Genetics · Study notes
Classical Mendelism handles single genes with two alleles, but real traits routinely bend those rules. This chapter covers the standard extensions: intermediate and co-expressed alleles, more than two alleles at one locus, genes that mask other genes, and alleles that alter survival. Every stated ratio is checked by summing its parts; the sums in this chapter are almost always 16, and the lethal ratio sums to 3.
In incomplete dominance, the heterozygote shows a phenotype intermediate between the two homozygotes. The classic example is flower color in snapdragons: a cross between a red-flowered plant and a white-flowered plant gives pink F1 offspring, and the F1 self cross produces red, pink, and white plants in the 1:2:1 ratio (sums to 4). The pink intermediate is a dosage effect: one functional allele makes about half as much pigment as two. The intermediate is not permanent, because crossing two pink plants recovers red and white plants, proving the alleles re-segregate cleanly.
Think of mixing one cup of red paint with one cup of white paint. The result is pink, a color halfway between the two. If you pour the pink paint into two jars and later use each jar alone, the two original colors come back. Incomplete dominance is like this, the two versions of a gene blend into a middle color, but the gene versions themselves are never destroyed.
In codominance, the heterozygote expresses both parental phenotypes at the same time instead of a blend. The ABO blood type system is the standard example: the IA and IB alleles are codominant, so a person with the IAIB genotype makes both the A sugar and the B sugar. Sickle cell trait shows the same logic: a person with one normal HBB allele and one sickle HBB allele (Glu6Val) makes both normal beta globin and sickle beta globin, so the alleles are codominant in expression even though the disease itself is autosomal recessive. A cross between an IAi and an IBi parent gives AB, A, B, and O offspring in the 1:1:1:1 ratio (sums to 4).
Common Mistake: Codominance is not incomplete dominance. The heterozygote IAIB shows both parental phenotypes at once and types as AB, while an incomplete dominance heterozygote would be intermediate. In an IAi x IBi cross the classes are AB, A, B, and O in the 1:1:1:1 ratio (sums to 4); treating AB as a blend collapses the count to three and misidentifies the ratio.
Picture a cow with some red hairs and some white hairs all over its body. No color blends into a new color, both colors stay fully visible at the same time. That is codominance, both versions of the gene are doing their own jobs. The red hairs stay red and the white hairs stay white.
Multiple alleles means that a single locus can hold three or more alleles in a population, even though each diploid individual carries at most two. The ABO locus is the classic system: the ABO gene (9q34.2) encodes the ABO glycosyltransferase, which adds the A or B sugar to the H antigen. The three alleles are IA and IB, which are codominant, and i, which is recessive to both; the i allele adds neither sugar, giving type O. Six genotypes therefore collapse into four phenotypes.
| Genotype | Blood type phenotype |
|---|---|
| IAIA, IAi | A |
| IBIB, IBi | B |
| IAIB | AB |
| ii | O |
Two different genotypes can give the same phenotype, so a type O child (ii) can come from several parental combinations; the 1:1:1:1 ratio (sums to 4) of an IAi x IBi cross follows from the gamete classes. Multiple alleles are not unique to ABO: the HBB gene (11p15.4) carries several disease alleles, including the sickle allele (Glu6Val in beta globin) and Beta thalassemia alleles that reduce beta globin synthesis.
CROSS: ABO blood type with three alleles at one locus
Parents: IAi x IBi
Assumptions: one locus with three alleles, IA and IB codominant, i recessive, equal viability
Gametes per parent: IA and i in 1:1, IB and i in 1:1
Genotypic outcome per locus: 1:1:1:1 (IAIB, IAi, IBi, ii)
Expected phenotypic ratio: 1:1:1:1 (sums to 4)
Probability requested: P(type O) = 1/2 x 1/2 = 1/4Imagine a hat with red, blue, and yellow tickets. Each player draws only two tickets, so no player holds more than two colors at once, but the hat keeps all three colors ready. Multiple alleles work the same way. The population carries many versions of a gene, and each person gets just two copies, one from each parent.
Epistasis is an interaction in which one gene masks the phenotypic effect of another gene at a different locus; the masking gene is epistatic to the masked gene. In recessive epistasis, a homozygous recessive genotype at one locus hides the other locus, so in a pigment pathway the cc class blocks all color regardless of the A locus, giving the 9:3:4 ratio (sums to 16). The color locus in such examples is often tyrosinase, the product of TYR (11q14.3) that makes melanin: when tyrosinase is lost, no pigment can form. In dominant epistasis, a single dominant allele at one locus masks the other locus, giving the 12:3:1 ratio (sums to 16). In duplicate gene action, either dominant allele alone produces the phenotype, so only the double homozygous recessive differs, giving the 15:1 ratio (sums to 16).
CROSS: Recessive epistasis in a two gene pigment pathway
Parents: CcAa x CcAa
Assumptions: genes unlinked, complete dominance at both loci, cc blocks all pigment, equal viability
Gametes per parent: CA, Ca, cA, ca in 1:1:1:1
Genotypic outcome per locus: 1:2:1
Expected phenotypic ratio: 9:3:4 (sums to 16)
Probability requested: P(pigmented) = 3/4 x 3/4 = 9/16A toy factory has a machine that makes wheels and a machine that paints the cars. If the wheel machine is broken, no car is ever finished, no matter how well the paint machine works. The broken machine hides the work of the whole factory. That is epistasis, one part of the system hiding the effect of another part.
All of the interactions above reorganize the same sixteen-cell dihybrid grid, so the ratios keep the total of 16, except the lethal case at 3. The catalog below is the complete list to memorize, matching the Modified Ratio Catalog in REFERENCE.md.
| Interaction type | Phenotypic ratio | Sum of parts | One line description |
|---|---|---|---|
| Independent assortment, complete dominance | 9:3:3:1 | 16 | Two unlinked genes, no interaction |
| Recessive epistasis | 9:3:4 | 16 | Homozygous recessive at one gene hides the other gene |
| Complementary gene action | 9:7 | 16 | Both dominant alleles needed for the phenotype |
| Dominant epistasis | 12:3:1 | 16 | One dominant allele at one gene hides the other gene |
| Dominant suppression | 13:3 | 16 | One dominant allele suppresses the other gene |
| Duplicate gene action | 15:1 | 16 | Either dominant allele produces the phenotype |
| Duplicate genes with cumulative effect | 9:6:1 | 16 | Two dominant alleles together give an additive phenotype |
| Lethal allele | 2:1 | 3 | Homozygous lethal class is missing |
In complementary gene action (9:7, sums to 16), both genes must supply a dominant allele, so only AB stands apart. In dominant suppression (13:3, sums to 16), a dominant allele at the suppressor locus silences the other gene, leaving the rare phenotype to 3 of 16 offspring. In duplicate genes with cumulative effect (9:6:1, sums to 16), two dominant alleles give an additive phenotype, one gives an intermediate, and none gives the baseline. The recognition strategy is to ask which classes collapse: the double recessive merging with one other class signals recessive epistasis (9:3:4, sums to 16); only AB standing apart signals complementary gene action (9:7, sums to 16); a total of 3 signals a lethal allele (2:1, sums to 3).
CROSS: Complementary gene action in flower color
Parents: AaBb x AaBb
Assumptions: genes unlinked, complete dominance at both loci, both dominant alleles needed for color, equal viability
Gametes per parent: AB, Ab, aB, ab in 1:1:1:1
Genotypic outcome per locus: 1:2:1
Expected phenotypic ratio: 9:7 (sums to 16)
Probability requested: P(colored) = 3/4 x 3/4 = 9/16Common Mistake: Do not memorize the number pattern without the mechanism. The 9:3:4 ratio (sums to 16) is recessive epistasis, not dominant epistasis, and the 12:3:1 ratio (sums to 16) is dominant epistasis, not duplicate gene action. Always verify that the parts add to 16 before naming the interaction.
A chocolate bar with 16 squares can be broken into piles of 9, 3, 3, and 1 squares. The same bar can also be broken into piles of 9 and 7, or 15 and 1. The bar never changes size, only the way the squares are grouped changes. Modified ratios are like this, the same 16 outcomes regrouped into different phenotype piles.
Pleiotropy is the situation in which one gene affects multiple, seemingly unrelated traits. Sickle cell disease is the textbook case: a single Glu6Val change in beta globin (HBB, 11p15.4) produces sickled red cells, anemia, and pain crises. Marfan syndrome shows the same logic: a defect in fibrillin (FBN1, 15q21.1) leads to tall stature, aortic dilation, and lens dislocation from one altered structural protein. Huntington disease (HTT, 4p16.3, CAG repeat expansion) likewise pairs chorea with cognitive decline.
One wall switch can turn on the ceiling light, the fan, and the heater all at once. Flipping that switch changes three things with a single action. Pleiotropy is the same idea, one gene with one job that touches many parts of the body. That is why one small change in a gene can bring several different problems together.
Penetrance is the proportion of individuals with a given genotype who show any phenotype at all, and incomplete penetrance means some carriers never express the trait. The index example is Hereditary breast and ovarian cancer syndrome, autosomal dominant with incomplete penetrance: some individuals with a disease allele never develop cancer. Expressivity is the degree to which the phenotype appears, and variable expressivity means affected individuals differ in severity. Neurofibromatosis type 1 (NF1, 17q11.2, autosomal dominant) illustrates the range: one affected individual may show a few cafe au lait spots while another shows many neurofibromas. Penetrance is the on-off switch, and expressivity is the dimmer. An autosomal recessive comparison is Hemochromatosis (HFE, 6p21.3, C282Y variant), whose features, iron overload and liver damage, require two disease alleles.
Some seeds planted in the same garden never sprout at all, even though they are the same kind of seed. That is incomplete penetrance, the genes are present but no plant shows. The seeds that do sprout grow to different heights, some tall and some short. That is variable expressivity, the same genes producing different strengths of the trait.
A lethal allele kills the organism when present in a genotype, usually early in development. When a dominant allele is lethal in homozygotes, an Aa x Aa cross produces zygotes in the 1:2:1 genotypic split (sums to 4), but the AA class dies, so the surviving offspring show a 2:1 phenotypic ratio (sums to 3). A recessive lethal is the mirror image: the aa class dies, all surviving offspring show the dominant phenotype, and the surviving genotypic split is 1:2 (sums to 3). The 2:1 ratio (sums to 3) appears only when the dead class is excluded from the count.
CROSS: Dominant allele that is lethal when homozygous
Parents: Aa x Aa
Assumptions: A dominant to a, AA lethal before birth, no other viability differences
Gametes per parent: A and a in 1:1
Genotypic outcome per locus: 1:2:1
Expected phenotypic ratio: 2:1 (sums to 3)
Probability requested: P(affected among live offspring) = (1/2)/(3/4) = 2/3A bakery bakes twelve muffins and always throws away the four that come out burnt. A customer counting the muffins on the shelf sees only eight, not twelve. The burnt muffins are real, they are just never counted. A lethal allele is like the burnt muffins, one class of offspring is removed before anyone counts the rest.
Gene interaction is the general term for any situation in which two or more genes jointly determine one phenotype, and epistasis is one special case. The catalog covers the spectrum: masking (epistasis), cooperation (complementary gene action), redundancy (duplicate gene action), and additivity (duplicate genes with cumulative effect). The corresponding ratios are the 9:7 ratio (sums to 16), the 15:1 ratio (sums to 16), and the 9:6:1 ratio (sums to 16). Modifier genes are a milder interaction, changing the strength or timing of another gene's effect without masking it. None of this requires linkage: the modified ratios all assume independent assortment.
Baking a cake needs flour and eggs, two different ingredients that both have to be there. If either one is missing, there is no cake at all. Some genes are like that, they must work together. Other recipes are more forgiving, sugar or honey both make the cake sweet, so either ingredient alone is enough.
The same genotype can produce different phenotypes in different environments. Temperature sensitive alleles are the cleanest example: the mutant protein works at a permissive temperature and fails at a restrictive one. The classic illustration uses tyrosinase (TYR, 11q14.3), the enzyme that synthesizes melanin: a temperature sensitive tyrosinase variant makes pigment only where the body is cool enough, so animals such as Siamese cats have dark extremities on a pale body. Diet is another environmental variable: Phenylketonuria (PAH, 12q23.2, autosomal recessive) follows from loss of phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine, and the index notes intellectual disability if untreated, so dietary control decides the outcome of the same genotype. A phenocopy is the mirror image: an environment that mimics the phenotype of a genotype without the genotype being present.
The same cookie dough makes flat cookies in a cool oven and puffy cookies in a hot oven. The dough did not change, the oven did. Genes work this way too, the same gene can give different results in different conditions. That is why the phenotype depends on both the genes and the environment.
Problem: A person with blood type A and genotype IAi has children with a person of blood type AB. What is the probability that a child has blood type A?
Given: Parent one genotype is IAi; parent two is IAIB; IA and IB are codominant, and i is recessive.
Plan: List the gametes of each parent, form the four zygote classes, then add the classes that express the A sugar.
Solution: Parent one gametes are IA and i, each with probability 1/2. Parent two gametes are IA and IB, each with probability 1/2. Zygote classes: IAIA = 1/2 x 1/2 = 1/4, IAIB = 1/2 x 1/2 = 1/4, IAi = 1/2 x 1/2 = 1/4, IBi = 1/2 x 1/2 = 1/4. Blood type A comes from IAIA or IAi, so P(A) = 1/4 + 1/4 = 1/2.
Answer: The probability of a type A child is 1/2.
Problem: Two Aa individuals mate, and the AA genotype is lethal before birth. What proportion of live offspring is affected?
Given: Zygote classes are 1/4 AA, 1/2 Aa, and 1/4 aa, and the AA class never survives.
Plan: Add the surviving classes, then express the affected class as a fraction of the survivors.
Solution: Survivors are Aa plus aa: 1/2 + 1/4 = 3/4 of all zygotes. P(affected given live) = (1/2)/(3/4) = 1/2 x 4/3 = 2/3. This is the 2:1 ratio (sums to 3) written as a probability.
Answer: 2/3 of live offspring are affected.
Problem: In the cross CcAa x CcAa, the cc genotype blocks all pigment (recessive epistasis). What is the probability that an offspring is pigmented?
Given: Pigmented requires C_ and A_; the cc class is unpigmented regardless of the A locus.
Plan: Use independent assortment: P(pigmented) = P(C_) x P(A_).
Solution: P(C_) = 3/4 and P(A_) = 3/4, so P(pigmented) = 3/4 x 3/4 = 9/16. The full partition checks: 9/16 pigmented, 3/16 for the C_aa class, and 4/16 for the cc class, and 9 + 3 + 4 = 16.
Answer: The probability of a pigmented offspring is 9/16.
High-Yield:
- Every modified dihybrid ratio adds to 16: the 9:3:3:1 ratio (sums to 16), the 9:3:4 ratio (sums to 16), the 9:7 ratio (sums to 16), the 12:3:1 ratio (sums to 16), the 13:3 ratio (sums to 16), the 15:1 ratio (sums to 16), and the 9:6:1 ratio (sums to 16); the lethal ratio 2:1 (sums to 3) is the exception.
- Recessive epistasis hides with a homozygous recessive pair, giving 9:3:4 (sums to 16); dominant epistasis hides with one dominant allele, giving 12:3:1 (sums to 16).
- ABO has three alleles at one locus, IA and IB codominant and i recessive; the ABO gene (9q34.2) encodes the ABO glycosyltransferase, which adds the A or B sugar to the H antigen.
- A lethal allele removes one zygote class, so a 1:2:1 genotypic split (sums to 4) becomes the 2:1 ratio (sums to 3) among survivors.
- Penetrance asks whether any phenotype appears at all; expressivity asks how strong the phenotype is.
Key terms are emphasized and defined within the main notes.
See the formulas, procedures, and process blocks in the main notes where applicable.
See the labeled common-mistake callouts in the main notes where present.
Use the quick-review or recap section in the main notes.
5 questions here, of 12 in this lesson’s practice set. Answers stay hidden until you check.
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