Chapter 7 -- Linkage, Recomination, and Eukaryotic Gene Mapping
These notes are provided to help direct your study from the textbook. They are not designed to explain all aspects of the material in great detail; that is what class time and the textbook is for. If you were to study only these notes, you would not learn enough genetics to do well in the course.
Soon after Mendel's rules were rediscovered, it was found that some loci
did not assort independently. (figures 7.3, 7.5, and 7.7)
While looking at figure 7.7, some points to consider are . . .
- The F2 generation does not fit the predicted ratio of 1:1:1:1 that we
would expect from a dihybrid testcross. Thus, this cross violates
the Rule of Independent Assortment
- Two phenotypes are in very high frequency have the same phenotypes as
the original parents (P1). These are called non-recombinants or
- Two phenotypes are in low frequency and combine the phenotypes of the
two original parents (P1). These are called recombinants or
The simplest explanation is that these two loci lie close to each other
on the same chromosome. They are linked on the same chromosome.
# of recombinants
100 x ----------------- = % of recombinants
# of offspring
1% recombination = 1 map unit, or 1 centimorgan, in honor of T.H. Morgan,
one of the first persons to propose this linkage, and first to win a
Nobel prize in genetics.
Mechanism for recombinant gametes: Recombinant gametes, which
eventually form recombinant offspring, result from crossing-over in
Prophase I (figure 7.6). In zygonema, bivalents are formed via the
synaptonemal complex and the homologues are paired. By pachynema, we
can see the tetrads, and by diplonema, we can see the chiasmata
where crossing-over occurred.
trans or replusion configuration = In the dihybrid, the mutant alleles are across from
each other on separate chromosomes (figure 6.8). (trans = repulsion)
cis or coupling configuration = In the dihybrid, the mutant alleles are on the same
chromosome (figure 6.8). (cis = coupling)
When crossing-over, we make the following three assumptions.
- It leads to recombinations of linked genes in reproduction, which
can be seen in the results of the crosses in figures
7.7 and 7.8.
- It takes place after chromosomes have replicated (in the four strand
stage). Each cross-over event involves only two of the four
- It is a process that involves an exchange of parts of homologous
Timing of crossing-over -- assumption #2
The tetrad analysis of Neurospora provided strong evidence that
crossing-over occurred after replication and during meiosis.
2 possibilities were originally proposed
The two theories predict different types of tetrads (or patterns of
spores) will be formed during meiosis in Neurospora.
- crossing-over occurred in G1 phase before replication of the
- crossing-over occurs after replication of the DNA, which occurs during S phase.
To address this question we can use an analysis of the pattern of spores
in the asci of fungi in the group Ascomycetes (Figure 6.19 and Table 6.4).
Most of these fungi are haploid and are only diploid for a brief period
in their life cycle.
- many fungi produce a tetrad of spores as products of a single
- may consist of ordered or unordered arrangement, where
ordered spores retain a fixed position with the ascus.
A cross between an ab strain and an a+b+ strain will yield a parental
ditype pattern of spores in the ascus, if no crossing-over occurred.
The same cross would yield only the nonparental ditype pattern, if
crossing-over occurred before DNA synthesis (2-strand stage).
Finally, the same cross would yield either the tetratype pattern if
crossing-over occurred after DNA synthesis (4-strand stage), or the
nonparental ditype pattern if 2 cross-overs occurred in the 4-strand stage.
Thus the presence of the tetratype pattern, which can only occur with a
cross-over in the 4-strand stage, indicates that crossing-over occurs
after DNA replication.
Physical Exchange of Genetic Material. -- assumption #3
Two experiments were conducted, one by Creighton and McClintock on corn,
and the other by Stern on Drosophila. The purpose was to relate the
physical exchange between homologous chromosomes (as seen by microscopy)
to the genetic recombination as seen in the phenotypes. Your book
presents the work of Crieghton and McClintock (page 170), while a very
similar experiment by Stern is summarized below.
Stern was able to isolate two strains of Drosophila, each having
different X chromosomes that could be distinguished under the
microscope. One strain had a piece of chromosome 4 attached to the X chromosome. The
other strain had a piece of the Y chromosome attached to the X. He now
had a system in which he could distinguish these chromosomes from each
other and from the normal X chromosomes. In addition, one of the
abnormal strains was homozygous recessive for carnation eye color and
homozygous for the dominant trait bar-eyes.
First, he mated two strains to produce F1 females that were
heterozygous for those chromosomes. The phenotypes of the
F1s were wildtype at carnation-eyed locus and bar-eyed at bar-eyed
locus (both dominant traits). Then he mated this heterozygous F1
to a male with carnation eyes (a testcross).
- He found that recombinant flies that were wild type at the carnation
locus carried at least part of the chromosome with the Y chromosome
- He also found that recombinant flies that were bar-eyed carried at
least part of the chromosome with chromosome 4 attached.
This was very powerful support for the concept that crossing-over
involved a physical exchange of material between non-sister chromatids.
Now that we have seen the consequences of crossing-over, and have
examined some of the physical properties of crossing-over, we can go
on to chromosome mapping. Most of the crosses we have dealt with
have been two-point crosses. Through a long and laborious breeding
experiment, we could produce maps of chromosomes. However, we can do
it much faster using a three-point cross. Also, the three-point
cross allows us to evaluate the effects of double cross-overs.
Be sure to study very carefully the problems in the text, especially figures
7.13 and 7.14. To work these problems, it is very important that you
approach them in a very consistent and systematic fashion. Failure to do
so will have you hopelessly mired in confusion.
Determining gene order
an example of a worked problem is in figure 7.13
Also note that as the parental phenotypes are composed of a
gamete from the female that is either b pr c or b+ pr+ c+, the alleles
are in the cis configuration.
- The pair of phenotypes with the highest frequency is always the
- The pair of phenotypes with the lowest frequency is always the
double cross-over group. The probability of a double cross-over
is approximately the product of the probability of the single
- compare the wild type class b+pr+c+ to the purple double cross-over
class b+prc+ and we can see that the purple locus
does not match indicating that the purple locus is in the middle.
Determining Map Distance
The next step is to set-up a table that is titled " number of recombinants
between". The percent
recombination is calculated as before.
# of recombinants
100 x ----------------- = % of recombinants
# of offspring
thus for the distance from b to pr
100 x ----------------- = 5.9% or 5.9 m.u.
The distance from b to c, the two outside loci, (25.4 m.u.) is the sum of
the distance from b to pr (5.9 m.u.) and the distance from pr to c
Coefficient of coincidence
Are cross-overs occurring independently of each other, or does one
cross-over affect the likelihood of a second cross-over in the same
This is addressed by asking Is the number of observed double
cross-overs equal to the expected number?
The product rule yields the probability of a double cross-over. Then multiplying the probability of a double cross-over times the number of offspring for the cross yields the expected number of double cross-overs. The ratio of the observed number of double cross-overs divided by the expected number of double cross-overs is the coefficient of coincidence.
Human chromosomal maps
The analysis of pedigrees can be used to map a human chromosome, and to
determine the percent recombination between two loci. This process
involves examining the percent of recombination between the locus of
interest and several markers that are scattered around the genome.
The likelihood of recombination is evaluated via the calculation of a
lod score, which is the logarithm of the probability of these offspring
given linkage divided by the probability of these offspring given
independent assortment. A lod score greater than 1 indicates linkage
and a lod score greater than 3 is seen as statistically significant.
Pedigree analysis is also used in genetic counseling.
Review the calculation of a lod score as shown on pages 182 and 183.
Evaluation of risk in genetic counseling
If the loci were not linked, we would be able to tell the woman that
her son would have a 50% chance of being hemophilic. When she becomes
pregnant . . .
- a female who is known to be a carrier of hemophilia (normal
father, carrier mother) wants to know if it is possible to
determine if the baby she is carrying has hemophilia
- determine that she is heterozygous at the GPD locus (GPD-A, GPD-a)
H = normal, h = hemophilia HhAa
- from previous mapping studies, it is known that these two loci lie
about 5 map units apart
- determine the genotype of her father at the GPD locus.
The father was not hemophiliac and thus had the H allele. If he has the
a allele at the GPD locus, then a allele at GPD is linked to the
H allele at the hemophilia locus in the mother and the A allele at
the GPD locus is linked to the h allele at the hemophilia locus
(h is linked to A and H is linked to a).
This is also used for other genetic traits such as Huntington's chorea,
though this trait occurs on an autosome and the process is somewhat more
- aspirate some amniotic fluid - this can be done at 3 months past
- examine the fetal cells in the fluid to determine the sex (by
karyotyping the cells)of the fetus - if the fetus
is female then there is no problem, assuming the father is normal
- if the fetus is male, examine the fetal cells in the fluid to
determine which form of GPD is present
- if GPD-A is present, then the fetus has a 95% chance of being
(crossing-over does not allow us to make 100% prediction)
- if GPD-a is present, then the fetus has a 95% chance of being normal
(this is much better than the 50/50 guess before this technique
Increased knowledge of the DNA sequences of genes and the advent of polymerase chain reaction techniques have allowed genetic counselors to examine directly the gene of interest to see if abnormalities occur. As these techniques are developed the above method is supplanted by this new ability.
Last update on 13 February 2005
Provide comments to Dwight Moore at email@example.com
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