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Sidana [21]
3 years ago
5

Iron rich deposits that formed about 2 billion years ago have red layers. Layers formed before then were not red. The red layers

formed when oxygen made during photosynthesis reacted with iron molecules in the ocean. The oxidized iron formed layers of sedimentary rock
. Which BEST summarizes this phenomenon?
A. The lithosphere changed the atmosphere. B. The hydrosphere change the atmosphere.
C. The atmosphere change the biosphere.
D. The biosphere change the lithosphere
Biology
1 answer:
Amanda [17]3 years ago
3 0

Answer:

D. The biosphere change the lithosphere

Explanation:

The red formed due to oxygen which is produced by plants in the process of photosynthesis. This oxygen then reacted with iron atom and formed oxidized iron layers of sedimentary rock. So here plants are responsible for the formation of these layers and we considered plants as part of biosphere so we can say that biosphere is responsible for the change of lithosphere.

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In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa
Dimas [21]

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 X^2 = \frac{(O - E)^2}{E}, where O = observed frequency and E = expected frequency.

Phenotype                O                  E                                           X^2

Wild type                 473          9/16 x 830 =  466.875     \frac{(473 - 466.875)^2}{466.875} = 0.08

wild w/ebony b      156          3/16 x 830 = 155.625       \frac{(156 - 155.625)^2}{155.625} =  0.0009

Dumpy w/wild b     149          3/16 x 830 = 155.625      \frac{(149 - 155.625)^2}{155.625} = 0.28

dumpy w/ebony b  52            1/16 x 830 = 51.875        \frac{(52 - 51.875)^2}{51.875} = 0.0003

Total X^2 = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of X^2 (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>

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