PCB 5065.
Recombination Lecture 5:  Mechanisms of crossing over

Instructor: Dean Gabriel


    In considering the Holliday Model, it is important to note two things:  1)  that the model is perfectly reciprocal (no information is gained or destroyed, except by mismatch correction) and 2) the default situation should yield aberrant 4:4 (ab4:4) asci (they are aberrant because they must result from heteroduplex DNA):

b        or         b
b                    b

b                    b
+                    +
b                    +
+                    b

+                    +
+                    +

Note that a single mismatch repair in the heteroduplex region gives 5:3 tetrads and two mismatch repairs can give 6:2 tetrads, or 4:4 tetrads, half of which should be aberrant.

The Holliday model faced three problems:  first, there were no ab4:4s found in yeast.  Holliday answered that was because yeast had such an efficient repair mechanism.  After all, other fungi did exhibit ab4:4s, and in yeast , the frequency of 6:2s  >> 5:3 > > ab 5:3 >>> ab 4:4.    However, there was another, similar problem that Holliday's model could not answer, and that was that an expected type of 5:3 tetrad was almost never found in either yeast and Ascobolus.  Since these were almost never found, they were called "aberrant" 5:3s.   "Normal" 5:3s were those 5:3 patterns that were normally observed.  Since Holliday proposed that two, symmetrical, heteroduplexes were found, one would expect the following 5:3 patterns, based on a single repair (in each case, correction from wild type to mutant is shown):

Aberrant 5:3s:              Normal 5:3s:
b            b                        b     b
b            b                        b     b

b            +                        b     b
+            b                        b     b
b            b                        b     +
b            b                        +     b

+            +                        +    +
+            +                        +     +

There is no explanation in the Holliday model for the failure to find equal numbers of ab5:3s and "normal" 5:3s..  When the model doesn't fit the facts, it is time to think about adjusting the model. The excess of 5:3s vs. ab 5:3s implied a lack of symmetry.

Note where the axis of reductional division lies (in the middle).  For 'b' to lie below this axis, a DNA strand had to donate the information from the mutant parent.  For '+' to lie above this axis, a DNA strand had to donate the information from the wild type parent.  Hence, ab5:3s demand a symmetrical pair of donations, resulting in two heteroduplexes.  To achieve the ab5:3 ratio, one of these had to be corrected.  On the other hand, to achieve an excess of normal 5:3s, only one heteroduplex is needed, and in the example given, the mutant DNA strand donated the information to the strand below.   Thus, asymetrical donations will result in normal 5:3s, but not ab5:3s, 


(A second way to detect symmetrical vs. nonsymmetrical heteroduplexes is to use unordered tetrads [for example in yeast], and look at the middle marker of three closely linked markers for gene conversions that do not result in recombination of the outside markers:  For example, in yeast, one would look for sectored colonies in tetrads at the b locus, and then check for recombination of the outside markers in those tetrads, in a cross abc X +++.  The various possibilities are:

a   b    c            a   b   c            a    b    c

a  b/+  c            a   b   c           a  +    c

+  b/+ +          +  b/+ +           +  b/+  +

+   +   +            +   +   +           +    +   +

ab4:4                   5:3                     ab5:3

Note that the ab5:3 shows up because of two things:  b/+, which will sector,  is now associated with A+ and C+.  This does not make it an ab5:3, since the same thing happens with normal 5:3s.  However, within the same tetrad, one must sees a + c colonies, which could only arise from a symmetrical heteroduplex.   In fact, ab5:3s were only rarely seen in yeast, which fit with the fact that ab4:4s were also only rarely seen.

This implied that there might not be two regions of heteroduplex DNA.  Why?
 
 

To overcome these problems, and still preserve the linear integrity of the chromosome during meiosis, Matthew Meselson and Charles Radding proposed a model that was very similar to Holliday's, but which added a note of asymetry and simplicity:  they proposed that recombination initiates with a nick on one chromatid only, and that a single Watson/Crick strand invades a homologous area on another chromatid, displacing a Watson/Strick strand on the recipient chromatid.  The displaced strand is destined to be destroyed, thusly:


 
 

 Now note the default situation:  "normal" 5:3.  As a homework assignment, please show why Meselson-Radding always gives normal 5:3 patterns and never ab5:3 (unless there is branch migration & mismatch repair).

The ab5:3 or ab4:4, which result from mismatch corrections involving two regions of heteroduplex DNA, can only occur in the symmetrical region around 'b', and by contrast with the Holliday model, this region is limited to regions where branch migration has occurred.  (Since ab5:3 and ab4:4 were rare, Meselson-Radding postulated very little of it).   Compare this to the Holliday model, where symmetrical heteroduplex DNA is all that you get.  By postulating a high level of repairs, Meselson-Radding explained the high frequency of 6:2s, and in fact, explained the frequency data observed generally of 6:2 > 5:3 >> ab 5:3 > ab 4:4, and this was what was observed, mostly...