2/16/16
BIOLOGY 207 - Dr.Locke
Lecture#15 - Mapping genes on chromosomes by
recombination
Required readings and problems:
Reading: Open Genetics, Chapter 7
Problems: Chapter 7
Optional
Griffiths (2008) 9th Ed. Readings: pp 134-145; 155-157
Problems: 9th Ed. Ch. 4: 10-16, 32, 35-37, 42; not 11b 12c
Campbell (2008) 8th Ed. Readings: Concept 15.3
Concepts:
How are gene loci mapped using recombination?
1. Three linked genes can serve as an example for the construction of a genetic map
using recombination frequencies derived from a 3-point test cross.
2. This method of determining genetic distances can be extended to loci spread along
a whole chromosome and throughout a genome.
3. Resolution on genetic maps is limited by the markers available and the number of
progeny that can be analysed, consequently genetic maps are more refined in
prokaryotes than in eukaryotes.
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 1
2/16/16
Genes are physically arranged in a linear order along a chromosome.
The order and relative distances between them can be mapped by using the frequency
of crossingover along a chromosome.
In previous lectures -> Independent Assortment vs Linkage
Linked loci are syntenic: don't show independent assortment
- not 9:3:3:1 ratio in F2 cross - not 1:1:1:1 ratios in test cross
Closely linked:
-------•--------------A-B----- e.g. 2%
-------•--------------a-b----Partially linked:
-------•---------A------B----- e.g. 30%
-------•---------a------b-----
Absolute linkage = 0%
Independent Assortment = 50%
Today: 3-point cross
First - a 3-point cross to identify two linked gene loci and a third that assorts
independently (unlinked).
Second - a 3-point cross to identify three linked gene loci.
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 2
2/16/16
First example: 3 loci in Drosophila
scute echinus
vg (vestigial)
-----|----|----|--5.6 m.u.
unlinked
sc scute - loss of thoracic bristles
ec echinus - roughened eye surface all recessive
vg vestigial - vestigial wings
1) Construct a triply recessive stock and cross with wild type to produce an F1
sc ec vg
x
sc+ ec+ vg+
P1
sc ec vg
sc+ ec+ vg+
sc ec vg
sc+ ec+ vg+
gametes
sc ec vg
sc+ ec+ vg+
F1
2) Cross F1 triple heterozygote to triply recessive stock
sc ec vg
sc+ ec+ vg+
8 combinations
Biol207 Dr. Locke section
x
sc ec vg
sc ec vg
sc ec vg
Lecture#15
gametes
Fall'11
page 3
2/16/16
Progeny: 8 gametic types / phenotypes are possible:
#
Parental
sc ec vg
sc+ ec+ vg+
Recombinant
sc ec vg+
sc+ ec+ vg
sc ec+ vg
sc+ ec vg+
sc ec+ vg+
sc+ ec vg
|sc-ec
|sc-vg
|ec-vg
|Par Rec.|Par. Rec.|Par. Rec.
233 |233
239 |239
241
231
12
14
14
16
1000
100 %
|241
|231
|
|
|
|
|944
|
|233
|239
12
14
14
16
56
|233
|239
|
241 |
|
231 |
| 12
|
| 14
|
|
14 | 14
|
16 | 16
| 498 502 |502
241
231
12
14
.
498
Note 1) considerable deviation from 1:1:1:1:1:1:1:1 ratio that would be expected for all
unlinked genes. Therefore -> linked
2) sc <--> ec Calculate distance # recomb./total = 56/1000 = 5.6% 2 loci linked
3) sc <--> vg = 502/1000 = 50.2% therefore 2 loci not-linked
4) ec <--> vg = 498/1000 = 49.8% therefore 2 loci not-linked
Therefore the map must be:
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 4
2/16/16
Second example - 3 point test cross:
3 loci with recessive mutant alleles:
v vermilion coloured eye
cv cross veinless wing
ct cut wing edge
Coupling
a- b- (x) a+ b+
a- ba+ b+
Repulsion
a- b+ (x) a+ ba- b+
a+ b-
a- ba+ b+
a- b+
a+ b-
(x)
a- ba- ba- b-
a- ba- ba- b-
a- ba+ b+
a+ ba- b+
a- b+
a+ ba+ b+
a- b-
Both mutants from the same parent
Biol207 Dr. Locke section
(x)
Lecture#15
Mutations from different parents
Fall'11
page 5
2/16/16
(1) Use v+/v+ cv/cv ct/ct (crossveinless, cut phenotype) cross to
v/v cv+/cv+ ct+/ct+ (vermilion)
v+ cv ct
v+ cv ct
x
v+ cv ct
v cv+ ct+
v cv+ ct+
P1
v cv+ ct+
gametes
v cv+ ct+
v+ cv ct
F1
2) Cross F1 triple heterozygote to triply recessive stock
v cv+ ct+
x
v cv ct
v+ cv ct
v cv ct
8 types
Biol207 Dr. Locke section
v cv ct
Lecture#15
gametes
Fall'11
page 6
2/16/16
Progeny: 8 gametic types / phenotypes are possible:
#
Parental
v
cv+ ct+
v+ cv
ct
Recombinant
v
cv
ct+
v+ cv+ ct
ct
ct+
400
409
31
28
v
v+
cv
cv+
61
65
v
v+
cv+ ct
2
cv
ct+
4
totals 1000
| v-cv
| v-ct
|
cv-ct |
|Par. Rec.|Par. Rec.|Par. Rec.|
|
|
|
|
|400
|400
|400
|
|409
|409
|409
|
|
|
|
|
|
31 | 31
|
31 |
|
28 | 28
|
28 |
|
|
|
|
|
61 |
61 | 61
|
|
65 |
65 | 65
|
|
|
|
|
| 2
|
2 |
2 |
| 4
|
4 |
4 |
|815
185 |868
132 |935
65 |
|
18.5%|
13.2% |
6.5%
|
Conclude: all three loci are linked and a map can be produced.
Note: the order of loci is changed to take into account the mapping data.
The order for the analysis above was v cv ct but the genetic order is v ct cv.
v
ct
cv
|<=-------- 13.2 --------=> | <=-------6.5------=>|
13.2 + 6.5 = 19.7
|<=-----------------------18.5----------------------=>| = 18.5?
Why is 18.5 not the same as 19.7?
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 7
2/16/16
Answer:
We need to correct for double crossovers!
Double crossovers are not yet taken into account
The last two (rare) classes are double cross over class
Parental Double recombinants phenotypic class above
v ct+ cv+ ----=> v ct cv+
(v cv+ ct class above)
X
X
v+ ct cv ----=> v+ ct+ cv
(v+ cv ct+ class above)
To correct for double crossovers we need to count this class twice
(once for each crossover event) as a recombinant (not as a parental).
Therefore 31+28+61+65+(2+4)+(2+4)=197 or 19.7%
See 3-locus “Cheet sheet”:
Use multiple 3-point crosses to map chromosomes
The 3-point cross produces a genetic map for 3 loci at a time. By using additional loci
spread along the chromosome, the whole chromosome can be mapped.
Example: First map 3-loci A B C
Next add D
A, B, D 3 point cross
Next add E
A, B, E 3 point cross
One can map A, B, C, D, E or D A B C E
Further extension will permit mapping of all loci along a chromosome and by extension
to all chromosomes.
Example: Mapping of the tomato genome.
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 8
2/16/16
Were do crossovers occur along a chromosome?
- can occur all along a chromosome, however,
- some regions differ in crossover frequency.
Centromere & telomere regions have reduced crossing over.
Recombination
Low
Frequency
High
In well mapped organisms (eg. Drosophila) there is a reduction in the frequency of
crossing over near the centromere and telomeres.
Message: Crossing over is not equally likely all along a chromosome.
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 9
2/16/16
Resolution of Genetic maps
The resolution of genetic maps depends upon two factors.
1) Number of marker loci
- Marker genes have to have a phenotype that permits the alleles to be distinguished.
- more loci make a better map
2) Number of progeny.
- More progeny make a better map.
- The ability to score recombinants among 100's, 1000's, 104 etc. means that one can
identify rare recombinants and thus map loci that are very close together.
4
1/100 = m.u. 1/10 = 0.01 map units apart etc.
Because of these two factors, the genetic maps of simple prokaryotes are more
refined than eukaryotes.
1) Use many of the auxotrophic marker loci.
6
2) Screen ~10 progeny on a plate if one uses selectable systems.
Biol207 Dr. Locke section
Lecture#15
Fall'11
page 10