7 Linkage, Recombination,

and Eukaryotic Gene Mapping

LINKED GENES
AND BALD HEADS

F or many, baldness is the curse of manhood. Twenty-five

percent of men begin balding by age 30 and almost half are
bald to some degree by age 50. In the United States, baldness
affects more than 40 million men, who spend hundreds of mil-
lions of dollars each year on hair-loss treatment. Baldness is not
just a matter of vanity: bald males are more likely to suffer from
heart disease, high blood pressure, and prostate cancer.
Baldness can arise for a number of different reasons,
including illness, injury, drugs, and heredity. The most-com-
mon type of baldness seen in men is pattern baldness—techni-
cally known as androgenic alopecia—in which hair is lost pre-
(a) (b) maturely from the front and top of the head. More than 95%
of hair loss in men is pattern baldness. Although pattern bald-
ness is also seen in women, it is usually expressed weakly as
mild thinning of the hair. The trait is clearly stimulated by male
Pattern baldness is a hereditary trait. sex hormones (androgens), as evidenced by the observation
Recent research demonstrated that a gene that males castrated at an early age rarely become bald (but this
for pattern baldness is linked to genetic
markers located on the X chromosome,
is not recommended as a preventive treatment for baldness).
leading to the discovery that pattern A strong hereditary influence on pattern baldness has long
baldness is influenced by variation in the been recognized, but the exact mode of inheritance has been
androgen-receptor gene. The trait is seen controversial. An early study suggested that it was autosomal
in three generations of the Adams family: dominant in males and recessive in females, an example of a sex
(a) John Adams (1735–1826), the second
president of the United States, was father
influenced trait (see Chapter 4). Other evidence and common
to (b) John Quincy Adams (1767–1848), folklore suggested that a man inherits baldness from his
sixth U.S. president, who was father to mother’s side of the family, exhibiting X-linked inheritance.
(c) Charles Francis Adams (1807–1886). In 2005, geneticist Axel Hillmer and his colleagues set out
[Part a: National Museum of American Art, to locate the gene that causes pattern baldness. They suspected
Washington, D.C./Art Resource, New York.
Part b: National Portrait Gallery, Washington,
that the gene for pattern baldness might be located on the X
D.C./Art Resource, New York. Part c: chromosome, but they had no idea where on the X chromo-
Bettmann/Corbis.] some the gene might reside. To identify the location of the
(c) gene, they conducted a linkage analysis study, in which they
looked for an association between the inheritance of pattern baldness and the inheritance of
genetic variants known to be located on the X chromosome. The genetic variants used in the
study were single-nucleotide polymorphisms (SNPs, pronounced “snips”), which are posi-
tions in the genome where individuals vary in a single nucleotide. The geneticists studied the
inheritance of pattern baldness and SNPs in 95 families in which at least two brothers devel-
oped pattern baldness at an early age.

161
162 Chapter 7

What Hillmer and his colleagues found was that pattern baldness and SNPS from the X
chromosome were not inherited independently, as predicted by Mendel’s principle of inde-
pendent assortment. Instead, they tended to be inherited together, which occurs when genes
are physically linked on the same chromosome and segregate together in meiosis.
As we will learn in this chapter, linkage between genes is broken down by recombination
(crossing over), and the amount of recombination between genes is directly related to the
distance between them. In 1911, Thomas Hunt Morgan and his student Alfred Sturtevant
demonstrated in fruit flies that genes can be mapped by determining the rates of recombina-
tion between the genes. Using this method for the families with pattern baldness, Hillmer and
his colleagues demonstrated that the gene for pattern baldness is closely linked to SNPs
located at position p12–22, a region that includes the androgen-receptor gene, on the X chro-
mosome. The androgen receptor gene encodes a protein that binds male sex hormones.
Given the clear involvement of male hormones in the development of pattern baldness, the
androgen-receptor gene seemed a likely candidate for causing pattern baldness. Further
analysis revealed that certain alleles of the androgen-receptor gene were closely associated
with the inheritance of pattern baldness, and the androgen-receptor gene is almost certainly
responsible for much of the differences in pattern baldness seen in the families examined.
Additional studies conducted in 2008 found that genes on chromosomes 3 and 20 also appear
to contribute to the expression of pattern baldness. TRY PROBLEM 13

T his chapter explores the inheritance of genes located

on the same chromosome. These linked genes do not
strictly obey Mendel’s principle of independent assortment;
tion: it tells us that, in the process of separation, the two alleles
at a locus act independently of alleles at other loci.
The independent separation of alleles results in recom-
rather, they tend to be inherited together. This tendency bination, the sorting of alleles into new combinations.
requires a new approach to understanding their inheritance Consider a cross between individuals homozygous for two
and predicting the types of offspring produced. A critical different pairs of alleles: AA BB × aa bb. The first parent,
piece of information necessary for predicting the results AA BB, produces gametes with the alleles A B, and the sec-
of these crosses is the arrangement of the genes on the ond parent, aa bb, produces gametes with the alleles a b,
chromosomes; thus, it will be necessary to think about the resulting in F1 progeny with genotype Aa Bb (Figure 7.1).
relation between genes and chromosomes. A key to under-
standing the inheritance of linked genes is to make the con-
P generation
ceptual connection between the genotypes in a cross and the
AA BB  aa bb
behavior of chromosomes in meiosis.
We will begin our exploration of linkage by first com- Gamete formation Gamete formation
paring the inheritance of two linked genes with the inheri-
tance of two genes that assort independently. We will then Gametes AB ab
examine how crossing over, or recombination, breaks up
linked genes. This knowledge of linkage and recombination
Fertilization
will be used for predicting the results of genetic crosses in
which genes are linked and for mapping genes. Later in the
chapter, we will focus on physical methods of determining F1 generation
Aa Bb
the chromosomal locations of genes. The final section exam-
ines variation in rates of recombination. Gamete formation

7.1 Linked Genes Do Not Assort

Gametes A B ab Ab aB
Independently
Original combinations of New combinations of
Chapter 3 introduced Mendel’s principles of segregation alleles (nonrecombinant alleles (recombinant
and independent assortment. Let’s take a moment to review gametes) gametes)
these two important concepts. The principle of segregation
states that each individual diploid organism possesses two Conclusion: Through recombination, gametes
contain new combinations of alleles.
alleles at a locus that separate in meiosis, with one allele going
into each gamete. The principle of independent assortment 7.1 Recombination is the sorting of alleles into new
provides additional information about the process of segrega- combinations.
 Linkage, Recombination, and Eukaryotic Gene Mapping 163

Recombination means that, when one of the F1 progeny Experiment

reproduces, the combination of alleles in its gametes may
Question: Do the genes for flower color and pollen shape
differ from the combinations in the gametes from its par-
in sweet peas assort independently?
ents. In other words, the F1 may produce gametes with alleles
A b or a B in addition to gametes with A B or a b. Methods Cross two strains homozygous for two
Mendel derived his principles of segregation and inde- different traits.
pendent assortment by observing the progeny of genetic
P generation
crosses, but he had no idea of what biological processes pro- Homozygous strains
duced these phenomena. In 1903, Walter Sutton proposed a
biological basis for Mendel’s principles, called the chromo- Purple flowers, Red flowers,
long pollen round pollen
some theory of heredity (see Chapter 3). This theory holds
that genes are found on chromosomes. Let’s restate Mendel’s
two principles in relation to the chromosome theory of 
heredity. The principle of segregation states that a diploid
Pollen
organism possesses two alleles for a trait, each of which is
located at the same position, or locus, on each of the two Fertilization
homologous chromosomes. These chromosomes segregate
in meiosis, with each gamete receiving one homolog. The
principle of independent assortment states that, in meiosis, F1 generation
Purple flowers,
each pair of homologous chromosomes assorts indepen-
long pollen
dently of other homologous pairs. With this new perspective,
it is easy to see that the number of chromosomes in most
organisms is limited and that there are certain to be more
genes than chromosomes; so some genes must be present on
the same chromosome and should not assort independently.
Genes located close together on the same chromosome are Self-fertilization
called linked genes and belong to the same linkage group.
Linked genes travel together in meiosis, eventually arriv- Results F2 generation
ing at the same destination (the same gamete), and are not
expected to assort independently.
All of the characteristics examined by Mendel in peas
did display independent assortment and, after the rediscov-
ery of Mendel’s work, the first genetic characteristics studied
in other organisms also seemed to assort independently.
How could genes be carried on a limited number of chro- 284 21 21 55
mosomes and yet assort independently? Purple Purple Red Red
flowers, flowers, flowers, flowers,
The apparent inconsistency between the principle of
long round long round
independent assortment and the chromosome theory of pollen pollen pollen pollen
heredity soon disappeared as biologists began finding genetic
characteristics that did not assort independently. One of the Conclusion: F2 progeny do not appear in the 9  3  3  1
first cases was reported in sweet peas by William Bateson, Edith ratio expected with independent assortment.
Rebecca Saunders, and Reginald C. Punnett in 1905. They
crossed a homozygous strain of peas having purple flowers 7.2 Nonindependent assortment of flower color and pollen
shape in sweet peas.
and long pollen grains with a homozygous strain having red
flowers and round pollen grains. All the F1 had purple flowers
and long pollen grains, indicating that purple was dominant
over red and long was dominant over round. When they 7.2 Linked Genes Segregate Together
intercrossed the F1, the resulting F2 progeny did not appear and Crossing Over Produces
in the 9 : 3 : 3 : 1 ratio expected with independent assortment
(Figure 7.2). An excess of F2 plants had purple flowers and
Recombination Between Them
long pollen or red flowers and round pollen (the parental Genes that are close together on the same chromosome
phenotypes). Although Bateson, Saunders, and Punnett were usually segregate as a unit and are therefore inher-
unable to explain these results, we now know that the two loci ited together. However, genes occasionally switch from
that they examined lie close together on the same chromo- one homologous chromosome to the other through the
some and therefore do not assort independently. process of crossing over (see Chapter 2), as illustrated
164 Chapter 7

Meiosis I Meiosis II

Late Prophase I Metaphase I Anaphase I Metaphase II Anaphase II Gametes

Crossing
over

Recombinant
chromosomes

Genes may switch from a chromosome to In meiosis II, genes that ...will then assort ...and end up in
its homolog by crossing over in meiosis I. are normally linked... independently... different gametes.

7.3 Crossing over takes place in meiosis and is responsible for recombination.

in Figure 7.3. Crossing over results in recombination; it only a single line, with the understanding that genes located on
breaks up the associations of genes that are close together the same side of the line lie on the same chromosome:
on the same chromosome. Linkage and crossing over can A B
be seen as processes that have opposite effects: linkage
a b
keeps particular genes together, and crossing over mixes
them up. In Chapter 5, we considered a number of excep- This notation can be simplified further by separating the
tions and extensions to Mendel’s principles of heredity. alleles on each chromosome with a slash: AB/ab.
The concept of linked genes adds a further complication to Remember that the two alleles at a locus are always
interpretations of the results of genetic crosses. However, located on different homologous chromosomes and there-
with an understanding of how linkage affects heredity, we fore must lie on opposite sides of the line. Consequently, we
can analyze crosses for linked genes and successfully pre- would never write the genotypes as
dict the types of progeny that will be produced. A a
B b
Notation for Crosses with Linkage
because the alleles A and a can never be on the same chromo-
In analyzing crosses with linked genes, we must know not some. It is also important to always keep the same order of the
only the genotypes of the individuals crossed, but also the genes on both sides of the line; thus, we should never write
arrangement of the genes on the chromosomes. To keep
A B
track of this arrangement, we introduce a new system of
notation for presenting crosses with linked genes. Consider b a
a cross between an individual homozygous for dominant because it would imply that alleles A and b are allelic (at the
alleles at two linked loci and another individual homozygous same locus).
for recessive alleles at those loci (AA BB × aa bb). For linked
genes, it’s necessary to write out the specific alleles as they
are arranged on each of the homologous chromosomes:
Complete Linkage Compared
with Independent Assortment
A B a b
We will first consider what happens to genes that exhibit com-
A B a b plete linkage, meaning that they are located very close together
on the same chromosome and do not exhibit crossing over.
In this notation, each line represents one of the two homolo-
Genes are rarely completely linked but, by assuming that no
gous chromosomes. Inheriting one chromosome from each
crossing over occurs, we can see the effect of linkage more
parent, the F1 progeny will have the following genotype:
clearly. We will then consider what happens when genes assort
A B independently. Finally, we will consider the results obtained if
a b the genes are linked but exhibit some crossing over.
A testcross reveals the effects of linkage. For example,
Here, the importance of designating the alleles on each chromo- if a heterozygous individual is test-crossed with a homozy-
some is clear. One chromosome has the two dominant alleles A gous recessive individual (Aa Bb × aa bb), the alleles that
and B, whereas the homologous chromosome has the two reces- are present in the gametes contributed by the heterozygous
sive alleles a and b. The notation can be simplified by drawing parent will be expressed in the phenotype of the offspring,
(a) If genes are completely linked (b) If genes are unlinked
(no crossing over) (assort independently)

Normal Mottled Normal Mottled

leaves, tall leaves, dwarf leaves, tall leaves, dwarf

 

M D m d Mm Dd mm dd

m d m d
Gamete formation Gamete formation Gamete formation Gamete formation

1/2 M D 1/2 m d m d 1/4 M D 1/4 m d 1/4 M d 1/4 m D md

Nonrecombinant Nonrecombinant Recombinant
gametes gametes gametes

Fertilization Fertilization

Normal Mottled Normal Mottled Normal Mottled

leaves, tall leaves, dwarf leaves, tall leaves, dwarf leaves, dwarf leaves, tall

M D m d 1/4 Mm Dd 1/4 mm dd 1/4 Mm dd 1/4 mm Dd

1/2 1/2

m d m d Nonrecombinant Recombinant
progeny progeny
All nonrecombinant progeny
Conclusion: With complete linkage, only Conclusion: With independent assortment, half the
nonrecombinant progeny are produced. progeny are recombinant and half the progeny are not.

7.4 A testcross reveals the effects of linkage. Results of a testcross for two loci in tomatoes that
determine leaf type and plant height.

because the homozygous parent could not contribute domi- M D m d

nant alleles that might mask them. Consequently, traits that P
M D m d
appear in the progeny reveal which alleles were transmitted
by the heterozygous parent.
Consider a pair of linked genes in tomato plants. One of
A
M D
the genes affects the type of leaf: an allele for mottled leaves F1
(m) is recessive to an allele that produces normal leaves (M). m d
Nearby on the same chromosome the other gene determines
The geneticist would then use these F1 heterozygotes in a
the height of the plant: an allele for dwarf (d) is recessive to
testcross, crossing them with plants homozygous for mot-
an allele for tall (D).
tled leaves and dwarf height:
Testing for linkage can be done with a testcross, which
requires a plant heterozygous for both characteristics. A M D m d
geneticist might produce this heterozygous plant by crossing m d m d
a variety of tomato that is homozygous for normal leaves
and tall height with a variety that is homozygous for mottled The results of this testcross are diagrammed in Figure 7.4a.
leaves and dwarf height: The heterozygote produces two types of gametes: some

165
166 Chapter 7

with the M D chromosome and others with the Crossing Over with Linked Genes
m d chromosome. Because no crossing over occurs, Usually, there is some crossing over between genes that lie
these gametes are the only types produced by the heterozy- on the same chromosome, producing new combinations
gote. Notice that these gametes contain only combinations of traits. Genes that exhibit crossing over are incompletely
of alleles that were present in the original parents: either the linked. Let’s see how it takes place.
allele for normal leaves together with the allele for tall height
(M and D) or the allele for mottled leaves together with the Theory The effect of crossing over on the inheritance
allele for dwarf height (m and d). Gametes that contain only of two linked genes is shown in Figure 7.5. Crossing over,
original combinations of alleles present in the parents are which takes place in prophase I of meiosis, is the exchange
nonrecombinant gametes, or parental gametes. of genetic material between nonsister chromatids (see Fig-
The homozygous parent in the testcross produces only ures 2.16 and 2.18). After a single crossover has taken place,
one type of gamete; it contains chromosome m d the two chromatids that did not participate in crossing over
and pairs with one of the two gametes generated by the are unchanged; gametes that receive these chromatids are
heterozygous parent (see Figure 7.4a). Two types of progeny nonrecombinants. The other two chromatids, which did
result: half have normal leaves and are tall: participate in crossing over, now contain new combinations
M D of alleles; gametes that receive these chromatids are recom-
m d binants. For each meiosis in which a single crossover takes
place, then, two nonrecombinant gametes and two recom-
and half have mottled leaves and are dwarf: binant gametes will be produced. This result is the same as
m d that produced by independent assortment (see Figure 7.4b);
m d so, when crossing over between two loci takes place in every
meiosis, it is impossible to determine whether the genes are
These progeny display the original combinations of traits on the same chromosome and crossing over took place or
present in the P generation and are nonrecombinant progeny, whether the genes are on different chromosomes.
or parental progeny. No new combinations of the two traits, For closely linked genes, crossing over does not take
such as normal leaves with dwarf or mottled leaves with tall, place in every meiosis. In meioses in which there is no cross-
appear in the offspring, because the genes affecting the two ing over, only nonrecombinant gametes are produced. In
traits are completely linked and are inherited together. New meioses in which there is a single crossover, half the gametes
combinations of traits could arise only if the physical connec- are recombinants and half are nonrecombinants (because a
tion between M and D or between m and d were broken. single crossover affects only two of the four chromatids); so
These results are distinctly different from the results that the total percentage of recombinant gametes is always half
are expected when genes assort independently (Figure 7.4b). the percentage of meioses in which crossing over takes place.
If the M and D loci assorted independently, the heterozygous Even if crossing over between two genes takes place in every
plant (Mm Dd) would produce four types of gametes: two meiosis, only 50% of the resulting gametes will be recombi-
nonrecombinant gametes containing the original combina- nants. Thus, the frequency of recombinant gametes is always
tions of alleles (M D and m d) and two gametes containing half the frequency of crossing over, and the maximum pro-
new combinations of alleles (M d and m D). Gametes with portion of recombinant gametes is 50%.
new combinations of alleles are called recombinant gametes.
With independent assortment, nonrecombinant and recom-
binant gametes are produced in equal proportions. These four
types of gametes join with the single type of gamete produced CONCEPTS
by the homozygous parent of the testcross to produce four Linkage between genes causes them to be inherited together and
kinds of progeny in equal proportions (see Figure 7.4b). The reduces recombination; crossing over breaks up the associations
progeny with new combinations of traits formed from recom- of such genes. In a testcross for two linked genes, each crossover
binant gametes are termed recombinant progeny. produces two recombinant gametes and two nonrecombinants.
The frequency of recombinant gametes is half the frequency of
CONCEPTS crossing over, and the maximum frequency of recombinant gam-
A testcross in which one of the plants is heterozygous for two etes is 50%.
completely linked genes yields two types of progeny, each type
displaying one of the original combinations of traits present in the ✔ CONCEPT CHECK 1
P generation. Independent assortment, in contrast, produces four For single crossovers, the frequency of recombinant gametes is half
types of progeny in a 1 : 1 : 1 : 1 ratio—two types of recombinant the frequency of crossing over because
progeny and two types of nonrecombinant progeny in equal
a. a test cross between a homozygote and heterozygote produces
proportions. 1
/2 heterozygous and 1/2 homozygous progeny.
 Linkage, Recombination, and Eukaryotic Gene Mapping 167

(a) No crossing over

1 Homologous chromosomes 2 If no crossing

pair in prophase I. over takes place,...

A B A B 3 …all resulting chromosomes in

A B gametes have original allele
A B Meiosis II
a b a combinations and are nonrecombinants.
b
a b a b

(b) Crossing over

1 A crossover may take 2 In this case, half of the resulting gametes will have
place in prophase I. unchanged chromosomes (nonrecombinants)…

A B A B Nonrecombinant 3 ….and half will have

A B A b Recombinant recombinant chromosomes.
Meiosis II
a b a B Recombinant
a b a b Nonrecombinant

7.5 A single crossover produces half nonrecombinant gametes and half recombinant gametes.

b. the frequency of recombination is always 50%. and dwarf, and 7 are mottle leaved and tall. These plants are
c. each crossover takes place between only two of the four the recombinant progeny.
chromatids of a homologous pair. The results of a cross such as the one illustrated in
d. crossovers occur in about 50% of meioses. Figure 7.6 reveal several things. A testcross for two indepen-
dently assorting genes is expected to produce a 1 : 1 : 1 : 1
phenotypic ratio in the progeny. The progeny of this cross
clearly do not exhibit such a ratio; so we might suspect that
Application Let’s apply what we have learned about link- the genes are not assorting independently. When linked
age and recombination to a cross between tomato plants genes undergo some crossing over, the result is mostly non-
that differ in the genes that encode leaf type and plant recombinant progeny and fewer recombinant progeny. This
height. Assume now that these genes are linked and that result is what we observe among the progeny of the testcross
some crossing over takes place between them. Suppose a illustrated in Figure 7.6; so we conclude that the two genes
geneticist carried out the testcross described earlier: show evidence of linkage with some crossing over.
M D m d
Calculating Recombination Frequency
m d m d
The percentage of recombinant progeny produced in a cross
When crossing over takes place in the genes for leaf is called the recombination frequency, which is calculated
type and height, two of the four gametes produced are as follows:
recombinants. When there is no crossing over, all four result-
ing gametes are nonrecombinants. Thus, over all meioses, recombinant number of recombinant progeny
= ×100%
the majority of gametes will be nonrecombinants (Figure frequency total number of progeny
7.6a). These gametes then unite with gametes produced by
the homozygous recessive parent, which contain only the In the testcross shown in Figure 7.6, 15 progeny exhibit new
recessive alleles, resulting in mostly nonrecombinant prog- combinations of traits; so the recombination frequency is:
eny and a few recombinant progeny (Figure 7.6b). In this 8+7 15
cross, we see that 55 of the testcross progeny have normal ×100% = ×100% = 12.2%
55 + 53 + 8 + 7 123
leaves and are tall and 53 have mottled leaves and are dwarf.
These plants are the nonrecombinant progeny, containing Thus, 12.2% of the progeny exhibit new combinations of
the original combinations of traits that were present in the traits resulting from crossing over. The recombination fre-
parents. Of the 123 progeny, 15 have new combinations of quency can also be expressed as a decimal fraction (0.122).
traits that were not seen in the parents: 8 are normal leaved TRY PROBLEM 15
168 Chapter 7

(a) Coupling and Repulsion

Normal Mottled In crosses for linked genes, the arrangement of alleles on
leaves, tall leaves, dwarf the homologous chromosomes is critical in determining the
outcome of the cross. For example, consider the inheritance
of two genes in the Australian blowfly, Lucilia cuprina. In
this species, one locus determines the color of the thorax:
a purple thorax (p) is recessive to the normal green thorax
(p+). A second locus determines the color of the puparium:
 a black puparium (b) is recessive to the normal brown
puparium (b+). The loci for thorax color and puparium
color are located close together on the chromosome. Sup-
pose we test-cross a fly that is heterozygous at both loci
Meioses with and M D m d with a fly that is homozygous recessive at both. Because
without crossing these genes are linked, there are two possible arrangements
over together result m d m d on the chromosomes of the heterozygous progeny fly. The
in less than 50% dominant alleles for green thorax (p+) and brown puparium
recombination Gamete Gamete
on average. (b+) might reside on one chromosome of the homologous
formation formation
pair, and the recessive alleles for purple thorax (p) and
black puparium (b) might reside on the other homologous
No crossing Crossing
over over chromosome:
p b
p b
MD m d MD m d M d m D m d

Nonrecombinant Nonrecombinant Recombinant This arrangement, in which wild-type alleles are found
gametes (100%) gametes (50%) gametes (50%) on one chromosome and mutant alleles are found on the
other chromosome, is referred to as the coupling, or cis,
Fertilization configuration. Alternatively, one chromosome might bear
the alleles for green thorax (p+) and black puparium (b), and
the other chromosome carries the alleles for purple thorax
(b) (p) and brown puparium (b+):
p b
Normal Mottled Normal Mottled
leaves, tall leaves, dwarf leaves, dwarf leaves, tall p b

This arrangement, in which each chromosome contains

one wild-type and one mutant allele, is called the repulsion, or
trans, configuration. Whether the alleles in the heterozygous
parent are in coupling or repulsion determines which pheno-
types will be most common among the progeny of a testcross.
When the alleles are in the coupling configuration,
the most numerous progeny types are those with a green
thorax and brown puparium and those with a purple
thorax and black puparium (Figure 7.7a); but, when the
M D m d M d m D
alleles of the heterozygous parent are in repulsion, the most
m d m d m d m d numerous progeny types are those with a green thorax and
Progeny black puparium and those with a purple thorax and brown
55 53 8 7
number puparium (Figure 7.7b). Notice that the genotypes of the
Nonrecombinant Recombinant parents in Figure 7.7a and b are the same (p+p b+b × pp bb)
progeny progeny
and that the dramatic difference in the phenotypic ratios
Conclusion: With linked genes and some crossing of the progeny in the two crosses results entirely from the
over, nonrecombinant progeny predominate.
configuration—coupling or repulsion—of the chromo-
7.6 Crossing over between linked genes produces somes. Knowledge of the arrangement of the alleles on the
nonrecombinant and recombinant offspring. In this testcross, chromosomes is essential to accurately predict the outcome
genes are linked and there is some crossing over. of crosses in which genes are linked.
 Linkage, Recombination, and Eukaryotic Gene Mapping 169

(a) Alleles in coupling configuration (b) Alleles in repulsion configuration

Green thorax, Purple thorax, Green thorax, Purple thorax,

brown puparium black puparium brown puparium black puparium

Testcross  Testcross 
p+ b+ p b p+ b p b

p b p b p b+ p b

Gamete formation Gamete formation Gamete formation Gamete formation

p+ b+ p b p+ b p b+ p b p+ b p b+ p+ b+ p b p b

Nonrecombinant Recombinant Nonrecombinant Recombinant

gametes gametes gametes gametes

Fertilization Fertilization

Green thorax, Purple thorax, Green thorax, Purple thorax, Green thorax, Purple thorax, Green thorax, Purple thorax,
brown black black brown black brown brown black
puparium puparium puparium puparium puparium puparium puparium puparium

p+ b+ p b p+ b p b+ p+ b p b+ p+ b+ p b

p b p b p b p b p b p b p b p b
Progeny 40 40 10 10 Progeny 40 40 10 10
number number
Nonrecombinant Recombinant Nonrecombinant Recombinant
progeny progeny progeny progeny
Conclusion: The phenotypes of the offspring are the same, but their numbers
differ, depending on whether alleles are in coupling or in repulsion.

7.7 The arrangement (coupling or repulsion) of linked genes on a chromosome affects the results of a testcross.
Linked loci in the Australian blowfly, Luciliá cuprina, determine the color of the thorax and that of the puparium.

somes; in this case, they exhibit independent assortment and

CONCEPTS combine randomly when gametes are formed. An individual hetero-
In a cross, the arrangement of linked alleles on the chromosomes zygous at two loci (Aa Bb) produces four types of gametes (A B, a b,
is critical for determining the outcome. When two wild-type A b, and a B) in equal proportions: two types of nonrecombinants
alleles are on one homologous chromosome and two mutant and two types of recombinants, In a testcross, these gametes will
alleles are on the other, they are in the coupling configuration; result in four types of progeny in equal proportions (Table 7.1).
when each chromosome contains one wild-type allele and one Second, the genes may be completely linked—meaning that
mutant allele, the alleles are in repulsion. they are on the same chromosome and lie so close together that
crossing over between them is rare. In this case, the genes do not
✔ CONCEPT CHECK 2 recombine. An individual heterozygous for two closely linked genes
The following testcross produces the progeny shown: Aa Bb × in the coupling configuration
aa bb → 10 Aa Bb, 40 Aa bb, 40 aa Bb, 10 aa bb. Were the genes in
A B
the Aa Bb parent in coupling or in repulsion?
a b
produces only the nonrecombinant gametes containing alleles A B or
CONNECTING CONCEPTS a b; the alleles do not assort into new combinations such as A b or a B.
In a testcross, completely linked genes will produce only two types of
Relating Independent Assortment, Linkage, progeny, both nonrecombinants, in equal proportions (see Table 7.1).
and Crossing Over The third situation, incomplete linkage, is intermediate
We have now considered three situations concerning genes at dif- between the two extremes of independent assortment and com-
ferent loci. First, the genes may be located on different chromo- plete linkage. Here, the genes are physically linked on the same
170 Chapter 7

Evidence for the Physical Basis

Table 7.1 Results of a testcross
(Aa Bb × aa bb) with complete
of Recombination
linkage, independent assortment, Walter Sutton’s chromosome theory of inheritance, which
and linkage with some stated that genes are physically located on chromosomes,
was supported by Nettie Stevens and Edmund Wilson’s
crossing over
discovery that sex was associated with a specific chromo-
some in insects (p. 75 in Chapter 4) and Calvin Bridges’s
Situation Progeny of Test Cross
demonstration that nondisjunction of X chromosomes
Independent Aa Bb (nonrecombinant) 25% was related to the inheritance of eye color in Drosophila
assortment aa bb (nonrecombinant) 25% (pp. 82–83). Further evidence for the chromosome the-
Aa bb (recombinant) 25% ory of heredity came in 1931, when Harriet Creighton
aa Bb (recombinant) 25% and Barbara McClintock (Figure 7.8) obtained evidence
Complete linkage Aa Bb (nonrecombinant) 50% that intrachromosomal recombination was the result of
(genes in coupling) aa bb (nonrecombinant) 50% physical exchange between chromosomes. Creighton and
Linkage with some Aa Bb (nonrecombinant)
McClintock discovered a strain of corn that had an abnor-
¶
more
crossing over aa bb (nonrecombinant) than 50% mal chromosome 9, containing a densely staining knob
(genes in coupling) Aa bb (recombinant) at one end and a small piece of another chromosome
¶
less
aa Bb (recombinant) than 50% attached to the other end. This aberrant chromosome
allowed them to visually distinguish the two members of a
homologous pair.
chromosome, which prevents independent assortment. However, They studied the inheritance of two traits in corn deter-
occasional crossovers break up the linkage and allow the genes to mined by genes on chromosome 9. At one locus, a dominant
recombine. With incomplete linkage, an individual heterozygous at allele (C) produced colored kernels, whereas a recessive allele
two loci produces four types of gametes—two types of recombi- (c) produced colorless kernels. At a second, linked locus, a
nants and two types of nonrecombinants—but the nonrecombi-
dominant allele (Wx) produced starchy kernels, whereas a
nants are produced more frequently than the recombinants because
recessive allele (wx) produced waxy kernels. Creighton and
crossing over does not take place in every meiosis. In the testcross,
these gametes result in four types of progeny, with the nonrecom- McClintock obtained a plant that was heterozygous at both
binants more frequent than the recombinant (see Table 7.1). loci in repulsion, with the alleles for colored and waxy on
Earlier in the chapter, the term recombination was defined as
the sorting of alleles into new combinations. We’ve now considered
two types of recombination that differ in the mechanism that gener-
ates these new combinations of alleles. Interchromosomal recom-
bination takes place between genes located on different chromo-
somes. It arises from independent assortment—the random segre-
gation of chromosomes in anaphase I of meiosis—and is the kind of
recombination that Mendel discovered while studying dihybrid
crosses. A second type of recombination, intrachromosomal
recombination, takes place between genes located on the same
chromosome. This recombination arises from crossing over—the
exchange of genetic material in prophase I of meiosis. Both types of
recombination produce new allele combinations in the gametes; so
they cannot be distinguished by examining the types of gametes
produced. Nevertheless, they can often be distinguished by the
frequencies of types of gametes: interchromosomal recombination
produces 50% nonrecombinant gametes and 50% recombinant
gametes, whereas intrachromosomal recombination frequently
produces less than 50% recombinant gametes. However, when the
genes are very far apart on the same chromosome, they assort
independently, as if they were on different chromosomes. In this
case, intrachromosomal recombination also produces 50% recombi-
nant gametes. Intrachromosomal recombination of genes that lie far
apart on the same chromosome and interchromosomal recombina- 7.8 Barbara McClintock (left) and Harriet Creighton (right)
tion are genetically indistinguishable. provided evidence that genes are located on chromosomes.
[Karl Maramorosch/Courtesy of Cold Spring Harbor Laboratory Archives.]
 Linkage, Recombination, and Eukaryotic Gene Mapping 171

the aberrant chromosome and the alleles for colorless and Predicting the Outcomes of Crosses
starchy on the normal chromosome: with Linked Genes
Knob Extra piece Knowing the arrangement of alleles on a chromosome allows
us to predict the types of progeny that will result from a cross
C wx entailing linked genes and to determine which of these types
will be the most numerous. Determining the proportions of
the types of offspring requires an additional piece of infor-
c Wx mation—the recombination frequency. The recombination
frequency provides us with information about how often the
They crossed this heterozygous plant with a plant that alleles in the gametes appear in new combinations and there-
was homozygous for colorless and heterozygous for waxy fore allows us to predict the proportions of offspring pheno-
(with both chromosomes normal): types that will result from a specific cross with linked genes.
C wx c Wx In cucumbers, smooth fruit (t) is recessive to warty
fruit (T) and glossy fruit (d) is recessive to dull fruit (D).
c Wx c wx Geneticists have determined that these two genes exhibit a
This cross will produce different combinations of traits recombination frequency of 16%. Suppose we cross a plant
in the progeny, but the only way that colorless and waxy homozygous for warty and dull fruit with a plant homozy-
progeny can arise is through crossing over in the doubly gous for smooth and glossy fruit and then carry out a test-
heterozygous parent: cross by using the F1:
T D t d
C wx
t d t d
c Wx
What types and proportions of progeny will result from this
Crossing over testcross?
Four types of gametes will be produced by the heterozy-
C Wx c Wx
gous parent, as shown in Figure 7.9: two types of nonrecom-
 binant gametes ( T D and t d ) and two types
c wx c wx of recombinant gametes ( T d and t D ). The
recombination frequency tells us that 16% of the gametes
produced by the heterozygous parent will be recombinants.
Because there are two types of recombinant gametes, each
should arise with a frequency of 16%/2 = 8%. This fre-
Some colored, C Wx quency can also be represented as a probability of 0.08. All
starchy progeny the other gametes will be nonrecombinants; so they should
c wx
arise with a frequency of 100% − 16% = 84%. Because there
c wx
are two types of nonrecombinant gametes, each should arise
Some colorless,
with a frequency of 84%/2 = 42% (or 0.42). The other par-
waxy progeny c wx ent in the testcross is homozygous and therefore produces
only a single type of gamete ( t d ) with a frequency
Note: Not all progeny genotypes are shown.
of 100% (or 1.00).
Four types of progeny result from the testcross (see
Notice that, if crossing over entails physical exchange
Figure 7.9). The expected proportion of each type can be
between the chromosomes, then the colorless, waxy progeny
determined by using the multiplication rule, multiplying
resulting from recombination should have a chromosome
together the probability of each gamete. Testcross progeny
with an extra piece but not a knob. Furthermore, some of the
with warty and dull fruit
colored, starchy progeny should possess a knob but not the
extra piece. This outcome is precisely what Creighton and T D
McClintock observed, confirming the chromosomal theory t d
of inheritance. Curt Stern provided a similar demonstration
by using chromosomal markers in Drosophila at about the appear with a frequency of 0.42 (the probability of inheriting
same time. We will examine the molecular basis of recombi- a gamete with chromosome T D from the heterozy-
nation in more detail in Chapter 12. gous parent) × 1.00 (the probability of inheriting a gamete
172 Chapter 7

Geneticists have determined that the recombination frequency with chromosome t d from the recessive parent)
between two genes in cucumbers is 16%. How can we use = 0.42. The proportions of the other types of F2 progeny
this information to predict the results of this cross? can be calculated in a similar manner (see Figure 7.9). This
Warty, dull Smooth, glossy method can be used for predicting the outcome of any cross
fruit fruit with linked genes for which the recombination frequency is
known. TRY PROBLEM 17
Testcross 
T D t d Testing for Independent Assortment
t d t d In some crosses, the genes are obviously linked because there
are clearly more nonrecombinant progeny than recom-
Gamete formation Gamete formation binant progeny. In other crosses, the difference between
independent assortment and linkage is not so obvious. For
example, suppose we did a testcross for two pairs of genes,
T D t d T d t D t d such as Aa Bb × aa bb, and observed the following numbers
of progeny: 54 Aa Bb, 56 aa bb, 42 Aa bb, and 48 aa Bb. Is
Nonrecombinant Recombinant Nonrecombinant this outcome the 1 : 1 : 1 : 1 ratio we would expect if A and
gametes gametes gametes
B assorted independently? Not exactly, but it’s pretty close.
Predicted 0.42 0.42 0.08 0.08 1.00
frequency Perhaps these genes assorted independently and chance pro-
of gametes duced the slight deviations between the observed numbers
Fertilization
Because the recombination and the expected 1 : 1 : 1 : 1 ratio. Alternatively, the genes
frequency is 16%, the total might be linked, with considerable crossing over taking place
proportion of recombinant between them, and so the number of nonrecombinants is
gametes is 0.16.
only slightly greater than the number of recombinants. How
do we distinguish between the role of chance and the role of
Predicted frequency linkage in producing deviations from the results expected
of progeny with independent assortment?
We encountered a similar problem in crosses in which
Warty, 0.42  1.00
dull fruit = 0.42 genes were unlinked—the problem of distinguishing between
T D deviations due to chance and those due to other factors. We
addressed this problem (in Chapter 3) with the goodness-
t d Non- of-fit chi-square test, which helps us evaluate the likelihood
recombinant that chance alone is responsible for deviations between the
Smooth, 0.42  1.00 progeny numbers of progeny that we observed and the numbers that
glossy fruit = 0.42 we expected by applying the principles of inheritance. Here,
t d we are interested in a different question: Is the inheritance of
alleles at one locus independent of the inheritance of alleles
t d
at a second locus? If the answer to this question is yes, then
the genes are assorting independently; if the answer is no,
Warty, 0.08  1.00
= 0.08
then the genes are probably linked.
glossy fruit
T d A possible way to test for independent assortment is
to calculate the expected probability of each progeny type,
t d assuming independent assortment, and then use the good-
Recombinant ness-of-fit chi-square test to evaluate whether the observed
Smooth, 0.08  1.00 progeny numbers deviate significantly from the expected numbers.
dull fruit = 0.08 With independent assortment, we expect 1/4 of each phe-
t D notype: 1/4 Aa Bb, 1/4 aa bb, 1/4 Aa bb, and 1/4 aa Bb. This
expected probability of each genotype is based on the multipli-
t d cation rule of probability, which we considered in Chapter 3.
The predicted frequency of progeny For example, if the probability of Aa is 1/2 and the probability
is obtained by multiplying the of Bb is 1/2, then the probability of Aa Bb is 1/2 × 1/2 = 1/4.
frequencies of the gametes. In this calculation, we are making two assumptions: (1)
the probability of each single-locus genotype is 1/2, and
7.9 The recombination frequency allows a prediction of the
proportions of offspring expected for a cross entailing linked (2) genotypes at the two loci are inherited independently
genes. (1/2 × 1/2 = 1/4).
 Linkage, Recombination, and Eukaryotic Gene Mapping 173

One problem with this approach is that a sig- (a)

nificant chi-square value can result from a violation
of either assumption. If the genes are linked, then Brown body, Yellow body,
straight wings  curved wings
the inheritance of genotypes at the two loci are not
independent (assumption 2), and we will get a sig- y + y cv + cv yy cvcv
nificant deviation between observed and expected 1 A testcross is carried
numbers. But we can also get a significant deviation out between
Cross
if the probability of each single-locus genotype is cockroaches differing
in two characteristics.
not 1/2 (assumption 1), even when the genotypes are
assorting independently. We may obtain a significant 63 y + y cv + cv brown body, straight wings
deviation, for example, if individuals with one geno- 28 y + y cvcv brown body, curved wings
type have a lower probability of surviving or the 33 yy cv + cv yellow body, straight wings
penetrance of a genotype is not 100%. We could test 77 yy cvcv yellow body, curved wings
both assumptions by conducting a series of chi-square
(b) Contingency table
tests, first testing the inheritance of genotypes at each
locus separately (assumption 1) and then testing for 2 To test for independent Segregation of y + and y 3 ...with genotypes
independent assortment (assumption 2). However, a assortment of alleles Row for one locus
faster method is to test for independence in genotypes encoding the two traits, y +y yy totals along the top...
a table is constructed...
with a chi-square test of independence. +
Segregation cv cv 63 33 96 5 Numbers of each
of cv + and cv genotype are
The chi-square test of independence The chi- cv cv 28 77 105 placed in the
square test of independence allows us to evaluate 4 ...and genotypes table cells, and
Column 91 110 201 the row totals,
whether the segregation of alleles at one locus is inde- for the other locus totals
along the left side. column totals,
pendent of the segregation of alleles at another locus, Grand total and grand total
without making any assumption about the probability are computed.
of single-locus genotypes. To illustrate this analysis, (c)
we will examine results from a cross between German Number expected
row total  column total
cockroaches, in which yellow body (y) is recessive to
brown body (y+) and curved wings (cv) are recessive to Genotype
Number
observed ( grand total (
straight wings (cv+). A testcross (y+y cv+cv × yy cvcv) 96  91
produced the progeny shown in Figure 7.10a. y + y cv + cv 63 = 43.46
201
To carry out the chi-square test of independence, 6 The expected
105  91
we first construct a table of the observed numbers of y + y cvcv 28 = 47.54 numbers of
201 progeny,
progeny, somewhat like a Punnett square, except, here, assuming
96 110
we put the genotypes that result from the segregation yy cv + cv 33 = 52.46 independent
201
of alleles at one locus along the top and the genotypes assortment,
that result from the segregation of alleles at the other 105  110 are calculated.
yy cvcv 77 = 57.46
201
locus along the side (Figure 7.10b). Next, we compute
the total for each row, the total for each column, and (d)
the grand total (the sum of all row totals or the sum 7 A chi-square
(observed– expected)2 value is
of all column totals, which should be the same). These 2 =  expected calculated.
totals will be used to compute the expected values for
the chi-square test of independence. (63 – 43.46 )2 (28 – 47.54 )2 (33 – 52.54 )2 (77 – 5 7. 4 6 )2
=
43.46
+ 47.54
+ 52.54
+ 57.46
Now, we compute the values expected if the seg-
regation of alleles at the y locus is independent of the = 8.79 + 8.03 + 7.27 + 6.64
segregation of alleles at the cv locus. If the segregation = 30.73
of alleles at each locus is independent, then the pro-
portion of progeny with y+y and yy genotypes should (e) 8 The probability
is less than 0.005,
be the same for cockroaches with genotype cv+cv and df = (number of rows – 1)  (number of columns – 1) indicating that the
for cockroaches with genotype cvcv. The converse is df = (2 – 1)  (2 – 1) = 1  1 = 1 difference
also true; the proportions of progeny with cv+cv and between numbers
P < 0.005 of observed and
expected progeny
is probably not
7.10 A chi-square test of independence can be used to Conclusion: The genes for body color and type of wing
due to chance.
determine if genes at two loci are assorting independently. are not assorting independently and must be linked.
174 Chapter 7

cvcv genotypes should be the same for cockroaches with probability less than 0.05 is significantly different from the
genotype y+y and for cockroaches with genotype yy. With expected values and is therefore evidence that the genes are
the assumption that the alleles at the two loci segregate inde- not assorting independently. TRY PROBLEM 16
pendently, the expected number for each cell of the table can
be computed by using the following formula: Gene Mapping with Recombination
row total × column total Frequencies
expected number = Thomas Hunt Morgan and his students developed the idea
grand total
that physical distances between genes on a chromosome are
For the cell of the table corresponding to genotype y+y cv+cv related to the rates of recombination. They hypothesized
(the upper-left-hand cell of the table in Figure 7.10b) the that crossover events occur more or less at random up and
expected number is: down the chromosome and that two genes that lie far apart
are more likely to undergo a crossover than are two genes
96 (row total) × 91 (column total) 8736
= = 43.46 that lie close together. They proposed that recombination
201 (grand total) 201 frequencies could provide a convenient way to determine
With the use of this method, the expected numbers for each the order of genes along a chromosome and would give
cell are given in Figure 7.10c. estimates of the relative distances between the genes. Chro-
We now calculate a chi-square value by using the same mosome maps calculated by using the genetic phenomenon
formula that we used for the goodness-of-fit chi-square test of recombination are called genetic maps. In contrast, chro-
in Chapter 3: mosome maps calculated by using physical distances along
the chromosome (often expressed as numbers of base pairs)
(observed − expected)2 are called physical maps.
x2 = ∑
expected Distances on genetic maps are measured in map units
(abbreviated m.u.); one map unit equals 1% recombination.
Recall that ∑ means “sum,” and that we are adding together Map units are also called centiMorgans (cM), in honor of
the (observed − expected)2/expected value for each type of Thomas Hunt Morgan; 100 centiMorgans equals 1 Morgan.
progeny. With the observed and expected numbers of cock- Genetic distances measured with recombination rates are
roaches from the testcross, the calculated chi-square value is approximately additive: if the distance from gene A to gene B
30.73 (Figure 7.10d). is 5 m.u., the distance from gene B to gene C is 10 m.u., and
To determine the probability associated with this chi- the distance from gene A to gene C is 15 m.u., then gene B
square value, we need the degrees of freedom. Recall from must be located between genes A and C. On the basis of the
Chapter 3 that the degrees of freedom are the number of map distances just given, we can draw a simple genetic map
ways in which the observed classes are free to vary from the for genes A, B, and C, as shown here:
expected values. In general, for the chi-square test of inde-
pendence, the degrees of freedom equal the number of rows bßßßßßßß 15 m.u. ßßßßßßßB
in the table minus 1 multiplied by the number of columns
in the table minus 1 (Figure 7.10e), or A b 5 m.u. B B bßßß 10 m.u. ßßßB C
df = (number of rows − 1) × (number of columns − 1) We could just as plausibly draw this map with C on the left
In our example, there were two rows and two columns, and and A on the right:
so the degrees of freedom are:
bßßßßßßßß 15 m.u. ßßßßßßßB
df = (2 − 1) × (2 − 1) = 1 × 1 = 1
Therefore, our calculated chi-square value is 30.73, with 1 C bßßß 10 m.u. ßßßB B b 5 m.u. B A
degree of freedom. We can use Table 3.5 to find the associ- Both maps are correct and equivalent because, with infor-
ated probability. Looking at Table 3.5, we find that our cal- mation about the relative positions of only three genes, the
culated chi-square value is larger than the largest chi-square most that we can determine is which gene lies in the middle.
value given for 1 degree of freedom, which has a probability If we obtained distances to an additional gene, then we
of 0.005. Thus, our calculated chi-square value has a prob- could position A and C relative to that gene. An additional
ability less than 0.005. This very small probability indicates gene D, examined through genetic crosses, might yield the
that the genotypes are not in the proportions that we would following recombination frequencies:
expect if independent assortment were taking place. Our
Gene pair Recombination frequency (%)
conclusion, then, is that these genes are not assorting inde-
pendently and must be linked. As is the case for the good- A and D 8
ness-of-fit chi-square test, geneticists generally consider that B and D 13
any chi-square value for the test of independence with a C and D 23
 Linkage, Recombination, and Eukaryotic Gene Mapping 175

A B
A B
a b
a b

Double crossover

1 A single crossover A B 2 ...but a second crossover will

A B
will switch the A B reverse the effects of the first,
alleles on a b restoring the original parental
homologous combination of alleles...
a b
chromosomes,...
Meiosis II

3 ...and producing only

A B
nonrecombinant genotypes
A B in the gametes, although 7.11 A two-strand double crossover between
a b parts of the chromosomes two linked genes produces only nonrecombinant
a b have recombined. gametes.

Notice that C and D exhibit the greatest amount of between the same two genes reverses the effects of the first,
recombination; therefore, C and D must be farthest apart, thus restoring the original parental combination of alleles
with genes A and B between them. Using the recombination (see Figure 7.11). We therefore cannot distinguish between the
frequencies and remembering that 1 m.u. = 1% recombina- progeny produced by two-strand double crossovers and the
tion, we can now add D to our map: progeny produced when there is no crossing over at all. As we
shall see in the next section, we can detect double crossovers if
bßßßßßßßßßß 23 m.u. ßßßßßßßßßßB we examine a third gene that lies between the two crossovers.
bßßßßß 13 m.u. ßßßßßB Because double crossovers between two genes go undetected,
bßßßßßßßß 15 m.u. ßßßB map distances will be underestimated whenever double
crossovers take place. Double crossovers are more frequent
D bß 8 m.u. ßB A b 5 m.u. B B bß 10 m.u. ßB C between genes that are far apart; therefore genetic maps based
on short distances are usually more accurate than those based
By doing a series of crosses between pairs of genes, we can on longer distances.
construct genetic maps showing the linkage arrangements
of a number of genes. CONCEPTS
Two points should be emphasized about constructing A genetic map provides the order of the genes on a chromosome
chromosome maps from recombination frequencies. First, and the approximate distances from one gene to another based
recall that we cannot distinguish between genes on different on recombination frequencies. In genetic maps, 1% recombina-
chromosomes and genes located far apart on the same chro- tion equals 1 map unit, or 1 centiMorgan. Double crossovers
mosome. If genes exhibit 50% recombination, the most that between two genes go undetected; so map distances between
can be said about them is that they belong to different groups distant genes tend to underestimate genetic distances.
of linked genes (different linkage groups), either on different
chromosomes or far apart on the same chromosome. ✔ CONCEPT CHECK 3
The second point is that a testcross for two genes that are How does a genetic map differ from a physical map?
far apart on the same chromosome tends to underestimate
the true physical distance, because the cross does not reveal
double crossovers that might take place between the two genes
(Figure 7.11). A double crossover arises when two separate Constructing a Genetic Map with the Use
crossover events take place between two loci. (For now, we will of Two-Point Testcrosses
consider only double crossovers that occur between two of the Genetic maps can be constructed by conducting a series
four chromatids of a homologus pair—a two-strand double of testcrosses. In each testcross, one of the parents is het-
crossover. Double crossovers that take place among three and erozygous for a different pair of genes, and recombination
four chromatids will be considered later.) Whereas a single frequencies are calculated between pairs of genes. A testcross
crossover produces combinations of alleles that were not pres- between two genes is called a two-point testcross or a two-
ent on the original parental chromosomes, a second crossover point cross for short. Suppose that we carried out a series of
176 Chapter 7

two-point crosses for four genes, a, b, c, and d, and obtained 10 m.u. The summed distances will be only approximate
the following recombination frequencies: because any double crossovers between the two genes will be
missed and the map distance will be underestimated.
Gene loci in testcross Recombination frequency (%) By examining the recombination frequency between c
a and b 50 and d, we can distinguish between these two possibilities.
a and c 50 The recombination frequency between c and d is 28%; so
gene d must lie to the left of gene b. Notice that the sum of
a and d 50
the recombination between d and b (10%) and between b
b and c 20 and c (20%) is greater than the recombination between d
b and d 10 and c (28%). As already discussed, this discrepancy arises
c and d 28 because double crossovers between the two outer genes go
undetected, causing an underestimation of the true map
We can begin constructing a genetic map for these genes distance. The genetic map of these genes is now complete:
by considering the recombination frequencies for each pair
Linkage group 1
of genes. The recombination frequency between a and b is
a
50%, which is the recombination frequency expected with
independent assortment. Therefore, genes a and b may
either be on different chromosomes or be very far apart on Linkage group 2
the same chromosome; so we will place them in different d b c
linkage groups with the understanding that they may or may
not be on the same chromosome: b 10 m.u. B
bßßß 20 m.u. ßßßB
Linkage group 1 bßßßßßß 30 m.u. ßßßßßßB
a TRY PROBLEM 26

Linkage group 2 7.3 A Three-Point Testcross Can Be

b
Used to Map Three Linked Genes
Genetic maps can be constructed from a series of testcrosses
The recombination frequency between a and c is 50%, for pairs of genes, but this approach is not particularly effi-
indicating that they, too, are in different linkage groups. The cient, because numerous two-point crosses must be carried
recombination frequency between b and c is 20%; so these out to establish the order of the genes and because double
genes are linked and separated by 20 map units: crossovers are missed. A more efficient mapping technique
is a testcross for three genes—a three-point testcross, or
Linkage group 1 three-point cross. With a three-point cross, the order of the
a three genes can be established in a single set of progeny and
some double crossovers can usually be detected, providing
more-accurate map distances.
Linkage group 2 Consider what happens when crossing over takes place
b c among three hypothetical linked genes. Figure 7.12 illus-
bßßß 20 m.u. ßßßB trates a pair of homologous chromosomes of an individual
that is heterozygous at three loci (Aa Bb Cc). Notice that the
The recombination frequency between a and d is 50%, genes are in the coupling configuration; that is, all the domi-
indicating that these genes belong to different linkage nant alleles are on one chromosome ( A B C )
groups, whereas genes b and d are linked, with a recombina- and all the recessive alleles are on the other chromosome
tion frequency of 10%. To decide whether gene d is 10 m.u. ( a b c ). Three types of crossover events can
to the left or to the right of gene b, we must consult the c-to- take place between these three genes: two types of single
d distance. If gene d is 10 m.u. to the left of gene b, then the crossovers (see Figure 7.12a and b) and a double crossover
distance between d and c should be approximately the sum (see Figure 7.12c). In each type of crossover, two of the
of the distance between b and c and between c and d: 20 m.u. resulting chromosomes are recombinants and two are non-
+ 10 m.u. = 30 m.u. If, on the other hand, gene d lies to the recombinants.
right of gene b, then the distance between gene d and gene Notice that, in the recombinant chromosomes result-
c will be much shorter, approximately 20 m.u. − 10 m.u. = ing from the double crossover, the outer two alleles are the
 Linkage, Recombination, and Eukaryotic Gene Mapping 177

Centromere A B C
A B C Pair of homologous
a b c chromosomes
a b c

(a) Single crossover (b) Single crossover (c) Double

between A and B between B and C crossover
A B C A B C A B C
A B C A B C A B C
a b c a b c a b c
a b c a b c a b c

Meiosis Meiosis Meiosis

A B C A B C A B C
A b c A B c A b C
a B C a b C a B c
a b c a b c a b c

Conclusion: Recombinant chromosomes resulting from

7.12 Three types of crossovers can take place
the double crossover have only the middle gene altered.
among three linked loci.

same as in the nonrecombinants, but the middle allele is loci with flies that are homozygous for recessive alleles at
different. This result provides us with an important clue all three loci:
about the order of the genes. In progeny that result from a
st e ss st e ss
double crossover, only the middle allele should differ from P
the alleles present in the nonrecombinant progeny. st e ss st e ss

Constructing a Genetic Map n

st e ss
with the Three-Point Testcross F
st e ss
To examine gene mapping with a three-point testcross,
we will consider three recessive mutations in the fruit fly The order of the genes has been arbitrarily assigned because,
Drosophila melanogaster. In this species, scarlet eyes (st) are at this point, we do not know which is the middle gene.
recessive to wild-type red eyes (st+), ebony body color (e) Additionally, the alleles in these heterozygotes are in cou-
is recessive to wild-type gray body color (e+), and spineless pling configuration (because all the wild-type dominant
(ss)—that is, the presence of small bristles—is recessive to alleles were inherited from one parent and all the recessive
wild-type normal bristles (ss+). The loci encoding these three mutations from the other parent), although the testcross can
characteristics are linked and located on chromosome 3. also be done with alleles in repulsion.
We will refer to these three loci as st, e, and ss, but keep In the three-point testcross, we cross the F1 heterozy-
in mind that either the recessive alleles (st, e, and ss) or the gotes with flies that are homozygous for all three recessive
dominant alleles (st+, e+, and ss+) may be present at each mutations. In many organisms, it makes no difference
locus. So, when we say that there are 10 m.u. between st and whether the heterozygous parent in the testcross is male
ss, we mean that there are 10 m.u. between the loci at which or female (provided that the genes are autosomal) but, in
mutations st and ss occur; we could just as easily say that Drosophila, no crossing over takes place in males. Because
there are 10 m.u. between st+ and ss+. crossing over in the heterozygous parent is essential for
To map these genes, we need to determine their order determining recombination frequencies, the heterozygous
on the chromosome and the genetic distances between flies in our testcross must be female. So we mate female F1
them. First, we must set up a three-point testcross, a cross flies that are heterozygous for all three traits with male flies
between a fly heterozygous at all three loci and a fly homo- that are homozygous for all the recessive traits:
zygous for recessive alleles at all three loci. To produce flies
heterozygous for all three loci, we might cross a stock of st e ss st e ss
Female Male
flies that are homozygous for wild-type alleles at all three st e ss st e ss
178 Chapter 7

Wild type Scarlet, ebony, ing the recessive trait. With two classes of progeny possible
spineless
for each of the three loci, there will be 23 = 8 classes of phe-
 notypes possible in the progeny. In this example, all eight
phenotypic classes are present but, in some three-point
st+ e+ ss+ st e ss
crosses, one or more of the phenotypes may be missing if the
st e ss st e ss number of progeny is limited. Nevertheless, the absence of
a particular class can provide important information about
which combination of traits is least frequent and, ultimately,
Testcross
about the order of the genes, as we will see.
To map the genes, we need information about where and
Progeny Progeny Progeny how often crossing over has taken place. In the homozygous
genotype phenotype number recessive parent, the two alleles at each locus are the same, and
st+ e+ ss+ so crossing over will have no effect on the types of gametes
Wild type 283 produced; with or without crossing over, all gametes from
st e ss this parent have a chromosome with three recessive alleles
st+ e+ ss+
( st e ss ). In contrast, the heterozygous parent has
st e ss
All mutant 278
different alleles on its two chromosomes, and so crossing over
st e ss can be detected. The information that we need for mapping,
st e ss therefore, comes entirely from the gametes produced by the
st+ e ss heterozygous parent. Because chromosomes contributed by
Ebony, spineless 50 the homozygous parent carry only recessive alleles, whatever
st e ss
st+ e ss alleles are present on the chromosome contributed by the
st e+ ss+
heterozygous parent will be expressed in the progeny.
Scarlet 52 As a shortcut, we often do not write out the complete
st e ss + +
genotypes of the testcross progeny, listing instead only the
st e ss alleles expressed in the phenotype, which are the alleles
st+ e+ ss inherited from the heterozygous parent. This convention is
Spineless 5
used in the discussion that follows.
st e ss
st+ e+ ss
st e ss+ CONCEPTS
Scarlet, ebony 3 To map genes, information about the location and number of
st e ss + crossovers in the gametes that produced the progeny of a cross is
st e ss
needed. An efficient way to obtain this information is to use a
st+ e ss+ three-point testcross, in which an individual heterozygous at three
Ebony 43
e ss linked loci is crossed with an individual that is homozygous reces-
st
st+ e ss+ sive at the three loci.
st e+ ss
Scarlet,
41 ✔ CONCEPT CHECK 4
e ss spineless
st Write the genotypes of all recombinant and nonrecombinant progeny
st e+ ss
expected from the following three-point cross:
Total 755
m p s m p s
7.13 The results of a three-point testcross can be used to m p s m p s
map linked genes. In this three-point testcross of Drosophila
melanogaster, the recessive mutations scarlet eyes (st), ebony body
color (e), and spineless bristles (ss) are at three linked loci. The order Determining the gene order The first task in mapping
of the loci has been designated arbitrarily, as has the sex of the the genes is to determine their order on the chromosome.
progeny flies.
In Figure 7.13, we arbitrarily listed the loci in the order st,
Animation 7.1 e, ss, but we had no way of knowing which of the three loci
 www.whfreeman.com/pierce4e
illustrates how to was between the other two. We can now identify the middle
determine the order of the three linked genes. locus by examining the double-crossover progeny.
First, determine which progeny are the nonrecom-
The progeny produced from this cross are listed in binants; they will be the two most-numerous classes of
Figure 7.13. For each locus, two classes of progeny are progeny. (Even if crossing over takes place in every meiosis,
produced: progeny that are heterozygous, displaying the the nonrecombinants will constitute at least 50% of the
dominant trait, and progeny that are homozygous, display- progeny.) Among the progeny of the testcross in Figure 7.13,
 Linkage, Recombination, and Eukaryotic Gene Mapping 179

the most numerous are those with all three dominant traits
( st
+
e+ ss+ ) and those with all three recessive Table 7.2 Steps in determining gene order
traits ( st e ss ). in a three-point cross
Next, identify the double-crossover progeny. These prog-
eny should always have the two least-numerous phenotypes, 1. Identify the nonrecombinant progeny (two most-
because the probability of a double crossover is always less than numerous phenotypes).
the probability of a single crossover. The least-common prog- 2. Indentify the double-crossover progeny (two least-
eny among those listed in Figure 7.13 are progeny with spineless numerous phenotypes).
+
bristles ( st e+ ss ) and progeny with scarlet eyes
3. Compare the phenotype of double-crossover progeny
and ebony body ( st e ss+ ); so they are the double-
with the phenotype of nonrecombinant progeny. They
crossover progeny. should be alike in two characteristics and differ in one.
Three orders of genes on the chromosome are possible: the
4. The characteristic that differs between the double
eye-color locus could be in the middle ( e st ss ), the crossover and the nonrecombinant progeny is encoded
body-color locus could be in the middle ( st e ss ), by the middle gene.
or the bristle locus could be in the middle ( st ss e ).
To determine which gene is in the middle, we can draw the
chromosomes of the heterozygous parent with all three pos-
sible gene orders and then see if a double crossover produces for red eyes (st+) and both have an allele for gray body (e+),
the combination of genes observed in the double-crossover but the nonrecombinants have an allele for normal bristles
progeny. The three possible gene orders and the types of prog- (ss+), whereas the double crossovers have an allele for spine-
eny produced by their double crossovers are: less bristles (ss). Because the bristle locus is the only one that
differs, it must lie in the middle. We would obtain the same
results if we compared the other class of double-crossover
Original Chromosomes ss+ ) with other nonrecombinant
progeny ( st e
chromosomes after crossing over
progeny ( st e ss ). Again, the only locus that dif-
fers is the one for bristles. Don’t forget that the nonrecom-
e⫹ st⫹ ss⫹ e⫹ st⫹ ss⫹ e⫹ st ss⫹
binants and the double crossovers should differ at only one
1. : :
locus; if they differ at two loci, the wrong classes of progeny
e st ss e st ss e st⫹ ss are being compared.
st⫹ e⫹ ss⫹ st⫹ e⫹ ss⫹ st⫹ e ss⫹
2. : :
st e ss st e ss st e⫹ ss CONCEPTS
To determine the middle locus in a three-point cross, compare the
st⫹ ss⫹ e⫹ st⫹ ss⫹ e⫹ st⫹ ss e⫹
double-crossover progeny with the nonrecombinant progeny. The
3. : : double crossovers will be the two least-common classes of pheno-
st ss e st ss e st ss⫹ e types; the nonrecombinants will be the two most-common classes
of phenotypes. The double-crossover progeny should have the
The only gene order that produces chromosomes same alleles as the nonrecombinant types at two loci and different
with the set of alleles observed in the least-numerous alleles at the locus in the middle.
+
progeny or double crossovers ( st e+ ss and
st e ss+ in Figure 7.13) is the one in which the ss ✔ CONCEPT CHECK 5
locus for bristles lies in the middle (gene-order 3). Therefore, A three-point test cross is carried out between three linked genes.
this order ( st ss e ) must be the correct sequence The resulting nonrecombinant progeny are s+ r+ c+ and s r c and the
of genes on the chromosome. double-crossover progeny are s r c+ and s+ r+ c. Which is the middle
With a little practice, we can quickly determine which locus?
locus is in the middle without writing out all the gene orders.
The phenotypes of the progeny are expressions of the alleles
inherited from the heterozygous parent. Recall that, when we Determining the locations of crossovers When we
looked at the results of double crossovers (see Figure 7.12), know the correct order of the loci on the chromosome, we
only the alleles at the middle locus differed from the non- should rewrite the phenotypes of the testcross progeny in
recombinants. If we compare the nonrecombinant progeny Figure 7.13 with the alleles in the correct order so that we can
with double-crossover progeny, they should differ only in determine where crossovers have taken place (Figure 7.14).
alleles of the middle locus (Table 7.2). Among the eight classes of progeny, we have already iden-
+
Let’s compare the alleles in the double-crossover prog- tified two classes as nonrecombinants ( st ss+ e+
+
eny st e+ ss with those in the nonrecombinant and st ss e ) and two classes as double crossovers
+
progeny st e+ ss+ . We see that both have an allele ( st
+
ss e+ and st ss+ e ). The other
180 Chapter 7

four classes include progeny that resulted from a chromo-

Wild type Scarlet, ebony,
spineless some that underwent a single crossover: two underwent
single crossovers between st and ss, and two underwent
 single crossovers between ss and e.
To determine where the crossovers took place in these
st+ ss+ e+ st ss e
progeny, compare the alleles found in the single-crossover
st ss e st ss e progeny with those found in the nonrecombinants, just as
we did for the double crossovers. For example, consider
+
progeny with chromosome st ss e . The first
Testcross allele (st+) came from the nonrecombinant chromosome
st+ ss+ e+ and the other two alleles (ss and e)
must have come from the other nonrecombinant chromo-
some st ss e through crossing over:
Progeny Progeny Progeny
genotype phenotype number st⫹ ss⫹ e⫹ st⫹ ss⫹ e⫹ st⫹ ss e
st+ ss+ e+ : :
Wild type 283 st ss e st ss e st ss⫹ e⫹
st ss e + + +
st ss e + +
This same crossover also produces the st ss e
st ss e Scarlet,
progeny.
ebony, 278
st ss e spineless This method can also be used to determine the loca-
st ss e tion of crossing over in the other two types of single-cross-
This symbol indicates the Nonrecombinants are
over progeny. Crossing over between ss and e produces
position of a crossover. produced most frequently.
st+ ss+ e and st ss e+ chromosomes:
st+ ss e
Spineless, 50 st⫹ ss⫹ e⫹ st⫹ ss⫹ e⫹ st⫹ ss⫹ e
ebony
st ss e : :
+ ss e
st
st ss+ e+
st ss e st ss e st ss e⫹
Scarlet 52
We now know the locations of all the crossovers; their
st ss e
st ss+ e+ locations are marked with a slash in Figure 7.14.
st+ ss+ e
Ebony 43 Calculating the recombination frequencies Next, we
st ss e + + can determine the map distances, which are based on the fre-
st ss e
quencies of recombination. We calculate recombination fre-
st ss e+ quency by adding up all of the recombinant progeny, dividing
Scarlet,
41
st ss e
spineless this number by the total number of progeny from the cross,
st ss e+ and multiplying the number obtained by 100%. To determine
Double-crossover recombinants the map distances accurately, we must include all crossovers
are produced least frequently. (both single and double) that take place between two genes.
st+ ss e+ Recombinant progeny that possess a chromosome
Spineless 5 that underwent crossing over between the eye-color locus
st ss e (st) and the bristle locus (ss) include the single cross-
st+ ss e+ +
overs ( st / ss e and st / ss+ e+ )
st ss+ e + +
Scarlet, 3
and the two double crossovers ( st / ss / e and
st ss e ebony st / ss+ / e ); see Figure 7.14. There are a total of
st + e
ss 755 progeny; so the recombination frequency between ss
Total 755 and st is:
7.14 Writing the results of a three-point testcross with the loci
st−ss recombination frequency =
in the correct order allows the locations of crossovers to be (50 + 52 + 5 + 3)
determined. These results are from the testcross illustrated in Figure ×100% = 14.6%
755
7.13, with the loci shown in the correct order. The location of a
crossover is indicated by a slash (/). The sex of the progeny flies has The distance between the st and ss loci can be expressed as
been designated arbitrarily. 14.6 m.u.
 Chromosome 1 (X) Chromosome 2 Chromosome 3 Chromosome 4
Bent wing
Yellow body Net veins Cubitus veins
0.0 Scute bristles 0.0 Aristaless antenna 0.0 Roughoid eyes 0.0 Shaven hairs
1.5 White eyes 1.3 Star eyes 0.2 Veinlet veins Grooveless scutellum
3.0 Facet eyes 4.0 Held-out wings Eyeless
5.5 Echinus eyes
7.5 Ruby eyes 13.0 Dumpy wings
13.7 Crossveinless wings 16.5 Clot eyes
20.0 Cut wings 19.2 Javelin bristles
21.0 Singed bristles
26.0 Sepia eyes
27.7 Lozenge eyes 26.5 Hairy body
33.0 Vermilion eyes
36.1 Miniature wings
41.0 Dichaete bristles
43.0 Sable body 43.2 Thread arista
44.0 Garnet eyes 48.5 Black body 44.0 Scarlet eyes
51.0 Reduced bristles 48.0 Pink eyes
56.7 Forked bristles 54.5 Purple eyes 50.0 Curled wings
57.0 Bar eyes 54.8 Short bristles 58.2 Stubble bristles
59.5 Fused veins 55.0 Light eyes 58.5 Spineless bristles
62.5 Carnation eyes 57.5 Cinnabar eyes 58.7 Bithorax body
66.0 Bobbed hairs 66.7 Scabrous eyes 62.0 Stripe body
67.0 Vestigial wings 63.0 Glass eyes
66.2 Delta veins
72.0 Lobe eyes 69.5 Hairless bristles
75.5 Curved wings 70.7 Ebony eyes
74.7 Cardinal eyes

91.1 Rough eyes

100.5 Plexus wings

100.7 Claret eyes
104.5 Brown eyes
107.0 Speck body 106.2 Minute bristles

7.15 Drosophila melanogaster has four linkage groups corresponding to its four pairs of
chromosomes. Distances between genes within a linkage group are in map units.

The map distance between the bristle locus (ss) and the body summing the map distances between st and ss and between
locus (e) is determined in the same manner. The recombinant ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.). We can now
progeny that possess a crossover between ss and e are the sin- use the map distances to draw a map of the three genes on
+
gle crossovers st ss+ / e and st ss / e+ the chromosome:
and the double crossovers st / ss / e+ and
st / ss +
/ e . The recombination frequency is: bßßßßßßßßßß 26.8 m.u. ßßßßßßB

ss−e recombination frequency = st bßß 14.6 m.u. ßßB ss bß 12.2 m.u. ßB e

(43 + 41 + 5 + 3)
×100% = 12.2%
755 A genetic map of D. melanogaster is illustrated in
Figure 7.15.
Thus, the map distance between ss and e is 12.2 m.u.
Finally, calculate the map distance between the outer Interference and the coefficient of coincidence Map
two loci, st and e. This map distance can be obtained by distances give us information not only about the distances

181
182 Chapter 7

that separate genes, but also about the proportions of So the interference for our three-point cross is:
recombinant and nonrecombinant gametes that will be
interference = 1 − 0.6 = 0.4
produced in a cross. For example, knowing that genes st
and ss on the third chromosome of D. melanogaster are This value of interference tells us that 40% of the
separated by a distance of 14.6 m.u. tells us that 14.6% double-crossover progeny expected will not be observed,
of the gametes produced by a fly heterozygous at these because of interference. When interference is complete and
two loci will be recombinants. Similarly, 12.2% of the no double-crossover progeny are observed, the coefficient of
gametes from a fly heterozygous for ss and e will be coincidence is 0 and the interference is 1.
recombinants. Sometimes a crossover increases the probability of
Theoretically, we should be able to calculate the pro- another crossover taking place nearby and we see more
portion of double-recombinant gametes by using the mul- double-crossover progeny than expected. In this case, the
tiplication rule of probability (see Chapter 3), which states coefficient of coincidence is greater than 1 and the interfer-
that the probability of two independent events occurring ence is negative. TRY PROBLEM 28
together is calculated by multiplying the probabilities of the
independent events. Applying this principle, we should find
that the proportion (probability) of gametes with double CONCEPTS
crossovers between st and e is equal to the probability of The coefficient of coincidence equals the number of double cross-
recombination between st and ss multiplied by the prob- overs observed divided by the number of double crossovers
ability of recombination between ss and e, or 0.146 × 0.122 expected on the basis of the single-crossover frequencies. The
= 0.0178. Multiplying this probability by the total number interference equals 1 − the coefficient of coincidence; it indicates
of progeny gives us the expected number of double-cross- the degree to which one crossover interferes with additional
over progeny from the cross: 0.0178 × 755 = 13.4. Only crossovers.
8 double crossovers—considerably fewer than the 13
expected—were observed in the progeny of the cross (see ✔ CONCEPT CHECK 6
Figure 7.14). In analyzing the results of a three-point testcross, a student deter-
This phenomenon is common in eukaryotic organ- mines that the interference is −0.23. What does this negative interfer-
isms. The calculation assumes that each crossover event is ence value indicate?
independent and that the occurrence of one crossover does a. Fewer double crossovers took place than expected on the basis
not influence the occurrence of another. But crossovers are of single-crossover frequencies.
frequently not independent events: the occurrence of one b. More double crossovers took place than expected on the basis of
crossover tends to inhibit additional crossovers in the same single-crossover frequencies.
region of the chromosome, and so double crossovers are less c. Fewer single crossovers took place than expected.
frequent than expected.
d. A crossover in one region interferes with additional crossovers in
The degree to which one crossover interferes with the same region.
additional crossovers in the same region is termed the
interference. To calculate the interference, we first deter-
mine the coefficient of coincidence, which is the ratio of
observed double crossovers to expected double crossovers:
CONNECTING CONCEPTS
coefficient of coincidence = Stepping Through the Three-Point Cross
number of observed double crossovers
We have now examined the three-point cross in considerable detail
number of exxpected double crossovers and have seen how the information derived from the cross can be
used to map a series of three linked genes. Let’s briefly review the
For the loci that we mapped on the third chromosome of D. steps required to map genes from a three-point cross.
melanogaster (see Figure 7.14), we find that the 1. Write out the phenotypes and numbers of progeny
produced in the three-point cross. The progeny phenotypes
coefficient of coincidence = will be easier to interpret if you use allelic symbols for the traits
5+3 8 (such as st+ e+ ss).
= = 0.6
0.146 × 0.122 × 755 13.4 2. Write out the genotypes of the original parents used to
produce the triply heterozygous individual in the testcross
which indicates that we are actually observing only 60% of and, if known, the arrangement (coupling or repulsion) of the
alleles on their chromosomes.
the double crossovers that we expected on the basis of the
single-crossover frequencies. The interference is calculated as 3. Determine which phenotypic classes among the progeny
are the nonrecombinants and which are the double
interference = 1 − coefficient of coincidence crossovers. The nonrecombinants will be the two most-
 Linkage, Recombination, and Eukaryotic Gene Mapping 183

common phenotypes; double crossovers will be the two • Solution

least-common phenotypes.
a. We can represent the crosses in this problem as follows:
4. Determine which locus lies in the middle. Compare the
alleles present in the double crossovers with those present in ch b cn ch b cn
the nonrecombinants; each class of double crossovers should
P
ch b cn ch b cn
be like one of the nonrecombinants for two loci and should
differ for one locus. The locus that differs is the middle one.
5. Rewrite the phenotypes with the genes in correct order.
n
ch b cn
6. Determine where crossovers must have taken place to F1
ch b cn
give rise to the progeny phenotypes. To do so, compare
each phenotype with the phenotype of the nonrecombinant ch b cn ch b cn
Testcross
progeny. ch b cn ch b cn
7. Determine the recombination frequencies. Add the Note that we do not know, at this point, the order of the
numbers of the progeny that possess a chromosome with a genes; we have arbitrarily put b in the middle.
crossover between a pair of loci. Add the double crossovers The next step is to determine which of the test-
to this number. Divide this sum by the total number of
cross progeny are nonrecombinants and which are
progeny from the cross, and multiply by 100%; the result is the
recombination frequency between the loci, which is the same
double crossovers. The nonrecombinants should be the
as the map distance. most-frequent phenotype; so they must be the progeny
+ + +
with phenotypes encoded by ch b cn and ch b cn.
8. Draw a map of the three loci. Indicate which locus lies in
the middle, and indicate the distances between them.
These genotypes are consistent with the genotypes of the
parents, given earlier. The double crossovers are the least-
9. Determine the coefficient of coincidence and the + +
frequent phenotypes and are encoded by ch b cn and
interference. The coefficient of coincidence is the number
ch b cn+.
of observed double-crossover progeny divided by the number
of expected double-crossover progeny. The expected number We can determine the gene order by comparing the
can be obtained by multiplying the product of the two single- alleles present in the double crossovers with those present in
recombination probabilities by the total number of progeny in the nonrecombinants. The double-crossover progeny should
the cross. TRY PROBLEM 31 be like one of the nonrecombinants at two loci and unlike it
at one locus; the allele that differs should be in the middle.
+
Compare the double-crossover progeny ch b cn with the
Worked Problem nonrecombinant ch b cn. Both have cherub wings (ch) and
black body (b), but the double-crossover progeny have wild-
In D. melanogaster, cherub wings (ch), black body (b), and type eyes (cn+), whereas the nonrecombinants have cinnabar
cinnabar eyes (cn) result from recessive alleles that are all eyes (cn). The locus that determines cinnabar eyes must be
located on chromosome 2. A homozygous wild-type fly in the middle.
was mated with a cherub, black, and cinnabar fly, and the b. To calculate the recombination frequencies among the
resulting F1 females were test-crossed with cherub, black, genes, we first write the phenotypes of the progeny with the
and cinnabar males. The following progeny were produced genes encoding them in the correct order. We have already
from the testcross: identified the nonrecombinant and double-crossover progeny;
ch b+ cn 105 so the other four progeny types must have resulted from single
ch+ b+ cn+ 750 crossovers. To determine where single crossovers took place,
ch+ b cn 40 we compare the alleles found in the single-crossover progeny
ch+ b+ cn 4 with those in the nonrecombinants. Crossing over must have
ch b cn 753 taken place where the alleles switch from those found in one
ch b+ cn+ 41 nonrecombinant to those found in the other nonrecombinant.
ch+ b cn+ 102 The locations of the crossovers are indicated with a slash:
ch b cn+ 5 ch cn / b 105 single crossover
Total 1800 ch cn b 750 nonrecombinant
ch / cn b 40 single crossover
a. Determine the linear order of the genes on the ch / cn / b 4 double crossover
chromosome (which gene is in the middle?). ch cn b 753 nonrecombinant
b. Calculate the recombinant distances between the three ch / cn b 41 single crossover
loci. ch cn / b 102 single crossover
c. Determine the coefficient of coincidence and the ch / cn / b 5 double crossover
interference for these three loci. Total 1800
184 Chapter 7

Next, we determine the recombination frequencies and draw

a genetic map: To increase your skill with three-point crosses, try
working Problem 29 at the end of this chapter.
ch-cn recombination frequency =
40 + 4 + 41 + 5
×100% = 5%
1800
Effect of Multiple Crossovers
cn-b recombination frequency = So far, we have examined the effects of double crossovers
105 + 4 + 102 + 5 taking place between only two of the four chromatids
×100% = 12%
1800 (strands) of a homologous pair. These crossovers are called
two-strand crossovers. Double crossovers including three
ch-b map distance = 5% + 12% = 17%
and even four of the chromatids of a homologous pair also
may take place (Figure 7.16). If we examine only the alleles
bßßßßßßßß 17 m.u. ßßßßßßßßB at loci on either side of both crossover events, two-strand
double crossovers result in no new combinations of alleles,
b 5 m.u. B bßßßß 12 m.u. ßßßßB and no recombinant gametes are produced (see Figure
ch cn b 7.16). Three-strand double crossovers result in two of the
four gametes being recombinant, and four-strand double
c. The coefficient of coincidence is the number of observed crossovers result in all four gametes being recombinant.
double crossovers divided by the number of expected Thus, two-strand double crossovers produce 0% recombi-
double crossovers. The number of expected double nation, three-strand double crossovers produce 50% recom-
crossovers is obtained by multiplying the probability of a bination, and four-strand double crossovers produce 100%
crossover between ch and cn (0.05) × the probability of a recombination. The overall result is that all types of double
crossover between cn and b (0.12) × the total number of crossovers, taken together, produce an average of 50%
progeny in the cross (1800): recombinant progeny.
4+5 As we have seen, two-strand double crossover cause
coefficient of coincidence = = 0.83 alleles on either side of the crossovers to remain the same
0.05 × 0.12 ×1800
and produce no recombinant progeny. Three-strand and
Finally, the interference is equal to 1 − the coefficient of four-strand crossovers produce recombinant progeny, but
coincidence: these progeny are the same types produced by single cross-
interference = 1 − 0.83 = 0.17 overs. Consequently, some multiple crossovers go unde-

Two-strand double crossover

A B A B
A B A B 0% detectable
a b a b recombinants
a b a b

Three-strand double crossover

A B A B
A B A b 50% detectable
a b a b recombinants
a b a B
A B A b
A B A B 50% detectable
a b a B recombinants
a b a b

Four-strand double crossover

A B A b
A B A b 100% detectable
a b a B recombinants
a b a B
7.16 Results of two-, three-, and
four-strand double crossovers on 50% average
recombination between two genes. detectable recombinants
 Linkage, Recombination, and Eukaryotic Gene Mapping 185

50 Mapping Human Genes

Efforts in mapping human genes are hampered by the
Recombination (%)

inability to perform desired crosses and the small number of

progeny in most human families. Geneticists are restricted to
25 analyses of pedigrees, which are often incomplete and provide
limited information. Nevertheless, a large number of human
traits have been successfully mapped with the use of pedigree
data to analyze linkage. Because the number of progeny from
0 any one mating is usually small, data from several families
0 20 40 60 80 and pedigrees are usually combined to test for independent
Actual map distance (m.u.) assortment. The methods used in these types of analysis are
7.17 Percent recombination underestimates the true physical beyond the scope of this book, but an example will illustrate
distance between genes at higher map distances. how linkage can be detected from pedigree data.
One of the first documented demonstrations of link-
age in humans was between the locus for nail–patella syn-
tected when the progeny of a genetic cross are observed. drome and the locus that determines the ABO blood types.
Therefore, map distances based on recombination rates will Nail–patella syndrome is an autosomal dominant disorder
underestimate the true physical distances between genes, characterized by abnormal fingernails and absent or rudi-
because some multiple crossovers are not detected among mentary kneecaps. The ABO blood types are determined
the progeny of a cross. When genes are very close together, by an autosomal locus with multiple alleles (see Chapter 5).
multiple crossovers are unlikely, and the distances based on Linkage between the genes encoding these traits was estab-
recombination rates accurately correspond to the physical lished in families in which both traits segregate. Part of one
distances on the chromosome. But, as the distance between such family is illustrated in Figure 7.18.
genes increases, more multiple crossovers are likely, and the Nail–patella syndrome is rare, and so we can assume
discrepancy between genetic distances (based on recombina- that people having this trait are heterozygous (Nn); unaf-
tion rates) and physical distances increases. To correct for this fected people are homozygous (nn). The ABO genotypes
discrepancy, geneticists have developed mathematical map- can be inferred from the phenotypes and the types of
ping functions, which relate recombination frequencies to offspring produced. Person I-2 in Figure 7.18, for example,
actual physical distances between genes (Figure 7.17). Most has blood-type B, which has two possible genotypes: IBIB
of these functions are based on the Poisson distribution, or IBi (see Figure 5.6). Because some of her offspring are
which predicts the probability of multiple rare events. With blood-type O (genotype ii) and must have therefore inher-
the use of such mapping functions, map distances based on ited an i allele from each parent, female I-2 must have gen-
recombination rates can be more accurately estimated. otype IBi. Similarly, the presence of blood-type O offspring

I
A B A I A I A or I A i
This person has 1 2
A
I n I B
n B I B I B or I B i
nail–patella syndrome (Nn). O ii
i N i n

II
Nail–patella syndrome
All children in generation II B A O B B A A O A A
1 2 3 4 5 6 7 8 9 10 does not develop in
with nail–patella syndrome
IB n IA n i n IB n IB n IA n IA n i n IA n IA n children of generation II
have either blood type B or
with blood type A.
blood type O, indicating that i N i n i N i N i N i n i n i N i n i n
genes for the syndrome and
for blood type are linked. III
A A O B B
1 2 3 4 5
IA n IA n i n IB n IB n
i N i n i n i n i n

These outcomes resulted

from crossover events.

7.18 Linkage between ABO blood types and nail–patella syndrome was established by
examining families in whom both traits segregate. The pedigree shown here is for one such family.
The ABO blood type is indicated in each circle or square. The genotype, inferred from phenotype, is given
below each circle or square.
186 Chapter 7

in generation II indicates that male I-1, with blood-type A, probability of obtaining a particular set of observations
also must carry an i allele and therefore has genotype IAi. with the assumption of linkage and a certain recombina-
The parents of this family are: tion frequency is 0.1 and that the probability of obtaining
the same observations with the assumption of independent
I Ai Nn I Bi nn
assortment is 0.0001. The ratio of these two probabilities is
From generation II, we can see that the genes for 0.1
/0.0001 = 1000, the logarithm of which (the lod score) is 3.
nail–patella syndrome and the blood types do not appear Thus, linkage with the specified recombination is 1000 times
to assort independently. All children in generation II with as likely as independent assortment to produce what was
nail–patella syndrome have either blood-type B or blood- observed. A lod score of 3 or higher is usually considered
type O; all those with blood-type A have normal nails and convincing evidence for linkage. TRY PROBLEM 33
kneecaps. If the genes encoding nail–patella syndrome and
the ABO blood types assorted independently, we would Mapping with Molecular Markers
expect that some children in generation II would have
For many years, gene mapping was limited in most organ-
blood-type A and nail–patella syndrome, inheriting both
isms by the availability of genetic markers—that is, variable
the IA and N alleles from their father. This outcome indicates
that the arrangements of the alleles on the chromosomes of genes with easily observable phenotypes for which inheri-
tance could be studied. Traditional genetic markers include
the crossed parents are:
genes that encode easily observable characteristics such
IA n IB n as flower color, seed shape, blood types, and biochemical
i N i n differences. The paucity of these types of characteristics in
many organisms limited mapping efforts.
The pedigree indicates that there is no recombination In the 1980s, new molecular techniques made it possible
among the offspring (generation II) of these parents, but there to examine variations in DNA itself, providing an almost
are two instances of recombination among the persons in gen- unlimited number of genetic markers that can be used for
eration III. Persons II-1 and II-2 have the following genotypes: creating genetic maps and studying linkage relations. The
iB n IA n earliest of these molecular markers consisted of restriction
fragment length polymorphisms (RFLPs), which are varia-
i N i n
tions in DNA sequence detected by cutting the DNA with
Their child III-2 has blood-type A and does not have restriction enzymes (see Chapter 19). Later, methods were
nail–patella syndrome; so he must have genotype developed for detecting variable numbers of short DNA
sequences repeated in tandem, called microsatellites. More
IA n recently, DNA sequencing allows the direct detection of
i n individual variations in the DNA nucleotides. All of these
methods have expanded the availability of genetic markers
and must have inherited both the i and the n alleles from
and greatly facilitated the creation of genetic maps.
his father. These alleles are on different chromosomes in the
Gene mapping with molecular markers is done essen-
father; so crossing over must have taken place. Crossing over
tially in the same manner as mapping performed with
also must have taken place to produce child III-3.
traditional phenotypic markers: the cosegregation of two
In the pedigree of Figure 7.18, 13 children are from
or more markers is studied, and map distances are based on
matings in which the genes encoding nail–patella syndrome
the rates of recombination between markers. These methods
and ABO blood types segregate; 2 of them are recombinants.
and their use in mapping are presented in more detail in
On this basis, we might assume that the loci for nail–patella
Chapters 19 and 20.
syndrome and ABO blood types are linked, with a recom-
bination frequency of 2/13 = 0.154. However, it is possible
that the genes are assorting independently and that the Locating Genes with Genomewide
small number of children just makes it seem as though the Association Studies
genes are linked. To determine the probability that genes are The traditional approach to mapping genes, which we have
actually linked, geneticists often calculate lod (logarithm of learned in this chapter, is to examine progeny phenotypes
odds) scores. in genetic crosses or among individuals in a pedigree, look-
To obtain a lod score, we calculate both the probability ing for associations between the inheritance of particular
of obtaining the observed results with the assumption that phenotype and the inheritance of alleles at other loci. This
the genes are linked with a specified degree of recombina- type of gene mapping is called linkage analysis, because it
tion and the probability of obtaining the observed results is based on the detection of physical linkage between genes,
with the assumption of independent assortment. We then as measured by the rate of recombination, in progeny from
determine the ratio of these two probabilities, and the a cross. Linkage analysis has been a powerful tool in the
logarithm of this ratio is the lod score. Suppose that the genetic analysis of many different types of organisms.
 Linkage, Recombination, and Eukaryotic Gene Mapping 187

Another alternative approach to mapping genes is to 1 A series of different

conduct genomewide association studies, looking for non- haplotypes occur
random associations between the presence of a trait and in a population.
alleles at many different loci scattered across the genome.
1 3 5 4 2 2 4 4 4 3 5 4
Unlike linkage analysis, this approach does not trace the A B C D A B C D A B C D

inheritance of genetic markers and a trait in a genetic cross

or family. Rather, it looks for associations between traits and

Mutation 2 When a mutation
particular suites of alleles in a population.
(D–) first arises, it is
Imagine that we are interested in finding genes that associated with a
contribute to bipolar disease, a psychiatric illness charac- 2 2 4 –
special haplotype of
A B C D alleles at other loci.
terized by severe depression and mania. When a mutation
that predisposes a person to bipolar disease first arises in a
population, it will occur on a particular chromosome and
be associated with a specific set of alleles on that chromo-
No Crossing
some (Figure 7.19). A specific set of linked alleles, such as crossing over
this, is called a haplotype, and the nonrandom association over
between alleles in a haplotype is called linkage disequilib- 2 2 4 –
A B C D
rium. Because of the physical linkage between the bipolar
mutation and the other alleles of the haplotype, bipolar A1 B3 C5 D4
illness and the haplotype will tend to be inherited together.
Crossing over, however, breaks up the association between
3 With no 4 With crossing over,
the alleles of the haplotype (see Figure 7.19), reducing the recombination, the (D –) mutation
linkage disequilibrium between them. How long the linkage the (D –) mutation is now also
disequilibrium persists depends on the amount of recom- remains associated associated with
bination between alleles at different loci. When the loci are with the A2 B2 C4 the A1 B3 C5
haplotype. haplotype.
far apart, linkage disequilibrium breaks down quickly; when
the loci are close together, crossing over is less common and
linkage disequilibrium will persist longer. The important 2
A B
2
C
4
D
–
A
1
B
3
C
5
D
–

point is that linkage disequilibrium—the nonrandom asso-

ciation between alleles—provides information about the 7.19 Genomewide association studies are based on the
distance between genes. A strong association between a trait nonrandom association of a mutation (D−) that produces a
such as bipolar illness and a set of linked genetic markers trait and closely linked genes that constitute a haplotype.
indicates that one or more genes contributing to bipolar ill-
ness are likely to be near the genetic markers. including bipolar disease, height, skin pigmentation, eye
In recent years, geneticists have mapped millions of color, body weight, coronary artery disease, blood-lipid
genetic variants called single-nucleotide polymorphisms concentrations, diabetes, heart attacks, bone density, and
(SNPs), which are positions in the genome where people glaucoma, among others.
vary in a single nucleotide base (see Chapter 20). It is now
possible to quickly and inexpensively genotype people for CONCEPTS
hundreds of thousands or millions of SNPs. This genotyp- The development of molecular techniques for examining variation
ing has provided genetic markers needed for conducting in DNA sequences has provided a large number of genetic mark-
genomewide association studies, in which SNP haplotypes ers that can be used to create genetic maps and study linkage
of people with a particular disease, such as bipolar illness, relations. Genomewide association studies examine the nonran-
are compared with the haplotypes of healthy persons. dom association of genetic markers and phenotypes to locate
Nonrandom associations between SNPs and the disease genes that contribute to the expression of traits.
suggest that one or more genes that contribute to the dis-
ease are closely linked to the SNPs. Genomewide associa-
tion studies do not usually locate specific genes, but rather
7.4 Physical-Mapping Methods
associate the inheritance of a trait or disease with a specific Are Used to Determine
chromosomal region. After such an association has been the Physical Positions of Genes
established, geneticists can examine the chromosomal
region for genes that might be responsible for the trait.
on Particular Chromosomes
Genomewide association studies have been instrumental in Genetic maps reveal the relative positions of genes on a
the discovery of genes or chromosomal regions that affect chromosome on the basis of frequencies of recombina-
a number of genetic diseases and important human traits, tion, but they do not provide information that allow us to
188 Chapter 7

place groups of linked genes on particular chromosomes. 7.20a). If the gene is not within the deleted region, all of the
Furthermore, the units of a genetic map do not always pre- progeny will be wild type (see Figure 7.20b).
cisely correspond to physical distances on the chromosome, Deletion mapping has been used to reveal the chromo-
because a number of factors other than physical distances somal locations of a number of human genes. For example,
between genes (such as the type and sex of the organism) Duchenne muscular dystrophy is a disease that causes pro-
can influence recombination. Because of these limitations, gressive weakening and degeneration of the muscles. From
physical-mapping methods that do not rely on recombina- its X-linked pattern of inheritance, the mutated allele caus-
tion frequencies have been developed. ing this disorder was known to be on the X chromosome,
but its precise location was uncertain. Examination of a
Deletion Mapping number of patients having Duchenne muscular dystrophy,
One method for determining the chromosomal location of who also possessed small deletions, allowed researchers to
a gene is deletion mapping. Special staining methods have position the gene on a small segment of the short arm of the
been developed that reveal characteristic banding patterns X chromosome. TRY PROBLEM 34
on the chromosomes (see Chapter 9). The absence of one
or more of the bands that are normally on a chromosome Somatic-Cell Hybridization
reveals the presence of a chromosome deletion, a mutation Another method used for positioning genes on chromo-
in which a part of a chromosome is missing. Genes can be somes is somatic-cell hybridization, which requires the
assigned to regions of chromosomes by studying the associ- fusion of different types of cells. Most mature somatic (non-
ation between a gene’s phenotype or product and particular sex) cells can undergo only a limited number of divisions
chromosome deletions. and therefore cannot be grown continuously. However, cells
In deletion mapping, an individual that is homozygous that have been altered by viruses or derived from tumors
for a recessive mutation in the gene of interest is crossed that have lost the normal constraints on cell division will
with an individual that is heterozygous for a deletion divide indefinitely; this type of cell can be cultured in the
(Figure 7.20). If the gene of interest is in the region of the laboratory to produce a cell line.
chromosome represented by the deletion (the red part of Cells from two different cell lines can be fused by treat-
the chromosomes in Figure 7.20), approximately half of ing them with polyethylene glycol or other agents that alter
the progeny will display the mutant phenotype (see Figure their plasma membranes. After fusion, the cell possesses

(a) (b)

P generation Region of deletion Chromosome

with deletion
a a A+ a a A+ A+

 
Region of Chromosome
deletion with deletion
Cross Cross

aa Mutant A+ Wild type aa Mutant A+A+ Wild type

F1 generation
If the gene of interest is in
the deletion region, half of
A+ a a the progeny will display the A+ a A+ a
mutant phenotype.

If the gene is not within

the deletion region, all
of the progeny will be
wild type.

A+a Wild type a Mutant A+a Wild type A+a Wild type

7.20 Deletion mapping can be used to determine the chromosomal location of a gene.
An individual homozygous for a recessive mutation in the gene of interest (aa) is crossed with
an individual heterozygous for a deletion.
 Linkage, Recombination, and Eukaryotic Gene Mapping 189

Human fibroblast Mouse tumor cell two nuclei and is called a heterokaryon. The two nuclei of
a heterokaryon eventually also fuse, generating a hybrid cell
that contains chromosomes from both cell lines. If human
1 Human fibroblasts and and mouse cells are mixed in the presence of polyethylene
mouse tumor cells are
glycol, fusion results in human–mouse somatic-cell hybrids
mixed in the presence
of polyethylene glycol, (Figure 7.21). The hybrid cells tend to lose chromosomes
which facilitates fusion as they divide and, for reasons that are not understood,
of their membranes,... chromosomes from one of the species are lost preferentially.
In human–mouse somatic-cell hybrids, the human chro-
mosomes tend to be lost, whereas the mouse chromosomes
are retained. Eventually, the chromosome number stabilizes
2 ...creating hybrid cells when all but a few of the human chromosomes have been
called heterokaryons. lost. Chromosome loss is random and differs among cell
lines. The presence of these “extra” human chromosomes in
the mouse genome makes it possible to assign human genes
Heterokaryon to specific chromosomes.
To map genes by using somatic-cell hybridization
requires a panel of different hybrid cell lines. Each cell line
is examined microscopically and the human chromosomes
3 Human and mouse nuclei
in some hybrid cells fuse.
that it contains are identified. The cell lines of the panel are
chosen so that they differ in the human chromosomes that
they have retained. For example, one cell line might pos-
Hybrid cell with sess human chromosomes 2, 4, 7, and 8, whereas another
fused nucleus might possess chromosomes 4, 19, and 20. Each cell line in
the panel is examined for evidence of a particular human
gene. The human gene can be detected by looking either
4 Different human for the for the gene itself (discussed in Chapter 19) or for
chromosomes are
lost in different
the protein that it produces. Correlation of the presence of
cell lines. the gene with the presence of specific human chromosomes
often allows the gene to be assigned to the correct chromo-
some. For example, if a gene is detected in both of the afore-
mentioned cell lines, the gene must be on chromosome 4,
A B C D E because it is the only human chromosome common to both
Cell lines cell lines (Figure 7.22).
7.21 Somatic-cell hybridization can be used to determine which Sometimes somatic-cell hybridization can be used to
chromosome contains a gene of interest. position a gene on a specific part of a chromosome. Some

Human chromosomes present

Gene product
Cell line 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 X
present

A + + + + +
B + + + + + + + + + +
C – + + + +
D + + + + + +
E – + +
F + + + +

7.22 Somatic-cell hybridization is used to assign a gene to a particular human chromosome. A panel of six
cell lines, each line containing a different subset of human chromosomes, is examined for the presence of the gene
product (such as an enzyme). Four of the cell lines (A, B, D, and F) have the gene product. The only chromosome
common to all four of these cell lines is chromosome 4, indicating that the gene is located on this chromosome.
190 Chapter 7

The gene product The gene product …or its short chromosomes 1 and 16. Next, examine all the cell lines that
is present when is absent when the arm is missing. possess either chromosomes 1 and 16 and determine wheth-
chromosome 4 is entire chromosome
er they produce haptoglobin. Chromosome 1 is found in
intact. 4 is absent…
cell lines A, B, C, and D. If the gene for human haptoglobin
were found on chromosome 1, human haptoglobin would
be present in all of these cell lines. However, lines A and D do
not produce human haptoglobin; so the gene cannot be on
chromosome 1. Chromosome 16 is found only in cell lines B
and C, and only these lines produce human haptoglobin; so
the gene for human haptoglobin lies on chromosome 16.

For more practice with somatic-cell hybridiza-

2 4 11 2 11 2 4 11 tions, work Problem 35 at the end of this chapter.
Cell line 1 Cell line 2 Cell line 3

Conclusion: If the gene product is present in a

cell line with an intact chromosome but missing
from a line with a chromosome deletion, the gene
Physical Chromosome Mapping Through
for that product must be located in the deleted region. Molecular Analysis
So far, we have explored methods to indirectly determine
7.23 Genes can be localized to a specific part of a chromosome
by using somatic-cell hybridization. the chromosomal location of a gene by deletion mapping
or by looking for gene products. Researchers now have the
information and technology to actually see where a gene lies.
hybrid cell lines carry a human chromosome with a chro-
Described in more detail in Chapter 19, in situ hybridization
mosome mutation such as a deletion or a translocation. If
is a method for determining the chromosomal location of
the gene is present in a cell line with the intact chromosome
a particular gene through molecular analysis. This method
but missing from a line with a chromosome deletion, the
requires the creation of a probe for the gene, which is a single-
gene must be located in the deleted region (Figure 7.23).
stranded DNA complement to the gene of interest. The probe
Similarly, if a gene is usually absent from a chromosome but
is radioactive or fluoresces under ultraviolet light so that it
consistently appears whenever a translocation (a piece of
can be visualized. The probe binds to the DNA sequence of
another chromosome that has broken off and attached itself
the gene on the chromosome. The presence of radioactivity or
to the chromosome in question) is present, it must be pres-
fluorescence from the bound probe reveals the location of the
ent on the translocated part of the chromosome.
gene on a particular chromosome (Figure 7.24).
In addition to allowing us to see where a gene is located
Worked Problem on a chromosome, modern laboratory techniques now
allow researchers to identify the precise location of a gene
A panel of cell lines was created from human–mouse somatic- at the nucleotide level. For example, with DNA sequencing
cell fusions. Each line was examined for the presence of (described fully in Chapter 19), physical distances between
human chromosomes and for the production of human hap- genes can be determined in numbers of base pairs.
toglobin (a protein). The following results were obtained:

Cell Human Human chromosomes

line haptoglobin 1 2 3 14 15 16 21
A − + − + − + − −
B + + − + − − + −
C + + − − − + + −
D − + + − − + − −

On the basis of these results, which human chromosome

carries the gene for haptoglobin?

• Solution
First, identify the cell lines that are positive for the protein 7.24 In situ hybridization is another technique for determining
the chromosomal location of a gene. The red fluorescence is
(human haptoglobin) and determine the chromosomes that produced by a probe for sequences on chromosome 9; the green
they have in common. Lines B and C produce human hapto- fluorescence is produced by a probe for sequences on chromosome
globin; the only chromosomes that they have in common are 22. [Courtesy of Applied Imaging Corp.]
 Linkage, Recombination, and Eukaryotic Gene Mapping 191

chromosomes of females undergo about 50% more recombi-

CONCEPTS
nation than do the autosomal chromosomes of males.
Physical-mapping methods determine the physical locations of
Geneticists have found numerous recombination hot-
genes on chromosomes and include deletion mapping, somatic-cell
spots, where recombination is at least 10 times as high as the
hybridization, in situ hybridization, and direct DNA sequencing.
average elsewhere in the genome. The human genome may
contain an estimated 25,000 to 50,000 such recombination
hotspots. Approximately 60% of all crossovers take place
7.5 Recombination Rates Exhibit in hotspots. For humans, recombination hotspots tend to
be found near but not within active genes. Recombination
Extensive Variation hotspots have been detected in the genomes of other organ-
In recent years, geneticists have studied variation in rates of isms as well. In comparing recombination hotspots in the
recombination and found that levels of recombination vary genomes of humans and chimpanzees (our closest living
widely—among species, among and along chromosomes of relative), geneticists have determined that, although the DNA
a single species, and even between males and females of the sequences of humans and chimpanzees are extremely similar,
same species. For example, about twice as much recombi- their recombination hotspots are at entirely different places.
nation takes place in humans as in the mouse and the rat.
Within the human genome, recombination varies among
chromosomes, with chromosomes 21 and 22 having the CONCEPTS
highest rates and chromosomes 2 and 4 having the lowest
Rates of recombination vary among species, among and along
rates. Researchers have also detected differences between male chromosomes, and even between males and females.
and female humans in rates of recombination: the autosomal

CONCEPTS SUMMARY
• Linked genes do not assort independently. In a testcross for unit. Maps based on recombination rates are called genetic
two completely linked genes (no crossing over), only maps; maps based on physical distances are called physical
nonrecombinant progeny are produced. When two genes maps.
assort independently, recombinant progeny and • Genetic maps can be constructed by examining
nonrecombinant progeny are produced in equal proportions. recombination rates from a series of two-point crosses or
When two genes are linked with some crossing over between by examining the progeny of a three-point testcross.
them, more nonrecombinant progeny than recombinant
• Some multiple crossovers go undetected; thus, genetic maps
progeny are produced.
based on recombination rates underestimate the true physical
• Recombination frequency is calculated by summing the distances between genes.
number of recombinant progeny, dividing by the total
number of progeny produced in the cross, and multiplying by • Human genes can be mapped by examining the cosegregation
100%. The recombination frequency is half the frequency of of traits in pedigrees.
crossing over, and the maximum frequency of recombinant • A lod score is the logarithm of the ratio of the probability of
gametes is 50%. obtaining the observed progeny with the assumption of
• Coupling and repulsion refer to the arrangement of alleles on linkage to the probability of obtaining the observed progeny
a chromosome. Whether genes are in coupling configuration with the assumption of independent assortment. A lod score
or in repulsion determines which combination of phenotypes of 3 or higher is usually considered evidence for linkage.
will be most frequent in the progeny of a testcross. • Molecular techniques that allow the detection of variable
• Interchromosomal recombination takes place among genes differences in DNA sequence have greatly facilitated gene
located on different chromosomes through the random mapping.
segregation of chromosomes in meiosis. Intrachromosomal • Genomewide association studies locate genes that affect
recombination takes place among genes located on the same particular traits by examining the nonrandom association
chromosome through crossing over. of a trait with genetic markers from across the genome.
• A chi-square test of independence can be used to determine • Nucleotide sequencing is another method of physically
if genes are linked. mapping genes.
• Recombination rates can be used to determine the relative • Rates of recombination vary widely, differing among species,
order of genes and distances between them on a among and along chromosomes within a single species, and
chromosome. One percent recombination equals one map even between males and females of the same species.
192 Chapter 7

IMPORTANT TERMS
linked genes (p. 163) intrachromosomal recombination genetic marker (p. 186)
linkage group (p. 163) (p. 170) linkage analysis (p. 186)
nonrecombinant (parental) gamete genetic map (p. 174) genomewide association studies (p. 187)
(p. 166) physical map (p. 174) haplotype (p. 187)
nonrecombinant (parental) progeny map unit (m.u.) (p. 174) linkage disequilibrium (p. 187)
(p. 166) centiMorgan (p. 174) single-nucleotide polymorphism (SNP)
recombinant gamete (p. 166) Morgan (p. 174) (p. 187)
recombinant progeny (p. 166) two-point testcross (p. 175) deletion mapping (p. 188)
recombination frequency (p. 167) three-point testcross (p. 176) somatic-cell hybridization (p. 188)
coupling (cis) configuration (p. 168) interference (p. 182) cell line (p. 188)
repulsion (trans) configuration (p. 168) coefficient of coincidence (p. 182) heterokaryon (p. 189)
interchromosomal recombination mapping function (p. 185)
(p. 170) lod (logarithm of odds) score (p. 186)

ANSWERS TO CONCEPT CHECKS

1. c 5. The c locus
2. Repulsion 6. b
3. Genetic maps are based on rates of recombination; physical
maps are based on physical distances.
m+ p+ s + m+ p s m p+ s + m+ p+ s m p s + m+ p s + m p+ s m p s
4.
m p s m ps m p s m p s m p s m p s m p s mps

WORKED PROBLEMS
1. In guinea pigs, white coat (w) is recessive to black coat (W) P ww vv WW VV
and wavy hair (v) is recessive to straight hair (V ). A breeder
crosses a guinea pig that is homozygous for white coat and wavy A
hair with a guinea pig that is black with straight hair. The F1 are F1 Ww Vv
then crossed with guinea pigs having white coats and wavy hair
in a series of testcrosses. The following progeny are produced A
from these testcrosses: Testcross Ww Vv ww vv

black, straight 30 A
₁ black, straight
black, wavy 10 Ww Vv ⁄₄
₁ black, wavy
white, straight 12
Ww vv ⁄₄
₁ white, straight
ww Vv ⁄₄
₁ white, wavy
white, wavy 31 ww vv ⁄₄
Total 83
Because a total of 83 progeny were produced in the testcrosses,
a. Are the genes that determine coat color and hair type we expect 1/4 × 83 = 20.75 of each. The observed numbers of
assorting independently? Carry out chi-square tests to test your progeny from the testcross (30, 10, 12, 31) do not appear to fit
hypothesis. the expected numbers (20.75, 20.75, 20.75, 20.75) well; so
b. If the genes are not assorting independently, what is the independent assortment may not have taken place.
recombination frequency between them? To test the hypothesis, carry out a chi-square test of
independence. Construct a table, with the genotypes of the first
• Solution locus along the top and the genotypes of the second locus along
a. Assuming independent assortment, outline the crosses the side. Compute the totals for the rows and columns and the
conducted by the breeder: grand total.
 Linkage, Recombination, and Eukaryotic Gene Mapping 193

Row b. To determine the recombination frequencies, identify the

Ww ww totals recombinant progeny. Using the notation for linked genes, write
the crosses:
Vv 30 12 42
w v W V
P
vv 10 31 41 w v W V

Column 40 43 83 Grand total

A

A
totals F1 W V
w v
The expected value for each cell of the table is calculated with
the following formula: W V w v
Testcross
w v w v
expected number =
row total × column total
grand total
A
W V 30 black, straight
Using this formula, we find the expected values (given in w v (nonrecombinant progeny)
parentheses) to be: w v 31 white, wavy
Row w v (nonrecombinant progeny)
Ww ww totals W v 10 black, wavy
Vv 30 12 42 w v (recombinant progeny)
(20.24) (21.76) w V 12 white, straight
w v (recombinant progeny)
vv 10 31 41
(19.76) (21.24) The recombination frequency is:
number of recombinant progeny
Column 40 43 83 Grand total ×100%
A

totals total number progenny

or
Using these observed and expected numbers, we find the recombinant frequency =
calculated chi-square value to be: 10 + 12 22
×100% = ×100% = 26.5
30 + 31 + 10 + 12 83
(observed − expected) 2
␹2 = ∑
expected
2. A series of two-point crosses were carried out among seven loci
(30 − 20.24)2 (10 − 19.76)2 (12 − 21.76)2 (31 − 21.24)2 (a, b, c, d, e, f, and g), producing the following recombination
= + + +
20.224 19.76 21.76 21.24 frequencies. Using these recombination frequencies, map the seven
= 4.71 + 4.82 + 4.38 + 4.48 = 18.39 loci, showing their linkage groups, the order of the loci in each
linkage group, and the distances between the loci of each group:
The degrees of freedom for the chi-square test of
independence are df = (number of rows − 1) × (number of Recombination Recombination
columns − 1). There are two rows and two columns, and so the Loci frequency (%) Loci frequency (%)
degrees of freedom are: a and b 10 c and d 50
a and c 50 c and e 8
df = (2 − 1) × (2 − 1) = 1 × 1 = 1 a and d 14 c and f 50
a and e 50 c and g 12
In Table 3.5, the probability associated with a chi-square a and f 50 d and e 50
value of 18.39 and 1 degree of freedom is less than 0.005, a and g 50 d and f 50
indicating that chance is very unlikely to be responsible for b and c 50 d and g 50
the differences between the observed numbers and the b and d 4 e and f 50
numbers expected with independent assortment. The genes b and e 50 e and g 18
for coat color and hair type have therefore not assorted b and f 50 f and g 50
independently. b and g 50
194 Chapter 7

• Solution There is 50% recombination between f and all the other

To work this problem, remember that 1% recombination equals genes; so f must belong to a third linkage group:
1 map unit and a recombination frequency of 50% means that Linkage group 1 a bßßßßßßßß 14 m.u. ßßßßßßßßB d
genes at the two loci are assorting independently (located in
different linkage groups). b
The recombination frequency between a and b is 10%; so
bßßßßß 10 m.u. ßßßßßB b 4 m.u. B
these two loci are in the same linkage group, approximately
10 m.u. apart.
Linkage group 2
a b c e
Linkage group 1
b 10 m.u. B bßßßß 8 m.u. ßßßßßB

The recombination frequency between a and c is 50%; so c must Linkage group 3

lie in a second linkage group.
f
a b
Linkage group 1
b 10 m.u. B Finally, position locus g with respect to the other genes. The
recombination frequencies between g and loci a, b, and d are
c all 50%; so g is not in linkage group 1. The recombination
Linkage group 2 frequency between g and c is 12 m.u.; so g is a part of linkage
group 2. To determine whether g is 12 m.u. to the right or left
The recombination frequency between a and d is 14%; so d is of c, consult the g–e recombination frequency. Because this
located in linkage group 1. Is locus d 14 m.u. to the right or to the recombination frequency is 18%, g must lie to the left of c:
left of gene a? If d is 14 m.u. to the left of a, then the b-to-d
distance should be 10 m.u. + 14 m.u. = 24 m.u. On the other Linkage group 1 a bßßßßßßßß 14 m.u. ßßßßßßßßB d
hand, if d is to the right of a, then the distance between b and d
b
should be 14 m.u. − 10 m.u. = 4 m.u. The b–d recombination
frequency is 4%; so d is 14 m.u. to the right of a. The updated bßßßßß 10 m.u. ßßßßßB b 4 m.u. B
map is:

Linkage group 1 a bßßßßßßßß 14 m.u. ßßßßßßßßB d Linkage group 2 g bßßßßßßßß 18 m.u. ßßßßßßßßB e

b c

bßßßßß 10 m.u. ßßßßßB b 4 m.u. B bßßßßß 12 m.u. ßßßßßB b 8 m.u. B

Linkage group 2 Linkage group 3 f

c

Note that the g-to-e distance (18 m.u.) is shorter than the
The recombination frequencies between each of loci a, b,
sum of the g-to-c (12 m.u.) and c-to-e distances (8 m.u.)
and d, and locus e are all 50%; so e is not in linkage group 1 with
because of undetectable double crossovers between g and e.
a, b, and d. The recombination frequency between e and c is 8
m.u.; so e is in linkage group 2: 3. Ebony body color (e), rough eyes (ro), and brevis bristles
(bv) are three recessive mutations that occur in fruit flies. The
Linkage group 1 a bßßßßßßßß 14 m.u. ßßßßßßßßB d loci for these mutations have been mapped and are separated by
the following map distances:
b
e ro bv
bßßßßß 10 m.u. ßßßßßB b 4 m.u. B
bßßßß 20 m.u. ßßßßB bßßß 12 m.u. ßßßB
Linkage group 2 The interference between these genes is 0.4.
c e A fly with ebony body, rough eyes, and brevis bristles is
crossed with a fly that is homozygous for the wild-type traits.
bßßßß 8 m.u. ßßßßßB The resulting F1 females are test-crossed with males that have
 Linkage, Recombination, and Eukaryotic Gene Mapping 195

ebony body, rough eyes, and brevis bristles; 1800 progeny are A total of 26 double crossovers should be observed. Because
+ +
produced. Give the expected numbers of phenotypes in the there are two classes of double crossovers ( e / ro / bv
+
progeny of the testcross. and e / ro / bv ), we expect to observe 13 of each class.
Next, we determine the number of single-crossover progeny.
• Solution The genetic map indicates that the distance between e and ro is
The crosses are: 20 m.u.; so 360 progeny (20% of 1800) are expected to have result-
e ro bv e ro bv ed from recombination between these two loci. Some of them will
P be single-crossover progeny and some will be double-crossover
e ro bv e ro bv progeny. We have already determined that the number of double-
A crossover progeny is 26; so the number of progeny resulting from
a single crossover between e and ro is 360 − 26 = 334, which will
e ro bv
F1 be divided equally between the two single-crossover phenotypes
+ + +
e ro bv ( e / ro / bv and e / ro bv ).

A The distance between ro and bv is 12 m.u.; so the number of

progeny resulting from recombination between these two genes
e ro bv e ro bv is 0.12 × 1800 = 216. Again, some of these recombinants will be
Testcross
e ro bv e ro bv single-crossover progeny and some will be double-crossover prog-
eny. To determine the number of progeny resulting from a single
In this case, we know that ro is the middle locus because the crossover, subtract the double crossovers: 216 − 26 = 190. These
genes have been mapped. Eight classes of progeny will be single-crossover progeny will be divided between the two single-
produced from this cross: + + +
crossover phenotypes ( e ro / bv and e ro / bv ); so there will
e ro bv nonrecombinant be /2 = 95 of each of these phenotypes. The remaining progeny
190

e ro bv nonrecombinant will be nonrecombinants, and they can be obtained by subtraction:

e / ro bv single crossover between e and ro 1800 − 26 − 334 − 190 = 1250; there are two nonrecombinants
+ + +
e / ro bv single crossover between e and ro ( e ro bv and e ro bv); so there will be 1250/2 = 625 of each.
e ro / bv single crossover between ro and bv The numbers of the various phenotypes are listed here:
e ro / bv single crossover between ro and bv
e / ro / bv double crossover e+ ro+ bv+ 625 nonrecombinant
e / ro / bv double crossover e ro bv 625 nonrecombinant
e+ / ro bv 167 single crossover between e and ro
To determine the numbers of each type, use the map distances, e / ro+ bv+ 167 single crossover between e and ro
starting with the double crossovers. The expected number of e+ ro+ / bv 95 single crossover between ro and bv
double crossovers is equal to the product of the single-crossover e ro / bv+ 95 single crossover between ro and bv
probabilities:
e+ / ro / bv+ 13 double crossover
expected number of double crossovers = 0.20 × 0.12 × 1800 e / ro+ / bv 13 double crossover
= 43.2 Total 1800
However, there is some interference; so the observed number
4. The locations of six deletions have been mapped to the
of double crossovers will be less than the expected. The
Drosophila chromosome as shown in the following diagram.
interference is 1 − coefficient of coincidence; so the
coefficient of coincidence is: Chromosome
coefficient of coincidence = 1 − interference
Deletion 1
The interference is given as 0.4; so the coefficient of coincidence Deletion 2
equals 1 − 0.4 = 0.6. Recall that the coefficient of coincidence is: Deletion 3
coefficient of coincidence = Deletion 4
number of observed double crossovers Deletion 5
number of exxpected double crossovers Deletion 6

Rearranging this equation, we obtain: Recessive mutations a, b, c, d, e, f, and g are known to be located
in the same regions as the deletions, but the order of the
number of observed double crossovers = mutations on the chromosome is not known. When flies
coefficient of coincidence × number of expected double crossovers homozygous for the recessive mutations are crossed with flies
homozygous for the deletions, the following results are obtained,
number of observed double crossovers = 0.6 × 43.2 = 26 where the letter “m” represents a mutant phenotype and a plus
196 Chapter 7

sign (+) represents the wild type. On the basis of these data, Mutation b is expressed only when deletion 1 is present; so b
determine the relative order of the seven mutant genes on the must be located in the region of the chromosome covered by
chromosome: deletion 1 and none of the other deletions:
Mutations b a
Deletion a b c d e f g
Deletion 1
1 + m m m + + + Deletion 2
2 + + m m + + + Deletion 3
3 + + + m m + + Deletion 4
4 m + + m m + + Deletion 5
5 m + + + + m m Deletion 6
6 m + + m m m + Using this procedure, we can map the remaining mutations.
For each mutation, we look for the areas of overlap among
• Solution deletions that express the mutations and exclude any areas of
The offspring of the cross will be heterozygous, possessing one overlap that are covered by other deletions that do not express
chromosome with the deletion and wild-type alleles and its the mutation:
homolog without the deletion and recessive mutant alleles. For
b c d e a f g
loci within the deleted region, only the recessive mutations will
be present in the offspring, which will exhibit the mutant Deletion 1
phenotype. The presence of a mutant trait in the offspring Deletion 2
therefore indicates that the locus for that trait is within the Deletion 3
region covered by the deletion. We can map the genes by Deletion 4
examining the expression of the recessive mutations in the flies Deletion 5
with different deletions. Deletion 6
Mutation a is expressed in flies with deletions 4, 5, and 6 but
not in flies with other deletions; so a must be in the area that is
unique to deletions 4, 5, and 6:
a

Deletion 1
Deletion 2
Deletion 3
Deletion 4
Deletion 5
Deletion 6

COMPREHENSION QUESTIONS
Section 7.1 *5. What is the difference between genes in coupling
*1. What does the term recombination mean? What are two configuration and genes in repulsion? What effect does the
causes of recombination? arrangement of linked genes (whether they are in coupling
configuration or in repulsion) have on the results of a
Section 7.2 cross?
*2. In a testcross for two genes, what types of gametes are 6. How would you test to see if two genes are linked?
produced with (a) complete linkage, (b) independent 7. What is the difference between a genetic map and a
assortment, and (c) incomplete linkage? physical map?
3. What effect does crossing over have on linkage? *8. Why do calculated recombination frequencies between
4. Why is the frequency of recombinant gametes always half pairs of loci that are located far apart underestimate the
the frequency of crossing over? true genetic distances between loci?
 Linkage, Recombination, and Eukaryotic Gene Mapping 197

Section 7.3 Section 7.4

9. Explain how to determine, using the numbers of progeny 11. What is a lod score and how is it calculated?
from a three-point cross, which of three linked loci is the 12. List some of the methods for physically mapping genes
middle locus. and explain how they are used to position genes on
*10. What does the interference tell us about the effect of one chromosomes.
crossover on another?

APPLICATION QUESTIONS AND PROBLEMS

Introduction white-banded wings in a testcross. The progeny of this

13. The introduction to this chapter described the search for testcross are:
genes that determine pattern baldness in humans. In wild-type eyes, wild-type wings 418
1916, Dorothy Osborn suggested that pattern baldness is red eyes, wild-type wings 19
a sex-influenced trait (see Chapter 5) that is dominant in wild-type eyes, white-banded wings 16
males and recessive in females. More research suggested red eyes, white-banded wings 426
that pattern baldness is an X-linked recessive trait. Would
you expect to see independent assortment between a. What phenotypic proportions would be expected if the
genetic markers on the X chromosome and pattern genes for red eyes and for white-banded wings were
baldness if (a) pattern baldness is sex-influenced and located on different chromosomes?
(b) if pattern baldness is X-linked recessive? Explain your b. What is the percent recombination between the genes for
answer. red eyes and those for white-banded wings?
Section 7.2 *16. A geneticist discovers a new mutation in Drosophila
14. In the snail Cepaea nemoralis, an autosomal allele causing melanogaster that causes the flies to shake and quiver. She
a banded shell (BB) is recessive to the allele for an calls this mutation spastic (sps) and determines that spastic
unbanded shell (BO). Genes at a different locus determine is due to an autosomal recessive gene. She wants to
the background color of the shell; here, yellow (CY) is determine if the gene encoding spastic is linked to the
recessive to brown (CBw). A banded, yellow snail is recessive gene for vestigial wings (vg). She crosses a fly
crossed with a homozygous brown, unbanded snail. homozygous for spastic and vestigial traits with a fly
The F1 are then crossed with banded, yellow snails homozygous for the wild-type traits and then uses the
(a testcross). resulting F1 females in a testcross. She obtains the
following flies from this testcross.
a. What will the results of the
testcross be if the loci that vg+ sps+ 230
control banding and color are vg sps 224
linked with no crossing over? vg sps+ 97
b. What will the results of the vg+ sps 99
testcross be if the loci assort
independently? Total 650
c. What will the results of the Are the genes that cause vestigial wings and the spastic
testcross be if the loci are [Otto Hahn/Photolibrary.] mutation linked? Do a chi-square test of independence to
linked and 20 m.u. apart? determine whether the genes have assorted
15. In silkmoths (Bombyx mori), red eyes (re) and white- independently.
banded wing (wb) are encoded by two mutant alleles that 17. In cucumbers, heart-shaped leaves (hl) are recessive to
are recessive to those that produce wild-type traits (re+ normal leaves (Hl) and having numerous fruit spines (ns)
and wb+); these two genes are on the same chromosome. is recessive to having few fruit spines (Ns). The genes for
A moth homozygous for red eyes and white-banded leaf shape and for number of spines are located on the
wings is crossed with a moth homozygous for the wild- same chromosome; findings from mapping experiments
type traits. The F1 have normal eyes and normal wings. indicate that they are 32.6 m.u. apart. A cucumber plant
The F1 are crossed with moths that have red eyes and having heart-shaped leaves and numerous spines is
198 Chapter 7

crossed with a plant that is homozygous for normal es St es St

leaves and few spines. The F1 are crossed with plants that es St Y

A
have heart-shaped leaves and numerous spines. What
phenotypes and proportions are expected in the progeny
of this cross? es St
1630
es St
*18. In tomatoes, tall (D) is dominant over dwarf (d) and
smooth fruit (P) is dominant over pubescent fruit (p), es St
1665
which is covered with fine hairs. A farmer has two tall and es St
smooth tomato plants, which we will call plant A and es St
plant B. The farmer crosses plants A and B with the same 935
es St
dwarf and pubescent plant and obtains the following
numbers of progeny: es St
1005
es St
Progeny of es St
1661
Plant A Plant B Y
Dd Pp 122 2
es St
Dd pp 6 82 1024
Y
dd Pp 4 82
dd pp 124 4 a. Which progeny are the recombinants and which progeny
are the nonrecombinants?
a. What are the genotypes of plant A and plant B? b. Calculate the recombination frequency between es and St.

b. Are the loci that determine height of the plant and c. Are some potential genotypes missing among the progeny
pubescence linked? If so, what is the percent of the cross? If so, which ones and why?
recombination between them?
c. Explain why different proportions of progeny are 21. Recombination rates between three loci in corn are
produced when plant A and plant B are crossed with the shown here.
same dwarf pubescent plant.
Loci Recombination rate
19. Alleles A and a are at a locus on the same chromosome as R and W2 17%
is a locus with alleles B and b. Aa Bb is crossed with aa bb R and L2 35%
and the following progeny are produced:
W2 and L2 18%
Aa Bb 5
What is the order of the genes on the chromosome?
Aa bb 45
aa Bb 45
22. In tomatoes, dwarf (d) is recessive to tall (D) and opaque
aa bb 5 (light-green) leaves (op) are recessive to green leaves (Op).
The loci that determine height and leaf color are linked
What conclusion can be made about the arrangement of and separated by a distance of 7 m.u. For each of the
the genes on the chromosome in the Aa Bb parent? following crosses, determine the phenotypes and
proportions of progeny produced.
20. Daniel McDonald and Nancy Peer determined that eyespot
DATA (a clear spot in the center of the eye) in flour beetles is D Op d op
a.
ANALYSIS
caused by an X-linked gene (es) that is recessive to the d op d op
allele for the absence of eyespot (es+). They conducted a
series of crosses to determine the distance between the D op d op
b.
gene for eyespot and a dominant X-linked gene for d Op d op
striped (St), which causes white stripes on females and D Op D Op
acts as a recessive lethal (is lethal when homozygous in c.
d op d op
females or hemizygous in males). The following cross was
carried out (D. J. McDonald and N. J. Peer. 1961. Journal D op D op
d.
of Heredity 52:261–264). d Op d Op
 Linkage, Recombination, and Eukaryotic Gene Mapping 199

23. In German cockroaches, *26. A series of two-point crosses were carried out among
bulging eyes (bu) are seven loci (a, b, c, d, e, f, and g), producing the
recessive to normal eyes following recombination frequencies. Map the seven
(bu+) and curved wings (cv) loci, showing their linkage groups, the order of the loci
are recessive to straight in each linkage group, and the distances between the
wings (cv+). Both traits are loci of each group.
encoded by autosomal genes
Percent Percent
that are linked. A cockroach [W. Willner/Photolibrary.]
Loci recombination Loci recombination
has genotype bu+bu cv+cv, and the genes are in replusion.
Which of the following sets of genes will be found in the a and b 50 c and d 50
most-common gametes produced by this cockroach? a and c 50 c and e 26
a. bu+ cv+ d. cv+ cv a and d 12 c and f 50
b. bu cv e. bu cv+ a and e 50 c and g 50
+
c. bu bu a and f 50 d and e 50
Explain your answer. a and g 4 d and f 50
*24. In Drosophila melanogaster, ebony body (e) and rough b and c 10 d and g 8
eyes (ro) are encoded by autosomal recessive genes b and d 50 e and f 50
found on chromosome 3; they are separated by 20 m.u. b and e 18 e and g 50
The gene that encodes forked bristles (f ) is X-linked
b and f 50 f and g 50
recessive and assorts independently of e and ro. Give the
phenotypes of progeny and their expected proportions b and g 50
when a female of each of the following genotypes is 27. R. W. Allard and W. M. Clement determined
test-crossed with a male. DATA
recombination rates for a series of genes in lima beans
(R. W. Allard and W. M. Clement. 1959. Journal of
e+ ro+ f+ ANALYSIS
a. Heredity 50:63–67). The following table lists paired
e ro f recombination rates for eight of the loci (D, Wl, R, S,
L1, Ms, C, and G) that they mapped. On the basis of
e+ ro f+
b. these data, draw a series of genetic maps for the
e ro+ f different linkage groups of the genes, indicating the
25. Honeybees have haplodiploid sex determination: females distances between the genes. Keep in mind that these
DATA
are diploid, developing from fertilized eggs, whereas males rates are estimates of the true recombination rates and
are haploid, developing from unfertilized eggs. Otto that some error is associated with each estimate. An
ANALYSIS

Mackensen studied linkage relations among eight asterisk beside a recombination frequency indicates that
mutations in honeybees (O. Mackensen. 1958. Journal of the recombination frequency is significantly different
Heredity 49:99–102). The following table gives the results from 50%.
of two of MacKensen’s crosses including three recessive Recombination Rates (%) among Seven Loci in Lima Beans
mutations: cd (cordovan body color), h (hairless), and ch
Wl R S L1 Ms C G
(chartreuse eye color).
D 2.1* 39.3* 52.4 48.1 53.1 51.4 49.8
Queen genotype Phenotypes of drone (male) progeny Wl 38.0* 47.3 47.7 48.8 50.3 50.4
cd h 294 cordovan, 236 hairless, 262 cordovan R 51.9 52.7 54.6 49.3 52.6
cd h and hairless, 289 wild .type S 26.9* 54.9 52.0 48.0
L1 48.2 45.3 50.4
h ch 3131 hairless, 3064 chartreuse, 96 Ms 14.7* 43.1
h ch chartreuse and hairless, 132 wild type C 52.0
a. Only the genotype of the queen is given. Why is the *Significantly different from 50%.
genotype of the male parent not needed for mapping these
genes? Would the genotype of the male parent be required Section 7.3
if we examined female progeny instead of male progeny? 28. Raymond Popp studied linkage among genes for pink eye
DATA
b. Determine the nonrecombinant and recombinant progeny (p), shaker-1 (sh-1, which causes circling behavior, head
for each cross and calculate the map distances between cd, ANALYSIS
tossing, and deafness), and hemoglobin (Hb) in mice
h, and ch. Draw a linkage map illustrating the linkage (R. A. Popp. 1962. Journal of Heredity 53:73–80). He
arrangements among these three genes. performed a series of testcrosses, in which mice
200 Chapter 7

heterozygous for pink eye, shaker-1, and hemoglobin 1 a. Determine the order of these genes on the chromosome.
and 2 were crossed with mice that were homozygous for b. Calculate the map distances between the genes.
pink eye, shaker-1, and hemoglobin 2.
c. Determine the coefficient of coincidence and the
P Sh-1 Hb 1 p sh-1 Hb 2 interference among these genes.
p sh-1 Hb 2 p sh-1 Hb 2
30. Priscilla Lane and Margaret Green studied the linkage
DATA
The following progeny were produced. relations of three genes affecting coat color in mice:
ANALYSIS
mahogany (mg), agouti (a), and ragged (Ra). They
Progeny genotype Number carried out a series of three-point crosses, mating mice
p sh-1 Hb 2 that were heterozygous at all three loci with mice that
274
p sh-1 Hb 2 were homozygous for the recessive alleles at these loci
(P W. Lane and M. C. Green. 1960. Journal of Heredity
P Sh-1 Hb 1
320 51:228–230). The following table lists the results of the
p sh-1 Hb 2
testcrosses.
P sh-1 Hb 2
57 Phenotype Number
p sh-1 Hb 2
a Ra + 1
p Sh-1 Hb 1 + + mg 1
45
p sh-1 Hb 2 a + + 15
p Sh-1 Hb 2 + Ra mg 9
6 + + + 16
p sh-1 Hb 2
a Ra mg 36
p sh-1 Hb 1 a + mg 76
5 + Ra + 69
p sh-1 Hb 2
Total 213
p Sh-1 Hb 2
0 Note: + represents a wild-type allele.
p sh-1 Hb 2
a. Determine the order of the loci that encode mahogany,
P sh-1 Hb 1
1 agouti, and ragged on the chromosome, the map distances
p sh-1 Hb 2 between them, and the interference and coefficient of
Total 708 coincidence for these genes.
b. Draw a picture of the two chromosomes in the triply
a. Determine the order of these genes on the chromosome. heterozygous mice used in the testcrosses, indicating which
b. Calculate the map distances between the genes. of the alleles are present on each chromosome.
31. Fine spines (s), smooth fruit (tu), and uniform fruit color
c. Determine the coefficient of coincidence and the (u) are three recessive traits in cucumbers, the genes of
interference among these genes. which are linked on the same chromosome. A cucumber
*29. Waxy endosperm (wx), shrunken endosperm (sh), and plant heterozygous for all three traits is used in a testcross,
yellow seedling (v) are encoded by three recessive genes in and the following progeny are produced from this
corn that are linked on chromosome 5. A corn plant testcross:
homozygous for all three recessive alleles is crossed with a S U Tu 2
plant homozygous for all the dominant alleles. The s u Tu 70
resulting F1 are then crossed with a plant homozygous for S u Tu 21
the recessive alleles in a three-point testcross. The progeny s u tu 4
of the testcross are: S U tu 82
wx sh V 87 s U tu 21
Wx Sh v 94 s U Tu 13
Wx Sh V 3,479 S u tu 17
wx sh v 3,478 Total 230
Wx sh V 1,515
wx Sh v 1,531 a. Determine the order of these genes on the chromosome.
wx Sh V 292 b. Calculate the map distances between the genes.
Wx sh v 280 c. Determine the coefficient of coincidence and the
Total 10,756 interference among these genes.
 Linkage, Recombination, and Eukaryotic Gene Mapping 201

d. List the genes found on each chromosome in the parents following results are obtained, in which “m” represents a
used in the testcross. mutant phenotype and a plus sign (+) represents the wild
*32. In Drosophila melanogaster, black body (b) is recessive to type. On the basis of these data, determine the relative
gray body (b+), purple eyes (pr) are recessive to red eyes order of the seven mutant genes on the chromosome:
(pr+), and vestigial wings (vg) are recessive to normal Mutations
wings (vg+). The loci encoding these traits are linked, with
the following map distances: Deletion a b c d e f
1 m + m + + m
b pr v g 2 m + + + + +
3 + m m m m +
bß 6 ßB bßßßß ßßßßßßßB 4 + + m m m +
5 + + + m m +
The interference among these genes is 0.5. A fly with a 6 + m + m + +
black body, purple eyes, and vestigial wings is crossed with
a fly homozygous for a gray body, red eyes, and normal 35. A panel of cell lines was created from human–mouse
wings. The female progeny are then crossed with males somatic-cell fusions. Each line was examined for the
that have a black body, purple eyes, and vestigial wings. If presence of human chromosomes and for the production
1000 progeny are produced from this testcross, what will of an enzyme. The following results were obtained:
be the phenotypes and proportions of the progeny?
33. A group of geneticists are interested in identifying genes Cell Human chromosomes
that may play a role in susceptibility to asthma. They study line Enzyme 1 2 3 4 5 6 7 8 9 10 17 22
the inheritance of genetic markers in a series of families that A − + − − − + − − − − − + −
have two or more asthmatic children. They find an B + + + − − − − − + − − + +
association between the presence or absence of asthma and
C − + − − − + − − − − − − +
a genetic marker on the short arm of chromosome 20 and
calculate a lod score of 2 for this association. What does this D − − − − + − − − − − − − −
lod score indicate about genes that may influence asthma? E + + − − − − − − + − + + −

Section 7.4 On the basis of these results, which chromosome has the
gene that encodes the enzyme?
*34. The locations of six deletions have been mapped to the
Drosophila chromosome, as shown in the following deletion *36. A panel of cell lines was created from human–mouse
map. Recessive mutations a, b, c, d, e, and f are known to be somatic-cell fusions. Each line was examined for the
located in the same region as the deletions, but the order of presence of human chromosomes and for the production
the mutations on the chromosome is not known. of three enzymes. The following results were obtained.

Chromosome Enzyme Human chromosomes

Cell
Deletion 1 line 1 2 3 4 8 9 12 15 16 17 22 X
Deletion 2 A + − + − − + − + + − − +
Deletion 3 B + − − − − + − − + + − −
Deletion 4
C − + + + − − − − − + − +
Deletion 5
Deletion 6 D − + + + + − − − + − − +

When flies homozygous for the recessive mutations are On the basis of these results, give the chromosome
crossed with flies homozygous for the deletions, the location of enzyme 1, enzyme 2, and enzyme 3.

CHALLENGE QUESTION
Section 7.5
37. Transferrin is a blood protein that is encoded by the Genotype Phenotype
DATA transferrin locus (Trf ). In house mice, the two alleles at Trf a/Trf a Trf-a
this locus (Trf a and Trf b) are codominant and encode Trf a/Trf b Trf-ab
ANALYSIS
three types of transferrin: Trf b/Trf b Trf-b
202 Chapter 7

The dilution locus, found on the same chromosome, determines a. Calculate the recombinant frequency between the Trf and
whether the color of a mouse is diluted or full; an allele for the d loci by using the pooled data from all the crosses.
dilution (d) is recessive to an allele for full color (d+): b. Which crosses represent recombination in male gamete
Genotype Phenotype formation and which crosses represent recombination in
d+d+ d+ (full color) female gamete formation?
+
d d d+ (full color) c. On the basis of your answer to part b, calculate the
dd d (dilution) frequency of recombination among male parents and
female parents separately.
Donald Shreffler conducted a series of crosses to determine d. Are the rates of recombination in males and females the
the map distance between the tranferrin locus and the same? If not, what might produce the difference?
dilution locus (D. C. Shreffler. 1963 Journal of Heredity
54:127–129). The following table presents a series of
crosses carried out by Shreffler and the progeny resulting
from these crosses.

Progeny phenotypes
d d d d
Cross Trf-ab Trf-b Trf-ab Trf-b Total A mouse with the dilution
d Trf a d Trf b trait. [L. Montoliu, et al., Color
1 32 3 6 21 62 Genes. (European Society for
d Trf b d Trf b Pigment Cell Research, 2010).]

d Trf b d Trf a
2 16 0 2 20 38
d Trf b d Trf b

d Trf a d Trf b
3 35 9 4 30 78
d Trf b d Trf b

d Trf b d Trf a
4 21 3 2 19 45
d Trf b d Trf b

d Trf b d Trf b
5 8 29 22 5 64
d Trf a d Trf b

d Trf b d Trf b
6 4 14 11 0 29
d Trf b d Trf a