THE INHERITANCE OF SEX-LIMITED BILATERAL
ASYMMETRY I N BRUCHUS
J. K. BREITENBECHER
Marine Biological Laboratory, Woods Hole, Massachusetts
Received January 28, 1925
TABLE OF CONTENTS
PAGE
The origin of the character. ......................................................
Materials and methods.. .........................................................
Formulae.. .....................................................................
The experimental results. ........................................................
The inheritance of the first piebald female. ......................................
Piebald BR female ( p p ) mated with a normal male (PP).
.........................
Piebald RB female ( p p ) mated with a normal male (PP)..
........................
The inheritance of a P p female and a P p male. ..................................
Backcross test: A PP femalexa Pp male. ......................................
Backcross test: A P p femaleXa PP male. ......................................
Backcross test: A BR piebald ( p p ) femalexa P p male.. ..........................
Backcross test: An RB piebald ( p p ) femalexa Pp male.. .........................
A p p femalexa p p male.. .....................................................
A summary of the RB and BR piebald females.. .................................
DISCUSSION..
...................................................................
CONCLUSIONS
...................................................................
LITERATURE
CITED...............................................................
261
262
262
263
264
266
266
266
269
270
270
271
272
273
275
276
277
THE ORIGIN OF THE CHARACTER
The wild-type female, described in detailin previous publications,
(BREITENBECHER
1921,
1922,
1923)
has two black spotsbilaterally
located on each elytrum, while the wild male has no spots but transmits
the genes (ss) for black spots. On October 23, 1922, there appeared in my
cultures of Bruchus quadrzmaculatus Fabr. an abnormal female with two
red spots on her right elytrum insteadof black spots, which are normally
found on both elytra of the wild-type female. This unilateral insect originated from a homozygous culture (RRss)pure for red elytra andblack spots.
I n cultures of this unilateral mutation, the abnormality, red spots on the
right elytrum, and black spots on the left, did not breed true for this
asymmetrical trait, but many females were found with the left elytrum
The major portion of this experiment was conducted a t the UNIVERSITY
OF OKLAHOMA.
I t was completed a t Woods Hole, Massachusetts, where the superior advantages of the MARINE
BIOLOGICAL
LABORATORY
were enjoyed through the kindness of Doctor FRANK
R. LILLIE.
GENETICS10: 261 My 1925
262
J. K. RREITENBECHER
black spotted and the right elytrum red spotted, while many manifested
the originaltype of spotting. The males, however, in the homozygous
mutant cultures, for eithertype of female, were alwaysnon-spotted.
The size of the spots varied from four minute patches to quite large ones.
For these reasons, “piebald” seemed the most appropriate name for this
mu tation.
MATERIALS AND METHODS
The problem here is concerned only with elytra color and its spotting;
hence, for each of the types described, only such traits will be discussed.
The wild-type females had tan elytra with
black spots, as previously
shown (BREITEWBECHER
1921, 1923). The piebald females were crossed
piebald
with males of another (RRSS) type in order that the unilateral
black spot might be more distinctly visible on a red background. Two
types of piebald females were discovered: One (RB) had red spots only
on her left elytrum with black spots on her right elytrum;while the other
(BR) had black spots on the left elytrum with red spots on the right elytrum. All piebald males, for both kinds of females. appeared alike, having
somatically tan non-spotted elytra. Single pair matings were used throughout this test. Virgin females were obtained by removing them with a knife
from the cowpeas a few hours before they emerged. The cowpeas were
sterilized inMasonjarskept
a t a temperature of 50°C for one hour.
Each pair was placed in a two-ounce bottle, filled with these cowpeas, and
plugged with cotton. Humidity was provided by placing several cultures
in glass battery jars filled with a layer of water.Eachcontainer
was
coveredwith a glass plate and sealed with vaseline. Each jar with its
insects was kept in a constant-temperature oven a t 40°C. A generation
would appear in about threeweeks.
FORMULAE
The formulae used take into account both the
sex-limited traits involved
as well as both types of piebald females. The genes, pp,Fstand for the
homozygous piebaId recessive mutation.
The two types used are as follows:
(1) The normal type
(a) Females. Two bilateral red spots on each elytrum
(PP).
(b) Males. Non-spotted with tan elytra (PP).
(2) The piebald type
(a) Females. Elytra spotting of two kinds ( p p ) .
RB elytra: left, red spots; right, black spots.
BR elytra: left, black spots; right, red spots.
(h) Males. All with non-spotted elytra ( p p ) .
~
263
SEX-LIMITED BILATERAL ASYMMETRY IN BRUCHUS
THE EXPERIMENTAL RESULTS
I n the tests that follow, each individual carried some combination of
the following factors: PP,Pp, PP and p p . The piebald cultures were
crossed with stocks having R R genes, homozygous for red elytra, and SS
factors, pure for red spots on the elytra, in order to contrast the black
TABLE
1
PI:A ptebald female having red spots on her right elytrwn and black spots on her left elytrumXa
non-spotted male homozygousfor red elytra color and red-spotted elytra.
FI: No piebald females,23females with nwmal
bilateral red-spotted elytra, and 25 non-spotted males.
PI ELYTBAL SPOTTING
-l
Females
PAIR NUMBER
Males
No spots
Piebald
Red spots
R B*
60.1
b
6
0
.
2
60.3
60.4
60.5
60.6
60.7
60.8
60.9
60.12
60.13
60 14
60.15
60.16
60.17
60.18
Totals 18 pairs:
29
0
37
3
40
4
0
0
20
2
34
1
18
2
16
2
18
2
46 60.10 5
37 60.11 5
51
1
19
2
14
0
6
1
29
1
19
2
9
1
442
34
Ratio
Red-spotted : piebald
B Rt
-
0
3
5
1
2
3
4
2
3
1
0
2
3
0
2
533
6.3:l
44
36
* R B indicates a femalehaving red spots on the left elytrum
30
25
24
40
36
37
21
17
8
30
21
11
14.5: 1
9.2 :1
6.7:l
0.o:o
4.0: 1
5.7:l
6.0:l
4.0: 1
3.6:l
5.1:l
5.3 : 1
12.7:l
6.3: 1
14.0:O
2.0:l
7.2:l
9.5:l
3.0:l
26
33
74
25
31
2
1
2
andblack spots on the right one.
andred spots on the right one.
t B R indicates a female having black spots on the left elytmm
piebald spotting on either elytrum with the background
of a pure-red
elytrum. The piebald mutation, having black spots on
either elytrum was
represented by the genes p p , while the normal type, having bilateral red
elytra spots, transmitted the factors, PP, Pp and PP. As a result of such
a cross, the actual genetic formulae for all of these traits were RRSSPP,
RRSSPP, RRSSpPand RRSSpp. Since every individual was homozygous
for the RRSS genes, these unnecessary combinations have been omitted
from all tables and from the discussion which follows.
GENETICS10: My 1925
264
J. K. BREITENBECHER
The inheritance of theJirst piebald female
The complete data for two generations of the descendants of the first
piebald female crossed withanormalnon-spotted
male, are given in
table 1. The piebald mother had two red spots on her right elytrum and
TABLE
2
PI:Piebald BR female ( p p )Xnon-spotted male ( P P ) pure for red spots.
S : Red-spotted females(Pp) and non-spotted males ( P p ) .
PI ELYTRAL SPOTTING
PAIR NUbiBER
Red-spotted females
Non-spotted males
9
14
6
1
6
15
13
3
12
4
2
10
~307
304
300
293
282.6
282.5
282.4
282.3
282.1
274
272
271
269
267
266
265
264
263
262
260
259
258
245
242
23 1
229
224
218
217
215
Totals. ..............
Expected. ............
2
15
13
3
7
11
12
14
8
15
11
19
7
5
8
10
11
18
20
11
21
8
16
15
18
13
20
16
16
26
13
20
18
20
397
402.5
15
6
19
8
9
15
21
22
0
17
13
33
29
23
I
408
402.5
two black ones on her left elytrum. She was bred to a normal, tan-elytra,
non-spotted male homozygous for red spots and tan elytra. Their F1 offspring consisted of 23 females with normal, bilateral red-spotted elytra,
SEX-LIMITED BILATERAL ASYMMETRY I N BRUCHUS
265
no piebald females, and 25 non-spotted males. This test indicated that
piebald was a recessive. Eighteen of these F1 pairs were mated singly.
The progeny of the above-mentioned pairstotaled 442 bilateral redTABLE
3
The same tests as enumerated in table 2.
P I : Piebald, BR female (##)Xnon-spotted male ( P P ) pure for red spots.
F,: Red-spotted females ( P p ) and non-spotted males ( P p ) .
F1 ELYTBAL SPOlTlNG
PAIP NUMBER
196
193
192
189
187
184
180
172
170
165
156
155
131
128
126
125
121
113
112
99
95
94
92.3
92.2
92.1
89
83
76
75
73
70
Totals, .............
Exbected. ............
Red-spotted females
Non-spotted males
23
26
24
3
13
31
16
3
9
14
8
10
1
12
7
12
16
11
12
15
3
15
30
14
13
17
13
7
10
17
36
8
14
7
0
15
32
9
441
471
456
456
4
23
0
10
12
4
13
4
12
17
8
25
19
5
28
37
27
13
25
21
12
16
19
32
spotted females, 70 piebald females (34 were of the RB type and 36 of the
BR kind), and 533 non-spotted males. This gives an actual ratio of 6 :1,
when the expected ratio should be 3 : 1 in the F2 generation. The discrepGENETICS 10:
My
1925
266
J. K. BREITENBECHER
ancy is due to the fact that the piebald character is in many instances
very minute. Therefore, during
the earlier stages of the experiment, the
spotsfrequently escaped observation and for that reason were not recorded. Hence, the datafrom table 1 is not added to those of the remaining
tables.
( p p ) mated with normal males ( P P )
testsin which homozygous piebald ( p p ) females
Piebald BR females
Accuratebreeding
were crossed with normal homozygous ( P P ) males, were also conducted.
The parents in thirty separate instances were piebald BR females ( p p ) ,
which were bred to non-spotted, normalmales (PP). The F1offspring were
normal, bilateral red-spotted females(Pp) and normal, non-spotted males,
in approximately 1: 1 sex-limited ratio, thus indicating that the piebald
character is recessive. This set of tests is recorded in table 2.
Table 3 is a continuation
of the same test given in table 2. In this
experimentlike-parents,
piebald BR females ( p p ) , were matedwith
normal, non-spotted males ( P P ) . The data included tests for 31 separate
P1 pairs. Their F1 progeny consisted of 441 bilateral red-spotted, heterozygous females ( P p ) ,and 471 non-spotted, heterozygous males (Pp). The
same sex-limited, 1: 1, ratio is therefore evident. Hence, the conclusions
from the data given in tables 2 and 3 prove that when a piebald BR
female isbred to a normal non-spotted male ( P P ) , all offspring are
heterozygous (Pp) with piebald as a recessive.
Piebald RB females ( p p ) bred to normal males ( P P )
The information relative to such crosses is given in table 4. The P,
parents consisted of apiebald RB female ( p p ) matedwithanormal
non-spotted male ( P P ) . The F1 progeny was heterozygous ( P p ) and
appearednormal.The
female offspring from these parentsappeared
red-spotted, while the males were non-spotted. Forty separate PI pairs
gave 558 heterozygous normal red-spotted (Pp) females and 629 heterozygous normal non-spotted (Pp) males. Again a l : 1 sex-limited ratio
is manifest. Therefore, the data from tables 2, 3 and 4 prove. that when
any piebald ( p p ) female, whether aBR or an RB one, enters the cross with a
normal non-spotted ( P P ) male, the F, offspring is always normal and no
piebald appears, thus indicating that the trait is recessive.
The inheritance of a Pp female and a Pp male
The offspring from heterozygous, Pp, individuals is tabulatedin table5.
Each F1pair consisted of a red-spotted normal (Pp) female and a normal
non-spotted (Pp) male. The thirty-two separate F1 pairs totaled in the
SEX-LIMITED BILATERAL ASYMMETRY I N BRUCHUS
TABLE
4
PI: Piebald, RB female ( p p )Xnon-spotted male ( P P ) pure for red spots.
S : Red-spotted females ( P p ) and non-spotted males (Pp).
FIELYTRAL SPOTTING
~"
PAIR NUMBER
306
296
294
281
2 68
257
248
239
237
232
214
211
200
199
198
197
183
181
179
177
176
175
174
168
163
157
127
123
120
119
107
103
97
96
91
90
84
80
71
67
Totals. .............
Expected. ............
GENETICS10: My 1925
Red-spotted females
Non-spotted males
1
3
8
9
13
22
13
9
21
16
9
28
24
25
6
19
24
3
5
18
20
27
11
14
12
17
8
12
10
9
16
21
18
11
23
10
21
26
7
10
0
3
19
5
10
11
13
12
15
16
7
24
19
38
588
608.5
629
608.5
7
28
8
5
19
18
15
15
11
10
0
15
9
12
10
11
29
29
35
18
20
12
19
41
13
16
267
268
J. K. BREITENBECHER
TABLE
5
F I : Red-spotted female (Pp)Xnon-spotted male (Pp).
”
Il -
FI ELYTRAL SPOTTING
’
I-
__
Females
Fz PAIR NUMBER
Males
Non-spatted
”
Piebald (pp)
Red-spatted
(PP, PP, PP)
309
299
297
295
292
291
290
289
284
281.3
281.2
281.1
261
243
240
234
227
205
195
162
159
154
153
133
130
124
117
110
104
87
86.1
86
85
84.1
82
81.2
81.1
81
80.1
79.3
79.1
79
l
13
12
15
21
11
9
5
12
2
13
17
12
12
27
20
3
17
19
20
19
l
15
2
16
18
21
5
11
8
13
4
14
17
23
11
24
21
20
21
13
16
11
( P P , PP, PP, P P )
RB
BR
2
2
6
4
2
1
1
2
0
0
3
3
2
3
7
1
4
1
4
0
1
2
1
3
3
3
1
2
2
3
2
4
2
3
2
3
7
5
4
0
4
4
3
2
0
3
2
2
1
2
1
4
4
2
2
6
2
0
4
2
1
8
2
2
0
2
3
2
2
1
0
0
0
2
3
4
1
2
2
1
2
4
0
0
19
21
23
25
12
16
l5
26
3
22
20
20
10
23
34
2
23
21
14
18
11
15
13
17
24
21
5
22
10
23
9
10
17
32
17
26
24
19
29
23
16
21
269
SEX-LIMITED BILATERAL ASYMMETRY IN BRUCHUS
TABLE
5 (continued)
PIELYTPAL SPOTTING
PIPAIR NUMBER
BR
77
25
75.3
75.2
14
74
2
73.7
72
69
68.2
68
65.1
62.9
62.7
62.5
62.4
62.2
62.1
12
16
18
3
31
6
22
30
5
3
2
3
2
1
2
2
2
3
2
3
0
5
6
3
2
2
3
' 4
4
3
1
1
16
21
17
21
18
5
38
9
38
46
33
51
24
25
22
25
7
1
19
40
7
24
3
3
3
18
18
13
4
1192
Piebalds. .. . .1192
. ..
Exfiected, 3 : 1 :4
-""-
127
290
300
I
I200
F2 generation, 411 red-spotted (PP, Pp, p p ) females,143 piebald
(PP)
females (75 were of the RB type and 68of the BR kind), and 547 nonspotted (PP, Pp, PP, p p ) males. In table 6, which is a continuation of
table 5 , twenty-six separate F1heterozygous pairs (Pp) produced in the
Fa generation, 486 red-spotted (PP, Pp, pp) females, 147 piebald (PP)
females (88 were of the RB kind and 59 of the BR type), and 645 nonspotted (PP, Pp, PP, p p ) males. The totals from table 5 are 897 redspotted females, 290 piebald females, of which163were
RB and 127
1:4
BR,and 1192 non-spotted males. Thisresultapproximatesa3:
sex-limited ratio, or actually a 3:1 ratio, since all the males appeared alike.
There were also a few more RB than BR piebald females-enumerated
in this experiment.
Backcross test: A PP jemaleXa Pp male
Homozygous normal bilateral red-spotted ( P P ) females were mated
to non-spotted heterozygous (Pp) males. From 16 PI separate pairs, a
total progeny of 291 normal red-spotted (PP, Pp)
females, and 286 normal,
GENETICS10: My 1925
270
J. K. BREITENBECHER
non-spotted ( P P , P p ) maleswasproduced, which is a 1: 1 sex-limited
ratio. This shows the complete dominance of the normal character over
the piebald. The data are given in table 6.
TABLE6
Backcross test.
P,:I Red-spotted female ( P P ) X l non-spotted male ( P p )
PlP*Ia NUMBER
-!
-~
Red-spotted females (PP, PpIlNon-spotted males ( P P , P # )
7
288
286
285
84.2
80.2
77.8
13
15
16
16
23
30
16
21
26
17
0
11
25
12
16
30
14
39
18
16
6
12
77.1
75.1
73.4
73.3
73.2
71.3
71.2
71.1
70.2
70. l
Totals. . . . . . . . . . . . . . .
FIELYTBAL SPOTTING
10
16
30
25
12
14
28
28
15
1
ExBected. . . . . . . . .
291
288.5
I
286
288.5
Backcross test: A Pp femaleXa PPmale
This test, the complete data of which are given in table 7 , was the
reciprocal of the previous one. The P, parents were heterozygous redspotted (Pp) females and homozygous non-spotted ( P P ) males. Nineteen
separate pairs mated produced 325 normal red-spotted ( P P , Pp) females
and 307 normal non-spotted ( P P , P p )males, which is a 1:l sex-limited
ratio. The results further confirm the previous tests, which showed that
the normal trait is dominant to the recessive piebald.
Backcross test: A B R piebald ( p p ) femaleXa Pp male
Homozygous ( p p ) BR piebald females bredtoheterozygousnonspotted (Pp) malesgavefrom
the 1 2 separatepairsmatedinthe
F,
generation, a total of 100 normal red-spotted (Pp) females, 37 R B piebald
( p p ) females, 51 BR piebald ( p p ) females, and 209 non-spotted (Pp, p p )
males. By adding thetwotypes
of piebald females, 88 piebald ( p p )
SEX-LIMITED BILATERAL
ASYMMETRY
I N BRUCHUS
271
females are obtained. The observed ratio is 100 :88: 209 which approximates a 1: 1: 2 sex-limited ratio. This experiment proves that the characters, normal and piebald, segregate as a 1: 1 ratio, if only the two types
of piebald females are added. It illustrates that the pied trait is arecessive
t o the normal. However, the RBpiebald females in this testdid not breed
true for the RB type but again produced two kinds, the RB and BR, in
about equal numbers. No male manifested the piebald character. The
results are given in table 8.
TABLE
7
Reciprocal backcross test.
PI:1 red-spotted (Pp)f e m a l e X l non-spotted (PP) male.
Pz ELYTBAL SPOTI'ING
PI PAIR N U m E P
69.1
62.6
62.3
61 .l
61.4
61 .5
61.6
61.9
61.10
61.11
61.12
61.13
61.14
61.15
61.16
61.52
61 .S1
61.42
61 .43
Totals. ..............
Expected. .............
Females
Red-spotted (PP, P p )
Males
Non-spotted (PP,Pp)
7
5
2
30
15
1
33
20
0
27
18
22
13
0
8
5
3
12
13
26
24
22
69
325
316
2
21
22
29
22
3
7
12
9
4
2
17
20
24
61
307
316
Backcross test: Am RB piebald ( p p ) femalexu Pp male
This test is the same as the previous one with the exception that an RB
piebald female was used instead of a BR female. The experiment is given
in table 9. The parents were homozygous RB ( p p ) females and heterozyproduced 176
gous, non-spotted (Pp) males. Twentyseparatepairs
normal red-spotted ( P p ) females, 105 RB piebald ( p p ) females, 59 BR
piebald ( p p ) females, and 356 non-spotted (Pp, p p ) males. Byadding
GENEITCS10: My 1925
272
J. K. BREITENBECHER
together the two kinds of piebald ( p p ) females, 164 were obtained. The
actual ratio was 176: 164: 356, approximating a 1: 1: 2 sex-limited ratio.
This test proves that piebald is a recessive homozygous trait when the
twotypes,dextralandsinistral,areadded.Lastly,thetwodifferent
piebald patterns, RB and BR, did not breed true for each different kind,
but each produced both types in about equal numbers.
TABLE
8
Backcross test.
PI:A piebald B R female ( p p ) X non-spotted
a
male ( P p ) .
F IELYTBAL SPOTTING
___-
-
Females
PIPAIR NUMBER
Piebald ( p p )
_____-
Red-spotted
282.2
247
246
244
216
210
186
79.2
78
79.9
77.2
66.1
(PP)
RB
BR
2
8
9
3
17
16
8
15
2
6
5
0
7
0
0
0
0
11
4
2
0
5
3
11
15
0
7
2
1
0
5
9
~
Totals.
. ... ..
i
Expected, 1: 1 : 2
100
99
1
7
4
3
"I
37
51
"
88
99
Males
Non-spotted
(PP.PP)
3
16
15
13
27
22
19
15
9
34
13
23
209
1 OR
A p p female X a p p male
The result of this test is given in table 10. Each pair mated consisted
of a piebald ( p p ) RB female and a non-spotted ( p p ) male, both of which
were homozygous for the piebald factor. Twelve separate pairs gave a
total of 109 RB piebald ( p p ) females, 100 BR piebald ( p p ) females and
213 non-spotted ( p p ) males. By adding the twotypes of piebald ( p p )
females, a total of209 piebald females is obtained, which approximates a
1 : 1 sex-limited ratio. The ratio is a sex-limited one because the piebald
trait is not visible in the male. It further demonstrates that this mutation
bred pure for piebald, when the two types of piebald females were added
together. Furthermore, the RB piebald females were about equal, though
2 73
SEX-LIMITED BILATERAL ASYMMETRY I N BRUCHUS
the BR pattern was slightly in excess. Lastly, this experiment proved
that the homozygous RB piebald female mated with a non-spotted male,
homozygous for piebald, did not breed true for the RBunilateral pattern,
but always produced two types, the RB and BR,in about equal numbers.
TABLE
9
Backcross test.
PI: A piebald RB female ( p p ) X a nowspotted male (Pp).
PIELYTRAL SPOTTINQ
?I
Females
PAIR NIJYBEP
"
Males
Non-spotted
Piebald ($9)
Rd-spotted
235
228
226
225
223
282
208
205
204
202
201
188
185
173
169
160
118
777
773
628
Totals.
(PP)
RB
3
9
5
19
4
8
12
9
13
24
11
6
1
5
5
12
4
6
11
9
4
5
3
8
5
0
BR
0
5
9
2 .
0
6
3
10
7
4
10
16
10
6
1
3
4
2
0
5
2
0
0
1
0
10
3
9
4
2
0
0
5
6
105
.. ... .
176
Expected, 1 :1:2
I74
(PP,PP)
8
27
2
26
13
16
27
24
14
36
22
25
7
7
15
23
5
14
18
27
59
164
174
356
348
A summary of the RB and BR piebald females
A summary of the two kinds of piebald females, the RB and the BR
types, is given in table 11. These totals show that for the piebald females,
448 manifested the RB spotting and 373 the BR pattern; that is, those
females having the red spots on the left elytrum and black ones on the
right were morefrequent thanthose having black spots on the leftelytrum
GENETICS10: My 1925
274
J. K. BREITENBECHER
and red spots on the right. The numbers for all tests were about equal.
The conclusions from this set of tests are, first, that piebald is a sexlimited recessive, visible only in the females, and second, that the character
is always seen on either the left or right elytrum, never on both at the
same time.
TABLE10
P I : A piebald R B female ( p p ) X a non-spotted male ( p p ) .
l
NUMBER
1
SPOTTING F1 ELYTBAL
Females
Piehald ($p)
PI PAIR
-
RB
BR
0
304
1
15 15 222
0
1
221
213
5
24 212
12
7
2 182
11
9
12
209
190
129
116
111
109
Totals. . . . . . . . .
.
3
18
12
14
25
24
19
2
23
0
5
11
0
6
17
11
13
18
109
100
~
;:
24
P
213
211
209
21l
TABLE
11
A w m m a r y of the piebald females (pp).
TOTALS FROM TABLENUMBER
1
1
5
6
9
10
11
Totals..
Males
Non-spotted
.. . . . . . . . . . . . . I
PIEBALD FEMALES
-
"
I
RB
-
l
BR
34
75
88
37
105
109
36
68
59
51
59
100
448
373
14
SEX-LIMITED BILATERAL ASYMMETRY IN BRUCHUS
275
DISCUSSION
Since the twobilateralpigmentedareas
of Bruchushavedefinite
boundaries, appearing always in the same position on the elytra of the
females, they form a pattern resembling, somewhat, the many piebald
traits in animals. Aside from the sex-differences, they suggest ALLEN’S
(‘centers of pigmentation” (ALLEN1904), especially since there are other
pigmentedareaslocated
elsewhere onthis weevil, which arenot sexlimited. Careful examination of the several available species of Bruchidae
show that the sexes are not easily distinguished,except in B. quadrimaculatus, though each species manifests a design peculiar to its kind.
I n Bruchus, symmetry is externally visible in the duplication of spots,
patterns, colors, elytra, antennae, legs, etc., while asymmetry is rarely
manifested externally. Inherited asymmetry is illustrated by this piebald
mutation, originating from purely symmetrical insects.
In order tostudytheinheritance
of thistrait successfully, it was
necessary to contrast the normal bilaterality with the abnormal
asymmetry. This was accomplished by injuring the pupae on one side, thus
producing asymmetrical elytra colors in the adult female. Such abnormalities were not inherited, though some were similar to the elytra1 spot1922).
color mosaics, a list of which has been published (BREITENBECHER
These were also non-inherited, but were interpreted, a t t h a t time, as
somaticmutations.Themostfrequent
mosaic type, red spotsonone
elytrum and black spots on the other, resembles this piebald mutation,
except for its heritability.
The actual cause of sex-limited differences in insects has not been coqclusively demonstrated. One of the theories advanced by GOLDSCHMIDT
(1922) in explanation of sex-controlled characters in Lepiand FISCHER
doptera seems to apply to these traits in Bruchus, since the male shows
himself a fraction of a day earlier than the female. However, only twenty
percent of the mutations in this weevil are sex-limited, although in every
mutation discovered the males issued a t the same time, whethersex-limited
or not; hence, some other explanation should account for the remaining
eighty percent, unless it is assumed that such characters develop more
rapidly. According to BROOKS(1922) the Y chromosome isabsentin
Bruchus;therefore, BRIDGES’S (1922) interpretation thatboth sexual
and sex-limited charactersdependupon
“modifiers” within the X
chromosome of Drosophila,upon further investigation, may be found
to be equally applicable to Bruchus.
The results of selection are negative. Since the males for both types
of pied females appear alike, it is impossible to distinguish them. It is not
GENETICS10: M y 1925
276
J. K. RREITENBECHER
unreasonable to suppose that in this experiment, comprising a great number of matings, an RB female will have been mated with an RB male at
least half the time, and thattheir offspring should breed true for this type;
but such pairs always produced,on theaverage. one dextral female,
one sinistral female and two normal males, regardless of the character of
the parents which entered the test.
The piebald trait for Bruchus is an instance of bilateral asymmetry.
The bilaterality is shown by the arrangement of the two spots on each
elytrum. The asymmetry is manifested by the color differences of these
spots. If this pied character is considered as unilateral, the black spots
representing this condition,then the red spotsmight be theresult of
normal genes. Or, if the bilateral red spots should be the trait, then a
unilateral inhibitor might make the color difference. There seems to be
some delicate adjusting mechanism which shifts theasymmetrical spotting
right aqd left according to chance, since equal numbers of dextrals and
sinistr als is produced.
CONCLUSIONS
1. The sexes in homozygous piebald cultures are dissimilar in appearance. The female has two black spots on one elytrum and two red spots
on the other, or vice versa, equal numbers dextral and sinistral. The male
has no elytra spots and is without dextral or sinistral differences.
2. This trait is sex-limited although it is transmitted equally by both
sexes.
3. The piebald mutation is a recessive, the typical recessive frequency
being obtained by adding together the dextral and sinistral females.
4. This character is called piebald, since it varies in the amount of
spotting and size of spots.
5 . Symmetry is normally manifested in this insect by the duplications
of spots, patterns, colors, antennae, elytra, legs, etc., while asymmetry is
rarely seen externally.
6 . This piebald-asymmetry is inherited, while asymmetries discovered
for mosaics and from experimental injuries were not.
7 . This trait hasa normal bilaterality and symmetry in the pupal state,
its asymmetry not being visible until the insect becomes adult.
8. Unilateral characters in animals are usually limited toone body side.
However, this type of unilaterality is not found in this piebald trait,
since dextral females always produce both dextral and sinistral insects
in equal numbers.
9. Probably some delicately adjusting mechanism shifts this asymmetry
right or left according to chance.
SEX-LIMITED BILATERAL ASYMMETRY IN BRUCHUS
277
10. Lastly, the piebald trait for Bruchus is an instance of inherited
bilateral asymmetry.
LITERATURE CITED
ALLEN, G . M., 1904 The heredity of coat colors in mice. Proc. Amer.Acad. Arts and Sci.
40: 61-163.
J. K., 1921 The genetic evidence of a multiple (triple) allelomorph system in
BREITENBECHER,
Bruchus and its relation to sex-limited inheritance. Genetics 6 : 55-90.
1922 Somatic mutations and elytra1 mosaics in Bruchus. Biol. Bull. 43: 10-23.
1923 A red-spotted sex-limited mutation in Bruchus. Amer. Nat. 57: 59-65.
BRIDGES,
C. B., 1922 The origin of variations in sexual and sex-limited characters. Amer. Nat.
56: 51-63.
FRANK
C., 1922 A preliminary note on the chromosome number in the spermatocytes
BROOKS,
of Bruchus. Proc. Oklahoma Acad. Sci.2: 49-52.
GOLDSCHMIDT,
R., and FISCHER,
E., 1922 Argynnis puphie-vulesinu, ein Fall geschlechtskontrollierter Vererbung bei Schmetterlingen. Genetica 4: 247-278.
GENETICS10: My 1925