Mendelian Genetics
AP Biology· AP Biology CED — Heredity· 14 min read
1. Core Concepts & Mendel's Laws of Inheritance★★☆☆☆⏱ 3 min
Mendelian genetics is the study of discrete single-gene trait inheritance, derived from Gregor Mendel’s pea plant experiments. It is the foundational framework for all modern genetics, making up 8-10% of AP Biology Unit 5 exam weight, appearing regularly in both MCQs and FRQs.
Mendelian Genetics
The study of trait inheritance that follows Mendel’s two core laws, applies to unlinked single-gene traits.
Example:
Seed shape in pea plants is a classic Mendelian trait.
Mendel derived two core laws, both directly rooted in homologous chromosome behavior during anaphase I of meiosis:
Law of Segregation: Two alleles for a single trait separate during gamete formation, so each gamete receives only one allele.
Law of Independent Assortment: Alleles of different genes assort independently of one another during gamete formation, only true for unlinked genes on separate non-homologous chromosomes.
Two core probability rules are used to apply Mendel's laws: the product rule for independent events, and the sum rule for mutually exclusive events.
For a monohybrid cross between two heterozygous plants (Aa × Aa), what is the probability an offspring will be heterozygous?
- 1
Apply the Law of Segregation: each parent produces 50% A gametes and 50% a gametes:
- 2
- 3
Identify two mutually exclusive ways to get a heterozygous offspring: (A from parent 1, a from parent 2) OR (a from parent 1, A from parent 2).
- 4
Apply the product rule to each outcome:
- 5
- 6
Apply the sum rule to get total probability:
- 7
Exam tip:
On FRQ questions asking why segregation or independent assortment occur, always connect the process to anaphase I of meiosis (separation of homologous chromosomes) — this is required for full credit.
2. Punnett Square Analysis★★☆☆☆⏱ 3 min
A Punnett square is a visual tool to organize all possible gamete combinations from two parents and calculate expected genotype and phenotype ratios of offspring. For a monohybrid cross (one gene, two alleles), the 2×2 Punnett square produces 4 equally likely outcomes, each with 25% probability.
Monohybrid Cross
A cross between two individuals that are heterozygous for a single gene of interest.
Example:
Aa × Aa is a standard monohybrid cross that produces a 3:1 phenotypic ratio with complete dominance.
3. Dihybrid Cross Calculations★★★☆☆⏱ 4 min
For dihybrid crosses (two unlinked genes), a double heterozygote (AaBb) produces four gamete types (AB, Ab, aB, ab) each at 25% frequency, which can be organized into a 4×4 Punnett square. A faster, more error-resistant shortcut is to split the cross into two separate monohybrid crosses, calculate the desired probability for each gene, then multiply using the product rule.
The well-known 9:3:3:1 phenotypic ratio only applies to dihybrid crosses between two double heterozygotes (AaBb × AaBb) with complete dominance and unlinked genes. Never assume this ratio applies to all dihybrid crosses.
In pea plants, round seeds (R) are dominant to wrinkled (r), and yellow seeds (Y) are dominant to green (y). The genes assort independently. Cross RrYy × rrYy. What is the probability of getting a wrinkled, yellow seed?
- 1
Split the dihybrid cross into two independent monohybrid crosses: seed shape (Rr × rr) and seed color (Yy × Yy).
- 2
Calculate the probability of each desired phenotype: Wrinkled is recessive, so from Rr × rr. Yellow is dominant, so from Yy × Yy.
- 3
Apply the product rule for independent events:
- 4
- 5
A full 4×4 Punnett square confirms 6 out of 16 boxes match the desired phenotype, equal to .
Exam tip:
Never assume a 9:3:3:1 ratio applies to any dihybrid cross. Always calculate ratios from scratch for any parental combination other than AaBb × AaBb.
4. Test Cross Design & Interpretation★★★☆☆⏱ 4 min
A test cross is a diagnostic cross used to determine the genotype of an individual with a dominant phenotype. A dominant phenotype can come from either a homozygous dominant (AA) or heterozygous (Aa) genotype, so phenotype alone cannot reveal genotype.
Test Cross
A cross between an individual with an unknown dominant genotype and a homozygous recessive tester individual for the same trait.
Example:
Crossing a red-flowered unknown snapdragon with a homozygous recessive white-flowered snapdragon.
The homozygous recessive tester can only pass a recessive allele to offspring, so offspring phenotype depends entirely on the allele inherited from the unknown parent. If any offspring show the recessive phenotype, the unknown parent must be heterozygous. If all offspring show the dominant phenotype, the unknown parent is almost certainly homozygous dominant. For dihybrid crosses, a test cross (AaBb × aabb) can test for independent assortment: a 1:1:1:1 ratio confirms independent assortment, while deviation indicates linkage.
A snapdragon with red flowers (dominant, R) is test-crossed with a white-flowered snapdragon (rr). Out of 120 total offspring, 57 have red flowers and 63 have white flowers. What is the genotype of the original red parent?
- 1
The test cross (white-flowered) parent can only contribute an r allele to all offspring, so offspring phenotype is determined entirely by the allele inherited from the red parent.
- 2
If the red parent were homozygous dominant (RR), all offspring would inherit an R allele, resulting in 100% red-flowered offspring. The observed result does not match this, so RR is eliminated.
- 3
If the red parent were heterozygous (Rr), half of its gametes are R and half are r, so half of offspring will be Rr (red) and half will be rr (white), a 1:1 expected ratio.
- 4
The observed ratio (57:63) is very close to 1:1, with small deviation from random chance, so the original red parent is Rr.
Exam tip:
To get full credit for test cross reasoning on FRQs, always explicitly state that the tester parent is homozygous recessive and only contributes recessive alleles to offspring.
5. AP-Style Practice Check★★★☆☆⏱ 3 min
Test your understanding of Mendelian genetics with these AP-style practice questions:
In rabbits, black fur (B) is dominant to brown fur (b), and long fur (L) is dominant to short fur (l). The two genes assort independently. You cross two BbLl rabbits. What is the probability that an offspring will have the genotype Bbll?
1/16
1/8
3/16
1/4
Reveal answer
1 —Split the cross into two monohybrid crosses: , . Multiply for independent events: .
Tay-Sachs disease is an autosomal recessive Mendelian disorder. Two heterozygous carriers (Tt) have two children. What is the probability that exactly one of their two children will have Tay-Sachs disease?
1/16
1/4
3/8
1/2
Reveal answer
2 —Probability of one child being affected is , unaffected is . Two mutually exclusive outcomes: (affected first, unaffected second) or (unaffected first, affected second). Each outcome is , sum to .
6. Common Pitfalls
Wrong move:
Assuming the 9:3:3:1 phenotypic ratio applies to any dihybrid cross, regardless of parent genotypes.
Why:
Students memorize this ratio for dihybrid crosses and forget it only arises from one specific parental combination (AaBb × AaBb, unlinked, complete dominance).
Correct move:
Always split dihybrid crosses into two separate monohybrid crosses and apply the product rule, instead of relying on memorized ratios, unless the cross is confirmed to be AaBb × AaBb.
Wrong move:
Forgetting to add probabilities for multiple mutually exclusive ways to get a genotype, e.g., calculating P(Aa) from Aa × Aa as 1/4 instead of 1/2.
Why:
Students confuse the product rule and sum rule, and forget heterozygous genotype can form two different ways (A from mom/a from dad, or a from mom/A from dad).
Correct move:
Always list all mutually exclusive ways to get the desired outcome, calculate each probability with the product rule, then add them with the sum rule.
Wrong move:
Stating that segregation or independent assortment occurs in anaphase II of meiosis.
Why:
Students mix up homologous chromosome separation (anaphase I) and sister chromatid separation (anaphase II).
Correct move:
Always associate both segregation and independent assortment with anaphase I of meiosis, when homologous chromosomes separate.
Wrong move:
Using a heterozygous individual as the tester in a test cross.
Why:
Students remember test crosses determine unknown genotypes but forget the required tester genotype is homozygous recessive.
Correct move:
Always use a homozygous recessive individual as the tester parent, because only it contributes exclusively recessive alleles to offspring.
Wrong move:
Applying independent assortment to two genes located close together on the same chromosome.
Why:
Students forget the core condition required for independent assortment to hold.
Correct move:
Only apply independent assortment to genes on separate non-homologous chromosomes or genes far apart on the same chromosome where crossing over regularly separates them.
7. Quick Reference Cheatsheet
Category | Rule / Ratio | Notes |
|---|---|---|
Law of Segregation | Two alleles separate during gamete formation; each gamete gets one allele | Occurs in anaphase I of meiosis, all autosomal genes |
Law of Independent Assortment | Alleles of different genes sort independently into gametes | Only applies to unlinked genes (separate non-homologous chromosomes) |
Product Rule | Used for independent events (both outcomes must occur) | |
Sum Rule | Used for mutually exclusive events (either outcome can occur) | |
Monohybrid Cross (Aa × Aa) | Genotype: 1 AA : 2 Aa : 1 aa | Only applies with complete dominance |
Dihybrid Cross (AaBb × AaBb) | Phenotype: 9:3:3:1 | Only applies to unlinked genes with complete dominance |
Test Cross | Cross unknown dominant × homozygous recessive | If any offspring are recessive, unknown is heterozygous |
When this came up on past exams
AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.
- 2024 · AP Biology
FRQ on dihybrid cross probability
- 2023 · AP Biology
MCQ on test cross interpretation
Going deeper
What's Next
Mastering Mendelian genetics is the non-negotiable foundation for all other inheritance topics in AP Biology Unit 5: Heredity. Immediately after this topic, you will learn non-Mendelian genetics, which covers exceptions to Mendel’s rules including linked genes, incomplete dominance, codominance, polygenic inheritance, and sex linkage. Without a solid understanding of Mendelian probability and ratio calculation, you will not be able to analyze or predict inheritance patterns for these non-Mendelian traits, which are commonly tested on AP Biology FRQs. Mendelian genetics also forms the base for population genetics in Unit 7, where Hardy-Weinberg calculations rely on the same probability rules you learned here.