Science Class 2027: Check Your Genetics, Chemistry and Statistics Homework

1. Punnett Squares (Biology 2e, Chapter 12)

Problem: cross two pea plants heterozygous for seed color, Yy × Yy, then two plants heterozygous for color and shape, YyRr × YyRr.

1. Gametes: Yy makes Y and y. YyRr makes four: YR, Yr, yR, yr.
2. Monohybrid: the four boxes are YY, Yy, Yy, yy, so the genotype ratio is 1 : 2 : 1 and the phenotype ratio 3 yellow : 1 green.
3. Dihybrid: 4 × 4 = 16 boxes, 9 genotypes, and phenotypes 9 : 3 : 3 : 1. Shortcut: each gene is 3/4 dominant and 1/4 recessive, so 3/4 × 3/4 = 9/16, 3/4 × 1/4 = 3/16 (twice) and 1/4 × 1/4 = 1/16.
4. Check against real data: Mendel counted 6,022 yellow and 2,001 green F2 seeds, a 3.01 : 1 ratio. A chi-square test against 3 : 1 gives χ² = 0.01 (p = 0.90), a near-perfect fit.

Common mistakes: putting two alleles of one gene in a gamete (YY instead of YR), mixing up genotype and phenotype ratios, and expecting exact ratios in small families. Check with the Punnett square calculator; the chi-square step is on the chi-square calculator.

Mendel’s pea traits behave like single genes, but many human traits don’t. Eye color is the classic example: one HERC2 variant explains most blue-versus-brown differences, other genes tune the shade, and in Sturm and colleagues’ 2008 data about a quarter of people with two blue-associated copies still had green or hazel eyes. The baby eye color calculator works from those published counts instead of a one-gene chart.

2. Hardy-Weinberg (Biology 2e, Section 19.1)

Problem: 500 pea plants: 245 YY, 210 Yy and 45 yy. Find p, q and the expected genotype frequencies.

1. Count alleles, not plants: 1,000 alleles; Y = 2 × 245 + 210 = 700, so p = 0.7; y = 300, so q = 0.3.
2. Expected: p² = 0.49, 2pq = 0.42, q² = 0.09, or 245, 210 and 45 plants: exactly what was observed.
3. Testing a real sample: Cornell’s statistics handout uses 725 AA, 123 AB and 7 BB. p = 0.9199, expected counts 723.49, 126.02 and 5.49, χ² = 0.49 with 1 degree of freedom, p-value 0.48: consistent with equilibrium. (The handout prints a p-value of 0.104; for χ² = 0.49 on 1 degree of freedom the correct upper-tail value is about 0.48. The conclusion doesn’t change.)

Common mistakes: treating the recessive phenotype frequency as q instead of q² (0.09 means q = 0.3), dividing by individuals instead of alleles, and using 2 degrees of freedom instead of 1. Check with the Hardy-Weinberg calculator.

3. Limiting Reactant and Percent Yield (Chemistry 2e, Section 4.4)

Problem A: 2.00 g of Si and 1.50 g of N₂ react: 3Si + 2N₂ → Si₃N₄. Which is limiting?

1. Moles: 2.00 ÷ 28.086 = 0.0712 mol Si; 1.50 ÷ 28.014 = 0.0535 mol N₂.
2. Product each could make: 0.0712 ÷ 3 = 0.0237 mol Si₃N₄; 0.0535 ÷ 2 = 0.0268 mol. Silicon makes less, so it is limiting.
3. Theoretical yield: 0.0237 mol × 140.286 g/mol = 3.33 g of Si₃N₄.

Problem B: 1.274 g of CuSO₄ with excess zinc gives 0.392 g of copper. Theoretical yield: 1.274 ÷ 159.602 × 63.546 = 0.5072 g; percent yield 0.392 ÷ 0.5072 × 100 = 77.3%, matching OpenStax’s 77.3%.

Common mistakes: choosing the limiting reactant by grams (in OpenStax’s hydrogen problem, 10.0 g of O₂ is only 0.313 mol and runs out before 5.00 g of H₂, which is 2.48 mol), skipping the mole ratio, and dividing theoretical by actual. Check with the percent yield calculator, which balances the equation first.

4. One-Way ANOVA (Introductory Statistics 2e, Section 13.3)

Problem: grade averages from five members of each of four sororities. Do the mean grades differ (α = 0.01)?

1. Grand mean of all 20 values: 2.738.
2. Between-group sum of squares: Σ n(group mean − grand mean)² = 2.887, with k − 1 = 3 degrees of freedom, so MS_between = 0.962.
3. Within-group sum of squares: Σ (value − group mean)² = 6.904, with N − k = 16 degrees of freedom, so MS_within = 0.432.
4. F = 0.962 ÷ 0.432 = 2.23. OpenStax gives F = 2.23 and p = 0.1241, more than 0.01, so there is not enough evidence that the mean grades differ.

Common mistakes: running six separate t-tests instead of one ANOVA (which inflates the false-positive rate), mixing up the two degrees of freedom, and reading a significant F as “every group differs” when it only says at least one does. Check with the ANOVA calculator; for two groups, the t-test calculator is the simpler tool.

5. The Poisson Distribution (Introductory Statistics 2e, Section 4.6)

Problem: Leah receives about six calls between 8 and 10 a.m. What is the probability of more than one call in the next 15 minutes?

1. Rescale the mean to the interval: 6 calls per 120 minutes is 6 × 15 ÷ 120 = 0.75 calls per 15 minutes.
2. P(0) = e^−0.75 = 0.4724; P(1) = 0.75 × e^−0.75 = 0.3543.
3. P(more than 1) = 1 − P(0) − P(1) = 0.1734, OpenStax’s 0.1734.

The bakery example in the same section (12 loaves per half hour, so 2 per five minutes) asks for exactly three loaves: P(3) = e⁻² × 2³ ÷ 3! = 0.1804.

Common mistakes: using the 2-hour mean (6) for a 15-minute question, computing P(x = 1) when the question says “more than one”, and using Poisson for events that aren’t independent or have a fixed number of trials (that is binomial). Check with the Poisson distribution calculator.

6. Empirical and Molecular Formulas (Chemistry 2e, Section 3.2)

Problem: nicotine is 74.02% C, 8.710% H and 17.27% N, and 40.57 g of it is 0.2500 mol. Find the empirical and molecular formulas.

1. Use a 100 g sample: 74.02 g C ÷ 12.011 = 6.1627 mol; 8.710 g H ÷ 1.008 = 8.6409 mol; 17.27 g N ÷ 14.007 = 1.2330 mol.
2. Divide by the smallest (1.2330): C 4.998, H 7.008, N 1.000, so the empirical formula is C₅H₇N (81.12 g/mol with these atomic masses; OpenStax gives 81.13).
3. Molar mass: 40.57 g ÷ 0.2500 mol = 162.3 g/mol; 162.3 ÷ 81.12 ≈ 2, so the molecular formula is C₁₀H₁₄N₂.

Common mistakes: dividing grams instead of moles, rounding a ratio like 1.5 to 2 instead of doubling everything (OpenStax’s hematite example, Fe 1 : O 1.5, is Fe₂O₃), and stopping at the empirical formula when a molar mass is given. Check with the empirical formula calculator.

7. Population Growth (Biology 2e, Section 45.3)

Problem: 1,000 bacteria double every hour with unlimited food. How many after a day? And how does a carrying capacity change things?

1. Exponential: N = N₀ × 2^(t ÷ doubling time) = 1,000 × 2²⁴ = 16,777,216,000, the “more than 16 billion” in the textbook. In continuous form, N = N₀e^(rt) with r = ln 2 ÷ 1 hour = 0.693.
2. Logistic: dN/dt = rN(K − N)/K. OpenStax Calculus Volume 2 models deer with N₀ = 900,000, r = 0.2311 and K = 1,072,764: after 3 years N = K N₀e^(rt) ÷ (K − N₀ + N₀e^(rt)) ≈ 978,830, against 1.8 million if the herd simply doubled.

Common mistakes: mixing time units between r and t, entering a percentage (23.11) where the formula wants a rate (0.2311), and letting a logistic population overshoot K. Check with the population growth calculator.

8. Polynomial Division (College Algebra 2e, Section 5.4)

Problem: divide 6x³ + 11x² − 31x + 15 by 3x − 2.

1. 6x³ ÷ 3x = 2x²; 2x²(3x − 2) = 6x³ − 4x²; subtract to get 15x² − 31x + 15.
2. 15x² ÷ 3x = 5x; 5x(3x − 2) = 15x² − 10x; subtract to get −21x + 15.
3. −21x ÷ 3x = −7; −7(3x − 2) = −21x + 14; subtract to get 1.
4. Answer: 2x² + 5x − 7 with remainder 1, or 2x² + 5x − 7 + 1/(3x − 2). Check by multiplying back.

Common mistakes: leaving out a 0 placeholder for a missing power, subtracting only the first term of each product, and using synthetic division with the wrong sign (for x + 2, use −2). Check with the polynomial division calculator.

A Quick Checklist Before You Hand It In

• Units and bases: alleles not individuals, moles not grams, the right time interval.
• Ratios: does the equation balance, do the genotype frequencies add to 1, do the probabilities add to 1?
• Degrees of freedom: categories minus one, minus one more for each parameter estimated from the data.
• Sanity: percent yield under 100%, probabilities between 0 and 1, F and χ² never negative.

Sources

• OpenStax, Biology 2e (2018), sections 12.1–12.3 and 19.1
• OpenStax, Chemistry 2e (2019), sections 3.2 and 4.4
• OpenStax, Biology 2e (2018), section 45.3; Calculus Volume 2 (2016), section 4.4; College Algebra 2e (2021), section 5.4
• OpenStax, Introductory Statistics 2e, sections 4.6 and 13.3 (web edition accessed October 2026)
• Cornell Statistical Consulting Unit, The Hardy-Weinberg Principle in Population Genetics (2013, updated April 2022)

Frequently asked questions

How do I check a Punnett square answer?

Count the boxes: a monohybrid cross of two heterozygotes gives 1 : 2 : 1 genotypes and 3 : 1 phenotypes; a dihybrid gives 9 : 3 : 3 : 1 phenotypes from 16 boxes.

How do you find the limiting reactant?

Convert each reactant to moles, divide by its coefficient (or use the mole ratio to the product); the one that makes the least product is limiting.

Why does Hardy-Weinberg use 1 degree of freedom?

Three genotype classes minus one for the total and one for the allele frequency estimated from the data leaves one.

What does the F statistic in ANOVA mean?

The ratio of variation between group means to variation within groups. A large F with a small p-value suggests at least one group mean differs.

How do you change the Poisson mean for a different time interval?

Scale it in proportion: 6 calls per 2 hours is 0.75 calls per 15 minutes.