This document discusses genetic linkage, detailing how genes located close together on the same chromosome fail to assort independently. It outlines the concept of recombination during meiosis, methods for testing gene linkage through test crosses, and the establishment of gene maps to assess gene distances based on recombination rates. Additionally, it covers complications such as double crossovers and interference in genetic mapping, providing an overview of methodologies and experimental findings related to gene linkage.

1. Mr. Rajendra singh Jr.
Biotechnology

2.  Linkage is defined genetically: the failure of two genes to
assort independently.
 Linkage occurs when two genes are close to each other on the
same chromosome.
 However, two genes on the same chromosome are called
syntenic.
 Linked genes are syntenic, but syntenic genes are not always
linked. Genes far apart on the same chromosome assort
independently: they are not linked.
 Linkage is based on the frequency of crossing over between
the two genes. Crossing over occurs in prophase of meiosis 1,
where homologous chromosomes break at identical locations
and rejoin with each other.

3.  In 1900, Mendel’s work was re-discovered, and
scientists were testing his theories with as many
different genes and organisms as possible.
 William Bateson and R.C. Punnett were working
with several traits in sweet peas, notably a gene for
purple (P) vs. red (p) flowers, and a gene for long
pollen grains (L) vs. round pollen grains (l).

4.  PP LL x pp ll
 selfed F1: Pp Ll
 F2 results in table
 Very significant deviation
from expected Mendelian
ratio: chi-square = 97.4,
with 3 d.f. Critical chi
square value = 7.815.
 The null hypothesis for chi
square test with 2 genes is
that the genes assort
independently. These
genes do not assort
independently.
phen
otype
obs exp
ratio
exp
num
P_ L_ 284 9/16 215
P_ ll 21 3/16 71
pp
L_
21 3/16 71
pp ll 55 1/16 24

5.  Purpose of a test cross: the offspring
phenotypes appear in the same ratio as
the gametes in the parent being tested.
 Here, we want to see how many gametes
are in the original parental configuration
(PL or pl) and how many are in the
recombinant configuration (Pl or pL).
The parental types have the same
combination of alleles that were in the
original parents, and the recombinant
types have a combination of the mother’s
and father’s alleles.
 Original parents: PP LL x pp ll
 F1 test cross: Pp Ll x pp ll
Pheno-
type
obs
purple
long
392
purple
round
116
red
long
127
red
round
365
total 1000

6.  Parentals: 392 PL + 365 pl = 757.
757/1000 total offspring = 75.7% parental
 Recombinant: 116 Pl + 127 pL = 243.
243 /1000 = 24.3% recombinant.
 If the genes were unlinked, 50% would be recombinant.
These genes are linked, with 24.3% recombination between
the P gene and the L gene.
 If the genes were right on top of each other, that is, the two
phenotypes were both caused by the same gene (pleiotropy),
then there would be 0% recombination between them.
 The percentage of recombinants is always between 0% and
50%, and the percentage of parentals is always between 50%
and 100%.

7.  We have been following Mendel’s tradition in writing the two alleles for each gene
together, as in PP LL x pp ll.
 Now we need to start paying attention to the fact that genes are on chromosomes.
 If one parent contributes a P L chromosome and the other parent contributes a p l
chromosome, we write the heterozygote as PL/pl.
 Homozygotes (for all genes on that chromosome) are written without the slash: the
pp ll homozygote used in the test cross is written as p l.

8.  The original test cross we did was PL/pl x p l. Among the
offspring, PL and pl were parental types, and pL and Pl were
the recombinant types. There was 24.3% recombination
between the genes.
 The condition of having the dominant alleles for both genes on
the same parental chromosome, with both recessives on the
other parental chromosome, is called “coupling”: the P and L
genes are “in coupling phase”.
 The opposite condition, having one dominant and one
recessive on each parental chromosome, is called “repulsion”.
Thus, if the original parents were P l x p L, their offspring
would have the genes in repulsion phase: Pl / pL.

9.  Now do the test cross in repulsion:
Pl / pL x p l
 Here, the parental types are P l and
p L, and the recombinant types are
P L and p l.
 The numbers of offspring in each
type are quite different from the
originals.
 However, the percentage of
recombinants is the same: 24.3%.
 123 P L + 120 p l = 243
recombinant offspring.
243/ 1000 total offspring = 24.3 %
 The percentage of recombination
depends on the distance between
the genes on the chromosome, and
NOT on which alleles are on which
chromosome.
phenotype obs
P L 123
P l 372
p L 385
p l 120
total 1000

10.  From an evolutionary point of view, the purpose of sex is to
re-shuffle the combinations of alleles so the offspring receive
a different set of alleles than their parents had. Natural
selection then causes offspring with good combinations to
survive and reproduce, while offspring with bad combinations
don’t pass them on.
 Genes are on chromosomes. Meiosis is a mechanism for re-
shuffling the chromosomes: each gamete gets a mixture of
paternal and maternal chromosomes.
 However, chromosomes are long and contain many genes. To
get individual genes re-shuffled, there needs to be a
mechanism of recombining genes that are on the same
chromosome. This mechanism is called “crossing over.

11.  Crossing over occurs in prophase of meiosis 1, when the
homologous chromosomes “synapse”, which means to pair
closely with each other. DNA strands from the two
chromosomes are matched with each other.
 During synapsis, an enzyme, “recombinase”, attaches to each
chromosome at several randomly chosen points. The
recombinase breaks both DNA molecules at the same point,
and re-attaches them to opposite partners.
 The result of crossing over can be seen in the microscope as
prophase continues, as X-shaped structures linking the
homologues.
 The genetic consequence of crossing over is that each
chromosome that goes into a gamete is a combination of
maternal and paternal chromosomes.

13. http://www.youtube.com/watch?v=BhJf9MHHmc4
Or, maybe this animated GIF image will work

14.  Each gene is found at a fixed position on a particular chromosome.
Making a map of their locations allows us to identify and study them
better. In modern times, we can use the locations to clone the genes so we
can better understand what they do and why they cause genetic diseases
when mutated.
 The basis of linkage mapping is that since crossing over occurs at random
locations, the closer two genes are to each other, the less likely it is that a
crossover will occur between them. Thus, the percentage of gametes that
had a crossover between two genes is a measure of how far apart those two
genes are.
 As pointed out by T. H. Morgan and Alfred Sturtevant, who produced the
first Drosophila gene map in 1913. Morgan was the founder of Drosophila
genetics, and in his honor a recombination map unit is called a
centiMorgan (cM).
 A map unit, or centiMorgan, is equal to crossing over between 2 genes in
1% of the gametes.

15.  The easiest way to map genes is to compare them in groups of
3. This allows both the distances between them and their
order to be determined. Further genes can be added to the
map by using overlapping groups of 3.
 Mapping is usually done in test crosses. One parent is
heterozygous for two versions of the chromosome being
mapped, and the other parent is homozygous for the recessive
mutants being mapped. Recombination in the heterozygous
parent gives different combinations of alleles, which are
counted.
 Note that recombination also occurs in the homozygous
recessive parent, but it has no effect on the alleles in the
offspring because it is homozygous.

16.  In corn, c gives a green plant
body, while its wildtype allele c+
gives a purple plant body.
 bz (bronze) gives brown seeds,
while the wildtype allele bz+
gives
purple seeds.
 wx (waxy) gives waxy
endosperm in the seeds; wx+
gives
starchy endosperm.
 The genes are arranged on the
chromosome in the order c-bz-
wx.
 The cross: c bz wx / + + + x c bz
wx.
 Note the +’s are the dominant
wildtype alleles of the
corresponding gene.
phenotype count
c bz wx 318
+ + + 324
c bz + 105
+ + wx 108
c + + 18
+ bz wx 20
c + wx 4
+ bz + 3
total 900

17.  Genes are arranged in reciprocal pairs: each pair has
1 copy of the mutant allele and the wildtype allele for
each gene. The counts are roughly equal for
reciprocal pairs, because they are both products of the
same crossing over events.
 Parentals are the largest groups: c bz wx and + + +.
 Double crossovers, one between c and bz and another
between bz and wx, are the smallest groups.

18.  Basic process: determine the percentage of offspring
that had a crossover between each pair of genes.
 1. Examine c and bz first. Parental configuration was
c bz and + +. Therefore, the recombinant
configurations are c + and + bz.
 Count recombinants, ignoring the other gene (wx): 18
c + + , 20 + bz wx, 4 c + wx, 3 + bz +. Total is 45
recombinants out of 900 total offspring. 45 / 900 =
0.05. Need to multiply by 100 to get percentage:
0.05 x 100 = 5.0 map units between c and bz.

19.  2. Next examine bz and wx. Parentals are bz wx and + +, so
recombinants are bz + and + wx.
 Ignoring c, the count of recombinants is: 105 c bz +, 108 + +
wx, 4 c + wx, 3 + bz +. Total = 220 recombinants. 220 / 900
= 0.244. 0.244 x 100 = 24.4 map units between bz and wx.
 3. Now do c and wx. Parentals are c wx and + +, so
recombinant offspring are c + and + wx.
 Ignoring bz, the recombinants are 105 c bz +, 108 + + wx,
18 c + + , 20 + bz wx. Total = 251. 251/ 900 x 100 = 27.9
map units.

20.  All 3 genes are in the
proper order, and all 3
distances between pairs
of genes are shown.
 Note that distances don’t
add up. This is due to
double crossovers, which
we will discuss next.

21.  A double crossover is two crossovers both occurring between the two
genes being examined. The first crossover changes the parental
configuration of alleles to the recombinant configuration. The second
crossover changes the recombinant configuration back to the parental. The
net result is that the genes are in the parental configuration, same as if no
crossovers had occurred.
 Thus, any even number of crossovers is the same as 0 crossovers, and any
odd number is the same as 1 crossover.
 Since you only see the offspring and not the actual crossovers, it is very
easy to undercount the number that occurred
 Consider the c bz wx cross. If you were just looking at c and wx, and
hadn’t examined bz, the c + wx and + bz + offspring would be parental
and count as 0 crossovers. The only way you know that 2 crossovers
occurred is by examining bz. Perhaps other crossovers also occurred that
we didn’t detect, since we didn’t examine any other genes in between bz
and wx.
 This is the main reason why the sum of the c--bz and bz--wx distances
didn’t add up: double crossovers were counted as parentals.

22.  The further apart 2 genes are, the more
likely it is that undetected double
crossovers will occur between them.
 For this reason, gene maps are created
using short intervals.
 One consequence: the maximum
percentage of recombinant offspring is
50%, but many chromosomes are several
hundred map units long. For instance, the
human chromosome 13 is 125 map units
long. Genes on opposite ends of this
chromosome would have a 50% frequency
of recombinant offspring.
 Mapping function: number of actual
crossovers on x-axis, frequency of
recombinant offspring on y-axis.
 In general,
◦ for less than 20% recombinants, the
number of recombinants is roughly
equal to the number of crossovers;
◦ for 20-50% recombinants, the number
of recombinants is significantly less
than the number of crossovers.
◦ for 50% recombinants, the number of
crossovers is not determinable.

23.  There is a second issue with double crossovers:
interference.
 Interference is the inability of 2 crossovers to occur
very close to each other. Think of the chromosome
as a thick rope: it is impossible to bend it too tightly.
 It is possible to measure the amount of interference,
by comparing the actual number of double crossovers
to the number that you would expect based on the
number of single crossovers that occurred.

24.  Consider our c--bz--wx example. The c--bz interval was 5.0 map units, or
5% recombinants between those 2 genes. The probability of a crossover
between c and bz is 0.05, which you get by dividing the map distance by
100 to put it on a 0-1 scale instead of a 0-100 scale. The bz--wx interval
was 24.4 map units, or 24.4% recombinants, or a 0.244 probability.
 If a crossover between c and bz had no effect on the possibility of a
crossover between bz and wx (i.e. no interference), then a chance of a c--
bz crossover AND and bz--wx crossover would be the product of their
individual probabilities.
 That is, the expected frequency of double crossovers would be 0.05 *
0.244 = 0.0122. Since there were 900 offspring, the expected number of
double crossover offspring would be 10.98.
 The observed number of double crossovers was 7. The coefficient of
coincidence is the ratio of observed double crossovers to expected: 7 /
10.98 = 0.638. The interference is 1 minus the coefficient of coincidence,
= 1 - 0.638 = 0.362. This means that about 36% of the expected double
crossovers are not occurring due to interference.

25. ( )offspringtotal
mapdistIImapdistI
CO _
100100
2exp 











=
CO
COobs
CofC
2exp_
2_
._. =
._.1 CofCI −=

26.  Another problem that can arise is that when mapping
genes you usually don’t know their order. You need
to infer the order from the data.
 Also, not all the alleles are necessarily going to be in
coupling. Sometimes some alleles are in repulsion.
 Getting the gene order is a matter of comparing the
alleles present on the parental chromosomes to those
on the double crossover chromosomes.

27.  Note that the offspring counts
are arranged in reciprocal pairs.
Each member of the pair has the
opposite alleles, and the counts
for the two members of the pair
are approximately equal.
 The parental class of offspring
is the largest: a + d, and + b
+. No crossovers have occurred
here: the original parents had
these combinations of alleles.
 The double crossovers class
(2CO) is the smallest class: + b
d and a + +.
 The other two pairs are the two
single crossover classes.
offspring
phenotype
count
a + d 510
+ b + 498
a + + 3
+ b d 1
a b + 61
+ + d 59
a b d 35
+ + + 37
total 1204

28.  Since we don’t know (or care about) the orientation of these
genes relative to the chromosome as a whole, there are only
3 possible orders. These are based on which gene is in the
middle.
 The genes could be a--b--d, b--a--d, or a--d--b.
 Note that a--b--d is identical to d--b--a, etc.
 The order in which the genes are listed in the offspring
counts has nothing to do with their order on the
chromosome!!! The gene order is unknown at this point.
 To determine gene order, set up the parental chromosomes in
the F1, then see what the resulting double crossover
offspring would look like. If the observed 2CO’s match the
expected, then you have the correct gene order. If not, try a
different order.

29.  1. Try a--b--d. The F1 chromosomes would be a + d/ + b +.
A double crossover would give a b d and + + + 2CO offspring.
This does not match the observed a + + and + b d.
 2. Try b--a--d. The F1 chromosomes are + a d / b + +. A
double crossover would give b a + and + + d 2CO offspring.
This does not match the observed b + d and + a +.
 3. Try a--d--b. The F1 chromosomes are a d + and + + b. A
double crossover gives a + + and + d b, which matches the
observed 2CO offspring. Therefore, a--d--b is the correct
order; d is in the middle.

30.  a--d. Parentals are a d and + +, so recombinants are a
+ and + d. There are 61 + 59 + 3 + 1 = 124
recombinant offspring. Total offspring = 1204. So
map distance = 124/1204 * 100 = 10.3 map units.
 d--b. Parentals are d + and + b, so recombinants are d
b and + +. There are 35 + 37 + 3 + 1 = 76 of them. 76
/ 1204 * 100 = 6.3 map units.
 a--b. Parentals are a + and + b, so recombinants are a
b and + +. There are 61 + 59 + 35 + 37 = 192 of
them. 192 / 1204 * 100 = 15.9 map units.

31.  There were 3 + 1 = 4 observed
2CO’s.
 Expected 2CO:
= (10.3/100) * (6.3/100) * 1204
= 7.81
 Coef. of Coincidence = obs 2CO /
exp 2CO
= 4 / 7.81
= 0.512
 Interference = 1 - C. of C.
= 1 - 0.512
= 0.487

32.  1. Organize the data into reciprocal pairs. For example, a + c and + b + are
a reciprocal pair, and so are a b c and + + +. The number of offspring for
each member of a pair will be similar.
 2. Determine which pair is the parental class: it is the LARGEST class.
 3. Determine which pair is the double crossover (2CO) class: the
SMALLEST class.
 4. Determine which gene is in the middle. If you compare the parentals
with the 2COs, the gene which switched partners is in the middle. If you
think two genes have switched sides, it is the other gene that is in the
middle.
 5. Determine map distances for all three pairs of genes. Count the number
of offspring that have had a crossover between the genes of interest, then
divide by the total offspring and multiply by 100.
 6. Figure the expected double crossovers: (map distance for interval I 100 ) *
(map distance for interval II / 100 ) * (total offspring) = exp 2CO.
 6. Figure the interference as: I = 1 - (obs 2CO / exp 2CO). The observed
2CO class comes from the data.

33.  Here are some test cross data for a 3 point cross.
Determine the order of the genes, make a map
showing all map distances, and determine the
interference value.
 pr = purple eyes
 bl = black body
 dp = dumpy bristles
 Answer will be given in class.

34. phenotype count
wild type 756
purple 59
dumpy 414
black 21
purple dumpy 23
purple black 417
dumpy black 60
purple dumpy black 750
total 2500

35. phenotype count
wild type 60
purple 756
dumpy 23
black 417
purple dumpy 414
purple black 21
dumpy black 750
purple dumpy black 59
total 2500