Mendelian & Non-Mendelian Genetics

Mendelian Genetics — Deep Dive

Part 1 of 4 — Mendel's Laws & the Probability Toolkit

The Two Laws

1. Law of Segregation: A diploid individual carries two alleles for each gene; these separate during gamete formation so each gamete receives exactly one. Physical basis: separation of homologous chromosomes in anaphase I of meiosis.
2. Law of Independent Assortment: Alleles of genes on different chromosomes (or far apart on the same chromosome) are distributed to gametes independently. Physical basis: random orientation of homologous pairs at metaphase I.

Vocabulary You Must Not Confuse

The Canonical Crosses

Monohybrid (Aa $\times$ Aa): genotypes $1\ AA : 2\ Aa : 1\ aa$; phenotypes $3 : 1$.

Dihybrid (AaBb $\times$ AaBb): phenotypes $9\ A\_B\_ : 3\ A\_bb : 3\ aaB\_ : 1\ aabb$ — but ONLY if the genes assort independently and both show complete dominance.

Test cross: unknown dominant $\times$ homozygous recessive (aa).

• All dominant offspring → tested parent AA
• $1:1$ dominant : recessive → tested parent Aa

A test cross works because the recessive parent contributes only recessive alleles, so offspring phenotypes read out the tested parent's gametes directly.

The Probability Toolkit (faster than Punnett squares)

• Product rule (AND): independent events multiply. $P(A \text{ and } B) = P(A) \times P(B)$
• Sum rule (OR): mutually exclusive events add. $P(A \text{ or } B) = P(A) + P(B)$
• Complement: $P(\text{at least one}) = 1 - P(\text{none})$

For any multi-gene cross with independent assortment, decompose into single-gene crosses and multiply. A gamete from AaBbCc carries a specific allele combination (say $abc$) with probability $\tfrac{1}{2} \times \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{8}$ — and such a parent makes $2^3 = 8$ gamete types.

How Many Gamete Types?

An individual heterozygous at $n$ independently assorting loci produces $2^n$ genetically distinct gamete classes (assuming the loci are unlinked). AaBBCc → heterozygous at 2 loci → $2^2 = 4$ gamete types.

Worked Example — Trihybrid Cross Without a 64-Box Grid

Problem. In a plant, purple flowers (P) are dominant to white (p), tall stems (T) to short (t), and smooth seeds (S) to rough (s). All three genes assort independently. Cross: $PpTtSs \times PpTtss$. What fraction of offspring are expected to show purple, short, smooth?

Step 1 — Split into three monohybrid problems.

• Flower color: $Pp \times Pp$ → $P(\text{purple, i.e., } P\_) = \tfrac{3}{4}$
• Height: $Tt \times Tt$ → $P(\text{short, } tt) = \tfrac{1}{4}$
• Seed coat: $Ss \times ss$ → this is a test-cross pattern: $\tfrac{1}{2}\ Ss$ (smooth), $\tfrac{1}{2}\ ss$ (rough). So $P(\text{smooth}) = \tfrac{1}{2}$

Step 2 — Product rule. $P = \frac{3}{4} \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{32}$

Why this matters on the MCAT. A full trihybrid Punnett square has $8 \times 8 = 64$ cells. The decompose-and-multiply strategy takes 20 seconds and generalizes to "at least one" questions via the complement rule. For example, the probability that an offspring shows at least one recessive phenotype is $1 - P(\text{dominant for all three}) = 1 - \left(\tfrac{3}{4} \times \tfrac{3}{4} \times \tfrac{1}{2}\right) = 1 - \tfrac{9}{32} = \tfrac{23}{32}$.

Mendel's Laws & Probability 🎯

Key Takeaways — Part 1

• Segregation = homologs separate (anaphase I); Independent Assortment = random metaphase I orientation of different pairs
• Monohybrid: 3:1 phenotype, 1:2:1 genotype. Dihybrid F2: 9:3:3:1 — only with independence and complete dominance
• Test cross ($\times$ aa) reads out the unknown parent's gametes: all dominant → AA; 1:1 → Aa
• Product rule for AND, sum rule for OR, complement for "at least one"
• Heterozygous at $n$ independent loci → $2^n$ gamete classes; only heterozygous loci count

Mendelian Genetics — Deep Dive

Part 2 of 4 — Linkage, Recombination & Gene Mapping

When Independent Assortment Fails

Mendel's second law assumes genes travel on different chromosomes. Genes on the same chromosome tend to be inherited together — they are linked. Linkage is the most common reason a dihybrid cross deviates from the expected $9:3:3:1$ (or a test cross from $1:1:1:1$).

Crossing Over Breaks Linkage

During prophase I, homologous chromosomes synapse and exchange segments (crossing over at chiasmata). A crossover between two linked genes produces recombinant gametes; no crossover between them leaves parental gametes.

• Parental (nonrecombinant) gametes carry the same allele combinations as the parent's own chromosomes
• Recombinant gametes carry new combinations
• Because a crossover between two genes is the exception rather than the rule when genes are close together, parental classes outnumber recombinant classes

Recombination Frequency (RF)

$RF = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%$

• $1\%$ recombination = 1 map unit = 1 centimorgan (cM)
• Map distances add (approximately): if A–C is 7 cM and C–B is 5 cM, A–B is about 12 cM, so C lies between A and B. Ordering three genes = finding the one whose two distances sum to the third.
• RF increases with the physical distance between genes: more room for a crossover
• Maximum observable RF is 50% — genes very far apart on the same chromosome (or on different chromosomes) assort as if independent. RF can never exceed 50% because even guaranteed crossovers produce at most half recombinant chromatids

Reading a Test Cross for Linkage

Cross a dihybrid $AB/ab$ (alleles A and B on one homolog, a and b on the other — this arrangement is called cis or coupling) to $ab/ab$:

If the parent were in trans (repulsion) configuration ($Ab/aB$), the identities flip: $Ab$ and $aB$ are the parental classes. Always identify parental classes as the two most frequent — do not assume.

Sex Linkage Basics

• X-linked genes: males are hemizygous (one allele = phenotype). No father-to-son transmission of X-linked alleles
• In fruit fly and human crosses, reciprocal crosses give different results for X-linked genes — a classic experimental signature (autosomal reciprocal crosses give identical results)
• Y-linked traits: strictly father-to-all-sons

• Y-linked inheritance is also called holandric inheritance.

Worked Example — Two-Point Mapping From Test-Cross Data

Problem. A fly heterozygous for body color (b, black recessive) and wing shape (vg, vestigial recessive) is test-crossed to a $b\,vg/b\,vg$ fly. Offspring:

Step 1 — Identify parental vs. recombinant classes. The two largest classes (415 + 405) are parental → the heterozygous parent's chromosomes were $b^+ vg^+ / b\ vg$ (cis configuration). The two smallest (92 + 88) are recombinant.

Step 2 — Compute RF. $RF = \frac{92 + 88}{415 + 405 + 92 + 88} = \frac{180}{1000} = 18\%$

Step 3 — Convert to map distance. $18\%$ recombination → the genes are 18 map units (cM) apart.

Interpretation check. If the genes were unlinked, all four classes would be about 250 each ($1:1:1:1$). The strong excess of two classes proves linkage; the identity of the big classes reveals the parental configuration. On the MCAT, expect exactly this move: spot the skewed test-cross table, name the two rare classes as recombinants, and compute RF.

Linkage & Recombination 🎯

Key Takeaways — Part 2

• Linked genes (same chromosome) violate independent assortment; parental classes outnumber recombinant classes in a test cross
• $RF = \text{recombinants}/\text{total} \times 100\%$; $1\% = 1$ map unit (cM); RF maxes out at 50%
• Both rare reciprocal classes count as recombinants — never just one
• Identify parental configuration (cis $AB/ab$ vs. trans $Ab/aB$) from the two most frequent classes
• Reciprocal-cross asymmetry is the experimental fingerprint of X-linkage; no father-to-son X transmission
• Y-linked traits pass from father to all sons

• Low-yield extras: Y-linked inheritance is also called holandric inheritance

Mendelian Genetics — Deep Dive

Part 3 of 4 — Pedigrees & Human Inheritance Patterns

The Five Patterns and Their Fingerprints

• Named examples: X-linked hypophosphatemic rickets (X-linked dominant); LHON (Leber hereditary optic neuropathy) and MELAS (mitochondrial).

The Decision Procedure

1. Male-to-male transmission present? → rules out ALL X-linked patterns (fathers give sons a Y).
2. Trait skips generations? → likely recessive. Appears in every generation with an affected parent → likely dominant.
3. Sex bias? Strong male excess → X-linked recessive. Affected father with all daughters affected and no sons → X-linked dominant.
4. All children of affected mothers affected, none of affected fathers? → mitochondrial.

Obligate Carriers

Some genotypes can be deduced with certainty:

• Parents of a child with an autosomal recessive disease: both must be carriers (Aa) if unaffected
• Daughters of a man affected by an X-linked recessive condition: all are carriers (he gives each daughter his only X)
• Phenotypically normal sons of that man: carry no disease allele from him (they got his Y)

Conditional Probability — the "2/3 Problem"

The unaffected sibling of a person with an autosomal recessive disease is NOT $\tfrac{1}{2}$ carrier. Among the surviving phenotype classes of $Aa \times Aa$ ($1\ AA : 2\ Aa : 1\ aa$), being unaffected removes the $aa$ class:

$P(\text{carrier} \mid \text{unaffected}) = \frac{2}{3}$

This conditioning step is one of the most frequently tested genetics moves on the MCAT.

Recurrence Risk Arithmetic

Chain the probabilities of each required event with the product rule: (parent 1 is a carrier) $\times$ (parent 2 is a carrier) $\times$ (two carriers have an affected child, $\tfrac{1}{4}$).

Worked Example — A Full Recurrence-Risk Calculation

Problem. Cystic fibrosis (CF) is autosomal recessive. A healthy woman has a brother with CF; her parents are healthy. She marries a healthy man from the general population, where the carrier frequency is $\tfrac{1}{25}$. What is the probability their first child has CF?

Step 1 — The woman's carrier probability. Her parents must both be $Aa$ (they are healthy with an $aa$ son). She is healthy, so among the unaffected outcomes of $Aa \times Aa$: $P(\text{she is } Aa) = \frac{2}{3}$

Step 2 — The husband's carrier probability. No family history, so use the population figure: $P(\text{he is } Aa) = \frac{1}{25}$

Step 3 — Probability that two carriers have an affected child. $P(aa \mid \text{both } Aa) = \frac{1}{4}$

Step 4 — Multiply. $P(\text{child has CF}) = \frac{2}{3} \times \frac{1}{25} \times \frac{1}{4} = \frac{2}{300} = \frac{1}{150}$

Trap check. Using $\tfrac{1}{2}$ instead of $\tfrac{2}{3}$ for the sister is the single most common error — it ignores the information that she is unaffected. Every probability in the chain must be conditioned on everything the problem tells you.

Pedigrees & Risk 🎯

Key Takeaways — Part 3

• Decision order: check male-to-male transmission (kills X-linkage) → generation skipping (recessive vs. dominant) → sex bias → maternal-line-only (mitochondrial)
• Obligate carriers: unaffected parents of an AR child; every daughter of an XR-affected father
• The unaffected sibling of an AR patient is a carrier with probability $2/3$, not $1/2$ — condition on the known phenotype
• Recurrence risk = product of each required event: (carrier?) $\times$ (carrier?) $\times$ $1/4$
• Two affected AR parents cannot have an unaffected child; late-onset dominant disease keeps risk at $1/2$ despite a healthy young adult phenotype

• Low-yield extras: X-linked hypophosphatemic rickets is the textbook X-linked dominant example; LHON and MELAS are the named mitochondrial examples

Mendelian Genetics — Deep Dive

Part 4 of 4 — MCAT Integration: When Ratios Go Wrong

The MCAT's Favorite Genetics Passage

A typical passage gives you a cross, an expected Mendelian ratio, and data that DON'T fit. Your job is to name the modifier. Memorize this diagnostic table:

Incomplete Dominance vs. Codominance

• Incomplete dominance: heterozygote is intermediate (red $\times$ white snapdragons → pink). $RR : Rr : rr$ = red : pink : white $= 1:2:1$
• Codominance: heterozygote expresses BOTH alleles fully (ABO blood type $I^AI^B$ = AB; sickle trait shows both HbA and HbS)
• ABO is doubly instructive: $I^A$ and $I^B$ are codominant with each other, and both are dominant to $i$ — a multiple allele system with 6 genotypes and 4 phenotypes

Testing Fit: The Chi-Square Idea

The MCAT will not make you look up critical values, but you must reason with the statistic:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

• $O$ = observed count, $E$ = expected count under the Mendelian hypothesis
• Degrees of freedom = (number of phenotype classes) $- 1$
• Large $\chi^2$ (p below 0.05) → reject the hypothesized ratio; small $\chi^2$ → data are consistent with it
• Rejecting $9:3:3:1$ does NOT tell you the correct model — it invites a follow-up hypothesis (linkage? epistasis? lethality?)

Passage-Reading Tactics

1. Count the classes first. Three phenotypes from a "simple dominant" cross means the model in the passage stem is wrong.
2. Convert counts to fractions and compare with the canonical ratios above.
3. Check totals. A missing quarter of the offspring is a lethality clue.
4. Sex-split the data. Any male/female difference points to the X chromosome.

Worked Example — Diagnosing a Modified Ratio

Passage-style problem. Researchers cross two true-breeding white-flowered plant lines (Line 1 $\times$ Line 2). All F1 are purple. Selfing the F1 yields an F2 of 178 purple : 142 white — approximately $9:7$.

Step 1 — Two white parents give purple F1: what does that mean? Each line must be blocked at a different step of the pigment pathway (Line 1 = $ccPP$, Line 2 = $CCpp$). The F1 ($CcPp$) has one functional allele of each gene, restoring the pathway. This is complementation — the classic evidence for TWO genes.

Step 2 — Explain $9:7$. In the F2 of $CcPp \times CcPp$:

• $C\_P\_$ (functional at both steps) $= \tfrac{9}{16}$ → purple
• $C\_pp$, $ccP\_$, and $ccpp$ $= \tfrac{3+3+1}{16} = \tfrac{7}{16}$ → white (pathway broken at either step)

So the $9:3:3:1$ collapses to $9:7$: duplicate recessive epistasis in a two-step biosynthetic pathway.

Step 3 — Predict a test cross. $CcPp \times ccpp$ → $\tfrac{1}{4} CcPp$ purple : $\tfrac{3}{4}$ white ($1:3$). Making a confirmable prediction from your model is exactly what an MCAT experimental follow-up question asks for.

Distinguish from recessive epistasis ($9:3:4$): there, one gene's recessive genotype masks the second gene (e.g., albino masking coat color), leaving three phenotype classes, not two.

MCAT Integration — Modified Ratios 🎯

Key Takeaways — Part 4

• Ratio diagnostics: $2:1$ = homozygous lethal; $9:3:4$ = recessive epistasis; $9:7$ = duplicate recessive/complementation; $12:3:1$ = dominant epistasis; $1:2:1$ phenotypes = incomplete dominance or codominance
• Two mutant parents + wild-type offspring = complementation → the mutations are in DIFFERENT genes
• ABO: codominant $I^A$/$I^B$, both dominant to $i$ — multiple alleles, 4 phenotypes
• Penetrance = whether the phenotype appears at all; expressivity = how strongly it appears
• $\chi^2 = \sum (O-E)^2/E$ with df = classes $- 1$: a significant value rejects the ratio but does not name the correct model — design the next cross