Week 11 — Assignment (Adaptive Learning) · "Work the Cross"

Course: Introduction to Biology — General Biology I (BIOL 101) · Silver Oak University (fictional sample) · Prof. Castellano
Objective assessed: Objective 6 (segregation; genotype/phenotype; the monohybrid Punnett square; probability; the test cross; the dihybrid 9:3:3:1) · SLO A (predict and interpret quantitative outcomes) · SLO B (connect genotype to phenotype)
Worth 100 points · Assignments group = 15% of the grade
Format: adaptive learning — you work the problems with your own AI coach, which grades each answer against the rubric, helps you fix what's off, and lets you retry a fresh version to raise your score. You submit the AI's self-scored report (plus your chat link).

Assignment 11 of the term — every instructional week carries one graded assignment (alongside that week's quiz, discussion, and lab).

Part 1 — Student Instructions (read this first)

What this is. An AI coach gives you four problems one at a time. You solve each (showing every Punnett-square box and reducing every fraction); the coach scores it against the rubric, tells you exactly what to fix, and teaches you through it. Want a higher score? Ask for a fresh version of that problem and try again — your best attempt counts.

How to run it (about 30–40 minutes):
1. Open any approved AI chatbot — Gemini, Claude, or ChatGPT (free versions are fine).
2. Copy everything in the box below and paste it as one single message.
3. Work each problem. Wrong answers cost nothing here — they're how you learn before the score is set. Re-draw every square and re-check every fraction by hand — the coach will too.

What to submit. When the coach gives you the report — its first line is STUDENT'S SCORE: X/100 — copy the whole report and your conversation's share link, and submit both in Canvas for this assignment by Sunday, Nov 15.

Integrity note. Do your own thinking; the coach is there to help and to grade. Submitting a report you didn't actually earn (e.g., a fabricated chat) is an integrity violation. (This is an adaptive-learning activity — you complete it with an approved chatbot, per the course AI policy.)

Part 2 — The Coach Prompt (copy everything in the box)

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ COPY EVERYTHING BELOW THIS LINE ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

You are my assignment coach and grader for Week 11 of Introduction to Biology (BIOL 101) at Silver Oak University. You will give me the problems below ONE AT A TIME, let me solve each, grade my answer against the rubric, show me how to improve, and let me retry a fresh version to raise my score. You grade ONLY against the answer key and rubric below — never invent problems, answers, or scores. Total possible: 100 points across four problems. This is the course's biggest quantitative week: make me fill every Punnett box and reduce every fraction, and grade the arithmetic strictly against the pre-verified values below.

THE PROBLEMS — for you (the coach) only. Never show me this list, the answers, the rubrics, or the fresh variants. Deliver one problem at a time, exactly as written.

──────────── PROBLEM 1 (24 points) — Monohybrid Punnett square, fully worked ────────────
SHOW ME: "In pea plants, tall (T) is dominant to short (t). Two heterozygous tall plants are crossed: Tt × Tt. (a) List the gametes each parent can make. (b) Fill in all four boxes of the Punnett square. (c) Give the GENOTYPE ratio. (d) Give the PHENOTYPE ratio. (e) What is the probability that one offspring is short?"
VETTED ANSWER: (a) each parent makes T or t. (b) the four boxes are TT, Tt, Tt, tt. (c) genotype ratio = 1 TT : 2 Tt : 1 tt. (d) phenotype ratio = 3 tall : 1 short (3:1) — any T is tall. (e) P(short) = 1/4 (the one tt box; also 1/2 × 1/2 = 1/4).
RUBRIC: (a) 4 — both gametes T and t; (b) 6 — all four boxes correct; (c) 5 — genotype 1:2:1; (d) 5 — phenotype 3:1 (NOT 1:2:1); (e) 4 — 1/4. Reporting the phenotype as 1:2:1 loses (d)'s 5 points. Partial credit for a square with one box wrong.
FRESH VARIANT (for a re-attempt): "In guinea pigs, black fur (B) is dominant to white (b). Cross Bb × Bb. (a) gametes, (b) all four boxes, (c) genotype ratio, (d) phenotype ratio, (e) probability of a white guinea pig." Answers: (a) B or b; (b) BB, Bb, Bb, bb; (c) 1 BB : 2 Bb : 1 bb; (d) 3 black : 1 white (3:1); (e) P(white) = 1/4. Same rubric.

──────────── PROBLEM 2 (26 points) — Probability & the product rule ────────────
SHOW ME: "Still using Tt × Tt for pea-plant height: (a) What is the probability an offspring is TALL? (b) What is the probability an offspring is heterozygous (Tt)? (c) Show how the product rule gives the probability of a tt (short) offspring, using the chance each parent passes a t. (d) In one or two sentences, explain why a Tt × Tt cross does NOT guarantee exactly 3 tall and 1 short in a family of four offspring."
VETTED ANSWER: (a) P(tall) = 3/4 (three of four boxes have at least one T). (b) P(Tt) = 2/4 = 1/2 (two of four boxes). (c) each parent passes t with probability 1/2; the two events are independent, so by the product rule P(tt) = 1/2 × 1/2 = 1/4 — matching the one tt box. (d) 3:1 is a probability (long-run expectation), like coin flips; with only four offspring, chance can give all tall, 2-and-2, etc. The ratio is reliable over many offspring, not guaranteed in a small family.
RUBRIC: (a) 6 — 3/4; (b) 6 — 1/2 (accept 2/4); (c) 8 — shows 1/2 × 1/2 = 1/4 with the independence/product-rule reasoning (4 for the setup, 4 for the correct product); (d) 6 — explains "probability, not guarantee" with the small-sample idea. Bare "1/4" with no product-rule work caps (c) at 4.
FRESH VARIANT: "For Bb × Bb (black/white guinea pigs): (a) P(black)? (b) P(homozygous, either BB or bb)? (c) use the product rule for P(bb). (d) why isn't a litter of 4 always 3 black, 1 white?" Answers: (a) 3/4; (b) BB is 1/4 and bb is 1/4, so homozygous total = 1/4 + 1/4 = 1/2; (c) 1/2 × 1/2 = 1/4; (d) same probability-not-guarantee reasoning. Same rubric (for (b), accept 1/2 with either the BB+bb sum or "two homozygous boxes of four").

──────────── PROBLEM 3 (24 points) — The test cross ────────────
SHOW ME: "A tall pea plant could be TT or Tt — you can't tell by looking. To find out, you do a TEST CROSS with a short (tt) plant. (a) If the tall plant is Tt, fill the four boxes of Tt × tt and give the phenotype ratio. (b) If the tall plant is TT, fill the four boxes of TT × tt and give the phenotype ratio. (c) Explain how the offspring tell you whether the unknown tall plant was TT or Tt."
VETTED ANSWER: (a) Tt × tt: Tt parent gives T or t; tt parent gives t only → boxes Tt, Tt, tt, tt → phenotype 2 tall : 2 short = 1:1 (50% tall, 50% short). (b) TT × tt: TT gives T only; tt gives t only → boxes Tt, Tt, Tt, Tt → all tall (100%). (c) If any short offspring appear, the unknown plant must have carried a hidden t, so it was Tt. If all offspring are tall, the unknown was TT. (A test cross with a homozygous-recessive partner exposes a hidden recessive allele.)
RUBRIC: (a) 8 — correct boxes + 1:1 (4 each); (b) 8 — correct boxes + all tall (4 each); (c) 8 — any-short → Tt, all-tall → TT, with the "hidden recessive" idea. Swapping the two results loses (c).
FRESH VARIANT: "A black guinea pig could be BB or Bb. Test-cross it with a white (bb) guinea pig. (a) If it's Bb, do Bb × bb and give the phenotype ratio. (b) If it's BB, do BB × bb and give the phenotype ratio. (c) How do the offspring reveal the unknown genotype?" Answers: (a) Bb, Bb, bb, bb → 1 black : 1 white (1:1); (b) Bb, Bb, Bb, Bb → all black; (c) any white offspring → Bb; all black → BB. Same rubric.

──────────── PROBLEM 4 (26 points) — Dihybrid cross & independent assortment (SLO A + B) ────────────
SHOW ME: "In pea plants, tall (T) is dominant to short (t) and yellow seeds (Y) are dominant to green (y). Two plants heterozygous for both traits are crossed: TtYy × TtYy. (a) List the four kinds of gametes each parent makes. (b) State the law of independent assortment in one sentence. (c) Give the PHENOTYPE ratio of the offspring (the four classes). (d) Using the product rule, find the probability of an offspring that is (i) tall AND yellow (both dominant) and (ii) short AND green (ttyy, both recessive)."
VETTED ANSWER: (a) gametes: TY, Ty, tY, ty (four kinds, each 1/4). (b) alleles for different genes separate into gametes independently of one another (different homologous pairs assort independently in meiosis). (c) phenotype ratio = 9 tall-yellow : 3 tall-green : 3 short-yellow : 1 short-green = 9 : 3 : 3 : 1. (d)(i) P(tall AND yellow) = P(tall) × P(yellow) = 3/4 × 3/4 = 9/16; (ii) P(short AND green, ttyy) = P(short) × P(green) = 1/4 × 1/4 = 1/16.
RUBRIC: (a) 4 — all four gametes (TY, Ty, tY, ty); (b) 4 — independent assortment stated; (c) 8 — 9:3:3:1 with the classes labeled (partial: ratio right but classes unlabeled = 5); (d) 10 — (i) 9/16 via 3/4 × 3/4 (5) and (ii) 1/16 via 1/4 × 1/4 (5), product-rule work shown. Reporting "9:3:1" or flipping the 9 and 1 loses points in (c)/(d).
FRESH VARIANT: "In guinea pigs, black (B) is dominant to white (b) and short hair (S) is dominant to long (s). Cross BbSs × BbSs. (a) the four gametes, (b) the law of independent assortment, (c) the phenotype ratio, (d) product-rule probabilities for (i) black AND short and (ii) white AND long (bbss)." Answers: (a) BS, Bs, bS, bs; (b) same statement; (c) 9 black-short : 3 black-long : 3 white-short : 1 white-long (9:3:3:1); (d)(i) 3/4 × 3/4 = 9/16; (ii) 1/4 × 1/4 = 1/16. Same rubric.

HOW TO RUN IT (with me, the student):
- Greet me in 1–2 sentences, ask my FIRST NAME, then give Problem 1 exactly as written. (NAME FALLBACK: if I answer without giving my name, keep going, but ask before the final report.)
- ONE problem at a time. Never show the whole set, the answers, the rubrics, or the variants.
- AFTER I ANSWER each problem:
• Grade my answer against that problem's rubric and state the score plainly ("That earns 20 of 24"). Judge MEANING for words, but be STRICT on the arithmetic — a wrong box, a wrong ratio, or an unreduced/incorrect fraction loses the points tied to it, against the pre-verified values above.
• Say specifically what I got right, then TEACH the gap — re-draw the boxes or re-do the fraction so I actually learn (full feedback is the point of this assignment).
• OFFER A RE-ATTEMPT: "Want to raise your score? I'll give you a similar problem." If I say yes, deliver the FRESH VARIANT (not the same problem), grade it, and set this problem's score to my BEST attempt (capped at full marks). I can retry as many times as I want.
• Move on when I'm satisfied.
- If I ask about the material, answer briefly, then return to the current problem. If I go off-topic, one friendly sentence, then — IN THE SAME MESSAGE — back to the problem.
- Until the final report, every message ends with a problem, a question, or a clear next step.
- Score HONESTLY against the rubric — don't inflate to be nice, and don't lowball; a wrong ratio scores low, a fully worked correct cross earns full marks. Grade only against the vetted key above.

COMPLETION + REPORT. After I've finished all four problems (and any re-attempts), produce the report in EXACTLY this format — the FIRST LINE is my score:
STUDENT'S SCORE: X/100
WEEK 11 ASSIGNMENT — Work the Cross
Student: [name] | Date: ___
Problem 1 (Monohybrid Punnett square): a/24 — [one line]
Problem 2 (Probability & the product rule): b/26 — [one line]
Problem 3 (The test cross): c/24 — [one line]
Problem 4 (Dihybrid cross & independent assortment): d/26 — [one line]
Strongest skill: ___
Worth another look: ___
(The four problem scores must add up to the number on line 1.) Then say, verbatim: "Copy this entire report AND your share link to this chat, and submit both in Canvas for this assignment." End with one genuine sentence of encouragement.

GETTING STARTED
Begin now: greet me, ask my first name, and give me Problem 1.

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ COPY EVERYTHING ABOVE THIS LINE ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Instructor grading note (Prof. Castellano)

• Record the STUDENT'S SCORE: X/100 from line 1 of the submitted report into the Assignments group.
• Spot-check a sample of chat share links against the reported scores; the embedded vetted key (with every Punnett square and fraction pre-computed) means the coach grades the same way for every student and every chatbot, so checks are quick. Glance especially at the dihybrid (Problem 4) — that's where a chatbot is most likely to mis-grade.
• The answer key + rubric live inside the student prompt (embed-don't-trust), and every genetics value (1:2:1, 3:1, 1/4, 3/4, 1/2, 9:3:3:1, 9/16, 1/16) was re-derived by an independent Python check (quantitative gate: PASS), so the score is consistent across Gemini / Claude / ChatGPT. Known weak point (H5/H7): an AI-self-scored grade submitted by share link is gameable; this is acceptable here as one assignment among many, but for high-stakes use pair it with an in-class or proctored check.

Canvas placement block

canvas_object    = Assignment
title            = "Week 11 Assignment — Work the Cross (adaptive)"
assignment_group = "Assignments"
points_possible  = 100
grading_type     = points
assignment_type  = adaptive
submission_types = [online_text_entry, online_url]   # paste the report (score on line 1) + the chat share link
due_offset_days  = 5
published        = true
provenance       = "~ Prof. Castellano's edition · Fall 2026 · built with thecoursemaker.com"

Traditional variant — for comparison. This sample course is configured adaptive learning, so its actual Week-11 assignment is the AI-coached, self-scored version in I-assignment-and-rubric-week-11.md. This file shows the same Week-11 skills built the traditional way — the student completes the work and submits it, and the instructor grades against the rubric — so you can see both formats side by side. (Choosing assignment_type = traditional at course setup generates this style instead.)

Course: Introduction to Biology — General Biology I (BIOL 101) · Silver Oak University (fictional sample) · Prof. Castellano
Objective assessed: Objective 6 (segregation; genotype/phenotype; the monohybrid Punnett square; probability; the test cross; the dihybrid 9:3:3:1) · SLO A (predict and interpret quantitative outcomes) · SLO B (connect genotype to phenotype)
Worth 100 points · Assignments group = 15% of the grade

The Assignment

This week's skill is turning parents into predictions: set up a Punnett square, fill every box, read the ratios, and use probability to find the odds. In four parts, you'll work a monohybrid cross, compute probabilities with the product rule, run a test cross, and finish with a dihybrid cross. Show every Punnett-square box and reduce every fraction. Submit your answers as a document upload or text entry in Canvas. You'll be graded on the rubric below — read it before you start.

Part 1 — Monohybrid Punnett square (24 pts). In pea plants, tall (T) is dominant to short (t). Two heterozygous tall plants are crossed: Tt × Tt. (a) List the gametes each parent can make. (b) Fill in all four boxes of the Punnett square. (c) Give the genotype ratio. (d) Give the phenotype ratio. (e) What is the probability that one offspring is short?

Part 2 — Probability & the product rule (26 pts). Still using Tt × Tt: (a) What is the probability an offspring is tall? (b) What is the probability an offspring is heterozygous (Tt)? (c) Show how the product rule gives the probability of a tt (short) offspring, using the chance each parent passes a t. (d) In one or two sentences, explain why a Tt × Tt cross does not guarantee exactly 3 tall and 1 short in a family of four.

Part 3 — The test cross (24 pts). A tall plant could be TT or Tt — you can't tell by looking, so you do a test cross with a short (tt) plant. (a) If the tall plant is Tt, fill the four boxes of Tt × tt and give the phenotype ratio. (b) If the tall plant is TT, fill the four boxes of TT × tt and give the phenotype ratio. (c) Explain how the offspring tell you whether the unknown tall plant was TT or Tt.

Part 4 — Dihybrid cross & independent assortment (26 pts). Tall (T) is dominant to short (t) and yellow seeds (Y) are dominant to green (y). Two plants heterozygous for both traits are crossed: TtYy × TtYy. (a) List the four kinds of gametes each parent makes. (b) State the law of independent assortment in one sentence. (c) Give the phenotype ratio of the offspring (the four classes). (d) Using the product rule, find the probability of an offspring that is (i) tall AND yellow (both dominant) and (ii) short AND green (ttyy, both recessive).

Integrity & AI note. This is your own work, submitted for grading. You may use an approved chatbot (Gemini, Claude, or ChatGPT) to help you think — but submitting AI-generated answers as your own is not allowed, and you should re-draw every Punnett square and re-check every fraction yourself (chatbots garble these). If AI helped you think, add a one-line note of which tool and how. (Note: this is the traditional format. In this course's actual adaptive assignment, you work the problems with the chatbot and submit its self-scored report — see I-assignment-and-rubric-week-11.md.)

Rubric — 100 points

Levels describe observable differences so grading stays fast and consistent. (This same rubric is what the adaptive variant embeds for the AI to grade against.)

Instructor answer key — REMOVE BEFORE PUBLISHING TO STUDENTS

(All Punnett squares fully worked; every number pre-computed and independently re-verified — quantitative gate: PASS.)

• Part 1 (Tt × Tt):
• (a) Each parent makes T or t.

• (b) Fill the grid:

  	T	t
  T	TT	Tt
  t	Tt	tt

• (c) Genotype ratio = 1 TT : 2 Tt : 1 tt.

• (d) Phenotype ratio = 3 tall : 1 short (3:1) — any T is tall. (1:2:1 is the genotype ratio, a common error.)
• (e) P(short) = 1/4 (the one tt box; equivalently 1/2 × 1/2 = 1/4).
• Part 2 (Tt × Tt):
• (a) P(tall) = 3/4 (three of four boxes have at least one T).
• (b) P(Tt) = 2/4 = 1/2.
• (c) Each parent passes t with probability 1/2; independent events multiply, so P(tt) = 1/2 × 1/2 = 1/4, matching the single tt box.
• (d) 3:1 is a probability (a long-run expectation), like coin flips. With only four offspring, chance can give all tall, 2-and-2, etc.; the ratio is reliable across many offspring, not guaranteed in a small family.
• Part 3 (test cross):

• (a) Tt × tt:

  	t	t
  T	Tt	Tt
  t	tt	tt

  → 2 tall : 2 short = 1:1 (50% / 50%).
  - (b) TT × tt:

  	t	t
  T	Tt	Tt
  T	Tt	Tt

  → all Tt, 100% tall.
  - (c) If any short offspring appear, the unknown carried a hidden t → it was Tt. If all offspring are tall → it was TT. A homozygous-recessive partner exposes a hidden recessive allele.
  - Part 4 (TtYy × TtYy):
  - (a) Gametes: TY, Ty, tY, ty (four kinds, each 1/4).
  - (b) Independent assortment: alleles for different genes separate into gametes independently (different homologous pairs assort independently in meiosis).
  - (c) Phenotype ratio = 9 tall-yellow : 3 tall-green : 3 short-yellow : 1 short-green = 9 : 3 : 3 : 1.
  - (d) (i) P(tall AND yellow) = 3/4 × 3/4 = 9/16; (ii) P(short AND green, ttyy) = 1/4 × 1/4 = 1/16. (The dihybrid 9:3:3:1 is two independent 3:1 crosses multiplied.)

Canvas placement block

canvas_object    = Assignment
title            = "Week 11 Assignment — Work the Cross (traditional)"
assignment_group = "Assignments"
points_possible  = 100
grading_type     = points
assignment_type  = traditional
submission_types = [online_upload, online_text_entry]
due_offset_days  = 5
published        = true
rubric_ref       = "week-11-assignment-rubric"
provenance       = "~ Prof. Castellano's edition · Fall 2026 · built with thecoursemaker.com"