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Ch. 12 - The Genetic Code and Transcription
Klug - Essentials of Genetics 10th Edition
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook
Chapter 12, Problem 20

Messenger RNA molecules are very difficult to isolate in bacteria because they are rather quickly degraded in the cell. Can you suggest a reason why this occurs? Eukaryotic mRNAs are more stable and exist longer in the cell than do bacterial mRNAs. Is this an advantage or a disadvantage for a pancreatic cell making large quantities of insulin?

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Understand that bacterial mRNAs are quickly degraded because bacteria lack certain protective features found in eukaryotic mRNAs, such as a 5' cap and a poly-A tail, which help stabilize the mRNA molecule.
Recognize that the rapid degradation of bacterial mRNA allows bacteria to quickly adjust protein synthesis in response to environmental changes, which is beneficial for their survival.
Note that eukaryotic mRNAs, including those in pancreatic cells, have a 5' cap and a poly-A tail that protect the mRNA from degradation, resulting in greater stability and longer lifespan within the cell.
Consider that for a pancreatic cell producing large quantities of insulin, having stable mRNA is an advantage because it allows sustained and efficient production of insulin without the need for constant new transcription.
Conclude that the increased stability of eukaryotic mRNA supports the high demand for insulin production by ensuring that the mRNA remains available for translation over a longer period, enhancing protein synthesis efficiency.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

mRNA Stability and Degradation

mRNA stability refers to how long an mRNA molecule persists before being degraded. In bacteria, mRNAs are rapidly degraded to allow quick adaptation to environmental changes, resulting in short half-lives. This rapid turnover makes bacterial mRNAs difficult to isolate. In contrast, eukaryotic mRNAs are generally more stable due to protective features like the 5' cap and poly-A tail.
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Differences Between Prokaryotic and Eukaryotic mRNA

Prokaryotic mRNAs lack a 5' cap and poly-A tail, making them more susceptible to degradation by ribonucleases. Eukaryotic mRNAs have these modifications, which protect them from rapid degradation and allow longer existence in the cytoplasm. These structural differences influence mRNA lifespan and gene expression regulation in the two domains.
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mRNA Stability and Protein Production in Specialized Cells

In specialized cells like pancreatic cells producing insulin, longer mRNA stability is advantageous because it allows sustained protein synthesis without constant transcription. Stable mRNAs enable efficient production of large protein quantities, reducing energy expenditure. Conversely, rapid mRNA turnover would require continuous transcription, which is less efficient for high-demand proteins.
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Related Practice
Textbook Question

Define the process of transcription. Where does this process fit into the central dogma of molecular biology (DNA makes RNA makes protein)?

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Textbook Question

Describe the structure of RNA polymerase in bacteria. What is the core enzyme? What is the role of the σ subunit?

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Textbook Question

Write a paragraph describing the abbreviated chemical reactions that summarize RNA polymerase-directed transcription.

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Textbook Question

One form of posttranscriptional modification of most eukaryotic pre-mRNAs is the addition of a poly-A sequence at the 3' end. The absence of a poly-A sequence leads to rapid degradation of the transcript. Poly-A sequences of various lengths are also added to many bacterial RNA transcripts where, instead of promoting stability, they enhance degradation. In both cases, RNA secondary structures, stabilizing proteins, or degrading enzymes interact with poly-A sequences. Considering the activities of RNAs, what might be general functions of 3'-polyadenylation?

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Textbook Question

In a mixed copolymer experiment, messages were created with either 4/5C:1/5A or 4/5A:1/5C. These messages yielded proteins with the following amino acid compositions.

Using these data, predict the most specific coding composition for each amino acid.

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Textbook Question

Shown here are the amino acid sequences of the wild-type and three mutant forms of a short protein.

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Wild-type: Met-Trp-Tyr-Arg-Gly-Ser-Pro-Thr

Mutant 1: Met-Trp

Mutant 2: Met-Trp-His-Arg-Gly-Ser-Pro-Thr

Mutant 3: Met-Cys-Ile-Val-Val-Val-Gln-His                 _


Use this information to answer the following questions:

Using the genetic coding dictionary, predict the type of mutation that led to each altered protein.

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