Lesson 9 Gene Mutation and DNA Repair

Question 1

Transition

Answer 1

A change of a pyramidine to another pyramidine (C to T) or a purine to another purine (A to G)

Question 2

Transversion

Answer 2

When a Purine is interchanged with a pyramidine

Question 3

What can point mutations include

Answer 3

1. Base substitution

2. A short sequence of DNA may be deleted or added to the chromosomal DNA

Question 4

Suppressors or suppressor mutations

Answer 4

Second mutation that affects the phenotypic expression of a first mutation. Acts to suppress the phenotypic effects of another mutation.

Question 5

What are the two common reasons for position effects?

Answer 5

1. Gene has been relocated next to regulatory sequences from another
2. Gene has been relocated to a chromosome that has a different packing structure.

Question 6

What three chemical changes can result in spontaneous mutations?

Answer 6

Depurination, damnation, and tautomeric shifts

Question 7

depurination

Answer 7

Covalent bond between a purine base breaks and creates an apurinistic site. It this occurs just prior to DNA replication, it can cause a mutation.

Question 8

demination

Answer 8

Primarily occurs on the cytosine base. If it occurs on a methylcytosine, it creates rhyming, making it difficult for DNA repair enzymes to distinguish the correct base from an altered base.

Question 9

tautomeric shifts

Answer 9

Changes between Kato and Enola, or between amino and imino forms of the bases. Though these shifts are rare and transient, they may cause mutations if they occur just prior to DNA replication.

Question 10

Experiment 16A: environmental agents as mutagens

hypothesis

Answer 10

The exposure of flies to X rays will increase the rate of mutation.

Question 11

Experiment 16A: environmental agents as mutagens

Starting material

Answer 11

Female flies contain one normal X chromosome and a CIB X chromosome. The male flies contain a normal X chromosome.

Question 12

Experiment 16A: environmental agents as mutagens

protocol

Answer 12

1. Expose male flies to X rays. Have control group that is not exposed to X rays.
2. Mate the male flies to female flies carrying one normal X chromosome and one CIB X chromosome
3. Save about 1000 daughters with bar eyes (these have CIB X from mother and X from father that may or may not have a recessive lethal mutation)
4. Mate each bar-eyed daughter with normal (nonirradiated) meals. This is done in 1,000 individual tubes.
5. Count the number of crosses that do not contain any male offspring. These crosses indicate that the bar-eyed female parent contained an X-linked lethal recessive mutation on the nonCIBX chromosome.

Question 13

Experiment 16A: environmental agents as mutagens

Interpreting the data

Answer 13

In the absence of X ray treatment, only 1 cross in approximately 1,000 was unable to produce male offspring. This means that the spontaneous rate for any X-linked lethal mutation was relatively low. By comparison, X-ray treatment of the fathers that gave rise to these CIB females produced 91 crosses without male offspring. These females inherited their non-CIB chromosome from irradiated fathers, the results indicate that X rays greatly increase the rate of X-linked, recessive lethal mutations.

Question 14

pericentric inversion

Answer 14

If the centromere lies within the inverted region of the chromosome, this is the inverted region

Question 15

Parametric inversion

Answer 15

If centromere is found outside the inverted region, the inverted region is parametric

Question 16

Two types of mutagens

Answer 16

Chemical and physical

Question 17

What can activate a mutagen inside the body?

Answer 17

Cellular enzymes such as oxidases

Question 18

Nitrous acid mutations

Answer 18

Nitrous acid replaces amino groups with keto groups (-NH2 to =O). This changes cytosine to uracil and adenine to hypoxanthine. When this mutated DNA replicates, the modified bases do not pair with the appropriate nucleotides in the newly made strand. Instead, uracil pairs with adenine, and hypoxanthine pairs with cytosine.

Question 19

What does nitrous acid change cytosine to?

Answer 19

Uracil

Question 20

What does uracil pair with?

Answer 20

adenine

Question 21

What does nitrous acid change adenine to?

Answer 21

hypoxanthine

Question 22

What does hypoxanthine pair with?

Answer 22

cytosine

Question 23

alkylating agents

Answer 23

alkylate bases within DNA. During alkylation, methyl or ethyl groups are covalently attached to the bases.

Question 24

acridine dyes

Answer 24

Interfere with DNA replication process. Contain flat planar structures that interchelate into the double helix by sandwiching between adjacent base pairs, thereby distorting the helical structure. When this DNA is replicated, single-nucleotide additions and/or deletions can be incorporated into the newly made daughter strand.

Question 25

5-bromouracil and 2-aminopurine

Answer 25

nucleotide base analogues that become incorporated into daughter strands during DNA replication. 5BU is a thymine analogue that can be incorporated into DNA instead of thymine. 5BU can base pair with adenine, but will also tautomerize and base-pair with guanine. This cause a mutation in which an AT base pair is changed to a G-5BU pair.

Question 26

UV light

Answer 26

nonionizing radiation. Penetrates only the surface of material such as skin. Known to cause DNA mutation. Causes formation of cross-linked thymine diners. A thymine diner within a DNA strand may cause a mutation when that DNA strand is replicated.

Question 27

How does Ames test work?

Answer 27

1. Suspected mutagen mixed with a rat liver extract and bacterial cells. Rat liver provides a mixture of enzymes that may potentially activate a mutagen.
2. After incubation period, large no of bacteria are plated on a minimal growth medium that does not contain histidine.
3. Salmonella is not expected to grow on these plates unless a mutation has occurred that allows a bacterium to synthesize histidine. To estimate the mutation rate, colonies that grown on the minimal media are counted and compared with the total no. Of bacterial cells that were originally streaked on the plate.

Question 28

photolyase

Answer 28

Yeast cells use this enzyme to repair thymine diners. It can split the diner to restore DNA to its original condition.

Question 29

O6-alkylguanine alkyltransferase

Answer 29

Removes the methyl or ethyl groups from guanine bases that have been mutagenized by agents such as nitrogen mustards and ethyl methane-suffocate. Alkyltransferase is permanently inactivated so it can only be used one.

Question 30

DNA-N-glycosylase

Answer 30

Involved in base excision repair. Can recognize an abnormal base and cleave the bond between it and the sugar in the DNA backbone. This releases base and leaves behind an apyrimidinic nucleotide.

Question 31

AP-endonuclease

Answer 31

Recognizes AP nucleotide and makes a cut on the 5’ side.

Question 32

Nucleotide Excision Repair

Answer 32

Can repair many different types of DNA damage. Several nucleotides in damaged strand are removed from the DNA, and the undamaged strand is used as a template to resynthesize a normal strand.

Question 33

What does nucleotide excision repair require in E. coli?

Answer 33

Four key proteins: UvrA, UvrB, UvrC, and UvrD, plus help of DNA polymerase and DNA ligase

Question 34

Steps of nucleotide excision repair in E. coli

Answer 34

1. UvrA/UvrB trimmer scans along DNA. When trimmer encounters a thymine diner in DNA, the UvrA diner is released.
2. UvrC binds to UvrB.
3. UvrC makes two cuts in the damaged DNA strand. One cut is made eight nucleotides 5’ to the site, the other approximately 5 nucleotides 3’ from the site.
4. UvrC is released. UvrD binds to the site
5. UvrD is a helicase that causes the damaged strand to be removed.
6. UvrB, C and D are released.
7. DNA polymerase resynthesize a complementary strand.
8. DNA ligase makes the final connection.

Question 35

Methyl-directed mismatch repair

Answer 35

Used by E. coli, this system detects the mismatch and specifically removes the segment from the newly made daughter strand.

Question 36

Steps of methyl-directed mismatch repair

Answer 36

1. MutS slides along the DNA and recognizes base mismatches in the double helix.
2. MutI binds to MutS and acts as a linker between MutS and MutH.
3. The DNA must loop for their interaction to occur.
4. MutH identifies the methylated strand of DNA, which is the nonmutated parental strand.
5. MutH makes a cut in the nonmethylated strand, and an exonuclease digests the strand until it passes the MutS/MutL region.
6. This leaves a gap, which is filled in by DNA polymerase and sealed by DNA ligase.

Question 37

What is the purpose of genetic recombination?

Answer 37

Repair gaps that can’t be replicated

Question 38

When does recombination happen?

Answer 38

While two DNA copies are being made

Question 39

Steps of recombination

Answer 39

1. Gap is replaced with the same region from the other parental DNA strand: a short segment of strand A replaces the corresponding segment of strand C. This removes the gap to strand C, but creates gap in strand A.
2. The next step is to fill in strand A gap, using normal strand B as a template.
   Net result: neither of the replicated DNA double helices contain gaps, but source of gap is still present in double helix.

Question 40

Why are actively transcribed genes in eukaryotes and prokaryotes more efficiently repaired following radiation damage compared with nontranscribed DNA?

Answer 40

There are proteins that target DNA repair systems to sites of damage in actively transcribing genes. The template strand rather than the coding strand is preferentially repaired.

Question 41

transcription-repair coupling factor (TRCF)

Answer 41

E coli. Protein. Responsible for targeting the excision repair system to actively transcribed genes that contain DNA damage. Recruits UvrA/UvrB complex to region for nucleotide excision repair.