Indentification and Mapping of Unknown Mutations in the Fruit Fly in Drosophila melanogaster. By Michael Tekin and Vincent Saraceno

Transcription

1 Indentification and Mapping of Unknown Mutations in the Fruit Fly in Drosophila melanogaster By Michael Tekin and Vincent Saraceno Bilology 332 Section 2 December 5, 2012

2 Abstract The code of the unknown mutations is Sarah/Steve and the mutant s phenotype is black eyes and black body. Both of these traits were caused by a darker pigmentation. The mode of inheritance is simple autosomal recessive. Two genes are involved to produce this mutant phenotype. There was no observed gene interaction among these two genes. The gene symbols assigned for these unknown genes were the following: be=black eyes and bb= black body. The following chromosome maps were produced from the F2 and Female Backcross data. Please note the marker genes in the chromosome map below are coded as the following: G=glued and S=stubble F2 1A data be 16.4 cm bb F2 1B data be 24.7 cm bb F2 average be 20.6 cm bb Female Backcross be 12.3 cm G 16.6 cm S 15.8 cm bb F1 Results The fruit flies were scored from the following parental crosses15 mutant male X 10 wild females cross(1a) and the 15 wild males with 10 mutant females cross(1b). 20 males and 20 females from each cross were scored. All fruit flies scored were wild type. This information indicates that

3 both mutant traits are autosomal recessive and not sex linked since both sexes express wild type in each culture. F2 Results The F2 Progeny was produced by crossing the 1A F1 progeny with each other and 1B F1 progeny with each other. The following results was obtained from this cross. Results for number of genes There were a total of four different classes of fruit flies scored (excluding gender): Wild-type, Black Eyes, Black body, and Black eyes/black body. This indicates a minimum of two genes are involved, one gene coding for black eyes, and one gene coding for black bodies. Results for allelic relationship of genes: dominant or recessive The F2 monohybrid ratios of the 1A and 1B are seen in the tables below Table 1 Gene Actual ratio Average Ratio Black bodies 1077 Wild-type bodies:119 black bodies 9.1 Wild-type bodies:1 black bodies Black eyes 1083 Wild-type eyes:113 black eyes 9.6 Wild-type eyes:1 black eyes Table 1 is the average ratios between wild-type and mutant traits of the F2 progeny of the 1A cross. The ratio were found by taking the number fruit flies with the wild-type trait and dividing it by the mutant trait(since in all situation Wild-type>Mutant). A sample calculation is provided below. Sample calculation /119 = 9.1 Table 2 Gene Actual ratio Average Ratio Black bodies 895 Wild-type bodies:215 black bodies 4.2 Wild-type bodies:1 black bodies Black eyes 908 Wild-type eyes: 202 black eyes 4.5 Wild-type eyes:1 black eyes

4 Table 2 is the average ratios between wild-type and mutant traits of the F2 progeny of the 1B cross. The ratio were found by taking the number fruit flies with the wild-type trait and dividing it by the mutant trait(since in all situation Wild-type>Mutant). In theory simple dominance, a monohybrid should produce a 3:1 ratio of the dominant phenotype to the recessive phenotype. This is not the case; all of the monohybrid ratios are higher than 3:1. However, even if table 1 and table 2 does not show a perfect 3:1 ratio, both tables indicate that black eyes are recessive, and black body are recessive. This is because in all cases the number of wild-type phenotypes observed versus mutant phenotypes observed is greater than the 3:1 ratio. Results for whether the genes are autosomal or sex linked Since both mutant genes were determined to be recessive, if the either of the genes were sexedlinked, the 1A data would still form a 3:1 ratio of the wild-type to mutant phenotype. However, the recessive mutant phenotype would only be expressed by males. Moreover there would be two wild-type females for every wild-type male. Thus, both the black eye and black body trait would only be observed in males, if both genes were sex linked. In the 1B there would be a 1:1 ratio formed between the mutant recessive and wild-type dominant phenotype. This is not the case in either bottle. The amount of males and females in every phenotypic class are relatively equal, which is not the ratios predicted by sex-linkage. When doing a chi-squared analysis, the null hypothesis that this slight deviation in gender is due to random chance cannot be rejected (P is between.05 and.1). Moreover, affected females are present. Sex linkage predicts no affected females. Also, as seen in table 2 above, a 1:1 ratio was not form between either set of phenotypic classes. Thus, the 1A or 1B data does not support sex-linkage for black eyes or black body. Thus, both of these genes are most likely autosomal. Epitasis and Gene interaction The F2 data did not indicate any gene interaction nor epitasis.

5 Three proposed hypothesizes There was three proprosed hypothesizes on how these two genes are related. First, these two genes independently assort due to being on separate chromosomes. Second, there are on the same chromosome, but are farther than 50 cm apart. Last, these two gens are located on the same chromosome, within 50 cm of each other. Hypothesis one: Independent assortment If the first hypothesis is true, 1A would produce a dihybrid ratio of Wild-type: 9.1 Black Eyed: 9.6 Black Bodied: 1 Black Bodied and Black Eyes and 1B would produce a dihybrid ratio 18.9 Wild-type: 4.2 Black eyed 4.5 Black bodied: 1 Black Bodied and Black Eyes. These ratios are derived from multiplying the monohybrid ratio of black eyes from the monohybrid ratio of black bodies seen in table 1 and table 2. A sample calculation is provided below. Sample calculation Wild-type Bodies: 1Black Bodies X 9.1 Wild-type Eyes: 1 Black Eyes Wild-type: 9.1 Black Eyed: 9.6 Black Bodied: 1 Black Bodied and Black Eyes However after doing a chi-squared analysis, this hypothesis is rejected. In both 1A and 1B P<<<<.05, so random chance cannot explain the variation from the expected dihybrid ratios. Hypothesis Two: located on same chromosome but are more than 50 map units apart. If second hypothesis was correct then we would observe a 5 wild-type:1 black body:1 black eye:1 ratio black eyes and black body because crossing over only happen in female. However, this ratio have to be modified to account for environmental factors. The modified ratios for 1A was 5 Wild-type: 39 Black Bodied:.38 Black Eyes:.15 Black Bodied and Black Eyes and 1B was 5 Wild-type:.78 Black Bodied:.73 Black Eyes:.57 Black Bodied and Black Eyes. These were calculated using correction factors. Below is a sample calculation for the 1A modified ratio. Sample calculation 3

6 Genotype Number Correction factor Ratio bb + be =1 5(1.00)=5 bb bb be + 52 bb = bb bb/[.25(total)] 1(119/299)=.39 bb + be be 46 be = be be/[.25(total)] 1(113/299)=.38 bb bb be be 67 (bb be) = bb x be 1(.41)(.40)=.15 Total=5.92 When doing a chi squared analysis comparing these ratios to the ratios observed in both 1A and 1B we reject this hypothesis since the observed ratios cannot be explained from random chance (p>>>.05 for both 1A and 1B). Hypothesis three: both genes are located on the same chromosome and within 50 map units of each other Last, the third hypothesis was tested. Since in this hypothesis, black bodies and black eyes gene are located on the same chromotid, fruit flies with only black bodies or black eyes would be produced by recombination. However, due to the black eye and black body allele being recessive, these phenotypes are only half the recombinant fruit flies produced. The other half are mixed in with the wild-type, since the wild-type allele is dominant for both genes. This information was used to make the tables below. Table 6 Phenotype Number observed Recombinant or Nonrecombinant Wild-type 1031 Mixture Black Bodied 52 Recombinant/2 Black eyed 46 Recombinant/2 Blacked Eyed and Black Bodied 67 Non-recombinant

7 Table 5 is the total number of each phenotypic class excluding gender counted in the F2 progeny of the 1A cross, and whether or not they were caused by recombination. Table 7 Phenotype Number observed Recombinant or Nonrecombinant Wild-type 833 Mixture Black Bodied 75 Recombinant/2 Black eyed 62 Recombinant/2 Blacked Eyed and Black Bodied 140 Non-recombinant Table 6 is the total number of each phenotypic class excluding gender counted in the F2 progeny of the 1b cross, and whether or not they were caused by recombination. Thus the map distance between the two genes can be determined by the following equation: {[(Black eyed fruit flies + Black bodied fruit flies)*2]/(total number observed)}*100. From the 1A data the calculated map distance is 16.4 cm and for 1B the calculated map distance is 24.7 cm. A sample calculation is provided below. Sample calculation 4 [(52+46)*2]/1196= 16.4 cm Since in both 1A and 1B the map distances calculated are both were under 50 map units, the third hypothesis cannot be rejected. From taking an average of these two map distance an average map distance of 20.6 cm was estimated. Male parental backcross Methods The male parental back cross used two marker stocks, Marker stock 2: Bristle, lobe, curly males, and Marker stock 3: glued, stubble, LVM. Both marker stocks are heterozygous for these traits

8 and were crossed with homozygous mutant female virgins. In marker stock 2, all of the traits are dominant, and the curly gene prevents recombination. Also the curly gene is located on a different chromatid then the bristle and lobed gene. Thus, since the females are homozygous recessive, the only two phenotypes produced are bristle/lobe flies and curly flies. Male bristle lobe flies were taken and back crossed with mutant female virgins. Marker stock 3 also is all dominant traits, except LVM which is just a marker that prevent recombination. LVM is located on a different chromatid then glued and stubble. Again, since the females are homozygous recessive only the following two phenotypes should be observed: glued/stubble flies, and Wildtype. The glued, stubble flies were backcrossed with female mutant virgins. In the following tables you can see the phenotypic classes counted. Table 7 Phenotype Number observed Black-eyed, black bodied 27 Black-eyed, black bodied, bristle, lobe flies 24 Wild-type 26 Bristle, Lobe 25 Table 7 is the phenotypic class scored from the marker 2 cross Table 8 Phenotype Number observed glued, stubble 29 Black eyed, black bodied 21 Table 8 is the phenotypic class scored from the marker 2 cross Based on the F2 results and black eyed and black bodied always being paired in fruit flies observed in the male parental backcross, the black eye and black body gene are located on the same chromosome. The markers prevent recombination in the first generation, and in the second generation males were crossed, which also cannot recombine, with homozygous females. Thus, only two ratios should be observed. A 1:1:1:1 ratio if the markers are located on a different gene then the black eye and black body genes or a 1:1 ratio if the markers are located on the same

9 chromosome as the black eye and black body genes. Marker 2 is located on chromosome two, and when the mutants were crossed with marker stock 2 they produced 1:1:1:1 ratio. Thus, as seen in table 7 both the black eyes, and black bodied genes are not located on chromosome 2 because a 1:1:1:1 ratio is form. Marker 3 is located on chromosome three, and when the mutants were crossed with marker stock 3 they produced 1:1ratio, as seen in Table 8. This indicates both the black-eyed, and black bodied genes are located on chromosome 3. A chi squared analysis was ran to test if this hypothesis was correct. The hypothesis could not be rejected because.7>p >.5 Female Parent Backcross The F1 generation of the female parent backcross was produced by the same procedure used for the male parental backcross with marker stock 3. Then the glued, stubble female virgins taken from the F1was backcrossed with the mutant males. The following tables represent the results pertaining to eye color. Table 9 Phenotypes Number Observed Class Glued/stubble and black eyes 526 Parental Wild-type and Glued 117 Single Cross Over /stubble/black eyes Stubble/Black eyes and glued 123 Single Cross Over Glued/ Black eyes and stubble 5 Double Cross Over Total 771 n/a Table 10 is a three point cross: Glued, Stubble, Black Eyes (black or wild-type body was ignored). The classes were determined based on amount observed. The Two lowest was a Double crossovers, the two highest was the parental, and the remaining were the single crossovers. Glued/Black eyes and Stubble are produced by a double crossover (DCO). This indicate that glued gene is between stubble gene and black eye gene, since glued gene switches its position in the double crossover. So the gene order is Black eyes, glued then stubble. Based on this gene order and the table above the following gene map was calculated. This is seen in the figure below

10 Figure 1. Stubble 16.6cM Glued 15.8 cm Black Eyes Figure 1 is a chromosome map of the stubble, glued and black eyed genes based on gene order provided above. The map distance in figure 1were calculated using this equation: map distance =[(SCO+DCO)/total)] *100 Sample calculations are provided below. Sample calculation 5 Stubble to glued [(117+5)/771] = 16.6 CM Moreover, from table 10 an coefficient of coincidence (CoC) of.25 from this equation: [observed/( Freq SCO1 x Freq SCO2 x total)]. The interference value was.75 and was calculated using this equation: 1-CoC. Sample calculations are provided below. Sample calculation 6 CoC = 5/[(.168*.158)*771]=.25 Sample calculation 7 I= =.75 Next, the Black body gene was compared to the markers. This is represented in the following table. Table 10 Phenotypes Number Observed Class Glued/stubble and black bodies 562 Parental Wild-type and Glued 81 Single Cross Over /stubble/black bodies Stubble/Black bodies and glued 14 Double Cross Over Glued/ Black bodies and stubble 114 Single Cross Over Total 771 n/a

11 Table 10 is a three point cross: glued, stubble, black body (black or wild-type eyes were ignored). The classes were determined based on amount observed. The Two lowest was a Double crossovers, the two highest was the parental, and the remaining were the single crossovers. Since the double crossover classes are stubble/black body and glued, stubble is located in the middle. This is because the stubble gene is the gene that moved in the double crossover. Thus the gene order is Black bodies, Stubble, and Glued. Figure 2 below is the chromosome map obtained from this gene order and table 10. Glued 16.6CM Stubble 12.3CM Black body Figure 2 is a chromosome map of the stubble, glued and black eyed genes based on gene order provided above. The map distance in figure 1were calculated using this equation: map distance =[(SCO+DCO)/total)] Moreover, from table 10 an coefficient of coincidence (CoC) of.9 from this equation: [observed/( Freq SCO1 x Freq SCO2 x total)]. The interference value was.1 and was calculated using this equation: 1-CoC. Last, the overall chromosome map is represented in figure 3 Figure 3 Black eyes 12.3CM Glued 16.6CM Stubble 15.8CM Black bodies Figure 3 is a chromosome mape of the two marker genes: glued and stubble, and the two mutant genes black body and black eyes. Since the two shared genes in figure 1 and figure 2 produced the same map distance figure 3 was produced by simply piecing them together. Discussions and Conclusions There were several sources of error when during the crossing and scoring of the fruit flies. Although there wasn t much room for error in the F1, since all the fruit flies were wild-type, the F2 data could have easily been skewed by observation bias. This refers to the one tending to

12 count a fly as wild-type, if they are having trouble scoring them. Another source of error is the fact that we were not able to count flies regularly for the F2 data due weather complications. In fact 61% of 1A data (734 fruit flies out of 1196), and 45% of 1B data (496 fruit flies out of 1110) was scored in one day. This caused our numbers to be skewed towards wild type, assuming they survives better. This explain also explain why the 1A data was skewed more toward wild-type then 1B, since there was 236 more fruit flies in the 1A bottles then the 1B data. Both of these sources of error also caused a smaller amount of recombinants observed than actual number of recombinants. This will cause the F2 map distance to be smaller than it actually is. The F1 data told us that the genes in question were autosomally recessive, while the F2 data told us that there were 2 autosomal recessive genes, that independent assortment was not occurring, and the map distance between the genes that were less than 50 cm on the same chromosome. The distance between the two genes was estimated to be 20.6 cm. The male parent backcross indicated both genes are located on chromosome 3. The female parent backcross gave the order and map distance of each mutant and the markers form marker stock 3 on chromosome 3. The map distances differed in the F2 and female parent backcross. This is for the reason as stated earlier with the lack of continuous counting due to weather complications. The interference values in the female backcross were.75 between black eye gene and stubble been and.1 between glued gene and black body gene. This inference indicates the observed number of recombinants were lower than the expect number of recombinants. This could simply be due to distance. Since they in both cases the total distance of the segment is higher than 20 cm, more than one crossover could contribute to this. Moreover, recombination seems to happen more readily between the glued gene and the black body gene then the stubble gene and the black eye gene. After fully analyzing the black eye and black body genes, it was concluded they were both located on

13 chromosome 3, they were indeed caused by 2 genes, and the map distances were also found. Interference values were analyzed and it is likely that these genes are about 44.7 cm apart. F1 20 Wild-type Female 20 Wild-type Males F2 1A Wild-type Female X Mutant Male F2 Progeny Appendix Raw data Phenotypic class Genotype Number observed Wild-type males bb + be Black bodied males bb bb be + 29 Black eyed males bb + be be 25 Black bodied black eyed males bb bb be be 30 Wild-type Females bb + be Black bodied females bb bb be + 23 Black eyed females bb + be be 21 Black eyed black body females bb bb be be 37 1B Mutant Females X Wild-type Males Phenotypic class Genotype Number Observed Wild-type males bb + be + 441

14 Black bodied males bb bb be + 31 Black eyed males bb + be be 25 Black bodied black eyed males bb bb be be 51 Wild-type Females bb + be Black bodied females bb bb be + 44 Black eyed females bb + be be 37 Black eyed black body females bb bb be be 89 Male Backcross F1 Bristle lobe male X mutant female virgins Phenotype Number observed Black-eyed, black bodied 27 Black-eyed, black bodied, bristle, lobe flies 24 Wild-type 26 Bristle, Lobe 25 F1 glued stubble males x mutant female virgins Phenotype Number observed glued, stubble 29 Black eyed, black bodied 21 Female backcross F1 Glued, stubble female virgins X mutant males Three point cross: Glued, Stubble, Black Eyes (black or wild-type body ignored) Phenotype Number Observed Class Glued/stubble 260 Parental Black eyes 266 Parental Wild-type 58 Single Cross Over Glued /stubble/black eyes 59 Single Cross Over Stubble/Black eyes 60 Single Cross Over Glued 63 Single Cross Over Glued/ Black eyes 2 Double Cross Over Stubble 3 Double Cross Over

15 Total 771 n/a F1 Glued stubble female virgin X mutant males Three point cross: Glued, Stubble, Black Bodies (Black or wild-type eyes ignored) Phenotype Number Observed Class Glued/stubble 280 Parental Black bodies 282 Parental Wild-type 39 Single Cross Over Glued /stubble/black bodies 42 Single Cross Over Stubble/Black bodies 9 Double Cross Over Glued 5 Double Cross Over Glued/ Black bodies 58 Single Cross Over Stubble 56 Single Cross Over total 771 n/a Chi squared analysis Sex linkage phenotype observe expected Chi-sqaured Wild-type males balck bodied males black eyed males black bodied black eyed males wild-type female black bodied females black eyed females

16 black eyed black body females total Class 1 = 7 degree of freedom.1>p>.05 so null hypothesis is not rejected. Male Parental Backcross data marker stock 3 phenotype observe expected Chi-squared Glued stubble Black eyed/ black body Total 1.28 Class 1= 1 degree of freedom.7>p>.5 null hypothesis is not rejected Female Parental back cross Other calculation Recombination Frequencies and map units for female parental backcross for the black eyes gene compared to the markers DCO frequency (2 Glued/black eyes + 3 Stubble)/771 =.0065 SCO Frequencies Stubble to Glued gene (60 Stubble/black eyes + 63 glued +5 DCO)/771 =.166

17 Glued to Black eyes (59 Glued/Stubble/Black eyes + 58 Wild-type + 5 DCO)/771=.158 Map units Stubble to Glued gene.166 X 100 = 16.6 MU Glued to Black eyes.158 X 100 = 15.8 Interference/coefficients of coincidence ExpectedDCO.166X.158=.026 Coefficient of coincidence = observed DCO/expected DCO Coefficient of coincidence =.0066/.026 =.25 Interference = 1- coefficient of coincidence Interference= =.75 Recombination Frequencies and map units DCO frequency (9 glued + 5 Stubble/Black bodies)/771 =.018 SCO Frequencies Glued to Stubble gene (58 Stubble + 56 Glued/Black Bodies + 14 DCO )/771 =.166 Glued to Black body gene (42 Glued/Stubble/Black eyes + 39 Wild-type + 14 DCO)/771=.123 Map units

18 Stubble to Glued gene.166 X 100 = 16.6 MU Glued to Black body gene.123 X 100 = 12.3 Map of three gens Interference/coefficients of coincidence Expected DCO.166X.123=.020 Coefficient of coincidence = observed DCO/expected DCO Coefficient of coincidence =.018/.020 =.9 Interference = 1- coefficient of coincidence Interference= 1-.9 =.1