AP Biology

AP Biology Unit 5 review: Heredity

Unit 5 explains how meiosis and fertilization pass alleles to offspring and generate variation. It is 8–11% of multiple choice. Expect chromosome models, inheritance probabilities and chi-square evidence, with a clear distinction between genotype and phenotype.

Unit 5 on the May 2027 exam

AP Biology Unit 5 exam facts
2027 exam
Weight8–11% of the multiple-choice section
Exam dateMonday, May 3, 2027, afternoon session
Multiple choice60 questions in 90 minutes, worth 50% of the exam score
Free response6 questions in 90 minutes, worth 50%: two long questions (9 points each) and four short questions (4 points each).
CalculatorBoth sections allow a four-function calculator with square root or a nongraphing scientific calculator. Bluebook includes Desmos scientific. Handheld graphing calculators are not allowed in 2027.
Class timeAbout 8–10 class periods in College Board's pacing guide
2026 results15% of students scored a 5 on the whole exam, and about 71% scored 3 or higher

Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.

What Unit 5 covers

College Board's course description splits the unit into 5 topics. The second column is what you should be able to do for each one.

AP Biology Unit 5 topics
TopicWhat you should be able to do
5.1 MeiosisTrack homologs and sister chromatids through two divisions after one round of DNA replication.
5.2 Meiosis and Genetic DiversityExplain how crossing over, independent assortment and random fertilization generate new allele combinations.
5.3 Mendelian GeneticsUse segregation and independent assortment to calculate offspring probabilities and test expected ratios.
5.4 Non-Mendelian GeneticsInterpret incomplete dominance, codominance, linked genes, sex linkage and non-nuclear inheritance.
5.5 Environmental Effects on PhenotypeUse data to explain how an environmental change can alter phenotype without changing genotype.

From inheritance model to predicted offspring

Identify the inheritance pattern before choosing a ratio. These predictions assume the stated parental genotypes.

From inheritance model to predicted offspring
ModelCross or evidencePrediction
Complete dominanceAa × AaGenotypes 1 AA : 2 Aa : 1 aa; phenotypes 3 dominant : 1 recessive.
Incomplete dominanceAa × Aa, with an intermediate heterozygoteBoth genotype and phenotype ratios are 1:2:1.
Two unlinked lociAaBb × aabbFour offspring classes occur in equal expected proportions.
Linked lociAB/ab × ab/abParental AB and ab combinations exceed each recombinant class when recombination is below 50%.
X-linked recessiveCarrier mother × unaffected fatherHalf of sons are expected to express the trait; no daughters express it under this model.
Chi-square comparisonSum (observed − expected)²/expectedCompare the statistic with the critical value using the appropriate degrees of freedom.

Common mistakes

Each of these loses points on questions you otherwise understand.

  • Confusing genotype and phenotype ratios. Aa × Aa gives 1:2:1 genotypes but 3:1 phenotypes only with complete dominance. The CED flags this error on printed p. 95.
  • Writing the alternative as the null hypothesis. For a Mendelian goodness-of-fit test, state the predicted ratio and that observed deviations can be explained by chance. CED p. 95.
  • Calling a nonsignificant chi-square result proof of the model. Failing to reject means the data do not provide enough evidence against it. CED p. 95.
  • Separating sister chromatids in meiosis I. Homologs separate first; sister chromatids normally separate in meiosis II. CED topic 5.1.

6 practice questions with answers

Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.

  1. Question 1 · 5.1

    A diploid organism has 12 chromosomes. After normal meiosis II, how many chromosomes are in each gamete?

    • A3
    • B6
    • C12
    • D24
    Show the answer

    B. Meiosis reduces the chromosome set from 2n = 12 to n = 6. Replication before meiosis doubles DNA amount, not the number of homologous pairs. Each final gamete has one member of each pair.

  2. Question 2 · 5.3

    Two heterozygous parents, Rr and Rr, have a trait with complete dominance. Among 160 offspring, how many are expected to show the recessive phenotype?

    • A20
    • B80
    • C120
    • D40
    Show the answer

    D. Each parent contributes r with probability 1/2. Thus P(rr) = 1/4 and 160 × 1/4 = 40. The 80 expected heterozygotes show the dominant phenotype.

  3. Question 3 · 5.3

    Genes A and B assort independently. In AaBb × Aabb, what is the probability of an aabb offspring?

    • A1/8
    • B1/4
    • C3/8
    • D1/2
    Show the answer

    A. The A cross gives aa with probability 1/4. The B cross gives bb with probability 1/2. Independence allows multiplication: 1/4 × 1/2 = 1/8.

  4. Question 4 · 5.4

    A testcross of AB/ab with ab/ab produces 82 AB, 78 ab, 21 Ab and 19 aB offspring. What is the recombination frequency?

    • A10%
    • B40%
    • C20%
    • D80%
    Show the answer

    C. The less frequent Ab and aB classes are recombinants. There are 40 among 200 offspring, so recombination frequency is 40/200 × 100 = 20%. Count both recombinant classes.

  5. Question 5 · 5.3

    A model predicts equal counts in two categories. The observed counts are 44 and 56 out of 100. What is χ²?

    • A0.72
    • B1.44
    • C6.00
    • D12.00
    Show the answer

    B. Expected counts are 50 and 50. χ² = (44 − 50)²/50 + (56 − 50)²/50 = 0.72 + 0.72 = 1.44. Differences are squared before dividing by expected counts.

  6. Question 6 · 5.5

    Genetically identical seedlings grow at two light levels. Leaf size differs consistently between groups. Which conclusion is supported?

    • ALight necessarily changed every seedling’s DNA
    • BThe groups must have different inherited alleles
    • CLeaf size cannot have a genetic component
    • DLight affected phenotype within one genotype
    Show the answer

    D. Holding genotype constant helps isolate an environmental effect. The result does not show that genes are irrelevant to leaf growth, and a phenotypic difference alone is not evidence of mutation.

How to study Unit 5

Draw a cell with two homologous pairs and follow each chromosome through meiosis. Use two colors for parental homologs. Mark a crossover between nonsister chromatids and identify which final chromosomes are recombinant.

For each cross, write the gametes and their probabilities before constructing a grid. Work one monohybrid, one independent two-gene and one linked-gene problem. For chi-square, write the null hypothesis, expected counts and degrees of freedom before calculating; then connect the decision to the proposed inheritance model.

Ashore's Biology course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.

Questions

Still have a question? support@tryashore.com

It is 8–11% of the multiple-choice section, which is half the exam score. This is not a fixed share of the whole exam: free-response questions can combine units. Suggested pacing is 8–10 class periods.

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Starting check

SAT Math · 1 of 5

Calculator allowed

A jacket's price is increased by 20% and then the new price is decreased by 20%. The final price is what percent of the original price?
  • A96%96\%
  • B100%100\%
  • C104%104\%
  • D80%80\%