Answer: Yes.
Explanation:
DNP and cyanide are both cellular respiration inhibitor.
Cyanide is a neurotoxin that can prevent cellular respiration by it's inactivation activity on mitochondria enzymes called cytochrome oxidase.
DNP (2,4-Dinitrophenol is a chemical that inhibit cellular respiration. It help to shuttle proton across cell membranes of living cells by spreading the proton along along the mitochondria and collapsing the force that produced energy for the cell activities.
Answer:
The correct answer is "the temperature is above 23 °C but not exceeding 40 °C".
Explanation:
The main function of virgin queens in a bee's hive is reproduction. Normally virgin queens do not exit the hive except to perform mating flight(s), which normally are done with 10–20 drone bees and then return to the hive as mated queen bees. Usually virgin queens perform her mating flight(s) in ideal temperature conditions, which are with a temperature above 23 °C but not exceeding 40 °C.
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:
The total blood given is 5.5 liters
Thus, total alcohol needed for BAC (blood alcohol content) to exceed 0.1 gram per 100 ml will be 0.1/100 × 5500 = 5.5 grams alcohol.
1 oz = 28.3 grams
28.3 gram water = 28.3 ml
It has 20 percent alcohol by volume,
Therefore, alcohol in one drink will be:
20/100 × 28.3
= 5.66 ml
= 5.66 × 0.79 (density of ethanol)
= 4.47 gram
Thus, no of drinks will be 5.5 g / 4.47 g = 1.2
Hence, 1.2 drinks will make the BAC to exceed 0.1 g / 100 ml in the blood.