<span>John could make some simple changes to his breakfast to increase his fiber intake. Instead of orange juice, he could drink juice with added fiber such as V8 juice or fiber added grape juice. Instead of corn flakes, he could eat a high fiber cereal such as Grape Nuts. Instead of white toast, he could eat a slice of high fiber whole wheat bread. He can make some simple substitutions that will really impact his diet and his health.</span>
I think that since it lives on the ocean floor, at the deepest of the ocean. So anyway, the fin on the Eurypterids is still necessary but not that much, since they live on the ocean floor. Their movements mostly on the ocean floor. So fin would not help them "stick" to the floor. Also there are still gravity acts on the Eurypterids and the feet would help them move faster. Hope this helps.
Crabgrass preventers can be categorized under Residual Herbicides simply because they can stay in the soil for a considerable amount of time and control germination for the weeds. It prevents growth by staying on the ground longer.
Answer:
They are building blocks for larger polysaccharide which are important in structural rigidity of cells.
Polysaccharides such as Starch and Cellulose which are instrumental in cell structure and rigidity are formed from chains of monosaccharides.
They are also useful in the provision of energy which is used in cell functions (metabolism).
Monosaccharides such as Glucose, Sucrose are useful in provision of energy for physical and metabolic activities of the body.
Answer:
Explanation:
<em>The probability of producing plants with white axial flowers would be 1/16.</em>
From the illustration, All F1 individuals had red, axial flowers. It thus means that red and axial genes are dominant over white and terminal genes in the pea plant.
Let us assume that the allele for flower color is A (red) and a (white); and the allele for flower location is B (axial) and b (terminal).
Pure-breeding red, axial flower = AABB
Pure breeding white, terminal flower = aabb
AABB x aabb
F1 genotype = AaBb - all red and axial
At F2:
AaBb x AaBb
Progeny
<em>9 A_B_ red/axial</em>
<em>3 A_bb red/terminal</em>
<em>3 aaB_ white/axial</em>
<em>1 aabb - white/terminal</em>
Hence, the probability of producing plants with white axial flowers in the F2 generation is 1/16.
