Mendelian Genetics Practice Problems for the MCAT with Worked Answers

Most MCAT genetics questions reduce to a few moves: write the genotypes, split the cross by gene, and multiply or add probabilities.

Rules you need

• Product rule: independent events multiply. P(A and B) = P(A) × P(B).
• Sum rule: mutually exclusive outcomes add.
• Independent assortment: genes on different chromosomes are inherited independently, so treat each gene separately.

Problem 1: monohybrid

Aa × Aa. Probability of an aa child? 1/4. The dominant phenotype appears in 3/4, giving a 3:1 phenotype ratio and a 1:2:1 genotype ratio.

Problem 2: dihybrid

AaBb × AaBb. Probability of aabb? 1/4 × 1/4 = 1/16. Probability of a child showing both dominant traits? 3/4 × 3/4 = 9/16. The phenotype ratio is 9:3:3:1.

Problem 3: multiple children

Two Aa parents have three children. Probability that exactly one has the recessive phenotype? There are 3 orders, each (1/4)(3/4)(3/4), so 3 × 9/64 = 27/64.

Problem 4: testcross

A plant with the dominant phenotype is crossed with an aa plant. All offspring show the dominant trait: the plant is most likely AA. Half dominant and half recessive: the plant is Aa.

Problem 5: X-linked recessive

A carrier mother (X^AX^a) and an unaffected father (X^AY). Sons: 1/2 affected. Daughters: none affected, 1/2 carriers. Males are affected more often because they have one X.

Common traps

• Using the dihybrid ratio when genes are linked. Linked genes do not assort independently.
• Forgetting to condition on a known phenotype. An unaffected child of two carriers is carrier with probability 2/3.
• Multiplying when you should add. Alternative ways to reach the same outcome add.

One-minute recap

• Split by gene, then multiply.
• Aa × Aa gives 3:1; dihybrid gives 9:3:3:1.
• X-linked recessive: affected sons from carrier mothers.