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Discuss the effects of point mutations on a DNA strand. a. Mutations can cause a single change in an amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins. b. Mutations can cause a single change in amino acid. A missense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins. c. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Substitution mutations can cause a frame shift. This can result in non-functional proteins. d. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in functional proteins.

Short Answer

Expert verified
Option a is correct.

Step by step solution

01

- Understand Point Mutations

Point mutations are changes to a single nucleotide base in the DNA sequence. They can result in various effects such as a change in a single amino acid, or more dramatic changes like frame shifts and the introduction of stop codons.
02

- Examine Different Types of Point Mutations

There are several types of point mutations: 1. Missense mutations lead to a change in one amino acid in a protein. 2. Nonsense mutations introduce a premature stop codon, which can halt protein synthesis. 3. Insertion or deletion mutations result in a frame shift, altering the reading frame of the gene and potentially leading to a non-functional protein.
03

- Evaluate Each Option

Option a: Correctly describes that nonsense mutations can halt replication or reading and that insertion or deletion can result in a frame shift yielding non-functional proteins.Option b: Incorrectly states that missense mutation can halt replication or reading.Option c: Incorrectly associates substitution mutations with frame shifts, which are typically caused by insertion or deletion.Option d: Incorrectly suggests that insertion or deletion mutations can result in functional proteins, though they are more likely to lead to non-functional ones.
04

- Choose the Correct Answer

Based on the evaluations, the correct choice is option a.

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Key Concepts

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

nonsense mutation
A nonsense mutation is a specific type of point mutation where a single nucleotide change results in a premature stop codon within the coding sequence of a gene.
This effectively halts the translation process, resulting in an incomplete and typically non-functional protein.
Nonsense mutations can be particularly severe because they can lead to truncated proteins that are often unable to perform their intended functions.
This can disrupt cellular processes and potentially lead to serious consequences, like genetic disorders or diseases. Examples include Cystic Fibrosis and Duchenne Muscular Dystrophy, which are often caused by nonsense mutations in their respective genes.
Understanding nonsense mutations is important for comprehending how genetic variations can alter protein function and for developing possible treatments.
missense mutation
A missense mutation occurs when a change in a single nucleotide base leads to the substitution of one amino acid for another in the protein product.
This type of mutation can have a variety of effects depending on the location and nature of the amino acid change.
In some cases, the function of the protein may not be significantly affected if the new amino acid is similar in properties to the original.
However, in other cases, even a small change can result in a dysfunctional protein that can lead to diseases.
For example, Sickle Cell Anemia is caused by a missense mutation in the HBB gene, where the amino acid glutamic acid is replaced by valine, drastically changing the behavior of hemoglobin.
It is important to understand that missense mutations can vary widely in their impact, from benign to highly detrimental.
frame shift
Frame shift mutations are caused by insertions or deletions of a number of nucleotides in a DNA sequence that is not divisible by three.
This alters the reading frame of the gene, which is the way nucleotides are grouped into codons (sets of three bases coding for amino acids).
Because the reading frame is shifted, every codon downstream of the mutation will be misread during translation, often leading to a completely different and usually non-functional protein.
Frame shift mutations are particularly severe because they impact not just one amino acid, but the entire sequence of amino acids post-mutation.
An example of a condition caused by a frame shift mutation is Tay-Sachs Disease, which arises from the insertion of four nucleotides in the HEXA gene.
Recognizing the potential for frame shifts to result in drastic genetic changes is crucial for understanding genetic diseases and for developing genetic engineering solutions.

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Most popular questions from this chapter

Explain the events taking place at the replication fork. If the gene for helicase is mutated, what part of replication will be affected? a. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be separated at the beginning of replication. b. Helicase joins the DNA strands together at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone after the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be joined together at the beginning of replication. c. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby increasing the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. d. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by RNA polymerase to form a parent strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication.

Discuss how the scientific community learned that DNA replication takes place in a semi conservative fashion. a. Meselson and Stahl experimented with E. coli. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N.\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N\) This supports the semi-conservative replication model. b. Meselson and Stahl experimented with S. pneumonia. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the 1 \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N\) This supports the semi-conservative replication model. c. Meselson and Stahl experimented with E. coli. DNA grown in \(^{14} N\) was heavier than DNA grown in \(^{15} N\) When DNA in \(^{15} N\) switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N.\) This supports the semi-conservative replication model. d. Meselson and Stahl experimented with S. pneumonia. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N.\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating complete presence of \(^{14} N.\) This supports the semi conservative replication model.

Which of the following would be a good application of plasmid transformation? a. to make copies of DNA b. to isolate a change in a single nucleotide c. to separate DNA fragments d. to sequence DNA

Describe the Sanger DNA sequencing method used for the human genome sequencing project. a. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPS’s. b. A DNA sample is denatured by heating and then put into four tubes. A primer, RNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPS’s. c. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is removed from the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPS’s. d. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four deoxynucleotides (dNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent dNTP is added the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different dNTPS’s.

What is bacterial transformation? a. The transformation of a bacterium occurs during replication. b. It is the transformation of a bacterium into a pathogenic form. c. Transformation of bacteria involves changes in its chromosome. d. Transformation is a process in which external DNA is taken up by a cell, thereby changing morphology and physiology.

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