Open with a real 2023 headline: the FDA approved Casgevy, the first CRISPR-based medicine, to treat sickle cell disease. Show the phenotype logic briefly: one wrong base in the HBB gene changes one amino acid (Glu → Val) in β-globin; that single amino-acid change makes red blood cells sickle and clog capillaries. Ask students the driving question of the day: 'How does changing ONE letter in a 3-billion-letter DNA sequence change what a cell looks like and how a person feels?' Take 3–4 quick student responses, don't correct yet — tell them by the end of the block they will trace that exact change from DNA to mRNA to protein, and see how CRISPR fixes it. Use this to set the stakes: DNA is not abstract; it is written instructions the cell reads every second.

1. 10m

   DNA structure: nucleotides, base pairing, and the antiparallel double helix

   Content

   DNA is a polymer built from nucleotide monomers. Each nucleotide has three parts: a deoxyribose sugar, a phosphate group, and one nitrogenous base — adenine (A), thymine (T), guanine (G), or cytosine (C). Nucleotides link into a strand by phosphodiester bonds between the phosphate of one and the 3′ carbon of the next sugar, giving every strand a direction: a 5′ end (free phosphate) and a 3′ end (free hydroxyl). Two strands wrap around each other into a double helix, and they run antiparallel — one strand runs 5′→3′, the other runs 3′→5′. The sugar–phosphate backbones sit on the outside; the bases point inward and pair by hydrogen bonds following strict base pairing rules: A pairs with T using 2 H-bonds, G pairs with C using 3 H-bonds. That is why a G–C-rich region is harder to pull apart than an A–T-rich region. Chargaff's rule falls out of this: in any DNA sample, %A = %T and %G = %C. If a top strand reads 5′-ATGCCA-3′, the bottom strand must read 3′-TACGGT-5′.

   Delivery

   Emphasize the three-part nucleotide and the base-pairing rules — these are the anchors for the rest of the day. Walk the class through pairing a short strand at the board using student volunteers reading off the complement. Pre-empt the misconception that A always pairs with G or that pairing is random — it is dictated by hydrogen-bond geometry (A–T = 2, G–C = 3). Do a quick check: 'If %A = 20%, what is %G?' (answer: 30%). Keep it snappy — they will build a physical model in Activity 1.

2. 10m

   Transcription: making an mRNA copy in the nucleus

   Content

   A gene is a segment of DNA that codes for a protein, but DNA never leaves the nucleus in a eukaryotic cell. Instead, the cell makes a portable copy — messenger RNA (mRNA) — through transcription. The enzyme RNA polymerase binds a promoter sequence upstream of the gene, unwinds the double helix, and reads the template strand 3′→5′ while building a complementary mRNA strand 5′→3′. RNA differs from DNA in three ways: sugar is ribose (not deoxyribose), it is single-stranded, and it uses uracil (U) instead of thymine (T). So during transcription A on DNA pairs with U on RNA. Example: DNA template 3′-TAC GGA TTA ACT-5′ becomes mRNA 5′-AUG CCU AAU UGA-3′. After transcription in eukaryotes, the mRNA is processed (5′ cap, poly-A tail, introns spliced out) and then exits through nuclear pores into the cytoplasm. Only THEN does translation start. This is the first half of the central dogma: DNA → mRNA.

   Delivery

   Hammer the location: transcription happens in the NUCLEUS. This directly attacks the misconception that DNA leaves the nucleus and gets read at the ribosome. Emphasize the T → U swap and that the mRNA sequence matches the coding strand (except U for T), because students often try to base-pair a codon chart directly to DNA. Do a live transcription of the sickle-cell region on the board: normal DNA template 3′-CTC-5′ → mRNA 5′-GAG-3′ (Glu); mutant DNA template 3′-CAC-5′ → mRNA 5′-GUG-3′ (Val — we will translate this in the next beat). Ask: 'Why does the cell bother making a copy instead of using DNA directly?' Expected answer: DNA is the master file; mRNA is a working copy that protects the original and can be made in many copies for many ribosomes.

3. 12m

   Translation: reading codons and building a polypeptide at the ribosome

   Content

   Translation happens in the cytoplasm at a ribosome. The ribosome reads mRNA in groups of three bases called codons, moving 5′→3′. Each codon specifies one amino acid, decoded using a codon chart. There are 64 possible codons for 20 amino acids, so the code is redundant (multiple codons per amino acid) — but it is not ambiguous (a codon codes for only one amino acid). Every mRNA starts translation at the start codon AUG (Methionine) and stops at a stop codon (UAA, UAG, or UGA — no amino acid). Transfer RNA (tRNA) is the adapter: each tRNA carries one specific amino acid on one end and has a three-base anticodon on the other end that pairs with the mRNA codon. The ribosome links amino acids together by peptide bonds into a polypeptide, which folds into a functional protein. Trace it: mRNA 5′-AUG-GAG-3′ → Met–Glu. In sickle cell, the mutation changes codon 6 from GAG (Glutamate, negatively charged, soluble) to GUG (Valine, hydrophobic). One amino acid, but valine on the outside of β-globin makes hemoglobin molecules stick together and distort the red blood cell into a sickle shape.

   Delivery

   This is the beat where you slow down. Practice using a codon chart out loud — read 5′→3′, find first base on the left column, second on the top, third on the right. Do the sickle cell codon step-by-step with them: GAG → Glu vs GUG → Val. Explicitly separate translation from transcription: 'Transcription copies DNA into mRNA — same language, different alphabet (U for T). Translation converts mRNA into protein — a NEW language, using the codon chart.' Attack the misconception that DNA sits in the ribosome. Ask students to check with a neighbor: 'What is the anticodon on the tRNA that brings valine for GUG?' (answer: 3′-CAC-5′ or written 5′-CAC-3′ depending on convention — accept CAC).

4. 8m

   Gene regulation and cellular differentiation: same DNA, different cells

   Content

   Every somatic cell in your body has the same ~3 billion base pairs of DNA — a neuron, a muscle cell, and a pancreatic β cell all have the identical genome. So why do they look and act so different? Gene expression is regulated: only a subset of genes is transcribed in any given cell at any given time. Transcription factors — regulatory proteins — bind to promoter and enhancer regions of DNA and turn transcription of specific genes on or off. In a pancreatic β cell, the insulin gene is transcribed and translated at high rates; in a neuron, the insulin gene sits silent while genes for neurotransmitter receptors are active. Cellular differentiation during development is essentially the stable switching-on of a cell-type-specific set of genes. Regulation can happen at many levels: which genes are transcribed (transcriptional control), which mRNAs are stable or spliced (post-transcriptional), and which proteins are activated after they are made (post-translational). This is why 'one gene → one protein → one trait' is too simple: the same gene can be silent in one cell and highly expressed in another, and one gene can produce multiple proteins through alternative splicing.

   Delivery

   Directly confront TWO misconceptions here: (1) every gene is expressed in every cell at all times — no, only a small fraction is active in a given cell; and (2) genes and proteins map one-to-one — no, regulation and splicing complicate that. Use the concrete question: 'Why doesn't your skin cell make insulin?' Expected answer: the insulin gene is present but not transcribed — transcription factors specific to β cells are missing. Tie back to the driving question: differentiation is why one fertilized egg becomes ~200 different cell types with the same DNA. Preview that this connects to next unit's cell cycle and stem-cell biology (AL.BIO.3).

1. 40m

   Lab 1 — Build, Transcribe, and Translate the Sickle Cell GeneLab

   Students work in pairs. Distribute the handout below and the nucleotide cutouts. Circulate and check base pairing (A–T, G–C — never A–G), correct direction (antiparallel), and correct T→U swap during transcription. Expect ~15 minutes on Part 1 (DNA model), ~15 minutes on Part 2 (transcribe & translate normal vs. sickle), and ~10 minutes on Part 3 (analysis). Debrief the sickle mutation as a class before moving to the CRISPR activity. Student handout — Sickle Cell Molecular Lab Part 1 — Build the DNA double helix (β-globin, codons 5–7) Using the colored paper nucleotides, build the following section of the normal HBB gene. Tape the sugar–phosphate backbones down the sides of your paper and place the bases in the middle so they pair. - Top (coding) strand, written 5′ → 3′: C - C - T - G - A - G - G - A - G - Bottom (template) strand: fill in the complement, and write the correct direction at each end. - Bottom strand 3′ → 5′: ______ - ______ - ______ - ______ - ______ - ______ - ______ - ______ - ______ Check: - Are your strands antiparallel? Label the 5′ and 3′ ends on BOTH strands. - Use 2 twist ties between A–T pairs and 3 twist ties between G–C pairs. Which pair is stronger, and why? ______ Part 2 — Transcribe the gene into mRNA RNA polymerase reads the TEMPLATE strand 3′ → 5′ and builds mRNA 5′ → 3′. Remember: in RNA, U replaces T. - Template strand (3′ → 5′): G - G - A - C - T - C - C - T - C - mRNA (5′ → 3′): ______ - ______ - ______ - ______ - ______ - ______ - ______ - ______ - ______ Part 3 — Translate the mRNA using the codon chart Break the mRNA into codons (groups of 3, reading 5′ → 3′) and look each one up on your codon chart. - Codon 5: ______ → amino acid: ______ - Codon 6: ______ → amino acid: ______ - Codon 7: ______ → amino acid: ______ String the amino acid tags together in order. This is a piece of the normal β-globin protein. Part 4 — Introduce the sickle cell mutation On the coding strand of your DNA, change codon 6 from GAG to GTG (change ONE base: A → T on the coding strand; T → A on the template strand). Now redo the transcription and translation of codon 6 only. - New mRNA codon 6: ______ → new amino acid: ______ Part 5 — Analysis (answer in complete sentences) 1. How many DNA bases changed? ______ 2. How many amino acids changed? ______ 3. Glutamate (Glu) is negatively charged and interacts with water. Valine (Val) is hydrophobic and avoids water. Predict what happens when hemoglobin molecules carrying valine on the outside encounter each other in a red blood cell. 4. Explain in your own words WHY a change of ONE DNA base can change the shape of an entire red blood cell. 5. If a section of DNA has 30% adenine, what percentage is thymine? Guanine? Cytosine? Show your reasoning using Chargaff's rule.

   Materials

   • Colored paper nucleotide cutouts (A = green, T = red, G = blue, C = yellow, U = orange) — precut, ~30 per pair
   • Pipe cleaners or twist ties (to represent H-bonds and sugar-phosphate backbone)
   • Codon chart (one per student — standard mRNA codon table)
   • Tape or glue sticks
   • Colored pencils
   • Amino acid bead kit OR paper amino acid tags (Met, Glu, Val, Leu, Thr, Pro, Ala, etc.)
   • Student handout (below) — one per student

   Example outputs

   • Part 1 bottom strand (3′→5′): G-G-A-C-T-C-C-T-C. G–C pairs are stronger than A–T pairs because G–C has 3 hydrogen bonds vs. 2 for A–T.
   • Part 2 mRNA (5′→3′): C-C-U-G-A-G-G-A-G. Part 3: CCU → Pro, GAG → Glu, GAG → Glu. Part 4 mutant codon 6: GUG → Val. Part 5: 1 DNA base changed → 1 amino acid changed; valine on the outside makes hemoglobin molecules stick together into fibers, distorting the red blood cell into a sickle shape. If %A = 30%, then %T = 30%, and %G = %C = 20% each.
   • presentation_text
2. 25m

   Lab 2 — CRISPR Case Study: Editing the Sickle Cell Gene

   Students work in pairs. This activity connects the molecular understanding from Lab 1 to a real 2023 medical breakthrough (Casgevy). Circulate to make sure students understand that CRISPR does NOT translate — it edits DNA in the nucleus BEFORE transcription happens. Optional: pairs may open the HHMI BioInteractive CRISPR-Cas9 Mechanism click-through at https://www.biointeractive.org/classroom-resources/crispr-cas-9-mechanism-applications while working through the case study. Debrief as a class: was this a good use of the technology? Why or why not? Student handout — CRISPR and the Sickle Cell Cure Background In December 2023, the U.S. FDA approved Casgevy, the first CRISPR-based medicine. It treats sickle cell disease, the disorder you modeled in Lab 1. Casgevy does not fix the mutated HBB gene directly — instead, it uses CRISPR-Cas9 to disable a different gene called BCL11A. When BCL11A is turned off, patients start producing fetal hemoglobin again, which does not sickle. How CRISPR-Cas9 works (short version) - Guide RNA (gRNA): a short RNA molecule designed by scientists to be complementary to a specific 20-base target DNA sequence. - Cas9: a protein enzyme that acts like molecular scissors. - The gRNA + Cas9 complex enters the nucleus, the gRNA base-pairs with the target DNA sequence, and Cas9 cuts both strands. - The cell tries to repair the cut but usually makes small errors that disable the gene. Part 1 — Design your guide RNA The target DNA sequence (coding strand, 5′ → 3′) in BCL11A is: 5′ - A T G G C C T G A C G T A A C T G C A A - 3′ Design a guide RNA (5′ → 3′) that will base-pair with the TEMPLATE strand of this sequence. Remember: RNA uses U instead of T. - Guide RNA (5′ → 3′): ______________________ Part 2 — Central dogma check Circle where in the central dogma CRISPR acts: DNA ——transcription——▶ mRNA ——translation——▶ protein 1. At which step does CRISPR-Cas9 act? ______ 2. Why can CRISPR change a trait for the rest of a cell's life, while a drug that blocks a protein only works while the drug is present? 3. Casgevy disables BCL11A instead of fixing HBB. Explain, using what you know about gene regulation, how disabling ONE gene (BCL11A) can turn ON a different gene (fetal hemoglobin). Part 3 — Evaluate the technology (write 3–5 sentences) Casgevy costs about $2.2 million per patient. It has cured sickle cell symptoms in the first patients treated, but it requires chemotherapy to prepare the bone marrow, and long-term effects are unknown. It also edits only somatic cells — the changes are NOT passed to children. A different, more controversial use of CRISPR would be editing embryos (germline editing), which WOULD pass to future generations. Write a short position: Should CRISPR be used for (a) treating sickle cell in patients, and (b) editing embryos to prevent sickle cell in future children? Support your answer with at least one biological reason AND one ethical reason. Part 4 — Extend Give one OTHER real-world application of CRISPR (e.g., mosquitoes engineered to resist malaria, hornless dairy cattle, disease-resistant crops, screening embryos for BRCA mutations). Explain how it uses the same DNA-editing mechanism you just described.

   Materials

   • Student handout (below) — one per student
   • Computers with internet access (1 per pair) for the optional simulation
   • Colored pencils

   Example outputs

   • Part 1 guide RNA (5′→3′): AUGGCCUGACGUAACUGCAA — students who write the complement instead should be corrected: the guide RNA matches the coding strand (with U for T) and pairs with the template.
   • Part 2: CRISPR acts on DNA, before transcription. It changes the gene itself, so every mRNA and every protein made from that gene from then on is affected — a drug only blocks proteins temporarily. Disabling BCL11A works because BCL11A normally acts as a transcription factor that SILENCES the fetal hemoglobin gene; remove BCL11A → fetal hemoglobin gene turns back ON → healthy red blood cells.

   No-equipment fallback

   Skip the HHMI simulation and run the case study from the handout only. The handout is self-contained.

1. The DNA template strand of a short gene reads: 3′-T A C G G A C T T A C T-5′. (a) Write the mRNA sequence produced by transcription (5′ → 3′). (b) Using the codon chart, translate the mRNA into a sequence of amino acids. (c) In WHICH part of the cell does each step (transcription, translation) happen?

   calculation(a) mRNA (5′→3′): A-U-G-C-C-U-G-A-A-U-G-A (b) Codons: AUG, CCU, GAA, UGA → Methionine – Proline – Glutamate – STOP. Final polypeptide: Met-Pro-Glu (translation terminates at UGA; no amino acid added for the stop codon). (c) Transcription happens in the nucleus. Translation happens at the ribosome in the cytoplasm. DNA itself does NOT leave the nucleus.

2. A student says: 'Since every cell in your body has the same DNA, every cell must be making the same proteins.' Explain what is wrong with this statement, using the terms transcription factors, gene expression, and cellular differentiation. Give one specific cell-type example.

   short answerEvery cell does share the same genome, but each cell type only EXPRESSES a subset of its genes. Transcription factors — regulatory proteins specific to each cell type — bind to DNA and switch specific genes on or off, so different mRNAs (and therefore different proteins) are made in different cells. This regulation of gene expression drives cellular differentiation: one fertilized egg becomes ~200 different cell types with the same DNA. Example: a pancreatic β cell expresses the insulin gene at high levels, while a skin cell has the insulin gene but keeps it silent — the skin cell lacks the transcription factors needed to turn it on.

3. In sickle cell disease, the HBB gene has ONE base change: on the coding strand, GAG becomes GTG at codon 6. Which statement BEST explains why this tiny change causes such a large effect on red blood cell shape?

   multiple choiceCorrect answer: C. A. The mutation deletes the entire β-globin protein, so no hemoglobin is made. — Wrong; only one base is substituted, protein is still made. B. The mutation changes many amino acids because of frameshift. — Wrong; a substitution keeps the reading frame; only ONE codon is altered. C. The mutation changes one codon from GAG (Glutamate, charged/soluble) to GUG (Valine, hydrophobic), so β-globin molecules stick together and distort the cell. — CORRECT. D. The mutation prevents mRNA from leaving the nucleus, so translation never happens. — Wrong; the mRNA is exported normally; the defect is in the protein sequence.

4. CRISPR-Cas9 uses a guide RNA to direct the Cas9 enzyme to a target DNA sequence. At which step of the central dogma does CRISPR act, and why does this allow a permanent change to a cell's protein output?

   short answerCRISPR acts on the DNA itself — BEFORE transcription. It edits or disables the gene in the nucleus. Because every mRNA is transcribed from that DNA, and every protein is translated from that mRNA, changing the DNA changes ALL future mRNAs and proteins from that gene for the life of the cell (and its daughter cells). A drug that blocks a protein only works while the drug is present; a DNA edit is essentially permanent.

• ☐Precut colored paper nucleotides (A=green, T=red, G=blue, C=yellow, U=orange) — ~30 per student pair
• ☐Pipe cleaners or twist ties
• ☐Tape and glue sticks
• ☐Colored pencils
• ☐Standard mRNA codon chart — one per student
• ☐Amino acid bead kit or paper amino acid tags (Met, Glu, Val, Pro, Leu, Thr, Ala, at minimum)
• ☐Student handout for Lab 1 (Sickle Cell Molecular Lab) — one per student
• ☐Student handout for Lab 2 (CRISPR Case Study) — one per student
• ☐Computers with internet access (1 per pair) — optional, for HHMI CRISPR simulation
• ☐Formative assessment printout — one per student