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Alex777 [14]
3 years ago
7

A group of small fish live in a lake with a uniformly light-brown sandy bottom. Most of the fish are light brown, but about 10%

are mottled. This fish species is often prey for large birds that live on the shore. A construction company dumps a load of gravel in the bottom of the lake, giving it a mottled appearance. Which of these statements presents the most accurate prediction of what will happen to this fish population?
Biology
1 answer:
Nuetrik [128]3 years ago
7 0
Due to the mass majority of the fish population becoming mottled The fish population will decrease because the birds pray more on mottled fish the brown fish
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Answer:

See the answer below

Explanation:

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

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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>

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2 years ago
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3 years ago
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