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Chromosomal Basis
of Inheritance

How genes on chromosomes drive the patterns of inheritance that Mendel first described: from meiotic segregation to linked genes and recombination.

MCAT Biology

What Are Chromosomes?

Chromosomes are tightly coiled structures of DNA wrapped around histone proteins. In humans, somatic cells contain 46 chromosomes organized as 23 homologous pairs: one set from each parent.

Autosomes
22 pairs that carry non-sex-determining genes shared by all individuals.
Sex Chromosomes
The 23rd pair: \(XX\) (female) or \(XY\) (male), determining biological sex.

The Chromosome Theory of Inheritance

In 1902, Sutton and Boveri independently proposed that Mendel's factors (genes) are carried on chromosomes. The behavior of chromosomes during meiosis perfectly explains Mendel's laws.

1
Genes are on chromosomes: Each gene occupies a specific locus on a specific chromosome.
2
Chromosomes segregate: Homologs separate during meiosis I, explaining Mendel's Law of Segregation.
3
Independent assortment: Non-homologous chromosomes sort independently into gametes during meiosis I.

Independent Assortment in Meiosis

Diploid Parent Cell
\(2n=4\), Genotype \(AaBb\)
 A a B b Meiosis I Possibility 1 Possibility 2 (Equally likely alignments at Metaphase I) Meiosis II Meiosis II Resulting Gametes AB ab Ab aB 25% 25% 25% 25% Four types of gametes are produced in equal proportions

Morgan's Drosophila Experiments

Thomas Hunt Morgan (1910) provided the first evidence that genes are physically located on chromosomes. He discovered a white-eyed male fruit fly among normal red-eyed flies and traced the trait to the X chromosome.

1
Cross: White-eyed male (\(X^w Y\)) x Red-eyed female (\(X^W X^W\))
2
F1: All red-eyed. Females were \(X^W X^w\) carriers, males were \(X^W Y\).
3
F2: All females red-eyed, but half the males were white-eyed: only males showed the recessive phenotype.

Sex-Linked Inheritance Patterns

X-linked Recessive Carrier
\(X^A X^a\)
 Normal
\(X^A Y\)
 x Normal
\(X^A X^A\)
 25% Carrier
\(X^A X^a\)
 25% Normal
\(X^A Y\)
 25% Affected
\(X^a Y\)
 25% Key Features More males affected Carrier females pass to sons No male-to-male transmission Affected father + carrier mother = affected daughters = Female = Male

Linked Genes and Recombination

Genes located on the same chromosome tend to be inherited together: they are linked. However, crossing over during meiosis I can shuffle linked alleles, producing recombinant offspring.

Before Crossing Over A B a b Crossing Over After Crossing Over A b Recombinant a B Recombinant Parental Parental chiasma
\(\text{Recombination Frequency} = \frac{\text{recombinant offspring}}{\text{total offspring}} \times 100\%\)

Recombination Frequency and Gene Mapping

The recombination frequency (RF) between two linked genes reflects their physical distance on a chromosome. Genes farther apart cross over more often. For unlinked genes, independent assortment produces \(2^n\) gamete combinations where \(n\) is the number of heterozygous gene pairs.

Close Together
Low RF (e.g., 5%): rarely recombine, mostly parental types.
Moderate Distance
Moderate RF (e.g., 20%): some recombinants observed.
Far Apart / Unlinked
RF near 50%: behave as if on separate chromosomes.

Linked vs Unlinked Genes

Linked Genes Unlinked Genes
Location Same chromosome Different chromosomes
Assortment Tend to co-segregate Independent assortment
Recombination \(RF < 50\%\) \(RF = 50\%\)
Offspring ratio (dihybrid) Deviates from \(9:3:3:1\) Approximates \(9:3:3:1\)
Gamete types (from dihybrid) Parental > Recombinant All 4 types are equal (25%)
Test cross result Unequal classes \(1:1:1:1\) ratio

Chromosomal Abnormalities

Errors in chromosome number or structure can cause genetic disorders. These arise from nondisjunction (failure of chromosomes to separate) or structural rearrangements.

Aneuploidy
Abnormal chromosome number: monosomy (\(2n-1\)) or trisomy (\(2n+1\)). Example: Trisomy 21 (Down syndrome).
Polyploidy
Entire extra set of chromosomes (\(3n, 4n\)). Common in plants, usually lethal in animals.
Deletions
A segment is lost. Example: Cri-du-chat (5p deletion).
Translocations
A segment moves to a non-homologous chromosome. Example: Philadelphia chromosome (CML).

How Nondisjunction Works

Normal Meiosis
\(2n\)
\(n\)
\(n\)
\(n\)
\(n\)
All gametes normal (\(n\))
 Nondisjunction (Meiosis I)
\(2n\)
 Both homologs Empty
\(n\!+\!1\)
\(n\!+\!1\)
\(n\!-\!1\)
\(n\!-\!1\)
 Trisomy Monosomy Nondisjunction in Meiosis I affects ALL four gametes Nondisjunction in Meiosis II affects only 2 of 4 gametes

How to Solve Linkage Problems

1
Identify the cross: Determine the genotypes of the parents and whether it is a test cross (one parent homozygous recessive).
2
Classify offspring: Group progeny into parental types (most frequent) and recombinant types (least frequent).
3
Calculate RF: \(\text{RF} = \frac{\text{recombinant offspring}}{\text{total offspring}} \times 100\%\)
4
Determine linkage: \(RF < 50\%\) = linked (distance in cM = RF%). \(RF = 50\%\) = unlinked or far apart.
5
Build the map: Use RF values between gene pairs to order genes and set distances along the chromosome.

Example: Test Cross for Linkage

A fruit fly heterozygous for body color (B/b) and wing shape (V/v) is test-crossed with a homozygous recessive fly. The offspring are:

Phenotype Genotype Count Type
Gray body, normal wings\(BbVv\)405Parental
Black body, vestigial wings\(bbvv\)395Parental
Gray body, vestigial wings\(Bbvv\)100Recombinant
Black body, normal wings\(bbVv\)100Recombinant

Example: Three-Point Gene Mapping

Three linked genes have the following pairwise recombination frequencies:

A ↔ B
12 cM
B ↔ C
7 cM
A ↔ C
19 cM
A B C 12 cM 7 cM 19 cM total

Example: X-Linked Pedigree Analysis

A woman with normal vision whose father was color-blind marries a man with normal vision. What fraction of their sons will be color-blind?

1
Mother's genotype: Her father was \(X^c Y\), so she received \(X^c\). She's normal, so her other X must be \(X^C\). Mother = \(X^C X^c\) (carrier).
2
Father's genotype: Normal vision = \(X^C Y\).
3
Cross: \(X^C X^c \times X^C Y\). Sons get Y from father and either \(X^C\) or \(X^c\) from mother.
4
Result: \(\frac{1}{2}\) of sons will be \(X^c Y\) = color-blind. Answer: 50% of sons.

Common Mistakes

✕ Confusing Linkage with Pleiotropy
Linked genes are near each other on a chromosome. Pleiotropy is one gene affecting multiple traits. They are distinct concepts.
✕ Assuming RF = 50% Means Unlinked
RF of 50% can also mean genes are very far apart on the same chromosome (multiple crossovers average out).
✕ Forgetting Hemizygosity
Males have only one X. A single recessive X-linked allele is expressed. Do not assign carrier status to males.

Summary of Key Concepts

Concept Definition Key Detail
Chromosome TheoryGenes reside on chromosomesSutton-Boveri, 1902
SegregationAlleles separate in meiosis IHomologs go to different gametes
Independent AssortmentNon-homologous pairs sort independently\(2^n\) gamete combinations
LinkageGenes on same chromosome co-segregate\(RF < 50\%\)
RecombinationCrossing over shuffles linked alleles1 cM = 1% RF
Sex LinkageX-linked genes show sex-biased inheritanceMales are hemizygous
NondisjunctionFailure of chromosome separationAneuploidy (trisomy/monosomy)

Question 1 of 5

The chromosome theory of inheritance states that Mendel's Law of Segregation is explained by which event during meiosis?

Question 2 of 5

In a test cross of a dihybrid fly (\(AaBb\)) with a double recessive (\(aabb\)), the offspring ratios are: 42% \(AaBb\), 42% \(aabb\), 8% \(Aabb\), 8% \(aaBb\). What is the recombination frequency?

Question 3 of 5

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

Question 4 of 5

Nondisjunction during meiosis I differs from nondisjunction during meiosis II in that meiosis I nondisjunction:

Question 5 of 5

Three genes (X, Y, Z) show the following recombination frequencies: \(X\text{-}Y = 8 \text{ cM}\), \(Y\text{-}Z = 12 \text{ cM}\), \(X\text{-}Z = 20 \text{ cM}\). What is the correct gene order?