This reduction in starch content occurred because starch was changed into simple sugars
<h3>Explanation:
</h3>
Starch is a polymeric carbohydrate consist of many glucose units linked by glycosidic bonds. The iodine test is used to test for the presence of starch. Starch will change color to an intense "blue-black" colour after the addition of aqueous solutions of the triiodide anion. To do it we can add Iodine-KI reagent to a solution or directly on a potato or other materials such as bread, crackers, or flour The reaction between amylose that present in lesser amounts and iodine is said to account for the intense color change seen.
An iodine test of a tomato plant leaf revealed that starch was present at 5:00 p.m. on a sunny afternoon in July. When a similar leaf from the same tomato plant was tested with iodine at 6:00 am the next morning, the test indicates that less starch was present in this leaf than in the leaf tested the day before. This reduction in starch content occurred because starch was
- 1. changed directly into proteins
- 2. transported out of the leaves through the guard cells
- 3. transported downward toward the roots through tubes
- 4. changed into simple sugars
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He was C Determined This is because he kept on trying Hope this helps
Answer:
c. Physical adaptations to the environment drive the distribution of all three species in the wild.
Explanation:
Options A and B refer to competition between species. The experiment was about the colonizing ability of the three species, and not the interaction among them. So these two options are not correct.
Option D states that species A is better adapted to the upper intertidal zone than the middle or lower. But the table shows that species A is equally adapted to colonize upper and middle intertidal zone, and less adapted to colonize lower zones.
The correct option is C. When these competing species coexist, this is because of niche partitioning or niche differentiation. If there is not any differentiation between them, the dominant species displaces the weak species. In the exposed example, the three species coexist in the middle and lower zones, which means that they probably have different niches and got adapted to living to their environments. This adaptation to different conditions is what leads to their distribution.
In the exposed example, species A and B can live in the upper intertidal zone, where species C can not live because they can not tolerate environmental conditions. The three species can live in the middle zone, but still, A and B are more adapted to this area than C. Among A and B, B is the most adapted to living in the upper and middle zones. Species C seems to be very adapted to live in the lower intertidal zone, where species A and B can also live, but are less adapted to this area, probably due to environmental conditions or due to their vulnerability to predation. In this last area, species A is less adapted.
Answer:
You can use either your answer or the sample answer you gave. You are so nice by the way!
Explanation:
THE FOLLOWING ARE TWO ANSWERS YOU CAN CHOOSE FROM
ANSWER 1:
The red crystals are compounds. This is because when the sample is heated, a GAS (element) is released and BLUE POWDER (element) is left. Two elements make up these red crystals. A compound can also be shown through color change - the RED crystals, when heated, turn into BLUE powder. Elements cannot be broken down into substances using physical force, like heating. So yes, this is a compound.
ANSWER 2:
The red crystals are compounds. This is because when the sample is heated, a GAS (element) is released and BLUE POWDER (element) is left. Two elements make up these red crystals. A compound can also be shown through color change - the RED crystals, when heated, turn into BLUE powder. Elements cannot be broken down into substances using physical force, like heating. So yes, this is a compound.
EITHER ONE YOU CHOOSE WILL BE CORRECT, JUST SAYING!
Answer:
0.05 mg/mL ( B )
Explanation:
Given data:
20 mg/ml starch
2% solution = 2g of solute is in 100g of solvent
<u>Determine the new concentration in mg/ml </u>
Dilution equation = C1V1 = C2V2
new concentration ; applying the dilution factor
dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every 100 mL
∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL
