In January 1914 a vermilion female from a lethal 3 culture when bred to a vermilion male gave 71 daughters and only 3 sons; 34 of these daughters were tested, and every one of them gave a 2:1 sex-ratio. The explanation advanced (Morgan 1914_c_) was that the mother of the high ratio was heterozygous for lethal 3, and also for another lethal that had arisen by mutation in the X chromosome brought in by the sperm. On this interpretation the few males that survived were those that had arisen through crossing-over. The rarity of the sons shows that the two lethals were in loci near together, although here of course in different chromosomes, except when one of them crossed over to the other. As explained in the section on lethal 1 and 1_a_ the distance between the two lethals can be found by taking twice the number of the surviving males (2+3) as the cross-overs and the number of the females as the non-cross-overs. But the 34 daughters tested were also non-cross-overs, since none of them failed to carry a lethal. The fractions (6+0)/(71+34) = 6/105 give 5.7 as the distance between the lethals in question. In the case of lethals 3 and 3_a_ another test was applied which showed graphically that two lethals were present. Each of the daughters tested showed, by the classes of her sons, the amount of crossing-over between white and that lethal of the two that she carried. These cross-over values were plotted and gave a bimodal curve with modes 7 units apart. It had already been shown that the locus of one of the two lethals was at 26.5, and since the higher of the two modes was at about 23, it corresponds to lethal 3. The data and the curve show that the lethals 3 and 3_a_ are about 7 units apart, _i. e._, lethal 3_a_ lies at about 19.5. {75} LETHAL 1b.