Open with a real, emotionally sticky case: "In 2019 a hospital in New Jersey transfused a type-O patient with type-A blood by mistake. The patient's immune system attacked the donor cells within minutes." Ask students, cold: why does an O patient react violently to type-A blood, but an AB patient can accept almost anything? Take 2-3 hands. Do NOT resolve it — write "Iᴬ, Iᴮ, i" on the board and tell them by the end of the block they will predict, for any two parents' blood types, the probability their child could safely donate to any sibling. Then pivot to a second phenomenon: show them that identical twins with the same DNA can end up with different diseases — one gets type 2 diabetes, the other doesn't. Ask: if their genes are identical, what else must matter? Bank the two questions on the board — they anchor the whole lesson.

1. 8m

   Probability rules that actually run genetics

   Content

   Genetics is applied probability. Two rules do almost all the work. The rule of product: for independent events, P(A and B) = P(A) × P(B). Meiosis makes each gamete an independent draw, so the chance of getting a specific allele from mom AND a specific allele from dad is the product of the two separate chances. Example: Tt × Tt. Each parent gives T with probability 1/2 and t with probability 1/2, so P(TT offspring) = 1/2 × 1/2 = 1/4. The rule of sum: for mutually exclusive outcomes, P(A or B) = P(A) + P(B). In the same cross, a heterozygote can arise two ways — T from mom + t from dad, OR t from mom + T from dad — so P(Tt) = 1/4 + 1/4 = 1/2. Add up the four cells and you get the classic 1:2:1 genotype ratio and, if T is dominant, the 3:1 phenotype ratio. Remind them that probability describes long-run expectation; a family of four kids from Tt × Tt is NOT guaranteed to be 3 dominant + 1 recessive — each birth is an independent 3/4 vs 1/4 draw.

   Delivery

   Anchor every future calculation to these two rules. Ask a fast check: "If two carriers Aa × Aa have three children, what is the probability ALL three are unaffected?" Walk them to (3/4)³ ≈ 0.42. Pre-empt the misconception that "we already had one affected kid so the next one is safe" — each birth is independent. Emphasize the phrase "expected ratio" so they stop expecting exactly 3:1 in tiny families.

2. 8m

   Monohybrid crosses and the genotype/phenotype distinction

   Content

   A Punnett square is just a bookkeeping device for the rule of product. Set the two possible gametes of one parent across the top and the two of the other down the side; each cell is the product of that row and column. Work Tt × tt live: gametes T, t across; t, t down; cells Tt, tt, Tt, tt — genotype ratio 1 Tt : 1 tt (which is 1:1), phenotype 1 tall : 1 short. Genotype is the allele pair (Tt). Phenotype is what you see (tall). Two different genotypes — TT and Tt — give the same phenotype when T is fully dominant. This is exactly why you can't read a genotype off a phenotype: a tall pea plant might be TT or Tt, and only a test cross with tt reveals which. Also drive home that "dominant" describes how the allele behaves in a heterozygote — it says nothing about how common the allele is in the population. Polydactyly (extra fingers) is dominant but rare; the allele for cystic fibrosis is recessive but far more common in some populations than the polydactyly allele.

   Delivery

   The slide shows a Tt × tt Punnett square — talk through each cell as a product-rule calculation, not a magic grid. Then explicitly contrast TT vs Tt: same phenotype, different genotype. Head off the misconception "dominant = common" with the polydactyly example — ask the class to raise a hand if they thought dominant meant common; most will. Quick check: "A pea plant is tall. What are its possible genotypes?" (TT or Tt.)

3. 10m

   Codominance vs incomplete dominance — the ABO system

   Content

   These two patterns are the most-confused pair in genetics, so define them by what the heterozygote LOOKS like. Incomplete dominance: heterozygote is an intermediate blend. Classic case: snapdragon flower color. RR is red, rr is white, Rr is pink. One functional pigment allele makes half as much pigment, so the color is diluted. Codominance: both alleles are fully expressed, side by side, with no blending. Classic case: human ABO blood type. The ABO gene has three alleles — Iᴬ, Iᴮ, and i. Iᴬ codes for an enzyme that puts the A-sugar on red blood cells; Iᴮ codes for an enzyme that puts the B-sugar on; i is a nonfunctional allele. Iᴬ and Iᴮ are codominant to each other (a person with Iᴬ Iᴮ has BOTH A and B antigens — type AB, not a blended AB"ish"), and both are dominant over i. So genotypes map to phenotypes as: Iᴬ Iᴬ or Iᴬi → type A; Iᴮ Iᴮ or Iᴮ i → type B; Iᴬ Iᴮ → type AB; ii → type O. Work Iᴬ i × Iᴮ i: cells Iᴬ Iᴮ, Iᴬ i, Iᴮ i, ii — four different blood types (AB, A, B, O), each 1/4. This is the answer to the hook: a type-O person makes no A or B antigens, so their immune system treats donor A antigens as foreign invaders and attacks. AB has both antigens already and treats neither as foreign — "universal recipient."

   Delivery

   The slide shows the Iᴬ Iᴮ cross with both antigens visible on the same cell surface — walk through it as "both flags flying," contrasted with the pink snapdragon where the two colors have blended. This visual contrast is the whole point. Directly name the misconception: "Codominance and incomplete dominance are NOT the same." Give a one-line memory hook: incomplete = blend, codominance = both. Cold-call the ABO genotype-to-phenotype mapping until it's automatic before moving on.

4. 8m

   Reading pedigrees like a detective

   Content

   A pedigree is a family tree drawn with rules: squares = males, circles = females, filled = affected, half-filled = known carrier, horizontal line between two symbols = mating, vertical line down = offspring, roman numerals I, II, III on the left = generations. You infer the inheritance pattern by looking for tells. Autosomal recessive: trait skips generations; two unaffected parents can produce an affected child (both were carriers, Aa × Aa gives 1/4 aa); affects males and females about equally. Autosomal dominant: trait appears in every generation; every affected child has at least one affected parent; affects males and females about equally. X-linked recessive: mostly males affected; an affected father cannot pass it to his sons (he gives them Y, not Xᵇ); daughters of affected fathers are obligate carriers. Once you name the pattern, you calculate probabilities. Example: two unaffected parents already have one affected child (aa) with cystic fibrosis. Both parents must be Aa. Probability the next child is affected = 1/4. Probability an unaffected sibling is a carrier = given they are unaffected, they are AA or Aa in a 1:2 ratio, so P(Aa | unaffected) = 2/3, NOT 1/2. That conditional probability catches nearly everyone.

   Delivery

   The slide shows a three-generation pedigree with an affected grandchild and unaffected parents — walk through the reasoning out loud: skips a generation → probably recessive; both parents must be carriers; now count. Emphasize the 2/3 conditional probability — most students say 1/2 because they forget to remove the aa box from the sample space. Ask: "Why isn't it 1/2?" and wait. This sets up Activity 2.

5. 7m

   Why DNA is not destiny: heritable risk, somatic mutation, environment

   Content

   Most human diseases are not one-gene / one-phenotype. Three inputs combine. First, heritable (germline) risk: alleles inherited from parents that raise or lower risk — e.g., BRCA1 pathogenic variants raise lifetime breast cancer risk from ~13% to 55-70%. That's a RISK increase, not a guarantee. Second, somatic mutations: DNA changes that arise in a body cell during life (UV damage in skin cells, replication errors in dividing gut cells). These are NOT inherited by offspring, but they can drive disease in the individual — most cancers require several somatic mutations to accumulate in one cell lineage. Third, environment and behavior: diet, exercise, smoking, sun exposure, infections, stress, and access to care. Return to the identical-twin hook: same germline DNA, but different somatic mutations accumulate over decades and different environments (one smokes, one doesn't; one develops insulin resistance from diet, one doesn't) produce different disease outcomes. This is why family history matters (heritable risk) AND why lifestyle matters (environment) AND why cancer can strike people with no family history (somatic).

   Delivery

   Explicitly kill the misconception "one gene = one disease." Use the phrase "genes load the gun, environment pulls the trigger" — then complicate it: somatic mutations can pull the trigger too, without any inherited risk. Ask the class to reconcile the twin hook using these three inputs before moving to the activities. Expected: same heritable risk, different somatic mutation history, different environments → different phenotypes.

1. 30m

   Bead-Allele Blood Bank: modeling codominance and incomplete dominanceLab

   Students work in pairs. Each pair gets two bead sets: an ABO set (red = Iᴬ, blue = Iᴮ, white = i) and a snapdragon set (red = R, white = r). They will simulate crosses by random draw from parental "gamete cups," tally 40 offspring, compute observed ratios, and compare to Punnett-square predictions. Walk the room and check that pairs are drawing WITH replacement (each draw is independent) and recording both alleles before deciding phenotype. Student handout: Part 1 — Set up your gamete cups (ABO cross Iᴬi × Iᴮi) - Cup A (mother, Iᴬi): put in 10 red beads (Iᴬ) and 10 white beads (i). - Cup B (father, Iᴮ i): put in 10 blue beads (Iᴮ) and 10 white beads (i). Draw with replacement — put the bead back after each draw so probabilities stay constant. Part 2 — Predict before you simulate Fill in the Punnett square for Iᴬi × Iᴮ i: - Iᴬ × Iᴮ → ______ - Iᴬ × i → ______ - i × Iᴮ → ______ - i × i → ______ Predicted phenotype ratio (AB : A : B : O) = ______ : ______ : ______ : ______ Out of 40 offspring, expected counts: - AB: ______ - A: ______ - B: ______ - O: ______ Part 3 — Simulate 40 offspring Draw one bead from Cup A and one from Cup B. Record the genotype, then the phenotype. Return both beads. Repeat 40 times. Tally: - Iᴬ Iᴮ (AB): ______ - Iᴬ Iᴬ or Iᴬi (A): ______ - Iᴮ Iᴮ or Iᴮ i (B): ______ - ii (O): ______ Part 4 — Compare and reason 1. Did your observed counts exactly match the predicted counts? Why or why not? 2. What would happen to the match if you did 400 draws instead of 40? 3. In your Iᴬ Iᴮ offspring, are the A and B antigens BLENDED into one "AB antigen"? Explain what codominance actually means at the cell surface. Part 5 — Now switch to snapdragons (incomplete dominance) Empty your cups. Refill: - Cup A (Rr): 10 red + 10 white - Cup B (Rr): 10 red + 10 white Predict the Punnett-square outcome for Rr × Rr. Record the phenotype rule for snapdragons: - RR = red - Rr = pink (blended) - rr = white Expected phenotype ratio (red : pink : white) = ______ : ______ : ______ Draw 40 times with replacement and tally: - Red: ______ - Pink: ______ - White: ______ Part 6 — The big idea (write 2-3 sentences) Compare Iᴬ Iᴮ (AB blood) with Rr (pink snapdragon). Both are heterozygotes. Why is one called codominant and the other incomplete dominant? Use the words antigen, blend, and expressed.

   Materials

   • Small cups or bags for each student pair
   • Colored beads or pony beads: red beads = Iᴬ, blue beads = Iᴮ, white beads = i (about 20 of each per pair)
   • A separate set with red beads = R (snapdragon red) and white beads = r (white) — 10 each per pair
   • Student handout (content below)
   • Calculators

   Example outputs

   • Part 2 Punnett square filled: Iᴬ Iᴮ, Iᴬi, Iᴮ i, ii; predicted phenotype ratio 1:1:1:1; expected counts 10 AB, 10 A, 10 B, 10 O out of 40.
   • Part 3 sample tally after 40 draws: 12 AB, 9 A, 11 B, 8 O — close to but not exactly 1:1:1:1 because of small-sample variation.
   • Part 4 answer 3: The A and B antigens are NOT blended; both types of antigen sit side by side on the same red blood cell surface. Codominance means both alleles are fully expressed, producing both products, not a mixture.
   • Part 5: Rr × Rr predicts 1 red : 2 pink : 1 white (1:2:1) because Rr blends to pink. Sample tally: 9 red, 22 pink, 9 white.
   • Part 6: Iᴬ Iᴮ makes both A and B antigens fully — both alleles are expressed side by side, so it is codominant. Rr snapdragons make only half the red pigment because one allele is nonfunctional, so the color blends to pink — that is incomplete dominance.
2. 30m

   The Alvarez Family Pedigree: cystic fibrosis case analysis

   Individual work for 15 minutes, then pair-share and full-class review. Students analyze a three-generation pedigree, determine the inheritance pattern, compute carrier probabilities, and apply conditional probability. This is the pedigree the slide shows — the handout reproduces it in text so students can annotate. Student handout: The Alvarez family and cystic fibrosis (CF) CF is caused by mutations in the CFTR gene on chromosome 7. Use A = unaffected allele, a = CF allele. Pedigree (three generations) - Generation I - I-1 (grandfather, unaffected square) — I-2 (grandmother, unaffected circle) - Generation II — children of I-1 and I-2, and their spouses - II-1 (unaffected square, married in) - II-2 (unaffected circle, daughter of I-1 and I-2) - II-3 (unaffected square, son of I-1 and I-2) - II-4 (unaffected circle, married in, no family history of CF) - Generation III — children - III-1 (affected square, filled — child of II-1 and II-2), has CF - III-2 (unaffected circle — sister of III-1) - III-3 (unaffected square — child of II-3 and II-4) Questions 1. Is CF autosomal dominant, autosomal recessive, or X-linked? Give TWO pieces of evidence from the pedigree. 2. What must the genotypes of II-1 and II-2 be? Explain your reasoning. 3. What must the genotypes of I-1 and I-2 be? Explain. 4. III-2 is unaffected. She wants to know the probability she is a carrier (Aa). - Before conditioning: what are the four equally likely genotype outcomes from Aa × Aa? - Given III-2 is unaffected (not aa), what is P(Aa | unaffected)? - Show your work. 5. II-3 is unaffected. What is the probability HE is a carrier? (Hint: think about I-1 and I-2's genotypes.) 6. II-3 married II-4, who has NO family history of CF. In the U.S. general population, roughly 1 in 25 people is a CF carrier. What is the probability that III-3 has CF? - Step A: P(II-3 is a carrier) = ______ - Step B: P(II-4 is a carrier) = ______ - Step C: If both are carriers, P(child is aa) = ______ - Step D: Multiply A × B × C = ______ 7. III-1 (the affected child) will grow up and eventually consider having children. His partner has no family history of CF (population carrier rate 1/25). What is the probability their first child has CF? 8. Beyond the pedigree. CF severity varies even among people with the same CFTR genotype. Name one non-genetic factor and one somatic/other factor that could influence how severe an individual's CF becomes.

   Materials

   • Student handout (content below)
   • Calculators
   • Pencils

   Example outputs

   • Q1: Autosomal recessive. Evidence: (a) the trait skips a generation — I-1 and I-2 are unaffected but their grandchild III-1 is affected; (b) both sexes appear equally at risk and two unaffected parents (II-1 and II-2) produced an affected child, which is the signature of a recessive allele.
   • Q2: II-1 and II-2 must both be Aa. They are unaffected (so not aa) but produced III-1 who is aa — each parent had to contribute an a allele.
   • Q4: Four outcomes from Aa × Aa are AA, Aa, Aa, aa (each 1/4). Given III-2 is unaffected, remove aa from the sample space, leaving AA, Aa, Aa. P(Aa | unaffected) = 2/3.
   • Q6: II-3's parents I-1 and I-2 must both be Aa (they produced Aa child II-2). So P(II-3 carrier | unaffected) = 2/3. P(II-4 carrier) = 1/25. If both are Aa × Aa, P(child aa) = 1/4. Total = (2/3)(1/25)(1/4) = 2/300 = 1/150 ≈ 0.67%.
   • Q7: III-1 is aa, so he passes a with probability 1. His partner has P(carrier) = 1/25; if she is Aa, P(child aa) = 1/2. Total = 1 × (1/25) × (1/2) = 1/50 = 2%.
   • Q8: Environmental factor — exposure to lung pathogens like Pseudomonas aeruginosa, air quality, access to treatments like CFTR modulators. Somatic factor — accumulated lung tissue damage from repeated infections; modifier genes at other loci also affect severity.

1. A pea plant with genotype Tt is crossed with a tt plant. (a) Complete the Punnett square. (b) State the genotype ratio and phenotype ratio. (c) If the cross produces 80 seeds, how many are expected to be short (tt)?

   calculation(a) Cells: Tt, tt, Tt, tt. (b) Genotype ratio 1 Tt : 1 tt (1:1). Phenotype ratio 1 tall : 1 short (1:1). (c) Expected short seeds = 1/2 × 80 = 40.

2. A woman with blood type A whose father was type O marries a man with blood type AB. What are the possible blood types of their children, and what is the probability of each?

   calculationThe woman is A but her father was ii, so she must have received i from him — her genotype is Iᴬi. The man is Iᴬ Iᴮ. Cross Iᴬi × Iᴬ Iᴮ: - Iᴬ Iᴬ (type A): 1/4 - Iᴬi (type A): 1/4 - Iᴬ Iᴮ (type AB): 1/4 - Iᴮ i (type B): 1/4 Combined: P(A) = 1/2, P(AB) = 1/4, P(B) = 1/4, P(O) = 0.

3. Which statement best distinguishes codominance from incomplete dominance? A. Codominance produces a blended intermediate phenotype; incomplete dominance shows both alleles fully. B. Codominance shows both alleles fully expressed in the heterozygote; incomplete dominance produces a blended intermediate phenotype. C. Codominance only occurs on the X chromosome; incomplete dominance only occurs on autosomes. D. Codominance and incomplete dominance are two names for the same pattern.

   multiple choiceB. In codominance, both alleles are fully and separately expressed in the heterozygote (e.g., Iᴬ Iᴮ shows both A and B antigens — type AB). In incomplete dominance, the heterozygote shows a blended intermediate phenotype (e.g., Rr snapdragons are pink).

4. Two unaffected parents have a daughter with an autosomal recessive disorder. They have a second child who is phenotypically unaffected. What is the probability that this second child is a carrier of the disease allele?

   multiple choice2/3. Both parents must be Aa (they produced an aa child). Aa × Aa gives AA, Aa, Aa, aa each with probability 1/4. Because the child is unaffected, remove aa from the sample space, leaving AA, Aa, Aa. P(Aa | unaffected) = 2/3, NOT 1/2.

5. A student says, "My grandfather smoked for 50 years and never got lung cancer, so lung cancer must be entirely genetic — you either have the gene or you don't." Using the ideas of heritable risk, somatic mutation, and environment, write a 3-4 sentence rebuttal.

   short answerLung cancer risk is a combination of three inputs, not a single gene. Heritable variants (e.g., in DNA repair genes) can raise or lower baseline risk, but they set a probability, not a guarantee — some people with high risk never develop cancer, and some with low risk do. Smoking causes somatic mutations in lung cells over decades; whether those mutations accumulate in the right combination to trigger cancer is partly chance. Environment (secondhand smoke exposure, air pollution, radon, diet) and access to screening also matter, which is why one heavy smoker can escape lung cancer while another non-smoker develops it.

• ☐Colored beads or pony beads — red (~30/pair), blue (~10/pair), white (~30/pair)
• ☐Small cups or bags labeled Cup A and Cup B (2 per pair)
• ☐Bead-Allele Blood Bank student handout (1 per student)
• ☐Alvarez Family Pedigree student handout (1 per student)
• ☐Calculators (1 per student)
• ☐Pencils
• ☐Whiteboard and markers for working the hook and worked examples
• ☐Projector for slide deck