big rule: Only ~10% of energy transfers between trophic levels. Producers (bottom) have the MOST energy; top consumers have the LEAST. Always.
1. The Trophic Levels
| Level | What | Energy |
|---|---|---|
| 4. Tertiary consumers | Top predators | LEAST (top of pyramid) |
| 3. Secondary consumers | Eat herbivores | |
| 2. Primary consumers | Herbivores | |
| 1. Producers (autotrophs) | Plants, algae — make own food | MOST (bottom) |
| Decomposers | Bacteria, fungi — recycle nutrients | Side path |
2. The 10% Rule
Each level up, ~90% of energy is lost as heat (metabolism). Only ~10% transfers. That's why:
- Pyramids narrow at the top
- Top predators are rare
- Energy/food chain are short (4-5 levels max)
3. Vocabulary You Need
| Term | Definition |
|---|---|
| Autotroph | Makes own food (producer) |
| Heterotroph | Eats other organisms (consumer) |
| Herbivore | Eats plants |
| Carnivore | Eats meat |
| Omnivore | Eats both |
| Predator | Hunts/kills prey |
| Decomposer | Breaks down dead matter |
| Biotic | Living component (predators, prey, disease) |
| Abiotic | Non-living component (rainfall, temperature, soil) |
4. Worked Example: Food Web Cascade
Scenario: A pesticide kills off the mouse population. Mice eat grass; snakes eat mice and rabbits; rabbits eat grass.
First effect: Snakes lose a food source → start eating more rabbits OR have less food overall.
And: Less grass eaten by mice → more grass available → rabbits' food supply increases.
Removing one species ripples through the whole web — that's why biodiversity matters for stability.
And: Less grass eaten by mice → more grass available → rabbits' food supply increases.
Removing one species ripples through the whole web — that's why biodiversity matters for stability.
5. Practice Problems
Practice 1
What % of energy transfers between trophic levels?
About 10%. The other 90% is lost as heat (mostly from metabolism).
Practice 2
Which level contains the most stored energy?
Producers (bottom). Always.
Practice 3
A scorpion hunts, kills, and eats a spider. Which terms describe the scorpion?
Predator (hunts), carnivore (eats meat), consumer (heterotroph). NOT producer or autotroph.
Practice 4
Is rainfall a biotic or abiotic factor? Predator populations?
Rainfall = abiotic (non-living). Predator populations = biotic (living).
Practice 5
Wrens are added to an ecosystem where they eat grasshoppers. What population decreases first?
Grasshoppers — their direct food source. Indirect effects (less competition for plants, etc.) come later.
Practice 6
Why does losing biodiversity destabilize an ecosystem?
Fewer species means each is more critical. If one fails, no backup species can fill its role. The whole system is more likely to collapse from a single disturbance.
Practice 7
Why are decomposers important?
They break down dead organisms and recycle nutrients back into the soil so producers can use them. Without decomposers, nutrients would stay locked up and the system would stall.
📐 Hardy-Weinberg
Master the math, the conditions, and what the equations mean.
Why H-W feels confusing: Mixes algebra with biology. Once you have a 3-step formula, every problem is the same.
1. What It Measures
H-W tells us frequencies of alleles and genotypes:
- p = probability of grabbing the dominant allele (A)
- q = probability of grabbing the recessive allele (a)
2. The Two Equations
Equation 1: p + q = 1
Equation 2: p² + 2pq + q² = 1
| Variable | What it represents | Genotype |
|---|---|---|
| p | Dominant allele freq | — |
| q | Recessive allele freq | — |
| p² | Homozygous dominant freq | AA |
| 2pq | Heterozygous freq | Aa |
| q² | Homozygous recessive freq | aa |
The big trick: The question usually gives you the % SHOWING the recessive trait = q². Always start by taking the square root of that.
3. The 3-Step Solver
1Find q: Take √ of % showing recessive trait.
2Find p: p = 1 − q.
3Plug into p², 2pq, q² as needed.
For number of individuals: multiply the frequency by total population size.
4. Zooming In: How to Find 2pq
2pq is the part that confuses most students. It's the frequency of heterozygous (Aa) individuals — the "carriers." Here's exactly how to find it.
Where the "2" Comes From
When two parents have a baby, the heterozygous combination (Aa) can happen TWO ways:
- A from mom × a from dad = p × q
- a from mom × A from dad = q × p
Adding both: p×q + q×p = 2pq. That's why it's doubled. (Same logic as Bb appearing twice in a Bb × Bb Punnett square.)
How to Calculate It — Three Sub-Steps
1Find q first (square-root the % showing the recessive trait).
2Find p (p = 1 − q).
3Multiply: 2pq = 2 × p × q.
Worked Example
Q: 16% of the population shows the recessive trait. What % is heterozygous?
| Step | Math | Result |
|---|---|---|
| 1. Find q | q² = 0.16 → q = √0.16 | q = 0.4 |
| 2. Find p | p = 1 − 0.4 | p = 0.6 |
| 3. Find 2pq | 2 × 0.6 × 0.4 | 2pq = 0.48 = 48% |
Sanity check: p² + 2pq + q² = 0.36 + 0.48 + 0.16 = 1.00 ✓
So 48% of the population is heterozygous (Aa) — they carry the recessive allele but don't show the trait.
Common mistake: Forgetting to multiply by 2. A lot of students compute p × q and stop there. The answer for heterozygotes is double that.
Finding the NUMBER (not just %) of Heterozygotes
Q: A population has 1,000 people; 9% show the recessive phenotype. How many are heterozygous?
q² = 0.09 → q = 0.3
p = 1 − 0.3 = 0.7
2pq = 2 × 0.7 × 0.3 = 0.42
Number of heterozygotes = 0.42 × 1,000 = 420 people
p = 1 − 0.3 = 0.7
2pq = 2 × 0.7 × 0.3 = 0.42
Number of heterozygotes = 0.42 × 1,000 = 420 people
Fun Fact: 2pq Has a Maximum
2pq can never exceed 0.5. The maximum happens when p = q = 0.5 (a perfectly balanced gene pool). As one allele becomes more common than the other, 2pq shrinks. That's why "half the population is heterozygous" is the upper limit — heterozygotes can never make up more than 50% of a population in H-W equilibrium.
5. Worked Examples
Example 1: q² = 0.25
q = √0.25 = 0.5 | p = 0.5 | p² = 0.25 | 2pq = 0.5 | q² = 0.25
Example 2: 9% show recessive phenotype
q² = 0.09 → q = 0.3 | p = 0.7 | p² = 0.49 | 2pq = 0.42 | q² = 0.09
Example 3: p = 0.8
q = 0.2 | p² = 0.64 | 2pq = 0.32 | q² = 0.04
Example 4: Population 500, q² = 0.16. How many homo recessive?
0.16 × 500 = 80 individuals
6. The 5 Conditions for Equilibrium
All 5 must hold or the population is evolving:
- Large population (otherwise genetic drift)
- No migration
- No mutation
- Random mating
- No natural selection
No real population perfectly meets all 5. H-W is a baseline to compare against.
7. Practice Problems
Practice 1
If q² = 0.25, calculate q, p, p², and 2pq.
q = 0.5, p = 0.5, p² = 0.25, 2pq = 0.5
Practice 2
9% recessive phenotype. All allele and genotype frequencies?
q = 0.3, p = 0.7, p² = 0.49 (49% AA), 2pq = 0.42 (42% Aa), q² = 0.09 (9% aa)
Practice 3
If p = 0.8, calculate q and all genotype frequencies.
q = 0.2, p² = 0.64 (64% AA), 2pq = 0.32 (32% Aa), q² = 0.04 (4% aa)
Practice 4
Population of 500, q² = 0.16. How many homozygous recessive?
0.16 × 500 = 80 individuals
Practice 5
1,000 birds, 420 heterozygous, p = 0.7. Calculate genotype numbers.
q = 0.3, p² = 0.49 → 490 AA, 2pq = 0.42 → 420 Aa ✓, q² = 0.09 → 90 aa
Practice 6
200 flowers, 32 are white (recessive). % heterozygous?
q² = 32/200 = 0.16, q = 0.4, p = 0.6, 2pq = 0.48 = 48%
Practice 7
Dominant allele freq is 0.9. % homozygous dominant?
p² = 0.81 = 81%
Practice 8
Large population, no migration/mutation/env change. What happens to allele frequencies?
They remain constant (Hardy-Weinberg equilibrium). This was Q30 on your practice test.
Practice 9
Allele frequency goes from 80% A / 20% a to 60% A / 40% a over 50 years. What does this mean?
The population is evolving. (Q34 on your practice test.)
Practice 10
2,000 people, 1,800 can roll their tongues (dominant). How many heterozygous?
Non-rollers (aa) = 200/2000 = 0.1 = q². q ≈ 0.316, p ≈ 0.684, 2pq ≈ 0.432. ≈ 865 individuals.
8. Summary
Equations: p + q = 1 | p² + 2pq + q² = 1
3-step solver:
- q = √(% showing recessive phenotype)
- p = 1 − q
- Plug into p², 2pq, q²
5 conditions: large pop, no migration, no mutation, random mating, no selection.
🐢 Mark & Recapture
The technique: Capture some animals, mark them, release. Later capture again. The ratio of marked-to-total in the second capture lets you estimate the whole population.
1. The Formula
N = (M × C) / R
- N = population estimate
- M = number marked in FIRST capture
- C = total CAUGHT in SECOND capture
- R = RECAPTURED marked individuals (the ones with marks in the second catch)
2. The Intuition
The ratio of marked-to-total in your second sample should equal the ratio of marked-to-total in the whole population:
R/C = M/N
Rearrange: N = (M × C) / R
3. Why Marked Organisms Must Mix Back In
Between captures, the marked individuals need to randomly redistribute among the whole population. If they don't, your second sample won't be representative — you'll get a bad estimate.
4. Limitations (Often Asked!)
- Marks can fall off (under-counts recaptures → overestimates population)
- Assumes no births, deaths, immigration, or emigration between captures
- Marked organisms might behave differently (trap-shy or trap-happy)
- Marking might affect survival
5. Practice Problems
Practice 1
40 turtles marked. Later catch 60, of which 12 are marked. Population?
N = (40 × 60) / 12 = 2400 / 12 = 200 turtles
Practice 2
25 rabbits marked. Later catch 50, of which 5 are marked. Population?
N = (25 × 50) / 5 = 250 rabbits
Practice 3
80 fish marked. Later catch 100, of which 20 are marked. Population?
N = (80 × 100) / 20 = 400 fish
Practice 4
100 birds marked. Later catch 75, of which 15 are marked. Population?
N = (100 × 75) / 15 = 500 birds
Practice 5
Why must marked organisms be allowed to mix back into the population before the second capture?
If they don't mix, the second sample won't be representative. You might catch them in a cluster or miss them entirely, leading to a wildly inaccurate population estimate.
Practice 6
Name one assumption of mark-recapture and one way it could be violated.
Assumes no births/deaths/migration between captures. Could be violated by: marks falling off, animals dying, immigration, marked animals avoiding traps after the first experience.
Practice 7 — Backwards
A scientist estimates a population of 600 deer. She marked 30. If she catches 40 in the second capture, how many should be marked?
600 = (30 × 40) / R → R = 1200 / 600 = 2 marked deer
📈 Population Growth
Two main patterns: Exponential growth (J-curve, unlimited) and logistic growth (S-curve, levels off at carrying capacity). Real populations follow logistic.
1. Exponential Growth (J-Curve)
Population doubles in a fixed time. The graph is a J-shape. Only happens when resources are unlimited — usually in:
- Newly introduced species in a new habitat
- Bacteria in fresh nutrient medium
- Small populations with no competition or predation
Can't continue forever — eventually something limits it.
2. Logistic Growth (S-Curve)
Population starts exponential but slows as it approaches the environment's limits, then levels off. The graph is an S-shape.
The level-off point = carrying capacity (K) — the max population the environment can sustain.
3. What Causes Growth to Slow
| Density-dependent (more impact when crowded) | Density-independent (no relation to crowding) |
|---|---|
| Competition for food, water, space | Weather (drought, frost) |
| Predation | Natural disasters (fire, flood) |
| Disease (spreads faster when crowded) | Climate |
| Waste accumulation |
4. Stable Populations
When a population is stable (at the top of the S-curve), births + immigration ≈ deaths + emigration. The population isn't growing or shrinking — it's at carrying capacity.
5. Human Impact
- Habitat destruction reduces carrying capacity
- Pollution adds to density-independent stress
- Climate change shifts which species can thrive where
- Overharvesting (fishing, hunting) acts like extra predation
6. Practice Problems
Practice 1
J-curve vs. S-curve?
J-curve = exponential growth (unlimited). S-curve = logistic growth (levels off at carrying capacity).
Practice 2
Define carrying capacity.
The maximum population size an environment can sustainably support, given the available resources.
Practice 3
Is drought a density-dependent or density-independent factor?
Density-independent. A drought affects populations the same regardless of how crowded they are.
Practice 4
A caribou population reaches a stable size on an S-curve. What does this mean about births and deaths?
Births equal deaths (approximately). When the population is at carrying capacity, no net growth occurs.
Practice 5
Name three things that can cause exponential growth to slow.
Limited food/water/space, predation, disease, competition, waste accumulation — any of these can slow growth as population approaches carrying capacity.
Practice 6
Why is disease often density-dependent?
Diseases spread through contact. In a crowded population, contact is more frequent, so diseases spread faster. A sparse population won't experience the same disease pressure.