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Chapter 22: Problem 44

(a) What mRNA sequence does the following DNA sequence produce: T A C G G G T T T A T C? (b) What message does that mRNA sequence produce?

Short Answer

Expert verified
The mRNA sequence is AUG CCC AAA UAG, which translates to Methionine-Proline-Lysine.

Step by step solution

01

- Identify DNA to mRNA transcription rules

In transcription, each DNA base pairs with a complementary RNA base: Adenine (A) pairs with Uracil (U), Thymine (T) pairs with Adenine (A), Cytosine (C) pairs with Guanine (G), and Guanine (G) pairs with Cytosine (C).
02

- Transcribe DNA to mRNA

Apply the transcription rules to each base of the given DNA sequence: T A C G G G T T T A T C.- T pairs with A- A pairs with U- C pairs with G- G pairs with C- G pairs with C- G pairs with C- T pairs with A- T pairs with A- T pairs with A- A pairs with U- T pairs with A- C pairs with GSo the mRNA sequence is: A U G C C C A A A U A G.
03

- Divide mRNA sequence into codons

mRNA is read in groups of three bases called codons. Divide the mRNA sequence A U G C C C A A A U A G into codons:- AUG- CCC- AAA- UAG
04

- Translate mRNA codons to amino acids

Each mRNA codon corresponds to an amino acid. Using a codon table:- AUG codes for Methionine (start codon)- CCC codes for Proline- AAA codes for Lysine- UAG is a stop codonSo the translated message is: Methionine-Proline-Lysine.

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Key Concepts

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

mRNA sequence
The process of creating an mRNA sequence from a DNA sequence is called transcription. During transcription, the enzyme RNA polymerase reads the DNA strand and builds a complementary mRNA strand. DNA consists of four bases: Adenine (A), Thymine (T), Cytosine (C), and Guanine (G). In mRNA, however, Thymine (T) is replaced by Uracil (U). Hence, when a DNA strand undergoes transcription, the base pairing rules are:
A pairs with U,
T pairs with A,
C pairs with G,
G pairs with C.
Using these rules, you can convert a given DNA sequence into its corresponding mRNA sequence.
DNA transcription rules
Understanding transcription rules is crucial for converting a DNA sequence to mRNA. Let's review the rule again:
A (Adenine) in DNA pairs with U (Uracil) in mRNA,
T (Thymine) in DNA pairs with A (Adenine) in mRNA,
C (Cytosine) in DNA pairs with G (Guanine) in mRNA,
G (Guanine) in DNA pairs with C (Cytosine) in mRNA.
For the given DNA sequence, T A C G G G T T A T C:
1. T pairs with A
2. A pairs with U
3. C pairs with G
4. G pairs with C
5. G pairs with C
6. G pairs with C
7. T pairs with A
8. T pairs with A
9. T pairs with A
10. A pairs with U
11. T pairs with A
12. C pairs with G.
Hence, the resulting mRNA sequence is: A U G C C C A A A U A G.
codons
Once you have the mRNA sequence, the next step is to divide it into codons. Codons are groups of three nucleotide bases in mRNA, and each codon corresponds to a specific amino acid. For example, AUG, CCC, AAA, and UAG are codons in our mRNA sequence. Codons are essential because they enable the mRNA sequence to be accurately translated into a chain of amino acids, forming proteins. The mRNA sequence A U G C C C A A A U A G is divided into codons as follows:
AUG, CCC, AAA, UAG.
Each of these codons will be translated into an amino acid in the next step.
amino acid translation
The final step in the process involves translating the mRNA codons into amino acids. This process is called translation and takes place in the ribosome, where tRNA molecules bring amino acids to match each mRNA codon. Here are the translations for our specific codons:
1. AUG codes for Methionine (start codon)
2. CCC codes for Proline
3. AAA codes for Lysine
4. UAG is a stop codon, signaling the end of translation.
Thus, the mRNA sequence A U G C C C A A A U A G translates to the amino acid sequence Methionine-Proline-Lysine, with UAG indicating the stop of the translation. These amino acids will then fold into a functional protein necessary for various cellular processes.

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Most popular questions from this chapter

A dodecapeptide \(\mathbf{1}\) is hydrolyzed in acid to give two Arg, Asp, Cys, and His, two Leu and Lys, two Phe, and two Val. The Edman procedure yields a phenylthiohydantoin that shows two \(3 \mathrm{H}\) doublets and a multiplet integrating for \(1 \mathrm{H}\) at about \(\delta 4.0 \mathrm{ppm}\) in the \({ }^{1} \mathrm{H}\) NMR spectrum. Treatment of 1 with trypsin leads to Val- His'Phe-Leu*Arg, Asp.Cys.Leu.Phe Lys, and Val-Arg. Treatment of 1 with chymotrypsin leads to Val-His.Phe, Leu.Arg.Asp* Cys.Leu.Phe, and Lys.Val-Arg. Deduce the structure of 1 and explain your reasoning.

(a) A peptide of unknown structure is treated with the Sanger reagent, 2,4 -dinitrofluorobenzene, followed by acid hydrolysis. The result is formation of a compound containing signals integrating for eight aromatic hydrogens in its \({ }^{1} \mathrm{H}\) NMR spectrum. What information can you derive about the structure of this peptide? (b) A second peptide, similarly treated, gives a compound in which a single hydrogen appears in the high-resolution \({ }^{1} \mathrm{H}\) NMR spectrum as an eight-line signal at about \(\delta 3.5 \mathrm{ppm}\). What information can you derive about the structure of this peptide?

Reaction of \(N\)-phenylalanine (1) with nitrous acid affords \(N\)-nitroso amino acid (2), which on treatment with acetic anhydride yields "sydnone" (3). Sydnones belong to a class of compounds known as mesoionic compounds. These compounds cannot be satisfactorily represented by Lewis structures not involving charge separation. The name sydnone derives from the University of Sydney where the first examples were prepared in 1935. Reaction of 3 and dimethyl acetylenedicarboxylate (DMAD) gives pyrazole (4). Propose mechanisms for the formations of 2,3 , and 4. Hint: For the formation of 4 , it may be helpful to consider the other possible Lewis structures for sydnone 3 .

(a) Amides protonate on oxygen rather than nitrogen. Why? (b) Amides \(\left(\mathrm{p} K_{\mathrm{b}} \sim 14\right)\) are weaker bases than amines \(\left(\mathrm{p} K_{\mathrm{b}} \sim 5\right)\). Why?

A tripeptide of unknown structure is hydrolyzed in acid to one molecule of Ala, one of Cys, and one of Met. The tripeptide reacts with phenyl isothiocyanate, followed by treatment with acid, to give a phenylthiohydantoin that shows only a three-hydrogen doublet at about \(\delta 2 \mathrm{ppm}\) in the \({ }^{1} \mathrm{H}\) NMR spectrum. The tripeptide is not cleaved by \(\mathrm{BrCN}\). Deduce the structure from these data.
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