Separate the alleles. Then choose the probability rule.
Move from segregation and testcross evidence to independent-event probability and inheritance extensions without confusing dominance with frequency or blending.
From discrete alleles to bounded probability claims.
Use gamete-level segregation, explicit sample spaces, cross evidence, and the level of genetic interaction instead of memorized ratios without assumptions.
01
LESSON 1 · 18 MIN
Study + retrieve
Segregation, dominance, and testcrosses
Predict monohybrid outcomes and use a testcross to distinguish a homozygous dominant genotype from a heterozygote.
ESSENTIAL QUESTIONWhich alleles enter each gamete, and what does the observed offspring pattern actually reveal?
STUDY DIAGRAM · TEXT DESCRIPTION AVAILABLE
01
Separate alleles into gametes
A diploid individual carries two alleles at a locus, but each gamete receives one. A heterozygote Aa therefore produces A-bearing and a-bearing gametes in equal expected proportions when segregation is unbiased.
Aa → A or a
Gametes carry one allele
Expected probability is not a guaranteed small-family count
02
Define dominance narrowly
Under complete dominance, Aa and AA share the dominant phenotype while aa has the recessive phenotype. Dominant describes the heterozygote’s phenotype; it does not mean common, beneficial, stronger, or evolutionarily favored.
Dominance concerns Aa
Phenotype can hide genotype
Dominant ≠ frequent or fit
03
Use the homozygous recessive tester
Cross an individual with a dominant phenotype but unknown genotype to aa. Recessive offspring prove the unknown parent supplied a, while a sufficiently informative all-dominant sample supports but does not logically guarantee AA.
Unknown A_ × aa
Any aa offspring → unknown was Aa
Small samples carry uncertainty
Worked example
A tall plant of unknown genotype is crossed with tt and produces 48 tall and 52 short offspring. What is the unknown genotype?
1
Short offspring must be tt and receive t from each parent.
2
The tester always supplies t, so the unknown tall parent must sometimes supply t.
3
A tall parent carrying t is Tt, and Tt × tt predicts an expected 1:1 phenotype ratio.
ConclusionThe unknown tall parent is heterozygous Tt; the near 1:1 offspring pattern is the expected testcross signature.
Close the notes first
Retrieve the evidence boundary.
01What alleles can an Aa individual place in a gamete?
A or a, each with expected probability one-half.
The two homolog-associated alleles segregate during meiosis.
02What does dominant mean?
The allele’s phenotype is expressed in the heterozygote under the stated conditions.
Dominance does not specify allele frequency, fitness, or physical strength.
03Why use aa in a testcross?
The tester always contributes a, so offspring phenotypes reveal which allele the unknown parent contributed.
The recessive tester removes ambiguity from its own gametes.
02
LESSON 2 · 20 MIN
Study + retrieve
Probability and independent assortment
Apply product, sum, complement, and conditional probability while checking whether loci and events may be treated as independent.
ESSENTIAL QUESTIONAre these events independent, mutually exclusive, complementary, or conditional?
STUDY DIAGRAM · TEXT DESCRIPTION AVAILABLE
01
Multiply independent joint events
For independent events that must both occur, multiply their probabilities. In AaBb × AaBb, P(aa and B_) is 1/4 × 3/4 = 3/16 only if the loci assort independently and complete dominance applies at B.
AND → product when independent
Check genotype and dominance assumptions
Linkage can break the simple product
02
Add routes and use complements
Add probabilities for mutually exclusive routes to the same result. For at least one success, it is often faster to calculate one minus the probability of no successes.
OR → sum for disjoint routes
At least one = 1 − none
State the sample and event
03
Condition on the information given
Conditional probability removes outcomes inconsistent with the condition. From Aa × Aa, among offspring with the dominant phenotype, AA:Aa occurs in a 1:2 ratio, so P(Aa | dominant phenotype) = 2/3.
Restrict the sample space
Renormalize remaining outcomes
Phenotype information changes genotype odds
Worked example
For AaBb × AaBb with independently assorting loci, what is the probability of aa and at least one B allele?
1
At locus A, Aa × Aa gives P(aa) = 1/4.
2
At locus B, Bb × Bb gives P(B_) = 3/4.
3
The loci are stated to assort independently, so multiply: 1/4 × 3/4.
ConclusionThe probability is 3/16; without the independence assumption, that product is not justified.
Close the notes first
Retrieve the evidence boundary.
01When does the product rule apply?
When all named events must occur and their probabilities are independent or appropriately conditional.
Multiplication represents a joint path.
02What is the shortcut for at least one?
One minus the probability of none.
The two events are complements that exhaust the sample space.
03From Aa × Aa, what is P(Aa | dominant phenotype)?
Two-thirds.
Among the three dominant-phenotype genotype parts, two are Aa and one is AA.
03
LESSON 3 · 20 MIN
Study + retrieve
Extensions of Mendelian inheritance
Distinguish incomplete dominance, codominance, multiple alleles, pleiotropy, and epistasis without erasing discrete alleles.
ESSENTIAL QUESTIONIs the interaction between alleles at one locus, genes at different loci, or one gene and several traits?
STUDY DIAGRAM · TEXT DESCRIPTION AVAILABLE
01
Keep alleles discrete
In incomplete dominance, a heterozygote has an intermediate phenotype, but its alleles remain distinct and segregate. In a heterozygote under codominance, both allele products are detectably expressed rather than blended into a new permanent allele.
Incomplete dominance → intermediate heterozygote
Codominance → both products
Neither model erases alleles
02
Separate population and individual counts
A locus can have many alleles in a population, but a diploid individual carries at most two alleles at that locus. Multiple alleles expand population-level genotype possibilities without giving one diploid person every allele.
Population may have >2 alleles
Diploid individual has ≤2 at one locus
Allele count ≠ chromosome count
03
Name the level of interaction
Epistasis occurs when a genotype at one locus masks or modifies expression at another locus. Pleiotropy means one gene influences multiple phenotypic traits. These are distinct from polygenic inheritance, in which multiple genes contribute to one trait.
Epistasis: locus modifies locus
Pleiotropy: one gene → several traits
Polygenic: several genes → one trait
Worked example
Red flowers crossed with white flowers produce pink F1 plants, and pink × pink produces red:pink:white near 1:2:1. Did the alleles blend permanently?
1
The F1 heterozygote has an intermediate phenotype.
2
The parental phenotypes reappear in the F2 generation.
3
Reappearance shows the alleles remained discrete and segregated.
ConclusionThis is incomplete dominance, not permanent blending; genotype and phenotype ratios are both 1:2:1 in this cross.
Close the notes first
Retrieve the evidence boundary.
01What F2 phenotype ratio is expected for a simple incomplete-dominance heterozygote cross?
1:2:1.
Each genotype class has a distinct phenotype.
02How does codominance differ from incomplete dominance?
Codominance detectably expresses both allele products; incomplete dominance produces an intermediate heterozygote phenotype.
Both preserve discrete alleles but differ in phenotype.
03Epistasis or pleiotropy: one locus masks another locus?
Epistasis.
Pleiotropy instead describes one gene influencing several traits.
Randomized retrieval set
Now identify the transmitted allele, event rule, or interaction level.
Segregation, dominance, testcrosses, products, complements, conditional probability, incomplete dominance, codominance, multiple alleles, and epistasis are interleaved.
12 PRACTICE QUESTIONS
Retrieve before you review.
Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.
Scope and score notice
Classical Genetics foundations, not a score prediction.
The ADA lists Classical Genetics within Genetics but does not publish a subtopic item quota. DAT TRAIN does not invent one.
Linkage mapping, detailed pedigrees, penetrance, population genetics, clinical risk interpretation, and named rare syndromes remain outside this route unless a prompt supplies the required model.
Use your results to choose what to review next—not as an official DAT score prediction.