Transposable elements, allelism tests and RFLP mapping

When an RFLP probe detects a single polymorphic band in different maize lines, the common assumption is that these bands are allelic. While this assumption is undoubtedly valid in most cases, the prevalence of transposable elements (TEs) in maize can complicate the issue, i.e. how can a low copy number TE present in different locations in different maize lines be distinguished from a fixed gene locus? As an example, Bs1, a retrovirus-like transposon, is present in 1-5 copies in most maize lines, and its hybridization patterns look similar to those of standard RFLP probes (Johns et al., EMBO J., 1985). Unfortunately, the standard tests for allelism, analysis of backcross and F2 self progeny, are remarkably insensitive to this phenomenon.

Let us call the hybridization band in one line "A", the band in another line "B", and the absence of a band "O". The F1 heterozygote between these lines can be represented as A/B if the bands are allelic, and A/O B/O if the bands are non-allelic and unlinked. On a Southern blot there are 4 possible conditions: A, B, AB, and O. It may be possible to distinguish AA from AO in some conditions, but this cannot be reasonably assumed.

In a backcross, the A/B heterozygote is crossed to the A line, resulting in 50% A progeny and 50% AB progeny. For unlinked TEs, the progeny will be 25% A/A B/O, 25% A/O B/O, 25%, A/A O/O, 25% A/O O/O; this reduces to 50% AB, 50% A. Thus, a backcross is completely incapable of distinguishing allelic bands from unlinked TEs.

Selfing the F1 to produce an F2 gives a 1:2:1 ratio of A:AB:B if the bands are allelic, and a 9:3:3:1 ratio of AB:A:B:O if the bands are unlinked. A closer look at the latter ratio reveals some problems. The diagnostic class is the O, or null class: if null progeny are detected, then A and B cannot be allelic. However, the nulls are only 1/16 of the total progeny, and it would take an examination of 46 progeny to be 95% sure that getting zero null progeny is not a chance variation from 1/16. The situation is worse if the TEs are linked: the frequency of the null class varies inversely with the square of the linkage distance. If two TEs were 25% linked, it would take an examination of 191 F2 progeny to be 95% sure that zero null progeny was not due to chance variation.

There is another problem with null progeny: if no band appears in a particular lane of a blot, it is easy to decide that this is due to experimental error and simply ignore these lanes. If this is done, the 9:3:3:1 ratio expected from unlinked TEs degenerates into a 3:1:1 ratio. It would take an examination of 150 progeny to be 95% sure that a 3:1:1 ratio was not a chance variation of the 2:1:1 ratio expected from alleles. The situation is, of course, much worse with linked TEs.

So, if neither backcrosses nor F2 selfs will reliably distinguish TEs from alleles, what does work? The best allelism test would seem to be crossing the F1 heterozygote to a third line with a different hybridization band "C". For alleles, the resulting progeny are 50% AC and 50% BC. For unlinked TEs, the progeny are 25% C, 25% AC, 25% BC, 25% ABC. The C and ABC progeny, half of the total, are diagnostic for non-allelism. Also, there is no null class: a lane with only the C band is clearly distinguishable from a completely blank lane. It thus takes an examination of only 5 progeny for there to be a greater than 95% chance that at least 1 ABC or C will be seen, if unlinked TEs are involved. If 25% linkage exists between the TEs, examination of 11 progeny gives a greater than 95% chance of seeing at least one diagnostic offspring. This method is clearly much more powerful in distinguishing TEs from alleles in RFLP mapping than the standard backcross or F2 self methods.

One other RFLP method should be discussed: recombinant inbred (RI) lines. Completely homozygous RI lines yield the major advantage of an outcross to a third line: there is a 1:1:1:1 segregation of AB:A:B:O progeny. The problem of null lanes being considered experimental error can be partially eliminated by testing more than one individual from a given line; however, there is no C band acting as an internal control. The major problem with RI lines is that they are not completely homozygous. Burr et al.'s (Genetics, 1988, in press) RI lines are estimated to be 7.5% heterozygous for all loci (they expected 3.125%, based on 5 generations of selfing). Also, their mapping method deliberately ignored null lanes. Thus, there is certainly room for a small number of TE probes in their collection, and an examination of these RI lines with a new probe will not necessarily distinguish between alleles and TEs in different locations (especially linked TEs). However, if a new probe detected only A and B types, and never AB or O types, among the RI lines, then one could be reasonably certain that A and B are allelic or at least tightly linked. The presence of heterozygous or null RI lines suggests the need for further allelism tests.

From the preceding discussion it can be seen that distinguishing low copy number TE from alleles by genetic tests is best done by outcrossing F1 heterozygotes to a third line; use of RI lines is also an acceptable method. Examination of F2 self progeny is unlikely to reliably distinguish between these alternatives, and backcrosses are useless. The importance of these observations depends on how prevalent low copy number TEs are. Bs1 certainly falls into this category: a number of lines containing a single Bs1 element exist, and only sophisticated genetic analysis or DNA sequence analysis would determine that Bs1 is a TE and not a fixed gene. It may well be that other low copy number TEs exist; they may even be common.

Mitrick A. Johns


Please Note: Notes submitted to the Maize Genetics Cooperation Newsletter may be cited only with consent of the authors.

Return to the MNL 62 On-Line Index
Return to the Maize Newsletter Index
Return to the MaizeGDB Homepage