The strand of RNA moves to the ribosome.
✔ 4
The DNA double helix unzips.
✔ 1
The strand of RNA leaves the nucleus.
✔ 3
A protein is produced.
✔ 5
A strand of RNA is made from a strand of DNA.
✔ 2
Answer:
4. Glucose, which reduces cAMP, is an example of positive control, and lactase, which breaks down lactose, is an example of negative control.
Explanation:
During positive control, the presence of Glucose results in the repression of expression of lac operon. This concept is known as catabolite repression. During negative control, the lac genes are switched off by repressor when the inducer is absent (indicating the unavailability of lactose). But when lactose is present, lac binds the repressor protein and modifies it in order to dissociate it from the operator. The removal of the combination of repressor and inducer helps lac to be transcribed and expressed.
B. Widespread ice sheets
C. The formation of Pangaea
Answer:
C
Explanation:
Answer:
Pepsin is an enzyme present in the gastric juice which digests the proteins of the food material.
These pepsin are synthesized as pre-proenzymes or inactive forms by chief cells in the gastric mucosa of the stomach. These inactive forms consist of a signal peptide, activation peptide and active enzyme which gets activated in the acidic environment of stomach by Hydrochloric acid.
The stomach secrets this enzyme in an inactive form to prevent the digestion of protective proteins present in the lining of the digestive tract.
was added to 9.9 ml of sterile buffer. After thorough mixing, this
suspension was further diluted by a 1/100 dilution followed by a
1/10 dilution. One-tenth of a ml of this final dilution was plated
on agar plates. After incubation, 52 colonies were present. How
many colony-forming units were present in the total 10 gram sample
of hamburger?
Answer:
5.2 × 10 ⁹ cfus
Explanation:
Using the dilution factors
0.1 ml of the final dilution has 52 colonies
1 ml will have approximately 520 colonies
10 ml of the final sample will have 5200 colonies
at 1 / 100 dilution
1 ml of the sample will have 5200 colonies
100 ml of the sample will have 520000 colonies
1 ml of the 0.1 ml + 9.9 ml has 520000 colonies
10 ml will have 5200000
at the second stage of the dilution
0.1 ml of the slurry had 5200000 colonies
1 ml will have 52000000 colonies
10 ml will have 520000000 colonies
100 ml of the initial sample ( 10 grams + 90 ml ) = 5200000000 colonies =
5.2 × 10 ⁹ cfu
The answer is stratus cloud, a blanket in the sky meaning bad weather mostly rain :)
Answer:
The answer is stratus cloud, a blanket in the sky meaning bad weather mostly rain
Explanation:
b. activation of protein kinase molecules.
c. activation of G protein-coupled receptors.
d. regulation of transcription by signaling molecules.
Protein phosphorylation is not involved in activation of G protein coupled receptor.
Further Explanation:
Roles of Protein phosphorylation are :
G- protein coupled receptor is activated through the external signals in the forms of ligand.
In this, ligand binds to the active site and cause conformation changes in receptor and active G- protein.
Protein phosphorylation is a key process in many biological functions but it is not directly involved in the activation of G protein-coupled receptors. Those receptors are activated by the binding to a specific signaling molecule, not phosphorylation.
Protein phosphorylation is a biochemical process that adds a phosphate group to a protein molecule, significantly affecting its function. Indeed, protein phosphorylation is involved in the activation of receptor tyrosine kinases, the activation of protein kinase molecules, and regulation of transcription by signaling molecules. However, phosphorylation is not directly involved in the activation of G protein-coupled receptors. Instead, these receptors are activated by binding to a specific signaling molecule or ligand, leading to a cascade of signaling events within the cell.
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