Download Biology 212 General Genetics

Biology 212 General Genetics
Lectures 11 and 12: Linkage Mapping II and III
Spring 2007
Reading Chapter 4 pp. 132-139
Lecture Outline:
1. Double crossovers
2. Three factor linkage
example 1 (Table 4.1)
3. interference
4. Three factor mapping in Drosophila
example 2 (Solved problem)
Lecture:
1. Double crossovers




When genes lie far apart on the chromosome, more than one crossover may occur
between them.
Probability of multiple crossovers increases as map distance between two genes
increases.
Double crossovers may not be detected, thus recombination frequencies (and
therefore genetic map distances) may be underestimated.
Use of a marker between two distant genes makes it easier to detect double
crossover types. See Fig. 4.13
prior to double crossovers:
a
a
+
+
c
c
+
+
b
b
+
+
after double crossovers:
a
a
+
+

c
+
c
+
b
b
b
+
Ability to identify double crossovers types is important to construction of three
factor linkage maps.
2. Three factor linkage (=three point linkage)
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
Maps of three genes (factors)
Break down problem to series of two-factor linkage problems
1

Use double crossovers to help determine gene order
Example 1: Three linked corn plant traits
trait 1
lz=lazy (growth habit)
Lz=wild type growth
trait 2
gl=glossy leaf
Gl=normal leaf
Trait 3
su=sugary endosperm (seed kernel)
Su=normal endosperm
Carry out a testcross of a heterozygote corn parent:
Lz Gl Su
lz gl su
x
lz gl su
lz gl su
Produce the progeny shown in Table 4.1
Phenotype of testcross
progeny
Wildtype
Lazy
Glossy
Sugary
Lazy, glossy
Lazy, sugary
Glossy, sugary
Lazy, glossy, sugary
Genotype of gamete from
hybrid parent
Lz Gl Su
lz Gl Su
Lz gl Su
Lz Gl su
lz gl Su
lz Gl su
Lz gl su
lz gl su
Number of progeny
286
33
59
4
2
44
40
272
Interpretation of results and construction of a linkage map:



The two most frequent class identify the non-recombinant (parental) gametes.
The two rarest classes identify the double crossover gametes.
The remaining classes are single crossover gametes.
a. What is the order of the three genes?
Compare the genotype of the most frequent class to the least frequent class and determine
which marker changes places.
2
Lz Gl Su
lz gl su
parental types


Lz Gl su
lz gl Su
double recombinant types
The Su/su alleles have exchanged places on the chromosomes in the double
recombinants.
Therefore the su gene is between the lz and gl genes:
lz
su
gl
b. What are the distances between the three genes?


break up into a series of two-factor linkage problems
compare the arrangement in the parental classes to the other classes
o if the same then they are non-recombinant
o if they are different, then they are recombinant and are used to compute
the recombination frequency
i) lz to su
Lz
lz
Su
su
parental types
Lz
lz
su
Su
recombinant types
calculate the recombination frequency between lz and su:
40 + 33 + 4 + 2/740 x 100 = 10.7% recombination
10.7 map units separate lz and su
ii) su to gl
Su
su
Gl
gl
parental types
Su
su
gl
Gl
recombinant types
calculate the recombination frequency between su and gl:
59 + 44 + 4 + 2/740 = 14.7% recombination
3
14.7 map units separate su and gl
lz
su
10.7 m.u.
gl
14.7 m.u.
3. Interference
interference = when crossing over in one region interferes with crossing over in a
neighboring region
can use probabilities to determine the expected numbers of double crossovers
coefficient of coincidence = observed number of double crossovers/expected number;
gives a quantitative measure of interference; Interference = 1 - (coefficient of
coincidence)


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Interference and graphs of mapping functions (Fig. 4.17) can be used to compute
the actual map distances from the recombination frequencies.
For short distances the recombination frequency = map distance.
For long distances, need to know the extent of interference to correctly estimate
map distance.
In this course, we will use the additive distance of close markers to estimate the distance
of more distal markers.
4. Three factor mapping in Drosophila
example 2 (Solved problem--handout)
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