The problem provides information about a gene with a total molecular mass of 360,000 carbon units. Adenine (A) constitutes 50% of the Cytosine (C) content. The average molecular mass of a nucleotide is 300 carbon units. The problem asks us to: a) Calculate the number of each type of nucleotide in the gene. b) Determine the maximum number of amino acids that can be synthesized from this gene. c) If the mRNA molecule is synthesized from the second strand of the gene with the end sequence AAA-GCT-TTG-CGA-TGC-ATT, determine the corresponding amino acid sequence at the end of the gene. We are given the mRNA codons for Asparagine (AAC), Alanine (GCU), Threonine (ACG), Phenylalanine (UUU), and Arginine (CGA).

1. Problem Description

The problem provides information about a gene with a total molecular mass of 360,000 carbon units. Adenine (A) constitutes 50% of the Cytosine (C) content. The average molecular mass of a nucleotide is 300 carbon units. The problem asks us to:

a) Calculate the number of each type of nucleotide in the gene.

b) Determine the maximum number of amino acids that can be synthesized from this gene.

c) If the mRNA molecule is synthesized from the second strand of the gene with the end sequence AAA-GCT-TTG-CGA-TGC-ATT, determine the corresponding amino acid sequence at the end of the gene. We are given the mRNA codons for Asparagine (AAC), Alanine (GCU), Threonine (ACG), Phenylalanine (UUU), and Arginine (CGA).

2. Solution Steps

a) Calculate the number of each type of nucleotide in the gene.

Total molecular mass of the gene = 360,000 carbon units

Average molecular mass of a nucleotide = 300 carbon units

Total number of nucleotides = Total molecular mass / Average molecular mass

$N = \frac{360000}{300} = 1200$ nucleotides.

We are given that $A = 0.5C$. In a double-stranded DNA, A pairs with T and C pairs with G. So the number of A nucleotides equals the number of T nucleotides, and the number of C nucleotides equals the number of G nucleotides. $A = T$ and $C = G$.

$A = 0.5C$, therefore $T = 0.5G$.

The total number of nucleotides can be expressed as: $A + T + C + G = 1200$.

Substituting $A = 0.5C$ and $T = 0.5G$, we get: $0.5C + 0.5G + C + G = 1200$

$1.5C + 1.5G = 1200$. Since $C = G$, we can rewrite this as $3C = 1200$ or $3G = 1200$.

Therefore, $C = \frac{1200}{3} = 400$, and $G = 400$.

$A = 0.5C = 0.5 * 400 = 200$, and $T = 200$.

So, we have: A = 200, T = 200, C = 400, G =

4

0

0.

b) Determine the maximum number of amino acids that can be synthesized from this gene.

Since there are 1200 nucleotides in the gene, one strand will have 600 nucleotides. The coding region of the mRNA will have a length less than or equal to

6

0

0. Each codon consists of 3 nucleotides. Therefore the maximum number of codons, and hence the maximum number of amino acids is:

Number of codons = $\frac{600}{3} = 200$.

c) Determine the corresponding amino acid sequence at the end of the gene.

The given sequence is from the template strand, which is: AAA - GCT - TTG - CGA - TGC - ATT.

The mRNA sequence would be the reverse complement of the template strand. Note that Thymine (T) is replaced with Uracil (U) in the mRNA sequence.

The mRNA sequence corresponding to the gene sequence is:

5' - UUU - CGA - AAC - GCU - ACG - UAA - 3'

Since the problem asks us to determine the sequence *at the end* of the gene, we consider the mRNA from left to right as that end.

Given mRNA codon - Amino acid relationship:

UUU - Phenylalanine

CGA - Arginine

AAC - Asparagine

GCU - Alanine

ACG - Threonine

Thus, the amino acid sequence is: Phenylalanine - Arginine - Asparagine - Alanine - Threonine.

3. Final Answer

a) The number of each type of nucleotide in the gene is: A = 200, T = 200, C = 400, G =

4

0

0. b) The maximum number of amino acids that can be synthesized from this gene is

2

0

0. c) The amino acid sequence at the end of the gene is: Phenylalanine - Arginine - Asparagine - Alanine - Threonine.