Which of the following are linear strands of DNA and proteins that carry transferable hereditary information? A. mutations B. ch
2 answers:
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Answer:
See the answer below
Explanation:
Using the formula for calculating Chi square ():
=
where O = observed frequency and E = expected frequency.
The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.
For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.
Hence, expected frequencies are calculated as:
phenotype (1) = 9/16 x 293 = 164.81
phenotype (2) = 3/16 x 293 = 54.94
phenotype (3) = 3/16 x 293 = 54.94
phenotype (4) = 1/16 x 293 = 18.31
The is calculated thus:
Phenotype O E
1 88 164.81 = 35.80
2 62 54.94 = 0.91
3 62 54.94 = 0.91
4 81 18.31 = 214.64
Total = 35.80 + 0.91 + 0.91 + 214.64 = 252.26
To the nearest tenth = 252
Degree of freedom = n - 1 = 4- 1 = 3
tabulated = 7.815
<em>The calculated </em><em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. </em>