Manoa BIOL 406 – What type of DNA repair is most

11 Jun Manoa BIOL 406 – What type of DNA repair is most

Question
1. What type of DNA repair is most likely to be used to repair a TG mispairing in DNA?

A. Base excision repair

B. Nucleotide excision repair

C. Mismatch excision repair

2. Choose the pair that does not match.

A. Homologous recombination: Holliday junction

B. NHEJ: double stranded break repair

C. Dominant mutation: loss of function

D. crossing over: meiosis

3. Which of these parts is/are required to be considered a virus?

A. capsid

B. envelope

C. genome

D. both A and B

E. both A and C

4. A group of volunteers is screened for mutations leading to a disease that appears at birth and persists through adulthood. 10 volunteers tested positive for the mutation but only 8 have the disease phenotype. Which term best explains this observation?

A. codominant

B. heterozygous

C. variable penetrance

D. recessive

5. Why are yeast tetrads particularly useful for examining recessive mutations?

A. the tetrads have 4 copies of the genome so even if tetrads have 2 defective alleles they can survive.

B. the tetrads are haploid so recessive mutations can be observed easily

C. the tetrads divide more rapidly and allow expansion of the colony in a shorter time

D. the tetrads have some chromosomes in 2 copies and some in 1 copy so you can simultaneously examine heterozygous and homozygous loci.

6. When would you need to generate a temperature sensitive mutation in yeast?

A. When the mutation is lethal.

B. When you are studying yeast cellular temperature regulation

C. Any time you are doing a tetrad analysis

D. When you want to transform the cells using heat shock

7. We perform a complementation test on mutations x, y, and z. Individuals with x and y mutations are normal, individuals with x and z mutations have the mutant phenotype, individuals with y and z mutations are normal. What can we conclude from this complementation test?

A. x, y, and z are mutations in the same gene

B. y and z are mutations in the same gene

C. x and y are mutations in the same gene

D. x and z are mutations in the same gene

8. Two different mutations (A and B) lead to inability to produce tryptophan. Mutation A leads to accumulation of biosynthetic intermediate 1, mutation B leads to accumulation of biosynthetic intermediate 2. When these mutations are combined we see accumulation of biosynthetic intermediate 2. What should we conclude about the epistatic relationship of these mutations?

A. mutation A is in a gene whose product acts earlier in the biosynthetic pathway

B. mutation B is in a gene whose product acts earlier in the biosynthetic pathway

9. You are trying to map a dominant mutation, T, with a phenotype of no tail and you find linkage to a dominant mutation that causes white coat color in mice, W. You perform the following cross:

T w t w

——– X ———

t W t w

You get the following phenotypes. How many centimorgans apart are the T and W loci?

24: no tail, black

3: tail, black

40: tail, white

5: no tail, white

A. 11.1 cM

B. 12.5 cM

C. 62.5 cM

D. 88 cM

10. When we use homologous recombination to do a gene deletion (knock-out), why do we include the Neo gene?

A. To select for cells that have integration of our construct (desired DNA)

B. To select against cells that have integrated in a random location

C. To mark the cells and allow screening for cells with integrations

D. To increase the rate of homologous recombination