Chapter 16_Gene Mutation And DNA Repair

Question 1

Mutation

Answer 1

A heritable change in the genetic material.

Question 2

Point Mutation

Answer 2

A change in a single base pair within the DNA.

Question 3

Base Substitution

Answer 3

One base is substituted for another.

Question 4

Transition vs. Transversion

Answer 4

• Transition: A change of a pyrimidine to another pyrimidine (C to T), or a purine to another purine (A to G).
• Transversion: Purines and pyrimidines are interchanged (T to G).

Question 5

Silent Mutations

Answer 5

Those that do not alter the amino acid sequence of the polypeptide even though the nucleotide sequence has changed.

Question 6

Missense Mutations

Answer 6

Base substitutions in which an amino acid change does occur.

Question 7

Nonsense Mutations

Answer 7

Involve a change from a normal codon to a stop codon. This terminates the translation of the polypeptide earlier than expected, producing a truncated polypeptide.

Question 8

When a nonsense mutation occurs in a bacterial operon…

Answer 8

…it may also inhibit the expression of downstream genes. This phenomenon is called polarity.

Question 9

Frameshift Mutations

Answer 9

Involve the addition or deletion of a number of nucleotides that is not divisible by three. Because the codons are read in multiples of three, this shifts the reading frame. The translation of the mRNA then results in a completely different amino acid sequence downstream from the mutation.

Question 10

Except for silent mutations, new mutations are more likely to…

Answer 10

…produce polypeptides that have reduced rather than enhanced function.

Question 11

Why are Missense mutations less likely to alter function?

Answer 11

They involve a change of a single amino acid within polypeptides that typically contain hundreds of amino acids. When a missense mutation has no detectable effect on protein function, it is referred to as a neutral mutation.

Question 12

Give an example of a neutral mutation.

Answer 12

A missense mutation that substitutes an amino acid with a similar chemistry as the original amino acid is likely to be neutral or nearly neutral.

Question 13

Are silent mutations considered neutral mutations?

Answer 13

Yes.

Question 14

Up Promoter Mutations

Answer 14

Promoter mutations that increase transcription. Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations.

Question 15

Down Promoter Mutations

Answer 15

Causes the promoter to become less like the consensus sequence, decreasing its affinity for regulatory factors and decreasing the transcription rate.

Question 16

Reversion

Answer 16

Changes a mutant allele back to a wild-type allele.

Question 17

What is the effect of a mutation on a regulatory element/operator site?

Answer 17

May disrupt the ability of the gene to be properly regulated.

Question 18

What is the effect of a mutation on 5’-UTR/3’-UTR (the untranslated region of mRNA)?

Answer 18

May alter the ability of mRNA to be translated; may alter mRNA stability.

Question 19

What is the effect of a mutation on a splice recognition sequence?

Answer 19

May alter the ability of pre-mRNA to be properly spliced.

Question 20

Suppressors (Suppressor Mutations)

Answer 20

A second mutation at another site in the organism’s DNA may restore the normal growth rate, converting the mutant back to the wild-type phenotype.

Question 21

How are suppressors different from reversions?

Answer 21

A suppressor mutation differs from a reversion, because it occurs at a DNA site that is distinct from the first mutation.

Question 22

Intragenic Suppressor

Answer 22

When the second mutant site is within the same gene as the first. Often involves a change in protein structure that compensates for an abnormality in protein structure caused by the first mutation.

Question 23

Intergenic Suppressor

Answer 23

Suppressor mutation is in a different gene from the first mutation. Usually involve a change in the expression of one gene that compensates for a loss-of-function mutation affecting another gene. Alternatively, intergenic suppressors may involve proteins that participate in a common cellular pathway.

Question 24

How do the use of intergenic and intragenic suppressors in research differ?

Answer 24

• Intragenic suppressors are used to obtain information about protein structure and function.
• Intergenic suppressors are used to gain information about proteins that have similar or redundant functional roles, proteins that participate in a common pathway, multimeric proteins with two or more subunits, and the regulation of protein expression by transcription factors.

Question 25

Suppressor Mutations: Common Pathway

Answer 25

Two or more different proteins may be involved in a common pathway. A mutation that causes a defect in one protein may be compensated for by a mutation that alters the function of a different protein in the same pathway.

Question 26

Suppressor Mutations: Multimeric Protein

Answer 26

A mutation in a gene encoding one protein subunit that inhibits function may be suppressed by a mutation in a gene that encodes a different subunit. The double mutant has restored function.

Question 27

Suppressor Mutations: Transcription Factor

Answer 27

A first mutation causes loss of function of a particular protein. A second mutation may alter a transcription factor and cause it to activate the expression of another gene. This other gene encodes a protein that can compensate for the loss of function caused by the first mutation.

Question 28

What is a less common way of intergenic suppression?

Answer 28

Involves mutations in nonstructural genes that alter the translation of particular codons.

Question 29

Chromosomal Breakpoint

Answer 29

The region where two chromosome pieces break and rejoin with other chromosome pieces.

Question 30

A breakpoint within the middle of a gene is….

Answer 30

…very likely to inhibit gene function because it separates the gene into two pieces.

Question 31

Position Effect

Answer 31

A gene is left intact, but its expression may be altered when it is moved to a new location.

Question 32

How do position effects alter gene expression?

Answer 32

• One possibility is that a gene may be moved next to regulatory sequences for a different gene, such as silencers or enhancers, that influence the expression of the relocated gene.
• Alternatively, a chromosomal rearrangement may reposition a gene from a less condensed or euchromatic chromosome to a very highly condensed or heterochromatic chromosome. When the gene is moved to a heterochromatic region, its expression may be turned off. This second type of position effect may produce a variegated phenotype in which the expression of the gene is variable.

Question 33

Germ line

Answer 33

Cells that give rise to the gametes such as eggs and sperm.

Question 34

Germ-line mutation

Answer 34

Can occur directly in a sperm egg or egg cell, or it can occur in a precursor cell that produces the gametes.

Question 35

If a mutant gamete participates in fertilization…

Answer 35

…all cells of the resulting offspring will contain the mutation.

Question 36

When an individual with a germ-line mutation produces gametes…

Answer 36

…the mutation may be passed along to future generations of offspring.

Question 37

Geneticists categorize the cause of mutation in one of two ways.

Answer 37

Spontaneous and induced mutations.

Question 38

Examples of spontaneous mutations.

Answer 38

• Aberrant recombination: Abnormal crossing over may cause deletions, duplications, translocations, and inversions.
• Aberrant segregation: Abnormal chromosomal segregation may cause aneuploidy or polyploidy.
• Errors in DNA replication: A mistake by DNA polymerase may cause a point mutation.
• Transposable elements: Can insert themselves into the sequence of a gene.
• Depurination: On rare occasions, the linkage between purines and deoxyribose can spontaneously break, leading to mutation if not repaired.
• Deamination: Cytosine and 5-methylcytosine can spontaneously deaminate to create uracil or thymine.
• Tautomeric shifts: Spontaneous changes in base structure can cause mutations if they occur immediately prior to DNA replication.
• Toxic metabolic products: The products of normal metabolic processess, such as reactive oxygen species, may be chemically reactive agents that can alter the structure of DNA.

Question 39

Mutagens

Answer 39

Agents known to alter the structure of DNA which lead to mutations.

Question 40

Hot Spots

Answer 40

Certain regions of a gene that are more likely to mutate than other regions.

Question 41

Depurination

Answer 41

Involves the removal of a purine (adenine or guanine) from the DNA. The covalent bond between deoxyribose and a purine base is somewhat unstable and occasionally undergoes a spontaneous reaction with water that releases the base form the sugar, thereby creating an “apurinic site”.

Question 42

Deamination

Answer 42

The removal of an amino group from the CYTOSINE base. This produces uracil.

Question 43

Can depurination and deamination be corrected by DNA base excision repair enzymes? What happens if they can’t?

Answer 43

• Yes they can.
• If not, in depurination, because a complementary base is not present to specific the incoming base for the new strand, any of the four bases are added to the new strand in the region that is opposite the apurinic site.
• In deamination, if a DNA template strand has uracil instead of cytosine, a newly made strand will incorporate adenine into the daughter strand instead of guanine.

Question 44

What’s the difference between the deamination of cytosine and 5-methylcytosine?

Answer 44

• Deamination of cytosine produces Uracil

- Deamination of 5-methylcytosine produces Thymine.

Question 45

Tautomeric Shift

Answer 45

A temporary change in base structure. The tautomers are bases.

Question 46

Tautomers.

Answer 46

The tautomers are bases, which exist in keto and enol or amino and imino forms. These forms can interconvert by a chemical reaction that involves the migration of a hydrogen atom and a switch of a single bond and an adjacent double bond.

Question 47

What are the common stable tautomers of A, T, G, and C?

Answer 47

• G and T is the keto form

- A, C is the amino form

Question 48

What happens if one of the bases is in the enol or imino form?

Answer 48

Then they won’t obey the AT/GC rule of base pairing. Instead, hydrogen bonding will promote TG and CA base pairs.

Question 49

How does a tautomeric shift cause a mutation?

Answer 49

It must occur immediately prior to DNA replication. Basically after the DNA strand unwinds, but before anything else.

Question 50

Trinucleotide Repeat Expansion

Answer 50

(TNRE) A repeated sequence of three nucleotides readily increase in number form one generation to the next.

Question 51

Dynamic Mutation

Answer 51

(Anticipation) a phenomenon in which TNRE disorders progressively worsen in severity in future generation.

Question 52

How are TNRE disorders related to sex of parents?

Answer 52

Genomic imprinting. Dynamic mutation depends on whether the disease was inherited from the father.

Question 53

What does nitrous acid do?

Answer 53

Causes deamination.

Question 54

Ames Test

Answer 54

Uses strains of a bacterium that cannot synthesize the amino acid histidine. These strains contain a point mutation within a gene that encodes an enzyme required for histidine biosynthesis. The mutation renders the enzyme inactive, and the bacteria cannot grown on petri plates unless histidine has been added. However, a second mutation (reversion) may occur that restores the ability to synthesize histidine. Can cause a reversion back to wild type.

Question 55

What does the Ames test monitor?

Answer 55

The rate at which this second mutation occurs, thereby indicating whether an agent increases the mutation rate above the spontaneous rate.

Question 56

Direct repair

Answer 56

• An enzyme recognizes an incorrect alteration in DNA structure and directly converts it back to a correct structure.
• EX: UV light causes formation of thymine dimers. Photolyase recognizes the dimers and splits them, returns the DNA to its original condition. The repair mechanism requires light and is known as photoreactivation. This restores structure of DNA.
• EX: Protein alkyltransferase can remove methyl or ethyl groups from guanine bases that have been mutagenized by alkylating agents such as nitrogen mustard and EMS.

Question 57

Base excision repair and nucleotide excision repair

Answer 57

• An abnormal base or nucleotide is first recognized and removed from the DNA. A segment of DNA is excised, and then the complementary DNA strand is used as a template to synthesize normal DNA.

Question 58

Example of Base Excision Repair

Answer 58

• EX: Involves function of enzymes DNA N-glycosylases. This can recognize an abnormal base and cleave the bond between it and the sugar in the DNA backbone, creating an apurinic or apyrimidinic site. N-glycosylase recognizes a uracil within the DNA and cleaves (nicks) the bond between the sugar and base. This releases the uracil base and leaves behind an apyrimidinic site. This abnormality is recognized by a second enzyme AP endonuclease, which makes a cut on the 5’ side.
  NOW, Following this, three things can happen:
  
  • 1.) In some species (bacteria) DNA polymerase I (which has 5’ to 3’ endonuclease activity) removes a DNA segment containing the abnormal region and, at the same time, replaces it with normal nucleotides. This is nick translation (although DNA replication, NOT mRNA translation occurs).
  • 2.) DNA polymerase beta has the enzymatic ability to remove a site, which is missing a base, and then insert a nucleotide with the correct base in its place.
  • 3.) DNA polymerase can synthesize a short segment of DNA, which generates a flap. The flap is then removed by flap endonuclease.
    In ALL three cases the final step is carried out by DNA ligase that closes a gap in the DNA backbone.

Question 59

Example of Nucleotide Excision Repair

Answer 59

• In E. Cloi, NER system requires four key proteins, UvrA - UvrD plus DNA polymerase and ligase.
• Protein complex consisting of two UvrA molecules and one UvrB molecule tracks along the DNA in search of damaged DNA.
• When a damaged segment is identified, the two UvrA proteins are released, and UvrC binds to the site. The UvrC protein makes cuts in the damaged strand on both sides of the damaged site.
• After this, UvrD, which is a helicase, recognizes the region and separates the two strands of DNA. This releases a short DNA segment that contains the damaged region, and UvrB and UvrC are also released.
• Following this, DNA polymerase fills in the gap, using the undamaged strand as a template.
• Finally DNA ligase makes the final covalent connection between the newly made DNA and the original DNA strand.

Question 60

Mismatch Repair

Answer 60

Similar to excision repair except the DNA defect is a base pair mismatch, not an abnormal nucleotide. The mismatch is recognized, and a segment of DNA is removed. The parental strand is used to synthesize a normal daughter strand of DNA.

Question 61

How does mismatch repair known which strand has the incorrect base?

Answer 61

• Hemimethylation. Prior to DNA replication, the parental DNA is already methylated. After DNA replication, some time must pass before a newly made strand is methylated. Therefore, newly replicated DNA is hemimethylated (only the parent DNA strand is methylated). Hemimethylation provides a way for a DNA repair system to distinguish between the parental DNA strand and the daughter.

Question 62

How does mismatch repair work?

Answer 62

• Three proteins, MutS, MutL, and MutH detect the mismatch and direct the removal of the mismatched base from the newly made strand.
• Role of MutS is to locate mismatches.
• ONce one is found, MutS forms a complex with MutL. MutL acts as a linker that binds to MutH by a looping mechanism.
• This stimulates MutH, which is bound to a hemimethylated site, to make a cut in the newly made, nonmethylated DNA strand.
• After the strand is cut, MutU, which functions as helicase, separates the strands, and an exonuclease then digests the nonmethylated DNA strand in the direction of the mismatch and proceeds beyond the mismatch site.
• This leaves a gap in the daughter strand that is repaired by DNA polymerase and DNA ligase.

Question 63

Homologous Recombination Repair

Answer 63

Occurs at double-strand breaks or when DNA damage causes a gap in synthesis during DNA replication. The strands of a normal chromatid are used to repair a damaged chromatid.

Question 64

Nonhomologous End Joining

Answer 64

Occurs at double-strand breaks. The broken ends are recognized by proteins that keep the ends together; the broken ends are eventually rejoined.

Question 65

Explain HRR

Answer 65

• DSB is processed by the short digestion of DNA strands at the break site. This processing event is followed by the exchange of DNA strands between the broken an unbroken sister chromatids.
• The unbroken strands are used as templates to synthesize DNA in the region where the break occurred.
• Finally, the crisscrossed strands are resolved, they are broken and rejoined in a way that produces separate chromatids.

Question 66

Explain NHEJ

Answer 66

• Two broken ends of DNA are simply pieced back together.
• This requires the participation of several proteins that play key roles in the process.
• First, the DSB is recognized by end binding proteins. These then recognize additional proteins that form a cross bridge that prevents theh two ends from drifting apart.
• Additional proteins are recruited to the region that may process the ends of the broken chromosome by digesting particular DNA strands. This processing may result in the deletion of a small amount of genetic material from the region
• Finally, any gaps are filled in via DNA polymerase, and the DNA ends are ligated together.

Question 67

Pros and Cons of HRR and NHEJ

Answer 67

• HRR: An advantage is that HRR can be an error free mechanism for repairing a DSB. A disadvantage is that sister chromatids are available only during the S and G2 phases of the cell cycle in eukaryotes or following DNA replication in bacteria.
  NHEJ: Advantage is that it doesn’t involve the participation of a sister chromatid, so it can occur at any stage of the cell cycle. A disadvantage is that NHEJ can result in small deletions in the region that has been repaired.

Question 68

Transcription coupled DNA repair

Answer 68

Actively transcribed genes in eukaryotes and prokaryotes are more efficiently repaired following radiation damage than is nontranscribed DNA.

Question 69

What are the advantages of targeting of DNA repair enzymes to actively transcribing genes?

Answer 69

• Active genes are loosely packed and may be more vulnerable to DNA damage.
• DNA regions that contain actively transcribed genes are more likely to be important for survival than nontranscribed regions.

Question 70

Translesion synthesis (TLS)

Answer 70

The synthesis of DNA over a template strand that harbors some type of DNA damage.

Question 71

TLS vs. DNA polymerases

Answer 71

• TLS: Low fidelity, highly error prone replication
• DNA Polymerase: High fidelity, highly intolerant of geometric distortions imposed on DNA by the incorporation of incorrect nucleotides, and consequently, they incorporate wrong nucleotides with a very low frequency.

Question 72

SOS Resposne

Answer 72

Triggered by environmental factors. The result is up regulation of several genes that function to repair the DNA lesions.

Question 73

IF TLS is so error prone and harmful, why is it used?

Answer 73

Allows bacteria to survive under conditions of extreme environmental stress.