Genetics & Evolution

Genetics & Evolution for the MCAT

Part 1 of 7 — Mendelian Genetics

Mendel's Laws

1. Law of Segregation: The two alleles for each gene separate during gamete formation, so each gamete carries only one allele.
2. 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

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 $\tfrac{3}{4}$ tall (T_) and $\tfrac{1}{4}$ short (tt). So $P(\text{tall}) = \tfrac{3}{4}$.

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.

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

Genetics & Evolution for the MCAT

Part 2 of 7 — Non-Mendelian Inheritance

Extensions to Mendel

ABO Blood Type (MCAT FAVORITE)

• $I^A$ 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: mitochondrial myopathies and other disorders of oxidative phosphorylation, which hit energy-hungry tissues (muscle, nerve) hardest

• Named mitochondrial disorders: Leber hereditary optic neuropathy (LHON) and MELAS.

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. Cannot be excluded.

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

• Low-yield extras: the named mitochondrial disorders LHON and MELAS

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.

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 (two X chromosomes give females two chances to inherit it).

MCAT Pedigree Strategy

1. Look for male-to-male transmission → if present, the trait is NOT X-linked.
2. Does the trait skip generations? → if yes, it is likely recessive.
3. Count affected males vs. females → a strong male excess suggests X-linked recessive.
4. 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).

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: $\tfrac{1}{2}\ \text{male} \times \tfrac{1}{2}\ \text{affected} = \tfrac{1}{4}$.)

Pedigrees 🎯

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

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:

$p + q = 1$

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

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

The Five Conditions for Hardy-Weinberg

1. No mutations
2. No migration (no gene flow)
3. No natural selection
4. Large population (no genetic drift)
5. Random mating

If any condition is violated, the population is evolving.

MCAT Shortcut

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

Genetics & Evolution for the MCAT

Part 5 of 7 — Evolution & Natural Selection

Mechanisms of Evolution

Types of Natural Selection

Genetic Drift

• 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)

Fitness

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

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

Sexual Selection

• 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.

Evolution 🎯

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

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.

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)
• Hybrid breakdown (the F2 generation is weak or sterile)

Phylogenetics

• 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 remnants that have lost their ancestral function (human appendix, whale hip bones)

Patterns and Rates

• 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.

Speciation 🎯

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)

Genetics & Evolution for the MCAT

Part 7 of 7 — Genetic Diseases & Chromosomal Abnormalities

Autosomal Dominant Diseases

• Achondroplasia's gene is FGFR3, a growth-factor receptor whose overactive form slows cartilage growth in the long bones.

Autosomal Recessive Diseases

Chromosomal Abnormalities

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

Heterozygote Advantage

Sickle cell carriers (HbAS) are protected against severe 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

• Low-yield extras: achondroplasia's gene is FGFR3 (an overactive growth-factor receptor)