Question

Drosophilia homozygous for the lobed eye mutation (L) were mated with flies homozygous for a wing mutation known as plexus (px) All the f1 progeny had lobed eyes and wild type wings. The F1 flies were test crossed resulting in the following progeny: Number 18 F2 phenotypes Wild type eyes and wings Lobed eyes Plexus wings Lobed eyes and plexus wings 35 33 14 Write the genotypes of the parental and f1 flies. Use proper drosophilia nomenclature. Be sure to distinguish between the two groups (label each) Parentals: F1: Write the genotype of each F2 fly in the order as shown in the table. Use drosophilia nomenclature. Use the chi square test to determine whether the genes for lobed eyes and plexus wings are linked. Include chi square values, degrees of freedom, and p values. Distinguish between each. TABLE 5.2 Critical Chi-Square Values p Values Cannot Reject the Null Hypothesis Null Hypothesis Rejected 0.99 Degrees of 0.90 0.50 0.10 0.05 0.01 0.001 Freedom x? Values 0.02 0.45 3.84 6.64 10.83 0.02 0.21 1.39 4.61 5. 999.21 13.82 0.11 0.58 2.37 6.25 7.81 11.35 16.27 0.30 1.06 3.36 7.78 9.49 13.28 18.47 0.55 1.61 4.35 9.24 11.07 15.09 20.52 Note: x values that lie in the yellow region of this table allow you to reject the null hypothesis with > 95% confidence, and for recombination experiments, to postulate linkage. Provide a null hypothesis for the analysis (involving loved eyes and plexus wings) Based on chi square analysis are the genes for lobed eyes and plexus wings linked? Briefly explain what led you to conclusion. Of the 100 f2 flies shown in the table. 60 were female and 40 were male. Is this a 1:1 ratio as expected? Use the chi square test to determine whether the deviation from the expected 1:! Sex ratio are due to chance alone. Include chi square value, degrees of freedom, p value, and final conclusion below. Distinguish each number.

Answer 1

Answer is given below:

Phenotype lobed eye (Homozygous) - pleaus wing( ") Genotype LL - Papa fi has lobed eyes and wild type wings when Drosophila homozygous for lobed eyes (LL) and Drosophila homozygous for plexus wings ( px px) are mated . lobed eyes are dominant. Pleals wings are recessive. . Parental I has genetype of LL pat pxt Parental a has genotype of el papa F has genotype of Ll pxt pa LL pxt pxt X lupx px - Parental A (LPxt) px - gametes Le pat px - bobed eyes and wireld type wings. Le pat px x el px px - Test cross 90 X 1px - gametes x apxt Apx LPxt) PN B) px px Upx px Lepxtpx - luptpx (F2) Le papa lobed eyes, lobed eyes, wild type wild type eyes wild wings plexus coings eyes and pleals wings callings 1 : 1 : 1 : 1 In mendelian test cross, the expected phenotypic ratio should be lililit

Phenotype would type eyes and wings observed (o) Expected (E) (O-E) 18 10 = 25 -7 (0-2 (0-62 49 1-96 Lobed eyes 35 25 10 4 Plexus wings 33 25 2.56 Lobed eyes and plexus wings 4.84 -11 Total 100 100 x2=13.36 X²= 13.36 Total phenotypes = 4 . Regree of freedom =4-3 plealue = 0.00K P <0.01 conclusion – If Prealue is less than 0.05, then the null hypothesis is rejected. Here Pealue is less than 0.01 ; so, the null hypothesis is rejected. The deviation of expected from the observed value is considered significant. Deviation is not attributed by chance. :. So, the hypothesis of Independent Assortment is rejected. This a genes are linked Mull hypothesis real difference between valle. states that there is no observed and expected

(2) Total population =100 Females = 60 ; males = 40 Phenotype obserwed() expected (C) (0-E) (6-4) 60 100 50 10 100 (6-E)? a Female 50 -10 100 male x²=4 loo 100 Total x²=4 degree of freedom =2-1=21 (as phenotypes=27 pvalle = 0.OKP <0.05 conclusion - Pivalue is less than 0.05. so null hypothesis is rejected. This means, deviation in se observed and expected value is considered significant So, the changes in sex ratio is not attributed by chance.