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The Francis Crick Papers

Letter from Cy Levinthal to Francis Crick pdf (17,467,127 Bytes) transcript of pdf
Letter from Cy Levinthal to Francis Crick
Item is handwritten.
Number of Image Pages:
6 (17,467,127 Bytes)
Date Supplied:
ca. March 1957
Levinthal, Cy
Crick, Francis
Original Repository: Wellcome Library for the History and Understanding of Medicine. Francis Harry Compton Crick Papers
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Exhibit Category:
Defining the Genetic Coding Problem, 1954-1957
Metadata Record Letter from Francis Crick to Cy Levinthal (April 5, 1957) pdf (186,538 Bytes) transcript of pdf
Metadata Record Letter from Francis Crick to Cy Levinthal (February 1, 1957) pdf (129,677 Bytes) transcript of pdf
Box Number: 10
Folder Number: PP/CRI/D/1/1/12
Unique Identifier:
Document Type:
Letters (correspondence)
Physical Condition:
Series: Correspondence
SubSeries: Alphabetical Correspondence
SubSubSeries: Correspondence 1
Folder: Correspondence L
Dear Francis,
Please excuse my delay in writing but somehow things involved with students got out of hand in the last few months. I'm still waiting for some photographs of poppit bead models but I think that I can drum [?] things sufficiently well so that the ideas are clear.
First pairing must take place between open regions at corresponding points on different (mamma and poppa) molecules. Then the interactions must take place which lead to the observed heterotzgotes. I'll forget the first question except to say that perhaps Drew Schwartz has a useful idea in the [diagram] and [diagram] tetrads.
At any rate the only new thing beyond the enclosed and what I said last summer is the following sequence of operations.
First set poppit beads to do the following:
Molecule A has two chains A plus and A minus drawn as [diagram] and B has [diagram] let the winding of the chains to form the double helix be indicated [diagram]. I assume that the starting time is long compared to the running time so that there will only be one start in the process. If the start takes place and the upper end of A and pairing takes place we will have: [diagram] The dotted line is the newly formed daughter which will be labelled [sic] with a prime
(I) The growing point C of A1 plus switches over and inserts itself between the chains of B and continues growing on B minus at this point it becomes B1 plus. B1 plus need only be wrapped about B minus for one or two turns and in this region B plus would have unsaturated hydrogen bonds. This process can continue down the chains with A1 minus growing on A plus and A1B1 plus growing on B minus but winding around A minus. If this were to continue to the end it would lead to a heterozygous region extending from the original switch point (C) to the end of the molecule and such
structures are not found (at least to the level of 10 to the minus 3 power in phase.). It also seems more likely that the growing point at A1 minus should switch before very long since it is constrained by the rest of the structure to be near other plus chains to which it could switch. The closeout plus to which it could transfer (and also the only one which leads to short heterozygotes) is the newly formed B1 plus. This then is taken (II) as the second switching operation i.e. A1 minus switches to growing about B1 plus and thus the newly formed minus is A1B1 minus. [diagrams]
After the operation (II) the structure which is circled (ie. B plus the small amount at B1 minus) is connected to the rest by only a single phosphate-ester link at D. Therefore free rotation can take place about D.
Any thermal motion which pulls B away will exert a tug on D and will cause A1B1 plus and A1B1 minus to pull away from A. These
two will then wind around each other and A plus A minus will rewind so that there will be no net change in the number of saturated Hydrogen bonds. No net energy is required except for the viscous drug and the gain is the free energy of separation. when this back winding gets to the top end of A one has: [diagram]
At C the A1 plus changes to B1 plus and at D A1 minus changes to B1 minus. The new structure then can continue to the end or the whole thing could start over again with another switch. The point E could continue as indicated or it could insert into the unsaturated B minus chain so that the structure would then be symmetric, If this continues to the end one ends up with one new molecule and it has all of the observed properties of the heterozygote.
Both Seymor [sic] and Maury Fox have seen this with beads in fact it was Maury's sister who asked the right questions to produce the scheme, obviously there is no reason to believe that it has anything to do with reality except that its the only one I've been able to find after a good deal of trying.
New experiments to the drum [?] the facts are in progress but nothing new to report yours
The apparent negative interference, that is, the high probability of switching near a switch came out of this a rather natural way. If after the switch (II) but before the backwinding [sic] there is a reverse switch of B minus1 back to A plus then one must start over with a type II switch before the separation. In this way the heterozygote could look like [diagram]
The left hand chain, that is the one which switched second, has three switches. The possibility of further switches is ended as soon as the backwinding [sic] starts so the switches are all clustered in the region between the first switch at C and the start of the backwinding [sic].
The testable predictions of this model are 1) that the length of the region continuing multiple switches i.e. length of the negative interference region should be the same as the observed region of the heterozygote. 2) All clusters of switches should contain an odd number of switches.
1) has recently been demonstrated by Doermann's lab but 2) has not yet been tested.
please let me know what you think of this.
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