/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Suppose a gene has the sequence ... [FREE SOLUTION] | 91影视

91影视

Suppose a gene has the sequence ATGCGTTATCGGGAGTAG. A point mutation changes the gene to read ATGCGTTATGGGGAGTAG. How would the polypeptide product of this gene change?

Short Answer

Expert verified
The polypeptide changes from Ala (A) to Gly (G) at the third amino acid.

Step by step solution

01

- Original Gene Sequence

Identify the original gene sequence: ATGCGTTATCGGGAGTAG.
02

- Mutated Gene Sequence

Identify the mutated gene sequence: ATGCGTTATGGGGAGTAG.
03

- Identify the Point Mutation

Locate the difference between the two sequences. The original gene has 'C' as the 9th nucleotide while the mutated gene has 'G' instead.
04

- Transcription

Transcribe the original and mutated DNA sequences into mRNA. Original mRNA: UACGCAAUAGCCCUCAUC. Mutated mRNA: UACGCAAUAGGCUCAUC.
05

- Translation

Translate the mRNA sequences into their corresponding amino acids using the genetic code table. Original mRNA translates to: Tyr-Ala-Ile-Ala-Leu-Ile. Mutated mRNA translates to: Tyr-Ala-Ile-Gly-Leu-Ile.
06

- Determine Polypeptide Change

Compare the polypeptides. The original has Ala (A) whereas the mutated has Gly (G) at the third amino acid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

DNA transcription
DNA transcription is the process by which a segment of DNA is copied into mRNA by the enzyme RNA polymerase. This process is essential for converting the genetic code found in DNA into a format that can be read and utilized to create proteins. During transcription, the DNA double helix unwinds, allowing RNA polymerase to read one strand and synthesize a complementary strand of mRNA.

Let's break down the steps of transcription:
  • Initiation: RNA polymerase binds to a specific sequence on the DNA known as the promoter.
  • Elongation: The RNA polymerase moves along the DNA template strand, adding complementary RNA nucleotides. For instance, if the DNA sequence is 'ATG', the RNA would be 'UAC'.
  • Termination: The process continues until the RNA polymerase reaches a termination sequence, signaling the end of transcription. The mRNA strand is then released.
In our example, the original gene sequence 'ATGCGTTATCGGGAGTAG' is transcribed into mRNA as 'UACGCAAUAGCCCUCAUC'. The mutated gene sequence 'ATGCGTTATGGGGAGTAG' is transcribed into 'UACGCAAUAGGCUCAUC'.

The critical alteration occurs at the ninth nucleotide where a 'C' in the original sequence is changed to a 'G', leading to a change in the mRNA sequence.
mRNA translation
Translation is the next phase where the mRNA sequence is used to build a polypeptide chain, which will eventually fold into a functional protein. This process takes place in the ribosome, a molecular machine found within all living cells.

Each group of three nucleotides in mRNA, called a codon, corresponds to a specific amino acid. For example, the codon 'AUG' codes for methionine. The ribosome reads the mRNA sequence three nucleotides at a time and links the corresponding amino acids together to form a polypeptide chain.

Steps involved in mRNA translation:
  • Initiation: The ribosome assembles around the mRNA and starts at the start codon (usually 'AUG').
  • Elongation: Transfer RNA (tRNA) molecules bring amino acids to the ribosome, where the codons of the mRNA are matched with their corresponding anticodon sequences on tRNA.
  • Termination: When the ribosome reaches one of the stop codons ('UAA', 'UAG', 'UGA'), translation ends, and the polypeptide chain is released.
In our scenario, the mRNA sequences 'UACGCAAUAGCCCUCAUC' and 'UACGCAAUAGGCUCAUC' are translated. The original mRNA sequence produces a polypeptide 'Tyr-Ala-Ile-Ala-Leu-Ile', whereas the mutated sequence results in 'Tyr-Ala-Ile-Gly-Leu-Ile'. The single nucleotide change leads to one amino acid difference in the polypeptide chain.
polypeptide synthesis
Polypeptide synthesis refers to the formation of a polypeptide chain from amino acids during translation. This process is crucial as polypeptides fold into the proteins which perform a plethora of functions in biological organisms, from catalyzing biochemical reactions to providing structural support.

The sequence of amino acids, determined by the mRNA sequence, dictates how the polypeptide will fold and function. Even a single change in the amino acid sequence can significantly alter the protein's structure and function, potentially leading to various impacts on the organism.

Understanding the importance of minute changes:
  • Point mutations, like the one in our exercise, can lead to a single amino acid substitution. In our example, the substitution of 'Ala' to 'Gly' could potentially impact the protein's function.
  • Not all mutations negatively impact protein function. Some can be neutral or even beneficial, depending on their role in the protein and the cellular environment.
  • Such mutations can be observed in genetic disorders or evolutionary adaptations.
Summarizing our case, the mutation observed changes the ninth nucleotide from 'C' to 'G'. This results in the amino acid 'Ala' being replaced by 'Gly' in the polypeptide chain. Understanding these processes highlights the delicate balance and intricate relationship between DNA, mRNA, and proteins.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If mRNA is complementary to the DNA template strand and the DNA template stand is complementary to the DNA non-template strand, why are base sequences of mRNA and the DNA non- template strand not identical? Could they ever be? a. No, they cannot be identical because the T nucleotide in DNA is replaced with U nucleotide in RNA and AUG is the start codon. b. No, they cannot be identical because the T nucleotide in RNA is replaced with U nucleotide in DNA. c. They can be identical if methylation of the U nucleotide in RNA occurs and gives T nucleotide. d. They can be identical if de-methylation of the U nucleotide in RNA occurs and gives T nucleotide.

What part of central dogma is not always followed in viruses? a. The flow of information in HIV is from RNA to DNA, then back to RNA to proteins. Influenza viruses never go through DNA. b. The flow of information is from protein to RNA in HIV virus, while the influenza virus converts DNA to RNA. c. The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses. d. The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza

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

What would happen if the 5鈥 methyl guanosine was not added to an mRNA? a. The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm. b. The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell. c. The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule. d. The mRNA molecule would not be able to add the poly-A tail on its strand at the 5鈥 end.

Structure and function in biology result from both the presence of genetic information and the expression of that information. Some genes are continually expressed, whereas the expression of most genes is regulated, commonly at the level of transcription. At the initiation of transcription, the TATA-binding protein (TBP) provides access to the DNA strand to be transcribed. The 5鈥橳ATAAA3鈥 sequence called the TATA box is found in prokaryotes, archaebacteria, and eukaryotes. Even among eukarya, when the TATA box is not present among eukaryotes, the initiation of transcription involves TBP. Scientists attribute this common characteristic to the relative thermostability of the A-T interaction. Hydrogen bonds hold the two strands of the DNA double helix together. This type of bond has the smallest interaction energy of all intermolecular forces; as temperature increases, these bonds are broken. A. Explain the advantage, in terms of the energy required, which is provided by an AT-rich region in the sequence where transcription is initiated. B. The fact that the TATA box or the associated TBP are common to all domains provides evidence of common ancestry among all life. Pose a scientific question that would need to be addressed by a valid alternative explanation of this fact. C. A whole-genome survey of prokaryotes (Zheng and Wu, BMC Bioinformatics, 2010) showed that the relative amounts of guanine and cytosine in DNA poorly predicted the temperature range conditions that are suitable for an organism. Refine the question posed in part B, taking this result into account.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.