Mendelian Genetics — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Mendel’s laws of segregation and independent assortment, Punnett square analysis, probability rules (product/sum), test cross design, and genotype/phenotype ratio calculation for monohybrid and dihybrid crosses.

You should already know: Chromosome behavior during meiosis, basic definitions of allele, genotype, and phenotype, gamete formation process.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Mendelian Genetics?

Mendelian genetics is the study of discrete, single-gene trait inheritance following the rules first experimentally derived by Gregor Mendel from his controlled pea plant breeding work. It is the foundational framework for all modern genetics, and per the AP Biology Course and Exam Description (CED), it makes up 8-10% of the total Unit 5 (Heredity) exam weight. This topic appears regularly in both MCQ sections (typically 2-4 questions per exam) and as a core component of longer FRQs, where it is often paired with non-Mendelian extensions or pedigree analysis. Standard AP Biology notation for Mendelian genetics uses uppercase letters for dominant alleles, lowercase for recessive alleles: homozygous dominant = AA, homozygous recessive = aa, heterozygous = Aa. Mendelian genetics is sometimes called "classical genetics" to contrast it with non-Mendelian inheritance patterns that violate its core rules, like linked genes or polygenic traits. Mastery of Mendelian probability and ratio rules is required for almost all other inheritance topics on the AP exam.

2. Mendel’s Core Laws of Inheritance

Mendel derived two foundational laws from his experiments, both directly rooted in the behavior of homologous chromosomes during meiosis (a connection not confirmed until decades after Mendel’s work). The Law of Segregation states that the two alleles for a single trait separate during gamete formation, so each gamete receives only one allele for each gene. This aligns with the separation of homologous chromosomes during anaphase I of meiosis: each homologous chromosome carries one allele, and they end up in separate daughter cells after meiosis I. The Law of Independent Assortment states that alleles of different genes assort independently of one another during gamete formation. This means the allele a gamete receives for gene A does not influence the allele it receives for gene B, and this only holds for genes on separate non-homologous chromosomes (or genes far apart on the same chromosome). To apply these laws, we use two core probability rules: the product rule (for independent events: the probability of two events occurring together equals the product of their individual probabilities) and the sum rule (for mutually exclusive events: the probability of either event occurring equals the sum of their individual probabilities).

Worked Example

Problem: 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 ($P(A)=1/2$) and 50% a gametes ($P(a)=1/2$).
2. Identify the 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).
3. Apply the product rule to each outcome: $P(A\text{ from 1 and }a\text{ from 2})=(1/2)(1/2)=1/4$, and $P(a\text{ from 1 and }A\text{ from 2})=(1/2)(1/2)=1/4$.
4. Apply the sum rule to get total probability: $1/4+1/4=1/2$.

Exam tip: On FRQ questions that ask why segregation/assortment occur, always connect the process to anaphase I of meiosis (separation of homologous chromosomes) — this is the reasoning AP graders require for full credit.

3. Punnett Square Analysis for Crosses

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 Punnett square has 2 rows (gametes from parent 1) and 2 columns (gametes from parent 2), for 4 total equally likely boxes, each representing 25% probability. For a dihybrid cross (two unlinked genes), the Law of Independent Assortment tells us that a double heterozygote (AaBb) produces four gamete types (AB, Ab, aB, ab) each at 25% frequency, leading to a 4×4 Punnett square with 16 boxes. A faster shortcut for dihybrid calculations is to split the cross into two separate monohybrid crosses, calculate the probability of the desired outcome for each gene, then multiply the probabilities with the product rule. This avoids the work of drawing 16 boxes and reduces error. Common memorized ratios for standard crosses are 3:1 phenotypic ratio for Aa × Aa, 1:2:1 genotypic ratio for Aa × Aa, and 9:3:3:1 phenotypic ratio for AaBb × AaBb (with complete dominance).

Worked Example

Problem: 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 $P(rr)=1/2$ from Rr × rr. Yellow is dominant, so $P(Y\_)=3/4$ from Yy × Yy.
3. Apply the product rule for independent events: $P(\text{wrinkled and yellow})=P(rr)\times P(Y\_)=(1/2)(3/4)=3/8$.
4. Verify: A full 4×4 Punnett square gives 6 out of 8 boxes with the desired phenotype, confirming the result.

Exam tip: Never assume a 9:3:3:1 ratio applies to any dihybrid cross. The ratio only comes from a cross of two double heterozygotes for unlinked genes with complete dominance. Always calculate ratios from scratch for any other parent combination.

4. The Test Cross

A test cross is a diagnostic cross used to determine the genotype of an individual with a dominant phenotype. Because a dominant phenotype can result from either a homozygous dominant (AA) or heterozygous (Aa) genotype, phenotype alone cannot reveal the underlying genotype. To resolve this uncertainty, we cross the unknown dominant individual with a homozygous recessive individual for the same trait. The homozygous recessive tester can only pass a recessive (a) allele to offspring, so the phenotype of offspring depends entirely on the allele passed from the unknown parent. If any offspring show the recessive phenotype, the unknown parent must be heterozygous: a homozygous recessive offspring must inherit one recessive allele from each parent, so the unknown parent must carry a recessive allele. If all offspring show the dominant phenotype, the unknown parent is almost certainly homozygous dominant. For two genes, a test cross (AaBb × aabb) is also used to test for independent assortment: a 1:1:1:1 phenotypic ratio confirms independent assortment, while a deviation from this ratio indicates linkage.

Worked Example

Problem: 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. Recall that the test cross (white-flowered) parent can only contribute an r allele to all offspring, so offspring phenotype is determined by the allele inherited from the red parent.
2. If the red parent were homozygous dominant (RR), all offspring would inherit an R from the red parent, resulting in all 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. This is the key justification that AP graders look for.

5. Common Pitfalls (and how to avoid them)

• 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. 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 Aa can form two different ways. Correct move: Always list all mutually exclusive ways to get the desired outcome, calculate each probability with the product rule, then add them.
• 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.
• 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. Correct move: Always use a homozygous recessive individual as the tester parent, because only it contributes exclusively recessive alleles.
• Wrong move: Applying independent assortment to two genes located close together on the same chromosome. Why: Students forget the core condition for independent assortment. 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.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

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? A) 1/16 B) 1/8 C) 3/16 D) 1/4

Worked Solution: First, split the cross into two independent monohybrid crosses: Bb × Bb for fur color and Ll × Ll for fur length. Calculate $P(Bb)$ from Bb × Bb: there are two mutually exclusive ways to get Bb, so $P(Bb)=(1/2\times 1/2)+(1/2\times 1/2)=1/2$. Next, calculate $P(ll)$ from Ll × Ll, which is 1/4. Apply the product rule for independent events: $1/2\times 1/4=1/8$. The correct answer is B.

Question 2 (Free Response)

In tomato plants, red fruit (R) is dominant to orange fruit (r), and hairy stems (H) is dominant to smooth stems (h). The two genes assort independently. A horticulturist crosses RrHh × rrHh. (a) Calculate the expected phenotypic ratio of offspring from this cross. Show your working. (b) The horticulturist grows 200 offspring from this cross, and counts 87 red hairy, 28 red smooth, 72 orange hairy, 13 orange smooth. Do these observed results match the expected Mendelian ratio? Justify your answer. (c) Explain why independent assortment applies to these two genes, and what must be true about their genomic location.

Worked Solution: (a) Split the cross into two monohybrid crosses:

• Rr × rr: $1/2R\_\text{ (red)}:1/2rr\text{ (orange)}$
• Hh × Hh: $3/4H\_\text{ (hairy)}:1/4hh\text{ (smooth)}$ Multiply probabilities to get expected proportions: Red hairy: $(1/2)(3/4)=3/8$, Red smooth: $(1/2)(1/4)=1/8$, Orange hairy: $(1/2)(3/4)=3/8$, Orange smooth: $(1/2)(1/4)=1/8$. Expected phenotypic ratio: 3 red hairy : 1 red smooth : 3 orange hairy : 1 orange smooth.

(b) Expected counts out of 200: 75 red hairy, 25 red smooth, 75 orange hairy, 25 orange smooth. The observed counts are within ~15 of expected counts, and small deviations are due to random chance in gamete fertilization. The observed results match the expected Mendelian ratio.

(c) Independent assortment applies because the two genes are located on separate non-homologous chromosomes. During anaphase I of meiosis, each pair of homologous chromosomes aligns and separates independently of other pairs, so the allele for fruit color and the allele for stem texture sort into gametes independently of one another.

Question 3 (Application / Real-World Style)

Tay-Sachs disease is an autosomal recessive disorder that follows Mendelian inheritance. Two heterozygous carriers (Tt) have two children. What is the probability that exactly one of their two children will have Tay-Sachs disease?

Worked Solution: First, from a Tt × Tt cross, the probability of any single child having Tay-Sachs (genotype tt) is $1/4$, and the probability of a child not having Tay-Sachs is $3/4$. There are two mutually exclusive outcomes for exactly one affected child: (first child affected, second unaffected) OR (first child unaffected, second affected). Calculate the probability of one outcome: $(1/4)(3/4)=3/16$. Add the probabilities of the two mutually exclusive outcomes: $3/16+3/16=6/16=3/8=0.375$. This means there is a 37.5% chance that exactly one out of two children born to two carrier parents will have Tay-Sachs disease.

7. Quick Reference Cheatsheet

8. What's Next

Mastering Mendelian genetics is the non-negotiable foundation for all other inheritance topics in 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.

• Non-Mendelian Genetics
• Pedigree Analysis
• Meiosis
• Population Genetics