1. Fruit Fly Genetics Drosophila melanogaster

2. Model organism for Genetic Studies Drosophila melanogaster
It is easily cultured in the lab
Has a short generation time
Can produce many offspring
Has only 4 chromosomes visible under a light microscope
Many mutants have been created

3. What sex is it? Xy XX
Sexing flies: Male and female fruit flies can be distinguished from each other in three ways:
1) Only males have a sex comb, a fringe of black bristles on the forelegs.
2) The tip of the abdomen is elongated and somewhat pointed in females and more rounded in males.
3) The abdomen of the female has seven segments, whereas that of the male has only five segments.
Images from:

4. Normal fly is called a "wild type" and any fly exhibiting a phenotypic mutation is called a "mutant".
Mutant flies are given names that generally denote the type of mutation the fly exhibits.
EXAMPLE:
Drosophila melanogaster Eye colors (clockwise): brown, cinnabar, sepia, vermilion, white, wild. Also, the wild-eyed fly has a yellow body, the sepia-eyed fly has an ebony body, and the brown-eyed fly has a black body.

5. Each mutation is also given a letter code.
Wild type is denoted by a superscript + over the mutant letter code.
EXAMPLE:
Ebony body color = e
Wild type body color = e+
So homozygous ebony body fly = e e heterozygous ebony body fly = e+ e

6. Xf Xf = forked bristle female
Image from:
Sex-linked genes are located on one of the sex chromosomes (usually the X chromosome). Thus, the genotypic notation for a mutant gene for FORKED BRISTLES on the X chromosome would look like:
Xf Xf = forked bristle female
Xf+ Xf = wild type bristle heterozygote female
Xf Y = forked bristle male
Xf+ Y = wild type bristle male
Xf+ Xf+ = wild type bristle female
Image from:

7. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x PURPLE EYED Male
639
454
147
604
428
163
These data for PURPLE EYES suggest this pattern is characteristic of a ________________________ __________________________ gene dominant recessive autosomal X-linked
Write a NULL hypothesis that describes the mode of inheritance for the PURPLE EYED trait. There is no difference between the observed data and the data expected if purple eyes is a(n) ______________________ _______________________ genetic trait.

8. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x PURPLE EYED Male
639
454
147
604
428
163
ONLY Wild type; No purple suggests purple eyes is recessive to Wild type
No differences in male:female ratios Suggests purple eyes is NOT a sex linked trait

9. p p p+ p+ p p+ p p+ p p+ p p+ p+ p+ p p+ p p p MENDEL’s 3:1 ratio
IF THIS IS AN ____________________ _____________________ trait.
I would expect this pattern
p p p+
p p p+ p
p+ p p+ p
p+ p+ p+ p
p+ p p p
F1 offspring F2 offspring
75% WILD TYPE EYES % PURPLE EYES
100% WILD TYPE EYES
1000 flies EXPECTATIONS: IF YOU IGNORE SEX:
750 to be wild type eyes to be purple eyes If you DON’T IGNORE SEX: WILD TYPE EYED MALES
375 WILD TYPE EYED FEMALES
125 purple eyed MALES purple eyed FEMALES
MENDEL’s 3:1 ratio

10. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x PURPLE EYED Male
639
454
604
147
428
163
These data for PURPLE EYES suggest this pattern is characteristic of a ________________________ __________________________ gene dominant recessive autosomal X-linked
Write a NULL hypothesis that describes the mode of inheritance for the PURPLE EYED trait. There is no difference between the observed data and the data expected if purple eyes is a(n) ______________________ _______________________ genetic trait.
autosomal recessive

15. WILD TYPE 882 1192 X 0.75 PURPLE EYES 310 1192 X 0.25 TOTAL 1192
Are the deviations for the phenotypic ratio of the F2 generation within the limits expected by chance? To answer this question, statistically analyze the data using the Chi-square analysis. Calculate the Chi-square statistic for the F2 generation in the chart below.
Chi-square (X2) = __________________ How many degrees of freedom are there? _____________
Referring to the critical values chart, what is the probability (p) value for these data? _______________
WILD TYPE
882
1192 X 0.75
PURPLE EYES
310
1192 X 0.25
TOTAL

16. H0-
There is no difference between the frequencies observed and the frequencies expected if PURPLE eyes is an autosomal recessive trait.
WT X.75=894
purple eyes X.25=298
= -12
= 12
(-12)2
(12)2
144/894 = .161
144/298 = .483
1192
0.664
2-1=1
3.84
WT female /2= 447
purple female /2= 149
WT male /2= 447
purple male /2= 149
= 7
= = -19
= 14
72 = 49
(-2)2 = 4
(-19)2 = 361
14 2 = 196
49/447=0.109
4/149=0.0268
361/447=0.808
196/149=1.315
1192
2.259
4-1=3
7.81

17. If the calculated X2 value is greater than or equal to the critical value from the table, then the null hypothesis is REJECTED.
According to the probability (p) value, do you accept or reject your null hypothesis for this cross? Explain.
 The Chi square value is LESS than the critical value of 3.84 so the null hypothesis is ACCEPTED. There is NO difference between observed and expected ratios Purple eyes IS an autosomal recessive trait.
What are the genotypes of the P1 flies ? FEMALE __p+ p+ ____ MALE ___ p p ____________ What are the genotypes of the F1 flies? FEMALE ___ p+ p _____________ MALE _ p+ p _____________
How is this trait inherited?
Is the mutation autosomal or sex linked? ____AUTOSOMAL_______
Is the mutation dominant or recessive? __RECESSIVE__________

18. Make 2 Punnett squares showing parents and F1 and F2 offspring for this trait.

19. p p p+ p p+ p+ p p+ p p+ p p+ p + p+ p p+ p p p F1 F2
Make 2 Punnett squares showing parents and F1 and F2 offspring for this trait.
p p p+
p p p+ p
Purple eye male
p+ p p+ p
p+ p + p+ p
p+ p p p
WT female F1 F2

20. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x YELLOW BODY Male
570
262
Yellow body males
311
616
606
Yellow body males
These data for YELLOW BODY suggest this pattern is characteristic of a ________________________ __________________________ gene dominant recessive autosomal X-linked
Write a NULL hypothesis that describes the mode of inheritance for the PURPLE EYED trait. There is no difference between the observed data and the data expected if purple eyes is a(n) ______________________ _______________________ genetic trait.

21. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x PURPLE EYED Male
ONLY Wild type; No yellow bodies suggests yellow body is recessive to Wild type
Differences in male:female ratios Suggests yellow body is a sex linked trait

22. Xm+ y Xm y Xm+ Xm+ Xm+Xm Xm+y Xm Xm+Xm+ Xm+y Xm+Xm Xmy
IF GENE is X-linked recessive
Different pattern if gene is inherited from mom or dad
WILD TYPE MOM X MUTANT type dad F1 F2
Xm y
Xm+ Xm
Xm y
Xm+
Xm+Xm Xm+y
Xm+Xm+ Xm+y
Xm+Xm Xmy
50% wild type females
50% wild type males
50% wild type females 25% wild type males % mutant males 0% wild type males

23. FRUIT FLY CROSS- Practice Go to Fly Genetics: http://www
Cross 1: Wild Type Female x YELLOW BODY Male
639
454
604
147
428
163
These data for YELLOW BODY suggest this pattern is characteristic of a ________________________ __________________________ gene dominant recessive autosomal X-linked
Write a NULL hypothesis that describes the mode of inheritance for the YELLOW BODY trait. There is no difference between the observed data and the data expected if purple eyes is a(n) ______________________ _______________________ genetic trait.
X-LINKED recessive

27. H0-
There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an X-LINKED recessive trait.
272.25/589.5=.462
/294.75=3.639
/294.75=.896
WT female X 0.5=589.5
WT male X.25=294.75
Yellow male X.25=294.75
=16.5
=-32.75
= 16.25
16.52 =
(32.75)2 =
(16.25)2 =
TOTAL 1,179
4.997
3-1=2
5.99

28. There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an autosomal recessive trait.
H0-
Expect: 75% WT 25% yellow body
1179 X 0.75 = female/ male 1179 X .25 = female/ male
WT female
Yellow female
WT male
Yellow male
60.74
80.9
1179
4-1=3
7.81
> 7.81 REJECT THE NULL YELLOW BODY is NOT AN AUTOSOMAL RECESSIVE
MAKE A NEW HYPOTHESIS!!!!

29. H0-
There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an AUTOSOMAL recessive trait.
272.25/589.5=.462
/294.75=3.639
/294.75=.896
WT female X 0.5=589.5
WT male X.25=294.75
Yellow male X.25=294.75
Yellow female 0
=16.5
=-32.75
= 16.25
16.52 =
(32.75)2 =
(16.25)2 =
TOTAL 1,179
4.997
3-1=2
5.99

30. If the calculated X2 value is greater than or equal to the critical value from the table, then the null hypothesis is REJECTED.
According to the probability (p) value, do you accept or reject your null hypothesis for this cross? Explain.
 The Chi square value is LESS than the critical value of 5.99 so the null hypothesis is ACCEPTED. There is NO difference between observed and expected ratios YELLOW BODY IS an X-LINKED recessive trait.
What are the genotypes of the P1 flies ? FEMALE __Xb+ Xb+ ____ MALE ___ Xb y ____________ What are the genotypes of the F1 flies? FEMALE ___ Xb+ Xb _____________ MALE _ Xb+ y _____________
How is this trait inherited?
Is the mutation autosomal or sex linked? __X-LINKED____
Is the mutation dominant or recessive? __RECESSIVE____

31. Make 2 Punnett squares showing parents and F1 and F2 offspring for this trait.

32. Xm+ y Xm y Xm+ Xm+ Xm+Xm+ Xm+y Xm+Xm Xm+y Xm XmXm+ Xmy
IF GENE is X-linked recessive
Different pattern if gene is inherited from mom or dad
WILD TYPE mom X mutant dad F1 F2
Xm y
Xm+ Xm
Xm y
Xm+
Xm+Xm+ Xm+y
XmXm+ Xmy
Xm+Xm Xm+y
50% normal females
50% normal males
50% normal females % mutant males 25% normal males
0% mutant females
When using Virtual fly lab Choose ignore sex and see if it changes the ratios

33. m m m+ m+ m m+ m+ m+ m+ m m m+ m m+ m F1 All = m+m (wildtype) m+ m m m
IF GENE is AUTOSOMAL and RECESSIVE TO +
MALES:FEMALES 1:1
m+ m m+ m
F1 All = m+m (wildtype)
m+ m m+ m F2
¼ = m+ m % wildtype
½ = m+ m
¼ = m m % mutant
3 WT : 1mutant
m+ m+ m+ m
m+ m m m

34. M M M+ M+M M+M M+ M M+ M+M+ M+M M M+M M M
IF GENE is AUTOSOMAL and DOMINANT TO +
MALES:FEMALES 1:1
M M M+
M+M M+M F1
All = M+M (mutant)
M M M+ M F2
¼ = M+ M % wildtype
½ = M+ M
¼ = M M % mutant
1WT :3 mutant
M+M+ M+M
M+M M M

35. Xm+ y Xm y Xm+ Xm+ Xm+Xm+ Xm+y Xm+Xm Xm+y Xm XmXm+ Xmy
IF GENE is X-linked recessive
Different pattern if gene is inherited from mom or dad
WILD TYPE mom X mutant dad F1 F2
Xm y
Xm+ Xm
Xm y
Xm+
Xm+Xm+ Xm+y
XmXm+ Xmy
Xm+Xm Xm+y
50% normal females
50% normal males
50% normal females % mutant males 25% normal males
0% mutant females
When using Virtual fly lab Choose ignore sex and see if it changes the ratios

36. Xm y Xm+ y Xm Xm+ Xm+Xm Xm+y Xm+Xm Xmy Xm XmXm Xmy
IF GENE is X-linked recessive
Different pattern if gene is inherited from mom or dad
Mutant mom X wild type dad F1 F2
Xm y
Xm+ Xm
Xm+ y Xm
Xm+Xm Xm+y
XmXm Xmy
Xm+Xm Xmy
50% wild type females
50% mutant males
25% wild type females 25% mutant females % mutant males 25% wild type males
When using Virtual fly lab Choose ignore sex and see if it changes the ratios

37. m m m+ m+ m m+ m+ m+ m+ m m m+ m m+ m F1 All = m+m (wildtype) m+ m m m
IF GENE is AUTOSOMAL and RECESSIVE TO +
MALES:FEMALES 1:1
m+ m m+ m
F1 All = m+m (wildtype)
m+ m m+ m F2
¼ = m+ m % wildtype
½ = m+ m
¼ = m m % mutant
3 WT : 1mutant
m+ m+ m+ m
m+ m m m