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
© 2011 by Steven
M. Carr