Chapter 9
Genetics of DNA Function
Key Concepts
The one-gene–one-enzyme hypothesis implies that genes control the structure of proteins.
Studies on hemoglobin demonstrated that a gene mutation causes an altered amino acid
sequence in a protein.
The linear sequence of nucleotides in a gene determines the linear sequence of amino acids in
a protein.
Fine structural analysis of the rII genes in phage T4 showed that the gene consists of a linear
array of subelements that can mutate and recombine with one another; these subelements were
later correlated with nucleotide pairs.
A gene can be defined as a unit of function by a complementation test.
Introduction
What is the nature of the gene, and how do genes control phenotypes? For example, how can
one allele of a gene produce a wrinkled pea and another produce a round, smooth pea? Today
we realize that all biochemical reactions of a cell are catalyzed by enzymes, which have a
specific three-dimensional configuration that is crucial to their function. We now know that
genes specify the structures of proteins, some of which are enzymes, and we can even relate
the structure of the genetic material to the structure of proteins. Table 9-1 summarizes our
current model of the relation between genotype and phenotype.
How did we arrive at this point? By following two lines of inquiry:
1. What is the physical structure of the genetic material?
2. How does the genetic material exert its effect? (Or, in detail, how does the structure
function?)
Chapter 8 recounted the demonstration that DNA is the genetic material and detailed the
unraveling of the structure of DNA. In Chapters 9 and 10, we examine how genes function.
How genes work
The first clues about the nature of primary gene function came from studies of humans. Early
in the twentieth century, Archibald Garrod, a physician, noted that several hereditary human
defects are produced by recessive mutations. Some of these defects can be traced directly to
metabolic defects that affect the basic body chemistry—an observation that led to the notion
of “inborn errors” in metabolism. For example, phenylketonuria, which is caused by an
autosomal recessive allele, results from an inability to convert phenylalanine into tyrosine.
Consequently, phenylalanine accumulates and is spontaneously converted into a toxic
compound, phenylpyruvic acid. In a different example, the inability to convert tyrosine into
the pigment melanin produces an albino. Garrod's observations focused attention on metabolic
control by genes.
One-gene–one-enzyme hypothesis
Clarification of the actual function of genes came from research in the 1940s on Neurospora
by George Beadle and Edward Tatum, who later received a Nobel Prize for their work. Before
we describe their actual experiments, let's jump ahead and examine some aspects of
biosynthetic pathways, on the basis of our current understanding. We now know that
molecules are synthesized as a series of steps, each one catalyzed by an enzyme. For instance,
a biosynthetic pathway might have four steps, where 1 is the starting material and 5 is the
final product:
Each step is catalyzed by an enzyme: A, B, C, or D. In turn, each enzyme is specified by a
particular gene. We might say that gene A specifies enzyme A, gene B specifies enzyme B,
and so forth. Therefore, if we inactivate the gene responsible for an enzyme, we eliminate one
required step and the pathway is interrupted.
In the following diagram, enzyme B is eliminated owing to a mutation in gene B:
Now the cell cannot carry out the reaction that converts compound 2 into compound 3. It is
blocked at compound 2, and cannot go further. But what happens if we add back different
intermediate compounds? Suppose that, for example, we feed the cell compound 3 or 4. Can
the cell then synthesize the final product, 5? Yes, with either compound 3 or 4 it can, because
subsequent steps are not blocked. What if we add more of compound 1? No, adding
compound 1 will not allow the synthesis of product 5, because one of the subsequent steps is
blocked.
Let's test our understanding of the concept of biosynthetic pathways by looking at a sample
problem that illustrates this principle:
Let's assume that we have isolated five mutants, 1 to 5, that cannot synthesize compound G
for growth. We know the five compounds, A to E, that are required in the biosynthetic
pathway, but we do not know the order in which they are synthesized by the wild-type cell.
We have tested each compound for its ability to support the growth of each mutant, with the
following results (a plus sign means growth and a minus sign means no growth):
1. What is the order of compounds A to E in the pathway? How do we approach this type of
problem? First, let's work out the underlying principle and then draw a diagram. The main
points are: a mutation blocks a biosynthetic pathway by canceling out one enzyme needed in
the pathway. The mutant therefore lacks one compound needed in the pathway and thus
cannot make any of the compounds that come after the blocked step. If we add a compound
that is normally synthesized before the block, it will not help matters: the mutant will still be
unable to synthesize the rest of the compounds in the pathway. However, the mutant does
have the enzymes to make these later compounds; thus, if we add either the blocked
compound or any compound that comes after the block, the mutant will grow. So, in our
testing of mutants with blocks at different steps in the pathway, the compounds that are used
latest in the pathway will support the growth of the most mutants, and compounds that are
used earliest in the pathway will allow the growth of the fewest mutants. A look at our table
shows us that the compound supporting the fewest mutants is compound E, with no mutants
supported, followed by A (one mutant), then by C, B, D, and G, the final product. Now we
can construct our diagram.
2. At which point in the pathway is each mutant blocked? Clearly, a mutant that is blocked
between E and A cannot be supported by E but can be supported by all the other compounds.
Thus, we see that mutant 5 must be blocked in the E–A conversion. We also see that mutant 4
cannot be supported by E or A, so it must be blocked in the A–C conversion. By similar logic,
we obtain the order 5–4–2–1–3, which we can insert into the diagram as follows:
Now we can comprehend how Beadle and Tatum first worked out this experiment, using one
particular biosynthetic pathway of Neurospora.
Experiments of Beadle and Tatum
Beadle and Tatum analyzed mutants of Neurospora, a fungus with a haploid genome. They
first irradiated Neurospora to produce mutations and then tested cultures from ascospores for
interesting mutant phenotypes. They detected numerous auxotrophs—strains that cannot
grow on a minimal medium unless the medium is supplemented with one or more specific
nutrients. In each case, the mutation that generated the auxotrophic requirement was inherited
as a single-gene mutation: each gave a 1:1 ratio when crossed with a wild type. Figure 9-1
depicts the procedure that Beadle and Tatum used.
One set of mutant strains required arginine to grow on a minimal medium. These strains
provided the focus for much of Beadle and Tatum's further work. First, they found that the
mutations mapped into three different locations on separate chromosomes, even though the
same supplement (arginine) satisfied the growth requirement for each mutant. Let's call the
three loci the arg-1, arg-2, and arg-3 genes. Beadle and Tatum discovered that the auxotrophs
for each of the three loci differed in their response to the chemical compounds ornithine and
citrulline, which are related to arginine (Figure 9-2). The arg-1 mutants grew when supplied
with ornithine, citrulline, or arginine in addition to the minimal medium. The arg-2 mutants
grew on either arginine or citrulline but not on ornithine. The arg-3 mutants grew only when
arginine was supplied. We can see this more easily by looking at Table 9-2.
It was already known that cellular enzymes often interconvert related compounds such as
these. On the basis of the properties of the arg mutants, Beadle and Tatum and their
colleagues proposed a biochemical model for such conversions in Neurospora:
Note how this relation easily explains the three classes ofmutants shown in Table 9-2. The
arg-1 mutants have a defective enzyme X, so they are unable to convert the precursor into
ornithine as the first step in producing arginine. However, they have normal enzymes Y and Z,
and so the arg-1 mutants are able to produce arginine if supplied with either ornithine or
citrulline. The arg-2 mutants lack enzyme Y, and the arg-3 mutants lack enzyme Z. Thus, a
mutation at a particular gene is assumed to interfere with the production of a single enzyme.
The defective enzyme, then, creates a block in some biosynthetic pathway. The block can be
circumvented by supplying to the cells any compound that normally comes after the block in
the pathway.
We can now diagram a more complete biochemical model:
Note that this entire model was inferred from the properties of the mutant classes detected
through genetic analysis. Only later were the existence of the biosynthetic pathway and the
presence of defective enzymes demonstrated through independent biochemical evidence.
This model, which has become known as the one-gene–one-enzyme hypothesis, was the
source of the first exciting insight into the functions of genes: genes somehow were
responsible for the function of enzymes, and each gene apparently controlled one specific
enzyme. Other researchers obtained similar results for other biosynthetic pathways, and the
hypothesis soon achieved general acceptance. It was subsequently refined, as we shall see
later in this chapter. Nevertheless, the one-gene–one-enzyme hypothesis became one of the
great unifying concepts in biology, because it provided a bridge that brought together the
concepts and research techniques of genetics and biochemistry.
MESSAGE
Genes control biochemical reactions by producing enzymes.
We should pause to ponder the significance of this discovery. Let's summarize what it
established:
1. Biochemical reactions in vivo (in the living cell) consist of a series of discrete, step-by-step
reactions.
2. Each reaction is specifically catalyzed by a single enzyme.
3. Each enzyme is specified by a single gene.
Gene-protein relations
The one-gene–one-enzyme hypothesis was an impressive step forward in our understanding
of gene function, but just how do genes control the functioning of enzymes? Virtually all
enzymes are proteins, and thus we must review the basic facts of protein structure to follow
the next step in the study of gene function.
Protein structure
In simple terms, a protein is a macromolecule composed of amino acids attached end to end
in a linear string. The general formula for an amino acid is H2N−CHR−COOH, in which the
side chain, or R (reactive) group, can be anything from a hydrogen atom (as in the amino acid
glycine) to a complex ring (as in the amino acid tryptophan). There are 20 common amino
acids in living organisms (Table 9-3), each having a different R group. Amino acids are
linked together in proteins by covalent (chemical) bonds called peptide bonds. A peptide bond
is formed through a condensation reaction that includes the removal of a water molecule
(Figure 9-3).
Several amino acids linked together by peptide bonds form a molecule called a polypeptide;
the proteins found in living organisms are large polypeptides. For instance, the α chain of
human hemoglobin contains 141 amino acids, and some proteins consist of more than 1000
amino acids.
The properties of the amino acid side chains are responsible for the structure and function of
each protein. These side chains vary in many chemical properties. A key property is the
hydrophobic, or water repelling, character of each amino acid. Hydrophobic amino acids tend
to avoid contact with water and are often turned toward the inside of the protein, whereas the
amino acids that are charged or that can form hydrogen bonds with water are excluded from
the interior of the protein and are turned toward the exterior or surface of the protein.
Proteins have a complex structure that is traditionally thought of as having four levels. The
linear sequence of the amino acids in a polypeptide chain is called the primary structure of
the protein. Figure 9-4 shows the linear sequence of tryptophan synthetase (an enzyme) and
beef insulin (a hormonal protein).
The secondary structure of a protein refers to the interrelations of amino acids that are close
together in the linear sequence. This spatial arrangement often results from the fact that
polypeptides can bend into regularly repeating (periodic) structures, created by hydrogen
bonds between the CO and NH groups of different residues. Two of the basic periodic
structures are the α helix (Figure 9-5) and the β pleated sheet (Figure 9-6).
A protein also has a three-dimensional architecture, termed the tertiary structure, which is
created by electrostatic, hydrogen, and Van der Waals bonds that form between the various
amino acid R groups, causing the protein chain to fold back on itself. In many cases, amino
acids that are far apart in the linear sequence are brought close together in the tertiary
structure. Often, two or more folded structures will bind together to form a quaternary
structure; this structure is multimeric because it is composed of several separate polypeptide
chains, or monomers.
Figure 9-7 depicts the four levels of protein structure. In Figure 9-7c, we can see the tertiary
structure of myoglobin. Note how the α helix is folded back on itself to generate the
three-dimensional shape of the protein. Figure 9-7d shows the combining of four subunits
(two α chains and two β chains) to form the quaternary structure of hemoglobin. Figure 9-8
shows the structure of myoglobin in more detail. The combining of subunits to form a
multimeric enzyme can be seen directly in the electron microscope in some cases (Figure
9-9).
Many proteins are compact structures; such proteins are called globular proteins. Enzymes
and antibodies are among the important globular proteins. Other, unfolded proteins, called
fibrous proteins, are important components of such structures as hair and muscle.
MESSAGE
The linear sequence of a protein folds up to yield a unique three-dimensional
configuration. This configuration creates specific sites to which substrates bind and at
which catalytic reactions take place. The three-dimensional structure of a protein,
which is crucial for its function, is determined solely by the primary structure (linear
sequence) of amino acids. Therefore, genes can control enzyme function by
controlling the primary structure of proteins.
Protein motifs
Often, several elements of secondary structure combine to produce a pattern, or motif, that is
found in numerous other proteins. We can recognize motifs sometimes by their amino acid
sequence pattern and other times by observing the three-dimensional structure. Figure 9-10
shows two examples. The helix-loop-helix motif is found in calcium binding proteins, and a
variant of it is found in regulatory proteins that bind DNA. The zinc-binding motif, also found
in DNA binding proteins, is termed the zinc finger, because of the way that the residues
protrude outward, like a finger.
Determining protein sequence
If we purify a particular protein, we find that we can specify a particular ratio of the various
amino acids that make up that specific protein. But the protein is not formed by a random
hookup of fixed amounts of the various amino acids; each protein has a unique, characteristic
sequence. For a small polypeptide, the amino acid sequence can be determined by clipping off
one amino acid at a time and identifying it. However, large polypeptides cannot be readily
“sequenced” in this way.
Frederick Sanger worked out a brilliant method for deducing the sequence of large
polypeptides. There are several different proteolytic enzymes—enzymes that can break
peptide bonds only between specific amino acids in proteins. Proteolytic enzymes can break a
large protein into a number of smaller fragments, which can then be separated according to
their migration speeds in a solvent on chromatographic paper. Because different fragments
will move at different speeds in various solvents, two-dimensional chromatography can be
used to enhance the separation of the fragments (Figure 9-11). In this technique, a mixture of
fragments is separated in one solvent; then the paper is turned 90° and another solvent is used.
When the paper is stained, the polypeptides appear as spots in a characteristic
chromatographic pattern called the fingerprint of the protein. Each of the spots can be cut out,
and the polypeptide fragments can be washed from the paper. Because each spot contains
only small polypeptides, their amino acid sequences can be easily determined.
Using different proteolytic enzymes to cleave the protein at different points, we can repeat the
experiment to obtain other sets of fragments. The fragments from the different treatments
overlap, because the breaks are made in different places with each treatment. The problem of
solving the overall sequence then becomes one of fitting together the small-fragment
sequences—almost like solving a tricky jigsaw or crossword puzzle (Figure 9-12).
Using this elegant technique, Sanger confirmed that the sequence of amino acids (as well as
the amounts of the various amino acids) is specific to a particular protein. In other words, the
amino acid sequence is what makes insulin insulin.
Relation between gene mutations and altered proteins
We now know that the change of just one amino acid is sometimes enough to alter protein
function. This was first shown in 1957 by Vernon Ingram, who studied the globular protein
hemoglobin—the molecule that transports oxygen in red blood cells. As shown in Figure 9-7d,
hemoglobin is made up of four polypeptide chains: two identical α chains, each containing
141 amino acids, and two identical β chains, each containing 146 amino acids.
Ingram compared hemoglobin A (HbA), the hemoglobin from normal adults, with
hemoglobin S (HbS), the protein from people homozygous for the mutant gene that causes
sickle-cell anemia, the disease in which red blood cells take on a sickle-cell shape (see Figure
4-2). Using Sanger's technique, Ingram found that the fingerprint of HbS differs from that of
HbA in only one spot. Sequencing that spot from the two kinds of hemoglobin, Ingram found
that only one amino acid in the fragment differs in the two kinds. Apparently, of all the amino
acids known to make up a hemoglobin molecule, a substitution of valine for glutamic acid at
just one point, position 6 in the β chain, is all that is needed to produce the defective
hemoglobin (Figure 9-13). Unless patients with HbS receive medical attention, this single
error in one amino acid in one protein will hasten their death. Figure 9-14 shows how this
gene mutation ultimately leads to the pattern of sickle-cell disease.
Notice what Ingram accomplished. A gene mutation that had been well established through
genetic studies was connected with an altered amino acid sequence in a protein. Subsequent
studies identified numerous changes in hemoglobin, and each one is the consequence of a
single amino acid difference. (Figure 9-15 shows a few examples.) We can conclude that one
mutation in a gene corresponds to a change of one amino acid in the sequence of a protein.
MESSAGE
Genes determine the specific primary sequences of amino acids in specific proteins.
Colinearity of gene and protein
Once the structure of DNA had been determined by Watson and Crick, it became apparent
that the structure of proteins must be encoded in the linear sequence of nucleotides in the
DNA. (We shall see in Chapter 10 how this genetic code was deciphered.) After Ingram's
demonstration that one mutation alters one amino acid in a protein, a relation was sought
between the linear sequence of mutant sites in a gene and the linear sequence of amino acids
in a protein. (It is possible to map mutational sites within a gene by high-resolution
recombination analysis, as we saw in the rII system, described in Chapter 7 and expanded
later in this chapter.)
Charles Yanofsky probed the relation between altered genes and altered proteins by studying
the enzyme tryptophan synthetase in E. coli. This enzyme catalyzes the conversion of indole
glycerol phosphate into tryptophan. Two genes, trpA and trpB, control the enzyme. Each gene
controls a separate polypeptide; after the A and B polypeptides are produced, they combine to
form the active enzyme (a multimeric protein). Yanofsky analyzed mutations in the trpA gene
that resulted in alterations of the tryptophan synthetase A subunit. He produced a detailed
map of the mutations, and he then determined the amino acid sequence of each respective
altered tryptophan synthetase. His results were similar to Ingram's for hemoglobin: each
mutant had a defective polypeptide associated with a specific amino acid substitution at a
specific point. However, Yanofsky was able to show an exciting correlation that Ingram was
not able to observe, owing to the limitations of his system. Yanofsky found an exact match
between the sequence of the mutational sites in the gene map of the trpA gene and the
location of the corresponding altered amino acids in the A polypeptide chain. The farther
apart two mutational sites were in map units, the more amino acids there were between the
corresponding substitutions in the polypeptide (Figure 9-16). Thus, Yanofsky demonstrated
colinearity—the correspondence between the linear sequence of the gene and that of the
polypeptide. Figure 9-17 shows the complete set of data.
MESSAGE
The linear sequence of nucleotides in a gene determines the linear sequence of amino
acids in a protein.
X-ray determination of threedimensional structure of proteins
X-ray crystallography is a powerful method for determining the three-dimensional structure
of a protein in atomic detail. John Kendrew first applied this method to a molecule as
complex as a protein to elucidate the structure of myoglobin in 1957, and Max Perutz
succeeded in unraveling the complexities of hemoglobin several years later. Now, the
structures of hundreds of proteins are known. In this technique, crystals of the protein are
obtained in a concentrated salt solution of the pure protein. Then, a narrow beam of X rays is
passed through the crystal. The repeating pattern of atoms in the protein complex scatters (or
diffracts) the X-ray beams, giving a pattern of spots on X-ray film, as depicted in Figure
9-18a (left); see also Figure 9-18b. Information about the electron density in different parts of
the protein is contained in the position and intensity of each spot.
Sophisticated mathematical analysis is used to generate electron-density maps (Figure 9-17a;
right), which are in turn used to derive contour maps of the protein. Ultimately, detailed
models of a protein can be built, as shown in the space-filling model of chymotrypsin (Figure
9-17c). Amazingly, this model stems from the pattern of spots seen in Figure 9-17b.
Enzyme function
How can a single amino acid substitution, such as that in sickle-cell hemoglobin (Figure 9-13),
have such an enormous effect on protein function and the phenotype of an organism? Take
enzymes, for example. Enzymes are known to do their job of catalysis by physically grappling
with their substrate molecules, twisting or bending the molecules to make or break chemical
bonds. Figure 9-19 shows the gastric digestion enzyme carboxypeptidase in its relaxed
position and after grappling with its substrate molecule, glycyltyrosine. The substrate
molecule fits into a notch in the enzyme structure; this notch is called the active site.
Figure 9-20 diagrams the general concept. Note that there are two basic types of reactions
performed by enzymes: (1) the breakdown of a substrate into simpler products and (2) the
synthesis of a complex product from one or more simpler substrates.
Much of the globular structure of an enzyme is nonreactive material that simply supports the
active site. So we might expect that amino acid substitutions throughout most of the structure
would have little effect, whereas very specific amino acids would be required for the part of
the enzyme molecule that gives the precise shape to the active site. Hence, the possibility
arises that a functional enzyme does not always require a unique amino acid sequence for the
entire polypeptide. This possibility has proved to be the case: in a number of systems,
numerous positions in a polypeptide can be filled by several alternative amino acids, and
enzyme function is retained. But, at certain other positions in the polypeptide, only the
wild-type amino acid will preserve activity; in all likelihood, these amino acids form critical
parts of the active sites. Some of these critical amino acids in carboxypeptidase are indicated
in red in Figure 9-19.
MESSAGE
Protein architecture is the key to gene function. A gene mutation typically results in
the substitution of a different amino acid into the polypeptide sequence of a protein.
The new amino acid may have chemical properties that are incompatible with the
proper protein architecture at that particular position; in such a case, the mutation will
lead to a nonfunctional protein.
Genes and cellular metabolism: genetic diseases
When we think of enzyme activity in relation to cellular metabolism, we realize that the
inactivation of one or more enzymes can have staggering consequences. Most of us have been
amazed by the charts on laboratory walls showing the myriad interlocking, branched, and
circular pathways along which the cell's chemical intermediates are shunted like parts on an
assembly line. Bonds are broken, molecules cleaved, molecules united, groups added or
removed, and so forth. The key fact is that almost every step, represented by an arrow on the
metabolic chart, is controlled (mediated) by an enzyme, and each of these enzymes is
produced under the direction of a gene that specifies its function. Change one critical gene,
and the entire assembly line can break down.
Humans provide some startling examples. The list in Table 9-4 gives some representative
examples and suggests the magnitude of genetic involvement in human disease. Figure 9-21
shows a corner of the human metabolic map to illustrate how a set of diseases, some of them
common and familiar to us, can stem from the blockage of adjacent steps in biosynthetic
pathways.
Genetic observations explained by enzyme structure
Armed with an understanding of the gene-protein relation and how enzymes function, we can
now reexamine some of the genetic findings presented in earlier chapters and look at them in
regard to the biochemistry.
A good example can be found in temperaturesensitive alleles. Recall that some mutants
appear to be wild type at normal temperatures but can be detected as mutants at high or low
temperatures. We now know that such mutations result from the substitution of an amino acid
that produces a protein that is functional at normal temperatures, called permissive
temperatures, but is distorted and nonfunctional at high or low temperatures, called
restrictive temperatures (Figure 9-22).
As we have seen, conditional mutations such as temperature-sensitive mutations can be very
useful to geneticists. Stocks of the mutant culture can be easily maintained under permissive
conditions, and the mutant phenotype can be studied intensively under restrictive conditions.
Such mutants can be very handy in the genetic dissection of biological systems. For example,
with a temperature-sensitive allele, we can shift to a restrictive temperature at various times in
the course of development to determine the time at which a gene is active.
Genetic fine structure
Until the 1950s, our genetic and cytological analysis led to the concept of the chromosome as
a linear (one-dimensional) array of genes, strung rather like beads on an unfastened necklace.
Indeed, this model is sometimes called the bead theory. According to the bead theory, the
existence of a gene as a unit of inheritance is recognized through its mutant alleles. A mutant
allele affects a single phenotypic character, maps to one chromosome locus, gives a mutant
phenotype when paired, and shows a Mendelian ratio when intercrossed. Several tenets of the
bead theory are worth emphasizing:
1. The gene is viewed as a fundamental unit of structure, indivisible by crossing-over.
Crossing-over takes place between genes (the beads in this model) but never within them.
2. The gene is viewed as the fundamental unit of change, or mutation. It changes in toto from
one allelic form into another; there are no smaller components within it that can change.
3. The gene is viewed as the fundamental unit of function (although the precise function of the
gene is not specified in this model). Parts of a gene, if they exist, cannot function.
Yet, how can we reconcile the fact that the gene consists of a series of nucleotides with the
view that the gene is the smallest unit of mutation and recombination? Seymour Benzer's
work in the 1950s showed that the bead theory was not correct. Benzer demonstrated that,
whereas a gene can be defined as a unit of function, a gene can be subdivided into a linear
array of sites that are mutable and that can be recombined. The smallest units of mutation and
recombination are now known to be correlated with single nucleotide pairs.
Intragenic recombination
Benzer used the rII system, described in Chapter 7, to detect very low levels of recombination
in phage T4. Benzer started with an initial sample of eight independently derived rII mutant
strains, crossing them in all possible combinations of pairs by the double infection of E. coli B
and subsequent plating onto a lawn of E. coli K. Using recombinant frequencies, he could
map the mutations unambiguously to the right or the left of each other to get what we now
call a gene map (in this case, the map units are the frequency of rII+ plaques):
Recombination within a gene, called intragenic recombination, seems to be the rule rather
than the exception. It can virtually always be found at any locus if a suitable selection system
is available to detect recombinants. In other words, a mutant allele can be pictured as a length
of genetic material (the gene) that contains a damaged or non-wild-type part, a mutational
site. This partial damage is what causes the non-wild-type phenotype. Different alleles
produce different phenotypic effects because damage is at different sites within the wild-type
allele.
Thus, an allele a1 can be represented as
where the asterisk (*) is the mutant site within an otherwise normal gene (denoted by the sites
marked +). A cross between a1 and another mutant allele, a2 , can be represented as
and it is easy to see how the normal, wild-type gene
could be generated by a simple crossover anywhere between the two mutational sites.
Benzer showed that, contrary to the classical view, genes were not indivisible but could be
subdivided by recombination. Extending his analysis to hundreds of rII alleles, Benzer found
that the minimal recombinant frequency in a cross between a pair of different mutant alleles
was 0.01 percent, even though his analytical system was capable of detecting recombinant
frequencies as low as 0.0001 percent if they occurred. This led to the idea that genes were
composed of small units and that recombination could take place between but not within these
units. We now know that the single nucleotide pair is the smallest unit of recombination.
MESSAGE
A gene is composed of subelements that can recombine. The smallest subunit not
divisible by recombination is the single nucleotide pair.
Recombination within genes permits us to construct detailed gene maps, such as the map for
the rII region shown in Figure 9-23. Making crosses between various mutations provides us
with the relative frequencies of intragenic recombinants, and these recombinants reveal the
order and relative positions of the mutational sites within a gene. It should be noted that
recombination within genes is the same as recombination between genes, except that the scale
is different.
Mutational sites
Benzer exploited the properties of partial deletions to rapidly map new point mutations to
specific segments within the rII gene, a procedure called deletion mapping.
Using deletions in mapping mutational sites
Deletions are mutations that result from the elimination of segments of DNA. Deletions can
be intercrossed and mapped just like point mutations. The deleted region is represented
diagrammatically by a bar. If no wild-type recombinants are produced in a cross between
different deletions, then the bars are shown as overlapping. A typical deletion map might be
as follows:
Such deletion maps are useful in delineating regions of the gene to which new point mutations
can be assigned. The logic is that a mutation cannot give wild-type recombinants in a cross to
deletion in which the DNA corresponding to the wild-type region for that particular mutation
is no longer present. Therefore, mutations can be ordered against a set of deletions by rapid
tests that do not depend on extensive quantitative measurements.
Deletion mapping of the rII region
Benzer carried out the analysis in the following way. For example, consider the following
gene map, showing 12 identifiable mutational sites:
Let us suppose that one mutant, D1 , fails to give rII+ recombinants when crossed with
mutants carrying altered sites 1, 2, 3, 4, 5, 6, 7, or 8; therefore, D1 behaves as if it has a
deletion of sites 1 to 8:
Another mutant, D2 , fails to give rII+ recombinants when crossed with 5, 6, 7, 8, 9, 10, 11, or
12; therefore, D2 behaves as if it has a deletion of sites 5 to 12:
These overlapping deletions now define three areas of the gene. Let's call them i, ii, and iii:
A new mutant that gives rII+ recombinants when crossed with D1 but not when crossed with
D2 must have its mutational site in area iii. One that gives rII+ recombinants when crossed
with D2 but not with D1 must have its mutational site in area i. A new mutant that does not
give rII+ recombinants with either D1 or D2 must have its mutational site in area ii. For
example, assume that a mutant in area iii is crossed with D1 . We would envision the cross
schematically as follows, where the mutant site is shown with a red bar:
and draw the actual pairing as follows:
The more deletions there are in the tester set, the more areas can be uniquely designated and
the more rapidly new mutational sites can be located (Figure 9-24). When assigned to a region,
a mutation can be mapped against other alleles in the same region to obtain an accurate
position. Figure 9-25 shows the complexity of Benzer's actual map.
Analysis of mutational sites
The use of deletions enabled Benzer to define the topology of the gene—the manner in which
the parts are interconnected. His genetic experiments showed the gene to consist of a linear
array of mutable subelements. Benzer's next step was to examine the topography of the
gene—differences in the properties of the subelements. Operationally, he determined these
differences by asking whether all the subelements or sites were equally mutable or whether
mutations were prevalent at some sites and rare at others. For this study, it was essential to
work with the smallest mutable subelements possible. Instead of multisite mutations
(deletions) that exhibited no reversion, Benzer used revertible mutations, because they
probably represented small alterations, or point mutations. Additionally, he discarded mutants
with high reversion rates, because high reversion interferes with recombination tests.
Benzer used his deletion mutants to rapidly map the set of point mutations. He first localized
each mutation into short deletion segments and then crossed all the point mutations within a
segment against one another; any two revertible mutations that failed to recombine with each
other represented mutations at the same site.
Figure 9-26 shows the distribution of 1612 spontaneous mutations in the rII locus. In Benzer's
own words, “That the distribution is not random leaps to the eye.” This extraordinary
nonrandom distribution demonstrates that all sites are not equally mutable. Benzer termed
sites that are more mutable than other sites hot spots. The most prominent hot spot was
represented by more than 500 repeated occurrences in the collection of 1612 mutations. By
examining a Poisson distribution calculated to fit the number of sites having only one or two
occurrences (Figure 9-27), Benzer could show that at least 60 sites were truly more mutable
than those with only one or two occurrences. Mutations were not observed at all (by chance)
in at least 129 sites in this collection, even though these sites were as mutable as those
represented by one or two occurrences. When Benzer extended the analysis to include
mutagen-induced mutations, results were similar to the spontaneous mutation analysis: there
were also hot spots, although often different ones from the spontaneous hot spots, among the
mutations generated by mutagenic agents (see Chapter 16).
The size of a mutational site also was of interest. The physical size of the rII region can be
used to calculate the approximate number of base pairs in the region. From this figure, the
number of mutational sites was determined to be approximately one-fifth of the number of
nucleotide pairs. In other words, the smallest mutable site was five nucleotide pairs or less.
The deciphering of the genetic code (Chapter 10), together with the demonstrations by Ingram,
Yanofsky, and others (described earlier in this chapter) that single amino acid substitutions
resulted from single mutations, allowed Benzer to conclude that a mutation could result from
the alteration of a single nucleotide pair. (The direct sequencing of DNA, described in
Chapter 12, has since confirmed these conclusions in many examples.)
MESSAGE
The gene can be divided into a linear array of mutable subelements that correspond to
individual nucleotide pairs.
Destruction of the bead theory
Let's look again at the bead theory in light of Benzer's work. With the aid of deletion mapping,
Benzer was able to map an extraordinary number of mutations in the rII locus against one
another. His experiments showed that mutations in the same gene can indeed recombine with
one another. This result contradicts the bead theory of classical genetics, which held that
recombination could take place between genes but not within genes.
Benzer's analysis of the fine structure of the gene demonstrated that each gene consists of a
linear array of subelements and that these sites within a gene can be altered by mutation and
can undergo recombination. This finding also contradicts the bead theory, one tenet of which
implies that only the gene as a whole is mutable, not parts of the gene.
Subsequent work by several investigators identified each genetic site as a base pair in
double-stranded DNA. Therefore, Benzer's contribution bridged the gulf between classical
genetics and the knowledge of the chemical structure of DNA revealed by Watson and Crick.
According to the bead theory, the Watson-Crick structure made no sense. However, Benzer's
demonstration that genes do indeed have fine structures that can be revealed solely by genetic
analysis allowed a fusion of the two disciplines and helped to launch the modern era of
molecular genetics. Figure 9-28 illustrates the fine structure analysis of the rII locus and its
correspondence with the DNA structure.
Complementation
In another part of his studies, Benzer carried out a series of experiments designed to define
the gene in regard to function. Benzer studied the concept of complementation; he wanted to
find out whether the entire rII region of phage T4 acts as a single functional unit or whether it
is made up of subunits that function independently. Therefore, he tested the mutations that he
had mapped in the rII region to see whether various combinations of pairs of the mutations
would restore the wild-type phenotype. In other words, Benzer looked for complementation in
E. coli host cells that were temporarily “diploid” for the T4 chromosome. To do so, he carried
out a mixed infection with different rII mutants (Figure 9-29). His criterion for the wild-type
phenotype was the ability to lyse E. coli K hosts. (Recall that rII mutants cannot do so but that
wild-type phages can.)
Complementation tests such as the one conducted by Benzer are carried out in one cycle of
infection; they do not require the multiple cycles of reinfection needed for plaque formation.
Samples of the two phages to be tested are spread over a strip of host bacteria on a section of
a petri plate at a high ratio of phages to bacteria to ensure that essentially every bacterium is
infected with both phages. After a period of incubation, the growth or the absence of growth
of the bacteria in the strip indicates whether the bacteria have lysed as a result of the phage
infection.
It is important to understand the difference between recombination and complementation.
Recombination represents the creation of new combinations of genes through the physical
breakage and rejoining of chromosomes. The progeny from a cross in which recombination
has occurred have new genotypes that are different from the parental genotypes.
Complementation, on the other hand, does not involve any change in the genotypes of
individual chromosomes; rather it represents the mixing of gene products. Complementation
occurs during the time that two chromosomes are in the same cell and can each supply a
function. Afterward, each respective chromosome remains unaltered. In the case of rII
mutants, complementation occurs when two different phage chromosomes, with mutations in
different rII genes, are in the same host cell. However, progeny that result from this
complementation carry only the parental genotypes.
Tests of many different paired mutants allowed Benzer to separate the mutations into two
groups, labeled A and B. All mutations in the A group complemented those in the B group,
whereas no mutations in the A group complemented any other mutations in the A group and
no mutations in the B group complemented any other mutations in the B group. Benzer found
that all mutations in group A mapped in one half of the rII locus and that all mutations in
group B mapped in the other half of the rII locus (Figure 9-30).
What did the results of Benzer's complementation test mean? We learned in Chapter 4 that
lack of complementation is the diagnostic for mutations being in the same gene. Hence it must
be concluded that the groupings rIIA and rIIB must represent different genes, located next to
each other on the chromosome. Benzer called the rIIA and rIIB complementation groups
cistrons. We now equate cistrons with genes and recognize that the functional units of
heredity are genes. The term cistron, which was popular for some years, is now rarely
used.
Summary
The work of Beadle and Tatum in the 1940s showed that one gene codes for one protein,
often an enzyme. When an enzyme fails to function normally owing to a mutation, a variant
phenotype results. These variant phenotypes are often the basis of genetic diseases in any
organism, including humans. To understand how abnormal enzymes can cause phenotypic
change, we need to understand the structure of proteins. Composed of a specific linear
sequence of amino acids connected through peptide bonds, the proteins assume specific
three-dimensional shapes as a result of the interaction of the 20 amino acids that in different
combinations constitute the polypeptide chain. Different areas of this folded chain are sites for
the attachment and interaction of substrates. Furthermore, many functional enzymes and other
proteins are built by combining several polypeptide chains into a multimeric form.
Specific amino acid changes can be detected in a protein by the technique of fingerprinting
and amino acid sequencing. Work of this type by Sanger, Ingram, Yanofsky, and others has
demonstrated colinearity between a mutational site on the genetic map and the position of an
altered amino acid in a protein.
Further work by Benzer and others illustrated that the gene could be dissected into smaller
and smaller pieces. A structural unit of mutation and one of recombination were identified
and equated with a single nucleotide pair. A gene (formerly termed cistron) is defined at the
phenotypic level as a genetic region within which there is no complementation between
mutations. This unit codes for the structure of a single functional polypeptide.
In the early days of genetics, genes were represented as indivisible beads on a chain. We now
have a far different picture of the gene. Multiple intragenic mutational sites exist, and
recombination may take place anywhere within a gene. In addition, a closer connection
between genotype and phenotype was realized when it was established that one gene is
responsible for the synthesis of one polypeptide.
Solved Problems
1. Various pairs of rII mutants of phage T4 are tested in E. coli in both the cis (= coupling)
and trans (= repulsion) conformations, and the “burst size” (the average number of phage
particles produced per bacterium) for each test pair is compared. Results for six different r
mutants—rM, rN, rO, rP, rR, and rS—are as follows (+ indicates the wild-type allele):
If we assign the mutation rO to the A gene, what are the locations of the other five
mutations with respect to the A and B genes? (NOTE: This problem is similar to Problem 21
in the problem set for this chapter.)
See answer
Solution
In solving problems such as this one, we first look for the underlying principle and then
attempt to draw a diagram to help us work out the solution. The key principle is that, in the
trans position, mutations in the same gene will not complement one another and thus will
yield no progeny phage (no burst), whereas mutations in different genes will complement and
thus will yield progeny phages of normal burst size. Because rO is the A gene, we start with
the following initial diagram:
We can now look at the test results in the problem and assign each mutation to a gene on the
basis of whether complementation occurs in a mixed infection with rO mutants. From the
table, it is evident that N and P complement O and must be in a different gene (the B gene)
whereas M does not complement O and must be in the same gene (the A gene). This leaves us
with R and S, which can be assigned to the A and B genes, respectively, on the basis of their
complementation tests with M, N, and P. Therefore, we have the arrangement
We can check this assignment by examining the other crosses to find out if the results are
consistent with our answer.
2. The following deletion map shows four deletions (1 to 4) in the rIIA gene of phage T4:
Five point mutations (a to e) are tested against these four deletion mutants for their ability
(+) or inability (−) to give wild-type (r+) recombinants; the results are
What is the order of the point mutations?
See answer
Solution
The key principle here is that point mutations can recombine with deletions that do not
extend past the mutation, but they cannot recombine to yield wild-type phages with deletions
that do extend past the mutation. For the test results given in the problem, any mutation that
recombines with deletion 1 must be to the right of the deletion, any mutation that recombines
with deletion 2 must be to the right of deletion 2, and so forth. Let's look at point mutation a.
It recombines with deletions 1, 2, and 3 but not with deletion 4. Therefore, it is to the right of
deletions 1, 2, and 3 but not to the right of deletion 4. We can therefore place point mutation
a in the interval between deletions 3 and 4. Point mutation b recombines with deletions 1 and
2 and must be to the right of them. It does not recombine with deletions 3 and 4, so it is in the
interval between deletion 2 and 3. Point mutation c recombines with all the deletions and is
to the right of all deletions, even deletion 4. Finally, point mutations d and e recombine only
with deletion 1 and must therefore be in the interval between deletions 1 and 2. The solution
that we have just derived can be summarized as follows:
Problem 25 in the following problem set is similar to this sample problem. See if you can
apply the reasoning set forth here when you solve Problem 25.
Problems
1. Describe the primary, secondary, tertiary, and quaternary structures of a protein. Give
examples of each type of structure, where possible.See answer
2. What is the one-gene–one-enzyme hypothesis?
3. What is an auxotroph?
See answer
4. What is the exact change in a protein that results in sickle-cell anemia?
5. What experiment did Yanofsky do to demonstrate colinearity between the gene and the
protein?
See answer
6. What is the difference between complementation and recombination?
7. Complete the following table by filling in the plaque types:
See answer
8. A common weed, St.-John's-wort, is toxic to albino animals. It also causes blisters on
animals that have white areas of fur. Suggest a possible genetic basis for this reaction.
9. In humans, the disease galactosemia causes mental retardation at an early age because
lactose in milk cannot be broken down, and this failure affects brain function. How would
you provide a secondary cure for galactosemia? Would you expect this phenotype to be
dominant or recessive?
See answer
10. Aminocentesis is a technique in which a hypodermic needle is inserted through the
abdominal wall of a pregnant woman and into the amnion, the sac that surrounds the
developing embryo, to withdraw a small amount of amniotic fluid. This fluid contains
cells that come from the embryo (not from the woman). The cells can be cultured; they
will divide and grow to form a population of cells on which enzyme analyses and
karyotypic analyses can be performed. Of what use would this technique be to a genetic
counselor? Name at least three specific conditions under which amniocentesis might be
useful. (Note: This technique entails a small but real risk to the health of both the woman
and the embryo; take this fact into account in your answer.)
11. The table below shows the ranges of enzymatic activity (in units that we need not worry
about) observed for enzymes having roles in two recessive human metabolic diseases.
Similar information is available for many metabolic genetic diseases.
a. Of what use is such information to a genetic counselor?
b. Indicate any possible sources of ambiguity in interpreting studies of an individual
patient.
c. Reevaluate the concept of dominance in the light of such data.
See answer
12. Two albinos marry and have a normal child. How is this possible? Suggest at least two
ways.
See answer
13. In humans, PKU (phenylketonuria) is a disease caused by an enzyme inefficiency at step
A in the following simplified reaction sequence, and AKU (alkaptonuria) is due to an
enzyme inefficiency in one of the steps summarized as step B here:
A person with PKU marries a person with AKU. What phenotypes do you expect for their
children? All normal, all having PKU only, all having AKU only, all having both PKU
and AKU, or some having AKU and some having PKU.
14. Three independently isolated tryptophan-requiring strains of yeast are called TrpB, TrpD,
and TrpE. Cell suspensions of each strain are streaked on a plate supplemented with just
enough tryptophan to permit weak growth for a Trp− strain. The streaks are arranged in a
triangular pattern so that they do not touch one another. Luxuriant growth is noted at both
ends of the TrpE streak and at one end of the TrpD streak.
*a.
Do you think that complementation is involved?
b. Briefly explain the patterns of luxuriant growth.
c. In what order in the tryptophan-synthesizing pathway are the enzymatic steps defective
in TrpB, TrpC, and TrpE?
d. Why was it necessary to add a small amount of tryptophan to the medium to
demonstrate such a growth pattern?
See answer
15. In Drosophila pupae, certain structures called imaginal disks can be detected as
thickenings of the skin; after metamorphosis, these imaginal disks develop into specific
organs of the adult fly. George Beadle and Boris Ephrussi devised a means of
transplanting eye imaginal disks from one larva into another larval host. When the host
metamorphoses into an adult, the transplant can be found as a colored eye located in its
abdomen. The researchers took two strains of flies that were phenotypically identical in
their bright scarlet eyes: one owing to the sex-linked mutant vermilion (v); the other owing
to cinnabar (cn) on chromosome 2. If v disks are transplanted into v hosts or cn disks into
cn hosts, then the transplants develop as mutant scarlet eyes. Transplanted cn or v disks in
wild-type hosts develop wild-type eye colors. A cn disk in a v host develops a mutant eye
color, but a v disk in a cn host develops wild-type eye color. Explain these results, and
outline the experiments that you would propose to test your explanation.
16. In Drosophila, the autosomal recessive bw causes a dark-brown eye, and the unlinked
autosomal recessive st causes a bright scarlet eye. A homozygote for both genes has a
white eye. Thus, we have the following correspondences between genotypes and
phenotypes:
Construct a hypothetical biosynthetic pathway showing how the gene products interact
and why the different mutant combinations have different phenotypes.
17. Several mutants are isolated, all of which require compound G for growth. The
compounds (A to E) in the biosynthetic pathway to G are known but their order in the
pathway is not known. Each compound is tested for its ability to support the growth of
each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign
indicates no growth:
a. What is the order of compounds A to E in the pathway?
b. At which point in the pathway is each mutant blocked?
c. Would a heterokaryon composed of double mutants 1,3 and 2,4 grow on a minimal
medium? 1,3 and 3,4? 1,2 and 2,4 and 1,4?
18. In Neurospora (a haploid), assume that two genes participate in the synthesis of valine.
Their mutant alleles are val-1 and val-2, and their wild-type alleles are val-1+ and val-2+.
The two genes are linked on the same chromosome, and a crossover takes place between
them, on average, in one of every two meioses.
a. In what proportion of meioses are there no crossovers between the genes?
b. Use the map function to determine the recombinant frequency between these two genes.
c. Progeny from the cross val-1 val-2+ × val-1+val-2 are plated on a medium containing no
valine. What proportion of the progeny will grow?
d. The val-1 val-2+ strains accumulate intermediate compound B, and the val-1+val-2
strains accumulate intermediate compound A. The val-1 val-2+ strains grow on valine or
A, but the val-1+val-2 strains grow only on valine and not on B. Show the pathway order
of A and B in relation to valine, and indicate which gene controls each conversion.
See answer
19. In a certain plant, the flower petals are normally purple. Two recessive mutations arise in
separate plants and are found to be on different chromosomes. Mutation 1 (m1) gives blue
petals when homozygous (m1m1). Mutation 2 (m2) gives red petals when homozygous
(m2m2). Biochemists working on the synthesis of flower pigments in this species have
already described the following pathway:
a. Which mutant would you expect to be deficient in enzyme A activity?
b. A plant has the genotype M1/m1 ; M2/m2. What would you expect its phenotype to be?
c. If the plant in part b is selfed, what colors of progeny would you expect, and in what
proportions?
d. Why are these mutants recessive?
20. In sweet peas, the synthesis of purple anthocyanin pigment in the petals is controlled by
two genes, B and D. The pathway is
a. What color petals would you expect in a pure-breeding plant unable to catalyze the first
reaction?
b. What color petals would you expect in a pure-breeding plant unable to catalyze the
second reaction?
c. If the plants in parts a and b are crossed, what color petals will the F1 plants have?
d. What ratio of purple:blue:white plants would you expect in the F2?
See answer
21. Various pairs of rII mutants of phage T4 are tested in E. coli in both the cis (coupling) and
trans (repulsion) conformations. Comparisons are made of the “burst size” (the average
number of phage particles produced per bacterium). Results for six different r
mutants—rU, rV, rW, rX, rY, and rZ—are as follows:
If we assign rV to the A gene, what are the locations of the other five rII mutations with
respect to the A and B genes? (Problem 21 from M. Strickberger, Genetics. Copyright
1968 by Monroe W. Strickberger. Reprinted with permission of Macmillan Publishing
Co., Inc.)
See answer
22. There is evidence that occasionally in meiosis either one or both homologous centromeres
will divide and segregate precociously at the first division rather than at the second
division (as is the normal situation). In Neurospora, pan2 alleles produce a pale ascospore,
aborted ascospores are completely colorless, and normal ascospores are black. In a cross
between two complementing mutations pan2x × pan2y, what ratios of black:pale:colorless
would you expect in asci resulting from the precocious division of (a) one centromere? (b)
both centromeres? (Assume that pan2 is near the centromere.)
2 Protozoon mirabilis is a hypothetical single-celled haploid green alga that orients to light
3. by means of a red eyespot. By the selection of cells that do not move toward the light, 14
white-eyespot mutants (eye−) are isolated after mutation. It is possible to fuse haploid cells
to make diploid individuals. The 14 eye− mutants are paired in all combinations, and the
color of the eyespot is scored in each. The table (see top of next page) shows the results,
where a plus sign indicates a red eyespot and a minus sign indicates a white eyespot.
a. Mutant 14 is obviously different from the rest. Why might this be?
b. Excluding mutant 14, how many complementation groups are there, and which mutants
are in which group?
c. Three crosses are made, with the following results:
Explain these genetic ratios with symbols.
d. How many genetic loci take part altogether, and which of the 14 mutants are at each
locus?
e. What is the linkage arrangement of the loci? (Draw a map.)
24. You have the following map of the rII locus:
You detect a new mutation, rW, and you find that it does not complement any of the
mutants in the A or B gene. You find that wild-type recombinants are obtained in crosses
with rX, rY, rE, and rF but not with rZ or rD. Suggest possible explanations for these
results. Describe tests that you would use to choose between the explanations.
25. The following map shows four deletions (1 to 4) in the rIIA gene of phage T4:
Five point mutations (a to e) in rIIA are tested against these four deletion mutants for their
ability (+) or inability (−) to give r+ recombinants, with the following results:
a. What is the order of the point mutants?
b. Another strain of T4 has a point mutation in the rIIB gene. This strain is mixed in turn
with each of the rIIA deletion mutants, and the mixtures are used to infect E. coli K at a
multiplicity of infection great enough that each host cell will be infected by at least one
rIIA and one rIIB mutant. A normal plaque is formed with deletions 1, 2, and 3, but no
plaque forms with deletion 4. Given that the B gene is to the right of A, explain the
behavior of deletion 4. Does your explanation affect your answer to part a?
26. In a phage, a set of deletions is intercrossed in paired combinations. The following results
are obtained (a + indicates that wild-type recombinants are obtained from that cross):
a. Construct a deletion map from this table.
b. The first geneticists to do a deletion-mapping analysis in the mythical schmoo phage
SH4 (which lyses schmoos) came up with this unique set of data:
Show why this is a unique result by drawing the only deletion map that is compatible with
this table. (Do not let your mind be shackled by conventional expectations.)
27. In a haploid eukaryote, four alleles of the cys-2 gene are obtained. Each allele requires
cysteine, and all alleles map to the same locus. The four strains bearing these mutant
alleles are crossed to wild types to obtain a set of eight cultures representing the four
mutant alleles in association with each mating type. Then the mutant alleles are
intercrossed in all possible combinations of pairs. The haploid meiotic products from each
cross are plated on a medium containing no cysteine. In some crosses, cys+ prototrophs are
observed at low frequencies. The results follow, where the numbers represent the
frequencies of cys+ colonies per 104 meiotic products plated:
a. Draw a map of the four mutant sites within the cys-2 gene. Provide a measurement of
the relative intersite distances.
b. Do you see any evidence that mutation might have a role in the production of the
prototrophs?
*28. In Neurospora, there is a gene controlling the production of adenine, and mutants in this
gene are called ad-3 mutants. The his-2 locus is 2 m.u. to the left, and the nic-2 locus is 3
m.u. to the right of the ad-3 locus (his-2 controls histidine, and nic-2 controls nicotinamide).
Thus, the genetic map is:
Three different ad-3 auxotrophs are detected: ad-3a, ad-3b, and ad-3c. (Use a, b, and c as their
labels.) The following crosses are made:
The ascospores are then plated on a minimal medium containing histidine and nicotinamide,
and ad-3+ prototrophs are picked up. The table shows the results obtained. What is the map
order of the ad-3 mutants and the genetic distance between them?
See answer
29. In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more
than 50 units of “square factor” per cell will produce a square phenotype and less than 50
units will produce a round phenotype. Allele sf is a functional gene that causes the
synthesis of the square factor. Each sf allele contributes 40 units of square factor; thus, sf/sf
homozygotes have 80 units and are phenotypically square. A mutant allele (sn) arises; it is
nonfunctional, contributing no square factor at all.
a. Which allele will show dominance, sf or sn?
b. Are functional alleles necessarily always dominant? Explain your answer.
c. In a system such as this one, how might a specific allele become changed in evolution
so that its phenotype shows recessive inheritance at generation 0 and dominant inheritance
at a later generation?
See answer
*30. Explain how Benzer was able to calculate the number of sites with zero occurrences, as
depicted in Figure 9-27. (Assume, as Benzer did, that the sites with 0, 1, and 2 occurrences
are a set of sites with equal mutability.)
31. In Collinsia parviflora, the petal color is normally purple. Four recessive mutations are
induced, each of which produces white petals. The four pure-breeding lines are then
intercrossed in the following combinations, with the results indicated:
a. Explain all these results clearly, using diagrams wherever possible.
b. What F1 and F2 do you predict from crosses of 2 × 3 and 2 × 4?
*32. A biologist is interested in the genetic control of leucine synthesis in the haploid
filamentous fungus Aspergillus. He treats spores with mutagen and obtains five point
mutations (a to e), all of which are leucine auxotrophs. He first makes heterokaryons between
them to check on their functional relations. He determines the following results, where a plus
sign indicates that the heterokaryon grew and a minus sign indicates that the heterokaryon
did not grow, on a medium lacking leucine:
The biologist then intercrosses the mutations in all possible combinations. From each cross,
he tests 500 ascospore progeny by inoculating them onto a medium lacking leucine. The
results follow (the numbers indicate the number of leucine prototrophs in the 500 progeny):
Explain both sets of data genetically. (Note that the two leucine prototrophs from the b × d
cross were found not to be due to reversion.)
33. In Drosophila, the eye phenotype “star” is caused by recessive mutations (s) mapping to
one location on the second chromosome. This region is flanked at the left by the A locus
(allele A or a) and at the right by the B locus (allele B or b):
Six independently induced star mutations are each made homozygous, with both A B/A B
and a b/a b constitutions, and the six are intercrossed to study complementation at the star
locus. The results follow, where “+” indicates wild eye and “s” indicates star eye, both
being phenotypes of the F1.
a. How many genes are at the star location, and which mutational sites are in each gene?
b. The heterozygotes in brackets are allowed to produce gametes. In both cases, s+
recombinant gametes are identified. The gametes are tested for the flanking marker
conformation, which is a B in both the 1 × 6 heterozygote and the 2 × 4 heterozygote.
Order the genes in relation to the A/a and B/b loci.
34. In Norway in 1934, a mother with two mentally retarded children consulted the physician
Asbjørn Følling. In the course of the interview, Følling learned that the urine of the
children had a curious odor. He later tested their urine with ferric chloride and found out
that, whereas normal urine gives a brownish color, the children's urine stained green. He
deduced that the chemical responsible must be phenylpyruvic acid. Because of chemical
similarity to phenylalanine, it seemed likely that this substance had been formed from
phenylalanine in the blood, but there was no assay for phenylalanine. However, a certain
bacterium could convert phenylalanine into phenylpyruvic acid so that the level of
phenylalanine could be measured by using the ferric chloride test. It was found that indeed
the children had high levels of phenylalanine in their blood, and it was probably the source
of the phenylpyruvic acid. Følling noted that, in general in families with retarded children
whose urine stained green with ferric chloride, there was a proportion of 1/4 affected
children and 3/4 unaffected. This disease, which came to be known as phenylketonuria,
seemed to have a genetic basis and was inherited as a simple Mendelian recessive.
It became clear that phenylalanine was the culprit and that this chemical built up in the
PKU patients and was converted into high levels of phenylpyruvic acid, which then
interfered with the normal development of nervous tissue. This finding led to the
formulation of a special diet low in phenylalanine, which could be fed to newborn babies
diagnosed with PKU and which allowed normal development to continue without
retardation. Indeed, it was found that, after the child's nervous system had developed, the
patient could be taken off the special diet. However, tragically, PKU women who had
developed normally with the special diet were found to have babies who were born
mentally retarded, and the special diet had no effect on these children.
a. Why do you think the babies of the PKU mothers were born retarded?
b. Explain the reason for the difference in the results of the PKU babies and those of the
babies of PKU mothers.
c. Propose a treatment that might allow PKU mothers to have unaffected children.
d. Write a short essay on PKU integrating concepts at the genetic, diagnostic, enzymatic,
physiological, pedigree, and population levels.
Unpacking the Problem
1. Draw the pedigree of the Norwegian family as far as possible. Do you think that some
siblings might have been unaffected?
2. Could the original children brought to Følling (assume that they were 10 years old)
have benefited from the special diet if started at that age?
3. Why were the bacteria needed by Følling?
4. Why was a blood test considered important?
5. Why was there phenylpyruvic acid in the urine?
6. What is the relation expected between blood concentrations and urine concentrations?
7. Was the green substance in the ferric chloride test itself important?
8. What phenylalanine and phenylpyruvic acid concentrations are expected in the blood
and urine of unaffected children?
9. What was the source of the odor in the children's urine?
10. What were the genotypes of the parents?
11. Would families with parents having these genotypes be common?
12. What genotypes of offspring are expected and in what proportions?
13. What would be the genotypes of parents in most families in the population?
14. Why was it inferred that the disease was inherited?
15. Why was it inferred to be inherited as a Mendelian recessive?
16. At which developmental stage(s) is most of the nervous system developing?
17. Why was it believed that PKU adults no longer needed to be on the special diet?
18. What are the relative volumes of the maternal and fetal blood circulation systems?
19. Which types of entities are able to pass the placental barrier (cells, macromolecules,
small molecules)?
20. Which PKU-associated substances might pass into the mother from the child? into the
child from the mother?
21. What is an essential amino acid?
22. Is phenylalanine an essential amino acid?
23. How does PKU illustrate the one-gene–one-enzyme concept?
24. Why is PKU recessive?
25. What is the relevance of biochemical pathways to PKU?
26. In the chapter, find a chart that shows the position of the PKU-associated chemicals in
the chemistry of the cell.
27. Draw a diagram relating chromosome, gene, mutant site, mRNA, enzyme, substrate,
and product in normal individuals and in PKU sufferers.
Chapter 9*
1. The primary structure of a protein is the sequence of amino acids along its length. It is held
together by covalent bonds. The secondary structure of a protein is caused by hydrogen
bonding between CO and NH groups on different amino acids. Frequently, α helices and β
pleated sheets result. The tertiary structure of a protein is caused by electrostatic, hydrogen,
and Van der Waals bonds between the R groups of amino acids. If the final, functional
protein is composed of more than one polypeptide, then the protein has a quaternary
structure.
3. An auxotroph is a strain that requires at least one nutrient for growth beyond that normally
required for the organism.
5. Yanofsky analyzed mutations in the trpA gene and ordered them by using transduction. He
also determined the amino acid sequence of the altered gene products. By this means, he
demonstrated an exact correlation between the sequence of mutation sites and the sequence
of altered amino acids.
7.
9. Lactose is composed of one molecule of galactose and one molecule of glucose. A
secondary cure would result if all galactose and lactose were removed from the diet. The
disorder would be expected not to be dominant, because one good copy of the gene should
allow for at least some, if not all, breakdown of galactose. In fact, the disorder is recessive.
11. a. The main use is in detecting carrier parents and in diagnosing a disorder in the fetus.
b. Because the values for normal individuals and carriers overlap for galactosemia, there is
ambiguity if a person has from 25 to 30 units. That person could be either a carrier or
normal.
c. These wild-type genes are phenotypically dominant but are incompletely dominant at
the molecular level. A minimal level of enzyme activity apparently is enough to ensure
normal function and phenotype.
12. One less likely possibility is a germ-line mutation. More likely is that each parent was
blocked at a different point in a metabolic pathway. If one were A/A · b/b and the other
were a/a · B/B, then the child would be A/a · B/b and would have sufficient levels of both
resulting enzymes to produce pigment.
14. a. Complementation refers to genes within a cell and is not what is happening here. Most
likely, what is known as cross-feeding is taking place, whereby a product made by one
strain diffuses to another strain and allows growth of the second strain. Cross-feeding is
equivalent to supplementing the medium. Because cross-feeding seems to be taking place,
the suggestion is that the strains are blocked at different points in the metabolic pathway.
b. For cross-feeding to take place, the growing strain must have a block that is earlier in
the metabolic pathway than the block in the strain from which it is obtaining the product
for growth.
c. E–D–B
d. Without some tryptophan, there would be no growth at all, and the cells would not have
lived long enough to produce a product that could diffuse.
18. a. Sixty percent of the meioses will not have a crossover.
b. There are 20 m.u. between the two genes.
c. 10 percent
d.
20. a. White
b. Blue
c. Purple
d.
21. The cis and trans burst size should be the same if the mutants are in different cistrons; if
they are in the same cistron, the trans burst size should be zero. Therefore, if rV is
assumed to be in A, rW also is in A, and rU, rX, rY, and rZ are in B.
28.
29. a. The allele sn will show dominance over sf because there will be only 40 units of square
factor in the heterozygote.
b. Here, the functional allele is recessive.
c. The allele sf may become dominant over time in two ways: (1) it could mutate slightly,
so that it produces more than 50 units, or (2) other modifying genes may mutate to
increase the production of sf.