GENET275 Assignment #2
Due: Wednesday March 3, 2006
1. a)
SM6, Cy / Roi ; TM3, Sb / Ser 
SM6, Cy;
TM3,Sb
SM6, Cy;
Ser
Roi;
TM3, Sb
Roi; Ser
SM6, Cy; TM3,Sb
SM6, Cy/SM6, Cy;
TM3, Sb/TM3, Sb
SM6, Cy/SM6, Cy;
TM3, Sb/Ser
SM6, Cy/Roi;
TM3, Sb/TM3, Sb
SM6, Cy/Roi;
TM3, Sb/Ser
Brian Chan 1074844
Andrew Kwan 1056818
Calvin Cheung 1077159
Cindy Lee 1041791
SM6, Cy / Roi ; TM3, Sb / Ser
SM6, Cy; Ser
Roi; TM3, Sb
Roi; Ser
SM6, Cy/SM6,Cy;
TM3, Sb/Ser
SM6, Cy/SM6, Cy;
Ser/Ser
SM6, Cy/Roi;
TM3, Sb/Ser
SM6, Cy/Roi;
Ser/Ser
SM6, Cy/Roi;
TM3, Sb/TM3, Sb
SM6, Cy/Roi;
TM3, Sb/Ser
Roi/Roi;
TM3, Sb/TM3, Sb
Roi/Roi;
TM3, Sb/Ser
SM6, Cy/Roi;
TM3, Sb/Ser
SM6, Cy/Roi;
Ser/Ser
Roi/Roi;
TM3, Sb/Ser
Roi/Roi;
Ser/Ser
b) The only progeny that will survive are those heterozygous for both the 2nd and 3rd
chromosomes. (ie. Those that have the genotype SM6, Cy / Roi ; TM3, Sb / Ser )
-text indicates dead
2.
Start with: SM6, Cy/ ku80; +/+
Start with: +/+; p53/p53
SM6,Cy/ku80; +/+  SM6,Cy/Roi; TM3,Sb/Ser
+/+; p53/p53  SM6, Cy/ Roi; TM3, Sb/ Ser


SM6, Cy/ku80; TM3, Sb/+
SM6, Cy/ +; TM3, Sb/p53, e
(progeny that is Cy and Sb, but not Roi or Ser)
(progeny that is Cy and Sb, but not Roi or Ser)


SM6, Cy/ku80; TM3, Sb/+
SM6, Cy/ +; TM3, Sb/p53, e


SM6,Cy/Roi; TM3,Sb/Ser
SM6, Cy/ Roi; TM3, Sb/ Ser


SM6, Cy/k80; TM3, Sb/Ser
SM6, Cy/ Roi; TM3, Sb/p53, e
(progeny that is Cy, Sb, and Ser, but not Roi)
(progeny that is Cy, Roi, and Sb, but not Ser)


SM6, Cy/k80; TM3, Sb/Ser

SM6, Cy/ Roi; TM3, Sb/p53, e

SM6, Cy/ku80; TM3, Sb/p53, e
(progeny that is Cy and Sb, but not Roi or Ser)

SM6, Cy/ku80; TM3, Sb/p53, e  SM6, Cy/ku80; TM3, Sb/p53, e

SM6, Cy/ku80; p53, e/p53, e
(progeny that is Cy and e, but not Sb)
GENET275 Assignment #2
Due: Wednesday March 3, 2006
Brian Chan 1074844
Andrew Kwan 1056818
Calvin Cheung 1077159
Cindy Lee 1041791
3. The stpy gene is on the 2nd chromosome because there is no phenotype of both curly wings
and stumpy. Stumpy is recessive so to be expressed it must be homozygous since curly wings
and stumpy do not show up together, they must be on the same chromosome. There also aren’t
any progeny that only have stubble bristles because there was no wild-type alleles for Cy and
stpy in the F1 cross so all progeny in the F2 generation will have curly wings and/or are stumpy.
The gene is not X-linked or Y-linked because the ratios are approximately the same for both
females and males.
P genotype:
+/+; stpy/stpy; +/+ 
F1 genotypes:
Females:
+/+; SM6, Cy/stpy; TM3, Sb/+
+/+; SM6, Cy/stpy; +/+
+/+; +/stpy; TM3, Sb/+
+/+; +/stpy; +/+
F1 cross:
+/Y; SM6, Cy/+; TM3, Sb/+
Males:
+/Y; SM6, Cy/stpy; TM3, Sb/+
+/Y; SM6, Cy/stpy; +/+
+/Y; +/stpy; TM3, Sb/+
+/Y; +/stpy; +/+
SM6, Cy/stpy; TM3, Sb/+ 
SM6, Cy/stpy; TM3, Sb/+
SM6, Cy; TM3, Sb
SM6, Cy; +
stpy; TM3, Sb
SM6, Cy; SM6, Cy/SM6, Cy;
SM6, Cy/SM6, Cy; SM6, Cy/stpy;
TM3, Sb
TM3, Sb/TM3, Sb
TM3, Sb/+
TM3, Sb/TM3, Sb
SM6, Cy; SM6, Cy/SM6, Cy;
SM6, Cy/SM6, Cy; SM6, Cy/stpy;
+
TM3, Sb/+
+/+
TM3, Sb/+
stpy;
SM6, Cy/stpy;
SM6, Cy/stpy;
stpy/stpy;
TM3, Sb
TM3, Sb/TM3, Sb
TM3, Sb/+
TM3, Sb/ TM3, Sb
stpy: +
SM6, Cy/stpy;
SM6, Cy/stpy;
stpy/stpy;
TM3, Sb/+
+/+
TM3, Sb/+
-text indicates dead
The phenotypic ratio of the progeny that survive are 4:2:2:1 for
Curly wings and Stubble: Curly wings: Stubble and Stumpy: Stumpy
which matches the ratio (446: 223: 221: 110) of the F2 generation
stpy;+
SM6, Cy/stpy;
TM3, Sb/+
SM6, Cy/stpy;
+/+
stpy/stpy;
TM3, Sb/+
stpy/stpy;
+/+
4. First we would need to mutate the male gametes of the wild-type allele stock with EMS to get
new mutant alleles of the car gene car*. We would then screen for male flies with a carnationeye phenotype. Next we would perform genetic crosses with the balancer x-chromosome to
analyze the type of mutation that EMS caused. (Assuming no crossing-over)
P: (car, sn)*/ Y  FM7, w, Bar, fs / car, sn
F1:
(car, sn)*/ car, sn
FM7, w, Bar, fs/ Y
and
and
(car, sn)*/FM7, w, Bar, fs
car, sn/ Y
GENET275 Assignment #2
Due: Wednesday March 3, 2006
Brian Chan 1074844
Andrew Kwan 1056818
Calvin Cheung 1077159
Cindy Lee 1041791
If there is a mutation in the car gene, the (car+, sn+)*/ cn, sn females (non-Bar) will not have
red (wild-type) eyes – with the exception of a hypermorphic car allele.
We can then cross (car, sn)*/FM7, w, Bar, fs females with FM7, w, Bar, fs/ Y males.
F1 cross: FM7, w, Bar, fs/ Y  (car, sn)*/FM7, w, Bar, fs
F2:
FM7, w, Bar, fs/(car, sn)*
(car, sn)*/Y
and
and
FM7, w, Bar, fs/ FM7, w, Bar, fs (sterile)
FM7, w, Bar, fs / Y
If the mutant car allele is…
Null: F1: females are either pink-eyed or red-eyed Bar-eyed; males are either white-eyed Bareyed, or pink-eyed and singed
F2: females are red-eyed Bar-eyed or white-eyed Bar-eyed and sterile; males are all white-eyed
Bar-eyed because those carrying the car null allele would be dead if it was lethal.
Hypomorphic: F1: females are either pink-eyed or red-eyed Bar-eyed; males are either whiteeyed Bar-eyed, or pink-eyed and singed
F2: females are red-eyed Bar-eyed or white-eyed Bar-eyed and sterile; males are either pinkeyed or white-eyed Bar-eyed
Hypermorphic: These alleles would be hard to distinguish from the wild-type allele because the
phenotype would be red-eyes, but may be possible if the intensity of the red differs depending on
the expression of the hypermorphic allele.
F1: females are either red-eyed or red-eyed Bar-eyed; males are either white-eyed Bar-eyed, or
pink-eyed and singed
F2: females are either red-eyed Bar-eyed or white-eyed Bar-eyed and sterile; males are either
red-eyed or white-eyed Bar-eyed
Antimorphic: These alleles would be dominant, however we do not know the structure of the
protein so we are unsure of the extend of change in phenotype. Assuming Caranti /+ gives pinkeyes…
F1: females are either pink-eyed (unless hypomorph/antimorph is lethal) or pink-eyed Bar-eyed;
males are either white-eyed Bar-eyed, or pink-eyed and singed
F2: females are either pink-eyed Bar-eyed or white-eyed Bar-eyed and sterile; males are all
white-eyed Bar-eyed (those with the antimorphic allele would not survive if the car gene is
essential)
Neomorphic: These alleles would be hard to identify because we will not be able to determine if
new “functions” are due to the neomorphic allele of car or mutations on other genes. Also if
gain of function means loss of normal function, the neomorph can behave similar to that of null
alleles.