Inheritance

Inheritance is the transmission of genetic information from parents to offspring. The AAMC MCAT content outline expects you to apply Mendel’s laws of segregation and independent assortment, work Punnett squares for monohybrid and dihybrid crosses, handle non-Mendelian patterns (incomplete dominance, codominance, multiple alleles), analyze pedigrees and X-linked traits, and apply the Hardy-Weinberg equation to populations. This is one of the most heavily tested areas of the Bio/Biochem section.

Key exam topics. Nail Mendel’s laws and Punnett squares, non-Mendelian inheritance (incomplete dominance, codominance, ABO), sex-linkage and pedigrees, gene linkage and recombination mapping, and the Hardy-Weinberg equilibrium. Numerical Punnett-square and Hardy-Weinberg questions are almost guaranteed.

Key Genetic Terms

Gene: A unit of inheritance — a length of DNA coding for a polypeptide or functional RNA.
Allele: An alternative form of a gene at a given locus (e.g. T for tall, t for short).
Genotype / Phenotype: Genotype is the genetic make-up (TT, Tt, tt). Phenotype is the observable trait (tall, short).
Homozygous / Heterozygous: Two identical alleles (TT or tt) vs two different alleles (Tt).
Dominant / Recessive: A dominant allele expresses its phenotype in the heterozygote; a recessive allele needs both copies to be expressed.
Test cross: Cross of an unknown phenotype with a homozygous recessive individual to reveal the unknown genotype.

Mendel’s Laws of Inheritance

Gregor Mendel worked with the garden pea (Pisum sativum) and seven contrasting traits. From thousands of crosses he deduced three statistical laws.

Law 1 — Law of Dominance

When two pure-breeding parents differing in one character are crossed, only one form (the dominant) appears in the F₁; the other (recessive) is hidden. Example: TT × tt → all Tt (tall).

Law 2 — Law of Segregation

The two alleles of a gene separate during gamete formation, so each gamete carries only one allele. They reunite at random in fertilization. A monohybrid F₁ × F₁ cross gives a phenotypic ratio of 3 : 1 and a genotypic ratio of 1 : 2 : 1. This is the single most important of Mendel’s laws.

Law 3 — Law of Independent Assortment

Alleles of different genes located on different chromosomes assort independently in gamete formation. A dihybrid F₁ × F₁ cross (RrYy × RrYy) gives a phenotypic ratio of 9 : 3 : 3 : 1. Note the important caveat: independent assortment holds only for genes on different chromosomes (or far apart on the same one).

Chromosomal basis of inheritance. Mendel’s abstract “factors” are genes carried on chromosomes (the Sutton-Boveri chromosome theory). Segregation reflects the separation of homologous chromosomes in anaphase I of meiosis; independent assortment reflects the random orientation of homologous pairs at the metaphase-I plate. Fertilization then restores the diploid number.

Worked monohybrid cross

Parental cross: TT × tt → F₁ = Tt (all tall). F₁ self-cross: Tt × Tt → gametes T and t from each side. Punnett square:

TT, Tt, Tt, tt → genotypic ratio 1 TT : 2 Tt : 1 tt
Tall : short = 3 : 1

Worked dihybrid cross

RrYy × RrYy (round/yellow vs wrinkled/green pea seeds). Each parent makes 4 gamete types: RY, Ry, rY, ry. The 16-cell Punnett square gives phenotypes:

9 round yellow
3 round green
3 wrinkled yellow
1 wrinkled green

Hence the famous 9 : 3 : 3 : 1 ratio.

Common Mendelian cross outcomes (F₂ generation)
Cross	Genotypic ratio	Phenotypic ratio	When to expect
Monohybrid (Aa × Aa)	1 AA : 2 Aa : 1 aa	3 : 1	One gene, complete dominance
Test cross (Aa × aa)	1 Aa : 1 aa	1 : 1	To determine an unknown genotype
Dihybrid (AaBb × AaBb)	9 : 3 : 3 : 1 phenotypes (16 genotypes)	9 : 3 : 3 : 1	Two genes, independent, complete dominance
Dihybrid test cross (AaBb × aabb)	1 : 1 : 1 : 1	1 : 1 : 1 : 1	Two genes, independent (no linkage)
Incomplete dominance (Aa × Aa)	1 : 2 : 1	1 : 2 : 1	Heterozygote shows intermediate phenotype
Codominance (e.g. ABO)	1 : 2 : 1	Both alleles fully expressed	I^AI^B = AB blood group

Common trap. The 9 : 3 : 3 : 1 ratio holds only when the two genes are on different chromosomes (truly independent). If they are linked on the same chromosome, ratios deviate predictably (more parental types, fewer recombinants).

Gene Linkage and Crossing Over

Genes located close together on the same chromosome tend to be inherited as a unit — they are linked. Linked genes violate Mendel’s law of independent assortment.

Linkage groups

All genes on a single chromosome form one linkage group. Humans have 23 linkage groups (one per pair). The closer two loci sit, the more tightly they are linked.

Crossing over

During prophase I of meiosis, homologous chromosomes pair up (synapsis) and exchange equivalent segments at chiasmata. This recombines maternal and paternal alleles, producing recombinant gametes. The frequency of recombination between two loci (recombination frequency, RF) is roughly proportional to the distance between them. 1% RF = 1 map unit (centimorgan).

Linked genes therefore produce more parental-type gametes than recombinant ones. If the two loci are far apart, RF approaches 50%, mimicking independent assortment.

X-linked Recessive Inheritance

Some genes are carried on the X chromosome. Because males have only one X (XY), a single recessive allele is enough to express the trait. Females (XX) need two copies. As a result, X-linked recessive disorders are far more common in males.

Classical examples

Hemophilia A — deficiency of clotting factor VIII; affected individuals bleed excessively. Famous in the European royal families descended from Queen Victoria.
Red-green color blindness — defective opsin gene; cannot distinguish reds and greens.
Duchenne muscular dystrophy — defective dystrophin gene; progressive muscle wasting.

Pedigree pattern

Use X^H for the normal allele and X^h for the recessive disease allele.

X^HX^H — normal female
X^HX^h — carrier female (clinically normal)
X^hX^h — affected female (rare)
X^HY — normal male
X^hY — affected male

Carrier mother × normal father (X^HX^h × X^HY) → offspring:

1/4 normal daughter (X^HX^H)
1/4 carrier daughter (X^HX^h)
1/4 normal son (X^HY)
1/4 affected son (X^hY) — 50% of sons are affected

Pedigree shortcut. If a trait skips a generation, affects mostly males, and there is no father-to-son transmission, suspect X-linked recessive. (Sons inherit Y, not X, from their fathers, so the dad cannot pass an X-linked allele to a son.)

Autosomal inheritance patterns

Autosomal recessive: sickle cell anemia, cystic fibrosis, phenylketonuria, Tay-Sachs, albinism. Two affected parents cannot have an unaffected child; trait can skip generations.
Autosomal dominant: Huntington’s disease, achondroplasia, Marfan syndrome, familial hypercholesterolemia. Every affected child has at least one affected parent; no skipping.

Non-Mendelian Inheritance

Many traits deviate from simple complete dominance. These variations are heavily tested because they change the expected Punnett-square ratios.

Incomplete dominance

The heterozygote shows an intermediate (blended) phenotype because one allele does not fully mask the other. Classic example: a red (RR) snapdragon × white (rr) gives all pink (Rr) F₁. Crossing the pink F₁ gives a 1 red : 2 pink : 1 white ratio — the phenotypic ratio equals the genotypic ratio (1 : 2 : 1).

Codominance

Both alleles are fully and simultaneously expressed in the heterozygote — not blended. Example: the MN blood group, and the AB blood type where both A and B antigens appear on red blood cells. Distinguish carefully from incomplete dominance: codominance shows both distinct phenotypes at once, whereas incomplete dominance shows a single intermediate.

Multiple alleles — the ABO blood group

A gene can have more than two alleles in a population (though any individual still carries only two). The ABO gene has three alleles: I^A and I^B (codominant to each other) and i (recessive to both).

ABO genotypes and phenotypes
Blood type	Genotype(s)	Antigens on RBC	Antibodies in plasma
A	I^AI^A or I^Ai	A	anti-B
B	I^BI^B or I^Bi	B	anti-A
AB (universal recipient)	I^AI^B	A and B	none
O (universal donor)	ii	none	anti-A and anti-B

Penetrance vs expressivity

Penetrance is the proportion of individuals with a given genotype who actually show the expected phenotype (all-or-none, measured across a population). A dominant allele with 80% penetrance is expressed in only 80% of carriers. Expressivity is how strongly or variably the phenotype is displayed among those who do express it (a matter of degree in one individual). Example: polydactyly is incompletely penetrant and variably expressive.

Hardy-Weinberg Equilibrium

The Hardy-Weinberg principle lets you predict allele and genotype frequencies in a non-evolving population — it is the null model of population genetics and a guaranteed calculation on the MCAT.

The two equations

For a gene with two alleles, let p = frequency of the dominant allele and q = frequency of the recessive allele:

p + q = 1 (allele frequencies sum to 1)
p² + 2pq + q² = 1 (genotype frequencies sum to 1)

where p² = homozygous dominant, 2pq = heterozygous (carriers), and q² = homozygous recessive. Test-taking tip: the recessive phenotype frequency gives you q² directly, so take its square root to get q, then find p = 1 − q.

The five assumptions

Allele frequencies stay constant only if there is: (1) no mutation, (2) random mating (no assortative mating), (3) no gene flow (no migration in or out), (4) no natural selection (all genotypes equally fit), and (5) a very large population (no genetic drift). If any is violated, the population is evolving.

Common trap. The frequency of the recessive allele (q) is not the same as the frequency of the recessive phenotype (q²). If 1 in 100 people show a recessive disorder, q² = 0.01, so q = 0.1 — and carriers (2pq = 2 × 0.9 × 0.1 = 0.18) make up 18%, far more than the affected.

Worked MCQs

Five MCQs covering numerical and conceptual patterns most often tested. Work the cross on paper before reading the explanation.

Q1. A heterozygous tall pea plant (Tt) is self-pollinated. What is the expected phenotypic ratio in the offspring?

1 : 1
3 : 1 (tall : short)
9 : 3 : 3 : 1
1 : 2 : 1

Tt × Tt gives genotypes 1 TT : 2 Tt : 1 tt. Both TT and Tt are tall (3 tall) while tt is short (1 short), giving a 3 : 1 phenotypic ratio — the classic monohybrid result and a direct demonstration of the law of segregation.

Q2. A 9 : 3 : 3 : 1 phenotypic ratio in the F₂ generation supports which of Mendel’s laws?

Law of dominance
Law of segregation
Law of independent assortment
Law of linkage

A 9 : 3 : 3 : 1 dihybrid ratio appears only when the two gene loci segregate independently — that is, when they sit on different chromosomes (or far apart on the same chromosome). This is Mendel’s third law.

Q3. A carrier mother for hemophilia (X^HX^h) marries a normal man (X^HY). What is the probability that any one son is hemophilic?

0%
25%
50%
100%

Sons receive Y from the father and either X^H or X^h from the carrier mother — each with 50% probability. Half of the sons are X^HY (normal) and half are X^hY (hemophilic). Note: among all children the affected fraction is 25%.

Q4. Crossing over during meiosis occurs at structures called:

Centromeres
Chiasmata
Kinetochores
Telomeres

Chiasmata are the visible X-shaped points where non-sister chromatids of homologous chromosomes have exchanged equivalent segments during prophase I (specifically the pachytene stage). They are the cytological evidence of crossing over.

Q5. In a population at Hardy-Weinberg equilibrium, 16% of individuals are homozygous recessive (aa) for a trait. What percentage of the population are heterozygous carriers?

q² = 0.16, so q = 0.4 and p = 1 − 0.4 = 0.6. Heterozygous carriers are 2pq = 2 × 0.6 × 0.4 = 0.48 = 48%. The 36% distractor is p² (homozygous dominant). Remember that the recessive phenotype gives you q² directly — take its square root first.

Quick Recap

Mendel’s core laws: segregation (allele separation in anaphase I) and independent assortment.
Monohybrid F₂ = 3 : 1 phenotypic; 1 : 2 : 1 genotypic. Dihybrid F₂ = 9 : 3 : 3 : 1 (only if loci unlinked).
Incomplete dominance → intermediate phenotype (1 : 2 : 1); codominance → both alleles expressed (ABO, MN).
ABO: I^A/I^B codominant, i recessive; AB = universal recipient, O = universal donor.
Penetrance = fraction of genotypes that show the trait; expressivity = how strongly it shows.
Linked genes → more parental than recombinant gametes; RF (1% = 1 cM) measures map distance; crossing over occurs at chiasmata in prophase I.
X-linked recessive: more common in males; no father-to-son transmission; carrier mothers transmit to half of sons (hemophilia, red-green color blindness, Duchenne).
Hardy-Weinberg: p + q = 1 and p² + 2pq + q² = 1; q² = recessive phenotype; holds only with no mutation/selection/migration/drift and random mating.

Test yourself. Take a timed practice test or browse topic-wise MCQs to lock these concepts in.