2. enzyme
3. esophagus
4. lower esophageal sphincter (LES)
5. throat (pharynx)
6. stomach
7. stomach
8. protein
9. HCL
10. small intestine
11. chemical
12. nutrient
13. cells
15. bacteria
16. anus
17. solid
Answer:
1. P120 is degraded in the 26S proteasome
2. The 26S proteasome has a major role in protein degradation and is critical for protein homeostasis
3. Cell cycle and DNA replication are cellular processes regulated by the Ras and NFkB pathways
Explanation:
The proliferation-associated nucleolar protein (p120) is a protein known to be expressed during the interphase of the cell cycle, specifically in G1 and early S phase, where any problem with DNA replication trigger a checkpoint, i.e., a molecular cascade of signaling events that suspend DNA replication until the problem is resolved. In mammalian cells, the 26S proteasome is responsible for catalyzing protein degradation of about 80% (or even more) of their proteins. The 26S proteasome acts to degrade rapidly misfolded and regulatory proteins involved in the cell cycle, thereby having a major role in protein homeostasis and in the control of cellular processes. It is for that reason that inhibitors that block 26S proteasome function have shown to be useful as therapeutic agents in diseases associated with the failure of protein degradation mechanisms (e.g., multiple myeloma). The NF-κB are highly conserved transcription factors capable of regulating different cellular processes including, among others, cellular growth, inflammatory responses and apoptosis. Moreover, the MAPK/ERK pathway is able to transduce different signals received on the cell surface to the nucleus. The MAPK/ERK pathway is activated when a singling molecule binds to a cell receptor which triggers a signaling cascade that ends when a transcription factor induces the expression of target genes, ultimately producing a response in the cell (for example, the progression through the cell cycle).
Answer:
B
Explanation:
The statement that best explains the result would be that <u>the rate of photosynthesis is greatest for direct sunlight and least for the infrared light.</u>
The DPIP will normally replace and play the role of NADPH in the light reaction of the process of photosynthesis. Hence, it will become colorless as a result of reduction and the rate of photosynthesis can be monitored based on the magnitude of the disappearance of the dark blue color.
It means that the more colorless the liquid in the illustration is, the more the rate of photosynthesis. <em>The color change moved from dark blue to clear colorless under direct sunlight, from dark blue to nearly colorless under indirect sunlight, and from dark blue to slightly lighter under the infrared light.</em> <u>This clearly indicates that the rate of photosynthesis is highest under direct sunlight and lowest under infrared light with the indirect sunlight having an intermediate rate. </u>
The correct option is B.
Answer:
2 moles of F
Explanation:
for every molecule of CaF2, without the two atoms of F, CaF2 cannot be formed.
Now we multiply both CaF2 and F by 6.02 X 10^23 to convert CaF2 and F to moles, it will be the same ratio since we multiplied both by the same number
