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Chapter 15: Problem 50

\(\begin{array}{|c|c|}\hline \text { Codon on mRNA } & {\text { Amino Acid }} \\\ \hline \mathbf{G C A} & {\text { alanine }} \\ \hline \mathbf{A A G} & {\text { lysine }} \\ \hline \mathbf{G U U} & {\text { valine }} \\ \text { AAU } & {\text { asparagine }} \\ \hline \mathbf{U G C} & {\text { cysteine }} \\ \hline \mathbf{U C G} & {\text { serine }} \\ \hline \mathbf{U C U} & {\text { serine }} \\ \hline \text { UUA } & {\text { leucine }} \\ \hline \text { UAA } & {\text { stop }} \\ \hline\end{array}\) You are given three mRNA sequences: 1\. 5鈥�-UCG-GCA- AAU-UUA -GUU-3鈥� 2\. 5鈥�-UCU-GCA- AAU-UUA -GUU-3鈥� 3\. 5鈥�-UCU-GCA- AAU-UAA -GUU-3鈥� Using the table, write the peptide encoded by each of the mRNA sequences. a. 1. Serine-alanine-asparagine-leucine-valine 2\. Serine-alanine-asparagine-leucine-valine 3\. Serine-alanine-asparagine(-stop) b. 1. Serine-phenylalanine-asparagine-leucinevaline 2\. Serine-alanine-asparagine-leucine-valine 3\. Serine-alanine-asparagine (-stop) c. 1. Serine-alanine-asparagine-leucine-valine 2\. Serine-alanine-asparagine (-stop) 3\. Serine-alanine-asparagine-leucine-valine d. 1. Serine-alanine-asparagine-leucine-valine 2\. Serine-arginine-asparagine-leucine-valine 3\. Serine-alanine-asparagine(-stop)

Short Answer

Expert verified
a. Serine-alanine-asparagine-leucine-valine Serine-alanine-asparagine-leucine-valine Serine-alanine-asparagine(-stop)

Step by step solution

01

Decode the first mRNA sequence

The first sequence is 5鈥�-UCG-GCA-AAU-UUA-GUU-3鈥�. Each codon represents an amino acid:- UCG: Serine- GCA: Alanine- AAU: Asparagine- UUA: Leucine- GUU: ValineThe peptide for the first sequence is: Serine-alanine-asparagine-leucine-valine.
02

Decode the second mRNA sequence

The second sequence is 5鈥�-UCU-GCA-AAU-UUA-GUU-3鈥�. Each codon represents an amino acid:- UCU: Serine- GCA: Alanine- AAU: Asparagine- UUA: Leucine- GUU: ValineThe peptide for the second sequence is: Serine-alanine-asparagine-leucine-valine.
03

Decode the third mRNA sequence

The third sequence is 5鈥�-UCU-GCA-AAU-UAA-GUU-3鈥�. Each codon represents an amino acid until a stop codon appears:- UCU: Serine- GCA: Alanine- AAU: Asparagine- UAA: StopThe peptide for the third sequence is: Serine-alanine-asparagine(-stop).

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Key Concepts

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

codons
Codons are sequences of three nucleotides in mRNA that correspond to specific amino acids or stop signals during translation.
In mRNA translation, the sequence of nucleotides in mRNA is read in groups of three, called codons.
Each codon specifies a particular amino acid or a stop signal.
This is crucial because it determines the order in which amino acids are added to the growing polypeptide chain.
For example, in the sequence 5鈥�-UCG-GCA-AAU-UUA-GUU-3鈥�, each set of three nucleotides represents a codon.
UCG codes for serine, GCA for alanine, AAU for asparagine, UUA for leucine, and GUU for valine.
  • The entire mRNA is read one codon at a time during translation.
  • Properly matching codons to their corresponding amino acids is essential for correct protein formation.
amino acids
Amino acids are the building blocks of proteins.
There are 20 different amino acids that make up proteins in the body.
Each amino acid has a unique side chain that determines its properties and role in protein structure and function.
When codons in mRNA specify amino acids, the sequence of codons affects the resulting protein's structure and function.
For example, the first mRNA sequence translates to the peptide: serine-alanine-asparagine-leucine-valine.
These amino acids are linked via peptide bonds, forming a polypeptide chain.
  • Each amino acid plays a specific role in the protein's final structure.
  • The combination and order of amino acids determine the polypeptide's properties and function.
genetic code
The genetic code is a set of rules that defines how the sequence of nucleotides in mRNA is translated into an amino acid sequence in proteins.
It is composed of codons, each of which specifies one of the 20 amino acids or a stop signal for translation.
The genetic code is universal, almost identical across all organisms.
During translation, ribosomes read mRNA codons and match them with the appropriate transfer RNA (tRNA) molecules that carry amino acids, forming proteins.
For instance, in the provided exercise:
  • UCG, UCU both encode serine.
  • GCA encodes alanine.
  • AAU encodes asparagine.
  • UUA encodes leucine.
  • UAA is a stop codon.
The genetic code is crucial for maintaining the flow of genetic information from DNA to RNA to proteins, ensuring that cells function correctly.

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Most popular questions from this chapter

Only a fraction of DNA encodes proteins. The noncoding portion of a gene is referred to as the intron. The intron fraction depends upon the gene. Introns are rare in prokaryotic and mitochondrial DNA; in human nuclear DNA, this fraction is about 95%. The intron is transcribed into mRNA, but this noncoding mRNA is edited out before translation of the coding portion, or exon, of a gene. The edited exon segments are then spliced together by a spliceosome, a very large and complex collection of RNAs and proteins. Although introns do not encode proteins, they have functions. In particular, they amplify expression of the exon, although the mechanism is unknown. When introns are very long, which is common among mammalian genes with roles in development, they can significantly extend the time required to complete transcription. Analysis of genes common to different plant and animal species shows many shared intronic positions and base sequences, although in some organisms, such as yeast, many introns have been deleted. Because introns do not encode proteins, mutations can remain silent and accumulate. A. As described above, introns are ancestral remnants that are replicated because they do not disadvantage the organism. Consider the claim that introns are 鈥渏unk DNA.鈥� Evaluate the claim with supporting evidence. B. Introns may be retained during transcription. Explain how the retention of a transcribed intron between two transcribed exons within a gene could do the following: 鈥� block expression of one polypeptide sequence 鈥� increase expression of a polypeptide 鈥� alter the polypeptide expressed

The yeast life cycle is usually dominated by haploid cells, each with a single set of unpaired chromosomes. The cell propagates asexually, and the genetic material is replicated through mitosis. Cell division occurs every 2鈥�4 hours, leading to 60鈥�100 generations in a single day. Yeast also reproduce sexually, particularly under adverse environmental conditions. When two haploid cells鈥攚ith DNA containing complementary mating-type alleles鈥攃onjugate, a diploid zygote results. The diploid zygote can then complete the sexual segment of the life cycle through meiosis. After meiosis, four haploid spores are produced, which can germinate. Researchers can grow yeast easily on nutrient-containing plates. Because both asexual and sexual reproduction is rapid, yeast has become an important organism for the experimental investigation of mutagenesis and evolution among eukaryotes. Environmental factors, such as chemicals or radiation, induce mutations. High-energy UV-c radiation of less than 1 minute in duration will result in many mutated yeast cells. UV-c can be used to mutate a strain of yeast in which the synthesis of adenine is blocked. This mutation is observable because the \(ade-2\) mutant has a red color when cultured on nutrientcontaining plates. Exposure to uv-c also can result in additional mutations. In particular, one mutant, \(ade-7\), changes the color of the ade-2 mutant to white. A. You have a uv-c lamp, culture plates, and growth chambers at \(23^{\circ} \mathrm{C}\) and \(37^{\circ} \mathrm{C}\). You also have available known haploid strains that are \((ade-2,+,+)\), where + denotes the wild type. Design a plan to determine the rate of uv-c-induced mutations in nutrient-containing plates inoculated with yeast. Earth's ozone layer removes high-energy ultraviolet radiation, uv-c, from the solar radiation received at the surface. Lower-energy ultraviolet radiation, uv-b, strikes Earth鈥檚 surface. Damage to DNA induced by ultraviolet radiation occurs with the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine and cytosine, on the same strand of DNA. A repair enzyme, photolyase, which is activated by visible light, is present in plants and most animals, but not in humans. In characterizing the relationship between environmental mutagens and cell damage, a useful assumption is often made and referred to as the linear hypothesis. This assumption states that the extent of damage is proportional to the amount of radiation received. Mutation rates for a strain (preac) that does not produce photolyase and a wild-type (+) strain were studied. Cultures of the two strains of yeast were diluted, and nutrient-containing plates were inoculated in triplicate at 23 掳C. The plates were exposed to bright sunlight for varying time intervals. After exposure, the plates were incubated in the dark at 23 掳C. After incubation between 1 and 8 hours, data shown in the table below were collected by counting the density of living cells relative to the control, and averaging these among replicates. B. Using the data table below, graph the average survival fraction, relative to the wild-type control. Predict the number of mutations in a sample of 1,000 cells of the preac type that are exposed to bright sunlight for 15 seconds. C. Based on these data, evaluate the merits of the alternative theories of the adaptive advantage provided by sexual reproduction.

A poly-A sequence is added at the: a. 5鈥� end of a transcript in the nucleus b. 3鈥�-end of a transcript in the nucleus c. 5鈥� end of a transcript in the cytoplasm d. 3鈥�-end of a transcript in the cytoplasm

In any given species, there are at least how many types of aminoacyl tRNA synthetases? a. 20 b. 40 c. 100 d. 200

The DNA of virus A is inserted into the protein coat of virus B. The combination virus is used to infect E. coli. The virus particles produced by the infection are analyzed for DNA and protein contents. What results would you expect? a. DNA and protein from B b. DNA and protein from A c. DNA from A and protein from B d. DNA from B and protein from A
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