AP Biology · Unit 5 · 8–11% of Exam ⚡ SPRINT MODE
Heredity & Genetics
Calculation-heavy and concept-dense. The biggest traps: what separates in Meiosis I vs II, chi-square language ("fail to reject" ≠ "accept"), and codominance vs incomplete dominance. Master these and the genetics questions become predictable.
Meiosis I vs IICrossing OverNondisjunction
3:1 / 9:3:3:1Chi-SquareCodominance
X-LinkageMaternal Inheritance
⚡ Quick Glance — All Topics at a Glance
| Topic | Priority | Exam Format | Key Trap / Must-Know |
|---|---|---|---|
| 5.1–2 Meiosis & Diversity | ★★★ | MCQFRQ | Meiosis I = homologs separate; Meiosis II = sister chromatids separate. After Meiosis I: n chromosomes but STILL duplicated |
| 5.3 Mendelian Genetics | ★★★ | MCQCalcFRQ | "Fail to reject H₀" — NEVER say "accept." Chi-square p = 0.05 critical value at df=3 is 7.815 |
| 5.4 Non-Mendelian Genetics | ★★★ | MCQFRQ | Incomplete dominance = NEW intermediate phenotype; Codominance = BOTH original phenotypes expressed. Mitochondrial = maternal only |
| 5.5 Environmental Effects | ★★ | MCQ | Phenotypic change ≠ genetic change; same genotype can produce different phenotypes in different environments |
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Topics 5.1–5.2
Meiosis & Genetic Diversity
Meiosis I vs II — The Critical Distinction
| Feature | Meiosis I (Reductional) | Meiosis II (Equational) |
|---|---|---|
| What separates? | Homologous chromosomes separate to opposite poles | Sister chromatids separate (like mitosis) |
| Key prophase event | Synapsis + crossing over (chiasmata) — UNIQUE to meiosis | No synapsis; no crossing over |
| Metaphase alignment | Bivalents (homolog pairs) line up — independent assortment here | Individual chromosomes line up (like mitosis) |
| Result (per cell) | 2 cells, each with n chromosomes — but each chromosome is still duplicated (2 sister chromatids) | 4 cells total, each with n unduplicated chromosomes |
| Ploidy change | 2n → n (chromosome number HALVED here) | n → n (chromosome number unchanged; chromatids separate) |
| DNA content | 4C → 2C (each cell has 2× DNA of final gamete) | 2C → 1C (each cell reaches final gamete DNA level) |
Mitosis vs. Meiosis — Exam Comparison Table
| Feature | Mitosis | Meiosis |
|---|---|---|
| Purpose | Growth, repair, asexual reproduction | Gamete production (sexual reproduction) |
| Divisions | 1 | 2 (Meiosis I + II) |
| Daughter cells | 2 diploid (2n) | 4 haploid (n) |
| Genetic outcome | Identical to parent | Genetically unique |
| Synapsis / crossing over | ❌ Never | ✅ Prophase I only |
| Homolog separation | ❌ Never | ✅ Anaphase I |
| Sister chromatid separation | ✅ Anaphase | ✅ Anaphase II |
| Where it occurs | All somatic cells | Gonads (testes/ovaries) |
Three Sources of Genetic Diversity
Source 1 — Prophase I
🔀 Crossing Over (Recombination)
- Non-sister chromatids of homologs exchange segments at chiasmata
- Creates recombinant chromosomes with new allele combinations
- Only way to shuffle alleles on the same chromosome (linked genes)
- More crossovers between distant genes → higher recombination frequency
- Recombination frequency (%) = map distance in centimorgans (cM)
- 50% recombination = genes act as if on separate chromosomes (unlinked)
Source 2 — Metaphase I
🎲 Independent Assortment
- Each bivalent orients randomly at metaphase plate
- Orientation of one homolog pair has NO effect on any other pair
- With n pairs: 2ⁿ possible gamete chromosome combinations
- Humans: 2²³ ≈ 8.4 million unique gametes from this alone
- Basis of Mendel's Law of Independent Assortment
- Only applies to genes on different chromosomes (or very far apart)
Source 3 — Fertilization
🥚 Random Fertilization
- Any sperm can fertilize any egg
- Combinations: (2²³)² ≈ 70 trillion unique zygote genotypes
- Guarantees that every individual (except identical twins) is genetically unique
Nondisjunction → Aneuploidy
| When | What Fails | Gametes Produced | If Fertilized by Normal Gamete |
|---|---|---|---|
| Meiosis I | Homologs don't separate in Anaphase I | 2 gametes with n+1 (extra chromosome), 2 with n−1 — ALL 4 are abnormal | Trisomy (2n+1) or Monosomy (2n−1) |
| Meiosis II | Sister chromatids don't separate in Anaphase II | 1 with n+1, 1 with n−1, 2 normal gametes | Trisomy or Monosomy (only 2 of 4 gametes abnormal) |
Key Aneuploidies — AP Examples
⚠️ Nondisjunction Outcomes
- Trisomy 21 (Down syndrome): 3 copies of chr. 21; most common viable trisomy; risk ↑ with maternal age
- Klinefelter (XXY): male phenotype, infertile, taller
- Turner (XO): female phenotype, infertile, short stature
- Triple X (XXX): usually fertile female, often no symptoms
- Most autosomal trisomies = lethal (embryo doesn't survive)
- Extra X chromosomes tolerated → inactivated as Barr bodies
🎯 Exam Sniper
- MCQ: "At the end of Meiosis I, how many chromosomes does each cell contain in a 2n=8 organism, and are they duplicated?" → 4 chromosomes (n=4), each still consisting of 2 sister chromatids (still duplicated). Sister chromatids don't separate until Meiosis II
- MCQ: "Which event in meiosis is responsible for shuffling alleles between homologous chromosomes?" → Crossing over in Prophase I (NOT independent assortment — that shuffles whole chromosomes, not alleles within chromosomes)
- FRQ: "Explain how meiosis contributes to genetic variation." → Must mention ALL THREE: (1) crossing over in Prophase I creates recombinant chromosomes, (2) independent assortment in Metaphase I produces 2ⁿ chromosome combinations, (3) random fertilization of unique gametes
- MCQ (nondisjunction): "Nondisjunction occurs in Meiosis I. How many of the resulting gametes have an abnormal chromosome number?" → All 4 gametes are abnormal (2 with n+1, 2 with n−1) — Meiosis I nondisjunction affects both subsequent cells
💣 Trap Alert
- ❌ After Meiosis I, cells have n chromosomes but are STILL duplicated — each chromosome has 2 sister chromatids. They are haploid in chromosome count but have 2× the DNA of a final gamete
- ❌ Crossing over occurs in Prophase I only — NOT in Meiosis II, NOT in mitosis
- ❌ Independent assortment shuffles whole chromosomes; crossing over shuffles alleles within chromosomes — these are different mechanisms creating different kinds of variation
- ❌ Meiosis I nondisjunction → ALL 4 gametes abnormal; Meiosis II nondisjunction → only 2 of 4 gametes abnormal
Topic 5.3
Mendelian Genetics
Mendel's Two Laws
📐 Core Principles
- Law of Segregation: each organism has 2 alleles for each gene; they separate (segregate) into different gametes → each gamete gets one allele
- Law of Independent Assortment: genes on different chromosomes (or very far apart) assort into gametes independently of each other
- Dominant allele: expressed when at least one copy is present (A_)
- Recessive allele: only expressed when homozygous (aa)
Probability Rules
🎲 Faster Than Punnett Squares
- AND rule (multiply): probability of A AND B = P(A) × P(B) — for independent events
- OR rule (add): probability of A OR B = P(A) + P(B) — for mutually exclusive events
- Dihybrid cross: treat each gene independently → multiply probabilities
- Example: AaBb × AaBb → P(A_B_) = P(A_) × P(B_) = ¾ × ¾ = 9/16
- Much faster than a 16-square Punnett for multi-gene problems
Key Ratios to Memorize
| Cross | Phenotypic Ratio | Genotypic Ratio | Shortcut |
|---|---|---|---|
| Monohybrid F₂ (Aa × Aa) | 3:1 (dominant:recessive) | 1AA : 2Aa : 1aa | ¾ dominant, ¼ recessive phenotype |
| Dihybrid F₂ (AaBb × AaBb) | 9:3:3:1 | — | 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb |
| Test cross (A_ × aa) | 1:1 (if Aa) or all dominant (if AA) | — | Reveals genotype of unknown parent |
| Monohybrid F₁ (AA × aa) | All dominant (Aa) | All Aa | F₁ are all heterozygous |
Punnett Square Quick Reference
Monohybrid: Aa × Aa
| A | a | |
|---|---|---|
| A | AA | Aa |
| a | Aa | aa |
Phenotype: 3 dom : 1 rec
Test Cross: Aa × aa
| A | a | |
|---|---|---|
| a | Aa | aa |
| a | Aa | aa |
Phenotype: 1 dom : 1 rec
🎯 Exam Sniper
- MCQ: "In a dihybrid cross (AaBb × AaBb), what fraction of offspring will show both dominant phenotypes?" → ¾ × ¾ = 9/16. Use probability multiplication, NOT a 16-box Punnett
- MCQ: "A plant with purple flowers is crossed with a white-flowered plant. All F₁ are purple. F₁ × F₁ gives 3 purple : 1 white. What are the genotypes?" → P: AA × aa → F₁: Aa → F₂: 3 A_ : 1 aa. Purple = dominant
- MCQ (test cross): "A tall pea plant crossed with a short (homozygous recessive) plant produces 50% tall and 50% short offspring. What is the genotype of the tall parent?" → Must be Tt (heterozygous) — homozygous TT would give all tall
In pea plants, yellow (Y) is dominant over green (y) and round (R) is dominant over wrinkled (r). Two plants that are heterozygous for both traits (YyRr) are crossed. What is the probability that an offspring will be homozygous recessive for both traits?
- (A) 1/16 — because there is only one yyrr box in a 16-square Punnett
- (B) 1/16 — calculated as P(yy) × P(rr) = ¼ × ¼ = 1/16
- (C) 1/4 — because only one gene needs to be homozygous recessive
- (D) 3/16 — because it appears in the 9:3:3:1 ratio as the last category
Answer: (B) — For each gene independently: P(yy from Yy × Yy) = ¼; P(rr from Rr × Rr) = ¼. Since the two genes are on different chromosomes (independent assortment), multiply: ¼ × ¼ = 1/16. Both (A) and (B) give 1/16 but only (B) shows the correct probability reasoning. The AP exam awards method points — always show the multiplication approach. The 9:3:3:1 ratio confirms: the last class (double recessive) = 1/16.
Chi-Square Analysis
Chi-Square Hypothesis Test
χ² = Σ [(O − E)² / E]
O = Observed frequency E = Expected frequency
Degrees of freedom (df) = number of categories − 1
Critical value at p = 0.05: df=1 → 3.841 df=3 → 7.815 df=7 → 14.067
O = Observed frequency E = Expected frequency
Degrees of freedom (df) = number of categories − 1
Critical value at p = 0.05: df=1 → 3.841 df=3 → 7.815 df=7 → 14.067
The 5-Step Chi-Square Process
- State H₀ (null hypothesis): "There is no significant difference between observed and expected ratios" (e.g., "the data fit a 9:3:3:1 ratio")
- Calculate expected values: E = total × expected fraction (e.g., for 9:3:3:1 with 160 total: E(A_B_) = 160 × 9/16 = 90)
- Calculate χ²: For each category, compute (O−E)²/E, then sum all categories
- Determine degrees of freedom: df = number of phenotypic classes − 1 (for 4 classes: df = 3)
- Compare to critical value and conclude: If χ² < critical value → fail to reject H₀ (data consistent with expected ratio). If χ² > critical value → reject H₀ (significant deviation)
⚠️ Critical Language — Costs Points Every Year
- ✅ CORRECT: "We fail to reject the null hypothesis" — the data do not provide sufficient evidence to reject the expected ratio
- ✅ CORRECT: "We reject the null hypothesis" — the deviation is statistically significant
- ❌ WRONG: "We accept the null hypothesis" — you NEVER accept H₀ in statistics. This phrasing loses points on every AP FRQ
- ❌ WRONG: "We prove the hypothesis" — statistical tests cannot prove anything; they assess evidence
- The p = 0.05 threshold means: if H₀ were true, there is only a 5% chance of seeing a deviation this large by random chance alone
🎯 Exam Sniper — Worked Example
Scenario: Expected 9:3:3:1 ratio from dihybrid cross. Observed (160 total): 90 round yellow, 30 round green, 32 wrinkled yellow, 8 wrinkled green. Expected: 90, 30, 30, 10.
χ² = (90−90)²/90 + (30−30)²/30 + (32−30)²/30 + (8−10)²/10 = 0 + 0 + 0.133 + 0.4 = 0.533
df = 4−1 = 3; critical value = 7.815
0.533 < 7.815 → Fail to reject H₀ — data are consistent with a 9:3:3:1 Mendelian dihybrid ratio. The deviation from expected is likely due to chance.
Topic 5.4
Non-Mendelian Genetics
Dominance Patterns — The Big Three
✅ Complete Dominance
- One allele fully masks the other
- Heterozygote = same as dominant homozygote phenotype
- Ratios: 3:1 (F₂); 1:1 (test cross)
- Example: TT tall = Tt tall ≠ tt short
🟣 Incomplete Dominance
- Heterozygote shows NEW intermediate phenotype
- Neither allele fully dominant
- Ratios: 1:2:1 (F₂ — 3 distinct phenotypes)
- Example: Red (RR) × White (WW) → Pink (RW)
- Alleles are NOT blended — still pass on separately
🔵 Codominance
- Both alleles fully expressed simultaneously
- Heterozygote shows BOTH original phenotypes
- Ratios: 1:2:1 (F₂ — like incomplete)
- Example: ABO blood type — type AB has both A and B antigens on RBCs
- Example: roan cattle — both red and white hairs present
X-Linked Inheritance — Pedigree Rules
| Genotype | Sex | Phenotype | Exam Notes |
|---|---|---|---|
| X^A X^A | Female | Normal (homozygous dominant) | Cannot be carrier |
| X^A X^a | Female | Normal phenotype, carrier | Passes X^a to 50% of offspring |
| X^a X^a | Female | Affected | Must have received X^a from BOTH parents — father must be affected |
| X^A Y | Male | Normal | — |
| X^a Y | Male | Affected (hemizygous) | Only one X → one allele determines phenotype; no carrier state possible in males |
X-Linked Recessive — Pedigree Clues
📊 How to Identify X-Linkage
- Trait appears more often in males (they only need one copy)
- Affected sons must get X^a from their mother (who is a carrier)
- Affected daughters are rare — need X^a from BOTH parents
- An affected father CANNOT pass trait directly to sons (gives Y to sons)
- Father passes X^a to ALL daughters (who become at least carriers)
- Common examples: colorblindness, hemophilia A, Duchenne muscular dystrophy
More Non-Mendelian Patterns
🧬 Other Inheritance Modes
- Pleiotropy: one gene affects MULTIPLE phenotypic traits (e.g., sickle cell — affects RBC shape, anemia, organ damage, malaria resistance)
- Polygenic inheritance: MULTIPLE genes control ONE trait → continuous bell-curve distribution (height, skin color). Not 3:1 ratios
- Gene linkage: genes on same chromosome tend to be inherited together; recombination frequency <50% → linked
- Epistasis: one gene masks expression of another gene (modified 9:3:3:1 ratios)
Organelle Inheritance
🥚 Maternal Inheritance
- Mitochondrial DNA: inherited exclusively from mother (through egg cytoplasm) in animals
- Sperm mitochondria destroyed after fertilization
- All children of an affected mother will inherit mtDNA mutations
- Affected father → NONE of his children inherit it through paternal line
- Chloroplast DNA: also maternally inherited in most plants (through ovule)
- Pedigree clue: trait passes through ALL children of affected mothers; NONE through affected fathers
🎯 Exam Sniper
- MCQ (top trap): "Snapdragon plants with red flowers are crossed with white-flowered plants. F₁ are all pink. F₁ × F₁ gives: 1 red : 2 pink : 1 white. This pattern is best explained by..." → Incomplete dominance (NOT codominance — there's a NEW pink phenotype, not both red and white simultaneously)
- MCQ: "A colorblind man and a woman with normal vision (whose father was colorblind) have children. What is the probability their son is colorblind?" → Mother is X^A X^a (carrier); P(colorblind son) = ½ × ½ = ¼ or use probability: sons get Y from dad and X from mom; 50% chance X^a from mom
- Pedigree FRQ: "A trait appears in 3 of 4 sons of a carrier female, but in no daughters. What inheritance pattern is this?" → X-linked recessive — males are hemizygous, so any son who receives X^a is affected
- MCQ: "A mitochondrial disease affects a woman. Which of her children will be affected?" → ALL children (both sons and daughters) — mitochondria are inherited through the egg, regardless of sex
💣 Trap Alert
- ❌ Incomplete dominance ≠ codominance: incomplete = new intermediate phenotype (pink); codominance = both original phenotypes present simultaneously (AB blood type). Both give 1:2:1 ratios but express differently
- ❌ X-linked recessive CAN appear in females — but only if homozygous (X^a X^a), which requires an affected father AND carrier (or affected) mother. It's less common but absolutely possible
- ❌ Mitochondrial inheritance = STRICTLY maternal in animals. If a father is affected with a mitochondrial disease, NONE of his children inherit it through him
- ❌ Polygenic traits show CONTINUOUS distributions — if you see a normal/bell-curve distribution, it is polygenic and NOT explained by simple Mendelian ratios
In a pedigree, a woman with normal vision has a colorblind father. She marries a man with normal vision. What is the probability that their first child will be a colorblind daughter?
- (A) 1/2 — because the mother is a carrier
- (B) 1/4 — because Mendelian genetics predicts 1/4 recessive offspring
- (C) 1/8 — probability of a daughter (1/2) × probability she gets X^a from mother (1/2) × probability she is X^a X^a (impossible without father's X^a, so actually: 1/2 chance daughter × 1/2 chance carrier daughter, but daughter can't be affected unless she gets X^a from dad too — which she can't since dad has normal vision)
- (D) 0 — daughters cannot be colorblind
Answer: (D) — probability = 0 — The mother is a carrier (X^A X^a) because her father was colorblind (X^a Y) — she must have received his X^a. The father has normal vision (X^A Y). For a daughter to be colorblind, she needs X^a X^a: one X^a from mom (possible, prob = ½) AND one X^a from dad (impossible — dad only has X^A to give daughters). Therefore, NO daughters can be colorblind. Sons have a 50% chance of being colorblind (get X^a from mom + Y from dad). The key: colorblind daughters require BOTH parents to contribute X^a — that means the father must be colorblind.
Topic 5.5
Environmental Effects on Phenotype
Core Formula
🌱 Genotype + Environment = Phenotype
- Same genotype CAN produce different phenotypes in different environments
- Phenotypic plasticity: ability of one genotype to express different phenotypes in response to environmental conditions
- Environmental change does NOT alter DNA sequence — no mutation occurs
- Classic examples: Himalayan rabbit (dark fur in cold regions); hydrangea color (soil pH); tanning in sun (UV → melanin production)
- Temperature-dependent sex determination in reptiles (not genetic — environmental)
- Norm of reaction: range of phenotypes a genotype can produce across environments
Exam Examples
🌡 Phenotypic Plasticity Cases
- Himalayan rabbit: same enzyme (tyrosinase) temperature-sensitive — cold extremities → dark fur; warm body → light fur. Same genotype, different expression
- Hydrangea: same genotype → blue flowers in acidic soil (Al³⁺ available), pink in alkaline — pigment production depends on soil chemistry
- Suntanning: UV light stimulates melanin production in skin — genotype determines potential, environment determines expression
- Nutrition and height: genes set potential height range; nutrition determines where in that range a person ends up
🎯 Exam Sniper
- MCQ: "Himalayan rabbits grown in warm conditions have white coats, but the same rabbits exposed to cold develop dark extremities. What explains this?" → Phenotypic plasticity — the genotype determines the enzyme (temperature-sensitive tyrosinase); the environment (temperature) determines which regions of the body produce dark pigment. The genotype is unchanged
- MCQ: "A gardener plants genetically identical cuttings of the same hydrangea in two soils with different pH. The plants produce flowers of different colors. What does this demonstrate?" → Same genotype → different phenotypes due to environmental conditions; supports the concept that phenotype = genotype + environment
Practice
Sprint Practice — Mixed Questions
A student crosses two pea plants heterozygous for seed color (Yy × Yy). Out of 80 offspring: 64 yellow and 16 green seeds. The student performs a chi-square test against an expected 3:1 ratio. Calculate χ² and interpret the result using p = 0.05 (critical value at df=1 is 3.841).
- (A) χ² = 1.07; accept the null hypothesis; data support a 3:1 ratio
- (B) χ² = 1.07; fail to reject the null hypothesis; the deviation is likely due to chance
- (C) χ² = 3.84; reject the null hypothesis; the data do not support a 3:1 ratio
- (D) χ² = 4.27; reject the null hypothesis; yellow is not truly dominant
Answer: (B) — Expected: 80 × ¾ = 60 yellow; 80 × ¼ = 20 green. χ² = (64−60)²/60 + (16−20)²/20 = 16/60 + 16/20 = 0.267 + 0.8 = 1.067. df = 2−1 = 1; critical value = 3.841. Since 1.067 < 3.841, we fail to reject H₀. The deviation from expected is likely due to random sampling error, not a violation of Mendelian inheritance. Never say "accept H₀" — that loses points every time.
Two genes, A and B, are located on the same chromosome and are 20 centimorgans (cM) apart. An individual heterozygous for both genes (AB/ab coupling) is test-crossed with an aabb individual. What percentage of offspring would be recombinant (Ab/ab and aB/ab)?
- (A) 50% — because genes on the same chromosome assort independently
- (B) 20% — because 1 cM = 1% recombination frequency, and the genes are 20 cM apart
- (C) 80% — because recombinant gametes are the majority when genes are linked
- (D) 0% — because linked genes never recombine
Answer: (B) — 1 cM = 1% recombination frequency by definition. Two genes 20 cM apart have a 20% chance of crossing over between them during Prophase I of meiosis. In a test cross, gametes from the heterozygous parent are directly revealed in offspring phenotypes. Parental gametes (AB and ab) make up 80% of offspring; recombinant gametes (Ab and aB) make up 20% total (10% each). If genes were >50 cM apart, they'd appear to assort independently (50% recombinant) — but they're still physically linked.
⚠ Trap Alert
Unit 5 High-Frequency Exam Traps
- 🔬Meiosis I: homologs separate. Meiosis II: sister chromatids separate.The most consistently tested meiosis concept. In Anaphase I, one homolog from each pair goes to each pole — sister chromatids stay connected. In Anaphase II (and mitotic anaphase), centromeric cohesins break → sister chromatids finally separate. After Meiosis I: cell is haploid in chromosome count but still has duplicated chromosomes (2 chromatids each).
- 📊"Fail to reject H₀" — NEVER say "accept H₀"This costs points on nearly every genetics FRQ. When χ² < critical value, the correct statement is "we fail to reject the null hypothesis" or "the data are consistent with the expected ratio." Statistics never prove or accept hypotheses — they only provide or lack evidence to reject them.
- 🌸Incomplete dominance ≠ codominance — know the phenotypic differenceIncomplete dominance: a NEW, intermediate phenotype appears (red × white → pink). Codominance: BOTH original phenotypes are expressed simultaneously (blood type AB shows both A and B antigens). Both produce 1:2:1 ratios in F₂ crosses, but the nature of the heterozygote phenotype is completely different.
- 🧬X-linked recessive traits CAN appear in females (homozygous X^a X^a)This requires the father to be affected (X^a Y) and the mother to be at least a carrier (X^A X^a). While less common than in males (who are hemizygous), it is biologically and statistically possible. Never say "females cannot have X-linked recessive conditions."
- 🥚Mitochondrial inheritance is STRICTLY maternal — affected father passes nothing to childrenAll mitochondrial DNA in animal offspring comes from the egg. An affected father with a mitochondrial disorder cannot pass it to any of his children. An affected mother passes it to ALL children. This is the pedigree pattern to recognize: all children of affected mothers affected; no children of affected fathers affected.
- 🎯Independent assortment only applies to genes on DIFFERENT chromosomes (or very far apart)Mendel's Law of Independent Assortment states that genes on different chromosomes assort independently. Genes on the SAME chromosome are linked and tend to be inherited together (with recombination creating some exceptions). The 9:3:3:1 ratio assumes the two genes are unlinked.
- 🌡Environmental change ≠ genetic changePhenotypic plasticity (Himalayan rabbit, tanning, hydrangea color) does NOT change the DNA sequence. The genotype is identical before and after environmental exposure. Only mutations change the genotype. Environmental effects on phenotype are reversible and heritable only in the sense that the alleles for phenotypic plasticity are inherited — not the specific phenotype itself.
✓ Last-Min Checklist
Pre-Exam 10-Minute Checklist
Click each item to confirm before exam day.
Meiosis & Diversity (5.1–5.2)
- Meiosis I = reductional (homologs separate); Meiosis II = equational (sister chromatids separate)
- After Meiosis I: n chromosomes per cell, but each chromosome is STILL duplicated (2 sister chromatids)
- Crossing over: Prophase I, at chiasmata, between non-sister chromatids of homologs
- Independent assortment: Metaphase I, 2ⁿ gamete chromosome combinations
- Nondisjunction in Meiosis I → all 4 gametes abnormal; Meiosis II → 2 of 4 abnormal
- Trisomy 21 = Down syndrome; XXY = Klinefelter; XO = Turner
Mendelian Genetics & Chi-Square (5.3)
- Monohybrid F₂ = 3:1 phenotypic; Dihybrid F₂ = 9:3:3:1
- Use probability rules (multiply for AND, add for OR) instead of large Punnett squares
- Chi-square formula: χ² = Σ[(O−E)²/E]; df = categories − 1
- Critical value at df=3, p=0.05 = 7.815
- Correct conclusion: "fail to reject H₀" (NEVER "accept H₀")
Non-Mendelian Genetics (5.4)
- Incomplete dominance = NEW intermediate phenotype (pink snapdragon) → 1:2:1 ratio
- Codominance = BOTH original phenotypes expressed (AB blood type, roan cattle) → 1:2:1 ratio
- X-linked recessive: males hemizygous (X^a Y = affected); females need X^a X^a to be affected
- Mitochondrial = maternal inheritance only; affected mother → all children affected; affected father → no children affected
- Polygenic traits = continuous bell-curve distribution (NOT 3:1 ratios)
- Gene linkage: recombination frequency (%) = map distance in cM; <50% recombination = linked genes
⚡ Final Sprint Strategy for Unit 5
- Top 5 must-master: (1) Meiosis I vs II what separates, (2) Chi-square calculation + "fail to reject" language, (3) Codominance vs incomplete dominance distinction, (4) X-linkage pedigree reading rules, (5) Maternal/mitochondrial inheritance pattern
- Calculation speed tips: For multi-gene problems, ALWAYS use probability multiplication (not Punnett squares with 16+ boxes). P(A_B_) = ¾ × ¾ = 9/16 is instant; drawing a 4×4 grid wastes 3 minutes
- Pedigree reading strategy: First determine: autosomal vs. sex-linked (check if males and females affected equally); dominant vs. recessive (can two unaffected parents have an affected child? If yes = recessive)
- Connections: Meiosis → Unit 4 (cell cycle comparison); Inheritance → Unit 6 (gene expression, mutations); Chi-square → used in Unit 7 (Hardy-Weinberg, population genetics)