Heredity & Genetics
How are traits passed from parent to offspring — and why do siblings look different? Unit 5 covers meiosis as the engine of genetic diversity, Mendel's laws as the framework for predicting inheritance, and the many ways that real genetics deviates from simple Mendelian patterns. Chi-square analysis and pedigree interpretation are core skills for this unit.
Meiosis
Meiosis is the specialized cell division that produces haploid gametes (sperm and eggs) from diploid parent cells. Where mitosis produces two genetically identical daughter cells (2n → 2n), meiosis produces four genetically unique daughter cells (2n → 4 × n). Meiosis consists of two sequential divisions: Meiosis I and Meiosis II, with no DNA replication between them.
Before Meiosis Begins — DNA Replication in S Phase
Like mitosis, meiosis is preceded by DNA replication during the S phase of interphase. After replication, each chromosome consists of two identical sister chromatids joined at the centromere. A diploid cell (2n = 46 in humans) enters meiosis with 46 chromosomes, each comprising 2 sister chromatids = 92 chromatids total.
Meiosis I — The Reductional Division
Meiosis I separates homologous chromosome pairs. This is the truly unique division — it halves chromosome number from 2n to n. Sister chromatids remain attached throughout Meiosis I.
| Stage | Key Events | What Makes It Unique vs. Mitosis |
|---|---|---|
| Prophase I | Chromosomes condense; homologous chromosomes pair up side-by-side (synapsis) forming a bivalent (tetrad); crossing over (recombination) occurs at points called chiasmata; spindle forms; nuclear envelope breaks down | Synapsis + crossing over — unique to meiosis. In mitosis, homologs NEVER pair up. |
| Metaphase I | Bivalents (homolog pairs) align at the metaphase plate; spindle fibers from opposite poles attach to kinetochores of each homolog (one per homolog, not one per chromatid); random orientation of each bivalent (independent assortment) | Homolog pairs align as units — not individual chromosomes as in mitosis. |
| Anaphase I | Homologous chromosomes separate and move to opposite poles; sister chromatids remain joined — cohesins on chromosome arms cleaved but centromeric cohesins protected | Homologs separate (not sister chromatids). Each pole gets 1 of each homolog pair. |
| Telophase I + Cytokinesis | Nuclear envelopes may re-form; cytokinesis produces 2 haploid cells. Each cell is haploid in chromosome number (n), but each chromosome still consists of 2 sister chromatids — so DNA amount is still doubled relative to a true haploid gamete. | Result: 2 haploid cells (n chromosomes each), not 2 diploid cells as in mitosis. |
Meiosis II — The Equational Division
Meiosis II resembles mitosis in haploid cells — it separates sister chromatids. There is no DNA replication between Meiosis I and II.
| Stage | Key Events |
|---|---|
| Prophase II | Spindle re-forms in both haploid cells; chromosomes condense (if they decondensed in telophase I) |
| Metaphase II | Individual chromosomes (sister chromatid pairs) align at the metaphase plate; each kinetochore attached to spindle fibers from opposite poles |
| Anaphase II | Centromeric cohesins cleaved → sister chromatids separate and move to opposite poles; now called individual chromosomes |
| Telophase II + Cytokinesis | Nuclear envelopes re-form; 4 haploid daughter cells produced; each cell has n chromosomes, each unduplicated (1 chromatid) |
Mitosis vs. Meiosis — Full Comparison
| Feature | Mitosis | Meiosis |
|---|---|---|
| Purpose | Growth, tissue repair, asexual reproduction | Sexual reproduction — gamete production |
| Number of divisions | 1 | 2 (Meiosis I + II) |
| Daughter cells | 2 | 4 |
| Ploidy of daughter cells | Diploid (2n → 2n) | Haploid (2n → n) |
| Genetic identity | Genetically identical to parent | Genetically unique |
| Synapsis | Never | Yes — in Prophase I |
| Crossing over | No | Yes — in Prophase I |
| Homolog separation | No — homologs never interact | Yes — in Anaphase I |
| Sister chromatid separation | Anaphase | Anaphase II |
| Occurs in | All somatic (body) cells | Gonads (testes, ovaries) |
A diploid organism (2n = 8) undergoes meiosis. At the END of Meiosis I, how many chromosomes does each daughter cell contain, and are they duplicated or unduplicated?
- (A) 8 unduplicated chromosomes
- (B) 4 unduplicated chromosomes
- (C) 4 chromosomes, each still consisting of 2 sister chromatids (duplicated)
- (D) 8 chromosomes, each consisting of 2 sister chromatids
Meiosis and Genetic Diversity
Sexual reproduction generates enormous genetic diversity — far more than asexual reproduction. Meiosis contributes to this diversity through three mechanisms: crossing over, independent assortment, and the subsequent random fertilization of gametes.
Mechanism 1 — Crossing Over (Recombination)
During Prophase I, homologous chromosomes synapse (pair up) and non-sister chromatids physically exchange segments at points called chiasmata. This shuffles alleles between homologs, creating recombinant chromosomes with new combinations of alleles that differ from either parent. The further apart two genes are on a chromosome, the more likely a chiasma will form between them — the basis of genetic mapping (map units / centimorgans). Crossing over is the only source of diversity for genes on the same chromosome (linked genes).
Mechanism 2 — Independent Assortment
During Metaphase I, each homologous pair (bivalent) lines up at the metaphase plate independently of all other pairs. The orientation of each bivalent (which homolog faces which pole) is random. For an organism with n pairs of homologs, there are 2ⁿ possible gamete combinations from independent assortment alone. For humans (n = 23): 2²³ ≈ 8.4 million possible chromosome combinations per gamete — before considering crossing over!
Mechanism 3 — Random Fertilization
When two gametes fuse at fertilization, the resulting zygote can be any combination of the ≈8.4 million possible gametes from each parent. The number of possible zygote genotypes: (2²³)² ≈ 70 trillion unique combinations — essentially guaranteeing every individual is genetically unique (identical twins excepted).
Nondisjunction — When Meiosis Goes Wrong
Nondisjunction is the failure of chromosomes to separate correctly during meiosis. This produces gametes with an abnormal number of chromosomes. When fertilized by a normal gamete, the resulting zygote will be aneuploid — having an abnormal chromosome number.
| When It Occurs | What Fails to Separate | Gametes Produced | Resulting Zygote |
|---|---|---|---|
| Meiosis I nondisjunction | Homologous chromosomes fail to separate in Anaphase I | Two gametes with n+1 (extra chromosome), two gametes with n−1 (missing chromosome) | Trisomy (2n+1) or Monosomy (2n−1) |
| Meiosis II nondisjunction | Sister chromatids fail to separate in Anaphase II | One gamete with n+1, one gamete with n−1, two normal gametes | Trisomy or Monosomy (but fewer abnormal gametes than Meiosis I nondisjunction) |
Trisomy 21 (Down syndrome): Three copies of chromosome 21. Most common viable trisomy in humans. Risk increases with maternal age (more meiotic errors with older oocytes).
Sex chromosome aneuploidies: XXY = Klinefelter syndrome (male phenotype, infertile); XO = Turner syndrome (female phenotype, short stature, infertile); XXX = Triple X (usually fertile female with mild effects); XYY = Jacob syndrome.
Most autosomal trisomies are lethal — embryos do not survive to term. The sex chromosomes are more tolerant of extra copies because one X is inactivated (Barr bodies) in cells with more than one X.
A cell undergoes nondisjunction during Meiosis I, such that both members of one homologous pair move to the same pole. Which of the following correctly describes the gametes produced from this cell?
- (A) All four gametes will be normal (haploid) because nondisjunction in Meiosis I is corrected by Meiosis II.
- (B) Two gametes will have an extra chromosome (n+1), and two gametes will be missing a chromosome (n−1).
- (C) One gamete will have an extra chromosome, one will be missing a chromosome, and two will be normal.
- (D) All four gametes will have an extra chromosome.
Mendelian Genetics
Gregor Mendel's experiments with pea plants (1860s) established two fundamental laws of inheritance that hold for genes on different chromosomes. These laws describe how alleles (alternate versions of a gene) are transmitted from parents to offspring.
Key Vocabulary
A gene is a specific DNA sequence that encodes a trait (e.g., flower color gene). An allele is a specific version of that gene (e.g., purple allele P or white allele p). Each diploid organism has two alleles per gene — one per homolog.
Genotype: the actual allele combination (e.g., Pp, PP, pp). Phenotype: the observable trait expression (e.g., purple or white). Same phenotype can arise from different genotypes (PP and Pp both → purple if P is dominant).
A dominant allele masks the recessive allele in a heterozygote. A recessive allele is only expressed in homozygous form (pp). Dominance is a relationship between alleles — not an intrinsic property of the allele itself.
Homozygous dominant (PP): two identical dominant alleles. Homozygous recessive (pp): two identical recessive alleles. Heterozygous (Pp): two different alleles. A carrier is a heterozygous individual for a recessive disease allele.
Mendel's Law of Segregation
During gamete formation, the two alleles of each gene segregate (separate) from each other so that each gamete carries only one allele for each gene. This is the direct consequence of homologs separating in Meiosis I. Each gamete gets one allele per gene — chosen randomly.
Mendel's Law of Independent Assortment
Alleles for genes on different (non-homologous) chromosomes assort into gametes independently of one another. This is a direct consequence of random bivalent orientation in Metaphase I. Genes on the same chromosome are linked and do NOT independently assort (unless far enough apart that crossing over makes them appear unlinked). The AP exam will specify whether genes are on same or different chromosomes.
Monohybrid Cross (1 gene, 2 alleles)
Cross of two heterozygotes (Pp × Pp):
Genotypic ratio: 1 PP : 2 Pp : 1 pp (1:2:1). Phenotypic ratio: 3 dominant : 1 recessive (3:1). The classic F2 ratio that Mendel observed.
Dihybrid Cross (2 independent genes)
Cross of two dihybrid parents (AaBb × AaBb) — genes on different chromosomes. Expected phenotypic ratio: 9:3:3:1 (9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb). This 9:3:3:1 ratio is the hallmark of independent assortment. If the observed ratio deviates significantly from 9:3:3:1, the genes may be on the same chromosome (linked).
Test Cross
A test cross crosses an organism of unknown genotype (but dominant phenotype) with a homozygous recessive organism. If all offspring show the dominant phenotype → unknown parent was homozygous dominant. If half show dominant, half recessive → unknown parent was heterozygous. This is the standard way to determine genotype in classical genetics.
Probability Rules in Genetics
| Rule | Formula | When to Use | Example |
|---|---|---|---|
| Product Rule (AND) | P(A and B) = P(A) × P(B) | Two independent events both occurring | P(two dominant offspring) from Pp × Pp: ¾ × ¾ = 9/16 |
| Sum Rule (OR) | P(A or B) = P(A) + P(B) [mutually exclusive] | Either of two mutually exclusive outcomes | P(PP or pp) from Pp × Pp: ¼ + ¼ = ½ |
Chi-Square Test (χ²) — Testing Genetic Hypotheses
The chi-square test determines whether observed results differ significantly from expected Mendelian ratios. A large χ² value means the difference is probably not due to chance — reject the null hypothesis (the data do not fit Mendelian expectations).
Sum across all phenotypic classes. Degrees of freedom = number of classes − 1
Null hypothesis (H₀): The observed ratios are consistent with Mendelian expectations (any deviation is due to chance alone).
p-value > 0.05: Fail to reject H₀ — data fit the expected Mendelian ratio. The deviation is likely due to random sampling error.
p-value ≤ 0.05: Reject H₀ — data do NOT fit the expected ratio. There is a statistically significant deviation. May indicate linked genes, selection, or a different inheritance pattern.
AP exam critical point: Students commonly write "accept H₀" — the correct phrasing is "fail to reject H₀". You never "prove" a null hypothesis — you only fail to find evidence against it.
Pedigree Analysis
Pedigrees trace inheritance patterns in families. Key rules for determining inheritance:
| Pattern | Key Observation in Pedigree |
|---|---|
| Autosomal recessive | Trait skips generations; affected individuals often have two unaffected carrier parents; both sexes equally affected; consanguineous couples (related) increase risk |
| Autosomal dominant | Trait appears in every generation (vertical pattern); every affected individual has at least one affected parent; both sexes equally affected |
| X-linked recessive | More common in males (XY — only one X copy needed); carrier mothers → affected sons; affected fathers cannot pass to sons (sons get Y from father) |
| X-linked dominant | Affected fathers pass to ALL daughters (never to sons); both sexes affected; more females affected than males |
In a monohybrid cross between two heterozygous tall plants (Tt × Tt), a student observes 78 tall and 22 short offspring out of 100 total. The student performs a chi-square test. The expected ratio is 3:1 (75 tall : 25 short). The calculated χ² = (78−75)²/75 + (22−25)²/25 = 0.12 + 0.36 = 0.48. With 1 degree of freedom, the critical value at p = 0.05 is 3.84. What is the correct conclusion?
- (A) Reject H₀; the data do not fit the 3:1 ratio because there are more tall plants than expected.
- (B) Fail to reject H₀; the data are consistent with the 3:1 Mendelian ratio because χ² = 0.48 < 3.84.
- (C) Accept H₀; the 3:1 ratio is proven correct because χ² is small.
- (D) Reject H₀; a value of 0.48 is too close to zero to support Mendelian inheritance.
Non-Mendelian Genetics
Many traits deviate from simple Mendelian patterns. These non-Mendelian inheritance patterns arise when dominance relationships are different, when genes are located on sex chromosomes, when genes are linked, or when genes are located outside the nucleus.
Codominance
Both alleles are fully and simultaneously expressed in the heterozygote — neither masks the other. The heterozygote has a distinct phenotype showing both alleles' products simultaneously, not a blend.
Example: ABO blood types — I^A and I^B alleles are codominant. A person with genotype I^A I^B has blood type AB: both A and B antigens are present on red blood cells. Phenotypic ratio in codominance crosses ≠ 3:1 but rather 1:2:1 (with the heterozygote showing a distinct phenotype).
Incomplete Dominance
Neither allele fully masks the other — the heterozygote shows an intermediate (blended) phenotype. This is NOT a mixing of alleles; both alleles are still present but neither fully expressed.
Example: Snapdragons — R (red) × W (white) → RW heterozygote is pink. Cross of two pink plants (RW × RW): 1 Red (RR) : 2 Pink (RW) : 1 White (WW) — a 1:2:1 phenotypic ratio (same as genotypic ratio), unlike the 3:1 of complete dominance.
Both produce a 1:2:1 phenotypic ratio in heterozygote × heterozygote crosses. The difference: codominance = both phenotypes expressed simultaneously (A and B antigens both on cell — not "mixed"); incomplete dominance = a new intermediate phenotype (pink, not both red AND white simultaneously).
Sex-Linked Inheritance (X-Linked)
Genes on the X chromosome are X-linked. Males (XY) are hemizygous for X-linked genes — they have only one copy, so even recessive alleles are expressed. Females (XX) can be carriers (heterozygous) for recessive X-linked traits without showing the phenotype.
| Cross | Offspring | Key Insight |
|---|---|---|
| Carrier female (X^A X^a) × normal male (X^A Y) | X^A X^A (normal ♀), X^A X^a (carrier ♀), X^A Y (normal ♂), X^a Y (affected ♂) | 50% of sons affected; daughters are carriers but unaffected (if A is dominant) |
| Affected male (X^a Y) × normal female (X^A X^A) | X^A X^a (all carrier daughters), X^A Y (all normal sons) | Affected father CANNOT pass X-linked allele to sons; all daughters are carriers |
| Carrier female (X^A X^a) × affected male (X^a Y) | X^A X^a (carrier ♀), X^a X^a (affected ♀), X^A Y (normal ♂), X^a Y (affected ♂) | 50% daughters affected — X-linked recessive CAN appear in females |
Classic X-linked recessive traits: red-green colorblindness, hemophilia A & B, Duchenne muscular dystrophy. Always more common in males because males only need one recessive allele.
Genetic Linkage and Gene Mapping
Genes on the same chromosome are linked — they tend to be inherited together (violating independent assortment). However, crossing over during Prophase I can separate linked alleles. The recombination frequency (percentage of recombinant offspring) is proportional to the distance between genes: 1% recombination frequency = 1 map unit (centimorgan, cM).
- Genes <50 cM apart → show linkage (fewer recombinants than expected for independent assortment)
- Genes ≥50 cM apart (or on different chromosomes) → appear to assort independently (recombination frequency approaches 50%). Note: genes on the same chromosome that are very far apart are not truly independent — they simply recombine so frequently that they behave statistically like unlinked genes.
- Map distance = (recombinant offspring / total offspring) × 100%
Pleiotropy
A single gene influences multiple, seemingly unrelated phenotypic traits. Because the traits are caused by the same gene, they are always inherited together. Example: sickle cell anemia — the HbS allele causes not just abnormal hemoglobin shape but also: anemia, pain crises, organ damage, increased malaria resistance (in heterozygotes), and susceptibility to bacterial infections. Another example: Marfan syndrome — single connective tissue gene mutation → tall stature, long fingers, aortic aneurysm, lens dislocation.
Polygenic Inheritance
A single trait is controlled by two or more genes (additive effect). Each gene contributes a small, often equal amount to the phenotype. Result: a continuous distribution (bell curve) of phenotypes rather than discrete categories. Examples: human height, skin color, birth weight, eye color. The more genes involved, the smoother the continuous distribution.
Non-Nuclear (Cytoplasmic / Maternal) Inheritance
Genes in chloroplasts and mitochondria are inherited differently from nuclear genes because these organelles are not distributed equally during meiosis. They are transmitted primarily through the egg cytoplasm (maternal inheritance):
- Animals: Mitochondria are transmitted almost exclusively through the egg (sperm contributes almost none). Mitochondrial DNA traits show maternal inheritance — affected mothers typically pass traits to all children; affected fathers pass to none. (The degree of expression can vary due to heteroplasmy — different proportions of mutant vs. normal mitochondria in different cells.)
- Plants: Both mitochondria and chloroplasts are transmitted through the ovule (egg), not the pollen. Therefore, both are also maternally inherited in most plants.
- The cytoplasmic organelles divide by binary fission and are distributed randomly to daughter cells — not by precise chromosomal mechanisms.
Example: Mitochondrial diseases (e.g., Leber's hereditary optic neuropathy, MERRF syndrome) show maternal inheritance — ALL children of an affected mother may be affected; children of an affected father are unaffected.
X-linked recessive: "If an affected father cannot pass the trait to his sons, but a carrier mother can pass it to 50% of her sons" — always trace the X chromosome specifically. Daughters get one X from dad and one from mom; sons get Y from dad and X from mom.
Deviation from 9:3:3:1: If a dihybrid cross gives a ratio other than 9:3:3:1 (e.g., 9:7, 12:3:1, 15:1), this indicates gene interaction (epistasis) or linkage. The exam may present unusual ratios and ask you to identify the modification.
Mitochondrial inheritance: If a pedigree shows a trait passing from ALL affected mothers to ALL their children (sons AND daughters), but from affected fathers to NONE of their children, this is maternal inheritance via mitochondrial DNA — not autosomal, not X-linked.
In snapdragons, flower color shows incomplete dominance: R¹R¹ = red, R¹R² = pink, R²R² = white. A pink-flowered plant is crossed with a white-flowered plant. (a) List the expected genotypic ratio among offspring. (b) List the expected phenotypic ratio. (c) If 200 offspring are produced, predict the expected number of each phenotype.
Cross: R¹R² × R²R² (pink × white).
Gametes from pink: R¹ or R² (each 50%). Gametes from white: R² only.
Offspring: R¹R² (pink) and R²R² (white) — 1:1 genotypic ratio.
(b) Phenotypic ratio:
Since R¹R² = pink and R²R² = white: 1 Pink : 1 White.
(c) Expected numbers from 200 offspring:
Pink (R¹R²): 200 × ½ = 100 plants
White (R²R²): 200 × ½ = 100 plants
Red (R¹R¹): 0 plants (impossible — white parent contributes only R² alleles)
Environmental Effects on Phenotype
Genotype alone does not fully determine phenotype. The environment in which an organism develops and lives can profoundly alter gene expression, leading to different phenotypes from the same genotype. This ability is called phenotypic plasticity.
Key Principle: Genotype × Environment = Phenotype
The same genotype can produce different phenotypes depending on the environmental conditions during development and life. Conversely, different genotypes can sometimes produce the same phenotype in the same environment (phenocopy). The norm of reaction describes the range of phenotypes a single genotype can express across different environments.
Examples of Environmental Influence on Phenotype
Hydrangeas produce blue flowers in acidic soil (aluminum ions absorbed) and pink flowers in alkaline soil — same genotype, different pigment expression. Flower color is directly affected by soil chemistry altering enzyme activity in the anthocyanin pathway.
Arctic hares and stoats (Himalayan rabbits) produce white fur in cold temperatures and darker fur in warm temperatures. The enzyme tyrosinase (required for dark pigment) is heat-sensitive — it is inactivated at body temperature in summer and active in cold extremities/winter. Same genotype; temperature determines phenotype.
In many reptiles (turtles, crocodilians, some lizards), sex is determined by the incubation temperature of eggs, not by sex chromosomes. There are no sex chromosomes — temperature-sensitive enzymes regulate sex hormone production during a critical developmental window. This is an entirely environmental determination of phenotype.
UV radiation exposure activates melanin synthesis in animals — the same individual produces more melanin (tanning) in response to UV. The MC1R gene affects baseline melanin, but UV exposure modulates actual melanin output in the same individual (same genotype → different phenotype with different UV exposure).
Both traits are polygenic (influenced by many genes) AND heavily influenced by environment: nutrition during development, disease history, stress hormones. Identical twins raised apart can have significantly different heights and weights despite identical genotypes.
The presence of mating-type pheromones (chemical environment) alters yeast gene expression — the same yeast cell expresses different pheromone pathways depending on whether it detects a potential mating partner. Signal = environmental cue; gene expression = response.
Phenotypic plasticity ≠ genetic change. When an organism changes phenotype in response to environment, its DNA sequence (genotype) has NOT changed. Only the pattern of gene expression changes. This is a critical distinction — often confused with mutation or evolution.
Norm of reaction: A graph showing how a single genotype responds to a range of environmental conditions. If different genotypes have different norms of reaction that cross each other, this means the ranking of genotypes changes depending on the environment — an important concept for evolution (genotype × environment interactions).
This concept bridges genetics (Unit 5) and evolution (Unit 7). Phenotypic variation produced by both genetic and environmental factors provides the raw material on which natural selection acts. Selection acts on phenotype, not directly on genotype.
Himalayan rabbits carry a gene for dark fur, but the enzyme encoded by this gene is only active at temperatures below 33°C. In warm body regions the fur is white; at cool extremities (ears, nose, paws) the fur is dark. Two Himalayan rabbits (same genotype) are raised in different environments: one at 20°C and one at 37°C. Which of the following best describes the expected outcome?
- (A) Both rabbits will have identical phenotypes because their genotypes are the same.
- (B) The rabbit at 37°C will develop more mutations to adapt to the warm temperature.
- (C) The rabbit at 20°C will likely develop dark fur over most of its body; the rabbit at 37°C will have mostly white fur because the enzyme is inactive at high temperatures (and will regain activity if the temperature drops).
- (D) The rabbit at 37°C will have darker fur because higher temperatures increase enzyme activity generally.
Mixed Practice Questions
A woman is a carrier for an X-linked recessive disorder (genotype X^A X^a). Her partner is unaffected (genotype X^A Y). What is the probability that their first child is an affected son?
- (A) 1/4, because half of sons are affected and half of children are sons
- (B) 1/2, because half of sons will be affected
- (C) 1/4 — probability of being a son (1/2) × probability that the son receives X^a from mother (1/2) = 1/4
- (D) 0, because the father is unaffected and cannot pass the disorder to sons
A student observes that a disease in a family always passes from affected mothers to all of their children, but never from affected fathers to any of their children. (a) Identify the inheritance pattern. (b) Explain the cellular and molecular basis of this inheritance pattern. (c) Predict the expected offspring of a cross between an affected woman and an unaffected man.
(b) Cellular and molecular basis: The gene causing this disease is located in mitochondrial DNA, not the nuclear genome. Mitochondria are organelles that contain their own circular DNA (remnant of the ancestral α-proteobacterium, per endosymbiotic theory). During reproduction, virtually all mitochondria in the zygote come from the egg's cytoplasm — sperm contribute almost no mitochondria to the zygote. Therefore, mitochondrial alleles are inherited exclusively through the mother. An affected mother's eggs contain her mitochondrial DNA, passing the mutation to ALL her children regardless of sex. An affected father's sperm contribute only nuclear DNA to the offspring; his mitochondria are excluded from the zygote.
(c) Expected offspring: All children of an affected mother (who carries the mitochondrial mutation) are expected to be affected, because ALL children receive their mitochondria from the mother's egg. There is no paternal contribution to the mitochondrial genome. Both sons and daughters would be affected.
High-Frequency Errors to Avoid
- 🔬Homologous chromosomes separate in Meiosis I; sister chromatids separate in Meiosis IIStudents frequently mix these up. In Anaphase I, homologs (one from mom, one from dad) separate. In Anaphase II (and mitotic Anaphase), sister chromatids separate. This distinction is critical for every genetics question involving chromosome number and ploidy.
- 🎲Saying "accept H₀" after chi-square analysisThe correct conclusion when χ² is less than the critical value is "fail to reject H₀." You never "accept" or "prove" the null hypothesis — you simply lack sufficient evidence to reject it. This phrasing error costs points on FRQs regularly.
- 🌸Confusing incomplete dominance with codominanceIncomplete dominance → NEW intermediate phenotype (pink snapdragon from red × white). Codominance → BOTH original phenotypes expressed simultaneously (AB blood type — both A and B antigens present). Neither is a blend of alleles; alleles remain intact in both cases.
- 🧬Saying X-linked recessive traits can NEVER appear in femalesX-linked recessive traits CAN appear in females — but only if she is homozygous recessive (X^a X^a). This is less common because she must receive X^a from both parents. It is possible if her father is affected (X^a Y) and her mother is at least a carrier (X^A X^a).
- 🌡Saying phenotypic change due to environment = genetic changePhenotypic plasticity (e.g., tanning in sun, fur color change with temperature) does NOT alter the DNA sequence. The genotype is unchanged; only gene expression patterns change. This is not a mutation or evolution — it is a physiological or developmental response.
- 🥚Forgetting that mitochondrial inheritance is STRICTLY maternal in animalsSperm do contribute a small number of mitochondria at fertilization in some organisms, but these are almost always destroyed. Functionally, ALL mitochondrial DNA in animal offspring comes from the egg. In plants, BOTH mitochondria AND chloroplasts are transmitted through the ovule — both are maternally inherited.
- 📊Treating polygenic traits as Mendelian single-gene traitsPolygenic traits (height, skin color) show a continuous, bell-curve distribution — not the discrete 3:1 or 9:3:3:1 ratios of Mendelian genetics. If a trait shows a normal distribution in the population, it is almost certainly polygenic (and likely also influenced by environment).
Unit 5 — Key Takeaways
2n → 4 haploid cells through 2 divisions. Meiosis I: homologs separate (reductional). Meiosis II: sister chromatids separate (equational). Prophase I = synapsis + crossing over. Mitosis vs. meiosis table critical.
3 sources: crossing over (Prophase I), independent assortment (Metaphase I — 2ⁿ gamete types), random fertilization. Nondisjunction: homologs fail to separate (Meiosis I) → n+1 or n−1 gametes → trisomy/monosomy.
Segregation: alleles separate into gametes. Independent assortment: genes on different chromosomes assort independently. Monohybrid F2: 3:1 phenotypic. Dihybrid F2: 9:3:3:1. Test cross with homozygous recessive. Chi-square: fail to reject vs. reject H₀ at p = 0.05.
Codominance (both alleles fully expressed — AB blood type). Incomplete dominance (intermediate phenotype — pink snapdragon). X-linkage (hemizygous males; carrier females). Linkage + gene mapping (% recombination = map units). Pleiotropy (one gene → many traits). Maternal inheritance (mitochondria/chloroplasts via egg).
Phenotypic plasticity: same genotype → different phenotypes in different environments. Examples: Himalayan rabbit fur, hydrangea color, temperature-dependent sex determination. Genotype + Environment = Phenotype. Environmental change ≠ genetic change.
Unit 5 = 8–11% of the AP Biology Exam. Highest-yield topics: Meiosis I vs. II events and outcomes (especially Anaphase I vs. Anaphase II), nondisjunction mechanics and resulting aneuploidies, monohybrid and dihybrid Punnett square calculations, chi-square hypothesis testing (calculation + correct interpretation), X-linked inheritance pedigrees, codominance vs. incomplete dominance distinction, and mitochondrial/maternal inheritance patterns. Always use probability rules systematically for multi-gene problems rather than drawing large Punnett squares — it is faster and less error-prone.