lecture 16- chapter 11

Question 1

Mutation rates

Answer 1

• DNA level: 10^-9 per replicated bp. Similar across different organisms
• Phenotypic level: 10^-6 to 10^-8. Varies across organisms
• Hot spots: genes with higher mutation rate mainly associated with large gene (i.e DYS gene)

Question 2

Fluctuation test

Answer 2

evaluated the nature of bacterial mutations that produced resistance to bacteriophage infection

for the RMH, the mutations occur spontaneously before exposure to T1, random fluctuation

for AMH, the environmental condition (T1 addition) triggers the mutation, similar number of resistant bacteria

Question 3

random mutation hypotheis

Answer 3

predicted that different bacterial cultures would develop resistance mutations at different times, yielding variable numbers of resistant bacteria

results showed great variation in the numbers of bacteriophage-resistant bacteria in the cultures, supporting this hypothesis

Question 4

Adaptive mutation hypothesis

Answer 4

predicts that all populations carry approx. the same proportion of phage-resistant cells. This would have been supported if the number of bacteriophage-resistant bacteria in each culture were about equal.

Question 5

germ-line mutations

Answer 5

• mutations that occur in germ-line cells, giving rise to sperm and egg
• can be passed from one generation to the next

Question 6

somatic mutations

Answer 6

cells not in the germline are somatic. Somatic cells divide by mitosis and only direct descendants of the original mutated cell will carry the mutation.

Question 7

Point mutations

Answer 7

occur at a specific, identifiable position in a gene or a specific location anywhere else in the genome

Question 8

Synonymous mutation

Answer 8

• coding sequence mutation

- a bp change that does not alter the resulting aa due to the redundancy of the genetic code

Question 9

Missense mutation

Answer 9

• coding sequence mutation

- a bp change that results in an aa change in the protein

Question 10

Nonsense mutation

Answer 10

• coding sequence mutation

- a bp change that creates a stop codon in a place of a codon specifying an aa

Question 11

Frameshift mutation

Answer 11

• coding sequence mutation
• insertion or deletion of one or more bp resulting in the addition or deletion of mRNA nucleotides, which may alter the reading frame of the message
• the wrong aa sequence is produced starting at the point of mutation; premature stop codons may also be produced

Question 12

Base pair substitution mutations

Answer 12

the replacement of one nucleotide bp by another

Question 13

transition mutations

Answer 13

one purine replaces another or one pyrimidine replaces another

Question 14

Transversion mutations

Answer 14

a pyrimidine is replaced by a purine or vice versa (exchanging a one ring and a two ring)

Question 15

Regulatory mutations

Answer 15

• some point mutations alter the amount (but not the aa) of protein product produced by a gene
• these regulatory mutations affect regions such as promoters, introns, and the regions coding for 5’- UTR and 3’-UTR

Question 16

Promoter mutation

Answer 16

• mutations that alter consensus sequence of nucleotides of promoters
• these interfere with efficient transcription initiation b/c they may not get identified and bound as they should
• some promoter mutations cause mild to moderate reductions in transcription level, whereas other may abolish transcription

Question 17

Splicing mutations

Answer 17

• efficient splicing of introns from mRNA requires specific sequences at either end of the intron
• these can result in splicing errors and the production of mutant proteins due to the retention of intron sequences in the mRNA

Question 18

Cryptic splice site mutations

Answer 18

• some bp substitution mutations produce new splice sites that replace or compete with authentic splice sites during mRNA processing.
• this leaves 19 additional nucleotides of the intron in the mRNA

Question 19

Polyadenylation mutations

Answer 19

• mutation of the polyadenylation signal sequence at the 3’ end of eukaryotic mRNA can block the proper processing of mRNA
• this occcurs in a rare mutant form of the human alpha-globin gene, leading to severe reduction in functional a-globin protein produces

Question 20

True reversion

Answer 20

• very rare

- WT DNA seq. is restored by a second mutation within the same codon, could be a change of a diff. nucletoide

Question 21

Intragenic reversion

Answer 21

very rare, occurs through second mutation elsewhere in the same gene to restore the reading frame

Question 22

Second site reversion

Answer 22

• occurs by mutation in a different gene that compensates for the original mutation, restoring the organism to WT. This happens often.
• also known as suppressor mutations because the second mutation “suppresses” the mutant phenotype caused by the first mutation
• good approach to map/discover genes involve in specific processes

Question 23

how do mutations occur?

Answer 23

arise in cells without exposure to agents capable of inducing mutation (mutagens). They arise primarily through errors in DNA replication or spontaneous changes in the chemical structure of a nucleotide base.

Question 24

DNA replication has very high fidelity due to

Answer 24

• the accuracy of DNA polymerases
• the proofreading ability of DNA polymerases
• the efficiency of mismatch repair

Question 25

Strand slippage

Answer 25

• causes alteration in number of DNA repeats
• The DNA polymerase of the replisome temporarily dissociates from the template and a portion of newly replicated DNA forms a temporary hairpin. Resumption of replication leads to re-replication of some of the repeats and an overall increase in the number of repeats on the daughter strand

Question 26

Trinucleotide repeat expansion disorders

Answer 26

involve strand slippage mutations that cause some hereditary diseases in humans and other organisms. WT alleles of the genes in question normally have a variable number of repeats, increases in the number of repeats beyond a certain threshold cause the disorders.

Question 27

Non-complementary base pairing

Answer 27

• similar to a third-base wobble, sometimes occurs during DNA replication and is called non-watson-and-crick bp.
• these include G with T pairing or A with C pairing; this type of mispairing is identified as incorporated error
• without repair, replication of the incorporated error converts it into a mutation in an event called replicated error

Question 28

Depurination

Answer 28

the loss of a purine (A,G) from a nucleo. by breaking the covalent bond linking the nucleo. base to the sugar. A lesion of this type is called an apurinic site (AP). Most AP sites are repaired before replication by BER; if left unrepaired, DNA pol will usually compensate by putting an A into the daughter strand site during replication

Question 29

Deamination

Answer 29

the loss of an amino group (NH2) from a nucleo. base. I.e when C is deaminated, an oxygen atom takes its place, converting the C into U. DNA mismatch repair removes the U from the DNA and replaces it with C, restoring the WT sequence.

Question 30

Deamination of methylated cytosine

Answer 30

when methylated C is deaminated, a T base is produced, which is now paired with G. If repair does not occur, the G-T pair remains and replication will produce two sister chromatids, one with the mutant A-T pair (b/c of the T strand as template) and one with the WT G-C pair.

Question 31

induce mutations

Answer 31

are produced by mutagens in an experimental setting to study the types of damage caused, the mutation process itself, or repair responses to damage

Question 32

chemical mutagens

Answer 32

• can increase the mutation rate >1000X
• can act as base analogs or base modifiers and lead to transitions
• intercalating agents (b/w two strands) modify the structure of the double helix and can induce frameshift or deletion mutations

Question 33

Photoproducts

Answer 33

• aberrant structures with additional bonds b/w nucleotides caused by UV radiation
• one common photoproduct is a thymine dimer, formed by covalent bonds b/w the 5 & 6 carbons of adjacent thymines
• a second is called a 6-4 photoproduct formed by a covalent bond b/w the 6 carbon on one thymine and the 4 carbon on the other

Question 34

consequences of photoproducts

Answer 34

• UV-induced damage can be addressed by photoreactive repair which can identify and correct most pyrimidine dimers. Those that are not repaired cause disruption of replication. Such disruptions leads to mutations, these are the primary cause of the stong association b/w excessive UV exposure and skin cancer

Question 35

Base excision repair (BER)

Answer 35

• multistep process leading to the removal of an incorrect or damaged DNA base and repair by synthesis of a new strand segment.
• DNA glycosylases recognize and remove modified bases, creating an AP site. AP endonucleose creates a SS “nick” near the AP site. Then the DNA pol remove and replaces several nucleo. of the nicked strand by nick translation

Question 36

Nucleotide excision repair (NER)

Answer 36

-removal of a strand segment containing DNA damage and replacement by new DNA synthesis (UV-repair in prokaryotes and eukaryotes)

i. e of NER
- UV-damage repair uses proteins encoded by the genes uvrA,uvrB,uvrC,uvrD. uvrA and uvrB proteins bind to the DNA strand opposite the photoproduct, then UVRA dissociates and uvrB denatures the DNA around the lesion. UVRC joins UVRB, forming the UVRBC complex; UVRC then cleaves the damaged DNA strand about 4 or 5 nucleo. to the 3’ & 5’ sides of phosphoproduct. UVRD helps remove the DNA fragment containing the photoproduct, then the DNA polymerase fills the gap and DNA ligase seals the sugar-phosphate backbone.

Question 37

Mismatch repair

Answer 37

• removal of a DNA bp mismatch by excision of a segment of the newly synthesized strand followed by resynthesis of the excised segment
• if mismatched nucleotides escapre the DNA pol proofreading then they may be detected and repaired by mismatch repair
• repair enzymes distinguish b/w the original, correct nucleo. and the new, mismatched nucleo. using the presence of methylation on the original strand

Question 38

Hemimethylated DNA region

Answer 38

consensus sequence recognized by MutH

Question 39

DNA mismatch repair by MutS and MutH in E.coli

Answer 39

1) the E.coli protein MutH binds to the hemimethylated DNA region
2) MutS locates and binds to the DNA mismatch and then forms a complex with MutL. The MutS/L complex binds to MutH
3) The MutH protein breaks a phosphodiester bond on the 5’ side of the guanine of a GATC sequence on the unmethylated daughter strand. Exonuclease enzymes digest nucleotides from the “nick” through the mismatched nucleotide
4) DNA pol fills the gap in the daughter strand. DNA ligase completes the repair. Dam methylase methylates the adenine of the GATC sequence on the daughter strand

Question 40

Which proteins do we have the equivalent of

Answer 40

MutL and MutS