Lecture 14, Chapter 18: Mutations (Dr. Bittel Study Guide) Flashcards

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1
Q

What is the difference between a transition and a transversion? Which type of base substitution is usually more common?

A

Transition mutations are base substitutions in which one purine (A or G) is changed to the other purine, or a pyrimidine (T or C) is changed to the other pyrimidine. Transversions are base substitutions in which a purine is changed into a pyrimidine or vice versa. Although transversions would seem to be statistically favored because there are eight possible transversions and only four possible transitions, about twice as many transition mutations are actually observed in the human genome.

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2
Q

Briefly describe expanding nucleotide repeats. How do they account for the phenomenon of anticipation?

A

Expanding nucleotide repeats occur when DNA insertion mutations result in an increase in the number of copies of a nucleotide repeat sequence. Such an increase in the number of copies of a nucleotide sequence may occur by errors in replication or unequal recombination. Within a given family, a particular type of nucleotide repeat may increase in number from generation to subsequent genereation, increasing the severity of the mutation in a process called anticipation.

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3
Q

What is the difference between a missense mutation and a nonsense mutation? Between a silent mutation and a neutral mutation?

A

A base substitution that changes the sequence and the meaning of a mRNA codon, resulting in a different amino acid being inserted into a protein, is called a missense mutation. Nonsense mutations occur when a mutation replaces a sense codon with a stop (or nonsense) codon.
A nucleotide substitution that changes the sequence of an mRNA codon, but not the meaning is called a silent mutation. In neutral mutations, the sequence and the meaning of an mRNA codon are changed. However, the amino acid substitution has little or no effect on protein function.

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4
Q

Briefly describe two different ways in which intragenic suppressors may reverse the effects of mutations.

A

Intragenic suppression is the result of second mutations within an already mutated gene that restore a wild-type phenotype. The supressor mutations are located at different sites within the gene from the orginal mutation. One type of suppressor mutation restores the original phenotype by reverting the meaning of a previously mutated codon to that of the original codon. The suppressor mutation occurs at a different position than the first mutation, which is still present within the codon. Intragenic suppression may also occur at two different locations within the same protein. If two regions of a protein interact, a mutation in one of these regions could disrupt that interaction. The suppressor mutation in the other region would restore the interaction. Finally, a frameshift mutation due to an insertion of deletion could be suppressed by a second insertion or deletion that restores the proper reading frame

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5
Q

How do insertions and deletions arise?

A

Strand slippage that occurs during DNA replication and unequal crossover events due to misalignment at repetitive sequences have been shown to cause deletions and additions of nucleitides to DNA molecules. Strand slippage results from the formation of small loops on either the template or the newly synthesized strand. If the loop forms on the template strand, then a deletion occurs. Loops formed on the newly syntesized strand result in insertions. If during crossing over, a misalignment of the two strands at repetitive sequence occurs, then the resuolution of the crossover will result in one DNA molecule containing an insertion and the other molecule containing a deletion.

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6
Q

How do base analogs lead to mutations?

A

Base analogs have structures similar to the nucleotides and if present, may be incorporated into the DNA during replication. Many analogs have an increased tendency for mispairing, which can lead to mutations. DNA replication is required for the base analog-induced mutations to be incorporated into the DNA.

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7
Q

How do alkylating agens, nitrous acid, and hydroxylamine produce mutations?

A

Alkylating agents donate alkyl groups (either methyl or ehtyl) to the nucleotide bases. The addition of the alkyl group results in mispairing of the alkylated base and typically leads to transition mutations. Nitrous acid treatment results in the deamination of cytosine, producing uracil, which pairs with adenine. During the next round of replication, a CG to AT transition will occur. The deamination of guanine by nitrous acid produces xanthine. Xanthine can pair with either cytosine or thymine. If paired with thymine, then a CG to TA transition occurs. Hydroxylamine works by adding a hydroxyl group to cytosine, producing hydroxylaminocytosine. The hydroxylaminocytosine has an increased tendency to undergo tautomeric shifts, which allow pairings with adenine, resulting in GC to AT transtions.

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8
Q

What is the purpose of the Ames test? How are his bacteria used in this test?

A

The Ames test allows for rapid and inexpensive detection of mutagenic and potentially carcinogenic compounds using bacteria. The majority of carcinogenic compounds result in DNA damage and are mutagens. The increase in reversion rate of his bacteria to his is used to detect the mutagenic potential of the compound being tested

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9
Q

What general characteristics are found in many transposable elements?

A

Most transposable elements have terminal inverted repeats and are flanked by short direct repeats that are generated at insertion sites during the transposition process. Many also contain a gene encoded one of the enzymes necessary for transposition (transposase or reverse transcriptase)

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10
Q

How does a retrotransposon move?

A

A retrotransposon relocated through an RNA intermediate. First, it is transcribed into RNA. A reverse transcriptase encoded by the retrotransposon then reverse the RNA template into a DNA copy of the transposon, which then integrates into a new location in the host genome

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11
Q

What are some differences between Class I and Class II transposable elements?

A

Class I transposable elements utilize replicative transposition through an RNA intermediate. Class II trasnposable elements utilize either replicative or conservative transposition and transpose through a DNA intermediate

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12
Q

Why are transposable elements often called genomic parasites?

A

Because they may not have an apparent benefit to a cell or organism and may be harmful to it. They exist because they are efficient at replicating and spreading. This does not preclude the possibility that transposable elements may serve an important role in genome plasticity, and evolution

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13
Q

List at least three different types of DNA repair and briefly explain how each is carried out.

A

Mismatch repair.

Direct repair:

Base exision repair:

Nucleotide exision repair:

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14
Q

Mismatch repair

A

Replication errors that are the result of base-pair mismatches are repaired. Mismatch-repair enzymes recognize distortions in the DNA structure due to mispairing and detect the newly synthesized strand by the lack of methylation on the new strand. The bulge is exised and DNA polymerase fills the gap and DNA ligase seals the repair

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15
Q

Direct Repair

A

DNA damage is repaired by directly changing the damaged nucleotide back to its original structure. As for example the repair done by a photolyase or methyl transferase.

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16
Q

Base exision repair

A

After the damaged base is removed by glycosylases, the phosphodiester bond is exised by AP endonucleases and other enzymes remove the deoxyribose sugar; then the entire nucleotide is replaced by DNA polymerase and the nick is sealed by DNA ligase

17
Q

Nucleotide exision repair

A

Repair enzymes recognize distortions of the DNA double helix. Damaged regions are exised by enzymes (nucleases and helicases), which separate the strands of DNA and cut phosphodiester bonds on either side of the damaged region. The gap generated by the exision step is filled in by DNA polymerase and sealed by DNA ligase

18
Q

What are the two major mechanisms for repair of double strand breaks? How do they differ?

A

Homologous recombination and non homologous end joining. Homologous recombination requires an identical DNA molecule (usually a sister chromatid) to repair a double strand break. Non-homologous end joining does not require a template

19
Q

A codon that specifies the amino acid Gly undergoes a single-base substitution to become a nonsense mutation. In accord with the genetic code given in Figure 15.10, is this mutation a transition or a transversion? At which position of the codon does this mutation occur?

A

By examining the four codons that encode Gly, CGU. GGC, GCA, and GGG and the three nonsense codons, UGA, UAA and UAG we can determine that only one of the Gly codons, GGA could be mutated to a nonsense codon by the single substitution of a U for a G at the first position:

GGA–>UGA
Because uracil is a pyrimidine and guanine is purine, the mutation is a transversion

20
Q

If a single transition occurs in a codon that specifies Phe, what amino acids can be specified by the mutated sequence?

CUC (Ser), UCC

A

Two codons can encode for Phe: UUU and UUC. A single transition could occur at each of the positions of the codon resulting in different meanings

Original codon: (Ser) UUU (Phe)

UUU CUU (Leu), UCU (Ser)m UUC (Phe)

UUC If a single transversion occurs in a codon that specifies Phe, what amino acids can be sepecified by the mutated sequence? Mutated codon (amino acid encoded) AUU (Ile), UAU (Tyr), UUA (Leu), GUU (Val), UGU (Cys)