Chromosomal Basis of Inheritance

Chromosome Theory of Inheritance

The chromosome theory of inheritance states that genes are located on chromosomes and that the behavior of chromosomes during meiosis accounts for Mendel's laws of segregation and independent assortment.

Key Principles

• Genes have specific loci (positions) on chromosomes
• Chromosomes undergo segregation during meiosis I → explains Mendel's Law of Segregation
• Chromosomes on different pairs assort independently → explains Mendel's Law of Independent Assortment
• Each chromosome carries hundreds to thousands of genes

The Evidence: Chromosomes Behave Like Mendel's Factors

Because chromosomes do exactly what Mendel's factors do, biologists concluded that genes are carried on chromosomes. Fruit-fly crosses (below) then tied one specific gene to one specific chromosome. The AP exam tests this reasoning, not the names of the scientists involved.

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Linkage Groups

A linkage group is a set of genes located on the same chromosome that tend to be inherited together.

Why Linkage Matters

• Mendel's Law of Independent Assortment applies only to genes on different chromosomes
• Genes on the same chromosome do NOT assort independently — they are linked
• The number of linkage groups equals the haploid chromosome number (n)

Linked vs. Unlinked Genes

• Unlinked genes (on different chromosomes): produce a 1:1:1:1 ratio in test cross offspring
• Completely linked genes: produce only parental types (no recombinants)
• Partially linked genes: produce mostly parental types with some recombinant types

💡 Crossing over during prophase I of meiosis can separate linked genes, producing recombinant chromosomes.

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Morgan's Drosophila Experiments

Thomas Hunt Morgan used the fruit fly Drosophila melanogaster as a model organism because of its advantages:

• Short generation time (~2 weeks)
• Prolific reproduction (hundreds of offspring)
• Only 4 pairs of chromosomes (easy to study)
• Many visible mutations (eye color, wing shape, body color)

The White-Eye Discovery (1910)

Morgan crossed a white-eyed male fly with a red-eyed (wild-type) female:

P cross: ♀ red-eyed × ♂ white-eyed $F_{1}$: All red-eyed (red is dominant)

$F_{1}$ × $F_{1}$ cross:

• $F_{2}$ females: all red-eyed
• $F_{2}$ males: ½ red-eyed, ½ white-eyed

Morgan's Conclusion

The white-eye gene must be on the X chromosome:

• Males have only one X → a single recessive allele is expressed
• Females have two X chromosomes → need two copies of the recessive allele

This was the first gene mapped to a specific chromosome, directly supporting the chromosome theory of inheritance.

🔬 Morgan won the Nobel Prize in Physiology or Medicine in 1933 for his discoveries concerning the role of chromosomes in heredity.

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Sex-Linked Traits

Sex-linked traits are controlled by genes located on the sex chromosomes (X or Y). Because males and females have different combinations of sex chromosomes, these traits show distinctive inheritance patterns.

Sex Determination

In the XX-XY system:

• The X chromosome is large and carries many genes unrelated to sex
• The Y chromosome is small and carries few genes, mostly involved in male development
• Because the Y lacks most X-linked genes, a male has only one copy of nearly every X-linked gene

X-Linked Recessive Traits

X-linked recessive traits are the most commonly tested sex-linked pattern on the AP exam.

Key Characteristics

• More common in males than females
• Males need only one copy of the recessive allele (hemizygous: $X^{a}$ Y)
• Females need two copies to express the trait $(X^{a}X^{a})$
• Carrier females $(X^{A} X^{a})$ are phenotypically normal but can pass the allele to sons
• An affected male gets the allele from his mother (never from father)
• An affected father passes the allele to all daughters (who become carriers) but no sons

Why Males Are Affected More Often

Red-green color blindness and hemophilia are familiar X-linked recessive traits. In a population, a male shows the trait whenever his single X carries the allele, but a female shows it only when both of her X chromosomes carry it. So if 1 in 12 X chromosomes carries a color-blindness allele, about 1 in 12 males is color-blind, while far fewer females are.

Punnett Square: Carrier Female × Normal Male

Result: 50% of sons affected, 50% of daughters are carriers

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X-Linked Dominant Traits

X-linked dominant traits are expressed in both heterozygous females and hemizygous males.

Characteristics

• Affected males pass the trait to all daughters and no sons
• Affected heterozygous females pass to 50% of all children regardless of sex
• Often more common in females (they have two X chromosomes, so two chances to inherit)
• Never passed from father to son, because a son receives his father's Y, not his X

Y-Linked Traits

Genes on the Y chromosome that have no counterpart on the X:

• Passed exclusively from father to son
• All sons of an affected father are affected
• No daughters ever inherit the trait
• Very few Y-linked genes affect phenotype beyond male development, so Y-linked traits are rare in pedigrees

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Concept Check 🎯

Linked Genes & Recombination

When genes are on the same chromosome, they tend to be inherited together — this is genetic linkage. However, crossing over during meiosis can shuffle linked alleles, producing recombinant offspring.

Parental vs. Recombinant Types

Consider two linked genes (A and B) on the same chromosome:

Example

If a parent has alleles AB on one chromosome and ab on the homolog:

• Parental gametes: AB, ab
• Recombinant gametes: Ab, aB (from crossing over)

💡 Key rule: If two genes are linked, the recombinant classes will always be less frequent than the parental classes. If they're unlinked, all four classes appear in roughly equal proportions (~25% each).

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Recombination Frequency & Map Distance

Recombination frequency (RF) measures how often crossing over separates two linked genes.

Calculating RF

$RF = \frac{\text{Number of recombinant offspring}}{\text{Total offspring}} \times 100\%$

Map Distance

• 1% recombination frequency = 1 centimorgan (cM) = 1 map unit (m.u.)
• The more often two genes recombine, the farther apart they are on the chromosome
• Calculating map distance from recombination data is called gene (genetic) mapping

Important Limits

⚠️ RF never exceeds 50% because at most 50% of the meiotic products from a single crossover are recombinant (only 2 of 4 chromatids participate in any single crossover event).

Example Calculation

A test cross produces:

• 412 parental type 1
• 388 parental type 2
• 97 recombinant type 1
• 103 recombinant type 2

$RF = \frac{97 + 103}{412 + 388 + 97 + 103} \times 100\% = \frac{200}{1000} \times 100\% = 20\%$

Map distance = 20 cM (the genes are 20 map units apart)

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Building a Map from Two-Point Distances

Each two-point test cross gives one distance. Because map distances between nearby genes are roughly additive, distances measured in pairs can place three genes in order on a chromosome.

Worked Example: Ordering Three Genes

Three separate two-point test crosses give these recombination frequencies:

Step 1 — Find the outer genes. The largest distance (A–C, 19 cM) must span the two genes at the ends.

Step 2 — Place the middle gene. That leaves B in the middle.

Step 3 — Check additivity. A–B + B–C = 12 + 7 = 19 cM = A–C. ✓

Map: A ——12—— B ——7—— C

💡 Genes very far apart recombine about 50% of the time, just like genes on different chromosomes, so a long distance is measured most accurately by adding up shorter intervals between genes in the middle.

Testing for Linkage with Chi-Square

Linkage is identified by quantitative analysis: if two genes assort independently, a dihybrid test cross predicts four classes in a 1:1:1:1 ratio. Using the counts from the Recombination Frequency example (412, 388, 97, 103; total 1,000), the expected count is 250 per class:

$\chi^2 = \frac{162^2}{250} + \frac{138^2}{250} + \frac{153^2}{250} + \frac{147^2}{250} \approx 361$

With 4 classes, df = 4 − 1 = 3, and the critical value at p = 0.05 is 7.81. Because 361 is far above 7.81, we reject the hypothesis of independent assortment: the excess of parental types means the genes are linked.

(Three-point crosses and interference calculations are beyond the scope of the AP exam.)

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Chromosomal Mutations — Changes in Chromosome Number

Chromosomal mutations involve changes in the number or structure of chromosomes. These large-scale alterations can have dramatic effects on phenotype and are a significant source of genetic disorders.

Nondisjunction

Nondisjunction is the failure of chromosomes (or chromatids) to separate properly during cell division.

When It Occurs

Consequences

After fertilization with a normal gamete:

• Gamete with extra chromosome → trisomy (2n + 1)
• Gamete with missing chromosome → monosomy (2n − 1)

⚠️ Most aneuploid embryos are spontaneously aborted. It is estimated that ~25% of all conceptions involve chromosomal abnormalities.

Concept Check 🎯

Why Changes in Chromosome Number Disrupt Development

The CED states that changes in chromosome number often result in new phenotypes and in disorders with developmental limitations. The reason is gene dosage.

Gene dosage

A single chromosome carries hundreds to thousands of genes. A diploid cell normally has two copies of each, and development depends on gene products being made in balanced amounts.

Because the change affects every gene on the chromosome at once, many proteins are out of balance together. That is why aneuploidy usually changes development far more than a mutation in one gene does.

Predictable patterns

• More genes, bigger effect. Trisomies of large, gene-rich chromosomes are almost always lethal early in development; the few that survive to birth involve small chromosomes carrying relatively few genes.
• Sex chromosomes are a special case. Extra or missing sex chromosomes are often less severe, partly because the Y chromosome carries few genes.
• Every cell is affected. An error in meiosis produces a gamete, so the resulting zygote, and every cell that descends from it, carries the abnormal number.

Worked Example: Tracing an aneuploid zygote

Problem. In a human, the two homologs of one chromosome pair fail to separate in meiosis I. One of the resulting gametes is fertilized by a normal gamete. Give the possible chromosome numbers of the zygote.

Step 1. Nondisjunction in meiosis I gives two gametes with 24 chromosomes (n + 1) and two with 22 (n − 1).

Step 2. A normal gamete carries 23 chromosomes.

Step 3. 24 + 23 = 47 (trisomic, 2n + 1), or 22 + 23 = 45 (monosomic, 2n − 1).

📝 Scope note: You should be able to predict and explain the effects of changes in chromosome number, but the names of specific disorders caused by them are not tested on the AP exam.

Fill in the Blanks 🔍

Polyploidy

Polyploidy involves having more than two complete sets of chromosomes. It can arise when an error in mitosis or meiosis leaves a whole set of chromosomes in one cell.

Polyploidy in Nature

• Rare in animals, where it is usually lethal because it disrupts development.
• Common in plants. Many crop plants, including wheat, potatoes and strawberries, are polyploid.
• Triploids are usually sterile. With three copies of each chromosome, homologs cannot pair and separate evenly in meiosis, so most gametes are unbalanced. Seedless bananas and watermelons are triploid.

Polyploidy and Speciation

Suppose a diploid plant produces a tetraploid offspring through an error in cell division. The tetraploid can often self-fertilize and produce fertile tetraploid offspring. If it crosses with a diploid plant, however, the offspring are triploid and sterile. The tetraploid is therefore reproductively isolated from its parent population after a single generation, even though the two grow side by side. This is an example of sympatric speciation (Topic 7.10).

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Structural Chromosomal Changes

Structural chromosomal abnormalities involve rearrangements of chromosome segments rather than changes in chromosome number. They can arise from chromosome breakage, from errors in DNA repair, or from mistakes in crossing over. The CED states that alterations in chromosome structure lead to genetic disorders.

Four Major Types

💡 These changes affect gene dosage (how many copies of a gene are present), gene regulation (which regulatory sequences sit next to a gene), and can disrupt genes at the points where the chromosome broke.

Deletions

A deletion occurs when a chromosomal segment is lost.

Effects

• Every gene in the deleted segment is left with only one copy, on the other homolog, so its product is made at about half the normal amount.
• Because many genes are lost at once, deletions are usually harmful, and large deletions are often lethal.

Duplications

A duplication occurs when a chromosomal segment is copied, resulting in extra genetic material.

Effects

• Generally less harmful than deletions (extra genes vs. missing genes), but extra copies can still upset gene dosage.
• Duplications are an important source of new genes in evolution. While one copy keeps the original function, the other copy is free to accumulate mutations, and some of those changes give the copy a new function.
• The globin gene family, which includes the genes for the different chains of hemoglobin, evolved through repeated duplications followed by divergence of the copies.

How a Mistake in Crossing Over Makes Both

Crossing over normally exchanges matching segments, so no genes are gained or lost. If two homologs pair slightly out of register, a crossover between them is unequal:

One crossover event produces a duplication on one chromatid and the matching deletion on the other. Gametes that receive these chromatids pass the structural change to the next generation.

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Inversions

An inversion occurs when a chromosome segment breaks at two points, flips 180°, and reinserts.

Effects

• All genes are still present, so the organism is often phenotypically normal.
• An inversion causes a problem when a break point falls inside a gene, cutting it in two, or when it separates a gene from the regulatory sequences that control its transcription.

Translocations

A translocation is the transfer of a chromosomal segment to a non-homologous chromosome.

Effects

• If no genes are lost or broken, the carrier is usually healthy, but some of the gametes the carrier makes receive a chromosome with a missing or extra segment. Offspring from those gametes have unbalanced gene dosage.
• A translocation can break a gene at the junction, or move a gene next to different regulatory sequences. A gene that promotes cell division, placed under the control of a strongly active regulatory region, may be expressed far more than normal, and that overexpression can contribute to cancer.

Worked Reasoning: Predicting the Effect of a Rearrangement

Name the Change 🔍

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Non-Nuclear (Extranuclear) Inheritance

Not every gene sits on a chromosome in the nucleus. Mitochondria and chloroplasts carry their own small, circular DNA, and traits determined by that DNA show a pattern of inheritance that does not follow Mendel's rules. The CED calls this non-nuclear inheritance.

Why Organelle Genes Break Mendel's Rules

💡 Organelle DNA is circular and resembles a bacterial chromosome, one line of evidence that mitochondria and chloroplasts descend from free-living prokaryotes (endosymbiosis).

(Other parent-of-origin effects you may see in textbooks, such as genomic imprinting and maternal-effect genes, are beyond the scope of the AP exam.)

Random Assortment of Organelles

A single cell can hold hundreds of mitochondria. When the cell divides, its mitochondria are split between the two daughter cells at random; nothing sorts them evenly, the way the spindle sorts chromosomes.

This matters when a cell contains a mixture of normal and mutant mitochondrial DNA:

Consequences

• Variable severity. Siblings who inherit mitochondria from the same mother can receive very different proportions of mutant mitochondria, so a mitochondrial disorder can be mild in one child and severe in another.
• Tissue differences. Mutant mitochondria cause the most harm in tissues with high energy demand, such as muscle and nerve, because mitochondria carry out most of the cell's ATP production by cellular respiration.
• No fixed ratio. Because the split is random, offspring do not fall into predictable Mendelian classes.

Concept Check 🎯

Maternal Inheritance in Animals and Plants

Animals

The egg contributes nearly all of the zygote's cytoplasm, including its mitochondria. Sperm mitochondria are not passed on. Mitochondrial traits are therefore maternally inherited:

• An affected mother can pass the trait to all of her children, sons and daughters alike.
• An affected father passes it to none of his children.

Plants

In plants, mitochondria and chloroplasts are transmitted in the ovule (which contains the egg) and not in the pollen. Traits determined by chloroplast or mitochondrial DNA are therefore also maternally inherited: offspring resemble the plant that produced the seed.

Worked Example: Reciprocal Crosses with a Variegated Plant

Some four-o'clock plants have green branches, white branches (whose chloroplasts cannot make chlorophyll) and variegated branches with both. Flowers on different branches are crossed, and the seedlings are scored.

Step 1 — Compare the reciprocal crosses. Green × white and white × green give opposite results, so the sex of the parent matters.

Step 2 — Identify which parent decides. In every row, the seedlings match the seed (egg) parent, whatever the pollen carried.

Step 3 — Explain the variegated row. A variegated branch's eggs receive random mixtures of normal and defective chloroplasts, so the seedlings vary with no fixed ratio.

Conclusion. Leaf color here is determined by chloroplast DNA, inherited through the egg's cytoplasm.

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Concept Check 🎯

AP Exam Practice — Chromosomal Inheritance Problems

This section contains multi-step genetics problems similar to those on the AP Biology exam. Work through each problem carefully, applying concepts from the entire unit.

Problem-Solving Strategy

1. Identify the inheritance pattern (autosomal vs. sex-linked, dominant vs. recessive, linked vs. unlinked)
2. Assign genotypes to all known individuals
3. Set up appropriate crosses (Punnett squares, branch diagrams)
4. Calculate expected ratios and compare to observed data
5. Check your work — do the ratios make biological sense?

Practice Problem Set 1: Sex-Linked Crosses 🎯

Gene Mapping Practice

Problem: Two-Point Test Cross

A test cross involving two linked genes produces the following offspring:

Step 1: Identify parental and recombinant classes

• Parental: AB (354) and ab (346) → total = 700
• Recombinant: Ab (52) and aB (48) → total = 100

Step 2: Calculate recombination frequency

$RF = \frac{100}{800} \times 100\% = 12.5\%$

Step 3: Map distance = 12.5 cM

The genes are 12.5 centimorgans apart on the same chromosome.

Practice Problem Set 2: Gene Mapping 🎯

Practice Problem 3: Chromosome Number and Structure 🔍

Practice Problem Set 3: Multi-Step Problems 🎯