Biology 212 General Genetics Spring 2007

Lectures 11 and 12: Linkage Mapping II and III

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 c b

a c b

+ + +

+ + +

after double crossovers:

a c b

a + b

+ c b

+ + +

·  Ability to identify double crossovers types is important to construction of three factor linkage maps.

2. Three factor linkage (=three point linkage)

·  Maps of three genes (factors)

·  Break down problem to series of two-factor linkage problems

·  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 x lz gl su

lz gl su lz gl su

Produce the progeny shown in Table 4.1

Phenotype of testcross progeny / Genotype of gamete from hybrid parent / Number of progeny
Wildtype / Lz Gl Su / 286
Lazy / lz Gl Su / 33
Glossy / Lz gl Su / 59
Sugary / Lz Gl su / 4
Lazy, glossy / lz gl Su / 2
Lazy, sugary / lz Gl su / 44
Glossy, sugary / Lz gl su / 40
Lazy, glossy, sugary / lz gl su / 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.

Lz Gl Su Lz Gl su

lz gl su lz gl Su

parental types 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 Su parental types

lz su

Lz su recombinant types

lz Su

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 Gl parental types

su gl

Su gl recombinant types

su Gl

calculate the recombination frequency between su and gl:

59 + 44 + 4 + 2/740 = 14.7% recombination

14.7 map units separate su and gl

lz su gl

10.7 m.u. 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)

·  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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