linkage

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Linkage& Crossing over
• Linkage occurs when genes are on the same chromosome. When genes occur on the
same chromosome, they are inherited as a single unit. Genes inherited in this way are
called Linked.
• Genes are located on specific regions of a certain chromosome, termed the gene locus
(plural: loci). A gene therefore is a specific segment of the DNA molecule.
• Linkage groups are invariably the same number as the pairs of homologous
chromosomes an organism possesses. Recombination occurs when crossing over has
broken linkage groups, as in the case of the genes for wing size and body color that
Morgan studied.
• Chromosome mapping was originally based on the frequencies of recombination
between alleles.
In dihybrid testcrosses for frizzle and white in chickens, Hutt (1931) obtained:
frizzled is dominant over normal (if one combines slightly and extremely frizzled). white is
dominant over colored. P1: White, Normal Colored, Frizzle F1: White, Frizzle
Testcross: White, Frizzle (F1) x Coloured, Normal
Note the marginal counts are in the 1:1 ratio we expect, but there is deviation in the main
table from 1:1:1:1. This deviation is due to linkage between the two genes. The percent
recombination is 100*(18+13)/157 = 19.7%. Under independent assortment the percent
recombination should be 50%. After mating another set of chickens of exactly the same
genotypes however, the following counts were made,
In the first testcross, the Frizzled and Coloured phenotypes seemed to co
segregate, but the reverse is seen in the second cross. This is what is referred to as
repulsion of the dominant traits (frizzled and white) in the first case, and coupling in the
second. The percentage deviation from 1:1:1:1 seems to be about the same in each table,
but in opposite directions. Actually, we always ignore the sign, and calculate the
recombination in this table as 100*(4+2)/33=18.2%. If one examines a large number of
genes in such a fashion in any organism, sets of genes are always linked together, while

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assorting independently (recombination 50%) with respect to members of other linkage
groups. It was realized in the 1920s that each linkage group corresponds to a
chromosome.
 Mapping
If one can arrange testcrosses for triple (or higher order) heterozygotes and recessives (a
three-point cross), the recombination can be calculated for the three pairs of genes. The
data will look like this example:
Trait A is controlled by a gene with alleles A and a, A dominant to a
Trait B is controlled by a gene with alleles B and b, B dominant to b
Trait C is controlled by a gene with alleles C and c, C dominant to c
Testcross is AaBbCc x abc/abc
Data from three-point cross of corn (colourless, shrunken, waxy) due to Stadler.
The table deviates drastically from the expected
1:1:1:1:1:1:1:1, so linkage is being observed.
ABC and abc are the two commonest phenotypes, and are "reciprocal classes", so the
heterozygote parent's phase was ABC/abc, rather than AbC/aBc etc. Recombination
events between A and B are calculated from the marginal AB table and so forth,
C AB = 100*(529+536)/45832 = 2.3% cAC cAB + cBC
C BC = 100*(4455+12+4654+20)/45832 = 19.9%
C AC = 100*(509+4455+524+4654)/45832 = 22.1%
When similar experiments are carried out involving larger numbers of loci from the same
linkage group, it becomes obvious that the set of pairwise recombination percentages
suggest strongly that the genes are ordered in a linear fashion, with recombination acting
as the distance between them.

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The linkage map that one constructs using recombination distance turns out to correspond
to the physical map of genes along the linear structure of the chromosome. Recombination
is the "phenotypic" effect of crossover or chiasma formation between homologous
chromosomes, whereby they exchange segments of DNA.
Positions on a linkage map are loci. Since "gene" can be taken to mean the different gene
forms (alleles), or the factor controlling a phenotype, geneticists often refer to the latter as
the locus sh1, rather than the gene sh1.
 Double Crossovers
If all three markers are in the same linkage group, that is, on the same chromosome, then
we can observe an ABC/abc undergoing two recombination events, one between A and B,
and another between B and C, to give AbC/aBc. This is what is going on in cells 7 and 8 of
the earlier example. If we had been looking only at dihybrid test cross data, then we would
not be able to detect these double recombinants. One notices that double recombinants
are not very common, so the effect on the estimates of the percent recombination is not
large. The corollary of this is that most chromosomes will experience only zero or one
recombinants. The estimated double recombination rate does add to our estimate of the
distance between the more distant loci (A and C in the example).
 Interference
The term interference refers to the fact that recombination seems to be suppressed close
to a first recombination event. The coincidence coefficient is the ratio of the observed
number of double recombinants to the expected number. For a given distance between
two loci, one can estimate the number of double recombinants that one would expect. At a
trivial level, imagine three loci, each 10% recombination distance apart. Then we would
expect in 1% of cases that a double recombinant would occur (one in each interval). The
rate of double recombinants is usually less than this expected value. The expected
frequency of double cross overs is thus is the product of the observed frquencies of the
ingle crosses overs. Interferences is calculated as I = 1-c where, I = Index of interference
c = Coefficent of coincidence c = Observed frequency of Double Cross overs
/Expected frequency of Double Cross overs
 Deriving Linkage Distance and Gene Order from Three-Point Crosses
By adding a third gene, we now have several different types of crossing over products that
can be obtained. The following figure shows the different recombinant products that are
possible.

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Now if we were to perform a testcross with F1, we would expect a 1:1:1:1:1:1:1:1 ratio. As
with the two point analyzes described above, deviation from this expected ratio indicates
that linkage is occurring.
The best way to become familiar with the analysis of three-point test cross data is to go
through an example. We will use the arbitrary example of genes A, B, and C. First make a
cross between individuals that are AABBCC and aabbcc. Next the F1 is testcrossed to an
individual that is aabbcc. We will use the following data to determine the gene order and
linkage distances. As with the two-point data, we will consider the F1 gamete composition.
The best way to solve these problems is to develop a systematic approach. First,
determine which of the the genotypes are the parental gentoypes. The genotypes found
most frequently are the parental genotypes. From the table it is clear that the ABC and abc
genotypes were the parental genotypes. Next we need to determine the order of the
genes. Once we have determined the parental genotypes, we use that information along
with the information obtained from the double-crossover.
The double crossover gametes are always in the lowest frequency. From the table
the ABc and abC genotypes are in the lowest frequency. The next important point is that a
double-crossover event moves the middle allele from one sister chromatid to the other.
This effectively places the non-parental allele of the middle gene onto a chromosome with
the parental alleles of the two flanking genes. We can see from the table that the C gene
must be in the middle because the recessive c allele is now on the same chromosome as
the A and B alleles, and the dominant C allele is on the same chromosome as the
recessive a and b alleles. Now that we know the gene order is ACB, we can go about
determining the linkage distances between A and C, and C and B.
The linkage distance is calculated by dividing the total number of recombinant gametes
into the total number of gametes. This is the same approach we used with the two-point
analyses that we performed earlier.
What is different is that we must now also consider the double-crossover events.
For these calculations we include those double-crossovers in the calculations of both
interval distances. So the distance between genes A and C is 17.9 cM
[100*((81+85+5+8)/1000)], and the distance between C and B is 7.0 cM
[100*((27+30+5+8)/1000)].
Now let's try a problem from Drosophila, by applying the principles we used in the above
example. The following table gives the results we will analyze.

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Step 1: Determine the parental genotypes.
The most abundant genotypes are the partenal types. These genotypes are v cv+ ct+ and
v+ cv ct. What is different from our first three-point cross is that one parent did not contain
all of the dominant alleles and the other all of the recessive alleles.
Step 2: Determine the gene order
To determine the gene order, we need the parental genotypes as well as the double
crossover geneotypes. As we mentioned above, the least frequent genotypes are the
double-crossover geneotypes. These geneotypes are v cv+ ct and v+ cv ct+. From this
information we can determine the order by asking the question: In the double-crossover
genotypes, which parental allele is not associated with the two parental alleles it was
associated with in the original parental cross. From the first double crossover, v cv+ ct, the
ct allele is associated with the v and cv+ alleles, two alleles it was not associated with in
the original cross. Therefore, ct is in the middle, and the gene order is v ct cv.
Step 3: Determing the linkage distances.
• v - ct distance caluculation. This distance is derived as follows: 100*((89+94+3+5)/1448)
= 13.2 cM • ct - cv distance calculation. This distance is derived as follows:
100*((45+40+3+5)/1448) = 6.4 cM
Step 4: Draw the map.
Three-point crosses also allows one to measure interference (I) among crossover events
within a given region of a chromosome. Specifically, the amount of double crossover gives
an indication if interference occurs. The concept is that given specific recombination rates
in two adjacent chromosomal intervals, the rate of double-crossovers in this region should
be equal to the product of the single crossovers. In the v ct cv example described above,
the recombination frequency was 0.132 between genes v and ct, and the recombination
frequency between ct andcv was 0.064. Therefore, we would expect 0.84% [100*(0.132 x
0.64)] double recombinants. With a sample size of 1448, this would amount to 12 double
recombinants. We actually only detected 8. To measure interference, we first calculate the

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coefficient of coincidence (c.o.c.) which is the ratio of observed to expected double
recombinants. Interference is then calculated as 1 - c.o.c. The formula is as follows:
For the v ct cvdata, the interference value is 33% [100*(8/12)]. Most often, interference
values fall between 0 and 1. Values less than one indicate that interference is occurring in
this region of the chromosome.
Note that if two genes are very far apart they will not be linked in practice even if they
reside on the same chromosome.
 Crossing over
The process in which interchange (exchange) of chromosomal segments between non-
sister chromatids of homologous chromosomes occurs resulting in the formation of
recombinants.
Crossing over and independent assortment are mechanisms that produce new
combination of genes. Crossing over is the mechanism for the formation of recombinants
for linked genes while the genes present on different chromosomes assort independently
and form recombinants along with parental types.
Crossing over takes place after the homologous chromosomes are paired resulting in the
formation of chiasmatic structures. After terminalization, recombinant chromatids are
produced. The frequency of crossing over between any two genes is directly proportional
to the distance between the genes. The maximum frequency of recombination that can
result from crossing over between two linked genes is 50%. Independent assortment of
genes also produces 50% of recombinant gametes. Therefore, recombination frequency
never exceeds 50%.
Essential features of crossing over
• Crossing over occurs at Pachytene stage after the synapsis of the homologous
chromosomes has occurred in prophase I to meiosis.
• Only two of the four chromatids are involved in crossing over.
• The crossing over resulting in recombinants occurs between the non-sister
chromatids of the homologous pair of chromosomes. Crossing over that involves

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sister chromatids also occurs, but it is seldom detectable genetically, since the
sister chromatids are genetically identical.
• The number of chiasmata per set of chromosomes depends upon the length of
chromosome. Longer chromosomes will have greater number of chiasmata than
short chromosomes.
• Chromosomes with recombinant combinations of linked genes are formed by the
occurrence of crossing over in the region between the two loci.
• Chances of crossing over are more between distantly placed genes while closely
linked genes will have less chance of crossing over (i.e.), the probability that
crossing over will occur between two loci increases with increasing distance
between the two loci on the chromosome.
Kinds of crossing over
 Single cross over: If only one chiasma is formed all along the length of
chromosome pair, it is known as single cross over. The chromatids of homologous
chromosomes contact and break only at one point.
 Double cross over (DCO): In DCO, chromatids break and rejoin at two points i.e.
two chiasmata are formed along the entire length of the chromosome. In D.C.O.,
the formation of each chiasma is independent of the other.
Mechanism of crossing over
o Synapsis and formation of Synaptinemal complex:
During prophase-I of meiosis the maternal and paternal chromosomes of homologous pair
come close together and pair at zygotene stage and lie side by side all along their lengths.
The paired chromosomes are known as bivalents. The pairing of homologous
chromosomes occurs due to the shortage of DNA and histones in chromosomes and some
amount of DNA and histone synthesis occurs during zygotene stage. Once the
homologous chromosomes pair, they form a complex structure called as synaptinemal
Complex
o Duplication of chromosomes
Each of the homologous chromosomes in a bivalent, splits logitudinally into two sister
chromatids. Thus the bivalent consists of four chromatids and is known as a tetrad. The
longitudinal splitting of chromosomes is achieved by the separation of already duplicated
DNA molecules.
o Chiasmata formation
When the paired chromosomes start separating, the chromatids remain in contact at one
or more points and thus establish one or more exchanges per bivalent and these points of
contact are known as chiasmata. At each chiasma two non-sister chromatids of the
bivalent break at the corresponding points and then rejoin with the exchange of segments.
The breakage of chromatids is brought about by a nuclear enzyme, endonuclease and the
fusion of broken segments takes place due to the action of enzyme ligase.
o Terminalization
After crossing over, the non-sister chromatids start repelling each other, because the
forces at attraction keeping them together lapse. The chromatids separate from the
centromere towards the tip and the chiasmata moves in zipper like fashion towards the
ends. The movement of chiasmata is known as terminalization.

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Why map
1) Gene isolation - through map-based cloning or positional cloning
2) To mark or tag genes of interest
Loci for diseases, especially in humans
Plant or animal breeding strategies
3) Basic genetic studies - constructing maps in individual organisms and comparing
maps between species
10% or less - close linkage
30% or greater - pushing it – underestimates recombination
50% - independent assortment (It doesn't make sense to have recombination fractions
greater than 50%).
Crossing over has to occur between the 2 loci being examined or it cannot be detected.
Also, if a double crossover occurs, recombination will not be detected unless an
intermediate locus can also be analyzed.
Linkage maps, or recombinational maps: based on gene and/or marker segregation.
Physical map
Cytogenetic maps: based on the physical characteristics of the chromosomes.
Others: ultimate physical map is DNA sequence.
How Do You Begin Mapping
1) Identify genes of interest and /or marker loci.
2) Construct or identify a segregating population.
Either 1 or 2 may be done first or they may be done in conjunction with each other.
3) Determine the genotype of each member of the population at each locus, i.e. make
hypotheses about loci and alleles.
4) Test for random segregation of alleles at individual loci
5) Test whether alleles at different loci segregate independently of each other.
6) If 2 genes do not assort independently, determine the recombination frequency
between the 2 genes.
7) Do this for all identified loci.
8) Use recombination frequencies to order linked genes into ordered groups.