Fruit Fly Genetics Drosophila melanogaster Slide show by
Fruit Fly Genetics Drosophila melanogaster Slide show by: Kelly Riedell/Brookings Biology http: //www. nasaexplores. com/show 2_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/Biol 114/Vfly 1. asp
What sex is it? XX Xy 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/Genetics 1. 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
Image from: http: //dev. biologists. org/content/develop/130/18/4427/F 7. 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+ = 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. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x PURPLE EYED Male 639 0 604 454 147 428 0 163 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. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x PURPLE EYED Male 639 0 604 454 147 428 0 163 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
IF THIS IS AN __________ I would expect this pattern p ___________ trait. p p+ p+ p p+ p+ p+ p p F 1 offspring p+ p p p F 2 offspring 100% WILD TYPE EYES MENDEL’s 3: 1 ratio 75% WILD TYPE EYES 25% PURPLE EYES 1000 flies EXPECTATIONS: IF YOU IGNORE SEX: 750 to be wild type eyes 250 to be purple eyes If you 375 125 DON’T IGNORE SEX: WILD TYPE EYED MALES WILD TYPE EYED FEMALES purple eyed FEMALES
FRUIT FLY CROSS- Practice Go to Fly Genetics: http: //www. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x PURPLE EYED Male 639 604 0 454 147 428 0 163 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
Are the deviations for the phenotypic ratio of the F 2 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 F 2 generation in the chart below. WILD TYPE PURPLE EYES 882 310 1192 X 0. 75 1192 X 0. 25 TOTAL 1192 Chi-square (X 2) = _________ How many degrees of freedom are there? _______ Referring to the critical values chart, what is the probability (p) value for these data? ________
H 0 - There is no difference between the frequencies observed and the frequencies expected if PURPLE eyes is an autosomal recessive trait. WT purple eyes 882 1192 X. 75=894 882 -894= -12 310 1192 X. 25=298 310 -298= 12 1192 (-12)2 (12)2 144/894 =. 161 144/298 =. 483 0. 664 2 -1=1 3. 84 454 purple female 147 WT male 428 purple male 163 1192 WT female 894/2= 447 298/2= 149 894/2= 447 149/2= 149 4 -1=3 7. 81 454 -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 2. 259
If the calculated X 2 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 P 1 flies ? FEMALE __p+ p+ ____ MALE ___ p p ______ What are the genotypes of the F 1 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 F 1 and F 2 offspring for this trait.
Make 2 Punnett squares showing parents and F 1 and F 2 offspring for this trait. p+ p+ p+ p p+ p p p+ p WT female F 1 Purple eye male p p+ p pp F 2
FRUIT FLY CROSS- Practice Go to Fly Genetics: http: //www. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x YELLOW BODY Male Yellow body males Yellow body females 570 0 616 0 262 311 606 0 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.
FRUIT FLY CROSS- Practice Go to Fly Genetics: http: //www. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x PURPLE EYED Male Yellow females 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
IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad WILD TYPE MOM X MUTANT type dad F 1 F 2 m+ X y m X y Xm+ X Xm+Xm Xm+y X m+ 50% wild type females 50% wild type males Xm+Xm+ Xm+y Xm Xm+Xm X my 50% wild type females 25% wild type males 25% mutant males 0% mutant females
FRUIT FLY CROSS- Practice Go to Fly Genetics: http: //www. sciencecourseware. org/vcise/drosophila/ Cross 1: Wild Type Female x YELLOW BODY Male 639 604 0 454 147 428 0 163 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) X-LINKED recessive _______________________ genetic trait.
H 0 - There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an X-LINKED recessive trait. WT female 606 1179 X 0. 5=589. 5 606 -589. 5=16. 52 = 272. 25 WT male 262 1179 X. 25=294. 75 262 -294. 75=-32. 75 (32. 75)2 =1072. 56 Yellow male 311 1179 X. 25=294. 75 311 -294. 75= 16. 25 (16. 25)2 = 264. 063 TOTAL 1, 179 272. 25/589. 5=. 462 1072. 56/294. 75=3. 639 264. 063/294. 75=. 896 4. 997 3 -1=2 5. 99
H 0 - There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an autosomal recessive trait. 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 606 Yellow female 0 WT male 262 Yellow male 311 WT female 442. 125 147. 375 163. 875 -147. 375 -189. 125 163. 625 1179 4 -1=3 26855. 016 21719. 391 35768. 266 26773. 141 60. 74 147. 375 80. 9 181. 667 470. 682 7. 81 470. 682 > 7. 81 REJECT THE NULL. . . YELLOW BODY is NOT AN AUTOSOMAL RECESSIVE MAKE A NEW HYPOTHESIS!!!!
H 0 - There is no difference between the frequencies observed and the frequencies expected if YELLOW BODY is an AUTOSOMAL recessive trait. WT female 606 1179 X 0. 5=589. 5 606 -589. 5=16. 52 = 272. 25 WT male 262 1179 X. 25=294. 75 262 -294. 75=-32. 75 (32. 75)2 =1072. 56 Yellow male 311 1179 X. 25=294. 75 311 -294. 75= 16. 25 (16. 25)2 = 264. 063 TOTAL 1, 179 272. 25/589. 5=. 462 1072. 56/294. 75=3. 639 264. 063/294. 75=. 896 4. 997 3 -1=2 5. 99
If the calculated X 2 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 P 1 flies ? FEMALE __Xb+ ____ MALE ___ Xb y ______ What are the genotypes of the F 1 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 F 1 and F 2 offspring for this trait.
IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad WILD TYPE mom X mutant dad F 1 F 2 m+ X y m X y Xm+ X Xm+Xm Xm+y X m+ 50% normal females 50% normal males Xm+Xm+ Xm+y Xm Xm. Xm+ X my 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+ IF GENE is AUTOSOMAL and RECESSIVE TO + m+ m MALES: FEMALES 1: 1 m m+ m+ m+ m m m+ m F 1 All = m+m (wildtype) mm F 2 ¼ = m + m+ ½ = m+ m ¼=mm 75% wildtype 25% mutant 3 WT : 1 mutant 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 IF GENE is AUTOSOMAL and DOMINANT TO + MALES: FEMALES 1: 1 F 1 All = M+M (mutant) M M +M MM F 2 ¼ = M + M+ ½ = M+ M ¼=MM 25% wildtype 75% mutant 1 WT : 3 mutant
IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad WILD TYPE mom X mutant dad F 1 F 2 m+ X y m X y Xm+ X Xm+Xm Xm+y X m+ 50% normal females 50% normal males Xm+Xm+ Xm+y Xm Xm. Xm+ X my 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
IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Mutant mom X wild type dad F 1 F 2 m+ Xm y Xm Xm+Xm Xmy 50% wild type females 50% mutant males Xm+Xm Xm+y Xm X my 25% wild type 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+ IF GENE is AUTOSOMAL and RECESSIVE TO + m+ m MALES: FEMALES 1: 1 m m+ m+ m+ m m m+ m F 1 All = m+m (wildtype) mm F 2 ¼ = m + m+ ½ = m+ m ¼=mm 75% wildtype 25% mutant 3 WT : 1 mutant http: //www. exploratorium. edu/exhibits/mutant_flies/curly-wings. gif
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