📖 ContentsChapter 2
Genetics · Chapter 2

Extensions of Mendel

Mendel's peas gave clean 3:1 and 9:3:3:1 ratios, but most real genes are not so tidy. Heterozygotes can look intermediate, both alleles can show at once, a gene can have dozens of alleles, some genotypes never survive to be counted, and one gene can hide the effects of another. This chapter shows that none of these complications overturn Mendel: they are his rules operating on the biochemistry of real gene products, and once you see the pathway behind a strange ratio, the ratio stops being strange.

By the end of this chapter you can

  • Distinguish complete dominance, incomplete dominance and codominance, and explain each in terms of gene product.
  • Analyze multiple allele series, including the ABO blood group and dominance hierarchies, and predict offspring of crosses involving them.
  • Recognize the 2:1 ratio produced by a recessive lethal allele and contrast recessive, dominant and conditional lethals.
  • Explain gene interaction that produces novel phenotypes while preserving a 9:3:3:1 ratio.
  • Identify the major forms of epistasis from their modified dihybrid ratios (9:3:4, 12:3:1, 9:7, 15:1, 13:3 and 9:6:1) and relate each to a biochemical pathway.
  • Define and calculate penetrance, define expressivity, and explain how environment, modifier genes and age alter phenotype.
  • Explain phenocopies, norms of reaction and temperature sensitive alleles.
  • Design and interpret a complementation test to decide whether two recessive mutations lie in the same gene, and sort mutants into complementation groups.

Section 1Incomplete dominance, codominance and multiple alleles

Dominance is a property of the heterozygote's phenotype

In Chapter 1, dominance meant that a heterozygote looked like one homozygote. That is complete dominance. But dominance is not built into an allele; it describes how the phenotype of the heterozygote compares with the two homozygotes, and the answer depends on the trait and on how closely you look. Three relationships are recognized:

Incomplete dominance

The classic example is flower color in snapdragons (Antirrhinum) and four o'clocks (Mirabilis). A true breeding red line crossed with a true breeding white line gives pink F1 plants. Intercrossing the pink plants gives an F2 of 1 red : 2 pink : 1 white. This looks at first like blending, but it is not: red and white reappear unchanged in the F2, proving that the alleles stayed discrete. The red allele CR encodes an enzyme in anthocyanin pigment synthesis, while CW makes no functional enzyme. Two doses give full red pigment, one dose gives half as much, which we see as pink, and none gives white.

With incomplete dominance the phenotypic ratio equals the genotypic ratio, 1:2:1, because each genotype has its own phenotype. Symbols change accordingly: since neither allele is dominant, geneticists use superscripts on a common base letter rather than upper and lower case letters.

P red CRCR × white CWCW F1 all pink CRCW F2 1 red 2 pink 1 white Pigment depends on dose CWCW CRCW CRCR 0 half full copies of CR: 0, 1, 2 pigment made
Figure 2.1 Incomplete dominance. The pink F1 is intermediate, yet pure red and pure white return in the F2, so the alleles never blended. The bar chart shows the underlying reason: pigment is proportional to the number of functional CR alleles.

Codominance

In codominance both alleles are expressed in the heterozygote, each detectable in its own right. The human MN blood group is a clean example. The GYPA gene encodes glycophorin A, a red cell surface protein, and two common alleles, LM and LN, make forms that differ at two amino acids. People with genotype LMLM have type M cells, LNLN have type N, and heterozygotes LMLN have type MN, with both antigens present on their cells. Two MN parents have children in the ratio 1 M : 2 MN : 1 N.

Whether a pair of alleles looks dominant, incompletely dominant or codominant often depends on the level of analysis. Sickle cell disease illustrates this. At the level of the whole person under ordinary conditions, the sickle allele HbS is recessive: heterozygotes (sickle cell trait) are generally healthy. At the level of red cell shape under low oxygen, heterozygote cells sickle somewhat, so the allele is incompletely dominant. At the level of the protein, heterozygotes make both normal hemoglobin A and hemoglobin S, which can be separated by electrophoresis, so the alleles are codominant. Same alleles, three answers.

Multiple alleles

Any one diploid individual carries at most two alleles of a gene, but a population can carry many. A gene with three or more alleles in the population has multiple alleles, and the alleles form an allelic series. With n alleles, the number of possible genotypes is n(n + 1)/2: n homozygotes plus n(n − 1)/2 heterozygotes. Three alleles give 6 genotypes; four give 10.

The ABO blood group combines multiple alleles with both dominance and codominance. The ABO gene encodes a glycosyltransferase that adds a sugar to a cell surface precursor called the H antigen. The IA allele's enzyme adds N acetylgalactosamine, making the A antigen; the IB enzyme adds galactose, making the B antigen; the i allele carries a frameshift and makes no active enzyme, so the H antigen is left unmodified (type O). IA and IB are each dominant to i and codominant with each other.

Blood typeGenotypesAntigens on red cellsAntibodies in plasmaCan receive red cells from
AIAIA, IAiAAnti BA, O
BIBIB, IBiBAnti AB, O
ABIAIBA and BNeitherA, B, AB, O (universal recipient)
OiiNeither (unmodified H)Anti A and anti BO only (O is the universal red cell donor)

Another kind of allelic series is a dominance hierarchy, in which each allele is dominant to those below it. Rabbit coat color is the textbook case, with four alleles of the C gene: C (full color, agouti) > cch (chinchilla, silvery gray) > ch (Himalayan, white with dark extremities) > c (albino). A cchch rabbit is chinchilla, a chc rabbit is Himalayan, and so on. The Himalayan allele is itself temperature sensitive, a point we return to in Section 4. Plant self incompatibility systems carry the idea to an extreme: the S locus of some species has dozens of alleles, and pollen cannot fertilize a plant that carries the same S allele, which enforces outcrossing.

Study tip. Worked problem: A woman with type A blood whose father was type O has children with a man of type B whose mother was type O. What blood types can their children have, and in what proportions? Step 1: Determine genotypes. A type O father can only transmit i, so the woman is IAi. Likewise the man is IBi. Step 2: Gametes: woman 1/2 IA, 1/2 i; man 1/2 IB, 1/2 i. Step 3: Combine: IAIB = 1/4 (AB); IAi = 1/4 (A); IBi = 1/4 (B); ii = 1/4 (O). Answer: 1 AB : 1 A : 1 B : 1 O. This single cross produces all four blood types, a favorite exam result. Note that ABO typing can exclude a man as a biological father (an AB man cannot father an O child) but cannot by itself prove paternity.
Connection. ABO compatibility is the first rule of transfusion medicine. Anti A and anti B antibodies appear in early infancy without any prior exposure to foreign blood, so giving type A red cells to a type O patient can trigger a rapid, sometimes fatal hemolytic reaction. A rare variant, the Bombay phenotype (genotype hh), lacks the H precursor entirely, so such a person types as O but makes anti H antibodies and can receive blood only from other Bombay donors. We return to this as an example of epistasis in Section 3.
Key point

Dominance describes the heterozygote's phenotype, not a property of the allele. Incomplete dominance gives intermediate heterozygotes and a 1:2:1 phenotypic ratio; codominance shows both allele products, as in the MN and AB blood types. Genes can have many alleles in a population, arranged in dominance hierarchies or mixed relationships like ABO, and n alleles give n(n + 1)/2 genotypes.

Section 2Lethal alleles and gene interaction

The yellow mouse puzzle

Soon after Mendel's work was rediscovered, the French geneticist Lucien Cuénot studied coat color in mice and hit a ratio that made no sense. Yellow mice never bred true. When yellow was crossed with yellow, the offspring were always about 2 yellow : 1 nonyellow (agouti), never 3:1. And every yellow mouse, when testcrossed, proved to be heterozygous. Where were the homozygous yellow mice?

The answer is that they die as embryos. The yellow allele, AY, is an allele of the agouti gene. It has a dominant effect on coat color (one copy makes the coat yellow) but a recessive lethal effect: AYAY embryos die shortly after implantation. A yellow by yellow cross (AYA × AYA) is expected to give 1/4 AYAY (dead), 1/2 AYA (yellow) and 1/4 AA (agouti). Among the survivors, yellow : agouti = 1/2 : 1/4 = 2:1. The mechanism was later found: AY carries a large deletion that fuses the agouti coding sequence to the promoter of a neighboring gene, Raly, which is essential for early development. The deletion destroys Raly function (lethal when homozygous) while driving agouti protein expression everywhere (yellow coat, plus obesity and diabetes in adult yellow mice).

yellow AYA × yellow AYA AY A AY A AYAY dies as embryo AYA yellow AYA yellow AA agouti Survivors: 2 yellow : 1 agouti One allele, two effects Coat color: one AY is enough DOMINANT (yellow) Survival: two AY needed to kill RECESSIVE (lethal) Expected 1:2:1, observed 2:1
Figure 2.2 A recessive lethal allele. The homozygous AYAY class is lost before birth, converting the expected 1:2:1 into 2:1 among living offspring. Note that dominance is assigned trait by trait: the same allele is dominant for coat color and recessive for lethality.

Kinds of lethal alleles

A lethal allele causes death of the organism, usually before reproductive age. Most are recessive lethals: heterozygotes survive (often appearing normal), and homozygotes die. Many essential genes, whose complete loss is incompatible with life, can mutate to recessive lethal alleles, and every human is estimated to carry a small number of them in the heterozygous state. A cross of two heterozygous carriers of a recessive lethal with no visible heterozygous phenotype simply produces about one quarter fewer offspring, with all survivors normal in appearance; only when the heterozygote has a visible phenotype, as with yellow mice, does the 2:1 ratio appear.

Other examples: the Manx allele in cats (M) causes taillessness in heterozygotes and is lethal early in development in homozygotes, so Manx by Manx crosses give 2 Manx : 1 tailed. Creeper chickens, with short legs, behave the same way. In humans, achondroplasia (caused by an activating mutation in FGFR3) is dominant for short limbed dwarfism, while homozygotes have a severe skeletal dysplasia that is usually lethal in infancy.

Dominant lethal alleles kill even in one copy. Those that act before reproduction are removed from the population in the same generation they arise, so they are seen only as new mutations. A dominant lethal can persist only if its effect appears after reproduction has begun. Huntington disease is the textbook example: symptoms typically begin between 30 and 50 years of age, after many affected people have had children, and the disease is eventually fatal.

Conditional lethal alleles are lethal only under certain conditions. Temperature sensitive (ts) mutants of bacteria, yeast and flies survive at a permissive temperature but die at a restrictive one, usually because the mutant protein misfolds at higher temperature. Geneticists prize ts alleles because they allow the study of essential genes: grow the organism at the permissive temperature, then shift it and see what fails. In humans, glucose 6 phosphate dehydrogenase deficiency is conditional in a different sense: carriers are usually healthy but can suffer severe hemolysis after eating fava beans or taking certain drugs.

Type of lethalGenotype that diesEffect on ratiosExample
Recessive lethal with dominant visible effectHomozygoteHeterozygote intercross gives 2:1Yellow mice (AY), Manx cats
Recessive lethal, no heterozygous phenotypeHomozygoteAll survivors look normal; litter size reduced by about 1/4Many essential gene loss of function alleles
Dominant lethal, early actingHeterozygote and homozygoteNot transmitted; seen only as new mutationsSevere developmental mutations
Dominant lethal, late actingHeterozygote (later in life)Transmitted 1:1 like any dominant traitHuntington disease
Conditional lethalDepends on environmentNormal ratios under permissive conditionsTemperature sensitive mutants, G6PD deficiency
Study tip. Worked problem: In mice, AY (yellow, recessive lethal) and A (agouti) are alleles at one locus, and an independently assorting gene has B (normal tail) dominant to b (kinked tail). What phenotypic ratio is expected among living offspring of AYA Bb × AYA Bb? Step 1: Solve each gene separately among survivors. Agouti locus: survivors are 2/3 yellow (AYA) and 1/3 agouti (AA). Tail locus: 3/4 normal (B_) and 1/4 kinked (bb). Step 2: Multiply: yellow normal = 2/3 × 3/4 = 6/12; yellow kinked = 2/3 × 1/4 = 2/12; agouti normal = 1/3 × 3/4 = 3/12; agouti kinked = 1/3 × 1/4 = 1/12. Step 3: Check: 6 + 2 + 3 + 1 = 12, so the fractions sum to 1. Answer: 6 yellow normal : 3 agouti normal : 2 yellow kinked : 1 agouti kinked, a 6:3:2:1 ratio. Spotting a 12 in the denominator is a strong hint that a lethal is involved, because 16 − 4 = 12 of the original Punnett cells survive.

Gene interaction that produces novel phenotypes

So far each trait has been controlled by one gene. But most traits are the products of several genes whose products interact, and when two genes affect the same trait, the dihybrid ratio may be modified or new phenotypes may appear. The term gene interaction covers all such cases; the simplest are those in which the 9:3:3:1 ratio is preserved but the four classes are four distinct forms of one trait.

The classic example, analyzed by William Bateson and Reginald Punnett in the early 1900s, is comb shape in chickens. Two genes, R (rose) and P (pea), act together:

A cross of true breeding rose (RRpp) with true breeding pea (rrPP) gives an F1 with walnut combs, a phenotype seen in neither parent. The F2 shows 9 walnut : 3 rose : 3 pea : 1 single. The ratio is Mendelian, but the walnut and single classes are novel phenotypes created by the combination of alleles at two loci. Similar interactions produce the red, brown, scarlet and white eyes that segregate when Drosophila carrying the brown and scarlet mutations are crossed and bell pepper colors (red, peach, brown, green, from two genes).

How can you tell gene interaction for a single trait from two separate traits? Look at the parents and the F1. If one trait (comb shape) appears in four forms in the F2, and the F1 shows a form not present in either parent, two genes are acting on that one trait. Testcrossing the F1 to rrpp confirms it: the four comb types appear 1:1:1:1.

Study tip. A one gene trait with four F2 phenotypes in a 9:3:3:1 ratio is impossible: one gene gives at most three genotypes in the F2. Whenever a single character shows ratios in sixteenths, assume two genes and write out the four genotype classes A_B_, A_bb, aaB_ and aabb before deciding which phenotype each class has.
Key point

Lethal alleles remove genotype classes and modify ratios: a recessive lethal with a visible heterozygote converts 1:2:1 into 2:1, and a lethal combined with a second gene gives ratios in twelfths. Late acting dominant lethals such as Huntington disease persist because they act after reproduction, and conditional lethals depend on the environment. When two genes shape one trait, the F2 can show 9:3:3:1 with novel phenotypes, as in chicken combs.

Section 3Epistasis and modified ratios

When one gene masks another

Epistasis is a form of gene interaction in which the genotype at one locus masks or modifies the phenotypic expression of a genotype at another locus. The gene doing the masking is epistatic; the gene whose effect is hidden is hypostatic. The word comes from Greek for "standing upon," and the image is apt: one gene stands on top of another. Epistasis is different from dominance. Dominance is an interaction between alleles at the same locus; epistasis is an interaction between different loci.

In a dihybrid intercross with epistasis, the 16 Punnett cells still occur in the usual proportions of 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Independent assortment has not changed. What changes is that some of these genotype classes share a phenotype, so the four classes merge into three or two phenotypic classes. Every epistatic ratio therefore still adds up to 16, and recognizing a ratio with a total of 16 is the first clue that two genes are involved.

Thinking in pathways

Epistasis is easiest to understand through biochemical pathways. Many traits are the end product of a series of enzyme catalyzed steps, each enzyme encoded by a different gene. A loss of function mutation in any step blocks everything downstream. If a mutation blocks an early step, it does not matter what alleles are present at genes acting later; the later enzymes have no substrate. The early acting gene is therefore epistatic to the later one.

A. Duplicate recessive epistasis (sweet pea flower color), 9:7 colorless precursor colorless intermediate purple pigment enzyme C enzyme P cc blocks pp blocks C_P_ purple (9) ; C_pp, ccP_, ccpp white (3 + 3 + 1 = 7) B. Recessive epistasis (Labrador coat color), 9:3:4 Gene B: pigment type B_ black, bb brown Gene E: deposit in hair E_ yes, ee no B_E_ black (9) bbE_ chocolate (3) __ee yellow (4) ee hides the B genotype: B_ee and bbee are both yellow
Figure 2.3 Epistasis seen as pathways. In A, two enzymes act in series, so a recessive block at either step gives the same white phenotype and 9:7 results. In B, the E gene controls whether pigment reaches the hair, so ee dogs are yellow whatever their B genotype, merging 3 + 1 into a class of 4.

Recessive epistasis: 9:3:4

In recessive epistasis, the homozygous recessive genotype at one locus masks the other locus. Labrador retriever coat color is the standard example. The B gene (TYRP1) determines whether the dark pigment eumelanin is black (B_) or brown (bb). The E gene (MC1R, the melanocortin 1 receptor) determines whether eumelanin is deposited in the hair at all; ee dogs make little eumelanin in the coat and are yellow, although their noses and lips still show the B genotype. A cross of two dogs heterozygous at both loci (BbEe × BbEe) gives 9 black (B_E_) : 3 chocolate (bbE_) : 4 yellow (3 B_ee + 1 bbee). Albinism in many animals works the same way: cc individuals lacking tyrosinase are white whatever their other coat color genes say, which is why albino mice are a frequent textbook example of 9:3:4.

The Bombay phenotype in humans is recessive epistasis too. The FUT1 gene (the H locus) makes the fucosyltransferase that builds the H antigen, the substrate on which the ABO enzymes act. People homozygous for a null allele (hh) make no H antigen, so even if they carry IA or IB, no A or B antigen can be made, and they type as O. The famous clue is a pedigree in which a "type O" mother has an AB child: she must carry an IB or IA allele masked by hh.

Dominant epistasis: 12:3:1

In dominant epistasis, a dominant allele at one locus masks the other locus. Fruit color in summer squash shows it. A dominant allele W blocks pigment formation, giving white fruit regardless of the second gene. Among ww plants, Y_ gives yellow and yy gives green. The F2 of WwYy × WwYy is 12 white (9 W_Y_ + 3 W_yy) : 3 yellow (wwY_) : 1 green (wwyy).

Duplicate recessive epistasis (complementary genes): 9:7

When two genes act in series in one pathway and a recessive loss of function at either one gives the same mutant phenotype, both genes must have a dominant allele for the wild type phenotype. Bateson and Punnett found this in sweet peas: two true breeding white flowered strains crossed to give purple F1 plants, and the F2 was 9 purple : 7 white. Purple requires C_P_; every genotype with cc or pp is white (Figure 2.3A). This is also called complementary gene action, because the two white parents each supply what the other lacks, which is exactly the logic of the complementation test in Section 5.

Duplicate dominant epistasis: 15:1

When either of two genes can supply the same function, as with duplicated genes, a dominant allele at either locus is enough for the wild type phenotype, and only the double homozygous recessive is mutant. Seed capsule shape in shepherd's purse (Capsella) segregates 15 triangular (any genotype with at least one A or one B) : 1 ovoid (aabb). Duplicate genes are common in plants with polyploid ancestry, and 15:1 ratios are frequent in crops such as wheat.

Dominant and recessive epistasis: 13:3

Sometimes the dominant allele of one gene and the recessive genotype of the other both produce the same phenotype. In chickens, C is required for color, and a dominant inhibitor I suppresses color. Only iiC_ birds are colored. Crossing two white breeds (white Leghorn, IICC, and white Wyandotte, iicc) gives white F1 (IiCc) and an F2 of 13 white (9 I_C_ + 3 I_cc + 1 iicc) : 3 colored (iiC_).

Duplicate genes with cumulative effect: 9:6:1

Fruit shape in another summer squash cross gives 9 disc (A_B_) : 6 sphere (A_bb or aaB_) : 1 long (aabb). Here a dominant allele at either locus has the same effect and the effects add: two loci with dominant alleles give disc, one gives sphere, and none gives long. Strictly this is gene interaction with additive effects rather than masking, but it is usually taught alongside epistatic ratios.

RatioNameClasses mergedPathway logicExample
9:3:3:1No epistasis (or novel phenotypes)NoneIndependent effects, or two genes jointly shaping one traitPea seed shape and color; chicken comb
9:3:4Recessive epistasisaaB_ + aabbaa blocks a step needed before B can showLabrador coat, albino mice, Bombay phenotype
12:3:1Dominant epistasisA_B_ + A_bbDominant A blocks or overrides the pathwaySummer squash fruit color
9:7Duplicate recessive (complementary genes)A_bb + aaB_ + aabbTwo required steps in seriesSweet pea flower color
15:1Duplicate dominantAll but aabbEither of two redundant genes sufficesShepherd's purse capsule shape
13:3Dominant and recessiveA_B_ + A_bb + aabbDominant inhibitor plus a required geneWhite Leghorn × white Wyandotte chickens
9:6:1Duplicate genes with cumulative effectA_bb + aaB_Each dominant allele adds to the phenotypeSummer squash fruit shape
Study tip. Worked problem: Two true breeding white sweet pea strains are crossed; the F1 are purple and the F2 contains 96 purple and 64 white plants. Test the 9:7 hypothesis and compare it with 3:1. Step 1: Total = 160. For 9:7, E(purple) = 160 × 9/16 = 90 and E(white) = 160 × 7/16 = 70. Step 2: χ2 = (96 − 90)2/90 + (64 − 70)2/70 = 36/90 + 36/70 = 0.400 + 0.514 = 0.914. df = 1. Step 3: 0.914 < 3.841, so we fail to reject 9:7 (P between 0.10 and 0.50). Step 4: For 3:1, E = 120 and 40. χ2 = (96 − 120)2/120 + (64 − 40)2/40 = 576/120 + 576/40 = 4.8 + 14.4 = 19.2, far above 6.635 (P < 0.01). Answer: The data fit 9:7 and decisively reject 3:1, supporting two complementary genes. The white parents must have been CCpp and ccPP.
Study tip. Worked problem: An F2 of squash gives 210 white, 54 yellow and 16 green fruits. Which ratio fits? Step 1: Total = 280. One sixteenth = 280/16 = 17.5 plants. Step 2: Divide each class by 17.5: 210/17.5 = 12.0; 54/17.5 = 3.09; 16/17.5 = 0.91. Step 3: The classes are close to 12 : 3 : 1. Answer: Dominant epistasis, 12:3:1. Dividing each class by one sixteenth of the total is the fastest way to recognize any modified dihybrid ratio; then confirm with chi square (df = number of classes − 1 = 2 here).
Connection. Epistasis matters in medicine because the effect of one gene variant often depends on others. The severity of sickle cell disease, for example, is reduced in people who also carry variants (in genes such as BCL11A) that keep fetal hemoglobin production high into adulthood. Drugs such as hydroxyurea raise fetal hemoglobin, and gene editing therapies that disable the BCL11A enhancer in blood stem cells deliberately recreate this protective interaction.
Key point

Epistasis is masking of one locus by another, and it changes how the nine, three, three and one classes are grouped rather than changing their frequencies. Learn the six modified ratios with their pathway logic: 9:3:4 (recessive epistasis), 12:3:1 (dominant epistasis), 9:7 (two steps in series), 15:1 (redundant genes), 13:3 (dominant inhibitor) and 9:6:1 (additive duplicate genes). Divide the data by one sixteenth of the total to recognize the ratio.

Section 4Penetrance, expressivity and the environment

Same genotype, different phenotypes

Mendel's traits behaved as if each genotype produced exactly one phenotype. Many genes are less predictable. Two people can carry the identical disease allele, yet one is severely affected, one mildly, and one not at all. Geneticists describe this variability with two terms that students often confuse: penetrance and expressivity.

Penetrance

Penetrance is the percentage of individuals with a particular genotype who show the expected phenotype at all. It answers a yes or no question across a group: does the person with the genotype have the trait? If 80 of 100 people carrying a dominant allele show the trait, penetrance is 80 percent. When every individual with the genotype shows the phenotype, the allele is fully penetrant; otherwise it shows incomplete (or reduced) penetrance. Incompletely penetrant dominant traits can appear to skip a generation: an unaffected parent may transmit the allele to an affected child, creating a pedigree that looks recessive at first glance.

Familiar examples include polydactyly (extra fingers or toes), which in some families shows clearly reduced penetrance, and inherited cancer predispositions. Women who carry a harmful BRCA1 variant have a lifetime breast cancer risk commonly cited at about 55 to 70 percent, far above the population risk of roughly 12 to 13 percent, but not 100 percent. Retinoblastoma, caused by RB1 mutations, shows about 90 percent penetrance in carriers of typical mutations.

Penetrance is often age dependent. Huntington disease alleles with 40 or more CAG repeats are essentially fully penetrant over a normal lifespan, but the probability of having symptoms rises with age, so a 25 year old carrier usually shows no signs. When counseling relatives, geneticists use age specific penetrance curves rather than a single number.

Expressivity

Expressivity is the degree or intensity with which a trait is expressed in those individuals who do show it. It answers a "how much" question about each affected person. An allele with variable expressivity produces a range of severity. Neurofibromatosis type 1 (NF1) is the standard example: almost all carriers show some sign by adulthood (penetrance near 100 percent), but one affected person may have only a few café au lait spots while a relative has hundreds of neurofibromas, learning difficulties and skeletal problems. Marfan syndrome likewise ranges from mild tall stature and joint laxity to life threatening aortic dilation within the same family.

Penetrance: does the carrier show the trait at all? All 10 carry the allele; 7 show the trait: penetrance = 7/10 = 70% Expressivity: how strongly is it shown? mild moderate severe
Figure 2.4 Penetrance and expressivity are different questions. Penetrance is a percentage of carriers who are affected (top). Expressivity is the range of severity among affected carriers (bottom). An allele can have full penetrance and highly variable expressivity, as in neurofibromatosis type 1.
Study tip. Worked problem: An autosomal dominant form of a skeletal disorder has 80 percent penetrance. A woman who is heterozygous (her genotype is known from testing) has children with an unaffected man who does not carry the allele. (a) What is the probability that their first child will be affected? (b) Their first child is born unaffected. What is the probability that this child carries the allele? Step 1: (a) P(child inherits the allele) = 1/2. P(affected | carries allele) = 0.8. Multiply: 1/2 × 0.8 = 0.40. Step 2: (b) Use conditional probability. P(carrier and unaffected) = 1/2 × 0.2 = 0.10. P(noncarrier and unaffected) = 1/2 × 1 = 0.50. P(unaffected) = 0.10 + 0.50 = 0.60. Step 3: P(carrier | unaffected) = 0.10 / 0.60 = 1/6, about 0.167. Answer: (a) 0.40, or 40 percent. (b) 1/6. The unaffected child is much less likely to be a carrier than the prior 1/2, but not zero, which is why an unaffected child of an affected parent may still pass on an incompletely penetrant allele.

Why penetrance and expressivity vary

Variation in penetrance and expressivity has three main sources. Modifier genes elsewhere in the genome alter the effect of the primary gene; differences in genetic background are why a mutation can produce a severe phenotype in one inbred mouse strain and a mild one in another. The environment, including diet, temperature, drugs, infections and toxins, changes how gene products work or how much they are needed. And chance events during development, such as which cells happen to acquire a second hit in a tumor suppressor gene or random variation in gene expression, can make even genetically identical twins differ.

Environmental effects on gene expression

Some of the clearest examples involve temperature. Himalayan rabbits and Siamese cats carry temperature sensitive alleles of the tyrosinase gene. The enzyme works at the cooler temperatures of the ears, nose, paws and tail, but is inactive at core body temperature, so pigment forms only at the extremities. Shave a patch of a Himalayan rabbit's back and keep an ice pack on it, and the new fur grows in dark. The genotype has not changed; its expression depends on the temperature.

Human disease offers equally clear cases. Phenylketonuria (PKU) results from homozygosity for loss of function alleles of phenylalanine hydroxylase. On an ordinary diet, phenylalanine accumulates and causes severe intellectual disability; on a low phenylalanine diet begun in the newborn period, affected children develop with normal or near normal intelligence. Newborn screening exists precisely so that the environment can be changed early enough to alter the phenotype. Lactase persistence and lactose intolerance, G6PD deficiency with fava bean exposure, and the effect of smoking on lung disease in people with alpha 1 antitrypsin deficiency are further gene by environment examples.

Plants show strong environmental effects too. Hydrangea flower color depends on soil pH, which controls how much aluminum is available to bind the pigment: acidic soil gives blue flowers and alkaline soil pink ones, in plants of the same genotype.

Phenocopies and norms of reaction

A phenocopy is a phenotype produced by environmental factors that mimics a phenotype normally produced by a specific genotype. Phenocopies are not inherited. Thalidomide, a sedative given to pregnant women around 1960, caused limb malformations that resembled those of rare inherited syndromes such as Roberts syndrome. Heat shock of Drosophila pupae can produce wing defects that look like specific mutations. In a pedigree, a phenocopy can mislead an analyst into seeing an inherited trait where there is none, or into assigning the wrong genotype.

A norm of reaction is the range of phenotypes a single genotype produces across a range of environments. It is usually drawn as a graph with environment on the horizontal axis and phenotype on the vertical axis, one line per genotype. When the lines are parallel, genotype and environment act additively; when they cross, the best genotype in one environment is not the best in another, a genotype by environment interaction. Classic studies of Drosophila eye facet number at different temperatures showed that genotypes respond to the same environmental change by different amounts, and studies of yarrow clones grown at different elevations showed that the ranking of genotypes can even reverse from one environment to another. Norms of reaction are a reminder that a phenotype belongs to a genotype in an environment, never to a genotype alone.

TermQuestion it answersMeasured asExample
PenetranceDoes a person with the genotype show the trait?Percent of carriers affectedBRCA1 breast cancer, polydactyly
Age dependent penetranceBy what age does the trait appear?Percent affected at each ageHuntington disease
ExpressivityHow severe is the trait in affected people?Range of severityNeurofibromatosis type 1, Marfan syndrome
Temperature sensitive alleleDoes the gene product work at this temperature?Phenotype at permissive versus restrictive temperatureSiamese cat and Himalayan rabbit coat
PhenocopyIs this genetic or environmental?Environmentally caused mimic of a mutant phenotypeThalidomide limb defects
Norm of reactionHow does one genotype respond across environments?Graph of phenotype against environmentPlant height at different elevations
Connection. Reduced penetrance is one of the biggest challenges in modern genetic testing. Population scale sequencing has shown that many variants once thought to cause disease turn up in healthy people at surprising rates, meaning their penetrance outside of affected families is much lower than first estimated. Laboratories now weigh this evidence carefully before calling a variant "pathogenic," and counselors explain risk as a probability rather than a certainty.
Key point

Penetrance is the percentage of carriers who show a trait; expressivity is how strongly it shows in those who do. Both are shaped by modifier genes, environment and chance. Temperature sensitive alleles, diet treatable disorders like PKU, phenocopies and norms of reaction all show that phenotype depends on genotype and environment together, and risk calculations must multiply transmission probabilities by penetrance.

Section 5Complementation tests

The question complementation answers

Suppose a geneticist screens thousands of fruit flies and recovers ten independent recessive mutants that all have white eyes, or a plant breeder finds five dwarf lines, or a clinic sees several families with the same recessive form of deafness. A basic question comes first: are these mutations in the same gene or in different genes? The phenotype alone cannot say, because, as Section 3 showed, mutations that block different steps of one pathway produce the same phenotype. The complementation test answers the question with a single cross.

How the test works

Cross two homozygous recessive mutants that have the same phenotype, and examine the F1:

The logic depends on the mutations being recessive loss of function alleles. In the F1 the two mutations sit on different homologs, a configuration called trans. Seymour Benzer, working on bacteriophage T4 in the 1950s, formalized this as the cis trans test and used the word cistron for a unit of function defined by complementation. A cistron corresponds, in nearly all cases, to what we now call a gene.

Notice how closely this mirrors the 9:7 ratio of Section 3. Two white sweet pea strains crossed to give purple F1 plants: that is complementation, and it told Bateson and Punnett that the strains carried mutations in different genes, C and P.

Mutations in different genes Mutations in the same gene gene A gene B from mutant 1 (top) and mutant 2 (bottom) one good A one good B F1 wild type complementation gene A gene B different sites, same gene no good A two good B F1 mutant no complementation
Figure 2.5 The logic of the complementation test. Two homologs in the F1 come one from each mutant parent (red crosses mark mutations). If the mutations are in different genes (left), each homolog supplies the function the other lacks. If they are in the same gene (right), neither homolog makes a working gene A product, even though the mutations are at different sites.

Complementation groups

With many mutants, every pair is crossed and the results are arranged in a complementation matrix. Mutants that fail to complement one another are placed in the same complementation group, and each group corresponds to one gene. The number of groups is therefore an estimate of the number of genes in which mutations can produce the phenotype. Complementation analysis of yeast cell division cycle (cdc) mutants, many of them temperature sensitive, identified dozens of genes needed for cell cycle progression, work honored with a Nobel Prize in 2001. Similar analyses defined the genes of the T4 phage rII region, the yeast mating pathway and many metabolic pathways.

Study tip. Worked problem: Five independently isolated recessive mutants of a plant all lack purple pigment. Each pair is crossed, and the F1 is scored as purple (+, complementation) or white (−, no complementation): 1 × 2: +; 1 × 3: −; 1 × 4: +; 1 × 5: +; 2 × 3: +; 2 × 4: +; 2 × 5: −; 3 × 4: +; 3 × 5: +; 4 × 5: +. How many genes are represented, and which mutants are in each? Step 1: Start with mutant 1. It fails to complement only mutant 3, so 1 and 3 are in the same gene: group I = {1, 3}. Step 2: Mutant 2 fails to complement only mutant 5: group II = {2, 5}. Step 3: Mutant 4 complements every other mutant, so it is alone: group III = {4}. Step 4: Check consistency: 3 complements 2, 4 and 5, as expected for a member of group I; 5 complements 1, 3 and 4, as expected for group II. Answer: Three genes. Group I contains mutants 1 and 3, group II contains 2 and 5, and group III contains 4. If these act in one pathway, an F2 from a cross between members of two different groups should give a 9:7 ratio.
Mutant12345
1−+−++
2−++−
3−++
4−+
5−

The matrix above is the same data in the conventional half matrix form. The diagonal is always −, because a mutant crossed with itself cannot complement.

Complementation in human families

Humans cannot be crossed deliberately, but nature sometimes performs the test. Congenital deafness is often autosomal recessive, and mutations in well over 50 different genes can cause it. Deaf people frequently marry one another, and the outcomes are informative. When two deaf parents with recessive deafness have only deaf children, they likely carry mutations in the same gene. When two deaf parents have children who all hear normally, the parents' mutations must be in different genes, and each child is heterozygous at both loci. The same reasoning applies to recessive forms of blindness such as retinitis pigmentosa, which can be caused by mutations in dozens of genes. Clinicians call this phenomenon locus heterogeneity: the same phenotype arising from mutations at different loci. Complementation can also be tested in cultured cells by fusing cells from two patients; if the hybrid cell recovers normal function, the patients have defects in different genes. This was how the complementation groups of xeroderma pigmentosum, a DNA repair disorder, were defined.

Limits and exceptions

F1 resultInterpretationUsual explanationExceptions to keep in mind
Wild typeComplementationMutations in different genesIntragenic complementation (same gene, multimeric product)
MutantNo complementationMutations in the same geneNonallelic noncomplementation; polar effects; a dominant mutation in either parent
Connection. Complementation thinking guides modern diagnosis. When an exome sequence shows that a child with a recessive disorder carries two different harmful variants in the same gene, a laboratory confirms that the variants are in trans (one inherited from each parent) by testing the parents. Two variants on the same homolog (cis) would leave one working copy, and the child would not be expected to be affected. The question is exactly Benzer's cis trans question, asked with sequencing instead of crosses.
Key point

A complementation test crosses two recessive mutants with the same phenotype. A wild type F1 means the mutations complement and lie in different genes; a mutant F1 means they fail to complement and lie in the same gene. Mutants that fail to complement form a complementation group, one per gene. The test works only for recessive mutations, and intragenic complementation and nonallelic noncomplementation are rare exceptions.

The chapter in one breath

  • Dominance describes the heterozygote's phenotype: complete, incomplete (intermediate, 1:2:1) or codominant (both products visible, as in MN and AB blood types).
  • The level of analysis matters: the sickle cell allele is recessive for disease, incompletely dominant for cell shape and codominant for the hemoglobin proteins.
  • Multiple alleles exist in populations; n alleles give n(n + 1)/2 genotypes. ABO combines dominance of IA and IB over i with codominance of IA and IB.
  • A recessive lethal with a visible heterozygous effect gives 2:1 (yellow mice, Manx cats); combined with a second gene, ratios fall in twelfths, such as 6:3:2:1.
  • Late acting dominant lethals like Huntington disease persist because they act after reproduction; conditional lethals depend on the environment.
  • Gene interaction can produce novel phenotypes in a 9:3:3:1 ratio, as in chicken comb shape.
  • Epistasis is masking of one locus by another and regroups the sixteenths: 9:3:4, 12:3:1, 9:7, 15:1, 13:3 and 9:6:1.
  • Epistatic ratios make sense as pathways: early blocks mask later steps, steps in series give 9:7, and redundant genes give 15:1.
  • Penetrance is the percent of carriers who show a trait; expressivity is the severity among those who do.
  • Environment changes phenotype: temperature sensitive tyrosinase in Siamese cats, diet in PKU, soil pH in hydrangeas; phenocopies mimic genetic traits, and norms of reaction chart one genotype across environments.
  • The complementation test asks whether two recessive mutations are in the same gene: wild type F1 means different genes, mutant F1 means the same gene.

Words worth knowing

TermWhat it means
ABO blood groupA human blood group determined by three main alleles, IA, IB and i, of a glycosyltransferase gene that modifies the H antigen.
Allelic seriesA set of multiple alleles of one gene, often arranged in order of dominance.
Bombay phenotypeA rare hh genotype that prevents H antigen formation, so no A or B antigen can be made regardless of ABO genotype.
Cis trans testBenzer's name for the complementation test, comparing two mutations on the same homolog (cis) with mutations on different homologs (trans).
CistronA unit of genetic function defined by the cis trans test; essentially a gene.
CodominanceA relationship in which the heterozygote shows the phenotypes of both alleles.
ComplementationRestoration of the wild type phenotype when two recessive mutations in different genes are combined in one individual.
Complementation groupA set of mutations that fail to complement one another and therefore lie in the same gene.
Complete dominanceA relationship in which the heterozygote is indistinguishable from the dominant homozygote.
Conditional lethalAn allele that kills only under certain conditions, such as a high (restrictive) temperature.
Dominant lethalAn allele that causes death even in a single copy; it persists only if it acts after reproduction.
Dominant epistasisMasking of one gene by a dominant allele at another gene, typically giving 12:3:1.
Duplicate recessive epistasisTwo genes in series where a recessive genotype at either gives the mutant phenotype, giving 9:7; also called complementary gene action.
EpistasisAn interaction in which the genotype at one locus masks or modifies the expression of another locus.
ExpressivityThe degree to which a trait is expressed in individuals who show it.
Gene interactionAny situation in which two or more genes together determine a single phenotype.
HaplosufficientDescribes a gene for which one functional copy produces enough product for a normal phenotype.
HypostaticDescribes a gene whose expression is masked by another, epistatic gene.
Incomplete dominanceA relationship in which the heterozygote phenotype is intermediate between the two homozygotes.
Incomplete penetranceA situation in which some individuals with a genotype do not show the expected phenotype.
Intragenic complementationRare complementation between two mutations in the same gene, usually when the protein is a multimer.
Lethal alleleAn allele that causes death of the organism, usually before reproductive age.
Locus heterogeneityThe production of the same phenotype by mutations in different genes.
Modifier geneA gene that alters the phenotypic effect of another gene.
Multiple allelesThe presence of three or more alleles of a gene in a population.
Norm of reactionThe range of phenotypes produced by one genotype across a range of environments.
PenetranceThe percentage of individuals with a given genotype who express the expected phenotype.
PhenocopyAn environmentally caused phenotype that mimics one produced by a particular genotype.
Recessive epistasisMasking of one gene by the homozygous recessive genotype at another gene, typically giving 9:3:4.
Recessive lethalAn allele that causes death only when homozygous.
Temperature sensitive alleleAn allele whose product functions at a permissive temperature but not at a restrictive one.

Check yourself

Pink snapdragons are crossed with white snapdragons. What phenotypes and proportions do you expect, and why can you never obtain a true breeding pink line?
Answer. CRCW × CWCW gives 1/2 pink and 1/2 white. Pink is always heterozygous, so pink plants crossed together always segregate 1 red : 2 pink : 1 white; a homozygous pink genotype does not exist.
A gene has 5 alleles in a population. How many different genotypes are possible, and how many of them are heterozygous?
Answer. n(n + 1)/2 = 5 × 6/2 = 15 genotypes. Of these, 5 are homozygous, so 10 are heterozygous (n(n − 1)/2 = 5 × 4/2 = 10).
A man with type AB blood is accused of fathering a child with type O blood. What can you conclude?
Answer. He can be excluded, barring a rare Bombay or other unusual genotype. An O child is ii and must receive i from each parent, but an AB man (IAIB) carries no i allele.
Two Manx cats are mated and produce 12 kittens. About how many would you expect to be Manx, and why is the ratio not 3:1?
Answer. About 8 Manx and 4 tailed. The MM quarter dies early in development, so among survivors Mm (Manx) : mm (tailed) = 2:1, and 2/3 of 12 is 8.
An F2 shows 9 black : 3 brown : 4 white. Name the type of interaction and give the genotypes of each class.
Answer. Recessive epistasis. Writing the masking gene as C: black = C_B_ (9), brown = C_bb (3), and white = ccB_ plus ccbb (3 + 1 = 4). The cc genotype hides whatever is present at the B locus, as in albino mice.
Two true breeding white flowered strains produce purple F1 plants. What does this tell you, and what F2 ratio do you predict?
Answer. The strains carry recessive mutations in different genes and complement each other (AAbb × aaBB gives AaBb). If both gene products are required in series for purple, the F2 is 9 purple : 7 white.
An F2 of 320 plants contains 299 with one phenotype and 21 with another. What ratio fits best, and what does it suggest about the genes?
Answer. One sixteenth of 320 is 20, so the classes are about 15 : 1 (299/20 = 14.95; 21/20 = 1.05). This suggests duplicate dominant epistasis: two redundant genes, either of which with a dominant allele gives the wild type phenotype.
A dominant disorder has 60 percent penetrance. An affected man (heterozygous) has a child with an unaffected woman who does not carry the allele. What is the chance the child will be affected?
Answer. P(inherit allele) × penetrance = 1/2 × 0.6 = 0.30, or 30 percent.
Explain the difference between penetrance and expressivity using neurofibromatosis type 1 as your example.
Answer. Penetrance asks whether carriers show any sign; for NF1 it is close to 100 percent by adulthood. Expressivity asks how severe the signs are; NF1 ranges from a few café au lait spots to many neurofibromas with learning and skeletal problems, even within one family, so its expressivity is highly variable.
Two parents who both have recessive congenital deafness have four children, all with normal hearing. Explain this result and give the children's likely genotype.
Answer. The parents carry mutations in different genes, so they complement: one parent is aaBB and the other AAbb. Every child is AaBb, with one working copy of each gene, and hears normally. This is locus heterogeneity revealed by a natural complementation test.