Fristensky Lab Protocols
Finding lines homozygous for a transgene
(update 12/2/98)
When a transgenic line contains a T-DNA insertion at a single locus,
the initial transformant plant T0, is hemizygous at that
locus.
For example, given a gene R present in hemizygous state in the T0
generation, selfing of any individual will give progeny
segregating
3:1 for the transgene. (The null allele is represented by "0").
T0 R0 | v T1 RR R0 00 1 : 2 : 1 ---------- 3 : 1
|
The 3:1 ratio should obtain for all subsequent generations if random
mating is allowed.
To identify individuals homozygous for a transgene,
-
Grow 10 - 20 plants until several true leaves are available, but well
before
flowering.
-
Extract DNA from each individual, and test for presence of the
transgene
by PCR. Use primers specific for the transgene, not the vector.
-
Discard null segregants, which do not give a band, and bag
positive
plants at flowering to prevent cross-pollination.
-
Collect seed from each plant and keep in separate envelopes.
-
For 5 positive parents, grow at least 15*
plants until several true leaves are available.
-
Extract DNA from each individual, and test for the presence of the
transgene
by PCR. (You may save some work by freezing leaf tissue for 5
sets
of progeny, but only testing one individual at a time, until you get a
true homozygote). If any line tests positive in all individuals, that
line
must be homozygous. If some individuals test negative, it may be
because
of bad DNA preps. It is sometimes worth re-doing the DNA preps and
doing
PCR on the new DNA.
All seed from that parent for which no null segregants were seen
should be homozygous, and it should be possible to propogate this line
indefinitely with no loss of the transgene due to segregation.
*How many plants need to be
tested
to ensure homozygosity?
We want to eliminate the possibility that a parent testing positive for
the transgene is heterozygous. When a single heterozygous plant is
tested,
the probability of a false positive (ie. seeing a + band rather than a
- band) is 0.75. Therefore, when N progeny
are tested the probability P that all N progeny will test positive is
0.75N.
If 2/3 of the parents that test positive are heterozygous, the
probability
P that a false positive will occur given N progeny testing positive is
P(heterozygous parent | all N
progeny
test positive) = 2/3 (0.75N)
The table illustrates the probability of a parent being heterozygous
given N progeny testing positive.
| N |
P |
| 1 |
0.500 |
| 5 |
0.158 |
| 10 |
0.037 |
| 15 |
0.009 |
| 20 |
0.002 |
Therefore, about 2 times out of 100 trials, if 10
progeny
all test positive, the parent was a heterozygote. If 15 progeny
test
positive, there is less than 1% probability that the parent was
heterozygous.
If 20 progeny test positive, there is only 0.2% probability that the
parent
was heterozygous.