Mendelian Genetics

Mendel's Laws

Part 1 of 7

In the 1860s, an Augustinian friar named Gregor Mendel quietly founded the science of genetics by counting peas. While other naturalists described inheritance in vague terms of "blending," Mendel treated heredity as a problem of discrete, countable units passed from parent to offspring. His genius was quantitative: he tracked thousands of plants across multiple generations and discovered ratios so clean that they pointed directly to the underlying mechanism — even though chromosomes and meiosis would not be understood for another forty years.

This part establishes the vocabulary and the three classical laws. Get these definitions exact, because every later part of the unit (Punnett squares, probability, pedigrees) is just an application of them.

Core vocabulary

Key distinction: Genotype is the cause; phenotype is the effect. Two organisms with different genotypes (YY and Yy) can share the same phenotype (yellow). This single fact is the source of more AP exam errors than any other in the unit.

Mendel's Pea Experiments: P, F1, and F2

Mendel chose the garden pea (Pisum sativum) deliberately. Peas have clearly contrasting traits (round vs. wrinkled seeds, tall vs. short stems), grow quickly, and — crucially — can be made to self-pollinate (controlled, predictable crosses) or be cross-pollinated by hand.

He began with true-breeding lines — plants that, when self-crossed, always produce offspring identical to the parent. A true-breeding tall plant is homozygous (TT); a true-breeding short plant is homozygous (tt).

The generations he tracked:

What Mendel observed:

1. In the F1 generation, every plant looked like one parent. Crossing tall $\times$ short gave 100% tall offspring — not medium-height plants. This disproved blending inheritance: the "short" factor had vanished from view but, as the next step showed, had not been destroyed.

2. In the F2 generation, the hidden trait reappeared. Self-crossing the tall F1 plants produced both tall and short offspring in a striking and repeatable 3:1 ratio (about 3 tall for every 1 short). The recessive "short" factor had been carried silently through the F1 and resurfaced.

This 3:1 ratio, appearing again and again across seven different traits, is the empirical fingerprint of Mendelian inheritance. The next sections explain why it appears.

The Three Laws

1. Law of Dominance

In a heterozygote, one allele (the dominant one) masks the phenotypic effect of the other (the recessive one). This is why F1 plants (Tt) are all tall: the T allele determines the phenotype even though t is present. Dominance describes the relationship between alleles within an individual — it says nothing about how common an allele is in a population. (A dominant allele can be rare; a recessive allele can be widespread.)

2. Law of Segregation

Each organism carries two alleles for each gene, and these two alleles separate (segregate) during gamete formation, so that each gamete receives only one allele. When fertilization unites two gametes, the offspring's two-allele pairing is restored.

This is the deepest of the three laws. A Tt parent produces two kinds of gametes in equal proportion — half carrying T, half carrying t. That 50/50 split of a heterozygote's alleles is what ultimately generates the 3:1 ratio.

3. Law of Independent Assortment

When forming gametes, the alleles of different genes assort independently of one another — the allele a gamete receives for seed color does not influence which allele it receives for seed shape. (This holds for genes on different chromosomes or far apart on the same chromosome. Genes close together are linked and violate it — covered in a later unit.) Independent assortment is the basis of the dihybrid cross and is explored fully in Part 3.

The Chromosomal Basis: Why Segregation Happens

Mendel did not know about chromosomes, but his laws are a direct consequence of meiosis:

• A diploid cell carries homologous chromosome pairs — one from each parent. The two alleles of a gene sit at the same locus on the two homologs.
• During anaphase I of meiosis, the homologous chromosomes are pulled to opposite poles. This physical separation of homologs is the law of segregation — the two alleles end up in different cells.
• Because each homologous pair lines up independently of every other pair at metaphase I, the alleles of different genes are distributed independently — the physical basis of independent assortment.

Connect the levels: "Alleles segregate" (Mendel's abstract rule) and "homologous chromosomes separate at anaphase I" (the cellular event) are the same phenomenon described at two scales. The AP exam loves to ask you to link them.

Checkpoint — Vocabulary in Action

Worked Example: Where Does the 3:1 Ratio Come From?

Problem. Mendel self-crossed heterozygous tall F1 pea plants (Tt $\times$ Tt). Predict the phenotypic ratio of the F2 offspring and explain why it arises.

Step 1 — Apply the law of segregation to find the gametes. Each Tt parent produces two gamete types in equal frequency: $\frac{1}{2}$ T and $\frac{1}{2}$ t.

Step 2 — Build a Punnett square. The Punnett square is a bookkeeping grid: parent gametes label the rows and columns, and each cell is one equally likely fertilization outcome.

Step 3 — Read the genotypes. The four equally likely cells are:

• 1 × TT (homozygous dominant)
• 2 × Tt (heterozygous)
• 1 × tt (homozygous recessive)

This is the genotypic ratio 1 : 2 : 1.

Step 4 — Translate genotype to phenotype using the law of dominance. Both TT and Tt show the dominant phenotype (tall) because the T allele masks t. Only tt shows the recessive phenotype (short):

• Tall = $\frac{1}{4} + \frac{2}{4} = \frac{3}{4}$ of offspring
• Short = $\frac{1}{4}$ of offspring

This gives the phenotypic ratio 3 : 1 — exactly what Mendel counted.

The big idea in one sentence: The 3:1 phenotypic ratio and the 1:2:1 genotypic ratio describe the same four offspring; dominance "collapses" three of the four genotypic outcomes into a single visible class. Keep these two ratios distinct — confusing them is a classic AP trap, and Part 2 drills it directly.

Checkpoint — Segregation and Meiosis

When Dominance Is Not Complete

Mendel happened to choose seven pea traits that each show complete dominance (the heterozygote looks exactly like the dominant homozygote). Many real traits do not, and the AP exam expects you to recognize the exceptions. Crucially, in all of these cases the law of segregation still holds — alleles still separate 50/50 into gametes; only the genotype-to-phenotype rule changes.

The diagnostic difference between 3:1 and 1:2:1. Under incomplete dominance and codominance, every genotype produces a distinct phenotype, so the phenotypic ratio equals the genotypic ratio (1:2:1). The "collapse" of three genotype classes into two phenotype classes — which gives 3:1 — happens only under complete dominance.

Worked micro-example (incomplete dominance). In snapdragons, red $(\cdot C^{R}\cdot )$ and white $(\cdot C^{W}\cdot )$ flowers cross to give all-pink F1 $(\cdot C^{R} C^{W}\cdot )$. Self-crossing the pink F1 gives an F2 of:

• $\frac{1}{4}$ red $(\cdot C^{R} C^{R}\cdot )$
• $\frac{1}{2}$ pink $(\cdot C^{R} C^{W}\cdot )$
• $\frac{1}{4}$ white $(\cdot C^{W} C^{W}\cdot )$

That is 1 red : 2 pink : 1 white — a 1:2:1 phenotypic ratio, the visual fingerprint of incomplete dominance. If you ever see a 1:2:1 phenotype ratio reported, suspect that dominance is incomplete (or codominant), not complete.

AP connection: these extensions are explored in a later unit, but Part 1's takeaway is conceptual: dominance is a statement about phenotype expression in heterozygotes, and it is logically separate from segregation. Segregation is universal; dominance relationships vary.

Exit Ticket — Part 1 Synthesis

Monohybrid Crosses

Part 2 of 7

A monohybrid cross follows the inheritance of a single gene (one trait). It is the foundational skill of Mendelian genetics: once you can set up, solve, and interpret a monohybrid cross fluently, every harder problem in this unit is built from the same moves.

This part walks through a reliable procedure, drills the all-important distinction between genotypic and phenotypic ratios, and introduces the test cross — the experimental trick geneticists use to figure out an unknown genotype.

A reliable five-step procedure

1. Assign symbols. Capital = dominant allele, lowercase = recessive (e.g., B = brown, b = blue).
2. Write the parent genotypes from the information given.
3. Determine each parent's gametes (apply the law of segregation).
4. Fill a Punnett square — each cell is one equally likely offspring.
5. Tally genotypic and phenotypic ratios separately.

Follow these steps every time. On the AP exam, partial credit is awarded for a correct Punnett square even when the final ratio is miscounted, so always show the grid.

Worked Example 1: A Complete Tt × Tt Cross

Problem. In pea plants, tall (T) is dominant to short (t). Two heterozygous tall plants are crossed. Give the genotypic ratio, the phenotypic ratio, and the probability that a randomly chosen offspring is a true-breeding tall plant.

Step 1–2 — Symbols and parents. T = tall (dominant), t = short (recessive). Both parents are Tt.

Step 3 — Gametes. Each Tt parent produces $\frac{1}{2}$ T and $\frac{1}{2}$ t gametes.

Step 4 — Punnett square.

Step 5 — Tally the ratios.

• Genotypic ratio = 1 TT : 2 Tt : 1 tt (i.e., 1 : 2 : 1)
• Phenotypic ratio = 3 Tall : 1 Short (i.e., 3 : 1)

Answering the final question. A true-breeding tall plant is homozygous dominant (TT) — only TT breeds true for tall, because Tt would yield some short offspring. From the square, exactly one of the four cells is TT, so:

$P(\text{true-breeding tall}) = \frac{1}{4}$

Trap watch: "Tall" is $\frac{3}{4}$ of offspring, but "true-breeding tall" is only $\frac{1}{4}$. The word true-breeding forces the genotype to be homozygous. Read genotype-vs-phenotype wording carefully.

Genotypic Ratio vs. Phenotypic Ratio

These two ratios describe the same offspring but answer different questions. Mixing them up is the single most common monohybrid error.

A useful sanity check: the phenotypic ratio is what you get by "collapsing" the genotypic ratio according to dominance. In a simple-dominance monohybrid cross, $1 + 2 = 3$ dominant-phenotype offspring and $1$ recessive — the 1:2:1 becomes 3:1. If a question gives you the phenotypic ratio and asks for genotypes, you must "un-collapse" it.

Why it matters for AP: A free-response prompt may ask for "the expected ratio of offspring genotypes" — that is 1:2:1, not 3:1. Read whether the question says genotype or phenotype and answer the one asked. Writing 3:1 when genotypes were requested loses the point.

Checkpoint — Reading the Ratios

The Test Cross: Revealing an Unknown Genotype

Here is the practical problem dominance creates: an organism showing the dominant phenotype could be either homozygous (TT) or heterozygous (Tt). You cannot tell which just by looking. So how does a breeder find out?

The solution is a test cross: cross the mystery individual (dominant phenotype, unknown genotype) with a homozygous recessive (tt). The recessive parent contributes only t gametes, so the offspring phenotypes act as a readout of the mystery parent's hidden allele.

Case A — the unknown is homozygous (TT):

All offspring are Tt → 100% dominant phenotype. No recessive offspring ever appear.

Case B — the unknown is heterozygous (Tt):

Offspring are $\frac{1}{2}$ Tt and $\frac{1}{2}$ tt → 1 dominant : 1 recessive phenotype.

The interpretation rule:

• If any recessive offspring appear, the unknown parent must be heterozygous (Tt) — it had to contribute a t allele.
• If all offspring show the dominant phenotype (especially over a large number), the unknown is most likely homozygous (TT).

Test cross vs. back-cross

A back-cross is any cross of an offspring back to one of its parents (or to an individual of the parental genotype). A test cross is the specific case where that cross is to a homozygous recessive individual in order to deduce a genotype. Every test cross to a recessive parent is a back-cross in form, but not every back-cross is a test cross (e.g., crossing back to a homozygous dominant parent reveals nothing about a hidden recessive allele).

Worked Example 2: Solving a Test Cross

Problem. A purple-flowered pea plant (P dominant to p) is test-crossed. Among 80 offspring, 38 are purple and 42 are white. What is the genotype of the purple parent?

Step 1 — Identify the cross. A test cross means the other parent is homozygous recessive (pp), contributing only p gametes.

Step 2 — Interpret the offspring. White offspring (pp) appeared. For an offspring to be pp, it must have received a p allele from each parent. Since white offspring exist, the purple parent must carry a p allele — so it cannot be PP.

Step 3 — Confirm with the expected ratio. A Pp $\times$ pp cross predicts a 1 purple : 1 white ratio. The observed 38 : 42 is statistically indistinguishable from 40 : 40 (sampling noise around the expected 1:1).

Conclusion: The purple parent is heterozygous (Pp).

Reasoning shortcut: The appearance of even one recessive offspring in a test cross is decisive proof of heterozygosity — you do not need the full 1:1 ratio to conclude it. The ratio is useful when zero recessives appear: the more dominant-only offspring you score, the more confident you are that the parent is homozygous (because the probability of getting all-dominant offspring by chance from a heterozygote shrinks rapidly — Part 4 quantifies this).

Checkpoint — Test Cross Reasoning

Dihybrid Crosses

Part 3 of 7

A dihybrid cross tracks two genes at once. This is where the law of independent assortment earns its keep: because the two genes assort independently, the famous 9 : 3 : 3 : 1 phenotypic ratio emerges in the F2 generation. Mendel's discovery of this ratio — by crossing peas that differed in both seed shape and seed color — was the experimental evidence for independent assortment.

We will use Mendel's classic example throughout:

• Seed shape: round (R) dominant to wrinkled (r)
• Seed color: yellow (Y) dominant to green (y)

A plant heterozygous for both genes is written RrYy (a dihybrid).

Counting gamete types: the $2^n$ rule

Before building any Punnett square, figure out how many gamete types each parent makes. For n heterozygous gene pairs, an individual produces $2^n$ genetically distinct gametes:

For an RrYy parent, the four gamete types are RY, Ry, rY, ry. A clean way to generate them is the FOIL method — distribute the alleles of gene 1 across the alleles of gene 2: $(R + r)(Y + y) = RY + Ry + rY + ry$. Each appears with probability $\frac{1}{4}$.

The 4×4 Punnett Square: RrYy × RrYy

With four gamete types from each parent, the Punnett square is a 4 × 4 grid (16 cells). List the gametes RY, Ry, rY, ry along the top and side, then combine.

Now classify all 16 offspring by phenotype (remember: any R gives round, any Y gives yellow):

This is the 9 : 3 : 3 : 1 phenotypic ratio — the signature of a dihybrid cross with two independently assorting genes, each showing simple dominance.

What the ratio means: the two single "outer" classes (round-green and wrinkled-yellow) are the recombinant phenotypes — new combinations not seen in the true-breeding parents. Their appearance is direct visual proof that the shape gene and the color gene assorted independently.

The Forked-Line Method: A Faster Route to 9:3:3:1

A 16-cell grid is slow and error-prone. Because the genes are independent, you can treat the dihybrid cross as two separate monohybrid crosses and multiply the results (this is the product rule, formalized in Part 4).

Step 1 — Solve each gene separately.

• Shape: Rr $\times$ Rr → $\frac{3}{4}$ round, $\frac{1}{4}$ wrinkled
• Color: Yy $\times$ Yy → $\frac{3}{4}$ yellow, $\frac{1}{4}$ green

Step 2 — Multiply across every combination (the "fork").

The four products are $\frac{9}{16}, \frac{3}{16}, \frac{3}{16}, \frac{1}{16}$ — the same 9:3:3:1 ratio, obtained without drawing 16 boxes. They sum to $\frac{16}{16} = 1$, a good check.

Why this works: independent assortment means the shape outcome and the color outcome are statistically independent events, so their probabilities multiply. The forked-line method is just the product rule applied phenotype-by-phenotype — and it scales to three, four, or more genes where a Punnett square becomes hopeless (a trihybrid cross would need a $64$-cell grid).

Checkpoint — Building the Dihybrid Cross

Worked Example: Probability of a Specific Genotype

Problem. From the cross RrYy × RrYy, what is the probability that an offspring has the exact genotype AABb... wait — using our symbols, what is the probability of genotype RRYy?

The Punnett-square route. Scan the 16-cell grid above for RRYy. It appears in 2 cells (row RY × column Ry, and row Ry × column RY). So $P(RRYy) = \frac{2}{16} = \frac{1}{8}$.

The faster product-rule route. Split into two independent monohybrid crosses and multiply the genotype probabilities:

• Gene 1 (Rr $\times$ Rr): $P(RR) = \frac{1}{4}$
• Gene 2 (Yy $\times$ Yy): $P(Yy) = \frac{2}{4} = \frac{1}{2}$

$P(RRYy) = P(RR)\times P(Yy) = \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}$

Both methods agree: $\frac{1}{8}$.

Try the variation: P(RrYy), the double heterozygote.

• $P(Rr) = \frac{1}{2}$ and $P(Yy) = \frac{1}{2}$
• $P(RrYy) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4}$ — consistent with the 4 RrYy cells out of 16 in the grid.

Strategy takeaway: For a specific genotype (not just a phenotype), apply the product rule to the per-gene genotypic probabilities (1:2:1 → $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$). For a specific phenotype, use the per-gene phenotypic probabilities ($\frac{3}{4}, \frac{1}{4}$). Picking the wrong per-gene fractions is the subtle mistake to avoid — and Part 4 makes the product/sum rules fully explicit.

Checkpoint — Dihybrid Probability

Why Punnett Squares Break Down — and Independent Assortment Saves You

The dihybrid grid has 16 cells. Add one more heterozygous gene and the picture gets ugly fast, because both the number of gamete types and the grid size grow as powers of 2.

A trihybrid self-cross would require you to draw and classify 64 boxes by hand — clearly impractical on a timed exam. This is exactly why the law of independent assortment is so powerful: because each gene assorts independently, you never need the full grid. You solve each gene as a one-line monohybrid cross and multiply the per-gene probabilities (the product rule of Part 4).

Worked example (trihybrid, no grid). From AaBbCc $\times$ AaBbCc, what fraction of offspring are A_ B_ cc (dominant for A and B, recessive for C)?

• $P(A\_) = \frac{3}{4}$
• $P(B\_) = \frac{3}{4}$
• $P(cc) = \frac{1}{4}$

$P(A\_ B\_ cc) = \frac{3}{4}\times\frac{3}{4}\times\frac{1}{4} = \frac{9}{64}$

That single multiplication replaces hunting through 64 boxes. The whole trihybrid phenotype distribution, by the way, is 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 — and every one of those eight numbers is just a product of $\frac{3}{4}$'s and $\frac{1}{4}$'s.

Takeaway: the Punnett square is a conceptual tool that shows why the ratios appear; the product rule is the computational tool you actually use once n exceeds 2. Part 4 formalizes this fully.

Checkpoint — Scaling Up

Probability in Genetics

Part 4 of 7

Punnett squares are wonderful teaching tools but terrible calculators once you have more than two genes — a trihybrid cross would need a 64-cell grid, and a question about "at least one recessive child out of four" can't be read off a square at all. The fix is to treat inheritance as what it really is: a problem in probability. Two simple rules handle almost everything on the AP exam.

The two fundamental rules

Product Rule (AND): The probability that two or more independent events both occur is the product of their individual probabilities.

$P(A \text{ and } B) = P(A)\times P(B)$

Use it when an outcome requires this AND that — e.g., "an offspring is aa and bb."

Sum Rule (OR): The probability that either of two mutually exclusive events occurs is the sum of their individual probabilities.

$P(A \text{ or } B) = P(A) + P(B)$

Use it when an outcome can be reached this way OR that way — e.g., "a heterozygous offspring, which can be formed as (mom's A, dad's a) or (mom's a, dad's A)."

Why these replace big squares: because genes assort independently, the outcome for each gene is an independent event — so the product rule lets you solve each gene as a tiny monohybrid cross and multiply. This is exactly the forked-line logic of Part 3, now stated as a general principle.

Worked Example 1: A Trihybrid Cross by Product Rule

Problem. Cross AaBbCc × AaBbCc (three independently assorting genes). What is the probability that an offspring is aabbcc — homozygous recessive for all three genes?

The square way is hopeless — each parent makes $2^3 = 8$ gamete types, so the grid is $8 \times 8 = 64$ cells. Use the product rule instead.

Step 1 — Solve each gene as a monohybrid cross. For Aa $\times$ Aa, the probability of the homozygous recessive aa is $\frac{1}{4}$. The same holds for each gene:

• $P(aa) = \frac{1}{4}$
• $P(bb) = \frac{1}{4}$
• $P(cc) = \frac{1}{4}$

Step 2 — Multiply (the offspring must be aa AND bb AND cc).

$P(aabbcc) = \frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \frac{1}{64}$

So 1 in 64 offspring is triple-recessive — the single corner cell you would have found after drawing all 64 boxes.

A phenotype variation. What fraction shows all three dominant phenotypes (A_ B_ C_)? Each gene gives $P(\text{dominant}) = \frac{3}{4}$, so:

$P(A\_ B\_ C\_) = \frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \frac{27}{64}$

(This is the "27" in the trihybrid 27:9:9:9:3:3:3:1 ratio — note that $\frac{27}{64}$ and $\frac{1}{64}$ are the two extreme classes.)

Worked Example 2: "At Least One" via the Complement

Problem. From the cross AaBbCc × AaBbCc, what is the probability that an offspring shows at least one recessive phenotype (i.e., is recessive for one or more of the three traits)?

Computing "at least one" directly would mean adding up many overlapping cases (recessive for A only, B only, A and B, all three, ...) — messy and error-prone. The clean trick is the complement rule:

$P(\text{at least one recessive}) = 1 - P(\text{no recessives})$

Step 1 — Find the complement (no recessive traits = dominant for ALL three). From Example 1: $P(\text{all three dominant}) = \frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \frac{27}{64}$

Step 2 — Subtract from 1. $P(\text{at least one recessive}) = 1 - \frac{27}{64} = \frac{37}{64}$

So about 58% of offspring will display at least one recessive trait.

General principle: Whenever a question says "at least one," compute $1 - P(\text{none})$. The "none" case is usually a single clean product, whereas the direct sum is a tangle of overlapping events. This complement strategy appears constantly on AP genetics and is worth making automatic.

Checkpoint — Product, Sum, and Complement

The Binomial: "Exactly k of n" Offspring

The product rule answers "all children affected." The complement answers "at least one." But what about "exactly 2 of 3 children are affected" — a specific count, in any birth order? For that, you need the binomial formula.

If each independent offspring has probability p of being affected and q = 1 − p of being unaffected, the probability of exactly k affected out of n children is:

$P = \binom{n}{k}\, p^{k}\, q^{\,n-k}$

where the binomial coefficient $\binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$ counts how many different birth orders give k affected children.

Why the coefficient is needed: "2 affected, 1 unaffected" can happen as (affected, affected, unaffected), (affected, unaffected, affected), or (unaffected, affected, affected) — 3 orderings. Each single ordering has probability $p^2 q$; multiplying by 3 accounts for all the ways it can occur. Forgetting $\binom{n}{k}$ (just writing $p^k q^{n-k}$) is the most common binomial mistake.

Worked Example 3: Exactly 2 of 3 Children Affected

Problem. Two carriers for an autosomal recessive disease (Cc $\times$ Cc) have 3 children. What is the probability that exactly 2 of the 3 are affected (cc)?

Step 1 — Per-child probabilities. From Cc $\times$ Cc:

• $p = P(\text{affected, } cc) = \frac{1}{4}$
• $q = P(\text{unaffected}) = \frac{3}{4}$

Step 2 — Identify n and k. $n = 3$ children, $k = 2$ affected.

Step 3 — Compute the coefficient. $\binom{3}{2} = \frac{3!}{2!\,1!} = 3$

Step 4 — Plug into the binomial. $P = \binom{3}{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{1} = 3 \times \frac{1}{16}\times\frac{3}{4} = 3\times\frac{3}{64} = \frac{9}{64}$

So $P(\text{exactly 2 of 3 affected}) = \frac{9}{64} \approx 0.14$.

Sanity check on the coefficient: if you had written only $p^2 q = \frac{3}{64}$, you would be computing the probability of one specific order (say, first two affected, third not). The factor $\binom{3}{2} = 3$ correctly scales that up to cover all three possible orders. Always ask: "in how many orders can this happen?"

Checkpoint — Binomial Problems

Using the Sum Rule Within a Single Gene

The product rule handles "AND across genes." The sum rule shines when a single outcome can be reached through several mutually exclusive genotypes — you compute each genotype's probability and add.

Worked example — "dominant phenotype" as a sum. In a Aa $\times$ Aa cross, the dominant phenotype arises from either AA or Aa. These are mutually exclusive (an offspring can't be both), so add:

$P(\text{dominant}) = P(AA) + P(Aa) = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$

That is where the familiar $\frac{3}{4}$ comes from — it is a sum of two genotype probabilities, not a single Punnett cell.

Worked example — combining sum and product. From AaBb $\times$ AaBb, what is the probability of an offspring that is either homozygous AA or homozygous aa at the first gene (i.e., not heterozygous), regardless of the B gene?

• At gene A, "homozygous" = AA OR aa (mutually exclusive): $P = P(AA) + P(aa) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
• The B gene is unrestricted, so it contributes a factor of $1$.

$P(\text{homozygous at A}) = \frac{1}{2}\times 1 = \frac{1}{2}$

Choosing the right rule — a quick test. Ask: can this outcome happen in more than one mutually exclusive way? If yes, add those ways (sum rule). Does the outcome require several independent conditions to all hold? If yes, multiply them (product rule). Many AP problems need both: add the genotype options within each gene, then multiply across genes.

The classic mistake: adding when you should multiply (or vice versa). "Recessive for trait 1 AND trait 2" must be multiplied ($\frac{1}{4}\times\frac{1}{4}$); "genotype AA OR aa" must be added ($\frac{1}{4}+\frac{1}{4}$). Always translate the question's "and"/"or" into the matching operation before computing.

Checkpoint — Sum vs. Product

Pedigree Analysis

Part 5 of 7

You cannot perform controlled crosses on humans, so geneticists trace traits through real families using a pedigree — a standardized family-tree diagram. The detective work of pedigree analysis is to read the pattern of affected and unaffected individuals across generations and deduce the mode of inheritance: is the trait autosomal or sex-linked? Dominant or recessive?

Reading the symbols

Generations are labeled with Roman numerals (I, II, III) from top to bottom; individuals within a generation are numbered left to right (I-1, I-2, II-1, ...).

The core logic: every individual's phenotype constrains their possible genotypes, and every parent-offspring link constrains what alleles were passed on. Pedigree analysis is just the rules of Mendel and probability applied backwards — from observed phenotypes to inferred genotypes.

The Three Common Modes of Inheritance

Most AP pedigree questions ask you to distinguish among these three patterns. Learn the diagnostic tells for each.

1. Autosomal Recessive (e.g., cystic fibrosis, albinism)

• The trait can skip generations — two unaffected parents can have an affected child (both parents are carriers, Aa $\times$ Aa).
• Affected individuals are aa; unaffected children of two carriers may be AA or Aa.
• Males and females are affected in roughly equal numbers.
• Tell: affected child from two unaffected parents → recessive. If those parents are Aa × Aa, watch for the trait reappearing after skipping a generation.

2. Autosomal Dominant (e.g., Huntington's disease, polydactyly)

• The trait appears in every generation — it does not skip (an affected child has at least one affected parent).
• Affected individuals carry at least one dominant allele (Aa or AA); unaffected individuals are aa.
• Males and females affected in roughly equal numbers.
• Tell: trait in every generation, no skipping, both sexes affected → dominant.

3. X-Linked Recessive (e.g., hemophilia, red-green color blindness)

• Far more males are affected than females (males are hemizygous: a single recessive allele on their one X is enough; written $X^a Y$).
• A trait can pass from an affected grandfather, through an unaffected carrier daughter ($X^A X^a$), to an affected grandson — the classic "skips to the maternal grandson" pattern.
• An affected female ($X^a X^a$) must have an affected father AND a carrier-or-affected mother.
• Tell: strong male bias, inheritance through unaffected mothers → X-linked recessive.

A quick decision tree

Worked Example: Deducing the Mode of Inheritance

Consider this small pedigree. Generation I is the grandparents; Generation II are their children (and spouses); Generation III are the grandchildren.

Step 1 — Recessive or dominant? Individuals I-1 and I-2 are both unaffected, yet their daughter II-1 is affected. Two unaffected parents producing an affected child is the signature of a recessive trait (the parents must each be carriers). A dominant trait could not skip from unaffected parents to an affected child.

Step 2 — Autosomal or X-linked? Affected individual II-1 is female. If the trait were X-linked recessive, she would be $X^a X^a$ and would need an affected father — but her father I-1 is unaffected. That contradiction rules out X-linked recessive. So the trait is autosomal recessive.

Step 3 — Assign genotypes (let the allele be a).

• II-1 is affected → aa.
• I-1 and I-2 are unaffected but produced an aa daughter, so each contributed an a → both are Aa (carriers).

Step 4 — A carrier-probability question. III-1 is an unaffected son of II-1 (aa) and II-2. The affected daughter III-2 (aa) proves II-2 carries a, and since II-2 is unaffected he is Aa. So the cross producing III-1 is aa × Aa.

The aa × Aa cross gives offspring $\frac{1}{2}$ Aa and $\frac{1}{2}$ aa. But we are told III-1 is unaffected, so he cannot be aa — he must be Aa.

$P(\text{III-1 is a carrier} \mid \text{unaffected}) = 1$

Every unaffected child of an aa parent is an obligate carrier (Aa), because the aa parent can only donate a. This conditional reasoning — restricting to the unaffected outcomes before computing the probability — is exactly the kind of step AP graders reward.

Checkpoint — Identifying the Pattern

Carriers, Conditional Probability, and a Note on Chi-Square

Carrier reasoning is conditional reasoning. A frequent AP twist: "An unaffected sibling of an affected (aa) child — what is the probability they are a carrier?" The parents must both be Aa (they produced an aa child). The Aa $\times$ Aa cross gives offspring $\frac{1}{4}$ AA, $\frac{1}{2}$ Aa, $\frac{1}{4}$ aa. But the sibling is unaffected, so we discard the aa outcome and renormalize over the surviving genotypes:

$P(Aa \mid \text{unaffected}) = \frac{P(Aa)}{P(AA) + P(Aa)} = \frac{\tfrac{1}{2}}{\tfrac{1}{4}+\tfrac{1}{2}} = \frac{\tfrac{2}{4}}{\tfrac{3}{4}} = \frac{2}{3}$

The answer is $\frac{2}{3}$, not $\frac{1}{2}$ — because knowing the sibling is unaffected removes the aa possibility and re-weights the rest. Forgetting to condition (and answering $\frac{1}{2}$) is a classic trap.

Chi-square ($\chi^2$) — testing whether data fit a Mendelian prediction. When real offspring counts don't exactly match an expected ratio, a chi-square goodness-of-fit test asks whether the deviation is small enough to be chance:

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

where O is the observed count and E the expected count for each class. A large $\chi^2$ means the data deviate more than chance comfortably allows (the genetic model may be wrong); a small $\chi^2$ means the data are consistent with the predicted ratio. (Part 7 walks through an actual chi-square computation — here just recognize the formula and its purpose.)

Checkpoint — Carrier and Conditional Probability

Two Rarer Patterns and a Key Inheritance Rule

Beyond the big three, AP pedigrees occasionally feature two less common modes. Knowing their tells lets you eliminate wrong answers quickly.

X-Linked Dominant (e.g., some forms of vitamin-D-resistant rickets)

• The trait appears in every generation (like autosomal dominant), but the transmission is sex-biased in a telltale way.
• An affected father passes the trait to ALL of his daughters and NONE of his sons — because he gives his X (carrying the dominant allele) to every daughter and his Y to every son.
• An affected heterozygous mother passes it to about half of each sex.
• Tell: affected father → all daughters affected, no sons affected. This father-to-all-daughters pattern is the unmistakable signature.

Y-Linked (holandric) (e.g., genes on the Y not shared with X)

• Only males are ever affected, and an affected father transmits to ALL of his sons and no daughters.
• Tell: strict father-to-son transmission, zero females affected, ever.

The unifying rule for sex-linked traits: a father gives his single X to every daughter and his Y to every son. A mother gives one of her two X's to each child regardless of sex. Internalizing this one sentence lets you predict any X-linked or Y-linked cross without memorizing separate cases.

Strategy: when a pedigree question gives answer choices spanning several modes, look first at an affected father's children. The split of his sons vs. daughters instantly separates autosomal, X-linked, and Y-linked patterns.

Checkpoint — Distinguishing Sex-Linked Patterns

Problem-Solving Workshop

Part 6 of 7

Real AP free-response questions rarely test one idea in isolation. They braid together crosses, probability rules, and pedigree reasoning into multi-part problems where each answer feeds the next. This workshop walks through three fully worked, multi-step problems that integrate the whole unit, then finishes with a checkpoint testing the methods.

The integrated problem-solving checklist

Before diving in, internalize this workflow — it converts a scary paragraph into a sequence of small, familiar moves:

1. Define every allele symbol and note which is dominant.
2. Extract genotypes from the phenotypes and family information given (work backwards from affected individuals, who pin down recessive genotypes).
3. Choose your tool per step: a Punnett square for one or two genes; the product rule for "AND across genes"; the complement for "at least one"; the binomial for "exactly k of n offspring."
4. Condition when told an outcome (e.g., "given the child is unaffected," remove impossible genotypes and renormalize).
5. Chain the steps: multiply independent stage-probabilities together to get the final answer.

Mindset: never reach for a 16- or 64-box Punnett square under time pressure. Decompose into per-gene monohybrid crosses and multiply. The grid is for learning; the probability rules are for solving.

Worked Problem 1: Cross + Probability Chain

Setup. In dogs, black coat (B) is dominant to brown (b), and a solid coat (S) is dominant to spotted (s); the two genes assort independently. A BbSs dog is crossed with a Bbss dog.

(a) What fraction of puppies are brown and spotted (bbss)?

Solve each gene separately, then multiply (product rule).

• Coat color Bb $\times$ Bb: $P(bb) = \frac{1}{4}$
• Coat pattern Ss $\times$ ss: $P(ss) = \frac{1}{2}$

$P(bbss) = \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}$

(b) What fraction are black and solid (B_ S_ )?

• $P(B\_) = \frac{3}{4}$ (from Bb $\times$ Bb)
• $P(S\_) = \frac{1}{2}$ (from Ss $\times$ ss: half Ss solid, half ss spotted)

$P(B\_ S\_) = \frac{3}{4}\times\frac{1}{2} = \frac{3}{8}$

(c) If this pair has 4 puppies, what is the probability that exactly 1 is brown and spotted?

This is a binomial with $p = P(bbss) = \frac{1}{8}$ (from part a), $q = \frac{7}{8}$, $n = 4$, $k = 1$:

$P = \binom{4}{1}\left(\frac{1}{8}\right)^{1}\left(\frac{7}{8}\right)^{3} = 4\times\frac{1}{8}\times\frac{343}{512} = \frac{1372}{4096} \approx 0.335$

Notice the chain: part (a)'s single-puppy probability ($\frac{1}{8}$) became the p plugged into part (c)'s binomial. Multi-part problems are designed so earlier answers feed later ones.

Worked Problem 2: Pedigree + Conditional Probability

Setup. Cystic fibrosis (CF) is autosomal recessive (f = recessive allele; affected = ff). A couple, Maria and Sam, are both unaffected. Maria's brother has CF. Sam was tested and is a known carrier (Ff).

(a) What is the probability that Maria is a carrier (Ff)?

Maria's brother is ff, so both of Maria's parents must be carriers (Ff $\times$ Ff). Maria herself is unaffected, so condition on the non-ff outcomes of that cross:

$P(\text{Maria is } Ff \mid \text{unaffected}) = \frac{\tfrac{1}{2}}{\tfrac{1}{4}+\tfrac{1}{2}} = \frac{2}{3}$

(b) What is the probability that Maria and Sam's first child has CF?

For the child to be ff, three things must all happen (product rule):

1. Maria is a carrier: $P = \frac{2}{3}$ (from part a).
2. Sam (known Ff) passes f: $P = \frac{1}{2}$ — this is certain to be possible since he IS a carrier.
3. Maria (if she is Ff) passes f: $P = \frac{1}{2}$.

$P(\text{child has CF}) = \underbrace{\frac{2}{3}}_{\text{Maria is } Ff}\times\underbrace{\frac{1}{2}}_{\text{Maria gives } f}\times\underbrace{\frac{1}{2}}_{\text{Sam gives } f} = \frac{2}{12} = \frac{1}{6}$

The subtlety: if you had wrongly assumed Maria was definitely a carrier, you would have gotten $\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}$. The correct answer $\frac{1}{6}$ properly weights by the $\frac{2}{3}$ chance that Maria carries the allele at all. Carrier uncertainty must be multiplied into the chain.

Worked Problem 3: Two Traits + "At Least One" Across Children

Setup. Two parents are each heterozygous for two independent recessive disorders, genotype AaBb × AaBb (disorder 1 = aa, disorder 2 = bb). They plan to have 2 children.

(a) What is the probability that a single child is affected by at least one of the two disorders?

Use the complement on a single child. "No disorder" means dominant for both genes (*A_ B_ *): $P(\text{healthy}) = P(A\_)\times P(B\_) = \frac{3}{4}\times\frac{3}{4} = \frac{9}{16}$ $P(\text{at least one disorder}) = 1 - \frac{9}{16} = \frac{7}{16}$

(b) What is the probability that BOTH children are completely healthy (neither disorder)?

Each child independently has $P(\text{healthy}) = \frac{9}{16}$. The two children are independent events, so multiply (product rule across children):

$P(\text{both healthy}) = \frac{9}{16}\times\frac{9}{16} = \frac{81}{256}$

(c) What is the probability that at least one of the two children has at least one disorder?

Complement of part (b): $P(\text{at least one child affected}) = 1 - \frac{81}{256} = \frac{175}{256}\approx 0.68$

Two layers of complement here: within a child ("at least one disorder" → $1 - P(\text{healthy})$) and across children ("at least one child affected" → $1 - P(\text{all healthy})$). Keeping the layers straight — which "at least one" you are computing — is the entire skill. Always name the event in words before you compute.

Worked Problem 4: Multiple Alleles + Binomial (ABO Blood Type)

Setup. Human ABO blood type uses three alleles: $I^{A}$ and $I^{B}$ are codominant to each other, and both are dominant to the recessive i. So genotype $I^{A}$ $I^{A}$ or $I^{A}$ i → type A; $I^{B}$ $I^{B}$ or $I^{B}$ i → type B; $I^{A}$ $I^{B}$ → type AB; ii → type O. A man of genotype $I^{A}$ i has children with a woman of genotype $I^{B}$ i.

(a) What blood-type ratio is expected among their children?

Treat it as a monohybrid cross with the four allele combinations. Gametes: man gives $\frac{1}{2}I^A,\ \frac{1}{2}i$; woman gives $\frac{1}{2}I^B,\ \frac{1}{2}i$.

The four equally likely children are AB : B : A : O = 1 : 1 : 1 : 1 — each blood type with probability $\frac{1}{4}$. (This 1:1:1:1 result is a hallmark of crossing two double-heterozygote-like parents who share no alleles.)

(b) What is the probability that a given child has type O blood?

Type O requires genotype ii — the child must inherit i from each parent: $\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}$. Consistent with the single O cell in the square.

(c) If the couple has 3 children, what is the probability that exactly 1 has type O blood?

Binomial with $p = P(\text{type O}) = \frac{1}{4}$, $q = \frac{3}{4}$, $n = 3$, $k = 1$:

$P = \binom{3}{1}\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{2} = 3\times\frac{1}{4}\times\frac{9}{16} = \frac{27}{64}\approx 0.42$

The integration: this problem chained a multiple-allele monohybrid cross (part a) → a single-genotype probability (part b) → a binomial across siblings (part c). Notice that codominance changed the phenotype labels (AB is its own class) but did not change the underlying segregation arithmetic — the gametes still split 50/50, exactly as in Mendel's peas.

Pitfalls That Wreck Multi-Step Problems

Integrated problems fail not because any single step is hard, but because a small early slip propagates. Here are the recurring failure points, each with the fix.

A two-line self-check before you commit an answer. First: is every probability between 0 and 1, and does a complete set of outcomes sum to 1? Second: did I answer the exact event asked (specific genotype vs. phenotype, "exactly k" vs. "at least one," carrier vs. affected)? These two checks catch the large majority of multi-step errors.

Habit to build: write the event in plain English first ("at least one of three children is aa"), then translate it into the operation ("$1 - (\frac{3}{4})^3$"). The translation step is where points are won or lost — never jump straight to numbers.

Checkpoint — Integrated Problem Solving

AP Review

Part 7 of 7

This final part synthesizes the unit, names the misconceptions that cost students the most points, and introduces the chi-square goodness-of-fit test — the quantitative skill the AP exam expects you to perform. Treat this as your pre-exam consolidation.

The unit in one map

The unifying insight: every quantitative answer in this unit comes from the fact that a heterozygote's two alleles segregate 50/50 into gametes, and the alleles of unlinked genes assort independently. Punnett squares, the 9:3:3:1 ratio, and the product rule are all consequences of those two facts.

The Five AP Traps That Cost the Most Points

Trap 1 — "Dominant means most common." Dominance describes which allele is expressed in a heterozygote, not how frequent it is in a population. A dominant allele can be vanishingly rare (e.g., Huntington's); a recessive allele can be widespread. Allele frequency is set by population genetics, not by dominance.

Trap 2 — Confusing genotypic and phenotypic ratios. A Tt $\times$ Tt cross is 1:2:1 by genotype but 3:1 by phenotype. Read whether the question asks for genotypes or phenotypes and answer the one asked. Writing 3:1 when genotypes were requested is an automatic miss.

Trap 3 — Assuming independent assortment always holds. The 9:3:3:1 ratio requires the two genes to be unlinked (on different chromosomes, or far apart on the same one). Genes that are linked stay together during meiosis and produce distorted ratios with excess parental types — independent assortment does not apply.

Trap 4 — Forgetting to condition on known outcomes. "An unaffected sibling of an aa child is a carrier with probability $\frac{2}{3}$, not $\frac{1}{2}$" — because being unaffected removes the aa possibility and you must renormalize over the surviving genotypes.

Trap 5 — Misapplying AND vs. OR. "Affected for trait 1 AND trait 2" → multiply. "Genotype can arise this way OR that way" → add. And for "at least one," use the complement $1 - P(\text{none})$ rather than summing overlapping cases.

Exam habit: when two answer choices both look plausible, pick the one that correctly distinguishes genotype from phenotype, or that conditions on the information given. The AP exam's distractors are built from exactly these five errors.

AP-Style Application Questions

Chi-Square Goodness-of-Fit: A Worked Computation

A chi-square test answers a precise question: are the deviations between my observed counts and the ratio I predicted small enough to be due to chance, or large enough to suggest my genetic model is wrong?

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

• O = observed count for each phenotype class
• E = expected count for each class (total offspring × predicted fraction)
• Sum the term over all classes.

Worked example. You predict a 3:1 ratio from a monohybrid cross and score 200 offspring: 160 dominant, 40 recessive. Do the data fit?

Step 1 — Compute expected counts (E) from the 3:1 prediction over 200 offspring:

• Dominant: $200 \times \frac{3}{4} = 150$
• Recessive: $200 \times \frac{1}{4} = 50$

Step 2 — Compute each class term $\frac{(O-E)^2}{E}$:

Step 3 — Sum to get $\chi^2$: $\chi^2 = 0.667 + 2.000 = 2.667$

Step 4 — Interpret with degrees of freedom. Degrees of freedom (df) = (number of classes − 1) = $2 - 1 = 1$. For df = 1, the critical value at $p = 0.05$ is 3.84. Our $\chi^2 = 2.667$ is less than 3.84, so we fail to reject the null hypothesis: the data are consistent with the predicted 3:1 ratio (the deviation is plausibly due to chance).

Decision rule to memorize: if $\chi^2 \geq$ critical value, reject the model (deviation too large to be chance — your predicted ratio is probably wrong). If $\chi^2 <$ critical value, fail to reject (data fit the model). A larger $\chi^2$ means a worse fit. Always pair $\chi^2$ with its degrees of freedom before comparing to the table.

Exit Ticket — Chi-Square and Synthesis

Rapid-Review Reference Card

Use this as a final pre-exam scan. Each row is a fact that, if recalled cleanly, prevents a common error.

The last-mile checklist for any genetics free-response

1. Define allele symbols explicitly (graders look for this).
2. State the cross in genotype form before computing.
3. Show the Punnett square or the probability multiplication — partial credit lives in the work.
4. Answer the exact question asked — genotype vs. phenotype, "at least one" vs. "exactly," carrier vs. affected.
5. Sanity-check probabilities: they must lie between 0 and 1, and a full set of mutually exclusive outcomes must sum to 1.

Final reminder: the deepest idea in this unit is that a heterozygote splits its alleles 50/50 into gametes and that unlinked genes do so independently. Almost every quantitative answer follows from those two sentences. Master them, and Mendelian genetics becomes arithmetic.

A Second Chi-Square: Testing a 9:3:3:1 Dihybrid

The monohybrid example used 2 classes (df = 1). Dihybrid crosses give 4 phenotype classes (df = 3), and the AP exam often pairs the larger table with a higher critical value. Here is the full computation.

Setup. A dihybrid cross is predicted to give 9:3:3:1. A total of 160 offspring are scored: 88, 34, 28, 10 in the four phenotype classes.

Step 1 — Expected counts (160 × each predicted fraction):

• $\frac{9}{16}\times 160 = 90$
• $\frac{3}{16}\times 160 = 30$
• $\frac{3}{16}\times 160 = 30$
• $\frac{1}{16}\times 160 = 10$

Step 2 — Per-class $\frac{(O-E)^2}{E}$:

Step 3 — Sum: $\chi^2 = 0.044 + 0.533 + 0.133 + 0.000 = 0.71$

Step 4 — Decide. df = (4 classes − 1) = 3; the critical value at $p = 0.05$ for df = 3 is 7.81. Since $\chi^2 = 0.71 < 7.81$, we fail to reject the null hypothesis — the observed counts fit the predicted 9:3:3:1 ratio comfortably (the genes appear to assort independently).

Contrast worth remembering: the same numerical $\chi^2$ can be "significant" or "not significant" depending on df, because the critical value rises with more classes (3.84 for df = 1, 7.81 for df = 3, 9.49 for df = 4). Always compute df = (classes − 1) before comparing to the table — judging $\chi^2$ without its df is the most common chi-square mistake.