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iogann1982 [59]
3 years ago
11

No matter what the cell type, prokaryotic or eukaryotic, plant or animal, the ____________ forms a protective barrier between th

e exterior environment and the interior of the cell. A) cell wall B) cytoskeleton Eliminate C) cell membrane D) glycoproteins
Biology
1 answer:
lidiya [134]3 years ago
7 0

Answer: C). cell membrane

The cell membrane is the membrane that can be found in both eukaryotic and prokaryotic cells. It is the cell that separates the interior of the cell from the extracellular environment. It is a protective layer that allows only the selected ions and organic molecules to enter inside the cells and go out of the cell, hence is called as selectively permeable or semi-permeable membrane. This phenomena maintains the concentration of ions and molecules inside the cell and prevents dehydration, cell rupture due to high concentration.

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The answer is D. Species
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Which of the following are contributing factors to the extinction of species?
meriva

Answer:

Deforestation

Climate Change

Invasive Species

Explanation:

Seventy percent of the Earth's plant and animals dwell in forests, and deforestation affects them directly. Once their habitat is lost, they are on their way to extinction.

Climate change also causes extinctions not only as a result of species intolerance to high temperature, but more commonly via a variety of related factors that alter a species’ interactions with other species.

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Whats used faster than theyre replaced
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White Blood cells Because they are produced in your body and they are constantly used more than they are replaced
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4 years ago
How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... C
fenix001 [56]

Answer:

There are 4 reading frames without stop codons.

Explanation:

1) To find the possible reading frames you need to separate your sequence in codons (trios)  for the 5'-> 3' sense, first, excluding the first nucleotide, then excluding 2 nucleotides and then not excluding nucleotides.

5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3'  NO STOP CODON

5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3'  NO STOP CODON

5'... CTT ACA GTT TAT <u>TGA</u> TAC GGA GAA GG... 3'

2)Then you find the complementary sequence (which you already have)

3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'

3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'

3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'

3) Next, you have to obtain the reverse complementary

CCT TCT CCG TAT CAA TAA ACT GTA AG NO STOP CODON

CC TTC TCC GTA TCA ATA AAC TGT AAG NO STOP CODON

C CTT CTC CGT ATC AAT AAA CTG <u>TAA</u> G

4)Now you need to find start (AGT in terms of DNA) and stop codons(<u>TAA</u>, <u>TAG,</u> <u>TGA</u>, also in terms of DNA), in 5->3 and de reverse complementary.

<em />

<em>As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.</em>

I added a table where you can find thestart and stop codons as well as the proteins that they translate to in terms of RNA.

I hope you find this information interesting and useful, good luck!

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How is the definition of "lipid" different from the types of definitions used for other biomolecules, such as amino acids, nucle
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