The Dutch pattern in rabbits is caused by a gene p which is rabbit in

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A REVISED MAP OF THE FIFTH CHROMOSOMEOF THE DOMESTIC RABBITOSAMA M. RIFAATDepartment of Genetics, University of Cairo, Giza, EgyptReceivedi5.ii.53I.INTRODUCTIONCASTLE(1926) described a three-point backcross experiment in therabbit involving the linked factors for Dutch spotting, English spotting,and Angora hair. He also considered that the three genes are locatedon the fifth chromosome of the rabbit in that order.The Dutch pattern in rabbits is caused by a gene pwhichisalmost completely recessive to self-colour P. The amount of whitein this type of rabbit is also influenced by at least three minor recessivegenes which hitherto are known to recombine freely with each other.English spotting En is an incompletely dominant mutation whichfinds its expression in a complete inhibition of pigment formation ina considerable part of the coat.Its dominance is also influencedby the presence or absence of minor modifiers.Angora hair 1 is the result of the increased length of the hair fibrefound in certain breeds of rabbits. It behaves as a recessive to normalshort hair.The present work is undertaken as an attempt to present accuraterecombination fractions between the three genes and to correct thehitherto known mapping of the fifth chromosome of the domesticrabbit in the light of a re-study and a re-calculation of the datapresented by Castle (1924-1926).2. DUTCH-ANGORALINKAGEInundertaking a comprehensive study of linkage in rabbits, Castlefound that the Dutch and Angora genes are linked together with arecombination fraction of 14.26 per cent. The evidence for thislinkage is summarised in table i.TABLE iDouble backeross data for (Dutch-Angora) linkage. Data from CastleNature of cross++p++113p172Total202Coupling...9720Repulsion...8843343852lOll1213r07

zo80. M. RIFAATSince the difference between the linkage values of the two sexesis not significant, all the data were combined together with the tn-coupling backcross data summarised in table 5.TABLE2Combined data for (Dutch-Angora) linkageNature of cross++p++1p1TotalCoupling.Repulsion..3308854433424383135273950111750Now we have a grand total of i 750 young of which 236 are cross-overs. This gives a recombination fraction of i 4857 I per cent.3.ENGLISH-ANGORA LINKAGECastlealso showed that the genes for English spotting and Angoiaare linked with a recombination fraction of 1359 per cent. Theevidence for this linkage is summarised in table 3.TABLE 3Double backcross data for (English-Angora) linkage. Data from CastleNature of crossEn+6oEni90++98+1617Total1407Coupling..Repulsion..546465501531560In connection with the original data of Castle, it should be notedthat the matings designated as repulsion should be considered ascoupling since all the heterozygous parents were of the constitutionTABLE 4Combined data for (English-Angora) linkageNature of crossEn+Eni+++1TotalCoupling.Repulsion8g1412064532658571019441532097En + 1+1, and those designated as coupling should be considered asrepulsion since the heterozygous parents were of the constitutionEnl/+ +.Evidentlythe use of the terms coupling and repulsion

THE FIFTH CHROMOSOME OF THE RABBIT109has been reversed. Coupling signifies that the two dominants arelocated in one chromosome and the two recessives in the other,irrespective of whether one or both mutants are dominant.Here again the difference between the linkage values in the twosexes is not significant and all the data were, therefore, combinedtogether with the tn-coupling backcross data summarised in table 5.In a grand total of 2097 young there are 276 crossovers or r3I6I66per cent. recombination fraction.4.ENGLISH-DUTCH LINKAGEItappears now that the two spotting genes p and En are linkedsince both are linked with Angora with almost equal recombinationfractions in each case. This indicates either that the genes for Dutchand English are close to the same locus or that they are on oppositesides of Angora and about equally distant from it. Castle determinedTABLE 5Tri-eoupling baciccross data for (English-Dutch-Angora) linkage.Data from CastleEn++f+plx +pl/+p1ShortDutchAngoraDutchShortEnglishAngoraEnglishAngoraDutchEnglishTotal3424023329I537the order by the introduction of Angora into a cross involving Englishand Dutch. Short-haired English rabbits have been crossed withtriple recessives, i.e. long-haired Dutch, resulting in the productionof triple heterozygotes En + + I+pl.Tn-coupling backcrosses werethen made and produced as a rule equal numbers of English andDutch individuals. In addition, very rare crossovers may also beproduced. These of course are of two kinds, viz. English-Dutchand self-coloured animals lacking both the Dutch and English factors.So far, Castle has obtained only one example of the first. Becauseof the rarity of crossovers Castle concluded that the two spotting genesare very closely linked, and located on the same side of Angora witha recombination fraction of oi86 per cent. The evidence for thisis summarised in table 5.These data indicate that there is one crossover in a total numberof 537 rabbits or o 186 per cent. recombination fraction.Castle says "Linkage tests English-Angora and Angora-Dutch,indicate that English is nearer to Angora than Dutch is, since thecrossover percentage in the former case is I3+o8 arid in the latterI42±o9. The data of table 17A (table 5 in this paper) indicate

iio0. M. RIFAATthe same relation. .. . Thisevidence is particularly convincingbecause it is based on matings in which both relations are simul-taneously being tested ".Now,examining the entire data of all the linkages we find thatthe data on Dutch-Angora give 236 recombinations out of 1750 or1348571 per cent.; the data on English-Angora give 276 recom-binations out of 2097 or i 3.16166 per cent. So far as these data gothey suggest that the Dutch gene is further from Angora than theEnglish gene; but this indication is very slight.The x2testof significance in this case is that of the two by twotable.TABLE 6Two by two tableCombinationsRecombinationsTotalsDutchEnglishTotals15141821236276'75°209733355123847Using Yates' correction for continuity we have for x2avalue ofo o6 100698 only, whence xiso •2469959. The probability of observinga xvalueas great or greater than o247, appropriate to the hypothesisthat the locus for English is in the middle is 04o25. On the oppositehypothesis that Dutch occupies the middle locus, we need the pro-bability of observing a disproportion less than that observed or inthe opposite direction. Consequently Yates' correction is reversed.now comes to o3423I5 and the probability to about o63394.The odds are, therefore, in the ratio o•63394 to 04025 in favour ofEnglish lying in the middle.But in the mating En + + I+plx +pl/ +pl the only observedcrossover between Dutch and English was Dutch-English-Angora,showing that the locus for English had in this case separated fromthe two loci for Dutch and Angora; if English were in the middlelocus, this would be a double crossover whereas if Dutch were inthe middle locus it would be a single crossover only.Now, if we accept Kosambi's formula for the relation betweenthe recombination fraction and the map distance it appears that ify and J2 are the recombination fractions in the two segments thenthe frequency of recombination in both these segments isso that of those gamete showing recombination in the second

THE FIFTH CHROMOSOME OF THE RABBITiiisegment the fraction which also show recombination in the firstmust be:2y1(y1+y2)Ifnow, as in our present problem, J2 is exceedingly small, thisfraction is reduced to 2312.Taking the two loci together there have been in all 512 recom-binations with Angora out of 3847 or I330907percent. fory. Hence2312iso'o35426 or i /28'2,hencesingle crossovers are nearly 292timesas frequent as double crossovers among gametes showing thecrossover between the English locus and the Dutch.Combining the two probabilities we have calculated, we see thatthe combined event of a single crossover with a sampling deviationexceeding O'247 of the standard error is about 18'3 times moreprobable than the combined event of a double crossover with adeviation exceeding - o '342.Inconsequence of these calculations I am inclined to believethat the Dutch gene lies in the middle between the English geneand the Angora gene. The recombination fractions on this hypothesiswill be as followsAngora-Dutch..13'216 per cent.Dutch-English..o'86Angora-English..13'402Aswe have determined the right order of the three genes forAngora, Dutch, and English, it remains to find the accurate recom-bination fractions between them.Since such problems, in this as in many other cases, lead totroublesome algebraic expressions, it is useful to apply the arithmeticalapproach supplied by Fisher's Scoring System (1946). As in all iterationprocesses of approximation the precision required in working out theconsequences of the trial values adopted is difficult to foresee, it is,therefore, usually convenient to carry out the work to the highestaccuracy readily obtainable; e.g. to use ten significant figures if aten figure machine is employed. This course was followed in thework set out below, in which, however, the numerical values have beencut down before publication usually to seven significant figures, sinceit appears from the results arrived at that this number would havebeen sufficient.Now, if we scorey1 andy2 at the trial values 13'23 and o'i86 percent. respectively we must therefore score 312 at31+32=13'40281per cent.'+4)1)2d312I 4)22moreover=_________ = 0'9980205d31(1+4,132)2dy12I_43,2=O• 281 8d32(1+4)132)29 5

1120. M. RIFAATAt 1323 per cent. recombination fraction the score for y (Angora-Dutch) will be173/)173/0.1323 - 13o76341040/(I - )) 1040/08677+II9857I - 1o9063and the amount of information will be:I2I3Jty)(I - ))) 1213/(01323)(o.8677) =1056651.At 13 4O recombination fraction the score for )2 (Angora-English)will be212/)212/0i340281 - I581758I61348/(1 - y)1348/0.8659719+155663245 - 25I257Iandthe amount of information will be:1560/(3))(I - )1) 156o/(01340281)(08659719) =134408oBenceFactorScoreInformationFor Angora-Dutch..o8oo - 25•o7597For English-Dutch.o928i587 - 23•32o65I338764I 157900The geometricmeanfor the two values equals:(I3o•8o)(o•998o2o5)(o928I587) =1245050In the case of the tn-coupling data, the expected frequencies ofthe four gametic types produced by the triple heterozygote could be•expressed in terms ofy1 and 3'2onlyas represented in (table a). Thetotal frequency is '+4)1)2 of which the two fractions, i.e. the originalcombinations and the recombinations, gives frequencies in thoseclasses (' - yi)('+4)1)2) gives old combinations fory1 andy,( '+4)LY2)gives the recombinations for the same segment.Similarly, fraction z )2 and .Y2ofthe total give frequencies forthe segment )2• Contingency table, simultaneous crossovers in bothsegments are represented by the following:)1+)'2)= 2)1)2Ly1±y2)since the frequency 2),y2(J1-l-)2) is involved twice in a sum of totalnumber of crossovers for both segments, 1.t. y +.y.

THE FIFTH CHROMOSOME OF THE RABBITi 13The other two tables (b and c) are constructed forwith respectto1 and2 respectively.TABLE aFrequencies of the four gametic types produced by the triple heteroygote

4Y2+4Y224)224Y2TABLE cDifferential coefficient with respect to J2 - 4)'i+4Yi)+2.Yi - I4Y1)2+2J1'4yJ_2j2 - 2J1(2y2+J1)the values for .Yi and Y2 are 01323 andthree tables may be represented afterTABLEa'00017958ooooo66oo868554I0132430209991225oooi86i81ooo9843TABLE b'+09935512+00009774+09945286 - 10009774 - 00009912 - Voo19686 - 00074262 - 00000138

TABLE bDifferential cofflcient with respectI - 4)P1+4J1y2+2y224Y1Y22J2' - I - 4yy+2)' - 2J2(2.Y1+Y2)to .Yi'+8y1y2 - y2 - 1 - 8)1)2

1+8)1)2 - 4)1 - i - 8yy4,'Now considering that000 186 respectively, thesubstitution as followso•866758301323642

- 00074400

1140. M. RIFAATTABLE c'+05067909 - 09659777 - 04591868 - 00340223 - 00359909 - 00700132+04727686 - 1ooI9686 - 05282000dmThe score equals a_/rn, but rn is equal to a/A where A is equal toi +4y,j' and a is equal to each fraction of the expectations. Therefore,log m =loga - log A.Hencex I/rn =x/axda/dy - i/AxdA/dyand this will give the score.Referring to the tn-coupling data illustrated in table 5, we findthat the old combinations and the new combinations of the twosegments, i.e.)1 andy2 are as follows:24746OldcombinationsNewcombinationsIo(Old combinations'I-iNew combinations....47363536I537The scores for,'1 are+''537'7+05517025 - 7554865 - 150107492Thescores fory2 are+I1133765 - 53738O72+o27I644I - 54478798The total score fory1 =+703o335.The total score fory2 =+635994.The amount of information for the two segments is obtained inthe form of a matrix consisting of i11 ,i12,i22.For i11 the amount of information equals:rn(i/mxdrn/dy1)2 =i/A{i/a(da/dy1)2 - i/A(dA/dy1)2}For i12 the amount of information equalsm(i/rnxdm/dy1)(i/rnxdrn/dy2) ..theproduct of the two informations=i/A{i/a(daldy )(da/dy2) - '/A(dA/dy1)(dA/dy2)}.

THE FIFTH CHROMOSOME OF THE RABBIT"5For i22 the amount of information equals:m(i/mxdm/dy2)2 =I/A{i/a(da/dy2)2_I/A(dA/dy2)2Information for,1Information for,12Information for,2I•s388927756969OOOO5320014880 - 00OOO05058092630.2572865 - 052574790.5403088 - 0003933402963Oo8745196076196182 - 0279889112680.848840353932974780645455379OO2893353Now the whole data and calculations should be combined asfollows for comparison:TestRabbitsScoresinuSj22A-DD-EAngora-Dutch .Angora-English.Tn-couplingii1560735 - iog.o63oo - 25O7597+7030335... - 2332065+6356641o566511338764478064...I24505045538.••i15790028933530 - 6383562 - 16964012873479129058830091430With these values we can write the equations:2873479dJ1+I29O588dy2 =+63835621290588dy1+3oogi43o4y2 =+I6964ozwhere dy1anddy2 stand for the adjustments required in the recom-bination fractions adopted. The solution of these equations is:=+00022394+022394 per cent.dy2 = - ooooo397 - 000397 per cent.The accurate recombination fractions for y, and y2 are I3454and oI82 per cent. respectively. Using Kosambi's formula, therecombination fraction fory12 will be I3623 per cent.The map lengths calculated from the recombination fractionsobtained (using table VII, Fisher and Yates) are given in the summary.5.SUMMARYAcorrection of the hitherto known mapping of the fifth chromosomeof the domestic rabbit has been made in the light of a re-study and are-calculation of the data presented by Castle (i 924- 1926).

xx60. M. RIFAATThe new recombination fractions and the corresponding mapdistances between the three genes involved are as follows : - Recombinationfraction per cent.Map distanceAngora-Dutch....Dutch-English....Angora-English....13454or8i3.62313.794 centimorganso182I3976Acknowledgment. - This work has been carried out in the Department of Genetics,University of Cambridge. I wish to express my gratitude to Professor R. A. Fisher,r.a.s., for his invaluable help and kind advice.6.REFERENCESCASTLE,W. E. 1924. Linkage of Dutch, English, and Angora in rabbits. P.X.A.S.,JO, 107-108.CASTLE, w. E. 1926. Studies of colour inheritance and of linkage in rabbits. CarnegieInst., Wash., 337, 1-47.cAsmE, w. E. 1930. The Genetics of Domestic Rabbits. Cambridge, Mass.CASTLE, W. E. 1934. Genetics of the Dutch coat pattern in rabbits. 3.Exp.Zool.,68, 377-391.FISHER, R. A. 1946. Statistical Methods for Research Workers. ioth Ed. Edinburgh:Oliver and Boyd.FISHER, R. A. 1946. A system of scoring linkage data, with special reference tothe pied factors in mice. Amer. Xat., 8o, 568-578.FISHER, H. A., AN1) YATES, F. 1948. Statistical Tables for Biological, Agricultura4 andMedical Research. 3rd Ed. Edinburgh: Oliver and Boyd.PUNNETr, H. C. 1912. Inheritance of coat-colour in rabbits. 3.Genet.,2, 221-238.PVNNETF, R. C.1920. The genetics of the Dutch rabbit - a criticism.3.Genei.9,303-317.PUNNZI-r, R. C., AND PEASE, H. 1925. On the pattern of the Dutch rabbit. 3.Genet.,55,375-412.PUNNETr, R. C. 1926. The Dutch rabbit - Castle, Pease, and Punnett. 3.Genet.,i6, 197-199.PIJNNETr, It. C. 1928. Further notes on Dutch and English rabbits. 3. Genet., 20,247-260.