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Non-Mendelian
Genetics

Explore inheritance patterns that go beyond Mendel's laws: incomplete dominance, codominance, polygenic traits, and epistasis.

AP Biology

What is Non-Mendelian Genetics?

Gregor Mendel described inheritance governed by dominant and recessive alleles. However, many traits do not follow these simple rules. Non-Mendelian genetics encompasses every pattern of inheritance where the phenotype cannot be predicted by straightforward dominance alone.

The Major Patterns of Non-Mendelian Inheritance

Incomplete Dominance
Heterozygote shows a blend of both homozygous phenotypes.
Codominance
Both alleles are fully expressed simultaneously in the heterozygote.
Polygenic
Multiple genes combine to produce a continuous range of phenotypes.
Epistasis
One gene masks or modifies the expression of another gene.

Why Non-Mendelian Genetics Matters

Most real-world traits are non-Mendelian. Human skin color, height, blood type, and many disease susceptibilities all involve inheritance patterns that deviate from simple dominance.

1
Medical genetics: Understanding codominance helps explain ABO blood typing and transfusion compatibility.
2
Agriculture: Plant breeders exploit incomplete dominance to produce flowers with novel colors.
3
Evolution: Polygenic traits create the continuous variation that natural selection acts upon.

Incomplete Dominance: The Blending Rule

RR Red WW White × RW Pink (Blend)

Codominance: Both Alleles Express Fully

\(C^R C^R\) Red coat \(C^W C^W\) White coat × Roan (Red + White patches) \(C^R C^W\)

Incomplete vs. Codominance

Incomplete Dominance Codominance
Heterozygote phenotype Intermediate blend Both phenotypes visible
Classic example Red × White = Pink Red × White = Red + White spots
Molecular basis One allele produces less product Both alleles produce full product
\( F_1 \) phenotypic ratio \( 1:2:1 \) \( 1:2:1 \)

Heterozygote phenotype diagnoses dominance relationship

Three comparable snapdragon plants with red, pink, and white flowers.The heterozygote phenotype distinguishes dominance relationships. Snapdragon heterozygotes retain both alleles while producing intermediate color; AB red cells display both A and B carbohydrate antigens.

Score the heterozygote phenotype

Score the heterozygote phenotype. The heterozygote phenotype distinguishes dominance relationships. Snapdragon heterozygotes retain both alleles while producing intermediate color; AB red cells display both A and B carbohydrate antigens.The heterozygote phenotype distinguishes dominance relationships. Snapdragon heterozygotes retain both alleles while producing intermediate color; AB red cells display both A and B carbohydrate antigens.

Codominance: both products are present

Codominance: both products are present. In the AB phenotype, both A and B carbohydrate antigen types occur on the same red cell because both allele-specific enzyme products are expressed.In the AB phenotype, both A and B carbohydrate antigen types occur on the same red cell because both allele-specific enzyme products are expressed.

Multiple Alleles: ABO Blood Type

A single gene can have more than two alleles in a population. The ABO blood group locus has three alleles: \( I^A \), \( I^B \), and \( i \). Each individual carries any two of them.

Type A \(I^A I^A\) or \(I^A i\) Type B \(I^B I^B\) or \(I^B i\) Type AB \(I^A I^B\) Type O \(ii\) Dominance relationships: \(I^A\) and \(I^B\) are codominant to each other Both \(I^A\) and \(I^B\) are dominant over \(i\)

Polygenic Inheritance: Many Genes, One Trait

When two or more genes contribute additively to a single phenotype, the result is a continuous distribution of phenotypes rather than distinct categories. Classic examples: human skin color, height, and eye color.

Phenotype (e.g., skin color intensity) Frequency Light Medium Dark

Epistasis: One Gene Masks Another

Gene 1: Pigment color B = black, b = brown Gene 2: Pigment deposit E = deposit, e = no deposit B_E_ Black bbE_ Brown _ _ ee Yellow (no pigment) ee is epistatic: masks Gene 1 entirely
Black, brown and yellow dogs illustrate three outcomes. B_E_ is black, bbE_ brown and __ee yellow in the selected model. An independently authored 4 by 4 cross shows the 9:3:4 expected ratio.
In this simplified coat-color model, pigment deposition can mask differences in pigment production. BbEe × BbEe with independent assortment predicts a 9:3:4 phenotype ratio; yellow does not identify the B genotype. Individual images do not establish genotype.
Sixteen gamete combinations
Sixteen gamete combinations

Pleiotropy and Linked Genes

Pleiotropy
One gene affects multiple phenotypic traits. Example: the sickle-cell allele (HbS) causes misshapen red blood cells, anemia, organ damage, and malaria resistance.
Linked Genes
Genes located close together on the same chromosome tend to be inherited together. They violate Mendel's Law of Independent Assortment unless separated by crossing over.

Sex-Linked Inheritance

Genes on the X chromosome follow a distinct pattern. Males (\( XY \)) express every X-linked allele because they have only one copy. Females (\( XX \)) can be carriers of recessive X-linked traits without showing symptoms.

Carrier \(X^H X^h\) Normal \(X^H Y\) × \(X^H X^H\) Normal \(X^H X^h\) Carrier \(X^H Y\) Normal \(X^h Y\) Affected Probability of affected son: 1/4 (or 1/2 of sons)

How to Solve Non-Mendelian Problems

1
Identify the pattern. Is the heterozygote a blend (incomplete dominance), both phenotypes (codominance), or something else?
2
Assign allele notation. Use superscripts for codominance (\( C^R C^W \)), separate letters for epistasis, or X-linkage notation (\( X^H Y \)).
3
Set up a Punnett square. For two-gene epistasis problems, use a \( 4 \times 4 \) dihybrid square.
4
Determine phenotypic ratios. Compare to expected Mendelian ratios to confirm the non-Mendelian pattern.
5
Check modified ratios. Epistasis: \( 9:3:4 \) or \( 9:7 \). Lethal alleles: \( 2:1 \). Multiple alleles: more genotypes than a simple cross.

Example 1: Snapdragon Flower Color

Cross two pink snapdragons (\( RW \times RW \)). What are the offspring phenotypes?

R W R W RR RW RW WW Ratio: 1 Red : 2 Pink : 1 White

Example 2: Labrador Coat Color Epistasis

Cross \( BbEe \times BbEe \). What is the phenotypic ratio?

1
Standard dihybrid ratio: \( 9 \; B\_E\_ : 3 \; bbE\_ : 3 \; B\_ee : 1 \; bbee \)
2
Apply epistasis: All \( ee \) genotypes become yellow regardless of B locus. Combine: \( 3 \; B\_ee + 1 \; bbee = 4 \) yellow.
3
Final ratio: 9 black : 3 brown : 4 yellow

Example 3: X-Linked Color Blindness

A carrier woman (\( X^C X^c \)) marries a normal man (\( X^C Y \)). What fraction of their sons will be color blind?

\(X^C\) Y \(X^C\) \(X^c\) \(X^C X^C\) Normal girl \(X^C Y\) Normal boy \(X^C X^c\) Carrier girl \(X^c Y\) Color blind boy

Common Mistakes to Avoid

✕ Confusing Incomplete Dominance with Codominance
A blend (pink from red + white) is incomplete dominance. Seeing both colors separately (red + white patches) is codominance.
✕ Forgetting X-Linkage in Males
Males have only one X. A single recessive allele is sufficient for expression: there is no "carrier" male for X-linked traits.
✕ Ignoring Epistatic Masking
When a modified dihybrid ratio appears (9:3:4, 9:7, 12:3:1), always check for an epistatic gene rather than assuming linkage or error.

Summary: Non-Mendelian Patterns at a Glance

Pattern Heterozygote Key Ratio Example
Incomplete Dom. Blend \( 1:2:1 \) Snapdragon color
Codominance Both visible \( 1:2:1 \) Roan cattle, ABO
Multiple Alleles Varies Multiple phenotypes ABO blood type
Polygenic Additive Bell curve Skin color, height
Epistasis Masked \( 9:3:4 \) etc. Lab coat color
Sex-Linked Carrier females Unequal by sex Color blindness

Question 1 of 5

A red flower is crossed with a white flower, and all \( F_1 \) offspring are pink. When two pink \( F_1 \) plants are crossed, the \( F_2 \) ratio is \( 1 \) red : \( 2 \) pink : \( 1 \) white. This is an example of:

Question 2 of 5

A person with blood type AB has the genotype \( I^A I^B \). The relationship between the \( I^A \) and \( I^B \) alleles is best described as:

Question 3 of 5

In Labrador retrievers, \( BbEe \times BbEe \) produces a 9:3:4 phenotypic ratio instead of the expected 9:3:3:1. This modified ratio indicates:

Question 4 of 5

A carrier woman (\( X^H X^h \)) and a normal man (\( X^H Y \)) have children. What is the probability that a randomly selected child will have hemophilia?

Question 5 of 5

Human skin color is determined by at least three genes, each with two alleles contributing additively. The phenotype distribution in a population forms a bell-shaped curve. This pattern is called: