Analysis of a tri-hybrid test cross
Theoretical:
Given three loci, linked on a single chromosome in the following
order
-----A----------B----------C-----
Then, a cross
between ABC//ABC
x abc//abc
tester involves
-----A----------B----------C-----
-----A----------B----------C-----
-----a----------b----------c-----
-----a----------b----------c-----
Recombination may not occur at all (parental type),
or may occur between A & B, or B
& C (single crossovers)
or between both: a double crossover
During Meiosis I,
single cross-over between the A
& B loci looks like:
-----A----------B--------C----- => ABC Parental
-----A---\ /---B--------C----- => aBC Single recombinant
X
-----a---/
\---b--------c----- => Abc Single recombinant
-----a----------b--------c----- => abc Parental
A single cross-over between the B
& C loci looks like:
-----A--------B----------C----- => ABC Parental
-----A--------B---\ /---C----- => abC Single recombinant
X
-----a--------b---/
\---c-----
=> ABc Single recombinant
-----a--------b----------c----- =>
abc Parental
Each crossover changes
the
phase of one locus with respect to (wrt) its
nearest neighbor.
The frequency of either single
crossover is proportional to the distance between loci,
and increases with distance
The frequency of a double
crossover is the product
of these frequencies:
therefore a
double-crossover is the rarest
type:
-----A----------B----------C-----
-----A---\ /---B---\ /---C-----
X
X
-----a---/
\---b---/ \---c-----
-----a----------b----------c-----
A double crossover will reverse the
phase of the middle
locus
-----A----------B----------C----- => ABC
-----A----------b----------C-----
=> AbC
-----a----------B----------c-----
=> aBc
-----a----------b----------c-----
=> abc
Thus, when you compare the
most common types (
the parental ABC & abc )
with the rarest (
the double-recombinants
AbC & aBc )
the cis /
trans phase
of the middle locus is reversed wrt both outside loci
Here: B is reversed wrt A & C (and b wrt a & c): B is the middle locus
Experimental
Consider three loci E, F
& G, linked on a
single chromosome, but in an
unknown order
The following cross is
constructed:
EEFFGG x eeffgg (P) => EeFfGg
x eeffgg (tester) (F1)
which produces the following counts among 1,000 F2 offspring.
Use of a tester means the genotype
of the offspring can be inferred directly from the phenotype
Genotype
|
Phenotype
|
Count
|
Class
|
E F G efg
|
'EFG'
|
370
|
P
|
| E F g efg |
'EFg'
|
8
|
D
|
| E f G efg |
'EfG'
|
37
|
II
|
| E f g efg |
'Efg'
|
95
|
I
|
| e F G efg |
'eFG'
|
85
|
I
|
| e F g efg |
'eFg'
|
43
|
II
|
| e f G efg |
'efG'
|
12
|
D
|
| e f g efg |
'efg'
|
350
|
P
|
(1) Identify the four pairs of phenotype
classes:
one parental (P) class, which is most
frequent, and
which
shows
the same phase relationships as the parents
two single recombinant (I & II) classes, between the middle locus and
each outside locus
one double
recombinant (D)
class, between the middle locus and both outside loci
As
this
requires two events, this is the rarest class
The two members of each recombinant pair should
occur in approximately equal numbers,
because each
crossover produces two products
(2) To determine gene locus
order:
Most common phenotypes EFG & efg are parental: no recombinants
Rarest phenotype EFg & efG will be the double-recombinants:
Compare phases of [EFG efg] vs [EFg efG]
*
The G
locus changes phase ("flips") wrt E & F, therefore G is in the middle *
-----E-----G-----F-----
(3) To determine the frequency
of recombination between locus pairs,
count all classes
in which the phase changes from the parental types
(i) for E
& G
EG
& eg are parental types
Eg
& eG are recombinant
types between E & G: (Class I):
count all classes
with this type
# (EFg + efG + eFG + Efg)
[(8 + 12) + (95 + 85)]
/ 1000 = 0.20 = 20% = 20
m.u. = 20 cM
(ii) for G & F
GF & gf are parental
types
Gf & gF are
recombinant types between F & G (Class II):
# (EFg + efG + EfG + eFg)
[(8 + 12) + (37 + 43)] / 1000 = 0.10 = 10% = 10 m.u. = 10 cM
(iii) for E
& F
EF & ef are parental
types
Ef & eF are
recombinant types between E & F
# (EfG + eFg + eFG + Efg)
[(37 + 43) + (95 + 85)] / 1000 = 0.26 = 26% = 26m.u. = 26 cM
(4) Map data then look like this:
The distance
between E & F is greatest, so these
are on the outside
E--------20cM--------G---10cM---F
E-----------26cM-----------F
but 0.26 < (0.20 + 0.10)
WHY?
In the double recombinants,
--A----B----C-- 
ABC
--A----b----C--
AbC
--a----B----c-- aBc
--a----b----c-- abc
so double
recombinants resemble parental type
but each represents 2 cross-overs between
outside ( E & F ) loci
(3.iii') Then
the corrected
count of crossovers between E & F
[(37 +
43) + (95 + 85) + (2)(8
+ 12)] = 0.3 = 30 m.u. = 30 cM
(5) Interference occurs where one cross-over decreases the
probability of others nearby:
Compare the
observed vs expected #
double-recombinants
Interference (I)
= 1 - (observed # D) /
(expected # D)
= 1 - [(8 + 12) / (0.20)(0.10)(1000)]
= 1 - f(Dobs) / f(Dexp)
= 1 - [(0.008 + 0.012) / [(0.20))(0.10)]
In these
data, I = 1 -
(8+12)/(20) = 0
: # expected = # observed no interference
All text
material © 2016 by Steven M. Carr