A particular eukaryotic protein is 300 amino acids long which of the folling could be the maximum number of nucleotides in the DNA template strand that codes for the amino acids in this protein?

Answer 1

there are 20!

They are:

Alanine

Arginine

Asparagine

Aspartic acid

Cysteine

Glutamic acid

Glutamine

Glycine

Histidine

Isoleucine

Leucine

Lysine

Methionine

Phenylalanine

Proline

Serine

Threonine

Tryptophan

Tyrosine

Valine

Hope this helps!

Answer 2

The likely answer is 99 amino acids.

The reasoning is as follows:

* since only one gene is mentioned, it sounds as if the protein has only one polypeptide (chain). * Also, there is no mention of introns, which are non-coding segments of DNA embedded in the gene, interrupting it like ads in a TV program.

* So, if the whole gene is coding for the polypeptide, every group of three adjacent nucleotides must be a codon, the unit of DNA that codes for one amino acid. Since there are 300 nucleotides, there must be 100 codons.

* However, the final codon must be a STOP codon, so that leaves 99 codons coding for 99 amino acids.

Answer 3

If the protein contains 300 amino acids, that means that there are 301 codons including the stop codon. Since each codon contains 3 nucleotides, This particular protein would have resulted from 903 nucleotides in the DNA template.

Answer 4

We use twenty amino acids and seven of these are taken in the diet (called essential ) while the others can be synthesized from precursor molecules.

Answer 5

there are three nucleotides per Condon and also 100 per amino acids so taking this into consideration you are looking at 99.

Answer 6

Each amino acid is coded for by 3 nucleotides (a 'triplet codon'):

3 x 300 = 900 nucleotides, not including the nucleotides involved with the start and stop codons.

Answer 7

Three nucleotides, one base, make up the code for one amino acid, so....,

3 * 331

= 993 nucleotides
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Answer 8

3 X 300 = 900..

Answer 9

9

Answer 10

110.