BSCI 410-Liu Homework#1 Spring 07. Due: Tuesday (Feb 13) at 11:00 AM in class

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1 BSCI 410-Liu Homework#1 Spring 07 Due: Tuesday (Feb 13) at 11:00 AM in class KEY 1. (4 points) Following is a list of mutational changes. For each of the specific mutations indicate which type of mutation best describes the change. Sometimes, more than one term could be used to describe each of the following statement (such as transition, transversion, insertion, deletion etc.). (1) The sequence of AAGCTTATCG is changed to AAGCTATCG. DELETION (AND/OR FRAMESHIFT) (1) (2) The sequence AACGTTATCG is changed to AATGTTATCG. TRANSITION (1) (3) The sequence AAGCTTATCG is changed to AAGCTTTATCG. INSERTION (AND/OR FRAMESHIFT (1) (4) The sequence AACGTCACACACACATCG is changed to AACGTCACATCG. DELETION (1) 2. (4 points) Mutate the AAG (Lys) codon one base at a time to illustrate missense, nonsense, silent, and neutral mutations. AAG=lysine MISSENSE: AUG (start), etc. (1) NONSENSE: UAG (1) SILENT: AAA (lysine) (1) NEUTRAL: AGG (arginine) (1) 3. (6 points) What effect on DNA could each of following mutagens cause? (1) UV irradiation thymine-thymine dimmer (1) (2) X-ray double stranded DNA break causing inversion, deletions or other rearrangements (1) (3) EMS chemical change (alkylating agent) causing transitions (1) 1

2 (4) Proflavin intercalation causing insertions and deletions (frameshifts) (1) (5) Nitrous acid deaminating agent causing transitions (1) (6) Transposon insertion inserts different nucleic acids into a gene and can disrupt frame, change encoding, introduce stops, alter splicing, etc. (1) 4. (5 points) Design a mutagenesis screen to isolate mutations involved in the biosynthesis of aspartic acid in yeast? 1. Mutagenize (1) 2. Grow in complete media (1) 3. Transfer (replica-plating) to minimal media (1) 4. Those that fail to grow in minimal media (MM) will be tested in the next step (1) 5. Test for growth in MM supplemented with aspartic acid to confirm. Those that fail to grow in MM but can now grow in MM supplemented with aspartic acid will be the mutants we want (1) 5. (8 points) Briefly compare and contrast following pairs of terms: (A) Auxotroph and Prototroph Auxotroph: strain requires nutrients/growth factors not normally needed by WT cells for viability (1) Prototrophs: strain which does not require any special growth factors (usually WT), can grow on MM (1) 2

3 (B) Complete medium and Minimal medium Complete medium: contains all nutrients necessary for growth (1) Minimal medium: contains only a carbon source (usually glucose), inorganic salt, and water. Can only support WT (autotrophy) (1) (C) Mutagenesis screen and Mutagenesis selection Mutagenesis screen: both WT and mutant organisms are present (i.e. survive) and require one to look through all of them to identify the mutant of interest (1) Mutagenesis selection: conditions set up so only the mutant of interest will survive (use death instead of growth to select) (1) (D) Retrotransposons and Transposons (list three differences) Retrotransposons: use an RNA intermediate to transpose to a new location; has a poly-a tail at one or LTR at both ends; still maintain the original (starting) retrotransposon (i.e. in crease in copy number rapidly); encodes reverse transcriptase (1) Transposons: require no RNA intermediate; has short inverted repeats at the end; may excise itself and insert into a new location (no net gain of the transposon); encode/require transposase (i.e. cut & paste (1) 6. (5 points) You have isolated five recessive maize mutants (d1-d5) which are defective in pigment biosynthesis. While wild type kernels are Red (R) in color; these d1-d5 mutant kernels are Yellow (Y) in color. To test if any of these mutations are allelic to each other, you did pairwise crosses among all five mutations and observed F1 phenotypes, which are summarized in the table below. How many genes are represented by these five mutations? How are these five mutations grouped into different genes? d1 d2 d3 d4 d5 d1 Y R R R R d2 Y R Y R d3 Y R Y d4 Y R d5 Y (R: red; Y: Yellow) Total of 3 genes (2) d1 (different gene from the rest) (1) d2=d4 (same gene) (1) d3=d5 (1) 3

4 7. (12 points) Wild type mouse with normal pigmentation is grey in color. Albino means white (no pigment at all). The following mutations (m1, m2, m3, and m4) affecting pigment synthesis in mouse are identified. They all affect the activity of the same gene. Given the following genotype as well as corresponding phenotype, could you describe the type of mutations using following terms: "recessive-null", "dominant-gain-offunction", "dominant-negative", "hypomorphic", "hypermorphic", and "neomorphic". Each situation listed below could sometimes be described by more than one term: (1) m1/+ grey mouse m1/m1 albino mouse recessive-null, hypomorph, or loss-of-function (3) (2) m2/+ albino mouse m2/m2 albino dominant negative or haploid-insufficient (3) (3) m3/+ Black (darker than normal) m3/m3 Black dominant gain-of-function or hypermorph (3) (4) m4/+ Redish brown (unusual color) m4/m4 Redish brown dominant gain-of-function or neomorph (3) 8. (10 points) Ethylene is a plant hormone that facilitate fruit ripening. Two types of tomato mutants were isolated that either could not ripen even when exposed to ethylene treatment or could ripen even without ethylene treatment. To determine the epistatic relationship among these mutations, double mutants were constructed and their phenotypes were observed and summarized below. Gene names never red (nrd): Single mutant phenotype: Form green tomatos even after ethylene treatment never ripen (nrp): Form green tomatos even after ethylene treatment always ripen (ar): Form red tomatos even without ethylene treatment (nrd, and nrp are defective in different genes). Gene names nrd ar: nrp ar: Double mutant phenotype: Form green tomatos even after ethylene treatment Form red tomatos even without ethylene treatment What is the regulatory relationship among these three genes? (Draw the regulatory pathway starting with Ethylene and ending with tomato ripening. Use arrows to indicate positive regulations and bars ( )to indicate negative regulations. 4

5 Ethylene NRP AR NRD tomato ripening 9. (8 points) Briefly explain what type of transposon each of the following is: (1). SINE from human Retrotransposon found in humans; require reverse transcriptase; leaves behind a footprint after hopping out; non-ltr-type retrotransposon (2) (2). P-element from drosophila Has inverted repeats which allows for transposase recognition; requires transposase; DNA-type transposon (2) (3). Ac/Ds from maize Ds (non-autonomous, dissociative) requires transposase from Ac to incorporate into genome, leaves behind a footprint after jumping out; DNA-type Ac (autonomous, activator) encodes transposase, DNA-type transposon, leaves footprint; cut & paste (2) (4). T-DNA from agrobacteria, stands for transfer DNA from agrobacteria. Has a piece of DNA residing in Ti (Tumor-inducing) plasmid in Agro. T-DNA can be transferred into plant cell and integrates into plant genome. It is an important vehicle/vector for genetic engineering in plants (2) 5

6 10. (8 points) Dr. Liu isolated five carnation flower mutants, which are named p1, p2, p3, p4, and p5. While the wild type flower has 5 petals, all these p1-p5 mutant flowers have 10 petals. (a) Dr. Liu first wants to determine if these p1-p5 mutations are recessive or dominant. How should she proceed with the cross? What is the likely outcome of the cross if these p1-p5 mutations are recessive? (You can use p1 as an example to illustrate the cross). Cross p1-p5 into WT to produce heterozygous plants (p1-p5/+) (2) If F1 (first) generation has a mutant phenotype (10 petals), then the mutation is dominant (2) p1/+ 10 petals = dominant p1/+ 5 petals = recessive (b) Dr. Liu then proceed to perform the complementation tests among these p1 to p5 mutations. Briefly describe the cross and the likely outcome if all five mutations reside in the same gene. Cross each homozygous mutant to one another to get F1 and see if the mutant phenotype is complemented in F1 (or saved). If they are, then they reside on different genes. If they fail to complement, then they are on different genes (2) If F1 progeny show WT (5 petal), they are in two different genes. If F1 progeny show 10 petals (mutant), then p1 & p2 reside on same gene (2) p1 X p2 p1/p2 or p1/+; p2/+ OR Cross these mutants to one another (pairwise). F1 show mutant phenotype (10 petals) they reside on the same gene. If F1 show WT phenotype (5 petals), they reside on 2 different genes. (4) 6