Answer:
The product of glycolysis is two molecules of pyruvate. It is a three-carbon compound. This pyruvate again undergoes oxidation in the cytoplasm. This process is called pyruvate oxidation which produces Acetyl CoA. The Acetyl CoA is a two-carbon molecule.
Acetyl CoA again used for the citric acid cycle. This is also called as Kreb's cycle / TCA cycle. Because citric acid has 3 carboxylic groups. The acetyl coenzyme produces NADH, FADH2, ATP. The citric acid cycle occurs in the mitochondrial membrane. This is an 8 step process. The first product is citric acid. The other products of each step are isocitrate, alpha-ketoglutarate, succinyl CoA, succinate, Fumarate, L - malate, and Oxaloacetate (OAA).
Another process of aerobic respiration is the electron transport chain ( ETS). Here the energy stored in NADH, FADH2 in the citric acid cycle are utilized. It is a chain of electron carriers. ETS occurs in the inner membrane of mitochondria.
In short, the glucose splits by glycolysis and produces ATP, NADPH, and final product pyruvate. The pyruvate is oxidized and forms acetyle coenzyme. This is used in the TCA / citric acid cycle. In this process also NADH, FADH2 which forms electrons are produced. Theses electrons are carried by different electron carriers and accepted by oxygen.
In the process of pyruvate oxidation 6 ATP, and in Kreb's cycle 18 ATPs, in ETS, 4 ATPs are produced. In addition to this in glycolysis produces 4 ATPs. The total number of ATP in aerobic respiration is 32 ATP.
Answer:
True
Explanation:
Serous membranes are found around organs to lubricate them and reduce friction
Q1. The answer is 1.
It can be calculated using the equation:
(1/2)ⁿ = x
x - decimal amount remaining,
n - a number of half-lives.
x = 50% = 50/100 = 0.5
n = ?
(1/2)ⁿ = 0.5
log((1/2)ⁿ) = log(0.5)
n * log(1/2) = log(0.5)
n * log(0.5) = log(0.5)
n = log(0.5)/log(0.5)
n = 1
Q10. The answer is 2.
It can be calculated using the equation:
(1/2)ⁿ = x
x - decimal amount remaining,
n - a number of half-lives.
Rhyolite #2 has 25% of the parent H remaining:
x = 25% = 25/100 = 0.25
n = ?
(1/2)ⁿ = 0.25
log((1/2)ⁿ) = log(0.25)
n * log(1/2) = log(0.25)
n * log(0.5) = log(0.25)
n = log(0.25)/log(0.5)
n = -0.602 / - 0.301
n = 2
Q3. The answer is 100 million years.
A number of half-lives (n) is a quotient of total time elapsed (t) and length of half-life (H):
n = t/H
n = 1
t = ?
H = 100 000 000 years
n = t/H
t = n * H
t = 1 * 100 000 000 years
t = 100 000 000 years<span>
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The difference is because there about different from one another, the reflexes will act different towards the speed
