Heredity
Unit 5 of AP Biology, worth 8–11% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Meiosis and genetic diversity, Mendelian genetics, non-Mendelian inheritance, chromosomal inheritance, probability and pedigrees.
How this unit is tested
Start by anchoring everything to meiosis: almost every heredity pattern in this unit traces back to how homologous chromosomes separate and how alleles get shuffled into gametes. Before doing any genetics problem, identify what kind of question it is: a straightforward Mendelian cross (use a Punnett square), a probability question (use the product and sum rules instead of drawing a huge square), a non-Mendelian inheritance pattern (identify which special case — incomplete dominance, codominance, epistasis, polygenic, or linkage — is being described), or a pedigree (work out dominant vs recessive and autosomal vs X-linked from the pattern of affected individuals).
For probability-heavy questions, resist the urge to build a 16-box Punnett square for every trait combination. Instead, calculate the probability for each gene separately using a simple 1-gene ratio, then combine the probabilities: multiply for 'and' (independent events happening together), and add for mutually exclusive 'or' outcomes, subtracting the overlap when the events can happen together.
For pedigrees, look for signature patterns: a trait that skips generations and needs two carrier parents is recessive; a trait appearing in every generation is usually dominant; a trait that appears almost exclusively in males suggests X-linked recessive; a trait passed from affected fathers to all daughters suggests X-linked dominant. Always test your hypothesis against every individual in the pedigree, not just the first couple of generations.
Finally, keep track of which mechanism is responsible for variation at each stage: crossing over and independent assortment happen during meiosis (before fertilization), random fertilization happens at fertilization, and nondisjunction is an error in meiosis that changes chromosome number rather than allele combinations.
What you have to know
Law of Segregation
Each organism carries two alleles for each gene, and these alleles segregate from each other during gamete formation so that each gamete receives only one allele for each gene.
Law of Independent Assortment
Alleles for genes located on different (non-homologous) chromosome pairs assort into gametes independently of one another during meiosis.
Product Rule (probability)
The probability that two or more independent events all occur together equals the product of their individual probabilities.
Sum Rule (probability)
The probability that either of two mutually exclusive events occurs equals the sum of their individual probabilities; if the events are not mutually exclusive, subtract the probability that both occur to avoid double-counting.
Chromosome Theory of Inheritance
Genes are located on chromosomes, and the behavior of chromosomes during meiosis (segregation and independent assortment) accounts for Mendel's laws of inheritance.
14 practice questions
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During meiosis I, which event directly increases genetic diversity by exchanging genetic material between homologous chromosomes?
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Answer. Crossing over between non-sister chromatids of homologous chromosomes during prophase I.
In prophase I, homologous chromosomes pair up (synapsis) to form a tetrad, and non-sister chromatids exchange segments of DNA at chiasmata. This physically shuffles alleles between maternal and paternal chromosomes, creating new allele combinations. Independent assortment and random fertilization also add diversity but do not involve physical exchange of DNA. -
In a cross between two dihybrid pea plants, YyRr x YyRr, where yellow (Y) is dominant to green (y) and round (R) is dominant to wrinkled (r), what is the probability that an offspring will be green and wrinkled (yyrr)?
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Answer. 1/16
Because the two genes assort independently, calculate each gene's probability separately using the product rule. For Yy x Yy, $P(yy)=1/4$. For Rr x Rr, $P(rr)=1/4$. Multiply the independent probabilities: $1/4 \times 1/4 = 1/16$. -
A pea plant with round seeds has an unknown genotype (RR or Rr). Describe a testcross that would determine its genotype, and state the expected results for each possibility.
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Answer. Cross the round-seeded plant with a homozygous recessive wrinkled plant (rr). If all offspring have round seeds, the unknown parent is RR; if about half the offspring are round and half are wrinkled, the unknown parent is Rr.
A testcross uses a homozygous recessive individual because it can only contribute a recessive allele, so any dominant phenotype appearing in offspring must come from the unknown parent's dominant allele. An RR parent produces only R gametes, giving all Rr (round) offspring, while an Rr parent produces R and r gametes equally, giving a 1:1 round-to-wrinkled ratio. -
In snapdragons, flower color shows incomplete dominance: red x white F1 crosses produce all pink offspring. If two pink F1 plants are crossed, what phenotypic ratio is expected in the F2 generation?
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Answer. 1 red : 2 pink : 1 white
Because neither allele is dominant, heterozygotes show an intermediate phenotype. Crossing the two pink heterozygotes gives genotype ratios of 1:2:1, which correspond directly to phenotypes since each genotype has its own distinct appearance: 1 red : 2 pink : 1 white. -
A man with blood type genotype I^A i and a woman with genotype I^B i have a child. What is the probability the child has type AB blood?
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Answer. 1/4
The I^A and I^B alleles are codominant, so both are expressed together as type AB when present in the same individual. The cross produces genotypes I^A I^B, I^A i, I^B i, and ii each with probability $1/4$, and only the I^A I^B genotype gives the AB phenotype. -
In Labrador retrievers, coat color depends on two genes: B/b (black vs brown pigment) and E/e, where the ee genotype prevents any pigment from being deposited in the coat, producing yellow fur regardless of the B genotype. A cross between two BbEe dogs produces offspring in what phenotypic ratio?
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Answer. 9 black : 3 brown : 4 yellow
This is recessive epistasis: the E gene determines whether the B gene's product is expressed at all. Of the standard 9:3:3:1 dihybrid classes, the two ee classes (3 B_ee and 1 bbee, totaling 4/16) all appear yellow because pigment isn't deposited, collapsing the ratio to 9 black : 3 brown : 4 yellow. -
Red-green color blindness is an X-linked recessive trait. A carrier woman has children with a man who has normal color vision. What proportion of their children overall are expected to be color blind?
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Answer. 1/4 of all children (half of the sons, none of the daughters)
The mother's gametes are half normal-allele X and half colorblind-allele X; the father contributes a normal X to daughters and a Y to sons. All daughters get his normal X, so none are colorblind. Half of sons inherit the colorblind X from mom and are colorblind. Since sons are half of all children, $1/2 \times 1/2 = 1/4$ of all children are expected to be colorblind. -
A geneticist examines a gamete with an extra chromosome resulting from nondisjunction. Karyotyping shows the extra chromosome consists of two genetically identical sister chromatids that failed to separate. In which meiotic division did the nondisjunction most likely occur, and why?
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Answer. Meiosis II, because the extra chromosome is made of two identical sister chromatids rather than two distinct recombined homologs.
If nondisjunction occurs in meiosis I, homologous chromosomes fail to separate, so the resulting gamete carries one copy each of two different (non-identical, recombined) homologs. If nondisjunction occurs in meiosis II after homologs have already separated correctly, sister chromatids fail to separate, producing a gamete with two identical copies of one chromosome — matching the scenario described. -
In a testcross of a dihybrid fly heterozygous for two linked genes, 940 and 935 offspring show parental phenotypes, while 65 and 60 offspring show recombinant phenotypes out of 2000 total offspring. Calculate the recombination frequency and express it as map units.
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Answer. 6.25 map units (6.25% recombination frequency)
Recombination frequency equals the number of recombinant offspring divided by the total, times 100. Recombinants total $65+60=125$; total offspring is 2000, so $RF = 125/2000 \times 100 = 6.25\%$. By convention, 1% recombination frequency corresponds to 1 map unit, so the genes are 6.25 map units apart. -
In a pedigree, an affected trait appears almost exclusively in males, affected fathers pass the trait's allele to all daughters (who become unaffected carriers, not affected), and affected sons always have carrier mothers rather than affected fathers. Which inheritance pattern best fits this pedigree?
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Answer. X-linked recessive
X-linked recessive traits appear predominantly in males because males are hemizygous, so a single recessive allele on their one X chromosome is enough to cause the phenotype. Affected fathers pass their X (carrying the recessive allele) to all daughters, making them carriers but not affected since they also get a normal allele from their mother, and sons are affected only if their mother is a carrier — matching the pattern described. -
For a dihybrid cross AaBb x AaBb, where A and B assort independently, what is the probability that an offspring is homozygous recessive for at least one of the two genes (aa or bb)?
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Answer. 7/16
P(aa) = 1/4 and P(bb) = 1/4 individually. Since these events are not mutually exclusive (a child could be both aa and bb), use the general addition rule: $P(aa \text{ or } bb) = P(aa)+P(bb)-P(aa \text{ and } bb)$. The overlap is $1/4 \times 1/4 = 1/16$ by the product rule, so $P(aa \text{ or } bb) = 1/4+1/4-1/16 = 7/16$. -
Human skin color varies continuously across a population rather than falling into a few discrete categories. Which mode of inheritance best explains this pattern, and why?
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Answer. Polygenic inheritance, because multiple genes each contribute a small additive effect to the phenotype, producing a continuous range of possible outcomes rather than a few discrete classes.
Traits controlled by many genes at different loci, each with additive alleles, generate a wide range of intermediate phenotypes rather than the sharp ratios seen with single-gene traits. The more genes involved, the finer the gradation of phenotypes, producing a bell-shaped distribution as seen with human height and skin color. -
Humans have a haploid number of $n = 23$ chromosomes. Ignoring crossing over, how many genetically distinct combinations of chromosomes are possible in a single gamete due to independent assortment alone?
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Answer. $2^{23}$, or about 8.4 million
During metaphase I, each of the 23 homologous chromosome pairs orients randomly and independently of the others, and each pair has two possible orientations determining which homolog goes into a given gamete. With 23 independent binary choices, the total number of possible combinations is $2^{23} \approx 8{,}388{,}608$. -
Name the three mechanisms occurring during meiosis and fertilization that together generate genetic variation among the offspring of sexually reproducing organisms.
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Answer. Crossing over (recombination) in prophase I, independent assortment of homologous chromosomes in metaphase I, and random fertilization of gametes.
Crossing over shuffles alleles between maternal and paternal chromatids within a chromosome; independent assortment randomly distributes whole maternal or paternal chromosomes into gametes; and random fertilization combines any one of millions of possible sperm with any one of millions of possible eggs. Together these ensure that, barring identical twins, no two offspring of the same parents are genetically identical.
What people get wrong
- Thinking crossing over happens between sister chromatids — it actually happens between non-sister chromatids of homologous chromosomes during prophase I; sister chromatids are genetically identical and swapping between them does nothing.
- Applying the standard 9:3:3:1 dihybrid ratio to an epistasis problem without adjusting it — instead, identify which genotype classes are masked by the epistatic gene and combine those categories (e.g., 9:3:4).
- Using the product rule ('and') when a question asks for 'either/or' outcomes, or vice versa — decide first whether the question wants both events to occur together (multiply) or one of several possible mutually exclusive outcomes (add).
- Assuming nondisjunction can only occur in meiosis I — it can also occur in meiosis II, and the two cases produce gametes with different chromosome compositions (non-identical homologs vs. identical sister chromatids).
- Reading a pedigree by only checking the first generation — always confirm a proposed inheritance pattern against every affected and unaffected individual in the whole pedigree, especially unaffected parents who produce affected offspring (a strong clue for recessive inheritance).
- Forgetting that males are hemizygous for X-linked genes — a male needs only one copy of a recessive X-linked allele to show the trait, so his phenotype directly reveals his single X-linked genotype.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Biology. Free, and no account needed to start.