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(a) Write the sequence of the mRNA molecule synthesized from a DNA template strand having the following sequence. $$5^{\prime}-\text { ATCGTACCGTTA-3 }^{\prime}$$ (b) What amino acid sequence is encoded by the following base sequence of an mRNA molecule? Assume that the reading frame starts at the \(5^{\prime}\) end. $$5^{\prime}-U U G C C U A G U G A U U G G A U G-3^{\prime}$$ (c) What is the sequence of the polypeptide formed on addition of poly(UUAC) to a cell-free protein-synthesizing system that does not require a start codon?

Short Answer

Expert verified
(a) 5'-UAAGGCAUGCGA-3'; (b) Leu-Pro-Ser-Asp; (c) (Leu-Leu-Thr)n.

Step by step solution

01

Transcribe DNA to mRNA

To transcribe the given DNA template to mRNA, substitute each DNA base with its complementary base. The DNA bases adenine (A), thymine (T), cytosine (C), and guanine (G) pair with uracil (U), adenine (A), guanine (G), and cytosine (C) respectively. The provided DNA sequence is 5'-ATCGTACCGTTA-3', which needs to be read in the 3'-to-5' direction to transcribe the complementary mRNA 5'-to-3'. Thus, the mRNA sequence will be 5'-UAAGGCAUGCGA-3'.
02

Translate mRNA to Amino Acids

The mRNA sequence is 5'-UUGCCUAGUGAUGG-3'. Use the genetic code table to translate this sequence. Split the mRNA into codons (triplets): UUG, CCU, AGU, GAU, G. Translate each codon: UUG codes for Leucine (Leu), CCU for Proline (Pro), AGU for Serine (Ser), GAU for Aspartic acid (Asp). Note that the sequence stops at the start of another codon which suggests the sequence ends with the question at hand. Therefore, the amino acid sequence is Leu-Pro-Ser-Asp.
03

Decode Poly(UUAC)

Poly(UUAC) can create repeating sequences of codons, but in a cell-free system without a start codon requirement, simply divide it into UUAC repeatedly: UUA, CUU, ACU,.... The codon UUA translates to Leucine (Leu), CUU to Leucine (Leu), and ACU to Threonine (Thr). Continuing this pattern will yield a repeated polypeptide sequence of Leu-Leu-Thr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

DNA to mRNA transcription
DNA to mRNA transcription is an essential step in the process of gene expression. It involves converting the nucleotide sequence of DNA into a complementary mRNA sequence. The mRNA, or messenger RNA, acts as a blueprint for protein synthesis. This process occurs in the nucleus of eukaryotic cells. The enzyme RNA polymerase plays a crucial role in this transcription process.

In transcription, the DNA double helix unwinds to expose the sequence of the template strand. The DNA template strand is read in the 3' to 5' direction. This ensures that the mRNA is synthesized in the 5' to 3' direction, which is necessary for subsequent translation.

To transcribe DNA into mRNA:
  • Adenine (A) pairs with uracil (U)
  • Thymine (T) pairs with adenine (A)
  • Cytosine (C) pairs with guanine (G)
  • Guanine (G) pairs with cytosine (C)
This complementary pairing ensures that the genetic information is accurately transcribed from the DNA template strand to produce the mRNA. For example, if a DNA sequence is 5'-ATCGTACCGTTA-3', the corresponding mRNA will be 5'-UAAGGCAUGCGA-3'.
genetic code translation
The genetic code translation occurs in the ribosome and is pivotal for protein synthesis. It involves translating the sequence of mRNA codons into an amino acid sequence, forming a protein. Each group of three nucleotides, termed a codon, corresponds to a specific amino acid or a stop signal.

The standard genetic code consists of 64 triplets, which include start and stop signals. Each codon specifies a particular amino acid.
  • For example, the mRNA sequence 5'-UUGCCUAGUGAUGG-3' consists of codons: UUG, CCU, AGU, GAU, and G
  • UUG translates to Leucine (Leu), CCU to Proline (Pro), AGU to Serine (Ser), and GAU to Aspartic acid (Asp)
Reading frames are critical in translation. The reading frame from the 5' end determines the grouping of the nucleotides into codons. Incorrect frames can lead to misinterpretation of the genetic information.

In the given sequence, the translation stops after decoding the known codons, resulting in the peptide Leu-Pro-Ser-Asp.
poly(UUAC) translation
In a protein-synthesizing system that does not require a start codon, the translation of synthetic nucleotide sequences, such as poly(UUAC), offers insights into codon decoding. Poly(UUAC) repeatedly segments into codons: UUA, CUU, and ACU.

The lack of initiation codons in cell-free systems allows continuous translation without the conventional start or stop sequences. For the sequence poly(UUAC):
  • UUA translates to Leucine (Leu)
  • CUU also translates to Leucine (Leu)
  • ACU translates to Threonine (Thr)
This results in a repeating polypeptide sequence of Leu-Leu-Thr. This pattern continues for as long as the synthetic mRNA is present. Understanding synthetic sequence translations helps in comprehending the flexibility and regulations of genetic code interpretation.

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Most popular questions from this chapter

A series of experiments were performed to establish the direction of chain growth in protein synthesis. Reticulocytes (young red blood cells) that were actively synthesizing hemoglobin were treated with \(\left[^{3} \mathrm{H}\right]\) leucine. In a period of time shorter than that required to synthesize a complete chain, samples of hemoglobin were taken, separated into \(\alpha\) and \(\beta\) chains, and analyzed for the distribution of \(^{3} \mathrm{H}\) within their sequences. In the earliest samples, only regions near the carboxyl ends contained radioactivity. In later samples, radioactivity was present closer to the amino terminus as well. Explain how these results determine the direction of chain growth in protein synthesis.

In a non-overlapping triplet code, each group of three bases in a sequence ABCDEF ... specifies only one amino acid-ABC specifies the first, DEF the second, and so forth-whereas, in a completely overlapping triplet code, ABC specifies the first amino acid, BCD the second, CDE the third, and so forth. Assume that you can mutate an individual nucleotide of a codon and detect the mutation in the amino acid sequence. Design an experiment that would establish whether the genetic code is overlapping or non-overlapping.

Why must tRNA molecules have both unique structural features and common structural features?

The energetic equivalent of two molecules of ATP is used to activate an amino acid, yet only one molecule of ATP is used. Explain.

Usually degeneracy should be avoided, except perhaps in well-controlled circumstances. However, it is advantageous for the genetic code to be degenerate. What does it mean to say that the code is degenerate, and explain why a degenerate code is valuable?

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