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luda_lava [24]
3 years ago
7

State the form in which carbohydrates are:1.transported in animals:plants​

Biology
1 answer:
Fittoniya [83]3 years ago
6 0
In animals the body absorbs & transports broken-down carbohydrates in the body. The monosaccharide units, glucose, galactose and fructose are transported through the wall of the small intestine and then into the portal vein which then takes these elements straight to the liver.

In plants Carbohydrates are mainly transported in the phloem in the form of sucrose. The cell membrane of the sieve cells contains pump proteins that actively convey sucrose into the vascular pathways.

Hope this helps !
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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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