Flylab Lab Notes for Franklin Xie Assignment 1

Franklin XieFebruary 11, 2013

Assignment 1

1.Flylab Lab Notes for Franklin Xie

-------------------------------------------------------------------------- Results of Cross #1

Parents

(Female: +) x (Male: SE)

Offspring

Phenotype Number Proportion Ratio Female: + 4978 0.5024 1.010 Male: + 4930 0.4976 1.000 Total 9908 These are the results of the F1 generation for my first monohybrid cross.-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #1

Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 4978 1.0000 4954.0000 0.1163Male: + 4930 1.0000 4954.0000 0.1163Total 9908 2.0000 9908.0000 0.2325

Chi-Squared Test Statistic = 0.2325Degrees of Freedom = 1Level of Significance = 0.6296Recommendation: Do not reject your hypothesisThis is the analysis of hypothesis of phenotype ratios from the F1 generation.-------------------------------------------------------------------------- Results of Cross #2

Parents

(Female: +) x (Male: +)

Offspring

Phenotype Number Proportion Ratio Female: + 3720 0.3675 2.993 Male: + 3858 0.3811 3.104 Female: SE 1301 0.1285 1.047 Male: SE 1243 0.1228 1.000 Total 10122 Observed phenotypic ratios of the F2 offspring of my first monohybrid cross.-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2

Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 3720 3.0000 3795.7500 1.5117Male: + 3858 3.0000 3795.7500 1.0209Female: SE 1301 1.0000 1265.2500 1.0101Male: SE 1243 1.0000 1265.2500 0.3913Total 10122 8.0000 10122.0000 3.9340

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Chi-Squared Test Statistic = 3.9340Degrees of Freedom = 3Level of Significance = 0.2687Recommendation: Do not reject your hypothesisThis is the analysis of hypothesis of phenotype ratios from the F2 generation.--------------------------------------------------------------------------

The correct phenotypic ratio for this experiment is 1:1 (wild type eyes to sepia eyes) in the F1 generation and 3:1 (wild type eyes to sepia eyes) in the F2 generation. This was what I expected because I assumed that incomplete dominance and codiminance were not factors that affected the trait for the flies' eye color. This meant that the flies that express the recessive trait must be homozygous recessive. I also assumed that the flies that express the dominant trait are true-bred and homozygous for the dominant alleles for eye type. The cross between the two true-bred fly types should, based on basic principles of dominance, produce only the dominant trait and heterozygous individuals. The parent cross, in this case, can be looked at as a test cross. Because my hypothesis for the F1 generation was analyzed and I was recommended to not reject it, I knew that I was most likely correct in assuming that all of the offspring were heterozygous. A cross between heterozygous individuals that express the dominant trait would produce offspring in a 3:1 ratio (dominant trait to trait recessive).

Because the F1 generation's population was completely heterozygous and only expressed one trait, I concluded that that trait is expressed by the dominant allele. In this case, the wild type eyes are expressed by the dominant trait. Further evidence is gathered from the F2 generation. My assumption that the F1 generation was completely heterozygous was proven as the F2 generation expressed traits in a 3:1 ratio (dominant to recessive). There is a 25% chance, in a monohybrid cross between heterozygous individuals, that an offspring with the recessive trait will be born. As about 75% of the population of flies expressed the dominant trait (wild type eyes) and 25% of the population expressed the recessive trait (sepia eyes), I arrived at the conclusion that the dominant allele codes for wild type eyes. I also conclude that the alleles for eye type in this species of flies do not function on the basis of codominance or incomplete dominance because the heterozygous flies of the F1 generation did not display both traits/a mix of both traits.

2. a.

Flylab Lab Notes for Franklin Xie--------------------------------------------------------------------------

Results of Cross #1

Parents

(Female: +) x (Male: BL)

Offspring

Phenotype Number Proportion Ratio Female: + 4934 0.4957 1.000 Male: + 5020 0.5043 1.017 Total 9954

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Chi Square Hypothesis Using Cross #1

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Chi-Squared Test Statistic = 0.7430Degrees of Freedom = 1Level of Significance = 0.3887Recommendation: Do not reject your hypothesis

-------------------------------------------------------------------------- Results of Cross #2

Parents


Offspring

Phenotype Number Proportion Ratio Female: + 371 0.3817 3.226 Male: + 361 0.3714 3.139 Female: BL 115 0.1183 1.000 Male: BL 125 0.1286 1.087 Total 972

-------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2

Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 371 3.0000 364.5000 0.1159Male: + 361 3.0000 364.5000 0.0336Female: BL 115 1.0000 121.5000 0.3477Male: BL 125 1.0000 121.5000 0.1008Total 972 8.0000 972.0000 0.5981


--------------------------------------------------------------------------Changing the number of offspring produced does not affect the results of the chi-square analysis on the data. The data is predictable for the crosses. Using the techniques described in my answer for question number 1, I conclude that the allele for wild type body color in this species of flies is dominant to the allele for black body color.

b.

Flylab Lab Notes for Franklin Xie -------------------------------------------------------------------------- Results of Cross #1

Parents

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(Female: +) x (Male: SN)

Offspring





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Results of Cross #2

Parents


Offspring

Phenotype Number Proportion Ratio Female: + 56 0.4828 2.435 Male: + 37 0.3190 1.609 Male: SN 23 0.1983 1.000 Total 116

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Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 56 3.0000 49.7143 0.7947Male: + 37 3.0000 49.7143 3.2516Male: SN 23 1.0000 16.5714 2.4938Total 116 7.0000 116.0000 6.5402

Chi-Squared Test Statistic = 6.5402Degrees of Freedom = 2Level of Significance = 0.0380Recommendation: Reject your hypothesis

--------------------------------------------------------------------------By restricting the number of offspring in a cross, the results become more unpredictable. The wild type bristles should be dominant over singed bristles

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on a 3 to 1 basis, assuming the parents are homozygous. The F2 generation should exhibit the dominant trait to recessive trait on a 3:1 ratio. However, because I restricted the F2 generation to approximately 100 offspring, the ratio became more unpredictable. This is because of the Law of Independent Assortment and the Law of Segregation. These laws account for the randomness of genetic crosses. Based on probability, the ratio of dominant to recessive traits should be 3:1, but because the outcome of genetic crosses is random in many cases, there will be a degree of unpredictability in the outcome of the crosses. This is more apparent when there are less offspring than more offspring because a large number of crosses, due to principles of probability, will cause the results to even out towards the expected outcome.

c. Flylab Lab Notes for Franklin Xie -------------------------------------------------------------------------- Results of Cross #1

Parents

(Female: +) x (Male: EY)

Offspring





-------------------------------------------------------------------------- Results of Cross #2

Parents


Offspring

Phenotype Number Proportion Ratio Female: + 3753 0.3711 2.990 Male: + 3783 0.3741 3.014 Female: EY 1321 0.1306 1.053 Male: EY 1255 0.1241 1.000 Total 10112

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Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 3753 3.0000 3792.0000 0.4011Male: + 3783 3.0000 3792.0000 0.0214Female: EY 1321 1.0000 1264.0000 2.5704Male: EY 1255 1.0000 1264.0000 0.0641Total 10112 8.0000 10112.0000 3.0570


--------------------------------------------------------------------------The F1 generation population consists of 104 flies. This generation is the outcome of the cross between homozygous individuals for variations of the same trait. I hypothesized that the offspring will express the dominant trait (wild type eye shape) on a ratio of 1:0. Even with such a limited number of offspring, the data is completely predictable because the cross between homozygous parents must produce heterozygous offspring at a 100% rate. In these different experiments (a, b, and c), the result of the different crosses is more predictable when the offspring population is larger because with more offspring, there is more opportunity for the result to even out towards what is probable.

3.

Flylab Lab Notes for Franklin Xie -------------------------------------------------------------------------- Results of Cross #1

Parents

(Female: E) x (Male: VG)

Offspring





--------------------------------------------------------------------------

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Results of Cross #2

Parents


Offspring

Phenotype Number Proportion Ratio Female: + 282 0.2803 10.444 Male: + 271 0.2694 10.037 Female: VG 98 0.0974 3.630 Male: VG 90 0.0895 3.333 Female: E 89 0.0885 3.296 Male: E 108 0.1074 4.000 Female: VG;E 41 0.0408 1.519 Male: VG;E 27 0.0268 1.000 Total 1006

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Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 282 9.0000 282.9375 0.0031Male: + 271 9.0000 282.9375 0.5037Female: VG 98 3.0000 94.3125 0.1442Male: VG 90 3.0000 94.3125 0.1972Female: E 89 3.0000 94.3125 0.2992Male: E 108 3.0000 94.3125 1.9865Female: VG;E 41 1.0000 31.4375 2.9087Male: VG;E 27 1.0000 31.4375 0.6264Total 1006 32.0000 1006.0000 6.6689

Chi-Sqaured Test Statistic = 6.6689Degrees of Freedom = 7Level of Significance = 0.4642Recomendation: Do not reject your hypothesis

--------------------------------------------------------------------------F1: wwNN (ebony, normal wings) x WWnn (wild type, vestigial wings)

wN wN wN wNWn WwNn WwNn WwNn WwNnWn WwNn WwNn WwNn WwNnWn WwNn WwNn WwNn WwNnWn WwNn WwNn WwNn WwNnOutcome: 100% WwNn (1:0 ratio)(All wild type bodies and wings)

F2: WwNn x WwNnWN Wn wN Wn

WN WWNN WWNn WwNN WwNnWn WWNn WWnn WwNn WwnnwN WwNN WwNn wwNN WwNnwn WwNn Wwnn wwNn WwnnWild Type, Normal: 9Ebony, Normal: 3Wild Type, Vestigial: 3Ebony, Vestigial: 19:3:3:1 ratio

The outcomes for my F1 and F2 crosses can be explained by the dihybrid punnett squares. The F1 punnett square shows that, by crossing two homozygous

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individuals, only one genotype and phenotype can be achieved within offspring (WwNn). When the offspring reproduce, however, They will provide an F2 generation with roughly the ratio of 9:3:3:1 [wild (color), normal (wild wings): ebony, normal: wild, vestigial: ebony, vestigial].

The actual ratio for the F2 generation was approximately 8.1:2.7:2.9:1. There is a small difference from the predicted and the actual outcome. This is due to the randomness of genetic crosses that occurs because, after the alleles segregate and assort independently during gamete formation, the flies’ sperm and egg randomly combine. It only likely, not certain, that the final result of the cross will produce offspring with phenotypes based on a 9:3:3:1 ratio. The actual ratio will even out due to principles of probability if more offspring are made from the F1 parents.

4. Flylab Lab Notes for Franklin Xie -------------------------------------------------------------------------- Results of Cross #1

Parents

(Female: +) x (Male: SV;DP;E)

Offspring





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Results of Cross #2

Parents


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Offspring

Phenotype Number Proportion Ratio Female: + 213 0.2105 26.625 Male: + 215 0.2125 26.875 Female: SV 57 0.0563 7.125 Male: SV 68 0.0672 8.500 Female: DP 69 0.0682 8.625 Male: DP 67 0.0662 8.375 Female: SV;DP 30 0.0296 3.750 Male: SV;DP 25 0.0247 3.125 Female: E 66 0.0652 8.250 Male: E 69 0.0682 8.625 Female: SV;E 28 0.0277 3.500 Male: SV;E 27 0.0267 3.375 Female: DP;E 31 0.0306 3.875 Male: DP;E 29 0.0287 3.625 Female: SV;DP;E 10 0.0099 1.250 Male: SV;DP;E 8 0.0079 1.000 Total 1012

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Chi Square Hypothesis Using Cross #2 Ignoring Sex

Phenotype Observed Hypothesis Expected Chi-Square Term+ 428 27.0000 426.9375 0.0026SV 125 9.0000 142.3125 2.1061DP 136 9.0000 142.3125 0.2800SV;DP 55 3.0000 47.4375 1.2056E 135 9.0000 142.3125 0.3757SV;E 55 3.0000 47.4375 1.2056DP;E 60 3.0000 47.4375 3.3268SV;DP;E 18 1.0000 15.8125 0.3026Total 1012 64.0000 1012.0000 8.8052

Chi-Sqaured Test Statistic = 8.8052Degrees of Freedom = 7Level of Significance = 0.2669Recomendation: Do not reject your hypothesis_____________________________________________________________________________

The F1 generation, as always, would result in offspring that exhibit the dominant traits at a 1:0 (100%) ratio. That is why my hypothesis for the Chi-square analysis of the F1 generation is correct.

The punnett square on the next page was used to predict the outcome of the F2 generation. A,B,C represent the dominant, or wild type, traits. a,b,c represent the traits, respectively, for dumpy wing shape, ebony body, and shaven bristles.

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ABC ABc AbC Abc aBC aBc abC abcABC AABBCC AABBCc AABbCC AABbCc AaBBCC AaBBCc AaBbCC AaBbCcABc AABBCc AABBcc AABbCc AABbcc AaBBCc AaBBcc AaBbCc AaBbccAbC AABbCC AABbCc AAbbCC AAbbCc AaBbCC AaBbCc AabbCC AabbCcAbc AABbCc AABbcc AAbbCc AAbbcc AaBbCc AaBbcc AabbCc AabbccaBC AaBBCC AaBBCc AaBbCC AaBbCc aaBBCC aaBBCc aaBbCC aaBbCcaBc AaBBCc AaBBcc AaBbCc AaBbcc aaBBCc aaBBcc aaBbCc aaBbccabC AaBbCC AaBbCc AabbCC AabbCc aaBbCC aaBbCc aabbCC aabbCcabc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc

The expected phenotypic ratio: All dominant traits:wild,wild,shave:wild,ebony,wild:dumpy,wild,wild:wild,ebony,shaven:

Expected Phenotype Ratio of F2 generationW=Wild (dominant) D=Dumpy Wings E=Ebony Body S=Shaven Bristles

WWW WWS WEW DWW WES DWS DEW DES27 9 9 9 3 3 3 1

The actual phenotypic ratio was approximately: 24:7:8:8:3:3:3:1. The randomness of genetic crosses occurs because, after the alleles segregate and assort independently during gamete formation, the flies’ sperm and egg randomly combine. It only likely, not certain, that the final result of the cross will produce offspring with phenotypes based on a 27:9:9:9:3:3:3:1 ratio. The actual ratio will even out due to principles of probability if more offspring are made from the F1 parents.

Assignment 2 Flylab Lab Notes for Franklin Xie -------------------------------------------------------------------------- Results of Cross #1

Parents

(Female: BW) x (Male: E)

Offspring




Chi-Squared Test Statistic = 0.0195Degrees of Freedom = 1Level of Significance = 0.8889

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Recommendation: Do not reject your hypothesis

-------------------------------------------------------------------------- Results of Cross #2

Parents

(Female: +) x (Male: BW;E)

Offspring

Phenotype Number Proportion Ratio Female: + 1237 0.1262 1.057 Male: + 1170 0.1194 1.000 Female: BW 1269 0.1295 1.085 Male: BW 1262 0.1287 1.079 Female: E 1253 0.1278 1.071 Male: E 1183 0.1207 1.011 Female: BW;E 1247 0.1272 1.066 Male: BW;E 1182 0.1206 1.010 Total 9803


Phenotype Observed Hypothesis Expected Chi-Square TermFemale: + 1237 1.0000 1225.3750 0.1103Male: + 1170 1.0000 1225.3750 2.5024Female: BW 1269 1.0000 1225.3750 1.5531Male: BW 1262 1.0000 1225.3750 1.0947Female: E 1253 1.0000 1225.3750 0.6228Male: E 1183 1.0000 1225.3750 1.4654Female: BW;E 1247 1.0000 1225.3750 0.3816Male: BW;E 1182 1.0000 1225.3750 1.5354Total 9803 8.0000 9803.0000 9.2656

Chi-Sqaured Test Statistic = 9.2656Degrees of Freedom = 7Level of Significance = 0.2341Recomendation: Do not reject your hypothesis

The F1 wild-type female is double-heterozygous. The female cannot be double homozygous, otherwise all of the offspring would have exhibited the dominant trait. A cross between double heterozygous females and double homozygous recessive males will produce a phenotype ratio of 1:1:1:1.

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Test Cross (aabb male) x (AaBb female)ab ab ab ab

AB AaBb AaBb AaBb

AaBb

Ab Aabb Aabb Aabb AabbaB aaBb aaBb aaBb aaBbab aabb aabb aabb aabb

This punnett square illustrates the outcome of the cross that I described. The actual outcome of the parent cross of the F1 generation created offspring on a 1:1:1:1 ratio (Wild eyes, Wild body: Brown eyes, wild body, Wild eyes: ebony body: Brown eyes, Ebony body). That is the predicted outcome of a test cross between a double homozygous recessive individual and double heterozygous individual (shown in the punnett square). The data proves that the female was almost certainly double-heterozygous.

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Flylab Lab Notes for Franklin Xie Assignment 1

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