Chapter 18 Mutations and DNA Repair

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

- Homologous recombination. If you have a break in one homolog, it can use the other as a template for repair - Non-Homologous joining: Two different chromosomes that have breaks shove chromosomes in any way possible to make some whole ones. This is a last ditch effort to protect the end of a chromosome, it is random and causes serious translocations and inversions.

How are breaks in the chromosomes repaired?

Base analogs have structures similar to 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

How do base analogs lead to mutations?

- Base Analogs: Chemicals with structures similar to one of the bases, that can be incorporated into DNA, causes misspairings - Alkylation, Deamination, Oxidation: Changes the chemistry of the wild type bases, making them similar to base analogs - Intercalation: Dyes that attach (shove there way into) to DNA bases, can cause insertions/frameshift mutations over time - Radiation: UV, Gamma

What are the causes of Induced mutations? (chemically induced only) (4)

-A transition is a base substitution in which a purine is replaced by a purine, or a pyrimidine is replaced by a pyrimidine. - A transversion is a base substitution in which a purine is replaced by a pyrimidine, and a pyrimidine is replaced by a purine. - The number of possible transversions is twice the number of possible transitions, but transitions are more common because its easier to transform a purine to another purine, rather than a purine to a pyrimidine.

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

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

What is the difference between class I and class II transposable elements?

The following nucleotide sequence is found in a short stretch of DNA. 5'- ATGT-3' 3'- TACA-5' If this sequence is treated with hydroxylamine, what sequences will result after replication?

5'-ATAT-3' 3'-TATA-5'

After observing the codons for Gly and the stop codons, we see that only one of the stop codons could be derived from the substitution of a single base in Gly. It is a transversion (G->U) of the first position of the Gly codon GGA to the stop codon UGA

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, is this mutation a transition or a transversion? At which position of the codon does the mutation occur?

a. 5'- ATTCGAACTGAC{Trans. element}TGACCGATCA-3' b. 5'- ATTCGAA{Trans. element}CGAACTGACCGATCA-3'

A particular transposable element generates, flanking direct repeats that are 4 bp long. Give the sequence that will be found on both sides of the transposable element if this transposable element inserts at the position, indicated on each of the following sequences. a. 5'- ATTCGAACTGAC{}CGATCA-3' b. 5'- ATTCGAA{}CTGACCGATCA-3'

- Expanding nucleotide repeats are mutations in which the number of copies of a set of nucleotides increases - There is an association between the number of repeats and the severity of the disease and the expansion of the repeats. Within a family a particular nucleotide repeat may cause increased nucleotide repeats from generation to generation, increasing the severity of the mutation as it gets passed on, this is anticipation.

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

No. Hydroxylamine cannot reverse nonsense mutations. Hydroxyl amine modifies cytosine-containing nucleotides and can only result in GC to AT transition mutations. In a stop codon, the GC to AT transition will only result in a different stop codon

Can nonsense mutations be reversed by hydroxylamine? Why or why not?

A mutation caused by insertion of either an Ac or Ds element in the pigment producing gene causes the corn kernel to be colorless. During development of the kernel, the Ac and Ds element may transpose out of the pigment gene, restoring a functional gene sequence. Therefore, pigment production in the cell and its descendants will be restored. - If the excision of the transposable element occurs early in the kernel development, the kernel will have a few large sectors of purple pigment. If the excision occurs later, the kernel will have relatively smaller sectors or specks of purple pigment.

Explain how Ac and Ds elements produce variegated corn kernels.

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 excised and DNA polymerase fills the gap and DNA ligase seals the repair.

How is mismatch repair carried out?

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

How is nucleotide excision repair carried out?

a. UUU= CUU(Leu), UCU (Ser), UUC (Phe) UUC= CUC(Leu), UCC (Ser), UUU (Phe) b. UUU= AUU (Ile), UAU (Tyr), UUA +UUG (Leu), GUU(Val), UGU(Cys) UUC= AUC (Ile), UAC (Tyr), UUA +UUG (Leu), GUC (Val), UGC (Cys)

Refer to the genetic code image and answer the following questions. a. If a single transition occurs in a codon that specifies Phe, what amino acids can be specified by the mutated sequence? b. If a single transversion occurs in a codon that specifies Phe, what amino acids can be specified by the mutated sequence?

Normal amino acid sequence: Met- Thr- Gly- Asn- Gln- Leu- Tyr- Stop a. Met- Thr- Gly- Ser- Gln- Leu- Tyr- Stop b. Met- Thr- Ala- Ile- Asn- Tyr- Ile c. Met- Thr- Gly- Asn- His- Leu- Tyr- Stop d. Met- Thr- Thr- Gly- Asn- Gln- leu- Tyr- Stop

The following nucleotide sequence is found on the template strand of DNA. First, determine the amino acids of the protein encoded by this sequence by using the genetic code provided. Then give the altered amino acid sequence of the protein that will be found in each of the following mutations. 3'-TAC,TGG,CCG,TTA,GTT,GAT,ATA,ACT-5' a. A transition at nucleotide 11 b. A one-nucleotide deletion at nucleotide 7 c. A T-> A transversion at nucleotide 15 d. An addition of TGG after nucleotide 6

a. Base substitution, missense mutation b. single base substitution of Arg, nonsense mutation c. A deletion caused a frameshift mutation d. deletion, caused a in-frame deletion e. Insertion, In-frame insertion

The polypeptide has the following amino acid sequence Met- Ser- Pro- Arg- Leu- Glu- Gly Given the following mutant sequences what possible mutations(single base substitution, insertion, deletion) could have caused the mutated sequence, and what is their phenotypic effects of the mutation?(nonsense, missense, frameshift) a. Met- Ser- Ser- Arg- Leu- Glu- Gly b. Met- Ser- Pro c. Met- Ser- Pro- Asp- Trp- Arg- Asp- Lys d. Met- Ser- Pro- Glu- Gly e. Met- Ser- Pro- Arg- Leu- Lue- Glu- Gly

- A Somatic mutations arise in somatic tissues which do not produce gametes. When they undergo mitosis, the mutation gets passed on to the daughter cells, leading to a population of genetically identical cells. The earlier the development of this mutation the larger the mutant clone of cells. - Germ-line mutations ultimately arise in cells that produce gametes, and can be passed to future generations, in their germ-line and somatic cells.

What is the difference between a somatic and germ line mutation?

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

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

D

Which of the following base changes in DNA is an example of a transition A. G-to-C B. A-to-C C. A-to-T D. A-to-G E. C-to-A

A

Which of the following enzyme activities is not a part of nucleotide-excision repair? A. Reverse transcriptase B. All of the above are part of nucleotide-excision repair. C. DNA helicase D. DNA polmerase E. DNA ligase

E

Which of the following types of mutations does NOT lead to a change in the amino acid sequence of the gene product? A. Loss-of-function B. Nonsense C. Missense D. Neutral E. All of the options lead to a change in amino acid.

C

__________ mutations produce new activities and are usually dominant. A. induced B. Lethal C. Gain-of-function D. Forward E. Spontaneous

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

How does a retrotransposon move?

- Breaks chromosomes - deletions - Duplications - Inversions - Translocations

What are the effects of Ionization radiation (X-rays) on DNA?

-The original mutation arose from a transversion of the second base of the codon UCA, C->G - The suppressor arose from another transversion of the third position of the codon UGA to either UGU or UGC

A gene has the following amino acid sequence Met- Lys- Ser- Pro- Ala- Thr- Pro A nonsense mutation caused by a single base pair substitution occurs, resulting in a protein with the amino acid sequence Met-Lys. An intergenic suppressor mutation allows the gene to produce the full-length protein. With the two mutations the gene now produces the protein with the following amino acid sequence Met- Lys- Cys- Pro- Ala- Thr- Pro Give the location and the nature of the original mutation and of the intergenic suppressor.

The transposon would only be able to transpose if another transposon of the same type were in the cell and able to recognize the inverted repeats and transpose its own element as well as other nonautonomous copies of the transposon with the same inverted repeats.

An insertion sequence contains a large deletion in its transposase gene. Under what circumstances would this insertion sequence be able to transpose?

C

Assume that a base-pair substitution mutation converts a DNA triplet (AAT) to another DNA triplet (AAA). A second mutation now changes the AAA triplet to the GAA triplet. (UUA and CUU code for leucine and UUU codes for phenylalanine.) This second mutation is an example of a(n) A. intergenic suppressor. B. loss-of-function mutation. C. intragenic suppressor. D. frameshift. E. transversion.

- An intragenic repressor is a mutation that takes place on the same gene as a previous mutation, undoing the effects of the first mutation. 1. The suppressor may may change a second nucleotide on the same codon as the first missense mutation, producing the amino acid that would have been produced had there been no mutation. 2. In the event of a deletion or insertion that causes a frameshift mutation, the intragenic suppressor may add or delete a base, restoring the reading frame.

Briefly describe two different ways in which intragenic repressors can reverse the effects of mutations?

A transversion in position two of the codon

Hemoglobin is a complex protein that contains four polypeptide chains. The normal hemoglobin found in adults- called adult hemoglobin- consists of two alpha and two beta polypeptide chains, which are encoded by different loci. Sickle cell hemoglobin, which causes sickle cell anemia, arises from a mutation in the beta chain of adult hemoglobin. Adult hemoglobin and sickle cell hemoglobin differ in a single amino acid: sixth amino acid from one end of adult hemoglobin is glutamic acid, whereas sickle cell hemoglobin has a Val in this position. After consulting the genetic code provided, indicate the type and location of the mutation that gave rise to sickle cell anemia.

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

How do insertion and deletions arise?

True. This is due the the exceptional repair and detection mechanisms of our DNA

True or False Spontaneous mutation rate is low

- Probability of mutation: induction (chemical/radiation) or spontaneous - Likelihood of repair - Likelihood of detection (actually observing the mutation)

What are mutation rates affected by? (Mutations per unit(cell division, gamete, genes replicated))

- mispairing (due to wobble) - Replication errors (due to incorporated mispaired base) - Strand slippage (essentially insertion, but also causes deletions) - Unequal crossing over (mispairings, cause unaligned homologs, causing unequal crossing over, resulting on one with an insertion and one with a deletion) Chemical changes: - Depurination( loss of a purine via bond break) - Deamination ( the loss of an amino group from a base )

What are the causes of spontaneous mutations? (6)

-Homologous recombination -Non-homologous recombination

What are the two major repair mechanisms for double stranded breaks?

- Alkylating agents donate alkyl groups (either methyl or ethyl) to the nucleotide bases. The addition of the alkyl group, results in mispairing of the alkylated bases and typically leads to transition mutations. - Nitrous acid treatment results in the deamination of cytosine, producing, producing a uracil, which pairs with adenine. During the next round of replication, a CG to AT will occur. The deamination, of guanine by nitrous acid produces a xanthine. Xanthine can pair with either cytosine or thymine, then a CG to TA transition can occur. - 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 transitions.

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

Some areas can sustain a mutation, like when it helps the organism survive in its environment. This may induce more mutations.

How do mutation rates change due to adaptive mutations?

After the damaged base is removed by the glycosylases, the phosphodiester bond is excised 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.

How is Base-excision repair carried out?

-Hydrolysis of cytosine ( misspairing during replication) - Thymine Dimers (2 bonds between T bases): This blocks transcription, and DNA replication (this is worse as the polymerase stalls, leading to apoptosis of the whole cell).

What are the effects of UV radiation on DNA?

they are reversible and complementary.

What characteristics do sequences that are found on the ends of an insertion sequence have?

- Terminal inverted repeats -flanked by short direct repeats that are generated at insertion sites during the transposition process - Many also contain a gene encoding one of the enzymes necessary for transposition.

What general characteristics are found in many transposable elements?

A mutation that occurs in the gene other than the one that had the original mutation, thus suppressing it. For example if a nonsense mutation is seen in the mRNA, a mutation in the tRNA that codes for a specific amino acid could make it so that it's anticodon is still able to bind to the stop codon, and instead of translation terminating the tRNA would translate it as a different amino acid.

What is an intergenic suppressor mutation? Give an example.

- A base substitution that results in a different amino acid in a protein is a missense mutation, while a nonsense mutation, is a base substitution that results in the coding of a nonsense codon, or a stop codon, thus terminating translation. - A silent mutation changes a codon to a synonymous codon that codes for the same amino acid. A neutral mutation is a missense mutation that alters the type of amino acid coded, but the amino acid has similar chemical properties as the other and does not affect the function of the protein

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

- Loss of function mutations causes a complete or partial absence of protein function, but they are recessive, the individual diploid organism must be homozygous for the mutation before the loss of function of the protein is exhibited. - Gain of function mutations, causes the cell to produce a protein or gene product whose function is not normally present. These are usually dominant, a single copy of the mutation leads to a new gene product.

What is the difference between loss of function mutations, and gain of function mutations?

- Lethal levels: significant DNA damage - Sub-lethal: Increased somatic mutations, effectively no germline mutations. (ex. Hiroshima)

What is the difference between the effects of lethal radiation levels and sub-lethal levels?

- Cumulative effect, as radiation dose increases so does the number of mutations. Whether it is one large dose at once, or an accumulation over time.

What is the relationship between radiation and mutation?

The following nucleotide sequence is found in a short stretch of DNA 5'-AG-3' 3'-TC-5' a. Give all the mutant sequences that can result from spontaneous depurination in this stretch of DNA b. Give all the mutant sequences that can result from spontaneous deamination in this stretch of DNA

a. The strand contains two purines (A,G), depurination results in the replacement of the purines by adenine 5'-AA-3' 3'-TT-5' b. If A is deaminated 5'-GG-3' 3'-CC-3' If C is deaminated 5'-AA-3; 3'-TT-5'


Ensembles d'études connexes

Unit 1 - Regulation of Investment Advisers

View Set

human nutri study guide for final

View Set

Eating Disorders Practice Questions (Test #2, Fall 2020)

View Set

Programming Languages: Final Exam Review

View Set

intro to public speaking mid term

View Set

An Update on Demineralization/Remineralization CE

View Set

Chapter 19: Documenting and Reporting

View Set