Fruit Fly Genetics Drosophila melanogaster http://www.nasaexplores.com/show2_articlea.php?id=04-006

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 http://bioweb.wku.edu/courses/Biol114/Vfly1.asp

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: http://math.hws.edu/javamath/ryan/Genetics1.html

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. http://www.answers.com/topic/drosophila-melanogaster-1

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 http://www.exploratorium.edu/exhibits/mutant_flies/ebony.gif

Xf Xf = forked bristle female Image from: http://dev.biologists.org/content/develop/130/18/4427/F7.large.jpg 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: http://www.exploratorium.edu/exhibits/mutant_flies/white-eyes.gif

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.

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

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 25% PURPLE EYES 100% WILD TYPE EYES 1000 flies EXPECTATIONS: IF YOU IGNORE SEX: 750 to be wild type eyes 250 to be purple eyes If you DON’T IGNORE SEX: 375 WILD TYPE EYED MALES 375 WILD TYPE EYED FEMALES 125 purple eyed MALES 125 purple eyed FEMALES MENDEL’s 3:1 ratio

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

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 1192

H0- There is no difference between the frequencies observed and the frequencies expected if PURPLE eyes is an autosomal recessive trait. WT 882 1192X.75=894 purple eyes 310 1192X.25=298 882-894= -12 310-298= 12 (-12)2 (12)2 144/894 = .161 144/298 = .483 1192 0.664 2-1=1 3.84 WT female 454 894/2= 447 purple female147 298/2= 149 WT male 428 894/2= 447 purple male 163 149/2= 149 454-447= 7 147-149= -2 428-447= -19 163-149= 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

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__________  

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

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

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.

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

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 25% mutant males 0% wild type males

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

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 1072.56/294.75=3.639 264.063/294.75=.896 WT female 606 1179X 0.5=589.5 WT male 262 1179X.25=294.75 Yellow male 311 1179X.25=294.75 606-589.5=16.5 262-294.75=-32.75 311-294.75= 16.25 16.52 = 272.25 (32.75)2 =1072.56 (16.25)2 = 264.063 TOTAL 1,179 4.997 3-1=2 5.99

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 = 884.25 442.125 female/442.125 male 1179 X .25 = 294.75 147.375 female/147.375 male WT female 606 442.125 Yellow female 0 147.375 WT male 262 442.125 Yellow male 311 147.375 163.875 -147.375 -189.125 163.625 26855.016 21719.391 35768.266 26773.141 60.74 147.375 80.9 181.667 1179 470.682 4-1=3 7.81 470.682 > 7.81 REJECT THE NULL . . . YELLOW BODY is NOT AN AUTOSOMAL RECESSIVE MAKE A NEW HYPOTHESIS!!!!

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 1072.56/294.75=3.639 264.063/294.75=.896 WT female 606 1179X 0.5=589.5 WT male 262 1179X.25=294.75 Yellow male 311 1179X.25=294.75 Yellow female 0 606-589.5=16.5 262-294.75=-32.75 311-294.75= 16.25 16.52 = 272.25 (32.75)2 =1072.56 (16.25)2 = 264.063 TOTAL 1,179 4.997 3-1=2 5.99

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____  

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

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 25% mutant males 25% normal males 0% mutant females When using Virtual fly lab Choose ignore sex and see if it changes the ratios

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+ 75% wildtype ½ = m+ m ¼ = m m 25% mutant 3 WT : 1mutant m+ m+ m+ m m+ m m m http://www.exploratorium.edu/exhibits/mutant_flies/curly-wings.gif

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+ 25% wildtype ½ = M+ M ¼ = M M 75% mutant 1WT :3 mutant M+M+ M+M M+M M M

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 25% mutant males 25% normal males 0% mutant females When using Virtual fly lab Choose ignore sex and see if it changes the ratios

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 25% mutant males 25% wild type males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

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+ 75% wildtype ½ = m+ m ¼ = m m 25% mutant 3 WT : 1mutant m+ m+ m+ m m+ m m m http://www.exploratorium.edu/exhibits/mutant_flies/curly-wings.gif