BSCI410-Liu/Spring 06 Exam #1 Feb. 23, 06

Size: px
Start display at page:

Download "BSCI410-Liu/Spring 06 Exam #1 Feb. 23, 06"


1 Your Name: Your UID# 1. (20 points) Match following mutations with corresponding mutagens (X-RAY, Ds transposon excision, UV, EMS, Proflavin) a) Thymidine dimmers b) Breakage of DNA backbone c) Frameshift d) Transition (G to A change) e) 8 bp insertion which is a repeat of adjacent endogenous sequence 2. Unlike human, plants grow outdoors and are under constant sunlight. Why don t plants get skin cancers? To answer this question, Dr. Liu has conducted an EMS mutagenesis screen and has identified several mutations in genes that control Arabidopsis plant s sensitivity to UV light. Two mutations (named uv1, uv2) caused increased sensitivity to UV irradiation, while two other mutations (named uv3, uv4) caused increased resistance to UV irradiation. (a) (4 points) What is the wild type function of UV1, UV2, UV3, and UV4? (Hint: simply state if each wild type gene promotes or inhibits UV resistance). (b) (4 points) Use uv1 as an example, describe a genetic cross to determine whether uv1 is recessive or dominant? (c) (4 points) How will you find out that uv1 and uv2 mutations are actually allelic to each other? Describe the genetic cross and the result from such a cross. 1

2 (c) (4 points) How will you find out that uv3 and uv4 are not allelic to each other (ie. reside in the same gene)? Describe the genetic cross and the result from such a cross. (d) (20 points) The four mutations, hence, represent three genes, UV1, UV3 and UV4. These three genes appear to act in a regulatory pathway that ultimately activate DNA repair. Such an increase in DNA repair could effectively repair UV-induced DNA damage, making plants more resistant to UV irradiation. The genetic epistatsis analysis was carried out and the results are summarized below: Gene name UV1: UV3: UV4: Single mutant phenotype More sensitive to UV irradiation More resistant to UV irradiation More resistant to UV irradiation Double mutant uv1 uv3: uv1 uv4: Phenotype More sensitive to UV irradiation More resistant to UV irradiation. What is the regulatory relationship among these three genes? Use arrow ( ) and bar ( ) to indicate their positive or negative regulatory activities. Start the pathway with UV irradiation and end the pathway with DNA repair. 3. (48 points) Multiple choices. Some questions may have more than one correct answer. (A) Which of the following vectors contain(s) the largest insert? (a) Cosmid; (b) YAC; (c) BAC; (d) plamid (B) Which of following are/is required for a cloning vector (a) DNA replication origin; (b) inverted repeats; (c) antibiotic resistance; (d) a-c 2

3 (C) During PCR amplifcation, which of following is/are NOT required (a) Reverse transcriptase; (b) dntp; (c) Taq polymerase; (d) primers (D) Sanger sequencing is based on (a) Chemical cleavage; (b) Nucleotide substitution; (c) ddntp-based termination (E) To construct cdna library, you will need to have: (a) A vector; (b) Reverse transcriptase; (c) mrna; (d) ligase; (e) a-d (F) Which of following mutations causes most severe damage to the gene function (a) Frameshift mutation; (b) silent mutation; (c) mis-sense mutation (G) Competent cells are cells that are (a) Capable of dividing; (b) infectious; (c) Able to take up DNA; (d) a-c (H) Transformation is a process when (a) Cells change shape; (b) plasmid vectors take up inserts; (c) foreign DNA is introduced into a host cell; (e) DNA integrates itself into the genome (I) An individual heterozygous for a dominant-negative mutation exhibits a phenotype similar to (a) An individual homozygous for a loss-of-function mutation in the same gene (b) An individual heterozygous for a gain-of-function mutation in the same gene (C) An individual heterozygous for a neomorphic mutation in the same gene (J) Allele-specific suppression describes the suppression of an existing mutation by a second site mutation (a) That changes a specific amino acid of an interacting protein (b) That changes any amino acid of an interacting protein (C) In the anticoden of a trna (if the existing mutation is a nonsense mutation) (K) A Contig is (a) Continuous string of DNA; (b) Overlapping pieces of DNA fragment; (c) random DNA pieces 3

4 (L) T-DNA is (a) A plant transposon; (b) a piece of DNA from agrobacteria Ti-plasmid; (c) a tool for introducing foreign DNA into plant cells 4. (21 points) Following is a sequencing gel. What is the DNA sequence of the synthesized strand? Label 5 and 3. dda ddg ddc ddt (added to each sequencing reaction) 5. (30 points) You are learning to utilize different molecular tools to accomplish different research goals. Pick one of these techniques (colony hybridization, PCR, restriction mapping, transformation, Northern analyses) to accomplish each of following goals: (a) Get enough DNA from a fossil (b) Introduce foreign DNA into E. coli (c) Identify the specific bacterial clone containing human insulin gene (d) Determine where an restriction enzyme cuts in a fragment of DNA (e) Determine the abundance and size of the mrna for the human insulin gene (f) Determine if someone has contracted HIV virus 4

5 6. (30 points) Contrast and compare (list 1 similarity and 1 major difference) (a) Shotgun sequencing vs. primer walking (b) cdna library vs. genomic DNA library (C) Southern blot vs. Northern blot (d) Ac vs. Ds transposon (e) Reverse Transcriptase vs. Transposase 5

6 (f) intragenic suppressor vs. extragenic suppressor 7. (15 points) Dr. Allen sequenced a piece of DNA and then entered the DNA sequence into a computer and asked the computer to analyze the restriction enzyme sites. The computer gave her following restriction map (numbers are the size of the fragments in KB). To verify the map made by the computer, she digested the DNA with EcoR1 (R1), then HindIII (H3), and then double digested with EcoRI (R1) and HindIII (H3). She then ran these digestions on a gel. Could you predict and draw the bands in the gel below? M is the size standard and the numbers next to the bands indicate the size of these bands in KB. R1 H3 R M R1 H3 R1+H