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Genetics & Evolution

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Genetics & Evolution - Complete Interactive Lesson

Part 1: Mendelian Genetics

Genetics & Evolution for the MCAT

Part 1 of 7 — Mendelian Genetics

Mendel's Laws

Law of Segregation: The two alleles for each gene separate during gamete formation, so each gamete carries only one allele.
Law of Independent Assortment: Genes located on different chromosomes (or far apart on the same chromosome) sort into gametes independently of one another.

Key Terminology

Term	Definition
Genotype	Genetic makeup (e.g., Aa)
Phenotype	Observable physical expression
Homozygous	Same alleles (AA or aa)
Heterozygous	Different alleles (Aa)
Dominant	Expressed in the heterozygote
Recessive	Only expressed when homozygous

Cross Types

Monohybrid cross (Aa $\times$ Aa):

Genotype ratio: 1 AA : 2 Aa : 1 aa
Phenotype ratio: 3 dominant : 1 recessive

Test cross: Cross an unknown dominant phenotype with a homozygous recessive (aa) individual.

If all offspring are dominant → the parent was AA
If 50% dominant, 50% recessive → the parent was Aa

MCAT Punnett Square Strategy

Always set up the cross systematically. For dihybrid crosses (AaBb $\times$ AaBb), use a 4 $\times$ 4 Punnett square or the shortcut phenotype ratio: $9\ \text{A\_B\_} : 3\ \text{A\_bb} : 3\ \text{aaB\_} : 1\ \text{aabb}$

Worked Example — Dihybrid Probability Without Drawing the Square

Problem. In pea plants, tall (T) is dominant to short (t), and round seeds (R) are dominant to wrinkled (r). Two plants heterozygous for both genes are crossed: $TtRr \times TtRr$ . What fraction of the offspring are expected to be tall AND wrinkled?

Strategy — use the product rule instead of the full 16-box grid. Because the two genes assort independently, you can treat each gene as a separate monohybrid cross and multiply the probabilities.

Step 1 — Probability of tall. $Tt \times Tt$ gives $\frac{3}{4}$ tall (T_) and short (tt). So .

Mendelian Genetics 🎯

Key Takeaways — Part 1

Mendelian ratios: monohybrid 3:1 (phenotype), 1:2:1 (genotype); dihybrid 9:3:3:1
Test cross with a homozygous recessive reveals an unknown genotype
Law of Segregation: alleles separate. Independent Assortment: genes on different chromosomes sort independently (cellular basis = random metaphase I orientation)
Use the product rule (multiply single-gene probabilities) to handle dihybrid questions quickly
Each fertilization is independent — past offspring don't change future probabilities

Part 2: Non-Mendelian Inheritance

Genetics & Evolution for the MCAT

Part 2 of 7 — Non-Mendelian Inheritance

Extensions to Mendel

Pattern	Description	Example
Incomplete dominance	Heterozygote = intermediate phenotype	Red $\times$ White → Pink flowers
Codominance	Both alleles fully expressed	Blood type AB (both A and B antigens)
Multiple alleles	>2 alleles exist in the population	ABO blood type ( $I^A$ , , i)

Part 3: Population Genetics

Genetics & Evolution for the MCAT

Part 3 of 7 — Sex-Linked Inheritance & Pedigrees

X-Linked Inheritance

Males (XY) have only ONE X chromosome → they are hemizygous for X-linked genes
X-linked recessive diseases affect males more often (no second X to mask the allele)
Carrier females ( $X^AX^a$ ) pass the recessive allele to ~50% of their sons
A father transmits his single X to all daughters and his Y to all sons → no male-to-male X-linked transmission

Common X-Linked Recessive Diseases

Color blindness
Hemophilia A and B
Duchenne muscular dystrophy
G6PD deficiency

Pedigree Analysis

Key patterns to recognize:

Autosomal Dominant: Affected individuals in every generation; males and females affected roughly equally; two unaffected parents do not produce affected children.

Part 4: Natural Selection

Genetics & Evolution for the MCAT

Part 4 of 7 — Population Genetics

Hardy-Weinberg Equilibrium

For a population that is NOT evolving, allele and genotype frequencies stay constant and obey:

$p + q = 1$

$p^2 + 2pq + q^2 = 1$

Part 5: Speciation & Phylogeny

Genetics & Evolution for the MCAT

Part 5 of 7 — Evolution & Natural Selection

Mechanisms of Evolution

Mechanism	Description	Direction
Natural selection	Differential survival/reproduction based on fitness	Adaptive
Genetic drift	Random changes in allele frequency	Random
Gene flow	Migration between populations	Reduces differences
Mutation	Introduces new alleles	Random, raw material

Types of Natural Selection

Type	Effect on Distribution	Example
Stabilizing	Narrows distribution (favors the average)	Human birth weight
Directional	Shifts the mean toward one extreme	Antibiotic resistance

Part 6: Immune System

Genetics & Evolution for the MCAT

Part 6 of 7 — Speciation & Phylogenetics

Speciation

Biological species concept: A species is a group of organisms that can interbreed and produce fertile offspring.

Type	Barrier	Example
Allopatric	Geographic isolation	A river divides a population
Sympatric	Reproductive isolation in the same location	Polyploidy in plants

Reproductive Barriers

Prezygotic (prevent mating or fertilization — no zygote forms):

Temporal isolation (different mating seasons or times of day)
Behavioral isolation (different courtship rituals or signals)
Habitat isolation (different microhabitats within the same region)
Mechanical isolation (incompatible reproductive anatomy)
Gametic isolation (sperm and egg cannot fuse)

Postzygotic (a hybrid forms but has reduced fitness):

Hybrid inviability (the embryo dies)
Hybrid sterility (the mule = horse $\times$ donkey is sterile)

Part 7: Review & MCAT Practice

Genetics & Evolution for the MCAT

Part 7 of 7 — Genetic Diseases & Chromosomal Abnormalities

Autosomal Dominant Diseases

Disease	Gene / Feature
Huntington's disease	HTT gene, CAG trinucleotide repeat expansion
Marfan syndrome	Fibrillin-1, connective tissue
Familial hypercholesterolemia	LDL receptor deficiency
Achondroplasia	FGFR3 mutation, short-limbed dwarfism

Autosomal Recessive Diseases

Disease	Feature	Population
Cystic fibrosis	CFTR chloride channel defect (thick mucus)	European descent
Sickle cell anemia	HbS (Glu→Val at position 6), pleiotropic	African descent
Phenylketonuria (PKU)	Cannot metabolize phenylalanine	Newborn screening
Tay-Sachs	Hexosaminidase A deficiency	Ashkenazi Jewish

Chromosomal Abnormalities

P(\text{tall}) = \tfrac{3}{4}

Step 2 — Probability of wrinkled. $Rr \times Rr$ gives $\tfrac{3}{4}$ round (R_) and $\tfrac{1}{4}$ wrinkled (rr). So $P(\text{wrinkled}) = \tfrac{1}{4}$ .

Step 3 — Multiply (product rule). $P(\text{tall and wrinkled}) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

This matches the $9{:}3{:}3{:}1$ ratio: the "tall, wrinkled" class (A_bb-style) is $\tfrac{3}{16}$ . On test day the product rule is faster and far less error-prone than drawing a 16-box square — reserve the grid for when you need the full distribution.

ABO Blood Type (MCAT FAVORITE)

Genotype	Blood Type	Antigens	Antibodies
$I^AI^A$ or $I^Ai$	A	A antigen	Anti-B
$I^BI^B$ or $I^Bi$	B	B antigen	Anti-A
$I^AI^B$	AB	Both A and B	Neither
ii	O	Neither	Anti-A and Anti-B

$I^A$ and $I^B$ are codominant to each other
Both are dominant over i
Type O = universal donor (no surface antigens to attack)
Type AB = universal recipient (no antibodies in plasma)

Mitochondrial (Cytoplasmic) Inheritance

mtDNA is inherited maternally (sperm mitochondria are excluded/degraded after fertilization)
An affected mother passes the trait to all of her children; affected fathers pass it to none
Examples: Leber hereditary optic neuropathy (LHON), MELAS, mitochondrial myopathies

Worked Example — A Blood-Type Paternity Question

Problem. A newborn has blood type O. The mother has blood type A. The hospital is trying to determine which of two men could be the father. Man 1 has type AB; man 2 has type B. Which man(men) could be the biological father?

Step 1 — Infer the genotypes from the phenotypes. The baby is type O, so its genotype must be ii — it received one i allele from each parent. Therefore every candidate parent must carry an i allele.

Step 2 — Check the mother. She is type A, but since the baby is ii, the mother must be $I^Ai$ (heterozygous). She contributed the i to the baby. Consistent.

Step 3 — Test each man for an i allele.

Man 1 (type AB): genotype is $I^AI^B$ . He has no i allele, so he cannot contribute the second i needed for an ii child. Excluded.
Man 2 (type B): he could be $I^BI^B$ or $I^Bi$ . Only $I^Bi$ can father a type O child, and that is genetically possible.

Conclusion. Man 2 is the only possible biological father. This is the core logic of ABO paternity/exclusion problems on the MCAT: a type O child requires an i allele from both parents, so any parent lacking an i (i.e., type AB) is ruled out.

Non-Mendelian 🎯

Key Takeaways — Part 2

Incomplete dominance: blended intermediate (1:2:1 phenotype = genotype). Codominance: both alleles fully expressed (AB blood).
ABO: $I^A$ and $I^B$ codominant, both dominant over i; type O child needs an i from BOTH parents
Pleiotropy = one gene, many effects. Epistasis = one gene masks another (different loci).
Polygenic traits show continuous variation (bell curve)
Mitochondrial DNA is inherited maternally → affected mother passes to all children

Autosomal Recessive: Can skip generations; often appears in consanguineous (related) matings; carrier $\times$ carrier gives a 25% affected risk.

X-Linked Recessive: Mostly males affected; a carrier mother → ~50% of sons affected; NO male-to-male transmission.

X-Linked Dominant: Affected fathers pass the trait to ALL daughters and NO sons; more females affected overall.

MCAT Pedigree Strategy

Look for male-to-male transmission → if present, the trait is NOT X-linked.
Does the trait skip generations? → if yes, it is likely recessive.
Count affected males vs. females → a strong male excess suggests X-linked recessive.
Two unaffected parents with an affected child → the trait is recessive (and the parents are carriers).

Worked Example — Carrier Risk for Hemophilia

Problem. Hemophilia A is X-linked recessive ( $X^h$ ). A phenotypically normal woman's brother and father both have hemophilia, but she herself does not. She marries an unaffected man ( $X^HY$ ). What is the probability that her first son has hemophilia?

Step 1 — Determine the woman's carrier status. Her father is affected ( $X^hY$ ). A father gives his only X to every daughter, so she must have received $X^h$ from him. Since she is unaffected, her other X carries the normal allele. Therefore she is an obligate carrier: genotype $X^HX^h$ , with probability 1 (not 1/2 — the affected father guarantees it).

Step 2 — Set up the cross. $X^HX^h$ (mother) $\times$ $X^HY$ (father).

	$X^H$ (from mom)	$X^h$ (from mom)
$X^H$ (dad)

Step 3 — Read off the sons. Among sons (the two cells in the Y row), half are $X^HY$ (normal) and half are $X^hY$ (affected). So $P(\text{son is affected}) = \tfrac{1}{2}$ .

Conclusion. Because the affected father makes the woman an obligate carrier, each son has a $\tfrac{1}{2}$ chance of hemophilia. (If the question had asked the chance that her first child is an affected boy, you would multiply: $\tfrac{1}{2}\ \text{male} \times \tfrac{1}{2}\ \text{affected} = \tfrac{1}{4}$ .)

Key Takeaways — Part 3

X-linked recessive: mainly males affected, NO male-to-male transmission, carrier females; affected father makes daughters obligate carriers
X-linked dominant: affected father → all daughters affected, no sons
Autosomal recessive: can skip generations, 25% risk from two carrier parents
Autosomal dominant: every generation, ~50% chance if one parent is affected
Pedigree strategy: check for male-to-male transmission, generation skipping, and sex ratios

Where $p$ = frequency of the dominant allele and $q$ = frequency of the recessive allele.

Term	Represents
$p^2$	Frequency of homozygous dominant individuals
$2pq$	Frequency of heterozygous individuals (carriers)
$q^2$	Frequency of homozygous recessive individuals

The Five Conditions for Hardy-Weinberg

No mutations
No migration (no gene flow)
No natural selection
Large population (no genetic drift)
Random mating

If any condition is violated, the population is evolving.

The recessive phenotype frequency equals $q^2$ (because recessive phenotype = homozygous recessive). So:

Start with $q^2$ = frequency of affected individuals
$q = \sqrt{q^2}$
$p = 1 - q$
Carrier (heterozygote) frequency $= 2pq$

Worked Example — Full Hardy-Weinberg Calculation

Problem. Cystic fibrosis is autosomal recessive. In a population, 1 in 2,500 newborns is affected. Assuming Hardy-Weinberg equilibrium, find (a) the recessive allele frequency $q$ , (b) the dominant allele frequency $p$ , (c) the carrier frequency, and (d) the fraction of the population that is homozygous dominant.

Step 1 — Identify $q^2$ . Affected = homozygous recessive = $q^2$ : $q^2 = \frac{1}{2500} = 0.0004$

Step 2 — Solve for $q$ . Take the square root: $q = \sqrt{0.0004} = 0.02$

Step 3 — Solve for $p$ . Since $p + q = 1$ : $p = 1 - q = 1 - 0.02 = 0.98$

Step 4 — Carrier frequency $2pq$ . $2pq = 2(0.98)(0.02) = 0.0392 \approx 0.04$ So about 1 in 25 people is a carrier — roughly 100 times more common than the disease itself.

Step 5 — Homozygous dominant $p^2$ . $p^2 = (0.98)^2 = 0.9604 \approx 96\%$

Sanity check. The three genotype frequencies must sum to 1: $0.9604 + 0.0392 + 0.0004 = 1.0000$ . ✓ The recurring MCAT insight: carriers ( $2pq$ ) vastly outnumber affected individuals ( $q^2$ ), because most copies of a rare recessive allele are "hidden" in heterozygotes.

Hardy-Weinberg 🎯

Key Takeaways — Part 4

Hardy-Weinberg: $p + q = 1$ and $p^2 + 2pq + q^2 = 1$ (only when the five conditions hold)
Start with $q^2$ (recessive phenotype frequency), take the square root to get $q$ , then $p = 1 - q$
Carrier frequency $2pq$ is always much higher than disease frequency $q^2$ for rare alleles
X-linked recessive: affected males occur at frequency $q$ , affected females at $q^2$
Any violation of the five conditions (especially drift in small populations) = evolution occurring

Bottleneck effect: A disaster sharply reduces population size → alleles lost at random
Founder effect: A small group colonizes a new area → reduced, non-representative genetic diversity
Both are random (unlike natural selection, which is adaptive)

$\text{Fitness} = \text{reproductive success (number of viable, fertile offspring)}$

Fitness is not about being the strongest or living the longest — it is about who leaves the most offspring that themselves reproduce.

A subset of natural selection acting on traits that improve mating success
Intersexual (mate choice): e.g., peacock tail favored by peahens
Intrasexual (competition): e.g., antler clashes between males
Can produce sexual dimorphism and traits that lower survival but raise mating success

Worked Example — Identifying the Mode of Selection

Problem. A bird species lays eggs whose color ranges from very pale to very dark. Over many generations on a particular island, researchers find that the average egg color stays the same, but the variance keeps shrinking — pale and dark eggs are picked off by predators, while medium-colored eggs survive best. What mode of selection is operating, and how would a frequency-vs-trait graph change over time?

Step 1 — Compare to the three modes.

Directional would shift the mean toward one extreme. The mean here is unchanged, so it is not directional.
Disruptive would favor the two extremes and hollow out the middle. Here the middle is favored, so it is the opposite of disruptive.
Stabilizing favors intermediate phenotypes and selects against both extremes. ✓

Step 2 — Name it. This is stabilizing selection — the classic example being a trait where intermediate values have the highest fitness.

Step 3 — Predict the graph. Plot frequency (y) against egg color (x). The distribution stays centered on the same mean but becomes taller and narrower over generations as extreme phenotypes are removed. Variance decreases; the mean is constant.

Contrast. If predators instead removed medium eggs (because they were easiest to see against a mottled background), you'd get disruptive selection — the curve would become bimodal (two peaks), and that mode can promote sympatric divergence and ultimately speciation.

Key Takeaways — Part 5

Natural selection is adaptive; genetic drift is random
Stabilizing: favors the average (variance shrinks). Directional: shifts the mean. Disruptive: favors extremes (becomes bimodal).
Bottleneck and founder effects reduce genetic diversity randomly
Fitness = relative reproductive success, not strength or survival alone
Sexual selection can favor survival-costly traits when they raise mating success; gene flow homogenizes populations

Hybrid breakdown (the F2 generation is weak or sterile)

Homologous structures: same evolutionary origin, possibly different function (human arm vs. whale flipper) → indicate common ancestry (divergent evolution)
Analogous structures: different origin, similar function (bird wing vs. insect wing) → result from convergent evolution
Vestigial structures: reduced, nonfunctional remnants (human appendix, whale hip bones)

Convergent evolution: unrelated lineages independently evolve similar traits (analogous structures)
Divergent evolution: related lineages accumulate differences (homologous structures)
Coevolution: two species reciprocally drive each other's evolution (e.g., flower and pollinator)
Gradualism (slow, steady change) vs. punctuated equilibrium (long stasis interrupted by rapid bursts)

Worked Example — Classifying a Reproductive Barrier and the Mode of Speciation

Problem. Two populations of a fly were once a single species. A mountain range rose between them, and after thousands of generations of separation they were brought back into contact. When researchers place them together, the flies court each other, but females of population A reject the courtship "song" of population B males, so no mating occurs. (a) Is the isolating barrier prezygotic or postzygotic, and which specific type is it? (b) What overall mode of speciation does this scenario illustrate?

Step 1 — Prezygotic vs. postzygotic. The barrier acts before fertilization — no zygote is ever formed because mating doesn't happen. So it is a prezygotic barrier.

Step 2 — Identify the specific type. The females reject the courtship song; this is a difference in mating behavior/signals. That is behavioral isolation (a prezygotic barrier).

Step 3 — Identify the mode of speciation. The divergence began while the populations were geographically separated by the mountain range. Speciation driven by an extrinsic geographic barrier is allopatric speciation. The reproductive isolation that we now observe on secondary contact evolved during that geographic separation.

Conclusion. (a) Prezygotic, specifically behavioral isolation. (b) Allopatric speciation. Note how the MCAT often layers concepts: the mode (allopatric) describes how populations were separated, while the barrier type (prezygotic/behavioral) describes the mechanism now preventing gene flow.

Key Takeaways — Part 6

Allopatric: geographic separation drives divergence. Sympatric: divergence in the same location (e.g., polyploidy).
Prezygotic barriers prevent mating/fertilization; postzygotic barriers reduce hybrid fitness (inviability, sterility, breakdown)
Homologous = same origin = common ancestry (divergent). Analogous = same function, different origin = convergent evolution.
Species concept: must interbreed AND produce FERTILE offspring
Gradualism (steady change) vs. punctuated equilibrium (stasis punctuated by rapid bursts)

Condition	Karyotype	Features
Down syndrome	Trisomy 21	Most common viable autosomal trisomy
Turner syndrome	45,X (monosomy X)	Female, short stature, infertile
Klinefelter syndrome	47,XXY	Male, tall, often infertile

Most aneuploidies arise from nondisjunction — the failure of chromosomes (meiosis I) or sister chromatids (meiosis II) to separate properly.

Heterozygote Advantage

Sickle cell carriers (HbAS) are resistant to malaria, which explains the high frequency of the sickle cell allele in malaria-endemic regions. This is balancing selection maintaining both alleles in the population.

Worked Example — Tracing Nondisjunction to a Trisomy

Problem. A child is born with Down syndrome (trisomy 21), having three copies of chromosome 21. Cytogenetic analysis shows the two maternally derived copies are non-identical (they are the two different homologs of chromosome 21 the mother carries). In which meiotic division did the nondisjunction most likely occur, and in which parent?

Step 1 — Recall the two types of nondisjunction.

Meiosis I nondisjunction: homologous chromosomes fail to separate, so a gamete receives both homologs (which are non-identical).
Meiosis II nondisjunction: sister chromatids fail to separate, so a gamete receives two identical copies of the same chromosome.

Step 2 — Use the clue. The two extra-from-mom copies are different homologs (non-identical). Receiving both homologs is the hallmark of meiosis I nondisjunction.

Step 3 — Assign the parent. The two extra copies trace to the mother, so the error occurred in maternal meiosis I. (This fits the epidemiology: most trisomy 21 cases arise from maternal meiosis I nondisjunction, and the risk rises with maternal age because oocytes are arrested in prophase I for years.)

Conclusion. Maternal meiosis I nondisjunction produced an egg with two non-identical copies of chromosome 21; fertilization by a normal sperm yielded a trisomic (2n + 1) zygote. Distinguishing "both homologs" (MI) from "two identical chromatids" (MII) is the key reasoning the MCAT tests.

Genetic Diseases 🎯

Genetics & Evolution — Complete! ✅

From Mendel to Hardy-Weinberg to natural selection to genetic diseases, genetics and evolution are heavily tested on the MCAT. Master Punnett squares, pedigree analysis, population-genetics calculations, and the meiosis-I-vs-II logic behind aneuploidy for test day.

High-yield reminders:

Late-onset dominant alleles (Huntington's) escape selection by acting after reproduction
Heterozygote advantage (sickle cell + malaria) is balancing selection
Aneuploidy comes from nondisjunction: MI = both homologs, MII = identical sister chromatids

Cannot be excluded.

Genetics & Evolution

ABO Blood Type (MCAT FAVORITE)

Mitochondrial (Cytoplasmic) Inheritance

Worked Example — A Blood-Type Paternity Question

Key Takeaways — Part 2

MCAT Pedigree Strategy

Worked Example — Carrier Risk for Hemophilia

Key Takeaways — Part 3

The Five Conditions for Hardy-Weinberg

MCAT Shortcut

Worked Example — Full Hardy-Weinberg Calculation

Key Takeaways — Part 4

Genetic Drift

Fitness

Sexual Selection

Worked Example — Identifying the Mode of Selection

Key Takeaways — Part 5

Phylogenetics

Patterns and Rates

Worked Example — Classifying a Reproductive Barrier and the Mode of Speciation

Key Takeaways — Part 6

Heterozygote Advantage

Worked Example — Tracing Nondisjunction to a Trisomy

Genetics & Evolution — Complete! ✅