Non-Mendelian Genetics — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Incomplete dominance, codominance, multiple alleles, polygenic inheritance, epistasis, pleiotropy, sex linkage, linked genes, and cytoplasmic (mitochondrial) inheritance, all core Non-Mendelian patterns tested on the AP Biology exam.

You should already know: Mendel’s laws of segregation and independent assortment, Punnett square construction, chromosomal basis of inheritance.

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 Non-Mendelian Genetics?

Non-Mendelian genetics describes any pattern of inheritance where phenotypes do not segregate according to Mendel’s original rules of complete dominant/recessive alleles, single-gene control of one trait, and autosomal location. Gregor Mendel’s pea plant experiments worked because he studied traits that fit simple Mendelian rules by chance, but most heritable traits in eukaryotes do not follow these patterns. In the AP Biology CED, Non-Mendelian genetics is part of Unit 5 (Heredity), which makes up 8–11% of the total exam score. Non-Mendelian topics account for roughly 3–4% of total exam points, and appear in both multiple choice (MCQ) and free response (FRQ) sections. MCQs typically ask you to identify an inheritance pattern from cross or pedigree data, while FRQs ask you to justify a pattern, calculate phenotype probabilities, or connect Non-Mendelian inheritance to evolution. Standard notation uses superscripts for multiple alleles or codominance (instead of simple uppercase/lowercase) to distinguish different alleles clearly. It is also called complex inheritance in some textbook frameworks.

2. Incomplete Dominance and Codominance

Incomplete dominance occurs when a heterozygote displays a phenotype that is intermediate between the two homozygous parental phenotypes. A common example is flower color: homozygous red × homozygous white produces all pink F1 offspring. A common misconception is that this supports old "blending inheritance" hypotheses, but that is not true: alleles remain distinct, so when F1 heterozygotes are selfed, the original parental phenotypes (red and white) reappear in the F2 generation. This produces a 1:2:1 genotypic ratio that matches the phenotypic ratio, unlike the 3:1 ratio seen in complete Mendelian dominance. Codominance occurs when both alleles are fully and simultaneously expressed in the heterozygote, with no intermediate blending. For example, heterozygous cattle with one red allele and one white allele have a roan coat with separate red and white hairs, not a pink or blended coat. Both incomplete dominance and codominance follow the law of segregation, because alleles still separate into gametes independently during meiosis.

Worked Example

A plant breeder crosses a true-breeding deep-blue flowering plant with a true-breeding white flowering plant. All F1 offspring have pale blue flowers. If F1 plants are self-pollinated, what proportion of F2 offspring will have pale blue flowers?

1. Step 1: Identify the inheritance pattern. The F1 phenotype is intermediate between the two homozygous parents, so this is incomplete dominance.
2. Step 2: Assign genotypes: Let $B^B{B}^B$ = deep blue, $B^W{B}^W$ = white, so heterozygous $B^B{B}^W$ = pale blue.
3. Step 3: F1 heterozygotes each produce 50% $B^B$ gametes and 50% $B^W$ gametes.
4. Step 4: A Punnett square for the self-cross gives a 1 deep blue : 2 pale blue : 1 white phenotypic ratio.
5. Step 5: The proportion of pale blue offspring is $2/4=1/2$.

Exam tip: Always distinguish phenotype blending from allele blending. If an exam question asks if this supports blending inheritance, the answer is no: alleles remain discrete and segregate to produce parental phenotypes in F2.

3. Multiple Alleles and Epistasis

Multiple alleles refers to a gene that has more than two alleles segregating in a population, even though each diploid individual still only carries two alleles per gene. The classic example is human ABO blood type, which has three alleles: $I^A$, $I^B$, and $i$, with $I^A$ and $I^B$ codominant to each other, and both dominant to $i$. Epistasis is an interaction between two different genes where one gene masks or modifies the phenotype of the second gene. This is different from dominance, which is an interaction between two alleles of the same gene. In the most common form of epistasis, homozygosity for a recessive allele of the masking gene blocks expression of the second gene entirely. For example, in mouse coat color, gene $B$ controls pigment color (B = black, b = brown), while gene $E$ controls pigment deposition: homozygous $ee$ blocks all pigment deposition, producing an albino mouse regardless of the genotype at $B$.

Worked Example

In mice, $B$ (black) is dominant to $b$ (brown), and $E$ (pigment deposition) is dominant to $e$ (no pigment). If two heterozygous black mice ($BbEe$) are crossed, what proportion of offspring will be brown?

1. Step 1: Confirm the epistatic interaction: $ee$ masks the $B/b$ gene, so only mice with at least one dominant $E$ allele will show coat color.
2. Step 2: Brown mice require two conditions: homozygous recessive $bb$ (for brown color) and at least one dominant $E$ (for pigment deposition).
3. Step 3: Calculate probability of each independent event: from $Bb\times Bb$, $P(bb)=1/4$. From $Ee\times Ee$, $P(E\_)=3/4$.
4. Step 4: Use the product rule for independent events: $P(\text{brown})=P(bb)\times P(E\_)=(1/4)(3/4)=3/16$. This matches the expected 9:3:4 (black:brown:albino) ratio for this cross.

Exam tip: Always calculate the probability of the masking gene first. If an individual has the masking genotype, you do not need to calculate the probability for the second gene, as it will not affect the phenotype.

4. Polygenic Inheritance and Pleiotropy

Polygenic inheritance occurs when multiple independent genes contribute additively to a single phenotypic trait. This produces continuous (quantitative) phenotypic variation, rather than the discrete phenotypic classes seen in single-gene Mendelian traits. Common examples include human height, skin color, and wheat kernel color. Each dominant additive allele contributes a small increment to the phenotype, so the more dominant alleles an individual carries, the more extreme the phenotype. Phenotypes of polygenic traits almost always form a bell-shaped normal distribution, because environment also contributes to variation. Pleiotropy occurs when a single gene affects multiple unrelated phenotypic traits. This is common for genes that code for proteins used in different tissues or with multiple cellular functions. For example, the recessive allele that causes sickle cell anemia affects hemoglobin structure, but also confers malaria resistance, and causes multiple downstream symptoms like organ damage and pain crises.

Worked Example

Wheat kernel color is controlled by three independently assorting additive genes: $A/a$, $B/b$, $C/c$. Each dominant allele adds one unit of red pigment, and recessive alleles add no pigment. How many distinct phenotypic classes of kernel color will be produced in a cross between two triple heterozygotes ($AaBbCc\times AaBbCc$)?

1. Step 1: For additive polygenic traits with $n$ heterozygous gene pairs, the number of distinct phenotypic classes is given by $2n+1$.
2. Step 2: In this cross, there are 3 heterozygous gene pairs, so $n=3$.
3. Step 3: Calculate: $2(3)+1=7$ distinct phenotypic classes.
4. Step 4: Verify: The number of dominant alleles can range from 0 (aabbcc, no red) to 6 (AABBCC, darkest red), giving 7 distinct values (0,1,2,3,4,5,6), which matches the calculation.

Exam tip: If an FRQ asks you to describe the distribution of a polygenic trait, always mention it forms a continuous bell-shaped (normal) curve, not discrete classes, because of additive contributions from multiple genes and environmental effects.

5. Sex-Linked and Cytoplasmic Inheritance

Sex-linked inheritance refers to genes located on the sex chromosomes (X or Y). In mammals, the X chromosome carries thousands of genes, while the Y chromosome carries very few, so most sex-linked traits are X-linked. Males are hemizygous for X-linked traits: they only have one X chromosome, so any allele on the X is expressed, even recessive alleles. This means X-linked recessive disorders are much more common in males than females. Y-linked traits are only passed from father to all sons, never to daughters, because fathers only pass the Y chromosome to sons. Cytoplasmic (mitochondrial) inheritance refers to genes located in the mitochondrial genome (or chloroplast genome in plants). Offspring inherit all their mitochondria from the mother’s egg, because sperm do not contribute cytoplasm to the zygote. This means mitochondrial mutations are passed exclusively from mother to all offspring, and never from father to offspring.

Worked Example

Red-green color blindness is an X-linked recessive disorder. A couple with normal vision has a son that is color-blind. What is the probability that their next child, which is a daughter, will be color-blind?

1. Step 1: Assign alleles: $X^C$ = normal vision, $X^c$ = color blindness. Males are $XY$, females are $XX$.
2. Step 2: The son inherits his X chromosome from his mother and Y from his father. Since the son is $X^{c}Y$ (color-blind), the mother must be a carrier: $X^C{X}^c$. The father has normal vision, so his genotype is $X^{C}Y$.
3. Step 3: Daughters inherit one X from the father (always $X^C$, since father is $X^{C}Y$) and one X from the mother (50% $X^C$, 50% $X^c$).
4. Step 4: Possible daughter genotypes are $X^C{X}^C$ (normal vision) and $X^C{X}^c$ (carrier, normal vision). There is no way for a daughter to get two $X^c$ alleles here, so the probability of a color-blind daughter is 0.

Exam tip: When tracing trait inheritance in a pedigree, if all offspring of an affected mother are affected and no offspring of an affected father are affected, the trait is mitochondrial, not nuclear.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Calling incomplete dominance "blending inheritance" and claiming alleles merge permanently in heterozygotes. Why: Students confuse the blended phenotype with the old incorrect blending inheritance hypothesis, forgetting alleles remain discrete. Correct move: Always specify incomplete dominance produces a blended phenotype in heterozygotes, but alleles segregate normally to produce parental phenotypes in the next generation.
• Wrong move: Confusing codominance with incomplete dominance, labeling a roan (red and white spotted) cow as incomplete dominance. Why: Students mix up intermediate phenotype vs simultaneous expression of both traits. Correct move: Ask: does the heterozygote show both full traits (codominance) or an intermediate between homozygotes (incomplete dominance) to sort them.
• Wrong move: Calling epistasis "dominance" between two different genes. Why: Students forget dominance is an interaction between alleles of the same gene, while epistasis is between two different genes. Correct move: Always check the level of interaction: same gene = dominance, different genes = epistasis.
• Wrong move: Assuming all X-linked traits are recessive, so any affected daughter must have an affected father. Why: Most tested X-linked traits are recessive, so students default to this assumption regardless of data. Correct move: Always test the pattern against the given data; for X-linked dominant traits, an affected father passes the trait to all daughters, no sons.
• Wrong move: Claiming Y-linked traits can be passed from father to daughter. Why: Students forget father gives Y only to sons and X only to daughters. Correct move: For Y-linked traits, always confirm transmission is exclusively father → all sons, no daughters.
• Wrong move: Counting only dominant alleles from one parent when calculating polygenic phenotype. Why: Students forget additive alleles from both parents contribute to the final phenotype. Correct move: Count all dominant alleles across all chromosomes from both parents to get the total phenotypic contribution.

7. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

In snapdragons, flower color is controlled by a single gene with two alleles: $C^R$ produces red pigment, $C^W$ produces no pigment. Heterozygotes have pink flowers (incomplete dominance). A pink-flowered snapdragon is crossed with a red-flowered snapdragon. What is the probability of getting a pink-flowered offspring from this cross? A) 0 B) 1/4 C) 1/2 D) 3/4

Worked Solution: First, identify the genotypes: the pink parent is heterozygous $C^R{C}^W$, and the red parent must be homozygous $C^R{C}^R$ (red is only seen in homozygotes for this trait). The pink parent produces 50% $C^R$ gametes and 50% $C^W$ gametes, while the red parent only produces $C^R$ gametes. All offspring inherit a $C^R$ from the red parent; 50% inherit $C^R$ from the pink parent (genotype $C^R{C}^R$, red) and 50% inherit $C^W$ from the pink parent (genotype $C^R{C}^W$, pink). The probability of a pink offspring is 1/2. Correct answer: C.

Question 2 (Free Response)

Coat color in Labrador retrievers is controlled by two interacting genes. Gene $B$ controls pigment color: $B$ = black, $b$ = brown. Gene $E$ controls pigment deposition: $E$ allows pigment deposition, and $ee$ produces yellow fur regardless of genotype at $B$. (a) Two heterozygous black Labradors ($BbEe\times BbEe$) are crossed. What is the expected phenotypic ratio of black : brown : yellow offspring? Show your work. (b) A breeder crosses a black Labrador and a yellow Labrador. The resulting litter has 3 black, 2 brown, and 5 yellow puppies. What are the most likely genotypes of the original parents? Justify your answer. (c) Explain why this interaction is an example of epistasis, not polygenic inheritance.

Worked Solution: (a) For independent genes, use the product rule:

• $Bb\times Bb$ gives $P(B\_)=3/4$, $P(bb)=1/4$
• $Ee\times Ee$ gives $P(E\_)=3/4$, $P(ee)=1/4$
• Black ($B\_E\_$) = $3/4\times 3/4=9/16$
• Brown ($bbE\_$) = $1/4\times 3/4=3/16$
• Yellow ($\_\_ee$) = $1\times 1/4=4/16$ Expected ratio = $\boxed{9:3:4}$ (black : brown : yellow)

(b) The presence of brown puppies ($bb$) means both parents carry the $b$ allele. The 5 yellow puppies ($ee$) means both parents carry the $e$ allele. The black parent must have at least one $B$ and one $E$, so it is $BbEe$. The yellow parent must be $ee$, and must carry $b$ to produce brown offspring, so it is $Bbee$. This cross produces an expected ratio of 3 black : 1 brown : 4 yellow, which matches the observed counts for the small litter. The most likely genotypes are black parent: $BbEe$, yellow parent: $Bbee$.

(c) Epistasis is an interaction where one gene masks the expression of a second different gene. Here, the homozygous recessive $ee$ genotype masks the expression of the $B/b$ gene, producing yellow fur regardless of $B/b$ genotype. Polygenic inheritance occurs when multiple genes contribute additively to a single trait to produce continuous variation. This interaction produces three discrete phenotypic classes with one gene masking the other, so it is epistasis, not polygenic.

Question 3 (Application / Real-World Style)

Leber's hereditary optic neuropathy (LHON) is a visual disorder caused by a deleterious mutation in the mitochondrial genome. A woman who carries the LHON mutation in all of her mitochondria marries a man who has no LHON mutation. If the couple plans to have four children, what is the expected number of their children that will inherit the LHON mutation? Justify your answer with rules of cytoplasmic inheritance.

Worked Solution: Mitochondrial DNA is exclusively maternally inherited in humans: all mitochondria in the zygote come from the mother’s egg, and the sperm contributes no mitochondrial DNA to the offspring. Because the mother carries the LHON mutation in all of her mitochondria, every one of her oocytes will contain the mutated mitochondrial genome, so all of her offspring will inherit the mutation. For four children, the expected number of children inheriting the LHON mutation is 4. This means all of the couple’s children are at risk of developing LHON, regardless of the father’s genotype.

8. Quick Reference Cheatsheet

9. What's Next

Non-Mendelian genetics describes the majority of heritable variation in real eukaryotic populations, and forms the foundation for all downstream study of genetic variation and evolution. Immediately after this topic, you will move to pedigree analysis in Unit 5, where you will use the inheritance patterns mastered here to identify trait modes and predict disease risk in families. Without correctly distinguishing between Non-Mendelian inheritance patterns, you will not be able to accurately interpret pedigree data or justify your conclusions on FRQs. This topic also feeds directly into population genetics in Unit 7, where you will calculate allele frequencies for multiple allele traits, and the study of complex human disease in modern genetics.

Pedigree Analysis Population Genetics Human Genetic Disorders Microevolution