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Inga [223]
3 years ago
10

Which of the following represents an activity within a population?​

Biology
1 answer:
elena55 [62]3 years ago
4 0

Explanation:a peacock spreading and shaking his feathers to attract a female

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4 0
3 years ago
22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

X^2  =  \frac{(O - E)^2}{E} 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 X^2 is calculated thus:

Phenotype            O               E                             X^2

  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

4 0
3 years ago
How does photosynthesis in a purple bacterium (Class Anoxyphotobacteria) differ from photosynthesis in an oak tree?
VladimirAG [237]
It differs due to the life cycle of each tree, and the very specific way the trees go.
7 0
3 years ago
Read 2 more answers
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