Non-Mendelian Inheritance and Chromosomal Genetics

Gregor Welcomes You!

Welcome back, investigators! In Chapter 11 you mastered Mendel's elegant laws — but nature is rarely that tidy. Many traits blend, overlap, or depend on multiple genes working in concert. Some alleles ride on sex chromosomes, and sometimes entire chromosomes fail to separate properly during meiosis. In this chapter, we venture beyond the 3:1 ratio into the rich, messy, fascinating reality of inheritance. Let's investigate!

Summary

Mendel's laws describe a simplified model of inheritance, but most traits do not follow simple dominant-recessive patterns. This chapter extends the Mendelian framework to incomplete dominance, codominance, multiple alleles, sex determination, and X-linked inheritance, then examines how polygenic inheritance and epistasis produce continuous variation and complex phenotypes. The chapter also covers genetic linkage and recombination frequency as tools for chromosomal mapping. It concludes with chromosomal genetics: nondisjunction, the aneuploidy conditions it produces, and the diagnostic power of pedigree analysis for determining inheritance patterns in human families.

Concepts Covered

This chapter covers the following 15 concepts from the learning graph:

Incomplete Dominance
Codominance
Multiple Alleles
Sex Determination
Sex-Linked Traits
X-Linked Inheritance
Polygenic Inheritance
Epistasis
Pleiotropy
Genetic Linkage
Recombination Frequency
Chromosomal Abnormalities
Nondisjunction
Aneuploidy
Pedigree Analysis

Prerequisites

This chapter builds on concepts from:

Chapter 11: Meiosis and Mendelian Genetics

Beyond Simple Dominance

In Chapter 11, every trait exhibited complete dominance — the heterozygote looked exactly like the homozygous dominant parent. But many allele interactions are more nuanced. Two important alternatives are incomplete dominance and codominance.

Incomplete Dominance

In incomplete dominance, the heterozygote displays a phenotype that is intermediate between the two homozygous phenotypes — neither allele fully masks the other.

Classic example: Flower color in snapdragons (Antirrhinum)

\(C^R C^R\) (homozygous) → red flowers
\(C^W C^W\) (homozygous) → white flowers
\(C^R C^W\) (heterozygous) → pink flowers

When two pink heterozygotes are crossed (\(C^R C^W \times C^R C^W\)):

	\(C^R\)	\(C^W\)
\(C^R\)	\(C^R C^R\) (red)	\(C^R C^W\) (pink)
\(C^W\)	\(C^R C^W\) (pink)	\(C^W C^W\) (white)

Phenotypic ratio: 1 red : 2 pink : 1 white

Genotypic ratio: 1 \(C^R C^R\) : 2 \(C^R C^W\) : 1 \(C^W C^W\)

Notice that in incomplete dominance, the phenotypic ratio equals the genotypic ratio (1:2:1) because every genotype produces a distinct phenotype.

Gregor's Tip

The AP exam loves to test the difference between incomplete dominance and codominance. The key: incomplete dominance produces a blended intermediate phenotype (pink from red + white), while codominance produces a phenotype where both alleles are fully expressed simultaneously. If you see a blended phenotype, it's incomplete dominance. If both traits appear side by side, it's codominance.

Codominance

In codominance, both alleles in the heterozygote are fully and simultaneously expressed — the phenotype is not intermediate but rather shows both parental traits distinctly.

Here is an example of how humans children inherit their blood type from their parents.

Diagram: Blood Type Explorer

Run the Blood Type Explorer Fullscreen

Classic example: ABO blood type — the \(I^A\) and \(I^B\) alleles

\(I^A I^A\) or \(I^A i\) → type A blood (A antigens on red blood cells)
\(I^B I^B\) or \(I^B i\) → type B blood (B antigens on red blood cells)
\(I^A I^B\) → type AB blood (both A and B antigens expressed)
\(ii\) → type O blood (neither antigen)

In the \(I^A I^B\) heterozygote, both the A antigen and the B antigen are produced — this is codominance because both alleles are fully expressed rather than blended.

Dominance Pattern	Heterozygote Phenotype	Example
Complete dominance	Same as homozygous dominant	Tall pea plants (Tt = TT phenotype)
Incomplete dominance	Intermediate blend	Pink snapdragons (\(C^R C^W\))
Codominance	Both alleles fully expressed	AB blood type (\(I^A I^B\))

Multiple Alleles

Mendel's crosses always involved genes with exactly two alleles. However, many genes exist as multiple alleles in a population — more than two allelic variants at a single locus. Any individual organism still carries only two alleles (one per homolog), but the population as a whole harbors more.

The ABO blood group illustrates both codominance and multiple alleles simultaneously: the gene has three alleles (\(I^A\), \(I^B\), and \(i\)), producing six possible genotypes and four phenotypes.

Genotype	Phenotype (Blood Type)	Antigens present
\(I^A I^A\) or \(I^A i\)	Type A	A
\(I^B I^B\) or \(I^B i\)	Type B	B
\(I^A I^B\)	Type AB	A and B
\(ii\)	Type O	Neither

Diagram: Blood Type Inheritance Explorer

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Blood Type Inheritance Explorer — Specification

Type: MicroSim (p5.js)
sim-id: blood-type-inheritance
Library: p5.js
Status: Specified

Learning objective: Students will be able to apply (Bloom's L3: Apply) knowledge of multiple alleles and codominance to predict offspring blood types from parental crosses, and solve blood type genetics problems.

Instructional Rationale: An interactive cross calculator lets students select parental blood types and immediately see the Punnett square and offspring probabilities. This builds fluency with multiple allele systems and reinforces the distinction between codominance and complete dominance (i is recessive to both \(I^A\) and \(I^B\)).

Canvas: 780 × 500 px, responsive.

Layout:

Top: Parent selectors — two dropdown menus for Parent 1 and Parent 2 blood type phenotypes (A, B, AB, O)
Below selectors: Possible genotype(s) for each parent shown as chips (e.g., Type A → could be \(I^A I^A\) or \(I^A i\))
Center: Punnett square grid (2×2 or combined if multiple genotypes possible)
Bottom: Results summary — offspring phenotype probabilities as colored bar chart segments

Interaction: - Select parent phenotypes → genotype possibilities appear - If a parent's genotype is ambiguous (e.g., Type A could be \(I^A I^A\) or \(I^A i\)), a toggle lets the user specify the exact genotype - Punnett square auto-fills with color-coded offspring - Hover over any cell to see genotype, phenotype, and antigen info - "Random Parents" button for quick practice

Colors: - Type A: red (#E74C3C) - Type B: blue (#3498DB) - Type AB: purple (#8E44AD) - Type O: gray (#95A5A6)

Responsive design: Grid and bar chart scale with container width.

Sex Determination and Sex-Linked Inheritance

Sex Determination

In humans and many other organisms, biological sex is determined by sex chromosomes. Humans have 22 pairs of autosomes plus one pair of sex chromosomes:

Females: XX (two X chromosomes)
Males: XY (one X, one Y chromosome)

The Y chromosome carries the SRY gene (Sex-determining Region Y), which triggers male development. In the absence of a functional SRY, the default developmental pathway produces female anatomy.

During meiosis in a male (XY), half the sperm receive an X chromosome and half receive a Y chromosome. Eggs from a female (XX) always carry one X. Therefore, the father's sperm determines the sex of the offspring:

Egg	X Sperm	Y Sperm
X	XX (female)	XY (male)

Expected sex ratio: 1 female : 1 male (50:50)

Sex-Linked Traits

Sex-linked traits are controlled by genes located on the sex chromosomes. Because the X chromosome is much larger than the Y chromosome and carries far more genes, most sex-linked traits are specifically X-linked.

X-Linked Inheritance

X-linked inheritance refers to the transmission of genes located on the X chromosome. Key features:

Males have only one X chromosome, so they are hemizygous for X-linked genes — a single copy of a recessive allele will produce the recessive phenotype
Females have two X chromosomes, so they can be homozygous dominant, heterozygous (carriers), or homozygous recessive
X-linked recessive conditions appear far more frequently in males than in females

Classic example: Red-green color blindness

\(X^B\) = normal vision (dominant)
\(X^b\) = color blindness (recessive)

Genotype	Sex	Phenotype
\(X^B X^B\)	Female	Normal vision
\(X^B X^b\)	Female	Normal vision (carrier)
\(X^b X^b\)	Female	Color blind
\(X^B Y\)	Male	Normal vision
\(X^b Y\)	Male	Color blind

Cross example: Carrier female × normal male (\(X^B X^b \times X^B Y\))

	\(X^B\)	\(Y\)
\(X^B\)	\(X^B X^B\) (normal female)	\(X^B Y\) (normal male)
\(X^b\)	\(X^B X^b\) (carrier female)	\(X^b Y\) (color-blind male)

Result: All daughters have normal vision (but half are carriers); half of sons are color blind.

Common Mistake

Students often write X-linked genotypes without the chromosome notation. Always use the format \(X^B X^b\) (not just Bb) for X-linked traits — the chromosome matters! Also remember: fathers cannot pass X-linked traits to their sons (they give sons their Y chromosome). An affected father passes his X-linked allele to all daughters, never to sons.

Diagram: X-Linked Inheritance Simulator

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X-Linked Inheritance Simulator — Specification

Type: MicroSim (p5.js)
sim-id: x-linked-inheritance
Library: p5.js
Status: Specified

Learning objective: Students will be able to apply (Bloom's L3: Apply) X-linked inheritance patterns to predict offspring genotypes and phenotypes, and analyze (Bloom's L4: Analyze) why X-linked recessive traits appear more frequently in males.

Instructional Rationale: An interactive cross tool that visually shows the X and Y chromosomes traveling from parents to offspring reinforces the physical basis for sex-linked inheritance patterns. Seeing the asymmetry between sons and daughters builds intuition.

Canvas: 780 × 500 px, responsive.

Layout:

Top: Parent genotype selectors
Mother: dropdown (\(X^B X^B\), \(X^B X^b\), \(X^b X^b\))
Father: dropdown (\(X^B Y\), \(X^b Y\))
Center: Animated cross diagram showing chromosomes passing from each parent
Mother's two X chromosomes shown as pink bars with allele labels
Father's X and Y shown as blue bars
Arrows trace which chromosome goes to each offspring
Bottom: Punnett square with results; offspring shown as icons with phenotype coloring

Interaction: - Select parental genotypes → cross auto-animates - "Step Through" mode: shows gamete formation, then fertilization one offspring at a time - Toggle between 3 X-linked traits: color blindness, hemophilia, Duchenne muscular dystrophy - Offspring tally: count of each phenotype/genotype class

Colors: Normal phenotype: green. Affected phenotype: red-orange. Carrier: green with orange border.

Responsive design: Chromosome diagrams and Punnett square scale proportionally.

Complex Inheritance Patterns

Polygenic Inheritance

Many traits are not controlled by a single gene but by the combined effects of multiple genes — this is polygenic inheritance. Polygenic traits show continuous variation (a bell-shaped distribution) rather than discrete categories.

Examples of polygenic traits:

Human skin color (at least 3–4 genes involved)
Height (hundreds of genes)
Eye color (multiple genes, including OCA2 and HERC2)
Milk production in cattle

Each contributing gene typically has a small additive effect. When you plot the phenotypes of a large population, the result is a smooth bell curve (normal distribution) rather than distinct classes.

Diagram: Polygenic Trait Distribution Visualizer

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Polygenic Trait Distribution Visualizer — Specification

Type: MicroSim (Chart.js)
sim-id: polygenic-distribution
Library: Chart.js
Status: Specified

Learning objective: Students will be able to explain (Bloom's L2: Understand) how increasing the number of contributing genes transforms a discrete phenotype distribution into a continuous bell curve, and compare (Bloom's L4: Analyze) polygenic inheritance with single-gene Mendelian patterns.

Instructional Rationale: A slider-controlled visualization showing how the phenotype distribution changes as the number of genes increases from 1 to 5+ builds an intuitive bridge between Mendelian ratios and continuous variation. Seeing the transition happen dynamically is far more instructive than a static figure.

Canvas: 760 × 440 px, responsive.

Layout:

Top: Slider labeled "Number of contributing genes" (range: 1 to 6)
Center: Bar chart / histogram showing phenotype distribution
X-axis: Phenotype value (e.g., number of "additive alleles" from 0 to 2n)
Y-axis: Frequency (proportion of offspring)
Bottom: Summary text: number of phenotypic classes, standard deviation, shape description

Data Visibility Requirements:

1 gene (2 alleles): 3 phenotypic classes in 1:2:1 ratio (Mendelian)
2 genes: 5 classes in 1:4:6:4:1 ratio
3 genes: 7 classes approximating a bell curve
4+ genes: smooth normal distribution
At each step, show the mathematical expansion (binomial coefficients) alongside the histogram

Interaction: - Slider adjusts gene number → histogram animates the transition - Hover over bars for exact frequency values - Toggle: overlay a normal distribution curve for comparison at 3+ genes - "Show Math" toggle: display binomial expansion beneath the chart

Colors: Bars gradient from light (few additive alleles) to dark (many additive alleles) in a green-to-brown skin color gradient (for the skin color example). A neutral blue palette is the default.

Responsive design: Chart width and bar counts scale with container.

Epistasis

Epistasis occurs when one gene modifies or masks the expression of a different gene at a separate locus. The "masking" gene is called the epistatic gene; the gene being masked is the hypostatic gene.

Classic example: Coat color in Labrador retrievers

Gene E/e controls pigment deposition (E = pigment deposited; ee = no pigment → yellow)
Gene B/b controls pigment color (B = black; bb = brown/chocolate)

Genotype at E locus	Genotype at B locus	Phenotype
E_ (at least one E)	B_ (at least one B)	Black
E_	bb	Chocolate
ee	B_ or bb	Yellow (pigment masked)

A dihybrid cross of BbEe × BbEe produces a 9:3:4 ratio instead of the standard 9:3:3:1:

9 B_E_ → black
3 bbE_ → chocolate
3 B_ee + 1 bbee → yellow (4 total)

The ee genotype is epistatic to the B/b locus — it masks whatever pigment gene is present.

Key Insight

Modified dihybrid ratios are a big signal on the AP exam that epistasis is at play. Whenever the F2 phenotypic ratio departs from 9:3:3:1 but the total still adds to 16, suspect epistasis. Common epistatic ratios include 9:3:4, 9:7, 12:3:1, and 15:1 — each reflects a different type of gene interaction.

Pleiotropy

Pleiotropy occurs when a single gene influences multiple, seemingly unrelated phenotypic traits. This is the opposite of polygenic inheritance (where many genes affect one trait) — in pleiotropy, one gene affects many traits.

Classic example: Sickle cell disease — a single point mutation in the \(\beta\)-globin gene (Glu → Val at position 6) produces:

Sickle-shaped red blood cells
Chronic anemia
Pain crises from blocked blood vessels
Increased resistance to malaria (in heterozygotes)
Organ damage (spleen, kidneys, lungs)

All of these diverse effects trace back to one gene mutation.

Inheritance pattern	Genes involved	Traits affected	Example
Mendelian (simple)	1 gene	1 trait	Pea seed shape
Polygenic	Many genes	1 trait	Human height
Pleiotropy	1 gene	Many traits	Sickle cell disease
Epistasis	2+ genes	1 trait (with masking)	Labrador coat color

Genetic Linkage and Recombination

Genetic Linkage

Mendel's Law of Independent Assortment works beautifully — when genes are on different chromosomes. But what happens when two genes are located on the same chromosome?

Genes on the same chromosome tend to be inherited together because they physically travel on the same DNA molecule during meiosis. This phenomenon is called genetic linkage. Linked genes do NOT assort independently — they violate Mendel's second law.

Thomas Hunt Morgan first demonstrated linkage in Drosophila by showing that certain gene combinations appeared in offspring far more often than the 1:1:1:1 ratio predicted by independent assortment.

Recombination Frequency

Although linked genes tend to travel together, crossing over during prophase I can separate them. The frequency with which crossing over occurs between two linked genes is called the recombination frequency (RF).

\[\text{Recombination frequency} = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%\]

Key principles:

RF ranges from 0% (genes very close together; virtually never separated by crossing over) to 50% (genes so far apart on the same chromosome that they behave as if unlinked)
RF of 50% is indistinguishable from genes on different chromosomes
RF is used as a measure of genetic map distance: 1% recombination = 1 centimorgan (cM) or 1 map unit
The closer two genes are on a chromosome, the lower the recombination frequency

Example: If a cross produces 400 offspring, of which 36 are recombinants:

\[RF = \frac{36}{400} \times 100\% = 9\%\]

These two genes are approximately 9 cM apart on the chromosome.

Diagram: Genetic Linkage and Recombination Mapper

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Genetic Linkage and Recombination Mapper — Specification

Type: MicroSim (p5.js)
sim-id: linkage-mapper
Library: p5.js
Status: Specified

Learning objective: Students will be able to calculate (Bloom's L3: Apply) recombination frequencies from offspring data and construct (Bloom's L6: Create) a genetic linkage map showing relative positions of genes on a chromosome.

Instructional Rationale: A hands-on mapping tool where students enter test cross data and see the resulting chromosome map builds both calculation skills and spatial understanding of gene arrangement. The ability to drag genes along the chromosome reinforces the relationship between RF and physical distance.

Canvas: 780 × 480 px, responsive.

Layout:

Top: Data entry panel — three gene pairs with text inputs for recombinant and total offspring counts
Gene pair A-B: recombinant count / total count
Gene pair B-C: recombinant count / total count
Gene pair A-C: recombinant count / total count
Center: Chromosome visualization — a horizontal bar with gene markers positioned according to calculated map distances
Bottom: Results table showing RF for each pair and resulting map distances in cM

Data Visibility Requirements:

Step 1: Enter offspring counts → RF calculated automatically
Step 2: Map distances displayed in cM
Step 3: Gene order determined (the largest RF = the outside genes)
Step 4: Genes positioned on chromosome bar proportionally

Interaction: - Enter test cross data → RFs auto-calculate - Genes animate into position on the chromosome map - "Load Example" button with 3 preset problems (Drosophila body color/wing shape/eye color) - Drag gene markers to test alternative orderings — correct order highlighted in green - Verify: does A-B + B-C ≈ A-C? Display consistency check

Preset data (Drosophila example): - body color – wing shape: RF = 17% - wing shape – eye color: RF = 9.5% - body color – eye color: RF = 26.5% - Map order: body color — 17 cM — wing shape — 9.5 cM — eye color

Colors: Genes: distinct colored markers. Chromosome: gray bar. RF labels: black text.

Responsive design: Chromosome bar length and font sizes scale proportionally.

Chromosomal Abnormalities

Sometimes errors occur during meiosis that alter the number or structure of entire chromosomes.

Nondisjunction

Nondisjunction is the failure of chromosomes to separate properly during cell division. It can occur during:

Meiosis I — homologous chromosomes fail to separate → both homologs go to one cell
Meiosis II — sister chromatids fail to separate → both chromatids go to one cell
Mitosis — sister chromatids fail to separate (affects only the individual, not offspring)

The result is gametes with an abnormal number of chromosomes — either one extra chromosome (n+1) or one missing (n-1).

Aneuploidy

When a gamete with an abnormal chromosome number is fertilized by a normal gamete, the resulting zygote has an abnormal chromosome count — this condition is called aneuploidy.

Trisomy (2n+1): three copies of a chromosome instead of two
Monosomy (2n-1): only one copy of a chromosome instead of two

Condition	Chromosome affected	Karyotype	Features
Down syndrome	Trisomy 21	47, +21	Intellectual disability, characteristic facial features, heart defects
Turner syndrome	Monosomy X	45, X	Female phenotype, short stature, infertility
Klinefelter syndrome	Extra X in male	47, XXY	Male phenotype, tall stature, infertility
Patau syndrome	Trisomy 13	47, +13	Severe intellectual disability, organ defects
Edwards syndrome	Trisomy 18	47, +18	Severe intellectual disability, often fatal in infancy

Most autosomal monosomies are lethal before birth, while some trisomies (particularly of smaller chromosomes) are survivable. Sex chromosome aneuploidies tend to be less severe than autosomal aneuploidies.

Diagram: Nondisjunction in Meiosis Visualizer

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Nondisjunction in Meiosis Visualizer — Specification

Type: MicroSim (p5.js)
sim-id: nondisjunction
Library: p5.js
Status: Specified

Learning objective: Students will be able to differentiate (Bloom's L4: Analyze) between nondisjunction in meiosis I versus meiosis II and predict the chromosomal composition of the resulting gametes.

Instructional Rationale: A side-by-side comparison of normal meiosis, nondisjunction in meiosis I, and nondisjunction in meiosis II — all shown step-by-step — allows students to pinpoint exactly where the error occurs and trace its consequences through both divisions.

Canvas: 800 × 500 px, responsive.

Layout: Three parallel columns showing the same cell progressing through meiosis:

Column 1: Normal meiosis (control)
Column 2: Nondisjunction in meiosis I — homologs fail to separate
Column 3: Nondisjunction in meiosis II — sister chromatids fail to separate

Data Visibility Requirements:

Start: All three columns show a diploid cell with 2n=4 (2 homologous pairs, color-coded)
After meiosis I: Column 1 shows two normal haploid cells; Column 2 shows one cell with both homologs and one empty; Column 3 shows two normal haploid cells (error hasn't happened yet)
After meiosis II: Column 1 shows four normal gametes (n=2 each); Column 2 shows two n+1 gametes and two n-1 gametes; Column 3 shows one n+1, one n-1, and two normal gametes
Chromosome counts displayed below each cell

Interaction: - Next/Previous buttons to step through stages (synchronized across columns) - Highlight the error step with a red flash - "Show Fertilization" button: shows what happens when each abnormal gamete is fertilized by a normal gamete (trisomy or monosomy) - Toggle: show a single chromosome pair or two pairs

Colors: Maternal chromosomes: pink. Paternal chromosomes: blue. Error highlight: red glow. Normal gamete: green border. Abnormal gamete: red border.

Responsive design: Three columns collapse to a tab interface on narrow screens.

Pedigree Analysis

A pedigree is a family tree diagram that traces the inheritance of a trait through multiple generations. Pedigrees are essential tools in human genetics because controlled crosses are impossible in humans — geneticists must observe patterns in existing families.

Pedigree Symbols

Circle = female
Square = male
Filled shape = affected individual
Half-filled shape = carrier (known or inferred)
Horizontal line connecting a circle and square = mating
Vertical line going down = offspring
Roman numerals (I, II, III...) = generations

Identifying Inheritance Patterns

By analyzing which individuals are affected and how the trait passes through generations, you can deduce the most likely mode of inheritance:

Pattern	Key clues
Autosomal dominant	Affected individuals in every generation; affected parent has at least one affected child; unaffected parents never have affected children
Autosomal recessive	Trait can skip generations; affected children often have two unaffected (carrier) parents; appears equally in males and females
X-linked recessive	Affects mostly males; affected males get allele from carrier mothers; affected fathers pass allele to all daughters (carriers) but no sons
X-linked dominant	Affected father passes trait to all daughters but no sons; affected mothers pass to ~50% of all children

Diagram: Interactive Pedigree Analyzer

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Interactive Pedigree Analyzer — Specification

Type: MicroSim (p5.js)
sim-id: pedigree-analyzer
Library: p5.js
Status: Specified

Learning objective: Students will be able to analyze (Bloom's L4: Analyze) pedigree diagrams to determine the most likely mode of inheritance, and evaluate (Bloom's L5: Evaluate) whether observed patterns are consistent with autosomal dominant, autosomal recessive, or X-linked recessive inheritance.

Instructional Rationale: Presenting multiple pedigrees that students must classify builds the diagnostic reasoning skills tested on the AP exam. The immediate feedback loop — guess the pattern, then check — reinforces the distinguishing clues for each inheritance mode.

Canvas: 800 × 520 px, responsive.

Layout:

Top: Pedigree display area — standard 3-generation pedigree drawn with standard symbols
Center: Multiple-choice selection panel — four buttons: Autosomal Dominant, Autosomal Recessive, X-Linked Recessive, X-Linked Dominant
Bottom: Explanation panel — after the student selects, shows why the answer is correct (or why it's incorrect with a hint)

Interaction: - "New Pedigree" button: loads one of 10+ pre-built pedigrees (randomized order) - Student clicks an inheritance pattern → immediate feedback - Correct: green highlight, explanation of the key clues - Incorrect: red highlight, hint pointing to a specific feature they should look at (e.g., "Notice that the trait skips generation II — what does that suggest?") - Hover over any individual in the pedigree: shows inferred genotype (after answer is revealed) - Score tracker: correct / total attempts - Difficulty selector: Easy (clear-cut patterns) / Hard (ambiguous with carrier inference required)

Pedigree data: 12 pre-built pedigrees stored as JSON: - 3 autosomal dominant - 3 autosomal recessive - 3 X-linked recessive - 3 X-linked dominant

Each pedigree object includes: nodes (generation, position, sex, affected status, genotype), edges (parent-child, mating), and correct inheritance pattern.

Colors: Unaffected: white fill. Affected: dark fill (#2C3E50). Carrier (revealed): half-fill. Correct answer: green glow. Incorrect: red glow.

Responsive design: Pedigree node sizes and spacing scale with container. On narrow screens, explanation panel appears below.

Solving a Pedigree: Worked Example

Consider a pedigree where two unaffected parents in generation I have one affected son and one unaffected daughter in generation II. The unaffected daughter marries an unaffected male, and they have one affected son in generation III.

Step 1: The trait skips generation II → likely recessive

Step 2: Both males and females can be affected → likely autosomal (not X-linked)

Step 3: Both parents in generation I must be carriers: Aa × Aa

Step 4: The unaffected daughter in generation II has a \(\frac{2}{3}\) probability of being Aa (since she is unaffected, she is either AA or Aa, in a 1:2 ratio)

Conclusion: Autosomal recessive inheritance

You've Got This!

Pedigree analysis combines everything you have learned about dominance, recessiveness, and sex linkage into a single detective challenge. On the AP exam, start by checking if the trait is dominant or recessive (does it skip generations?), then check if it's autosomal or X-linked (is it much more common in one sex?). With practice, you will be able to read pedigrees as fluently as you read Punnett squares.

Key Takeaways

Incomplete dominance produces an intermediate phenotype in heterozygotes (1:2:1 phenotypic ratio). Codominance produces a phenotype where both alleles are fully expressed (e.g., AB blood type).
Multiple alleles exist at a single locus in a population (e.g., \(I^A\), \(I^B\), \(i\) for ABO blood type), though each individual carries only two.
Sex determination in humans uses the XX/XY system; the SRY gene on the Y chromosome triggers male development.
X-linked recessive traits affect males disproportionately because males are hemizygous (only one X). Carrier mothers can pass the allele to affected sons.
Polygenic inheritance produces continuous variation from the additive effects of multiple genes (bell curve distribution).
Epistasis occurs when one gene masks or modifies the expression of another gene at a different locus, producing modified dihybrid ratios (e.g., 9:3:4).
Pleiotropy occurs when a single gene affects multiple phenotypic traits (e.g., sickle cell disease).
Genetic linkage — genes on the same chromosome tend to be inherited together, violating independent assortment.
Recombination frequency measures the rate of crossing over between linked genes: RF (%) = (recombinants / total) × 100. RF = 1% corresponds to 1 cM of map distance.
Nondisjunction — failure of chromosomes to separate during meiosis — produces aneuploid gametes, leading to conditions such as trisomy 21 (Down syndrome).
Pedigree analysis uses family inheritance patterns to deduce whether a trait is autosomal vs. X-linked and dominant vs. recessive.

AP Practice: Test Your Understanding

Question 1: In a cross between a red snapdragon (\(C^R C^R\)) and a white snapdragon (\(C^W C^W\)), all F1 offspring are pink. If two F1 plants are crossed, what phenotypic ratio is expected in the F2 generation?

Answer: 1 red (\(C^R C^R\)) : 2 pink (\(C^R C^W\)) : 1 white (\(C^W C^W\)). This 1:2:1 ratio is characteristic of incomplete dominance.

Question 2: A woman who is a carrier for hemophilia (\(X^H X^h\)) marries an unaffected man (\(X^H Y\)). What proportion of their sons will have hemophilia? What proportion of their daughters will be carriers?

Answer: Half (\(\frac{1}{2}\)) of their sons will have hemophilia (\(X^h Y\)). Half (\(\frac{1}{2}\)) of their daughters will be carriers (\(X^H X^h\)). No daughters will have hemophilia.

Question 3: In Labrador retrievers, a cross of BbEe × BbEe yields 91 black, 30 chocolate, and 39 yellow puppies. Is this consistent with a 9:3:4 epistatic ratio? How would you test this statistically?

Answer: Expected 9:3:4 ratio predicts 90 black : 30 chocolate : 40 yellow out of 160 total. The observed values (91:30:39) are very close. To test statistically, you would apply a chi-square test comparing observed to expected values. The chi-square value here would be very small, indicating no significant deviation from the 9:3:4 ratio.