Chapter 14

14

1

An albino giraffe with three normally pigmented companions.

Definitions of the Gene

Sir Archibald Garrod and Human Inborn Errors of Metabolism

In 1902, just two years after the discovery of Mendel’s work, Sir Archibald E.Garrod, a physician at the Hospital for Sick Children in London, published a pa-per entitled “The Incidence of Alkaptonuria: A Study in Chemical Individuality.”Like Mendel’s work, the concepts Garrod presented in this paper remainedlargely unknown by the scientists of the world for 40 years—until the conceptswere independently formulated by George W. Beadle and Edward L. Tatum in1941. Garrod’s paper described the results of his studies of individuals with al-kaptonuria, an innocuous, but easily detected, disorder. Because of the presenceof a chemical called homogentisic acid (formerly called alkapton), the urine ofaffected individuals turns black when exposed to air. Garrod clearly recognizedthat alkaptonuria was inherited. In his 1902 paper, he wrote: “There are good rea-sons for thinking that alkaptonuria is not the manifestation of a disease but israther the nature of an alternative course of metabolism, harmless and usuallycongenital and lifelong” (Lancet ii:1616).

In addition to alkaptonuria, Garrod studied albinism, cystinuria, pentosuria,and porphyrinurea in humans (See Human Genetics Sidelight: Human Inborn Er-rors of Metabolism). Garrod summarized the results of these studies in a treatiseentitled “Inborn Errors of Metabolism,” first presented as the Croonian Lecturesto the Royal College of Physicians in London in 1908 and published in bookform in 1909. In this book, Garrod clearly articulates the view that the metabolicdefects observed in individuals with these disorders are caused by recessive mu-tant genes. Garrod probably did not develop his amazing insight into metabolicprocesses totally by himself. His father, Alfred B. Garrod, was also a physician

CHAPTER OUTLINEEvolution of the Concept of the Gene:

SummaryEvolution of the Concept of the Gene:

FunctionEvolution of the Concept of the Gene:

StructureA Genetic Definition of the GeneComplex Gene-Protein Relationships

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and was the first to demonstrate the accumulation of the chemical uric acid in pa-tients with gout, a painful inflammation of the joints in the hands and feet. In anycase, Archibald E. Garrod was the first scientist to relate defects in genes toblocks in metabolic pathways. In this chapter, we focus on how Garrod’s conceptevolved into the current understanding of this basic unit of genetic information:the GENE.

The gene is to genetics what the atom is to chemistry. Thus throughout thistext we have focused our attention on the gene and the alternate forms of a geneor alleles. In the preceding chapters, we have examined the patterns of transmis-sion of independently assorting and linked genes, the chromosomal location ofgenes, the chemical composition of genes and chromosomes, the mechanism ofreplication of genes, mutational events in genes, and the mechanisms by whichgenes exert their effects on the phenotype of the organism. What is this unit ofgenetic information that we call the gene? As we will see, the concept of a gene isnot static; it has evolved through several phases since Wilhelm Johannsen intro-duced the term in 1909, and it will undoubtedly evolve through additional refine-ments in the future.

The gene has been defined as the unit of genetic in-formation that controls a specific aspect of the pheno-type. Such a description, though accurate, does notprovide a precise, unambiguous definition that can beused to identify a gene at the molecular level. At amore fundamental level, the gene has been defined asthe unit of genetic information that specifies the syn-thesis of one polypeptide. However, it is not a verygood operational definition. An operational defini-tion spells out an operation or experiment that can becarried out to define or delimit something. One genespecifying one polypeptide is a poor operational defi-nition because experimentally relating all the seg-ments of DNA that represent genes with all thepolypeptides is not feasible in a complex organism.Moreover, it is preferable to define the gene by usinggenetic approaches rather than biochemical experi-ments. Thus, in this chapter, we focus on the comple-mentation test as an operational definition of thegene. We will also consider the limitations of the com-plementation test, along with the unique structuralfeatures of selected genes.

EVOLUTION OF THE CONCEPT OF THEGENE: SUMMARY

Before discussing evidence supporting the variousconcepts, let’s summarize the important stages in theevolution of the gene concept. The gene theory of in-heritance began with the publication of Mendel’s clas-sic paper in 1866 but did not become an accepted part

of scientific knowledge until after the discovery ofMendel’s work in 1900. Mendel’s gene (not so-named)was the “character” or “constant factor” that con-trolled one specific phenotypic trait such as flowercolor in peas. At the time of the discovery of Mendel’swork, the English physician Sir Archibald E. Garrodwas studying several inherited diseases in humans.Garrod first recognized that homozygosity for reces-sive mutant alleles can cause defects in the normalprocesses of metabolism. His concept of the gene isprobably stated most accurately as one mutant gene-one metabolic block, which over 30 years later was re-fined to the one gene-one enzyme concept enunciatedby George W. Beadle and Edward L. Tatum. Sincemany enzymes contain two or more different poly-peptides, each encoded by a separate gene, the onegene-one enzyme concept subsequently was modifiedto one gene-one polypeptide.

Prior to 1940, genes were considered analogous tobeads on a string; recombination occurred between,but not within, genes. The gene was both the basicfunctional unit, which controlled one phenotypic trait,and the elementary structural unit, which could not besubdivided by recombination or mutation. ClarenceOliver’s 1940 report that recombination had occurredwithin the lozenge gene of Drosphila stimulated bothexcitement and much debate about its significance.When the debate ended, the nucleotide pair had re-placed the gene as the basic unit of structure, the unitof genetic material not subdivisible by mutation or re-combination.

In the early 1940s, Edward B. Lewis developed thecomplementation, or cis-trans, test for functional al-

2 CHAPTER 14 Definitions of the Gene

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lelism in Drosophila. This test subsequently was ex-ploited by Seymour Benzer to define experimentallythe gene in bacteriophage T4. The complementationtest provides an operational definition of the gene.This test allows a geneticist to determine whether twoindependent mutations occurred in the same gene orin two different genes. Moreover, the gene as definedby the complementation test is a perfect fit to the onegene–one polypeptide concept.

In the 1960s, elegant experiments by CharlesYanofsky, Sydney Brenner, and their collaboratorsshowed that the gene and its polypeptide productwere colinear structures, with a direct correlation be-tween the sequence of nucleotide pairs in the gene andthe sequence of amino acids in the polypeptide. How-ever, this simple concept of the gene as a continuoussequence of nucleotide pairs specifying a colinear se-quence of amino acids in the polypeptide gene prod-

Evolution of the Concept of the Gene: Summary 3

HUMAN GENETICS SIDELIGHT

Human Inborn Errors of Metabolism

As discussed in the text, Garrod’s concept of one mutantgene–one metabolic block was based on his studies of a fewinherited human disorders. Alkaptonuria, on which much ofGarrod’s information was based, is described in the text.However, Garrod also studied familial cases of cystinuria,pentosuria, porphyrinuria—inherited disorders characterizedby elevated levels of the amino acid cystine, five-carbon sug-ars, and the iron-binding porphyrin component of hemoglo-bin, respectively, in urine. The most common disorder thatGarrod studied was albinism, an autosomal recessive traitthat occurs at a frequency of about 1 in 20,000 newborns in theUnited States. The red eyes and white skin and hair of albinosresult from the absence of the black pigment melanin. Themost common forms of albinism are caused by mutations thatblock the biosynthesis of melanin from the amino acid tyro-sine. Of course, the pathway of melanin biosynthesis was un-known in 1909 when Garrod published his book Inborn Errorsof Metabolism. Nevertheless, he clearly understood that muta-tions caused specific blocks in metabolic pathways.

Today, over 4000 inherited human disorders have beendescribed, and the number of such hereditary abnormalitiesis constantly increasing. These disorders range from rela-tively innocuous disorders like alkaptonuria to those such asTay-Sachs disease (characterized by rapid degeneration ofthe central nervous system) that are lethal in early childhood.Sickle-cell anemia and phenylketonuria (PKU) are perhapsthe best known of the human inborn errors of metabolism.The severe anemia in individuals homozygous for the sickle-cell mutation is caused by a single amino acid substitution intheir b-globin (Chapter 13). Children with PKU lack the en-zyme phenylalanine hydroxylase. This enzyme deficiency re-sults in the accumulation of phenylpyruvic acid, which ishighly toxic to the central nervous system. If untreated, chil-dren with PKU develop severe mental retardation. However,if they are placed on a diet low in phenylalanine, they de-velop normal mental abilities (Chapter 20). Other inheritedhuman diseases are discussed throughout the book.

In the United States alone, over 120,000 children withinherited defects are born each year. Long-term medical care

of these birth defects is estimated to cost $10 to $20 billion.But the cost is even higher in terms of human suffering. Forexample, consider the tragic degeneration of the central ner-vous system that occurs in Huntington disease or the loss ofbody control resulting from progressive muscle degenera-tion associated with Duchenne muscular dystrophy (bothdiscussed in Chapter 20) or the consequences of the loss ofmemory that occurs in individuals with Alzheimer’s disease(Chapter 26). What are the causes of these inherited dis-eases? What can humans do, if anything, to keep the fre-quency of these disorders as low as possible? How manydifferent human inborn errors of metabolism are there? Canthe deleterious symptoms of these inherited disorders beeliminated by treatment? Can any of these inherited diseasesbe cured so that they won’t be transmitted from parents totheir children?

Human inherited diseases are caused by all of the typesof mutations discussed in Chapter 13 and by changes inchromosome structure and number as discussed in Chapter6. Because mutations are required for evolution, they cannever be completely eliminated. However, humans can min-imize the frequency of new mutations by avoiding contactwith highly mutagenic agents such as irradiation and chem-icals that damage DNA. All genes can mutate to nonfunc-tional states. Therefore, the number of human inborn errorsof metabolism is almost certainly equal to the number ofgenes required to develop and maintain a normal humanphenotype. Several inherited disorders such as PKU haveproven treatable once their molecular bases are known. Oth-ers are currently being treated by somatic cell gene therapy(Chapter 20). In the future, some inherited human diseasesmay be cured by germ-line gene therapy; but serious ethicalissues must be addressed before germ-line gene therapy canbe performed on humans. Given the rapidly increasingnumber of known inherited human diseases and the identi-fication and characterization of the genes responsible forthese disorders (Chapter 20), the tools of molecular geneticswill play an increasingly important role in the practice ofmedicine in the future.

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uct was short-lived. Overlapping genes and genes-within-genes were discovered in the late 1960s, andthe coding sequences of eukaryotic genes were shownto be interrupted by noncoding intron sequences inthe late 1970s. Moreover, some genes, for example,genes encoding immunoglobulins, were shown to bestored in germ-line chromosomes as short “gene seg-ments,” which are assembled into mature, functionalgenes during development.

Thus the definition of the gene needs to remainsomewhat pliable if it is to encompass all of the differ-ent structure/function relationships that occur in dif-ferent organisms. Here, we define the gene as the unitof genetic information that controls the synthesis ofone polypeptide or one structural RNA molecule. Asjust defined, the gene can be identified operationallyby the complementation test. As such, the gene in-cludes the 5' and 3' noncoding regions that are in-volved in regulating the transcription and translationof the gene and all noncoding sequences or introns

within the gene (Figure 14.1). The structural generefers to the portion that is transcribed to produce theRNA product. In the case of overlapping genes, thisdefinition requires that some nucleotide-pair se-quences be considered components of two or moregenes. For those cases where exons are spliced to-gether in various combinations to make related butdifferent proteins, the gene may be defined as a DNAsequence that is a single unit of transcription and en-codes a set of closely related polypeptides, sometimescalled “protein isoforms.” In germ-line chromosomes,the DNA sequences that encode segments of antibodychains probably should be called “gene segments,” be-cause this genetic information is not organized intounits that fit any of the standard definitions of thegene (Chapter 24).

Key Points: The concept of the gene has undergonemany refinements since its discovery by Mendel in


Transcription

Translation

Primary transcript: 5' 3'

Regulation of transcription

Regulation of translation

Signals for termination

of transcriptionCoding region

Typical prokaryotic gene.(a)

(b)

Polypeptide: N C

N CPolypeptide:

Typical eukaryotic gene.

Signals for termination

of transcriptionExon 3Intron 2Exon 2Intron 1Exon 1

Regulation of transcription

Regulation of translation

Transcription

Translation

5' 3'Primary transcript:

mRNA:

Removal of introns

Initiation codon Termination codon

A n

AUG

AUG

AUG UAA

AUG UAA

UAA

UAAInitiation codon Termination codon

Cap

1

1

3

2

2

Figure 14.1 Gene struc-ture. Prokaryotic genes usu-ally contain uninterruptedcoding sequences (a), whereasthe coding sequences of eu-karyotic genes are commonlyinterrupted by noncoding se-quences (b).

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1866. Most genes encode one polypeptide and can beoperationally defined by the complementation test.

EVOLUTION OF THE CONCEPT OF THEGENE: FUNCTION

In the preceding section, we briefly examined the evo-lution of the concept of the gene, the basic unit of ge-netic information. That summary included both func-tional and structural aspects of the gene. Now, let’stake a closer look at the gene as a unit of function.

Mendel: Constant Factors ControllingPhenotypic Traits

The law of combination of different characters, which gov-erns the development of the hybrids, finds therefore its foun-dation and explanation in the principle enunciated, that thehybrids produce egg cells and pollen cells which in equalnumbers represent all constant forms which result from thecombinations of the characters brought together in fertilisa-tion. (Mendel, 1866; translation by William Bateson)

Mendel’s characters or factors, which are now calledgenes, controlled specific phenotypic traits such asflower color, seed color, and seed shape. They werethe basic units of function, the units of genetic infor-mation that governed one specific aspect of the pheno-type. This definition of the gene as the basic unit offunction is almost universally favored by present-dayscientists. There has been no change in the concept of

the gene as the basic unit of function since its discov-ery by Mendel in 1866. However, the discovery of thechemical nature of the genetic material raised ques-tions about the structure of the gene, and the conceptof the molecular structure of this basic unit of functionhas undergone several changes and refinements sincethe discovery of Mendel’s work in 1900.

If we examine what is known about how genescontrol phenotypic traits, the need for a more precisedefinition of the gene will be obvious. The pathway bywhich a gene exerts its effect on the phenotype of anorganism is often very complex (Figure 14.2). Severalgenes may have similar effects on the same pheno-typic trait, making it difficult to sort out the effects ofindividual genes. All the genes of an organism are lo-cated in the same nuclei, and they do not all functionindependently. The phenotype of an organism is theproduct of the action of all the genes acting within therestrictions imposed by the environment. Each genealso has an effect on the population to which the or-ganism carrying the gene belongs. Ultimately, eachgene has a potential effect, small though it may be, onthe cumulative phenotype of the biosphere, for eachgene may affect the ability of the organism, or the pop-ulation, or the species to compete for an ecologicalniche in the biosphere (Chapter 28).

Garrod: One Mutant Gene–One MetabolicBlockAt the time of the discovery of Mendel’s work in 1900,Sir Archibald Garrod was studying several congenitalmetabolic diseases in humans. One of these was the

Evolution of the Concept of the Gene: Function 5

Gene Phenotypic effect

Regulatory genes Hormones,

gene-product interactions

Gene (DNA)

Cell-cell interactions

RNA Protein Effect on cell

Effect on tissue

Effect on organism

Environmental effects

Effect on population

Effect on phenotype

of biosphere

Figure 14.2 The com-plex pathway by which agene exerts its effect on thephenotype of an organism,a population, or the bios-phere.

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inherited disease alkaptonuria, which is easily de-tected because of the blackening of the urine upon exposure to air. The substance responsible for thisblackening is alkapton (or homogentisic acid), an in-termediate in the degradation of the aromatic aminoacids tyrosine and phenylalanine (Figure 14.3). Garrodbelieved that the presence of homogentisic acid in theurine was due to a block in the normal pathway of me-tabolism of this compound. Moreover, on the basis ofthe family pedigree studies, Garrod proposed thatalkaptonuria was inherited as a single recessive gene.The results of Garrod’s studies of alkaptonuria and afew other congenital diseases in humans, such as al-binism, were presented in detail in his book, Inborn Er-rors of Metabolism. Although the details of the bio-chemical pathway affected by the recessive mutationsthat cause alkaptonuria were not worked out until

many years later, Garrod clearly understood the rela-tionship between genes and metabolism. His conceptmight be best stated as one mutant gene–one meta-bolic block.

Early Evidence That Enzymes AreControlled by GenesIn proposing that metabolic reactions were controlledby genes in 1902, Sir Archibald E. Garrod displayedgreat insight, because there was little direct evidenceat the time to support his proposal. Some of the firstevidence showing that genes control the enzyme-cat-alyzed reactions of metabolic pathways was obtainedfrom studies of Drosophila eye color mutants. In 1935,George Beadle and Boris Ephrussi suggested that mu-tations in two genes controlling eye color resulted inblocks at two different steps in the biosynthesis of thebrown eye pigment in Drosophila. Wild-type fruit flieshave dark red eyes resulting from the presence of twoeye pigments, one bright red and the other brown. Thevermilion (v) and cinnabar (cn) mutant flies have brightred eyes owing to the absence of the brown pigment.

Like other insects, fruit flies undergo metamor-phosis; fertilized eggs develop into larvae, then pupae,and finally adult flies (Chapter 2). The larval stagescontain groups of embryonic cells called imaginaldisks that will develop into the specific adult struc-tures, such as eyes, wings, or legs, of the adult flies.Beadle and Ephrussi discovered that if they surgicallytransplanted eye imaginal disks from one larva intothe abdomen of another larva, the transplanted diskwould develop into a third eye, albeit nonfunctional,in the adult fly (Figure 14.4). When they transplantedeye disks from wild-type larvae into vermilion orcinnabar larvae, the transplanted disks developed intodark red wild-type eyes (Figure 14.4a). In these twocases, the genotype of the transplanted eye disk con-trolled its own phenotype; the phenotype of the thirdeye was not influenced by the genotype of the larvalhost. However, this was not always the case.

When Beadle and Ephrussi transplanted eye disksfrom vermilion or cinnabar larvae into the abdomens ofwild-type larvae, the disks developed into dark redwild-type eyes (Figure 14.4b). Beadle and Ephrussireasoned that the wild-type larval hosts must haveprovided some diffusible substance to the trans-planted mutant disks that allowed them to bypass themetabolic block and synthesize the brown eye pig-ment. In these two cases, the development of thetransplanted disks into adult eye structures was influ-enced by the genotype of the host organism.

Beadle and Ephrussi next transplanted eye disksfrom vermilion larvae into cinnabar hosts and eye disksfrom cinnabar larvae into vermilion hosts. The recipro-


NH2

PhenylalanineHydrolysis of

dietary proteins

CH2CHCOOH

NH2

CH2CHCOOHHO

Tyrosine

CH2COOH

Homogentisic acid

HO

OH

Homogentisic acid oxidase Alkaptonuria

CH3CCH2COOH + HOOCHC CHCOOH

O

Acetoacetic acid Fumaric acid

CO2 + H2O

Figure 14.3 Alkaptonuria in humans results from a blockin phenylalanine catabolism caused by a mutation in thegene encoding the enzyme homogentisic acid oxidase.When this enzyme is absent or inactive, its substrate, ho-mogentisic acid, accumulates in tissues and in urine.

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cal transplants yielded quite different results! The ver-milion disks transplanted into cinnabar larvae devel-oped into dark red wild-type eyes, whereas thecinnabar disks transplanted into vermilion larvae devel-oped into bright red cinnabar eyes (Figure 14.4c). Toexplain these results, Beadle and Ephrussi proposedthat the cinnabar hosts supplied a diffusible substancethat the implanted vermilion disks converted to thebrown eye pigment. They further suggested that thevermilion hosts did not produce a metabolite thatcould be converted to the brown pigment by thecinnabar disks.

In fact, the results of Beadle and Ephrussi’s recip-rocal transplant experiments can be easily explained if

the brown eye pigment is synthesized by a series ofenzyme-catalyzed reactions and the v1 and cn1 genescontrol different steps in the biosynthetic pathway(Figure 14.5). Beadle and Ephrussi proposed that (1) aprecursor molecule X is converted to a diffusible inter-mediate substance Y by the product of the v1 gene,and (2) intermediate Y is converted to the brown eyepigment by the product of the cn1 gene. If their pro-posal was correct, the cinnabar larval hosts, whichwere v1, would have provided diffusible substance Yto the transplanted vermilion disks. Because the vermil-ion disks were cn1, they would have converted sub-stance Y to the brown eye pigment and would havedeveloped into dark red wild-type eyes. The vermilion


Metamorphosis

Mutant eye disk

Wild-type eye disk

Wild-type disks are transplanted into vermilion or cinnabar larvae.(a)

(b)

(c)

Wild-type larva

Vermilion or cinnabar

larva

Mutant adult fly

Dark red wild-type eye color

Red-orange mutant eye color

Wild-type eye disk

Vermilion or cinnabar eye disk

Vermilion or cinnabar disks are transplanted into wild-type larvae.

Vermilion or cinnabar

larva

Wild-type larva

Wild-type adult fly



Vermilion eye disk

Vermilion disks are transplanted into cinnabar larvae and cinnabar disks are transplanted into vermilion larvae.

Vermilion larva

Cinnabar larva

Cinnabar adult fly


Red-orange cinnabar eye color

Cinnabar eye disk

Cinnabar larva

Vermilion larva

Vermilion adult fly

Red-orange cinnabar eye color

Red-orange vermilion eye color

Figure 14.4 Transplant experiments ofDrosophila eye imaginal disks by Beadle andEphrussi.

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larval hosts would not have provided anything com-parable to substance Y to the transplanted cinnabardisks. Since the cinnabar disks were v1, they wouldhave converted precursor X to intermediate Y. How-ever, in the absence of cn1, intermediate Y could notbe converted to the brown pigment. As a result,cinnabar disks transplanted into vermilion larvae devel-oped into bright red mutant eyes.

During the 1940s, Beadle and Ephrussi’s proposalwas proven correct, albeit with the addition of a fewsteps to the pathway. The brown eye pigment, xan-thommatin, was shown to be synthesized from theamino acid tryptophan (precursor X) by a sequence ofenzyme-catalyzed reactions (Figure 14.6). The v1 genecontrols the synthesis of an enzyme called tryptophanpyrrolase, which converts tryptophan to N-formyl-kynurenine. Then, N-formylkynurenine is convertedto kynurenine (intermediate Y) by the enzyme kynure-nine formylase. The cn1 gene controls the synthesis ofthe enzyme kynurenine hydroxylase, which convertskynurenine (substance Y) to 3-hydroxylkynurenine.Two additional enzyme-catalyzed reactions then con-vert 3-hydroxylkynurenine to the brown pigment xan-thommatin.

These early studies of Drosophila eye color mu-tants and the pathways of eye pigment biosynthesisprovided direct evidence for the genetic control of en-zyme-catalyzed metabolic pathways. The results ofsubsequent investigations demonstrated the geneticcontrol of metabolism in all types of living organisms,from viruses to humans.

Beadle and Tatum: One Gene–One EnzymeGeorge Beadle’s collaboration with Ephrussi in stud-ies of the synthesis of the brown eye pigment inDrosophila led him to search for the ideal organism touse in extending this work. The pink bread mold Neu-rospora crassa can grow on medium containing only (1)inorganic salts, (2) a simple sugar, and (3) one vitamin,biotin. Neurospora growth medium containing only

these components is called “minimal medium.”George Beadle and Edward Tatum reasoned that Neu-rospora must be capable of synthesizing all the otheressential metabolites, such as the purines, pyrim-idines, amino acids, and other vitamins, de novo. Fur-thermore, they reasoned that the biosynthesis of thesegrowth factors must be under genetic control. If so,mutations in genes whose products are involved inthe biosynthesis of essential metabolites would be ex-pected to produce mutant strains with additionalgrowth-factor requirements.

Beadle and Tatum tested this prediction by irradi-ating asexual spores (conidia) of wild-type Neurosporawith X rays or ultraviolet light, and screening theclones produced by the mutagenized spores for newgrowth-factor requirements (Figure 14.7). In order toselect strains with a mutation in only one gene, theystudied only mutant strains that yielded a 1:1 mutantto wild-type progeny ratio when crossed with wild-type. They identified mutants that grew on mediumsupplemented with all the amino acids, purines,pyrimidines, and vitamins (called “complete medi-um”), but could not grow on minimal medium. Theyanalyzed the ability of these mutants to grow onmedium supplemented with just amino acids, or justvitamins, and so on (Figure 14.7a). For example, Beadle

Precursor X Intermediate Y

Wild-type v+ gene

v+ enzyme

Wild-type cn+ gene

cn+ enzyme

Brown eye pigment

Figure 14.5 Beadle and Ephrussi’s interpretation of theresults of their reciprocal transplants of vermilion andcinnabar eye disks in Drosophila. They proposed that the v+

and cn+ genes control two different steps in the synthesis ofthe brown eye pigment and that the v+ gene product actsbefore the cn+ gene product.

Tryptophan (precursor X)

v+ enzyme Tryptophan pyrrolase

N-formylkynurenine

Kynurenine formylase

Kynurenine (intermediate Y)

cn+ enzyme Kynurenine hydroxylase

3–Hydroxykynurenine

Phenoxazinone synthetase

Phenoxazinone

Xanthommatin (brown eye pigment)

v+ gene

cn+ gene

Figure 14.6 Biosynthesis of the brown eye pigment xan-thommatin in Drosphila. The wild-type vermilion andcinnabar genes encode enzymes that catalyze two of the re-actions in this pathway. Note the consistency of Beadle andEphrussi’s proposed sequence of reactions (Figure 14.5)with the actual biosynthetic pathway.

Chpt 14 9/10/96 3:16 PM Page 8

and Tatum identified mutant strains that grew in thepresence of vitamins but could not grow in mediumsupplemented with amino acids or other growth fac-tors. They next investigated the ability of these vita-

min-requiring strains to grow on media supplementedwith each of the vitamins separately (Figure 14.7b).

In this way, Beadle and Tatum demonstrated thateach mutation resulted in a requirement for one


MinimalNucleosidesFolic acid

Choline

Vitamins

Growth only when vitamins added

Minimal medium

Amino acids

Purines and

pyrimidines

Complete medium

Complete medium

Wild type

Crossed with wild type of opposite sex

Complete medium (with vitamins, amino acids, etc.)

Conidia (asexual spores)

Fruiting body

X rays or ultraviolet light

Wild-type spores are irradiated, and the resulting strains are crossed with wild type.

Individual ascospores are tested for general growth requirements.

Individual strains are tested for specific growth requirements.

Sexual spore

Inositolp-Aminobenzoic acid

NiacinPantothenic acid

Pyridoxin

Growth only when the vitamin pantothenic acid is added

RiboflavinThiamin

1

2

3

Figure 14.7 Diagram of Beadle and Tatum’s experiment with Neurospora that led to theone gene–one enzyme hypothesis.

Chpt 14 9/10/96 3:16 PM Page 9

growth factor. By correlating their genetic analyseswith biochemical studies of the mutant strains, theydemonstrated in several cases that one mutation re-sulted in the loss of one enzyme activity. This work,for which Beadle and Tatum received a Nobel Prize in1958, was soon verified by similar studies of manyother organisms in many laboratories. The onegene–one enzyme concept thus became a central tenetof molecular genetics.

Appropriately, in his Noble Prize acceptancespeech, Beadle stated:

In this long, roundabout way, we had discovered what Gar-rod had seen so clearly so many years before. By now weknew of his work and were aware that we had added little ifanything new in principle. . . . Thus we were able to demon-strate that what Garrod had shown for a few genes and afew chemical reactions in man was true for many genes andmany reactions in Neurospora.

One Gene–One PolypeptideSubsequent to the work of Beadle and Tatum, manyenzymes and structural proteins were shown to beheteromultimeric, that is, to contain two or more dif-ferent polypeptide chains, with each polypeptide en-coded by a separate gene. For example, in E. coli, theenzyme tryptophan synthetase is a heterotetramercomposed of two a polypeptides encoded by the trpAgene and two b polypeptides encoded by the trpBgene. Similarly, the hemoglobins, which transportoxygen from our lungs to all other tissues of our bod-ies, are tetrameric proteins that contain two a-globinchains and two b-globin chains, as well as four oxy-gen-binding heme groups (see Figure 12.4). In hu-mans, the major form of adult hemoglobin containstwo a-globin polypeptides encoded by the HbA

a geneon chromosome 16 and two b-globin polypeptides en-coded by the HbA

b gene on chromosome 11. Other en-zymes, for example, E. coli DNA polymerase III(Chapter 10) and RNA polymerase II (Chapter 11),contain many different polypeptide subunits, each en-coded by a separate gene. Thus the one gene–oneenzyme concept was modified to one gene–onepolypeptide.

Key Points: The existence of a basic genetic ele-ment, the gene, that controlled a specific phenotypictrait was established by Mendel’s work in 1866. Sincethe discovery of Mendel’s results in 1900, the con-cept of the gene has evolved from the unit that canmutate to cause a specific block in metabolism, to theunit specifying one enzyme, to the sequence of nu-cleotide pairs in DNA encoding one polypeptidechain.

EVOLUTION OF THE CONCEPT OF THEGENE: STRUCTURE

In the preceding section, we examined the evolutionof the concept of the gene as the basic functional com-ponent of the genetic material. In this section, we ex-amine the gene from a structural perspective. What isthe structure of the gene? Do all genes have the samestructure?

The Pre–1940 Beads-on-a-String ConceptPrior to 1940, the genes in a chromosome were consid-ered analogous to beads on a string. Recombinationwas believed to occur only between the beads orgenes, not within genes. The gene was believed to beindivisible. According to this beads-on-a-string con-cept, the gene was the basic unit of genetic informa-tion defined by three criteria: (1) function, (2) recombi-nation, and (3) mutation. More specifically, the genewas

1. The unit of function, the unit of genetic material thatcontrolled the inheritance of one “character” or oneattribute of phenotype.

2. The unit of structure, operationally defined in twoways:a. By recombination: as the unit of genetic informa-

tion not subdivisible by recombination.b. By mutation: as the smallest unit of genetic mate-

rial capable of independent mutation.

Geneticists initially thought that all three criteria de-fined the same basic unit of inheritance, namely, thegene.

Geneticists now know that these criteria definetwo different units of inheritance. According to thecurrent molecular concept, the gene is the unit of func-tion, the unit of genetic information controlling thesynthesis of one polypeptide chain or, in some cases,one RNA molecule. The unit of structure is simply thestructural unit in DNA, the nucleotide pair. Because itclearly does not make sense to call each nucleotidepair a gene, geneticists have focused on the originaldefinition of the gene as the unit of function and havediscarded the beads-on-a-string view that the gene isnot subdivisible by recombination or mutation. This isclearly appropriate since the emphasis in Mendel’swork was on the Merkmal (or gene, as it is now called)controlling one phenotypic characteristic.

Discovery of Recombination Within the GeneIn 1940, Clarence P. Oliver published the first evi-dence indicating that recombination could occur


Chpt 14 9/10/96 3:16 PM Page 10

Evolution of the Concept of the Gene: Structure 11

within a gene. Oliver was studying mutations at thelozenge locus on the X chromosome of Drosophilamelanogaster. Two mutations, lzs (“spectacle” eye) andlzg (“glassy” eye), were thought to be alleles, that is,different forms of the same gene. The data availableprior to 1940 indicated that they mapped at the sameposition on the X chromosome. They had similar ef-fects on the phenotype of the eye, and heterozygouslzs/lzg females had lozenge rather than wild-type eyes.However, when lzs/lzg females were crossed with ei-ther lzs or lzg males and large numbers of progenywere examined, wild-type progeny occurred with afrequency of about 0.2 percent.

These rare wild-type progeny could be explainedby reversion of either the lzs or the lzg mutation. Butthere were two strong arguments against the rever-sion explanation. (1) The frequency of reversion of lzs

or lzg to wild-type in hemizygous lozenge males wasmuch lower than 0.2 percent. (2) When the lzs/lzg het-erozygotes carried genetic markers bracketing thelozenge locus, the rare progeny with wild-type eyes al-ways carried an X chromosome with lz+ that wasflanked by recombinant outside markers. Moreover,the same combination of outside markers always oc-curred, as though the sites of lzs and lzg were fixed rel-ative to each other and crossing over was occurringbetween them. Different sets of outside markers wereused and yielded the same result. If the lzs/lzg het-erozygous female carried X chromosomes of the type

the rare progeny with wild-type eyes all (with one ex-ception) contained an X chromosome with the follow-ing composition:

Among progeny of these matings, the reciprocal com-bination of outside markers (x+-y+) never appeared incombination with lz+. This result strongly suggestedthat the lzs and lzg mutations were located at distinctsites in the lozenge locus, and that the lz+ chromosomewas produced by crossing over between the two sites,as shown in the following diagram.

Definitive evidence for the involvement of recombina-tion required the recovery and identification of the lzs-lzg double mutant with the reciprocal combination ofoutside markers—that is,

Oliver was not able to identify this double mutant be-cause of the inability to distinguish it from theparental single-mutant phenotypes. The identificationof both products, the wild-type and double mutantchromosomes, produced by crossing over within thelozenge gene, was first accomplished by Melvin M.Green, one of Oliver’s students.

The results of these pioneering studies first indi-cated that the gene was more complex than a bead ona string. They showed that the gene was divisible, con-taining sites that were separable by crossing over.Oliver’s and Green’s results were the first step towardthe present concept of the gene as a long sequence ofnucleotide pairs, capable of mutating and recombin-ing at many different sites along its length.

Recombination Between AdjacentNucleotide PairsThe results obtained by Oliver, Lewis, and Green intheir studies of Drosophila genes all indicated that mu-table sites that are separable by recombination can ex-ist within a single gene. Seymour Benzer extended thispicture of the gene by demonstrating the existence of199 distinct sites of mutation that were separable byrecombination within the rIIA gene of bacteriophageT4 (Chapter 15). Benzer’s picture of the gene as a se-quence of nucleotide pairs capable of mutating atmany distinct sites was soon verified by the results ofmany researchers investigating gene structure in sev-eral different organisms, both prokaryotes and eu-karyotes. Given this information about the structure ofgenes and the known structure of DNA, it followedthat the smallest unit of genetic material capable ofmutation might be the single nucleotide pair and thatrecombination might occur between adjacent nu-cleotide pairs, whether between or within genes. Re-combination between adjacent nucleotide pairs of agene was first demonstrated by Charles Yanofsky inhis studies of the trpA gene encoding the a polypep-tide of tryptophan synthetase in E. coli. This enzyme, atetramer containing two a polypeptides and two bpolypeptides, catalyzes the final step in the biosynthe-sis of the amino acid tryptophan.

Yanofsky and colleagues isolated and character-ized a large number of tryptophan auxotrophs withmutations in the trpA gene. The wild-type trpA geneencodes an a polypeptide that is 268 amino acids long.

x((

x+ y

y+lzs

lzg

))

x+(

y+lzg

)lzs

x(

ylz+

)

x(

lzs

Crossover

)y+

( )x+ ylzg

x(

lz+)

y

Chpt 14 9/10/96 3:16 PM Page 11


Yanofsky and associates used the laborious techniquesof protein sequencing to determine the completeamino acid sequence of the wild-type a polypeptide.They also determined the amino acid substitutionsthat had occurred in several mutant forms of the tryp-tophan synthetase a polypeptide. They mapped the

mutations within the trpA gene by two- and three-fac-tor crosses, and compared the map positions with thelocations of the amino acid substitutions in the mutantpolypeptides.

Yanofsky’s correlated genetic and biochemicaldata for the trpA gene and the tryptophan synthetasea polypeptide showed that recombination can occurbetween mutations that alter the same amino acid.Mutations trpA23 and trpA46 both result in the substi-tution of another amino acid (arginine in the case ofA23, glutamic acid in the case of A46) for the glycinepresent at position 211 of the wild-type tryptophansynthetase a polypeptide. However, these two muta-tions occur at different mutable sites; that is, the A23and A46 sites are separable by recombination. Yanof-sky and colleagues determined the amino acids pre-sent at position 211 of the a polypeptide in other mu-tants as well as in revertants and partial revertants ofthe trpA23 and trpA46 mutants. By using this informa-tion and the known codon assignments, they wereable to determine which of the glycine, arginine, andglutamic acid codons were present in the trpA mRNAat the position encoding amino acid residue 211 of thea polypeptides present in trp1, trpA23, and trpA46cells, respectively (Figure 14.8).

Once the specific codons in mRNA are known, thecorresponding base-pair sequences in the structuralgene from which the mRNA is transcribed are alsoknown. One strand of DNA will be complementary tothe mRNA, and the second strand of DNA will becomplementary to the first strand. Therefore, Yanof-sky’s data demonstrated that the mutational eventsthat produced the trpA23 and trpA46 alleles were G:Cto A:T transitions at adjacent nucleotide pairs. Thetrp1 cells produced by recombination between chro-mosomes carrying mutations A23 and A46 demon-strated that recombination had occurred between ad-jacent nucleotide pairs in the trpA gene as shown inFigure 14.9. Yanofsky’s results clearly showed that the

Glu (GAA) Arg (AGA)Val (GUA)

Gly (GGA)

Ala (GCA)

Gly (GGA)

Val (GUA)

Gly (GGA)

Ala (GCA)

lle (AUA)

Thr (ACA)

Ser Gly (GGA)

7 1 15

14 25 8 2 2 14 16

1 2 2

AGC or

AGU

ArgAGA or

GGC or

CGU

Mutant

Wild-type

Mutants

Revertants or

partial revertants

Full revertants

or mutants

Figure 14.8 Pedigree of amino acid residue 211 (from theNH2 terminus) of the a polypeptide of tryptophan syn-thetase of E. coli. Amino acid 211 is altered in trpA23 andtrpA46 mutants (see Figure 14.11). The triplet codonsshown in parentheses are the only codons specific to the in-dicated amino acids that will permit all of the observedamino acid replacements to occur by single base-pair substi-tutions. The number beside each arrow indicates the num-ber of times that particular substitution was observed.These results indicate that the arginine and glutamic acidcodons encoding amino acid 211 of the a polypeptides intrpA23 and trpA46 mutant cells are AGA and GAA, respec-tively. Thus these two mutations alter adjacent nucleotidepairs in the trpA gene.

C •••

T ••

G A

C •••

G

T ••

C •••

T ••

A G A

T ••

T ••

A A

C •••

G

Recombination

Amino acid in polypeptide:

Codon in mRNA:

Base pairs in DNA:

Arginine

AGA

Base pairs in DNA:

Codon in mRNA: GAA

Amino acid in polypeptide:

Glutamic acid

Rare ( 1/100,000) wild-type recombinants

Base pairs in DNA

G G A

Glycine Amino acid in polypeptide

Codon in mRNA

~ =

Mutant trpA23

Mutant trpA46

Figure 14.9 Recombinationbetween mutations at adja-cent nucleotide pairs in thetrpA gene of E. coli. Muta-tions A23 and A46 both resultin an amino acid substitutionat position 211 of the tryto-phan synthetase a polypep-tide. Wild-type E. coli has aglycine residue at this posi-tion of the a polypeptide.A23 causes a glycine to argi-nine substitution; A46 causesa glycine to glutamic acidsubstitution (see Figure 14.8).

Chpt 14 9/10/96 3:16 PM Page 12

Evolution of the Concept of the Gene: Structure 13

unit of genetic material not divisible by recombinationis the single nucleotide pair.

The pre–1940 concept of the gene as (1) the small-est unit of genetic material that could undergo muta-tion and (2) the unit of genetic information that couldnot be subdivided by recombination was clearlywrong. Each nucleotide pair of a gene can change ormutate independently, and recombination can occurbetween adjacent nucleotide pairs. The results ofOliver, Lewis, Green, Benzer, Yanofsky, and manyothers have compelled geneticists to focus on the geneas the unit of function, the sequence of nucleotidepairs controlling the synthesis and structure of onepolypeptide or one RNA molecule.

Colinearity Between the Coding Sequence ofa Gene and Its Polypeptide ProductThe genetic information is stored in linear sequencesof nucleotide pairs in DNA (or nucleotides in RNA, insome cases). Transcription and translation convert this

genetic information into linear sequences of aminoacids in polypeptides, which function as the key inter-mediaries in the genetic control of the phenotype.

It is now known that the nucleotide-pair se-quences of the coding regions of the structural genesand the amino acid sequences of the polypeptides thatthey encode are colinear. That is, the first three basepairs of the coding sequence of a gene specify the firstamino acid of the polypeptide, the next three basepairs (four to six) specify the second amino acid, andso on, in a colinear fashion (Figure 14.10a). It is alsoknown that the coding regions of most of the genes inhigher eukaryotes are interrupted by noncoding in-trons (Chapter 12). However, the presence of intronsin genes does not invalidate the concept of colinearity.The presence of introns in genes simply means thatthere is no direct correlation in physical distances be-tween the positions of base-pair coding triplets in agene and the positions of amino acids in the polypep-tide specified by that gene (Figure 14.10b).

The first strong evidence for colinearity between agene and its polypeptide product resulted from stud-

1 2 3 4 5 6 7 8 9 10 11Base-pair triplets in coding region

Transcription

Coding region of typical uninterrupted prokaryotic gene.(a)

(b)

12 13 14 15 301 302 303 304

Translation

aa1 aa2 aa3 aa4 aa5 aa6 aa7 aa8 aa9 aa10 aa11

Amino acids in polypeptide gene product

aa12 aa13 aa14 aa15 aa301aa302aa303aa304

aa1 aa2 aa3 aa4 aa5 aa6 aa7 aa8 aa9 aa10

Amino acids in polypeptide gene product

aa301aa302aa303aa304

1 2 3 4 5 6 7 8 9 10Base-pair triplets in coding region

Transcription

Exon 2Exon 1

Intron 1

Intron 1

Exon n

Coding region of typical interrupted eukaryotic gene.

301 302 303 304

1 2 3 4 5 6 7 8 9 10

Processing of gene transcript

301 302 303 304Codons in primary transcript

1 2 3 4 5 6 7 8 9 10Codons in mRNA

Translation

301 302 303 304

1

2

1

2

3

Figure 14.10 Colinearitybetween the coding regions ofgenes and their polypeptideproducts.

Chpt 14 9/10/96 3:16 PM Page 13

ies of Charles Yanofsky and colleagues on the E. coligene encoding the a subunit of the enzyme trypto-phan synthetase. Tryptophan synthetase catalyzes thefinal step in the biosynthesis of the amino acid trypto-phan. As mentioned earlier, this enzyme contains twoa polypeptides encoded by the trpA gene and two bpolypeptides encoded by the trpB gene. Yanofsky andcoworkers performed a detailed genetic analysis ofmutations in the trpA gene and correlated the geneticdata with biochemical data on the sequences of thewild-type and mutant tryptophan synthetase apolypeptides. They demonstrated that there was a di-rect correlation between the map positions of muta-tions in the trpA gene and the positions of the resul-tant amino acid substitutions in the tryptophansynthetase a polypeptide (Figure 14.11).

About the same time, Sydney Brenner and associ-ates demonstrated a similar colinearity between thepositions of mutations in the gene of bacteriophage T4that encodes the major structural protein of the phagehead and the positions in the polypeptide affected bythese mutations. Brenner and colleagues studied am-ber (UAG chain-termination) mutations and demon-strated a direct correlation between the length of thepolypeptide fragment produced and the position ofthe mutation within the gene.

In the yeast S. cerevisiae, Fred Sherman and col-leagues demonstrated an uninterrupted colinear rela-tionship between the map positions of mutations inthe CYCI gene and the amino acid substitutions in themutant forms of iso-1-cytochrome c, the polypeptidespecified by the CYC1 gene. More recently, colinearityhas been documented for many interrupted genes ineukaryotes. The only difference observed with inter-rupted genes is that the linear sequence of nucleotidepairs encoding a colinear polypeptide is not one con-tinuous sequence of nucleotide pairs. Instead, noncod-ing sequences (introns) intervene between coding se-quences (exons).

Definitive evidence for colinearity has been pro-vided by direct comparisons of the nucleotide se-quences of genes and the amino acid sequences oftheir polypeptide products. One of the first caseswhere the amino acid sequence of a polypeptide andthe nucleotide sequence of the gene encoding it wereboth determined experimentally involved the coatprotein of bacteriophage MS2 and the gene that en-codes it. This small virus has an RNA genome that en-codes only four proteins, one being the coat proteinthat encapsulates the RNA. When the genetic codewas used to compare the nucleotide sequence of thecoat protein gene with the amino acid sequence of thecoat polypeptide, the sequences exhibited perfect col-inearity (Figure 14.12). Since then, similar results haveestablished colinearity between many genes and theirprotein products in organisms ranging from viruses tohumans. Thus colinearity between the coding regionsof genes and the amino acid sequences of theirpolypeptide products is a universal or nearly univer-sal feature of gene-protein relationships.

Key Points: The concept of the gene has evolvedfrom a bead on a string, not divisible by recombina-tion or mutation, to a sequence of nucleotide pairs inDNA encoding one polypeptide chain. The unit of ge-netic material not divisible by recombination or mu-tation is the single nucleotide pair.

A GENETIC DEFINITION OF THE GENE

With the emergence of the one gene–one polypeptideconcept, scientists could define the gene biochemi-cally, but they had no genetic tool to use in determin-ing whether two mutations were in the same or differ-ent genes. This deficiency was resolved when EdwardLewis developed the complementation test for func-


0 1.4 0.04 0.3 0.4 0.001 0.6 0.5 0.001 0.02 0.3

trpA geneA3 A33 A446 A487 A223 A23 A46 A187 A78 A58 A169 A96

Glu Glu Tyr Leu Thr Gly Gly Gly Gly Gly Ser Gln

Val Met Cys Arg I I e Arg Glu Val Cys Asp Leu (term.)

H2N—1—49 — 49 — 175—177—183—211—211—213—234—234—235—243—268—COOH

Genetic map Map distances (not to scale):

Amino acid in wild-type polypeptide:

Amino acid in mutant polypeptide:

Position of amino acid change in polypeptide:

Figure 14.11 Colinearity between the E. coli trpA gene and its polypeptide product, the apolypeptide of tryptophan synthetase. The map positions of mutations in the trpA gene areshown at the top, and the locations of the amino acid substitutions produced by these mutationsare shown below the map.

Chpt 14 9/10/96 3:16 PM Page 14

tional allelism in 1942. Lewis was studying the Star-as-teroid (small, rough eyes) locus in Drosophila and ob-served that flies of genotype S ast1/S1 ast had a moreextreme mutant eye phenotype than flies of genotypeS ast/S1 ast1. His results were complicated by the par-tial dominance of the Star (S) mutation. Lewis subse-quently performed similar experiments with two eye-color mutations, white (w) and apricot (apr), andobtained results that were easier to interpret.

The Complementation Test as anOperational Definition of the GeneFruit flies that are homozygous for the X-linked muta-tions apr (now wa) and w have apricot-colored eyesand white eyes, respectively, in contrast to the redeyes of wild-type Drosophila.

Lewis reported that heterozygous apr/w femaleshad light apricot-colored eyes and produced rare red-eyed progeny carrying recombinant apr1w1 chromo-somes. In addition, Lewis was able to identifyprogeny flies that carried X chromosomes with thereciprocal apr w recombinant genoypte. The observedfrequency of recombination between the apr and w

mutations was 0.03 percent. Clearly, the apr and w mu-tations were separable by recombination, and apr andw were not alleles according to the pre–1940 conceptof the gene. The two mutations appeared to be in thesame unit of function but in two different units ofstructure. When Lewis produced flies of genotype aprw/apr1 w1, they had red eyes just like those of wild-type flies. When he constructed flies of genotype aprw1/apr1 w, they had light apricot-colored eyes. Bothgenotypes contain the same mutant and wild-type ge-netic information, but in different arrangements. Thepresence of different phenotypes in organisms thatcontain the same genetic markers, but with the mark-ers present in different arrangements, is called a posi-tion effect, and the type of position effect observed byLewis is referred to as a cis-trans position effect.

Before we analyze Lewis’s results in greater detail,we need to define some terms. A double heterozygote,which carries two mutations and their wild-type alle-les, that is, m1 and m1

1 plus m2 and m21, can exist in ei-

ther of two arrangements (Figure 14.13). When the twomutations are on the same chromosome, the arrange-ment is called the coupling or cis configuration; a het-erozygote with this genotype is called a cis heterozy-gote (Figure 14.13a). When the two mutations are on

A Genetic Definition of the Gene 15

GCU – UCU – AAC – UUU – ACU – CAG – UUC – GUU – CUC – GUC – GAC – AAU – GGC – GGA – ACU – GGC – GAC – GUG – ACU – GUC – GCC – CCA – AGC – AAC – UUC –

GCU – AAC – GGG – GUC – GCU – GAA – UGG – AUC – AGC – UCU – AAC – UCG – CGU – UCA – CAG – GCU – UAC – AAA – GUA – ACC – UGU – AGC – GUU – CGU – CAG –

AGC – UCU – GCG – CAG – AAU – CGC – AAA – UAC – ACC – AUC – AAA – GUC – GAG – GUG – CCU – AAA – GUG – GCA – ACC – CAG – ACU – GUU – GGU – GGU – GUA –

GAG – CUU – CCU – GUA – GCC – GCA – UGG – CGU – UCG – UAC – UUA – AAU – AUG – GAA – CUA – ACC – AUU – CCA – AUU – UUC – GCU – ACG – AAU – UCC – GAC –

UGC – GAG – CUU – AUU – GUU – AAG – GCA – AUG – CAA – GGU – CUC – CUA – AAA – GUA – GGA – AAC – CCG – AUU – CCC – UCA – GCA – AUC – GCA – GCA – AAC –

UCC – GGC – AUC – UAC – UAA – UAG – ACG – CCG – GCC – AUU – CAA – ACA – UGA – GGA –

Ala Ser Asn Phe Thr Gln Phe Val Leu Val Asp Asn Gly Gly Thr Gly Asp Val Thr Val Ala Pro Ser Asn Phe

Ala Asn Gly Val Ala Glu Trp Ile Ser Ser Asn Ser Arg Ser Gln Ala Tyr Lys Val Thr Cys Ser Val Arg Gln

Ser Ser Ala Gln Asn Arg Lys Try Thr Ile Lys Val Glu Val Pro Lys Val Ala Thr Gln Thr Val Gly Gly Val

Glu Leu Pro Val Ala Ala Trp Arg Ser Tyr Leu Asn Met Glu Leu Thr Ile Pro Ile Phe Ala Thr Asn Ser Asp

Cys Glu Leu Ile Val Lys Ala Met Gln Gly Leu Leu Lys Asp Gly Asn Pro Ile Pro Ser Ala Ile Ala Ala Asn

Ser Gly IIe Tyr129

1 5 10 15 20 25

30 35 40 45 50

55 60 65 70 75

80 85 90 95 100

105 110 115 120 125

Tandem ochre

and amber termination

codons

Opal termination

codon

Coat protein initiation codon

– AGC – AUG–

Figure 14.12 Colinearity between the nucleotide se-quence of the bacteriophage MS2 coat protein gene and theamino acid sequence of the coat polypeptide that it encodes.Note that the amino acid sequence of this protein is pre-

cisely that predicted from the nucleotide sequence based onthe genetic code. In addition, note that all three terminationcodons are present between the coat gene and the genedownstream from it on the MS2 chromosome.

Chpt 14 9/10/96 3:16 PM Page 15

different chromosomes, the arrangement is called therepulsion or trans configuration. An organism withthis genotype is a trans heterozygote (Figure 14.13b).

Recall that apr w/apr1 w1 cis heterozygotes havewild-type red eyes, whereas apr w1/apr1 w trans het-

erozygotes have light apricot eyes. This is preciselythe result that would be expected if apr and w are mu-tations at different sites in the same gene, the unit ofgenetic information encoding a single polypeptide(Figure 14.14). Thus apr and w are considered alleles ofthe same gene, and apr is now designated wa in recog-nition of this relationship. If apr and w had been muta-tions in two different units of function, two differentgenes, both the cis and the trans heterozygotes shouldhave expressed the wild-type phenotype, namely, redeyes. In the trans heterozygote, the apr1 gene productwould be produced by the apr1 gene on the chromo-some carrying the w mutation, and the w1 gene prod-uct would be specified by the w1 gene on the chromo-some harboring the apr mutation .

Lewis’s discovery of the cis-trans position effectswith Star-asteroid and wa-w led to the development ofthe complementation test or trans test for functionalallelism. The complementation test allows geneticiststo determine whether mutations that produce thesame or similar phenotypes are in the same gene or indifferent genes. They must test mutations pairwise bydetermining the phenotypes of trans heterozygotes.That is, they must construct trans heterozygotes witheach pair of mutations to be analyzed and determinewhether these heterozygotes have mutant or wild-type phenotypes.


m2+m1

m2m1+

m2m1

m2+m1

+

trans heterozygote.

cis heterozygote.

(b)

(a)

Figure 14.13 The arrangement of genetic markers in cisand trans heterozygotes.

X chromosome 1Nonfunctional mutant white gene-product

White gene

apr w

X chromosome 2 Wild-type white gene product

Therefore, the cis heterozygote has wild-type red eyes.

apr+ w+

X chromosome 1Nonfunctional apr and w

White gene

apr w+

X chromosome 2 mutant white gene products

Therefore, the trans heterozygote has light apricot-colored eyes.

apr+ w

trans heterozygote.

cis heterozygote.

(b)

(a)

Figure 14.14 The cis-trans positioneffect observed by Edward Lewis withthe apr and w mutations of Drosophila.

Chpt 14 9/10/96 3:16 PM Page 16

Ideally, the complementation or trans test shouldbe done in conjunction with the cis test—a control thatis often omitted. Cis tests are performed by construct-ing cis heterozygotes for each pair of mutations to beanalyzed and determining whether they have mutantor wild-type phenotypes. Together, the complementa-tion or trans test and the cis test are referred to as thecis-trans test. Each cis heterozygote, which containsone wild-type chromosome, should have the wild-type phenotype whether the mutations are in the samegene or in two different genes (Figure 14.15). Indeed,the cis heterozygote must have the wild-type pheno-type for the results of the trans test to be valid. If the cisheterozygote has the mutant phenotype, the trans testcannot be used to determine whether the two muta-tions are in the same gene. Thus the trans test cannotbe used with dominant mutations. Because cis het-erozygotes contain one chromosome with wild-typecopies of all relevant genes, these genes should specifyfunctional products and a wild-type phenotype.

Whether two mutations are in the same gene ortwo different genes is determined by the results of thecomplementation or trans test. With diploid organ-isms, the trans heterozygote is produced simply bycrossing organisms that are homozygous for each ofthe mutations of interest. With viruses, trans heterozy-gotes are produced by simultaneously infecting host

cells with two different mutants. Regardless of howthe two mutations are placed in a common proto-plasm in the trans configuration, the results of thetrans or complementation test provide the same infor-mation.

1. If the trans heterozygote has the mutant phenotype(the phenotype of organisms or cells homozygousfor either one of the two mutations), then the twomutations are in the same unit of function, the samegene (Figure 14.16a).

2. If the trans heterozygote has the wild-type pheno-type, then the two mutations are in two differentunits of function, two different genes (Figure14.16b).

When the two mutations present in a trans het-erozygote are both in the same gene, as shown for mu-tations m1 and m2 in Figure 14.16a, both chromosomeswill carry defective copies of that gene. As a result, thetrans heterozygote will contain only nonfunctionalproducts of the gene involved and will have a mutantphenotype.

When a trans heterozygote has the wild-type phe-notype, the two mutations are said to exhibit comple-mentation or to complement each other and are locatedin different genes. In the example illustrated in Figure


Homologous chromosome

copy 1

cis heterozygote: mutations in one gene.(a)

(b)


copy 2

Therefore, the cis heterozygote will have the wild-type phenotype.

gene 1

+ +

m1

m2

Functional product of wild-type gene 1.

Nonfunctional product of gene 1 carrying m

1 and m

2 mutations.


copy 1

cis heterozygote: mutations in two different genes.


copy 2

Therefore, the cis heterozygote will have the wild-type phenotype.

gene 1

+

m1

m3

Functional products of wild-type genes

1 and 2.

Nonfunctional products of m

1 and m

3 mutant

alleles of genes 1 and 2.

gene 2

+

Figure 14.15 The cis test. The cisheterozygote should have the wild-type phenotype whether the mutationsare in the same gene (a) or in two dif-ferent genes (b).

Chpt 14 9/10/96 3:16 PM Page 17

14.16b, the chromosome carrying mutation m1 in gene 1has a wild-type copy of gene 2, which specifies func-tional gene 2 product, and the chromosome carryingmutation m3 in gene 2 has a wild-type copy of gene 1,which encodes functional gene 1 product. Thus thetrans heterozygote shown contains functional productsof both genes and has a wild-type phenotype.

Only the complementation test, or the trans part ofthe cis-trans test, is included in most genetic analyses.Constructing a chromosome that carries both muta-tions for the cis test is often difficult, especially witheukaryotes. Moreover, the cis heterozygotes almost al-ways have wild-type phenotypes if the mutations be-ing analyzed are recessive. Thus, in most instances,the results of complementation or trans tests can be in-terpreted correctly without carrying out the laboriouscis tests.

Seymour Benzer introduced the term cistron to re-fer to the unit of function operationally defined by thecis-trans test. However, today, most geneticists con-sider the terms gene and cistron to be synonyms. Thuswe will use gene, rather than cistron, throughout thistext.

The gene is operationally defined as the unit offunction by the complementation or trans test, which

is used to determine whether mutations are in thesame gene or different genes. The complementationtest is one of the three basic tools of genetics. The othertwo genetic tools are recombination (Chapter 7) andmutation (Chapter 13).

The information provided by complementation tests istotally distinct from that obtained from recombinationanalyses.

The results of complementation tests indicatewhether mutations are allelic, whereas the results of re-combination analyses indicate whether mutations arelinked and, if so, provide estimates of how far apartthey are on a chromosome. Nevertheless, studentssometimes confuse complementation and recombina-tion. Thus we will contrast complementation and re-combination by illustrating both phenomena with thesame three mutations in bacteriophage T4. Phage T4 isvery similar to phage T2, which we discussed in Chap-ter 10. Because of the simple structure of the virus andthe direct relationship between specific gene productsand phenotypes, the phage T4 system provides an ex-cellent mnemonic visualization of the difference be-tween complementation and recombination.

We discussed the amber mutations of bacterio-phage T4 in Chapter 12 (see Figure 12.23) and the



copy 1

trans heterozygote: mutations in one gene.(a)

(b)


copy 2

No functional gene 1 product is synthesized in the trans heterozygote; therefore, it will have a mutant phenotype.

gene 1

+

+

m1

m2

Nonfunctional m1

mutant product of gene 1

Nonfunctional m2

mutant product of gene 1


copy 1

trans heterozygote: mutations in two different genes.


copy 2

Functional products of both genes are synthesized in the trans heterozygote; therefore, it will have the wild-type phenotype.

gene 1

+

m1

m3

Functional wild-type product of gene 2

Functional wild-type product of gene 1

gene 2

+

Nonfunctional m1 mutant

product of gene 1

Nonfunctional m3 mutant

product of gene 2 Figure 14.16 The trans test. Thetrans heterozygote should have (a) themutant phenotype if the two muta-tions are in the same gene, and (b) thewild-type phenotype if the mutationsare in two different genes.

Chpt 14 9/10/96 3:16 PM Page 18

pathway of morphogenesis of phage T4 in Chapter 13(see Figure 13.11) Amber mutations produce transla-tion–termination triplets within the coding regions ofgenes. As a result, the products of the mutant genesare truncated polypeptides, which are almost alwaystotally nonfunctional. Therefore, complementationtests performed with amber mutations are usually un-ambiguous. When amber mutations occur in essentialgenes, the mutant phenotype is lethality—that is, noprogeny are produced when a restrictive host cell isinfected; and the wild-type phenotype is a normalyield (about 300 phage per cell) of progeny phage ineach infected restrictive host cell. In Chapter 13, wecalled such mutations conditional lethals and empha-sized their utility in genetic analysis. With conditionallethals, the distinction between the mutant and wild-type phenotypes is maximal: lethality versus normalgrowth.

Two of the three amber mutations that we will con-sider (amB17 and amH32) are in gene 23, which en-codes the major structural protein of the phage head;the other mutation (amE18) is in gene 18, which speci-fies the major structural protein of the phage tail. Wecan see from Figure 14.17 why complementation oc-curs between mutations amB17 (head gene) and amE18(tail gene) and why complementation does not occurbetween mutations amB17 and amH32 (both in headgene). Complementation is the result of the interactionof the gene products specified by chromosomes carry-ing two different mutations when they are present in acommon protoplasm. Complementation does not de-pend on recombination of the two chromosomes or in-volve any direct interaction between the chromo-somes. Complementation, or the lack of it, is assessed bythe phenotype (wild-type or mutant) of each trans heterozy-gote.

Figure 14.18 illustrates the occurrence of recombi-nation between the amber mutations used to illustratecomplementation in Figure 14.17. Although wehaven’t discussed viral genetics in detail (Chapter 15),we have covered the essential concepts of recombina-tion (Chapter 7). Recombination of phage genes occursby a process analogous to crossing over in eukaryotes,with linkage distances measured in map units, just asin eukaryotes. Recombination frequencies are mea-sured by infecting permissive host cells with two mu-tants so that the mutant chromosomes can replicateand participate in crossing over. Then, the progeny arescreened for wild-type recombinants by plating themon lawns of restrictive host cells (E. coli cells in whichonly the wild-type phage can grow). In the exampleshown in Figure 14.18, recombination is observed inboth crosses: (1) amB17 3 amE18, mutations in two dif-ferent genes, and (2) amB17 3 amH32, mutations in thesame gene. The only difference is that more recombi-nants are produced in cross 1, which involves two am-

ber mutations that are relatively far apart on the phageT4 chromosome, than in cross 2, which involves twomutations located near one another in the same gene.Recombination involves direct interactions between thechromosomes carrying the mutations, the actual breakage ofchromosomes, and reunion of parts to produce wild-typeand double-mutant chromosomes.

Complementation should occur in every trans het-erozygote containing mutations in two differentgenes. Recombination is detected by examining thegenotypes of the progeny of heterozygotes, not thephenotypes of the trans heterozygotes themselves.Moreover, a trans heterozygote will never producemore than 25 percent gametes (or progeny for hap-loids) with wild-type chromosomes. If the mutationsare closely linked, like the amber mutations shown inFigure 14.18b, the frequency of wild-type recombinantchromosomes will be much lower than 25 percent.

Structural allelism is the occurrence of two ormore different mutations at the same site and is deter-mined by the recombination test. Two mutations thatdo not recombine are structurally allelic; the muta-tions either occur at the same site or overlap a com-mon site. Functional allelism is determined by thecomplementation test as just described; two mutationsthat do not complement are in the same unit of func-tion, the same gene. Mutations that are both struc-turally and functionally allelic are called homoalleles;they do not complement or recombine with eachother. Mutant homoalleles have defects at the samesite or overlap a common site in the same gene. Muta-tions that are functionally allelic, but structurally non-allelic, are called heteroalleles; they recombine witheach other but do not complement one another. Mu-tant heteroalleles occur at different sites but within thesame gene.

Intragenic ComplementationThe results of complementation tests are usually un-ambiguous when mutations that result in the synthe-sis of no gene product, partial gene products, or to-tally defective gene products are used—for example,deletions of segments of genes, frameshift mutations,or polypeptide chain-terminating mutations. Ofcourse, the mutations must be recessive. When muta-tions causing amino acid substitutions are used, theresults of complementation tests are sometimes am-biguous because of the occurrence of a phenomenoncalled intragenic complementation.

The functional forms of some proteins are dimersor higher multimers consisting of two or morepolypeptides. These polypeptides may be either ho-mologous, the products of a single gene, or nonhomol-ogous, the products of two or more distinct genes.


Chpt 14 9/10/96 3:16 PM Page 19


Chromosome of mutant amB17 Chromosome of mutant amE18

Gene 18 Gene 23Gene 18 Gene 23

DNA:

mRNA:

Protein:

Mutations

E. coli B cell: restrictive host for amber mutants

Lysis yields infective progeny phage; therefore, the trans heterozygote has the wild-type phenotype.

Complementation between mutations amB17 and amE18.

Lack of complementation between mutations amB17 and amH32.(b)

(a)

Chromosome of mutant amB17 Chromosome of mutant amH32

Gene 18 Gene 23Gene 18 Gene 23

DNA:

mRNA:

Protein:

Mutation Mutation

E. coli B cell: restrictive host for amber mutants

No phage heads and, thus, no infective progeny phage are produced in the infected cell; therefore, the trans heterozygote has the mutant phenotype.

Figure 14.17 Complementation and noncomplementa-tion in trans heterozygotes. (a) Complementation betweenmutation amB17 in gene 23, which encodes the major struc-tural protein of the phage T4 head, and mutation amE18 ingene 18, which encodes the major structural protein of thephage tail. Both heads and tails are synthesized in the cell,

with the result that infective progeny phage are produced.(b) When the trans heterozygote contains two mutations(amB17 and amH32) in gene 23, no heads are produced, andno infective progeny phage can be assembled. Comparewith Figure 14.18.

Chpt 14 9/10/96 3:16 PM Page 20

When the active form of the protein contains two ormore homologous polypeptides (it may or may notalso contain nonhomologous polypeptides), intrageniccomplementation may occur. Intergenic complemen-tation (discussed in the preceding section) and intra-

genic complementation (discribed below) are two dis-tinct phenomena.

Let us consider an enzyme that functions as a homodimer, that is, a protein containing two copies ofa specific gene product (Figure 14.19). In organisms


amB17am+

Replication

Recombination between phage T4 chromosomes carrying mutations amB17 and amE18. (a)

Lysis yields progeny phage of four genotypes: Parental genotypes : ~_ 40% amB17 and ~_ 40% amE18; Recombinant genotypes: ~_ 10% wild-type (am+) and ~_ 10% double mutant (amE18-amB17).

Chromosome carrying mutation amB17

Chromosome carrying mutation amE18

Replication

am+amE18

am+am+

amB17amE18(40%) (40%)

(10%)

(10%)

Breakage and exchange

of parts

E. coli CR63 cell: Permissive host for amber mutants

amB17 am+

Replication

Recombination between phage T4 chromosomes carrying mutations amB17 and amH32. (b)

Chromosome carrying mutation amB17

Chromosome carrying mutation amH32

Replication

am+ amH32

amB17 amH32

am+ am+

(49%) (49%)

(1%)

(1%)

Breakage and exchange of parts

E. coli CR63 cell: Permissive host for amber mutants

Lysis yields progeny phage of four genotypes: Parental genotypes : ~_ 49% amB17 and ~_ 49% amH32; Recombinant genotypes: ~_ 1% wild-type (am+) and ~_ 1% double mutant (amB17-amH32).

Figure 14.18 Recombination between (a) the comple-menting mutations amB17 (gene 23) and amE18 (gene 18),and (b) the noncomplementing mutations amB17 and amH32(both in gene 23). Recombination occurs in both cases; how-

ever, fewer recombinants are produced in cells infectedwith amB17 and amH32 because the two mutations are lo-cated closer together on the phage T4 chromosome. Com-pare with Fig. 14.17.

Chpt 14 9/10/96 3:16 PM Page 21

that are homozygous for the wild-type allele of thegene, all the protein dimers will contain identicalwild-type polypeptides. Similarly, organisms that arehomozygous for any mutation in the gene will containdimers with two mutant polypeptides. An organismthat is heterozygous for two different mutations in thegene will produce some dimers that contain the twodifferent mutant polypeptides. We call these het-erodimers. Such heterodimers may have partial orcomplete (wild-type) function. If they do, intrageniccomplementation has occurred, and the trans het-erozygote has a wild-type phenotype or a phenotypeintermediate between mutant and wild-type (Figure14.19, bottom). In the case of noncomplementing mu-tations in a gene encoding a multimeric protein, theheteromultimers are nonfunctional, just like the mu-

tant homomultimers (protein multimers composed oftwo or more identical mutant polypeptides).

In several known cases of intragenic complemen-tation, the active form of the protein in the heterozy-gote has been purified and shown to be a heterodimeror heterotetramer containing two different mutantpolypeptides. Why such heteromultimers should beactive when the two corresponding homomultimersare inactive is not clear. Apparently, the wild-type se-quence of amino acids in the nonmutant segment ofone mutant polypeptide somehow compensates forthe mutant segment of the polypeptide encoded bythe second mutant allele, and vice versa (Figure 14.19).However, most proteins have complex three-dimen-sional structures, and until the exact structures of awild-type homomultimer, two mutant homomulti-mers, and an active heteromultimer composed of thetwo mutant polypeptides have been determined, themolecular basis of intragenic complementation willcontinue to be subject to speculation.

Limitations on the Use of theComplementation TestThe complementation test has been very useful in oper-ationally delimiting genes. Usually, two or more muta-tions that produce the same phenotype can be assignedto one or more genes based on the results of comple-mentation tests. However, in some cases, the results ofcomplementation tests cannot be used to delimit genes.As previously mentioned, complementation tests arenot informative in studies of dominant or codominantmutations or in cases where intragenic complementa-tion occurs. In addition, complementation tests aresometimes uninformative because of epistatic interac-tions between the mutant gene products. If the cis test isdone, such interactions are readily detected because thecis heterozygotes will have mutant phenotypes ratherthan the required wild-type phenotype.

Another limitation of the complementation test isencountered in working with so-called polar muta-tions. A polar mutation is a mutation that not only re-sults in a defective product of the gene in which it islocated, but also interferes with the expression of oneor more adjacent genes. The adjacent genes are alwayslocated on one side of the gene carrying the mutation(thus the term polar mutation). Such polar mutationsare frequently observed in prokaryotes in coordi-nately regulated sets of genes called operons (Chapter21). They usually are mutations resulting in polypep-tide chain-termination signals (nucleotide-pair tripletsyielding UAA, UAG, and UGA codons in mRNA)within genes. These polar mutations interfere with theexpression of genes located downstream (relative tothe direction of transcription) of the mutant gene. As


Active

Inactive

Inactive

Inactive

+

+

Inactive

Active

Active site

Wild-type

Mutant

Mutant

Wild-type or

intermediate

Wild-type

Gene

Genotype Protein Phenotype

Mutant 1

Mutation

Mutant 2

trans heterozygote

Mutation

Active site

Figure 14.19 Intragenic complementation sometimes oc-curs when the active form of an enzyme or structural pro-tein is a multimer that contains at least two copies of anyone gene product. Here, the functional form of the enzymeis a dimer composed of two polypeptides encoded by onegene. The amino acids altered by the mutations are shownas red circles in the polypeptide chains.

Chpt 14 9/10/96 3:16 PM Page 22

a result, polar mutations fail to complement mutationsin genes subject to the polar effect (Figure 14.20). Thusthe results of complementation tests performed withpolar mutations are often ambiguous.

Key Points: The complementation or trans test pro-vides an operational definition of the gene; it is usedto determine whether mutations are in the same geneor different genes. Intragenic complementation mayoccur when a protein is a multimer containing at leasttwo copies of one gene product.

COMPLEX GENE–PROTEINRELATIONSHIPS

Most prokaryotic genes consist of continuous se-quences of nucleotide pairs, which specify colinear se-quences of amino acids in the polypeptide gene prod-ucts. As we discussed in Chapter 11, most eukaryotic

genes are split into coding sequences (exons) and non-coding sequences (introns). However, because thespliceosomes usually excise introns from primarytranscripts by cis-splicing mechanisms (processes thatjoin exons from the same RNA molecule), the presenceof introns in genes does not invalidate the comple-mentation test as an operational definition of the gene.Nevertheless, in some cases, transcripts of split genesmay undergo several different types of splicing, mak-ing the relationships between genes and proteinsmore complex than the usual one gene–one polypep-tide. In other cases, expressed genes are assembledfrom “gene pieces” during the development of thespecialized cells in which they are expressed.

Alternate Pathways of Transcript Splicing:Protein IsoformsMany interrupted eukaryotic genes, such as the mam-malian hemoglobin genes and the chicken ovalbumin

Complex Gene-Protein Relationships 23

Chromosome

Mini-chromosome

Trans heterozygote: E. coli partial diploid.

Little or no functional β-galactoside permease is produced. Therefore, the trans heterozygote has a mutant phenotype — that is, it is unable to utilize lactose as an energy source.

Little or no β-galactoside permease due to polar mutation in Z gene.

lac Z

+Z – polar mutation

lac Y

lac Z lac Y

Transcription

mRNA

mRNA

Nonfunctional mutant β-galactosidase

Nonfunctional mutant β-galactoside permease

Functional wild-type β-galactosidase

Translation

Transcription

Translation

Y – mutation+

1

2

2

1

Figure 14.20 Lack of complementation between a polarmutation and a mutation in a downstream gene in the sametranscription unit. The ability of E. coli cells to utilize lac-tose as an energy source depends on the products of two co-transcribed genes: lacY, which encodes b-galactosidepermease, and lacZ, which encodes b-galactosidase. Tran-scription of the lacY and lacZ genes produces a multigenic

mRNA, which is translated to provide the two proteins. Atranslation–termination mutation near the translation–startsite in lacZ has a polar effect on translation of the lacY gene,reducing its translation efficiency to 1–2% of the normallevel. Thus the polar mutation in lacZ will not complementa null mutation in lacY, even though the two point muta-tions are in two different genes.

Chpt 14 9/10/96 3:16 PM Page 23

and 1a2 collagen genes discussed in Chapter 11, eachencode a single polypeptide chain with a specific func-tion. In these cases, the mRNA produced from a givengene contains all the exons of the gene joined togetherin the same order as they occur in the gene. However,the transcripts of some interrupted genes undergo al-ternate pathways of transcript splicing. That is, differ-ent exons of a gene may be joined to produce a relatedset of mRNAs encoding a small family of closely re-lated polypeptides called protein isoforms (Figure14.21). The alternate splicing pathways are often tis-sue-specific, producing related proteins that carry outsimilar, but not necessarily identical, functions in dif-ferent types of cells. The mammalian tropomyosingenes provide striking examples of genes which eachproduce a family of protein isoforms. Tropomyosinsare proteins involved in the regulation of muscle con-traction in animals. Because the various organs of ananimal contain different muscle types, all of whichneed to be regulated, the availability of a family of re-lated tropomyosins might be beneficial. In any case,one mouse tropomyosin gene is known to produce atleast 10 different tropomyosin polypeptides as a resultof alternate pathways of transcript splicing. Genes ofthis type obviously do not fit the one gene–one

polypeptide concept very well. For such genes, wherealternate splicing pathways give rise to two or moredifferent polypeptides, the gene can be defined as aDNA sequence that is a single unit of transcriptionand encodes a set of protein isoforms.

Assembly of Genes During Development:Human Antibody ChainsGenetic information is not always organized intogenes of the type described in the preceding sectionsof this chapter. In rare cases, genes are assembledfrom a storehouse of gene segments during the devel-opment of an organism. The immune system of verte-brate animals depends on the synthesis of proteinscalled antibodies to provide protection against infec-tions by viruses, bacteria, toxins, and other foreignsubstances. Each antibody contains four polypeptides,two identical heavy chains, and two identical lightchains. The light chains are of two types: kappa andlambda. Each antibody chain contains a variable re-gion, which exhibits extensive diversity from antibodyto antibody, and a constant region, which is largelythe same in all antibodies. In germ-line chromosomes,


Promoter

Primary transcript

Family of mRNAs

Polypeptide isoforms

Terminator

Gene

Transcription

Translation

Exon 5

Intron 4

Exon 4

Intron 3

Exon 3

Intron 2

Exon 2

Intron 1

Exon 1

Exon 1 Exon 2 Exon 3 Exon 4 Exon 5

Exon 4

Exon 4

Exon 1 Exon 2 Exon 3

Exon 1 Exon 2 Exon 3 Exon 5

Exon 1 Exon 2

Exon 1 Exon 3

Exon 5

Exon 4 Exon 5

Alternate splicing events

1

2

3

Figure 14.21 A single genemay produce a family of closelyrelated polypeptides by using alternate pathways of exon splicing.

Chpt 14 9/10/96 3:16 PM Page 24

the DNA sequences encoding these antibody chainsare present in gene segments, and the gene segmentsare joined together to produce genes during the differ-entiation of the antibody-producing cells (B lympho-cytes) from progenitor cells.

To illustrate this process of gene assembly duringdevelopment, let us briefly consider the DNA se-quences encoding kappa light chains in humans. (Wediscuss this topic in detail in Chapter 24.) A kappalight chain gene is assembled from three gene seg-ments: Vk (V for variable region), Jk (J for joining seg-ment), and Ck (C for constant region), during B lym-phocyte development. Together, the Vk and Jk genesegments encode the variable region of the kappa lightchain, whereas the Ck gene segment encodes the con-stant region. No functional Vk-Jk-Ck kappa light chaingene is present in any human germ-line chromosome.Instead, human chromosome 2 contains a cluster ofabout 300 Vk gene segments, another cluster of fiveJk gene segments, and a single Ck gene segment (Figure

14.22). During the differentiation of each B lympho-cyte, recombination joins one of the Vk gene segmentsto one of the Jk gene segments. Any Jk segments re-maining between the newly formed Vk-Jk exon and theCk gene segment become part of an intron that is re-moved during the processing of the primary tran-script. Similar somatic recombination events are re-sponsible for the assembly of the genes encodingantibody heavy chains, lambda light chains, and T-lymphocyte receptor proteins (Chapter 24).

Key Points: The transcripts of some genes undergoalternate pathways of splicing to produce mRNAswith different exons joined together. Translation ofthese mRNAs produces closely related polypeptidescalled protein isoforms. Other genes, such as thoseencoding antibody chains, are assembled from genesegments during development by regulated processesof somatic recombination.

Testing Your Knowledge 25

V1 V5

Germ-line chromosome 2

V10 V78 V300 J1 J5 C

Somatic cell recombination during B lymphocyte developmentRearranged chromosome 2

in B lymphocyte

V1 V5 V10 V78 J4 J5 C

V78 J4 J5 C

Assembled gene

Transcription

Primary transcript

V78 J4 C

RNA processing

TranslationKappa light antibody chain

mRNA

1

2

3

4

Figure 14.22 Assembly ofa gene encoding an antibodykappa light chain from genesegments during the develop-ment of a B lymphocyte inhumans.

TESTING YOUR KNOWLEDGE

1. In Drosophila, white, cherry, and vermilion are all sex-linkedmutations affecting eye color. All three mutations are reces-sive to their wild-type allele(s) for red eyes. A white-eyed fe-male crossed with a vermilion-eyed male produces white-eyed male offspring and red-eyed (wild-type) female

offspring. A white-eyed female crossed with a cherry-eyedmale produces white-eyed sons and light cherry-eyeddaughters. Do these results indicate whether or not any ofthe three mutations affecting eye color are located in thesame gene? If so, which mutations?

Chpt 14 9/10/96 3:16 PM Page 25


ANSWER The complementation test for allelism involves placing mu-tations pairwise in a common protoplasm in the trans con-figuration and determining whether the resulting trans het-erozygotes have mutant or wild-type phenotypes. If the twomutations are in the same gene, both copies of the gene inthe trans heterozygote will produce defective gene products,resulting in a mutant phenotype (see Figure 14.16a). How-ever, if the two mutations are in different genes, the twomutations will complement each other, because the wild-type copies of each gene will produce functional gene prod-ucts (see Figure 14.16b). When complementation occurs, thetrans heterozygote will have the wild-type phenotype. Thusthe complementation test allows one to determine whetherany two recessive mutations are located in the same gene orin different genes.

If the trans heterozygote has the mutant phenotype, thetwo mutations are in the same gene. If the trans heterozy-gote has the wild-type phenotype, the two mutations are intwo different genes. Because the mutations of interest aresex-linked, all the male progeny will have the same pheno-type as the female parent. They are hemizygous, with one Xchromosome obtained from their mother. In contrast, the fe-male progeny are trans heterozygotes. In the cross betweenthe white-eyed female and the vermilion-eyed male, the fe-male progeny have red eyes, the wild-type phenotype. Thusthe white and vermilion mutations are in different genes, asillustrated in the following diagram:

2. Suppressor-sensitive (sus) mutants of bacteriophage f29can grow on Bacillus subtilis strain L15, but cannot grow(that is, are lethal) on B. subtilis strain 12A. Wild-type (sus+)f29 phage can reproduce on both strains, L15 and 12A.Thus the f29 sus mutants are conditional lethal mutants likethe amber mutants of bacteriophage T4 (see Figure 12.23).Seven different sus mutants of phage f29 were analyzed forcomplementation by simultaneously infecting the restrictivehost (B. subtilis strain 12A) with each possible pair of mu-tants. Single infections with each of the mutants and withwild-type f29 were also done as controls. The results ofthese complementation or trans tests and the controls aregiven as progeny phage per infected cell in the accompany-ing table. Several infections performed with wild-type f29phage yielded 300 to 400 progeny phage per infected cell.The results of the cis controls are not given, but assume thatall of the cis heterozygote controls yielded over 300 progenyphage per infected cell. Also assume that no intragenic com-plementation occurs between any of the sus mutants stud-ied.

Phage f29 Progeny per Infected Bacterium

Mutant: 1 2 3 4 5 6 7

7 365 384 344 371 347 333 0.016 341 301 351 369 329 0.15 386 326 322 0.04 < 0.014 327 398 374 0.063 354 387 <0.012 0.01 <0.011 0.02

(a) Based on these data, how many genes are identified bythe seven sus mutants? (b) Which sus mutations are locatedin the same gene(s)?

ANSWERThe seven sus mutants yielded from <0.01 to 0.1 progenyphage per infected B. subtilis strain 12A cell; those data de-fine the mutant phenotype (basically no progeny). Infectionsof strain 12A cells with wild-type f29 produced 300 to 400progeny phage per infected cell, defining the wild-type phe-notype. We then examine the phenotypes of the trans het-erozygotes to determine whether any of the mutations arelocated in the same gene(s). In each case, we must askwhether the trans heterozygote has the mutant or the wild-type phenotype. If a trans heterozygote has the mutant phe-notype, the two sus mutations are in the same gene. If it hasthe wild-type phenotype, the two sus mutations are in dif-ferent genes. If you are unsure of why this is true, reviewFigure 14.16. Of the 21 trans heterozygotes examined, 19 ex-hibited the wild-type phenotype, indicating that in each casethe two mutations are in different genes. Two trans het-erozygotes, (1) sus1 on one chromosome and sus2 on a sec-ond chromosome and (2) sus4 on one chromosome and sus5on the another, had the mutant phenotype. Therefore, (a)the seven sus mutations are located in five different genes,with (b) mutations sus1 and sus2 in one gene and mutationssus4 and sus5 in another gene.

X chromosome from parent

trans heterozygote


Complementation yields wild-type phenotype; both v+ and w+

gene products are produced in the trans heterozygote.

v

v+

w+

v+ gene product

w+ gene product

w


trans heterozygote


No w+ gene product; therefore, mutant phenotype.

wch

No active (w+) gene product

w

In the cross between a white-eyed female and a cherry-eyedmale, the female progeny have light cherry-colored eyes (amutant phenotype), not wild-type red eyes as in the firstcross. Since the trans heterozygote has a mutant phenotype,the two mutations, white and cherry, are in the same gene:

Chpt 14 9/10/96 3:16 PM Page 26

Questions and Problems 27

QUESTIONS AND PROBLEMS14.1 In what ways does our present concept of the genediffer from the pre–1940 or classical concept of the gene?

14.2 What was the first evidence that indicated that theunit of function and the unit of structure of genetic materialwere not the same?

14.3 What is the currently accepted operational definitionof the gene?

14.4 Of what value are conditional lethal mutations for ge-netic fine structure analysis?

14.5 Eight independently isolated mutants of E. coli, all ofwhich are unable to grow in the absence of histidine (his-),were examined in all possible cis and trans heterozygotes(partial diploids). All of the cis heterozygotes were able togrow in the absence of histidine. The trans heterozygotesyielded two different responses: some of them grew in theabsence of histidine; others did not. The experimental re-sults, using + to indicate growth and 0 to indicate nogrowth, are given in the accompanying table. How manygenes are defined by these eight mutations? Which mutantstrains carry mutations in the same gene(s)?

Growth of Trans Heterozygotes (without Histidine)

Mutant: 1 2 3 4 5 6 7 8

8 0 0 0 0 0 0 1 07 1 1 1 1 1 1 06 0 0 0 0 0 05 0 0 0 0 04 0 0 0 03 0 0 02 0 01 0

14.6 Assume that the mutants described in Problem 14.5yielded the following results. How many genes would theyhave defined? Which mutations would have been in thesame gene(s)?

Growth of Trans Heterozygotes ( without Histidine)

Mutant: 1 2 3 4 5 6 7 8

8 1 1 1 1 1 1 0 07 1 1 1 1 1 1 06 1 1 1 1 0 05 1 1 1 1 04 1 1 0 03 1 1 02 0 01 0

14.7 What determines the maximum number of differentalleles that can exist for a given gene?

14.8 What is the difference between a pair of homoallelesand a pair of heteroalleles?

14.9 Two different inbred varieties of a particular plantspecies have white flowers. All other varieties of this specieshave red flowers. What experiments might be done to obtainevidence to determine whether the difference in flower colorin these varieties is the result of different alleles of a singlegene or the result of genetic variation in two or more genes?

14.10 The amber mutants of phage T4 are conditional lethalmutants. They grow on E. coli strain CR63 but are lethal onE. coli strain B. An amber mutant almost never exhibits intra-genic complementation with any other amber mutant; forthis problem, assume that no intragenic complementationoccurs between any of the mutants involved. The followingresults were obtained when eight amber mutants were ana-lyzed for complementation by infecting the restrictive host(E. coli strain B) with each possible pair of mutants. The re-sults of mixed infections by pairs of mutants are shown as 0if no progeny are produced and as 1 if progeny phage re-sulted from the infection with that particular pair of mu-tants.

Mutant: 1 2 3 4 5 6 7 8

8 1 1 1 1 1 1 0 07 1 1 1 1 1 1 06 1 1 1 1 1 05 0 1 0 1 04 1 1 1 03 0 1 02 1 01 0

(a) These data indicate that the eight amber mutations arelocated in how many different genes?

(b) Which mutations are located in the same gene orgenes?

14.11 Considering only base-pair substitutions, how manydifferent mutant homoalleles can occur at one site in a gene?

14.12 Are the following statements concerning the geneticelement referred to as the gene true or false?

(a) The classical (pre–1940) conception of the gene wasthat it was (1) a unit of physiological function or ex-pression, (2) the smallest unit that could undergo mu-tation, and (3) a unit not subdivisible by recombina-tion.

(b) In bacteria, the cis-trans test provides an operationaldefinition by which we usually can identify a gene asthe unit that specifies one mRNA molecule.

(c) Our present knowledge of the structure of the gene in-dicates that the units defined by criteria (2) and (3) instatement (a) above are both equivalent to a single nu-cleotide pair.

(d) Studies in the 1940s demonstrated the existence of het-eroalleles, clearly indicating that many mutations thatwere allelic by the functional criterion could be sepa-rated by recombination, and thereby indicating that the

Chpt 14 9/10/96 3:16 PM Page 27


units of function, mutation, and recombination are notequivalent.

(e) Homoalleles are functionally and structurally allelic;heteroalleles are functionally allelic but structurallynonallelic.

14.13 The rosy (ry) gene of Drosophila encodes the enzymexanthine dehydrogenase; the active form of xanthine dehy-drogenase is a dimer containing two copies of the rosy geneproduct. Mutations ry2 and ry42 are both located within theregion of the rosy gene that encodes the rosy polypeptidegene product. However, ry2/ry42 trans heterozygotes havewild-type eye color. How can the observed complementa-tion between ry2 and ry42 be explained given that these twomutations are located in the same gene?

14.14 Both temperature-sensitive (ts) mutant alleles and am-ber (am) mutant alleles have been identified and studied formany of the genes of bacteriophage T4. Different ts muta-tions within the same gene are frequently found to comple-ment each other, whereas different am mutations within thesame gene practically never complement one another. Whyis this difference to be expected?

14.15 Suppressor-sensitive (sus) mutants of bacteriophagelambda can grow on E. coli strain C600 but cannot grow (thatis, are lethal) on E. coli strain W3350. In other words, sus mu-tants are conditional-lethal mutants. Seven sus mutants wereanalyzed for complementation by simultaneously infectingthe restrictive host (E. coli strain W3350) with each possiblepair of mutants. Single infections with each mutant and withwild-type lambda were also done as controls. The results ofthese complementation or trans tests and the controls aregiven as progeny per infected cell in the accompanyingtable. Several infections with wild-type lambda yielded 120to 150 progeny phage per infected cell. The results of the cisheterozygote controls are not given, but assume that all ofthe cis heterozygotes yielded over 100 progeny phage per in-fected cell. Also assume that no intragenic complementationoccurs between any of these sus mutants.

Lambda Progeny per Infected Cell

Mutant: 1 2 3 4 5 6 7

7 0.01 133 0.01 146 134 128 0.016 131 142 161 0.06 0.1 0.15 120 126 134 0.05 <0.014 147 129 134 0.063 <0.01 147 <0.012 170 <0.011 0.02

(a) Based on the above data, how many genes are definedby the seven sus mutants?

(b) Which sus muations are located in the same gene(s)?

14.16 Is the number of potential alleles of a gene directlyrelated to the number of nucleotide pairs in the gene? Is sucha relationship more likely to occur in prokaryotes or in eu-karyotes? Why?

14.17 In Drosophila, white, eosin, and carnation are all sex-

linked recessive mutations affecting eye color. A white-eyedfemale crossed with a carnation-eyed male produced white-eyed male progeny and red-eyed (wild-type) female off-spring. A white-eyed female crossed with an eosin-eyedmale produced white-eyed sons and light eosin-eyed daugh-ters. Based on these data, which of the three mutations(white, eosin, and carnation), if any, are located in the samegene(s)?

14.18 Suppressor-sensitive (sus) mutants of bacteriophagef29 can grow on Bacillus subtilis strain L15 but cannot grow(that is, are lethal) on B. subtilis strain 12A. Wild-type (sus+)f29 phage can reproduce on both strains, L15 and 12A. Thusthe f29 sus mutants are conditional lethal mutants like theamber mutants of bacteriophage T4. Seven different sus mu-tants of phage f29 were analyzed for complementation bysimultaneously infecting the restrictive host (B. subtilis strain12A) with each possible pair of mutants. Single infectionswith each of the mutants and with wild-type f29 were alsodone as controls. The results of these complementation ortrans tests and the controls are given as progeny phage perinfected cell in the accompanying table. Several infectionsperformed with wild-type f29 phage yielded 300 to 400progeny phage per infected cell. The results of the cis con-trols are not given, but assume that all of the cis heterozy-gote controls yielded over 300 progeny phage per infectedcell. Also assume that no intragenic complementation occursbetween any of the sus mutants studied.


Mutant 1 2 3 4 5 6 7

7 0.01 384 0.01 371 347 333 0.016 341 301 351 0.06 329 0.15 386 326 322 367 <0.014 327 398 374 0.063 <0.01 387 <0.012 354 <0.011 0.02

(a) Based on these data, how many genes are identified bythe seven sus mutants?

(b) Which sus mutations are located in the same gene(s)?

14.19 Assume that the mutants described in Problem 14.18had yielded the following results.


Mutant 1 2 3 4 5 6 7

7 0.01 0.01 0.01 0.03 <0.01 0.01 0.016 0.08 0.09 0.05 0.06 0.1 0.15 0.02 <0.01 <0.01 0.04 <0.014 0.05 0.06 0.03 0.063 <0.01 <0.01 <0.012 0.01 <0.011 0.02

How many genes would they have defined? Which muta-tions would have been in the same gene(s)?

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14.20 The recessive mutations bl (black) and e (ebony) inDrosophila both produce flies with black bodies rather thangray bodies like wild-type flies. Mapping studies showedthat bl is located on chromosome 2, whereas e is on chromo-some 3. When homozygous bl/bl flies are crossed with ho-mozygous e/e flies, the heterozygous bl/e progeny havegray bodies. The observed complementation indicates thatthe two mutations are in two different genes. Was it neces-sary to perform a complementation test to conclude that thebl and e mutations were located in two different genes? If so,why? If not, why not?

14.21 Why was it necessary to modify Beadle and Tatum’sone gene–one enzyme concept of the gene to one gene–onepolypeptide?

14.22 In their analysis of gene function, Beadle and Tatumused Neurospora as an experimental organism, whereas Gar-rod had studied gene function in humans. What advantagesdoes Neurospora have over humans for such studies?

14.23 Based on the information provided in Figure 14.11,(a) are mutations trpA3 and trpA33 heteroalleles or homoal-leles? (b) Are mutations trpA78 and trpA78 heteroalleles orhomoalleles?

14.24 Based on the information given in Figure 14.11, whatis the maximum number of nucleotide pairs separating mu-tations trpA78 and trpA78?

14.25 Arthur Chovnick and colleagues have mapped alarge number of recessive mutations that produce fruit flieswith rose-colored eyes in the homozygous state. They alsohave performed complementation tests on these ry (rosy)mutations. Heterozygotes that carried mutations ry42 andry406 in the trans configuration had wild-type eyes, whereastrans heterozygotes that harbored ry5 and ry41 had rose-col-ored eyes. The results of two- and three-factor crosses unam-biguously demonstrated that mutations ry42 and ry406 bothmap between mutations ry5 and ry41. How can these resultsbe explained?

14.26 The sequences of nucleotide pairs that encode hu-man antibody chains are usually referred to as gene seg-ments rather than genes. Why?

14.27 Tropomyosins are proteins that mediate the interac-

tions between actin and troponin and regulate muscle con-tractions. In Drosophila, six different tropomyosins that havesome amino acid sequences in common, but differ in othersequences, are encoded by two tropomyosin genes (TmI andTmII). How can two genes encode six different polypep-tides?

14.28 In Drosophila, car (carnation) and g (garnet) are sex-linked mutations that produce brown eyes, in contrast to thedark red eyes of wild-type flies. The g and car mutationsmap at positions 44.4 and 62.5, respectively, on the linkagemap of the X chromosome. Is a complementation testneeded to determine whether these two mutations are in thesame gene or two different genes? If so, why? If not, whynot?

14.29 The loz (lethal on Z) mutants of bacteriophage X areconditional lethal mutants that can grow on E. coli strain Ybut cannot grow on E. coli strain Z. The results shown in thefollowing table were obtained when seven loz mutants wereanalyzed for complementation by infecting E. coli strain Zwith each possible pair of mutants. A + indicates that prog-eny phage were produced in the infected cells, and a 0 indi-cates that no progeny phage were produced. All possible cistests were also done, and all cis heterozygotes producedwild-type yields of progeny phage.

Mutant 1 2 3 4 5 6 7

7 1 1 0 1 0 0 06 1 1 1 1 1 05 1 1 0 1 04 0 0 1 03 1 1 02 0 01 0

Given that intragenic complementation does not occur be-tween any of the seven loz mutants analyzed here, (a) pro-pose four plausible explanations for the apparently anom-alous complementation behavior of loz mutant number 7. (b)What simple genetic experiments can be used to distinguishbetween the four possible explanations? (c) Explain whyspecific outcomes of the proposed experiments will distin-guish between the four possible explanations.

BIBLIOGRAPHYBEADLE, G. W., AND E. L. TATUM. 1942. Genetic control of bio-

chemical reactions in Neurospora. Proc. Natl. Acad. Sci.USA 27:499–506.

CARLSON, E. A. 1966. The Gene: A Critical History. W. B. Saun-ders, Philadelphia.

FINCHAM, J.R.S. 1966. Genetic Complementation. Benjamin,Menlo Park, CA.

GARROD, A. E. 1909. Inborn Errors of Metabolism. Oxford Uni-versity Press, New York. (Reprinted in H. Harris, 1963,Garrod’s Inborn Errors of Metabolism, Oxford Mono-graphs on Medical Genetics, Oxford University Press,London.)

OLIVER, C. P. 1940. A reversion to wild type associated withcrossing over in Drosophila melanogaster. Proc. Natl. Acad.Sci. USA 26:452–454.

STURTEVANT, A. H. 1965. A History of Genetics. Harper & Row,New York.

YANOFSKY, C., AND V. HORN. 1972. “Tryptophan synthetase achain positions affected by mutations near the ends ofthe genetic map of trpA of Escherichia coli.” J. Biol. Chem.247:4494–4498.

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