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VashaNatasha [74]
3 years ago
8

Which of the following cellular environments is conducive to the formation of disulfide bonds within or between proteins? Choose

all that apply. Which of the following cellular environments is conducive to the formation of disulfide bonds within or between proteins? Choose all that apply. Rough ER Cytosol Nucleus Mitochondria
Biology
1 answer:
STatiana [176]3 years ago
7 0

Answer:

Rough endoplasmic reticulum and Mitochondria.

Explanation:

Disulfide bonds are known as covalent bonds. They are formed by the oxidation of 2 cysteines and these bonds can provide stability to proteins. These bonds mainly formed in intermembrane space of mitochondria and cellular compartments outside the cytoplasm endoplasmic reticulum. Both of these organelles present in an oxidation state providing an atmosphere for disulfide bond formation.

Cytoplasm and Nuclei mostly exit in reducing state because of the existence of disulfide reductase which is reducing the disulfide bonds between the cysteine residue to thiolate state. So, the disulfide bond formation will not happen.

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Explanation:

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3 years ago
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Answer:

The correct answer is option e) "Over time, we could see an increase in the H1 allele frequency".

Explanation:

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Answer:

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5 0
3 years ago
Urgent please help!!!!!
antiseptic1488 [7]
I think the answer is most likely be J.

The first (F) one the population of the predator increases hugely while the population of the prey was neutral. And so both population didn’t seem to have any connection. Same goes for H. Graph G doesn’t make sense at all the population of the prey didn’t exist throughout the time in the graph but only exist in one single point of time and then just vanish again so that shouldn’t be the answer either.
In graph J, you can see the correlation between the two populations as the predator goes up and so does the prey.
You can search up on google predator-prey relationship graph to get better understanding.
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