GENETICS EXPERIMENTS FLYLAB - Essay Example

Fly Lab Experiment Teacher               Fly Lab Experiment PART MONOHYBRID CROSS a-d. Based on what I know about the principles of Mendelian genetics, the phenotypic ratio for crossing two parents that are homozygous for the genes is 1:1 or 100% because if the first gene were WW for wild type and the second were SS for sepia eyes, and one parent was WW while the other was SS, then all offspring would be W and S for both genes, naturally with wild type and sepia eye phenotypes. The result of the data is 4947 wild-type females with sepia eyes and 5048 wild type males with sepia eyes.

This is a 1:1.02 ratio. Thus, somehow it follows a 1:1 ratio although the difference of almost 100 offspring may actually have some significance. -------------------------------------------------------------------------- Results of Cross #1 Parents (Female: +) x (Male: SE) Offspring Phenotype Number Proportion Ratio Female: + 4947 0.4949 1.000 Male: + 5048 0.5051 1.020 Total 9995 -------------------------------------------------------------------------- PART 1 MONOHYBRID CROSS d-f. For the F2 generation, my expected phenotypic ratio was 1:1 still because the male and female flies involved were homozygous for the gene and since the phenotypes were wild type and sepia eyes, then both traits were dominant.

When the F1 male and female flies were mated and this resulted in 3697 and 3717. With my expected phenotype ratio of 1:1, the Chi-Square result recommended rejecting my hypothesis. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 3697 1.0000 2447.7500 637.5756 Male: + 3717 1.0000 2447.7500 658.1536 Female: SE 1194 1.0000 2447.7500 642.1771 Male: SE 1183 1.0000 2447.7500 653.

4951 Total 9791 4.0000 9791.0000 2591.4014 Chi-Squared Test Statistic = 2591.4014 Degrees of Freedom = 3 Level of Significance = 0.0000 Recommendation: Reject your hypothesis ----------------------------------------------------------------------- Another hypothesis was tested at 3:1, and this time, chi-square results favored the hypothesis. This means that in the F2 generation, there are 3 wild types for every 1 sepia-eyed fly. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 3697 3.0000 3671.6250 0.

1754 Male: + 3717 3.0000 3671.6250 0.5608 Female: SE 1194 1.0000 1223.8750 0.7293 Male: SE 1183 1.0000 1223.8750 1.3651 Total 9791 8.0000 9791.0000 2.8305 Chi-Squared Test Statistic = 2.8305 Degrees of Freedom = 3 Level of Significance = 0.4185 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- PART 1 MONOHYBRID CROSSES For the cross of the wild-type male and the ebony-bodied female fly, the expected phenotypic ratio is 1:1 for the F1 generation and 3:1 for the F2 generation, based on the results of previous experiments.

The results were the same with the actual ratios for the F1 and F2 generation experiments. For the F1 generation, there were equal number of wild types and ebony bodies. For the F2 generation, there are 3 wild types for every ebony bodies. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #3 (F1 Generation) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 501 1.0000 496.5000 0.0408 Male: + 492 1.0000 496.5000 0.0408 Total 993 2.0000 993.0000 0.

0816 Chi-Squared Test Statistic = 0.0816 Degrees of Freedom = 1 Level of Significance = 0.7752 Recommendation: Do not reject your hypothesis (1:1) -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #4 (F2 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 383 3.0000 376.1250 0.1257 Male: + 387 3.0000 376.1250 0.3144 Female: E 109 1.0000 125.3750 2.1387 Male: E 124 1.0000 125.3750 0.0151 Total 1003 8.0000 1003.0000 2.

5939 Chi-Squared Test Statistic = 2.5939 Degrees of Freedom = 3 Level of Significance = 0.4586 Recommendation: Do not reject your hypothesis (3:1) -------------------------------------------------------------------------- PART 2 DIHYBRID CROSSES -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #5 (F1 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 484 1.0000 488.0000 0.0328 Male: + 492 1.0000 488.0000 0.0328 Total 976 2.0000 976.0000 0.

0656 Chi-Squared Test Statistic = 0.0656 Degrees of Freedom = 1 Level of Significance = 0.7979 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #6 (F2 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 298 10.0000 295.5882 0.0197 Male: + 288 10.0000 295.5882 0.1948 Female: VG 98 3.0000 88.6765 0.9803 Male: VG 84 3.0000 88.6765 0.2466 Female: E 87 3.0000 88.6765 0.

0317 Male: E 91 3.0000 88.6765 0.0609 Female: VG;E 29 1.0000 29.5588 0.0106 Male: VG;E 30 1.0000 29.5588 0.0066 Total 1005 34.0000 1005.0000 1.5511 Chi-Squared Test Statistic = 1.5511 Degrees of Freedom = 7 Level of Significance = 0.9805 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- My predicted phenotypic ratio of 1:1 was accepted by the Chi-Square results. The phenotype for all F1 is only the wild type and the vestigial wing size did not appear, which means that it is a recessive gene.

My predicted ratio for the F2 generation was 9:3:3:1, which is based on the results if the other trait is recessive. Most flies of the F2 generation were mostly wild types. The others were only vestigial-winged types or only ebony-bodied types. The ones with the least number include the ones that are both vestigial-winged and ebony-bodied. The dominant trait was the wild type trait, represented by the part of the ratio which is 9. The heterozygous offspring were the vestigial and ebony characteristics, making up the 3.

When vestigial is dominant, the ebony is not expressed. Where the ebony is expressed, the vestigial is not. Lastly, the recessive trait is actually the combination of the vestigial and ebony characteristics, and since both recessive alleles are present here, then they both express themselves, which represent 1. PART 3 TRIHYBRID CROSSES The cross between the specified male and female flies resulted in a ratio which is different from the usual Mendelian trihybrid cross F2 generation ratio, which is 27:9:9:9:3:3:3:1.

-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #8 (F2 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 184 27.0000 213.8906 4.1771 Male: + 227 27.0000 213.8906 0.8035 Female: SV 72 9.0000 71.2969 0.0069 Male: SV 82 9.0000 71.2969 1.6068 Female: DP 70 9.0000 71.2969 0.0236 Male: DP 79 9.0000 71.2969 0.8323 Female: SV;DP 23 9.0000 71.2969 32.7166 Male: SV;DP 22 9.0000 71.2969 34.0854 Female: E 74 3.0000 23.7656 106.1825 Male: E 73 3.0000 23.7656 101.

9971 Female: SV;E 36 3.0000 23.7656 6.2982 Male: SV;E 24 3.0000 23.7656 0.0023 Female: DP;E 16 3.0000 23.7656 2.5375 Male: DP;E 17 3.0000 23.7656 1.9260 Female: SV;DP;E 6 1.0000 7.9219 0.4663 Male: SV;DP;E 9 1.0000 7.9219 0.1467 Total 1014 128.0000 1014.0000 293.8086 Chi-Squared Test Statistic = 293.8086 Degrees of Freedom = 15 Level of Significance = 0.0000 Recommendation: Reject your hypothesis -------------------------------------------------------------------------- Finally, I was able to come up with: 27:10:10:3:10:4:2:1, and this was accepted.

Chi Square Hypothesis Using Cross #8 Ignoring Sex Phenotype Observed Hypothesis Expected Chi-Square Term + 411 27.0000 408.6269 0.0138 SV 154 10.0000 151.3433 0.0466 DP 149 10.0000 151.3433 0.0363 SV;DP 45 3.0000 45.4030 0.0036 E 147 10.0000 151.3433 0.1246 SV;E 60 4.0000 60.5373 0.0048 DP;E 33 2.0000 30.2687 0.2465 SV;DP;E 15 1.0000 15.1343 0.0012 Total 1014 67.0000 1014.0000 0.4774 Chi-Squared Test Statistic = 0.4774 Degrees of Freedom = 7 Level of Significance = 0.9995 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- SEX-LINKED GENES The expected phenotypic ratio of 1:1 for the F1 generation was accepted by the Chi-Square test.

-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #9 (F1 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 502 1.0000 507.0000 0.0493 Male: T 512 1.0000 507.0000 0.0493 Total 1014 2.0000 1014.0000 0.0986 Chi-Squared Test Statistic = 0.0986 Degrees of Freedom = 1 Level of Significance = 0.7535 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- However, in the F2 generation, the Mendelian phenotypic ratio of 9:3:3:1 is not followed.

Instead, it was still 1:1:1:1. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #10 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 249 1.0000 258.7500 0.3674 Male: + 239 1.0000 258.7500 1.5075 Female: T 284 1.0000 258.7500 2.4640 Male: T 263 1.0000 258.7500 0.0698 Total 1035 4.0000 1035.0000 4.4087 Chi-Squared Test Statistic = 4.4087 Degrees of Freedom = 3 Level of Significance = 0.2206 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- With the next experiment on the vestigial-winged female and the white-eyed male, the ratio was 7:3:4:2:1:1, with the white eye gene being expressed only in the male fly.

-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #12 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 385 7.0000 391.6111 0.1116 Male: + 186 3.0000 167.8333 1.9664 Male: W 217 4.0000 223.7778 0.2053 Female: VG 109 2.0000 111.8889 0.0746 Male: VG 56 1.0000 55.9444 0.0001 Male: W;VG 54 1.0000 55.9444 0.0676 Total 1007 18.0000 1007.0000 2.4255 Chi-Squared Test Statistic = 2.4255 Degrees of Freedom = 5 Level of Significance = 0.

7877 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- RECOMBINATION AND CHROMOSOME MAPPING -------------------------------------------------------------------------- Results of Cross #16 Parents (Female: +) x (Male: SV;EY) Offspring Phenotype Number Proportion Ratio Female: + 2 0.0020 1.000 Male: + 4 0.0040 2.000 Female: SV 255 0.2560 127.500 Male: SV 246 0.2470 123.000 Female: EY 251 0.2520 125.500 Male: EY 233 0.2339 116.

500 Female: SV;EY 2 0.0020 1.000 Male: SV;EY 3 0.0030 1.500 Total 996 -------------------------------------------------------------------------- Based on the results above, there are 2 females for every 3 male flies with both shaven bristles and eyeless mutation. The percentage is 66.67%. This translates as 66.67 cM as the map distance between the two genes. SV EY 66.67 cM References Welcome to Fly Lab. (2013). Retrieved 13 Jun 2013 from Biology Labs Online: http://www.biologylabsonline.com/protected/FlyLab/