Why does it make more sense for the hydrophilic sugar-phosphates to be on the outside of the DNA molecule and the hydrophobic ni
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Answer:
See the answers below
Explanation:
Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):
(a) <em>9/16 D_E_ wild type </em>
<em> 3/16 D_ee wild wing, ebony body</em>
<em> 3/16 ddE_ dumpy wing, wild body</em>
<em> 1/16 ddee dumpy wing, ebony body</em>
(b) Chi square =
, where O = observed frequency and E = expected frequency.
Phenotype O E
Wild type 473 9/16 x 830 = 466.875 = 0.08
wild w/ebony b 156 3/16 x 830 = 155.625 = 0.0009
Dumpy w/wild b 149 3/16 x 830 = 155.625 = 0.28
dumpy w/ebony b 52 1/16 x 830 = 51.875 = 0.0003
Total = 0.3612
Degree of freedom = 4 - 1 = 3
Tabulated value of (0.05)= 7.815
(c) <em>Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.</em>
