Complementation and Gene Function: From Beadle and Tatum to Benzer

Cross two pure-breeding fruit fly lines that both look white-eyed, and the offspring can have ordinary red eyes if the whiteness has a different cause in each.

If whiteness were one thing, that could not happen. It happens because a red eye needs pigment, pigment is made in several steps by the products of several genes, and losing any one step leaves the eye pale. Two white-eyed strains may have lost different steps. This lesson asks two questions that follow. When many mutants share one phenotype, how many genes have been hit? And when the genes act in a chain, in what order?

The idea: one phenotype can hide many genes

Think of an assembly line with five stations. A broken station anywhere stops the product, so every breakdown looks the same from the end of the line. If one line has a broken station 2 and another a broken station 4, borrowing working stations from each gives a working line. If both broke station 2, borrowing helps nothing.

A few terms first. A locus is a gene’s place on a chromosome; an allele is one version of a gene. A recessive allele shows its effect only when no working allele is present, so a recessive mutant must be homozygous (two copies of the mutant allele) in a diploid. A mutation that inactivates a gene is usually recessive, because one working copy makes enough product.

Mapping, from linkage and mapping and three-point mapping, tells you where mutations sit. It does not tell you whether two nearby mutations damage the same function. For that you need a test of function.

The complementation test: put two mutations in one cell

Combine the two mutations in one diploid, one on each homologous chromosome. This arrangement is called trans. Then look at the phenotype.

If the mutations are in different genes, a and b, the diploid is a/+ ; b/+. Each chromosome set supplies the working allele the other lacks, and the phenotype is wild type. The mutations complement. If they are alleles of one gene, the diploid is a1/a2, with no working copy of that gene, and it stays mutant. The two white sweet peas in extending Mendel that gave purple hybrids are complementation in action.

Mutations that fail to complement one another form a complementation group, which in the simple model is one gene.

The test has conditions. Both mutations must be recessive, which you check first by crossing each to wild type. A dominant allele gives a mutant phenotype opposite any partner, so the trans result says nothing; a dominant-negative allele, whose product spoils the wild-type product, will look like “same gene” falsely. There are also exceptions with recessive alleles. In proteins built of several identical subunits, two differently damaged subunits can sometimes rebuild a working enzyme, called intragenic complementation. In human argininosuccinate lyase each active site holds residues from three subunits, which explains how this can happen. And mutations in different genes sometimes fail to complement.

Worked problem: how many genes? Six recessive white-eyed mutants of an idealized fly (1 to 6) are crossed in all fifteen pairs, and the F1 is scored (+ is wild-type F1, − is white). Assume each mutation is recessive, autosomal and fully penetrant.

1x2 +, 1x3 −, 1x4 +, 1x5 −, 1x6 +, 2x3 +, 2x4 +, 2x5 +, 2x6 −, 3x4 +, 3x5 −, 3x6 +, 4x5 +, 4x6 +, 5x6 +.

1. List the failures only: 1x3, 1x5, 2x6, 3x5.
2. Start with mutant 1. It fails with 3 and 5, so {1, 3, 5} is a group. Check the inside pair: 3x5 fails, as it must if all three are in one gene.
3. Mutant 2 fails only with 6: group {2, 6}. Check that neither 2 nor 6 fails with 1, 3 or 5. Neither does.
4. Mutant 4 complements all five others: group {4}.

Three complementation groups, so at least three genes. “At least”, because other genes needed for red pigment may simply not have been hit in these six mutants.

Beadle and Tatum: one mutant, one blocked step

The idea that a gene acts by controlling a chemical step is older than the experiments that proved it. In 1902, in the Lancet, the physician Archibald Garrod described alkaptonuria, in which urine darkens on standing because it holds homogentisic acid. He noted that it often appeared in children of parents who were cousins, the pattern expected for a recessive Mendelian trait, and he described it as an alternative course of metabolism. The recessive Mendelian reading of the cousin data came from William Bateson. The missing enzyme, homogentisate dioxygenase, was identified later.

George Beadle and Edward Tatum, at Stanford, turned the question around. Rather than start with a trait and hunt its chemistry, they started with known chemistry. The bread mold Neurospora grows on a simple minimal medium and makes its amino acids and most vitamins itself. In its growing phase it is haploid, one copy of each chromosome, so a recessive mutation shows at once. The design had four steps.

1. Irradiate the mold with X-rays.
2. Grow single-spore cultures on a complete medium, rich in amino acids and vitamins, so that a mutant unable to make one of them stays alive.
3. Test each culture on minimal medium. Keep the ones that grow on complete medium but not on minimal.
4. Add supplements one at a time to minimal medium. The single compound that restores growth names the synthesis that was lost. Then cross the mutant to wild type and check that the defect segregates as one gene.

Their 1941 paper reported three mutants among about 2,000 strains, needing pyridoxine (vitamin B6), the thiazole half of thiamine or para-aminobenzoic acid. The pyridoxine and thiamine mutants were Neurospora sitophila, and the para-aminobenzoic acid mutant N. crassa. Crosses of the pyridoxine mutant fit a single-gene difference, on what the authors called limited data. Beadle and Tatum shared the 1958 Nobel Prize with Joshua Lederberg.

The summary “one gene, one enzyme” came later; it is not the wording of the 1941 paper. Today you would say that a gene encodes a polypeptide or a functional RNA. Some genes encode only RNA, alternative splicing gives one gene several proteins, and an enzyme of several different subunits needs several genes.

Ordering a pathway from rescue data

In what order do the steps run? Adrian Srb and Norman Horowitz answered it for arginine in 1944. The compounds involved are the ones you met in the urea cycle in amino acid catabolism: ornithine gains a carbamoyl group to become citrulline, and citrulline is converted to arginine.

The rule is simple. A supplement rescues a mutant only if it enters the pathway after the block, because the mutant can still carry out every later step (see enzyme action for why one missing enzyme stops a pathway).

Worked problem: put the steps in order. Three classes of arginine-requiring Neurospora mutants are tested on minimal medium plus one supplement. Assume each mutant blocks one step and that uptake of every supplement is normal.

• Class A grows with ornithine, citrulline or arginine.
• Class B grows with citrulline or arginine.
• Class C grows only with arginine.

1. Arginine rescues every class, so it is the end product.
2. Citrulline rescues A and B but not C, so C’s block lies after citrulline. Ornithine rescues only A, so ornithine comes earlier than citrulline.
3. The order is precursor → ornithine → citrulline → arginine.
4. A is blocked before ornithine, B between ornithine and citrulline, and C between citrulline and arginine.
5. A blocked step leaves its substrate with nowhere to go, so class C is expected to accumulate citrulline.

The real data were less tidy. Srb and Horowitz had fifteen arginine-requiring strains. Heterokaryon tests and crosses showed at least seven were genetically different, yet the seven they described fell into these same three classes, so one rescue class need not be one gene. Under their one-gene, one-step model they inferred at least seven steps from sugar and ammonia to arginine. In double mutants, the block nearer arginine set the requirement, just as an assembly line broken at two stations needs parts that enter after the later one.

Benzer: function versus fine structure

A complementation test asks whether two mutations damage the same function. A recombination test asks whether the two sites can be separated by crossing over. Seymour Benzer, working at Purdue in the 1950s, showed how differently the two can answer.

His system was the rII region of the phage T4. rII mutants form large plaques on E. coli strain B but cannot grow on K12S(λ), a K-12 strain that carries the phage lambda. That host difference gives two tests.

• Function. Coinfect K12S(λ) with two rII mutants. If they lie in different functional units, each phage supplies what the other lacks and the cells release progeny. Benzer’s 1955 paper found that this test split the rII region into two independent segments. By 1959 he called such a unit, defined by the cis-trans test, a cistron.
• Recombination. Coinfect B, where all types grow, then plate the progeny on B to count everything and on K12S(λ) to count only wild-type recombinants. Each crossover between the two sites gives one wild-type and one double-mutant chromosome, and the double mutant cannot be seen on K12S(λ), so you double the wild-type fraction.

recombination frequency = 2 × (wild-type recombinants per mL) / (total progeny per mL)

Because only recombinants grow on the selective host, Benzer could count very rare ones and map many sites inside one cistron. Two mutations can fail to complement, because they are in one gene, and still recombine, because they sit at different points within it. A zero count has limits, though: finding no recombinants in a finite sample sets only an upper limit on the distance, and Benzer noted that fine-scale frequencies were not always additive.

Design problem

Two T4 rII mutants, rX and rY, each fail to grow on E. coli K12S(λ). Coinfecting K12S(λ) with rX and rY gives no growth. The cis control (an rX rY double mutant plus wild type) grows. Crossing rX and rY on E. coli B and plating the progeny gives 40 plaques per 0.1 mL on K12S(λ) at a 10^-3 dilution, and 200 plaques per 0.1 mL on B at a 10^-6 dilution. Each parent alone gives no plaques on K12S(λ). Interpret the results and give the recombination frequency.