The nutrients from the soil is taken up by the roots
to be distributed to the entire plant. The plant absorbs the nutrients into its
cells and utilized by the leaves, stem, trunk. They are also used to grow
flowers and to bare fruits.
Moreover, the nutrients are divided into two groups;
first the macronutrients which is needed by the plants in large amount such as
nitrogen, phosphorus, potassium, calcium, magnesium and sulfur. Second group is
the micronutrients, needed in tiny amounts that include zinc, copper,
manganese, boron, iron, sodium and cobalt. These two groups are essential for
normal plant growth and development.
Answer:
In trees, most savanna adaptations are to drought--long tap roots to reach the deep water table, thick bark for resistance to annual fires (thus palms are prominent in many areas), deciduousness to avoid moisture loss during the dry season, and use of the trunk as a water-storage organ (as in baobab).
Answer:
anwser is C
Explanation:
they both have 6 carbon atoms so A is wrong
both are monosacchride so B is wrong
both have same molecular formula which is (C₆H₁₂O₆) so D is wrong
in glucose the anomeric carbon is the first carbon, whereas in fructose, the anomeric carbon is the second carbon. The anomeric carbon is the one containing the carbonyl group (carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O)
Answer:
Alleles for feather colour exhibit incomplete dominance or co-dominance.
50% gray offspring + 50% black offspring
Explanation:
<em>It means that the alleles for feather colour in the hen exhibit incomplete dominance or co-dominance over one another.</em>
Assuming the allele for white colour is B, white colour will be b while the heterozygote Bb gives the gray phenotype.
Gray rooster + gray hen = 15 gray chicks, 6 black chicks and 8 white chicks.
15:6:8 is roughly 2:1:1 which is phenotypic ratio obtainable from crossing two heterozygous individuals as pointed out by Mendel.
Bb x Bb = 1BB, 2Bb, and 1bb
Crossing the gray rooster (Bb) with a black hen (bb):
Bb x bb = Bb, Bb, bb, and bb
= 2Bb (gray):2bb (black)
50% of the offspring will be gray while the remaining 50% will be black.
