The rough endoplasmic reticulum (RER) is is a series of flattened sacs connected to ribosomes. The Golgi Apparatus is an organelle composed of flattened stacked pouches known as <em>cisternae</em>.
In this case, the missing words indicate:
- Specifically, this organelle is attached on the ROUGH ENDOPLASMIC RETICULUM (RER) which will help to modify and package the polypeptide product.
- It puts it into a transport vesicle to be sent over to the GOLGI APPARATUS to add molecular tags and sort it, like a post office.
- The RER is required for the production and secretion of different hormones and proteins.
- The Golgi apparatus plays fundamental roles associated with the transport, modification, and packaging of proteins and lipids into vesicles, which are subsequently delivered to specific cellular destinations.
- Both the RER and Golgi apparatus form part of the endomembrane system.
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Option A is correct. Your olfactory nerve is the first cranial nerve (CN I). It's also section of your autonomic apprehensive system, which regulates physique functions. This nerve enables your feel of smell.
<h3>What are olfactory signals?</h3>
Listen to pronunciation. A sequence of occasions in which cells in the nose bind to scent-bearing molecules and send electrical indicators to the talent where they are perceived as smells.
<h3>What is an example of olfactory?</h3>
The excellent smell of spring flowers, for example, may be considered an "olfactory delight." A associated word, olfaction, is a noun referring to the experience of scent or the act or procedure of smelling.
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Im gonna answer the ones i stufied in grade5 and back soo question 2 a food web question 4 they break down dead and decaying organic material question7 omnivores
Answer:
The presence of a stop codon—UAA, UAG or UGA—in the A site of the ribosome is generally a signal to terminate protein synthesis. This process constitutes the last essential stage of translation, as it ensures the formation of full-sized proteins
Answer: 48.93 mL of sprite
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
Explanation:
