IB Biology How to Solve: IB Biology – Genetics (Monohybrid/Dihybrid Crosses, Sex-Linked, Pedigree, Chi-Squared Test)

How to Solve: IB Biology – Genetics (Monohybrid/Dihybrid Crosses, Sex-Linked, Pedigree, Chi-Squared Test)

Complete Guide

Introduction

"Mastering genetics unlocks 15–20% of your IB Biology Paper 2—including pedigrees that look like spiderwebs, dihybrid crosses with 16 boxes, and chi-squared tests that decide if your data is ‘good enough’ for full marks. One wrong allele, and your entire Punnett square collapses. Let’s fix that."

WHAT YOU NEED TO KNOW FIRST

1. Basic Mendelian genetics: Dominant vs. recessive alleles, homozygous vs. heterozygous.
2. Probability rules: Multiplication rule for independent events (e.g., and = multiply), addition rule for mutually exclusive events (e.g., or = add).
3. Meiosis basics: How gametes form (haploid, random assortment).

KEY TERMS & FORMULAS

Terms

• Allele: Version of a gene (e.g., A or a).
• Genotype: Genetic makeup (e.g., AaBb).
• Phenotype: Observable trait (e.g., "purple flowers").
• Autosomal: Gene on non-sex chromosomes (1–22 in humans).
• Sex-linked: Gene on X or Y chromosome.
• Carrier: Heterozygous for a recessive disorder (e.g., XᴬXᵃ).
• Pedigree symbols:
• Square = male, circle = female.
• Shaded = affected, unshaded = unaffected.
• Half-shaded = carrier (if sex-linked).
• Line through = deceased.

Formulas

1. Chi-squared (χ²) test – MEMORISE THIS [ \chi^2 = \sum \frac{(O - E)^2}{E} ]
2. O = observed count (your data).
3. E = expected count (from Punnett square).

4. ∑ = sum for all categories.

5. Degrees of freedom (df) – MEMORISE THIS [ df = \text{number of categories} - 1 ]

6. For monohybrid crosses: 2 phenotypes → df = 1.

7. For dihybrid crosses: 4 phenotypes → df = 3.

8. Critical value – Given on exam sheet

9. Compare χ² to critical value at p = 0.05 (5% significance).
10. If χ² > critical value → reject null hypothesis (data doesn’t fit expected ratio).
11. If χ² ≤ critical value → fail to reject (data fits).

STEP-BY-STEP METHOD

A. Monohybrid Crosses (1 Gene)

1. Identify parental genotypes (e.g., Aa × Aa).
2. Write gametes (e.g., A, a for both parents).
3. Draw Punnett square (4 boxes).
4. Fill in offspring genotypes (e.g., AA, Aa, Aa, aa).
5. Count phenotypes (e.g., 3 dominant : 1 recessive).
6. Calculate probabilities (e.g., 75% dominant, 25% recessive).

B. Dihybrid Crosses (2 Genes)

1. Identify parental genotypes (e.g., AaBb × AaBb).
2. Use FOIL method to list gametes (e.g., AB, Ab, aB, ab for both parents).
3. Draw 16-box Punnett square.
4. Fill in offspring genotypes (e.g., AABB, AaBb, etc.).
5. Count phenotypes (e.g., 9:3:3:1 ratio for unlinked genes).
6. Check for linkage (if ratio ≠ 9:3:3:1, genes may be linked).

C. Sex-Linked Crosses

1. Identify gene location (X or Y chromosome).
2. Write parental genotypes with sex chromosomes (e.g., XᴬXᵃ × XᴬY).
3. Write gametes (e.g., Xᴬ, Xᵃ for female; Xᴬ, Y for male).
4. Draw Punnett square (4 boxes).
5. Fill in offspring genotypes and sexes (e.g., XᴬXᴬ, XᴬXᵃ, XᴬY, XᵃY).
6. Count phenotypes by sex (e.g., 50% carrier females, 50% affected males).

D. Pedigree Analysis

1. Label generations (I, II, III) and individuals (1, 2, 3…).
2. Determine inheritance pattern:
3. Autosomal dominant: Affected individuals in every generation; no carriers.
4. Autosomal recessive: Skips generations; carriers possible.
5. Sex-linked recessive: More males affected; females can be carriers.
6. Assign genotypes (e.g., aa for affected, Aa for carriers).
7. Check for consistency (e.g., two unaffected parents can’t have an affected child if dominant).

E. Chi-Squared Test

1. State null hypothesis (H₀): "The observed data fits the expected ratio."
2. Calculate expected counts (e.g., 9:3:3:1 → total offspring = 160 → E = 90, 30, 30, 10).
3. Plug into χ² formula: [ \chi^2 = \frac{(O_1 - E_1)^2}{E_1} + \frac{(O_2 - E_2)^2}{E_2} + \dots ]
4. Find degrees of freedom (df = categories - 1).
5. Compare χ² to critical value (from exam sheet).
6. Conclude:
7. χ² > critical value → reject H₀ (data doesn’t fit).
8. χ² ≤ critical value → fail to reject H₀ (data fits).

WORKED EXAMPLES

Example 1 – Monohybrid Cross (Basic)

Question: In pea plants, tall (T) is dominant to short (t). Cross two heterozygous tall plants. What is the phenotypic ratio of the offspring?

Steps:
1. Parental genotypes: Tt × Tt.
2. Gametes: T, t for both.
3. Punnett square: | | T | t | |-------|-------|-------| | T | TT | Tt | | t | Tt | tt |
4. Offspring genotypes: 1 TT : 2 Tt : 1 tt.
5. Phenotypes: 3 tall : 1 short.

What we did and why: We used a Punnett square to predict offspring ratios by combining parental gametes. The 3:1 ratio confirms Mendel’s first law.

Example 2 – Dihybrid Cross (Medium)

Question: In fruit flies, gray body (G) is dominant to black (g), and long wings (L) are dominant to vestigial (l). Cross two flies heterozygous for both traits. What is the phenotypic ratio?

Steps:
1. Parental genotypes: GgLl × GgLl.
2. Gametes (FOIL): GL, Gl, gL, gl for both.
3. Punnett square (16 boxes): - Example offspring: GGLL, GgLl, ggll, etc.
4. Phenotypes: - Gray, long: 9 - Gray, vestigial: 3 - Black, long: 3 - Black, vestigial: 1
5. Ratio: 9:3:3:1.

What we did and why: We used the FOIL method to list all possible gametes, then combined them in a 16-box Punnett square. The 9:3:3:1 ratio shows independent assortment.

Example 3 – Sex-Linked Cross (Exam-Style)

Question: Hemophilia is a sex-linked recessive disorder (Xᴴ = normal, Xʰ = hemophilia). A carrier female marries a normal male. What is the probability their son will have hemophilia?

Steps:
1. Parental genotypes: XᴴXʰ × XᴴY.
2. Gametes: - Female: Xᴴ, Xʰ. - Male: Xᴴ, Y.
3. Punnett square: | | Xᴴ | Y | |-------|--------|--------| | Xᴴ | XᴴXᴴ | XᴴY | | Xʰ | XᴴXʰ | XʰY |
4. Offspring: - Sons: XᴴY (normal), XʰY (hemophiliac).
5. Probability: 1/2 (50%).

What we did and why: We tracked the X and Y chromosomes separately, focusing only on male offspring (XY). The 50% probability comes from the mother’s Xʰ gamete.

Example 4 – Pedigree Analysis (Exam-Style)

Question: The pedigree below shows a family with cystic fibrosis (autosomal recessive). What is the genotype of individual II-3?

I: [◻]──[●]
     |
II: [◻] [◻] [●] [◻]

Steps:
1. Label generations (I, II) and individuals (1–4).
2. Determine inheritance: Autosomal recessive (skips generations, unaffected parents have affected child).
3. Assign genotypes: - Affected (●) = aa. - Unaffected parents (I-1, I-2) must be Aa (carriers).
4. II-3 is affected → aa.

What we did and why: We used the pedigree to deduce genotypes by working backward from affected individuals. The key was recognizing the recessive pattern.

Example 5 – Chi-Squared Test (Exam-Style)

Question: A dihybrid cross (AaBb × AaBb) produces 160 offspring with the following phenotypes: - 85 AB, 35 Ab, 30 aB, 10 ab. Does this fit the expected 9:3:3:1 ratio? Use χ² at p = 0.05.

Steps:
1. Null hypothesis (H₀): Data fits 9:3:3:1.
2. Expected counts (total = 160): - AB: 9/16 × 160 = 90 - Ab: 3/16 × 160 = 30 - aB: 3/16 × 160 = 30 - ab: 1/16 × 160 = 10
3. χ² calculation: [ \chi^2 = \frac{(85-90)^2}{90} + \frac{(35-30)^2}{30} + \frac{(30-30)^2}{30} + \frac{(10-10)^2}{10} = 0.278 + 0.833 + 0 + 0 = 1.111 ]
4. Degrees of freedom: 4 categories - 1 = 3.
5. Critical value (from table, df = 3, p = 0.05): 7.815.
6. Compare: 1.111 < 7.815 → fail to reject H₀.

What we did and why: We calculated χ² to test if the observed data matched the expected ratio. Since χ² was less than the critical value, the data fits the 9:3:3:1 ratio.

COMMON MISTAKES

1. MISTAKE: Forgetting to write all possible gametes in dihybrid crosses. WHY IT HAPPENS: Students stop at AB and ab, missing Ab and aB. CORRECT APPROACH: Use FOIL (First, Outer, Inner, Last) to list all 4 gametes.

2. MISTAKE: Mixing up autosomal and sex-linked inheritance in pedigrees. WHY IT HAPPENS: Assuming all disorders are autosomal recessive. CORRECT APPROACH: Check if more males are affected (sex-linked) or if it skips generations (recessive).

3. MISTAKE: Calculating χ² with percentages instead of counts. WHY IT HAPPENS: Students use 9:3:3:1 directly instead of converting to expected counts. CORRECT APPROACH: Multiply ratios by total offspring to get E values.

4. MISTAKE: Ignoring degrees of freedom in χ². WHY IT HAPPENS: Students compare χ² to the wrong critical value. CORRECT APPROACH: df = categories - 1 (e.g., 4 phenotypes → df = 3).

5. MISTAKE: Mislabeling carriers in sex-linked pedigrees. WHY IT HAPPENS: Assuming females can’t be carriers for dominant disorders. CORRECT APPROACH: Only recessive sex-linked traits have female carriers (XᴬXᵃ).

EXAM TRAPS

1. TRAP: "Linked genes" disguised as dihybrid crosses. HOW TO SPOT IT: Phenotypic ratio ≠ 9:3:3:1 (e.g., 3:1 or 1:1). HOW TO AVOID IT: Check if the question mentions "linked genes" or gives a non-9:3:3:1 ratio.

2. TRAP: Chi-squared with incomplete data. HOW TO SPOT IT: Question gives observed counts but no expected ratio. HOW TO AVOID IT: Assume 3:1 (monohybrid) or 9:3:3:1 (dihybrid) unless stated otherwise.

3. TRAP: Pedigrees with "de novo" mutations. HOW TO SPOT IT: Two unaffected parents have an affected child in a dominant disorder. HOW TO AVOID IT: If the disorder is dominant, the child must have an affected parent. If not, it’s a mutation (rare, but possible).

1-MINUTE RECAP

"Listen up—this is your genetics cheat sheet for exam day. For monohybrid crosses, draw a 4-box Punnett square and count phenotypes. For dihybrids, use FOIL to list gametes, then fill 16 boxes—9:3:3:1 is your golden ratio. Sex-linked? Track X and Y separately; males can’t be carriers. Pedigrees: shade = affected, half-shade = carrier (if sex-linked). Chi-squared? Calculate expected counts, plug into the formula, compare to the critical value. Degrees of freedom = categories minus one. Common traps? Linked genes mess up the 9:3:3:1 ratio, and always use counts, not percentages, for χ². Now go crush that exam."