dt,

Kaun A

 

YALE UNIVERSITY
OSBORN BOTANICAL LABORATORY
NEW HAVEN, CONNECTICUT

July 4, 1947.

Dear Mather:

In mulling over the statement of the problem which I had previous-
ly stated in too abbreviated and confusing fashion, I think I have
achieved a partial solution; at least I hope it may serve to convey
the nature of mg difficulties. I hope you will have time to go over
this and detect any fallacies that may have crept in.

Assume the parental configurations: ABCd and abeD. Given the rela-
tive frequencies of the only detectable recombinatiog classes (A..D
or prototrophs) to estimate the absolute map distance A..D, assuming
linearity and no interference. The regions are as follows: :

A BC D and for the sake of convenience, let us call
1 2 3 the three "single-interchange" classes,
AbcD; ABcD, and ABCD, ji » and the "triple-interchange" class AbCD, k.

The grand-total accumulation of data.in the case where ABCD are

BM, Lac, V,TL respectively, is as follows:

de

1... 82 27 3% 27 8%

2.++1389 46.3 47 6

Bese 729 ~ 24.3 2.20005 §©24.9

ke... 6 2el 2935. .00.00
2998 100.0%

we aN

The k's are separated in the third column to give the re2z&tive dis-
tances.

In your analysis of the problem, you suggested that the abso-
lute distances might be given by the e ressions:

tt = e ' = = =

a = P1do94d3 5 “2 pP2a,43 and °3 Pz4,4, and k = p)Pops.
where p,, etc., was the chance of a crossover in region one, i.e.,-
the. distance from A to B. It has just occurred to me that the analysis
which I had just completed is congruous with yours if a more accurate

oA? meaning is given to these symbols, namely that Py is the probability
h

that there is an interchange in region 13 i.e., at there is an odd
number of crossovers in this region. Thecmap distance which can,of
course exceed 1.0¢7 while a true probability cannot, should be defined

as: no
\p dx For infinitesimal segments, of course, p
by can only refer to single crossovers.

According to this definition, the absolute distance A..D ‘is the
mean number of crossovers per chromatid pair.

As a first approximation, consider the case of two-strand crossin,

over, with the distances 1,2,and 3 equal to each other, and estimate

what the total distance must be to allow the class k of given size

(2.1%). There is assumed to be no interference. If the distance then

   

 

8s x, ‘the proportions: of x
YALE UNIVERSITY
OSBORN BOTANICAL LABORATORY
NEW HAVEN, CONNECTICUT

  

troph will be obtained. For purposes of orientation, it may be pointed.
out that the limiting probability of a k is 1/4: If the tatal (odd)°
number of crossovers is large, the chance that there will be an edd number
in region 1 is = to the chance of an even number, 1.e, = 1/2. ae
Similarly for the chances of region 2. This determines region 3 as odd,
so the overall probability is 0.5 x 0.5 = 0.25. This is to be expected,
since it means that with a lrge number of crossovers, each class has
an equal expectation. Z me
The chances of 1 crossover in each reggion, for x=3 can be estimated — ._
from the binomial distribution, and turns out to be 2/9 = .22. The Jimit ~~
of .25 is very nearly reached for n=5, but thecalculations are too space-
consuming to be worthwhile writing down.
The value of x correspomding:to k* .021 can now be determined. oa
Each single-crossover will yield only j. Triple-crossovers will yield
.22 k and .78 j, while odd n-ple crossovers will yield even more closely
.25 k and .75 j. The weight to be given to each ecrossover-class is given
by the Poisson series. To simplify the calculations summa somewhat the
yield for n=3 can be approximated as alss -25. The odd terms of the_
Posssoh distribution are, of course, sjnh x. The first term is ne :
(n=1) e7* (x). The proportion of single-crossovers to total odd crossovers .
is then x/sinh x . This will be the same as j-3k/ j+k.= 1 - 4k/j+k.
In the present case, k/jt+k = 2.1% or .021; x/sinh x = 916. 3
x= .73 or 73 morgans. This turns out not to be very different~

‘cr your estimate. x is of course 4 very sensitive function of k, e.g.

  
 

or xx k= .03, x= 90 morgans, so that the estimate can be regarded only  -
as a rough measure. With mmixx x only .7, the class n=5 can be disregarded
we compared to n=3 ( .012 : 1) so that the yield can be calculated on the
‘/$, basis of a comparison of the N=1 and n= 3 classes only. We can also
modify the calculation to include the found relative distances, neglec-
‘f ting the possible slight modification that the triples among them might
P have introduced. The chance that three crossovers will be distributed
yx Y” one each in the reBAtive distances: 27.8, 47.3 and 24.9 is: 6( .2e78x.473
A x.249) =.196. The proportion of J n=3 to n=l is simply x-/6. Thus we have

y 096 x°/6 = k/j+k = .021. : x= .72 or 72 morgans as a more exact
+ Xx ;

: estimate. About 8% will be the
’ . actual frequency of n=3.

So much for the hypothetical two strand case. | -

If crossing over occurs at a four-strand stage, the calculations are
much more involved. Let n be the number of crossovers per tetrad. If n=l).
the situation is much as in two-strand, since only the prototroph will.
be recovered. If n=2, however, 3 tetrdds out of 47(the digressive and . oe
the progressive) will yield j rototrophs. If n=3; 4 tetrads out of |
16 will contain k prototrophs, .2 will contain no prototraphs, while - ae
12 will contain J's. (One overlap, with 1 j; 1 k). For n=4, the situation -
is clearly even more complicated, The n=3 caée mentioned is for those
where there is one crossover in each region.(.196 of total). In general, ..
the following will have to be calculated: ; 8

a) for any n the BARRE chance that there will be crossovers in each re-
gion. also, the proportion:of the disproportionate types which yield
prototrophs (all j). :

b) the proportions of j amd k where there are crossovers in each region.

A

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#

YALE UNIVERSITY
OSBORN BOTANICAL LABORATORY
NEW HAVEN, CONNECTICUT K. Mather

: 3 AE

The problem seems so complex, phat I have begun by making some

 
  
  
    
   
  
      
   

simplifying approximations. Firstly, we will count the total number
of recombinants, rather than zygotes, in making our estimations, to: ;
avoid the difficulty of enumerating zygotes containing two kinds o
protrophs (digressive multiple exchanges). On this basis’ ‘then, ‘bot,
single~- and double- crossover types. (n=1,2) wil] be counted-as. =
giving all g prototrophs, while+the type where n=3, and there is*
one crossover in each region will be considered as yielding 3j:1k
I have not completed thedenalysis, but it seems very likely on “thi
basis that for any value of n, there will be this yation for 3
set of crossover classes in which there is one or @ crossovers:.
in each region. The difference between this and the two strand syste
is that the even-numbered types: must be counted also.
Theoretically, it should not be difficult to find an expression :
for p=f(n) where p is the chancé that if n marbles are thrown at. ‘random «
into three equal (or more precisely,somewhat unequal) boxes, that .
none of these boxes shall be empty. I have not been able to find it”
however, and have had to rely upon the binomial expansion, summing
all the appropriate cases, to gptain values of p from n=3 to n=8.
p must be known, since k/j+k sa P/4 x7 /n} 02)

 

ae

‘ eX -1
The effect of increasing x, then, is to augment the proportions ©

of higher values of n, which ink tarn increases the proportions of — |

the types in which there is at Jeast one exchange in each region. ot
Using this formula, and the calculated values of p, the following

trials were made as enumerated jn the table:

   

p F(x) \R(1) /4.F(1) F(2) p/4.F(2) {F(1.6) D/A. FL. 6)
L 0 =| 2. 0 1.6 Oo
05 0 q 2 Oo. . Lees oO - 8
17 .008 = 1.33 065 068 = .033;
04 O01 * 075 083 e027) = 03502
Ol ~- * 2 , O41 209 . 004.

~ - “ 209 OLT ° 02 , 2004.

: - ¢ "02.004 -— =

- - ¢ .005 - - a8

1.72 .009 ¥ 6.46 «210 5.93 O81:
re 4005 ¢ r=,032 r=,021

 

 

Thus, n=1.6 gives r=.021 in agreement with the expbotea 5/j+ke 2s ig.
1.6 crossovers per tetrad is,of: course equivalent to 80 morgans betweeh™
A..D, which is surprisingly close to the 2-strand estimate. Is this =
more than an accident?? &

- In the course of all these ‘elaculations, I became rather tired- of F
drawing four chromatids and spotting innumerable combinations of « :
srossovers on them to determine the proportions of various types. _. zi
Therefore I set myself to formulating an operational treatment of — |:
multiple crossovers; I should he ane erested to hear whether it would °

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q ve
4
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of any use to anyone, and whether anything similar has already been
published. 4 7 ore

A crossover is written as a3 for the exchange indicated, in region
involving strands 1 and 5. £13 is an operator, the operands of::
which are the 4 strands- sj, Sor Sz) and Sye5 &

For the purpose of developing the arithmatic of these operators,.there °‘
are two parts to the operation: a,3.s which have to be considered,

@ Oo
— ew ae Se ee

oa Fh

To combine operators, BoE 215 +d55+04 498) write the operands in
rate columns: 2

Then perform the operation indicated by the penultimate symbol, keeping

in ming the shift in ranks of the previous operator.And so forth.

The tetrad then consists of: ¢,8,; a1bz 89.3; boS,5 azo] 2 or
451) soda? aba, andch, as

is the correct answer. The generalization of this ‘method to any number?
of crossovers or regions is,of course,obvious. I am going, now, ti ee
\ work on derived rules for the production of. the various crossover, ¥pes,!
~ based on the relationships between subscripts. E.g., a triple exca Be.

tive frequencies of various tY¥
caloylate, <-m Ww, wor tour 3
bee

/ Tove | ;
ra PT nope I have made this le
of brevity.) Please accept my.

must be abo (ac), This can.
and would Zena be a prototroph,
t

ARPA RA eR

sy," . § 8 S43 and perform the operation -
ne C4 . (f) (13 c indicated by the righthand-
o (1) | bs b- (t) most term.(c,4). The change
ue _ (1) ay (1? Az in rank is very readily:. -
OO s symbolized by using thes:

  
   
 
 
  
   
   
  
 
   
      
          
  
 
 

~§-

tinge

The method is: :
///k hay B (b) @ (ce) D*
evr, :

@

a b Cc ad

 

 

> Uhr

 

on s,; it substitutes a for A . “nis may be written. (4) or more’ sit
(a).~ on 83 it substitutes Z for a. This has the sams: notation.

it converts s, from 'rank' I to rank 5 OS

it acts,of course, in similar fashtion on’ss. ny

an operator has no effect, @og. (l)s ; on operands of different rank
the operators are written from left to right, in the sequence of.”
the oroscoversi the order of. action is from right to left. a
aa = B

the combinations of operators are:

wy

eSniei atin aay

a = (a) ab = (b) (* (a)(ab))

b = (ab) ac = (be) | oe

¢ = (abe) po = (6)... abo =(a-boj=(ab)
patios ised ass PSS Doha shana me Boag L288 Po .

   

cee pon eR aioe Sime 0 SAT a SER ; pee ns ue
“Tae Sa EE SSA a eis 5 bee al age ce

sepa-”

subscript.

abeD; AbOD;: ABcda3 ch,°as You can -vé

* :
Sten

-only be produced if: 8) )+DyyeSny 82 :
i Me From suc rulés, the rela- .
es shpuld not be too aifficult to =. =

’ The milied 1 teach Rear eevee TY

ter somewhat olearer (if at the expense “
ratitude four-your-continued interest.

 

    
   
 
 
 

 

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