Since each glucose molecule produces two acetyl-CoA molecules, the Krebs cycle must be completed twice to produce the four CO2, six NADH, two FADH2, and two ATPs.
- Catabolic reactions occur within cells during cellular respiration. It is a biochemical process by which waste materials are removed and nutrients are broken down to generate energy, which is then stored in the form of ATP. The process of aerobic respiration needs oxygen.
- The Krebs cycle, also known as the citric acid cycle, is the last step of oxidation for amino acids, lipids, and glucose.
- Other than glucose, many animals rely on other substances for energy.
- Protein's metabolic byproduct, amino acids, are deaminated and converted to pyruvate and other Krebs cycle intermediates.
- They begin the cycle and are broken down, for example. On deamination, alanine turns into pyruvate, glutamate into -ketoglutarate, and aspartate into oxaloacetate.
- Acetyl CoA is created when fatty acids are -oxidized and enters the Krebs cycle. It is the primary mechanism through which cells produce ATP. Complete nutrient oxidation results in the production of a significant amount of energy.
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Answer:
Pylorus
Explanation:
The stomach is a J-shaped organ which is a part of the digestive system. The digestive system on one end is attached to the oesophagus and on the end to the small intestine.
The pylorus structure of the stomach is composed of thick muscles which help to mix or agitating the food in the stomach and controls the movement of the food to the duodenum which is the first part of the small intestine. The movement is controlled by the ring of muscles in the pylorus called pyloric sphincter.
Thus, Pylorus is the correct answer.
Answer:
2% of the progeny will be double crossovers for the trihybrid test cross
Explanation:
By knowing the positions of genes, we can estimate the distances in MU between them per region.
- Genes A and B are 10 map units apart (Region I)
- Genes B and C are 20 map units apart (Region II)
- Genes A and C are 30 map units apart
----A-------10MU--------B-------------20MU-------------C---
Region I Region II
We can estimate the recombination frequencies by dividing each distance by 100.
• recombination frequency of A-B region = 10MU / 100 = 0.10
• recombination frequency of B-C region = 20MU / 100 = 0.20
Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:
EDRF = recombination frequency in region I x recombination frequency in region II.
EDRF = 0.10 x 0.20 = 0.02
2% of the progeny will be double crossovers for the trihybrid test cross
