Name_____________________________________&&Student&ID_____________________________________&Page&1&General Genetics, Biol 283 Prof Walker Fall 2014 Homework 1 (50 points) 1. (3 points) Consider the experiments carried out by Griffith using Streptococcus pneumoniae. When Griffith injected mice with the virulent strain of S. pneumoniae, the mice died. When he injected them with the avirulent strain, they lived. When he heated the virulent strain to kill it, and then injected it, the mice lived. We can tabulate these data as follows: Strain Treatment Procedure Result Virulent none inject into mice dead mice Avirulent none inject into mice live mice Virulent heat treatment inject into mice live mice a.) (1 points) The key experiment performed by Griffith was to mix heat-treated virulent cells with untreated avirulent cells. What did Griffith probably think would result when he injected the mixture into mice? Explain briefly—use 1-2 sentences. (NOTE—I do NOT want to know what actually happened. I want to know what Griffith thought would happen.) b.) (1 points) What hypothesis did Griffith make to explain what he actually observed as a result from the experiment in part a? Be brief: use 2-4 sentences to explain the hypothesis. c.) (1 points) In a SINGLE sentence, explain why Griffith’s experiment does not demonstrate that DNA is the genetic materialName_____________________________________&&Student&ID_____________________________________&Page&2&2. (4 points) Hershey and Chase used radiolabeled compounds to physically trace the genetic material as it moved from bacteriophage into E. coli cells. a. (1 point) BRIEFLY outline how a phage infects E. coli. A diagram may work best. Be sure to point out the place/time where/when the genetic material is transferred. b. (2 points) In Hershey and Chase’s experiment, what was the evidence that genetic material had moved from the phage into the E. coli cells? Be specific. You can correctly answer in one-two sentences. c. (1 point) Hershey and Chase used a centrifuge to separate the phage (or at least their empty husks) from the E. coli cells. The E. coli formed a ‘pellet’ at the bottom of the tube, while the phage remained in the supernatant (the liquid part). If the phage were labeled with 35S, which would be radioactive in H&C’s experiment—the supernatant or the pellet?Name_____________________________________&&Student&ID_____________________________________&Page&3&3. (5 points)Name_____________________________________&&Student&ID_____________________________________&Page&4&4. (8 points)To study DNA repair mechanisms, geneticists isolated Neurospora crassa mutants that were sensitive to various types of radiation; for example, mutants that were more sensitive than the WT strain to UV light. Ten haploid mutants were isolated. a. (3 points)How would you do the complementation tests for these mutants? Please be specific about what you would do to determine phenotypes. b. (3 points) Based on the complementation chart below, how many genes are defined by these mutations? “+“ indicates complementation, “-“ indicates failure to complement. 1 2 3 4 5 6 1 - - + - + + 2 - + - + + 3 - + - + 4 - + + 5 - + 6 - c. (2 points) State which mutant(s) belong to each complementation group that you observed. NOTE: every mutant belongs to a complementation group, so tell me about all six mutant strains.Name_____________________________________&&Student&ID_____________________________________&Page&5& 5. (8 points) A mutant screen is carried out in N. crassa for mutants that are unable to synthesize the amino acid valine (VAL). A number of mutants are isolated and classified into four groups according to their ability to grow (+) or not grow (-) in minimal medium supplemented with possible intermediates 3-hydroxy-3-methyl-2-oxobutanoate (HMOB), 2,3-dihydroxy-3-methylbutanoate (DMB), 2-acetolactate (AL), 2-oxisovalerate (OV), and valine (VAL). The data are shown in the table below. HMOB DMB AL OV VAL Strain 1 - - - - + Strain 2 - - - + + Strain 3 - - - - + Strain 4 + + - + + Strain 5 - + - + + Strain 6 - + - + + Strain 7 + + + + + Strain 8 + + - + + Complete the diagram shown below. Each arrow indicates one or more biochemical reactions. Within each circle write the class(es) of mutants (1-8) whose products contribute to the reactions symbolized by the arrow, and in the squares write the name of the intermediate (HMOB, DMB, AL, OV or precursor) at that position in the pathway. HINT: This problem is very like 1.28 in your textbook. Check that problem and its answer for practice.Name_____________________________________&&Student&ID_____________________________________&Page&6&6. (6 points) The picture below shows the restriction map of a DNA fragment that is 1500 bp in length. Draw bands in the appropriate places in gel lanes 2-5 below: Lane 1 Molecular weight standards: 2 kb, 1.5 kb, 1 kb, 800 bp, 600 bp, 400 bp, 200 bp, 100 bp. Lane 2 Undigested fragment Feldmann T-DNA left border Lane 3 Fragment digested with EcoRV Lane 4 Fragment digested with EcoRV and NruI Lane 5 Fragment digested with AluIName_____________________________________&&Student&ID_____________________________________&Page&7&7. (8 points) Below is the sequence of a small gene. The start and stop codons in this gene are indicated in bold face. There is a small intron in this gene, shown as underlined sequence. As you can see, there is a polymorphism in the intron that distinguishes the two alleles shown. Allele 1: 5’ GGTCAAAGCA ACCAAACACA TAAAAGAGAG ATTTAATACA AAAGAAAGAG AAAAAAGAAA GATATGGCAG GACTCATCAA CAAGATCGGA GACGCACTCC ACATTGGAGG AGGCAACAAG GAAGGTGAGC ACAAGAAGGA AGAGGAACAC AAGAAACACG TTGACGACCA CAATCAAAGA CAAGATCCAC GGTGGTGAAG GTAAAAGCCA CGACGGAGAA GGCAAAAGCC ACGACGGTGA GAAGAAAAAG AAGAAGGACA AGAAGGAGAA GAAACATCAT GATGATGGTC ACCACAGCAG CAGCAGTGAC AGCGACAGCG ATTAAGGTGA GGAAGTGAGG AGGATCGCTT GAATAAAACA GATCTGGTTC TGGCTATTAT TAATTAATGT TGCTGTATGT TCTTATCATC TTAGAGAGAG GTTAAAGACA GGAGAACCG 3’ Allele 2: 5’ GGTCAAAGCA ACCAAACACA TAAAAGAGAG ATTTAATACA AAAGAAAGAG AAAAAAGAAA GATATGGCAG GACTCATCAA CAAGATCGGA GACGCACTCC ACATTGGAGG AGGCAACAAG GAAGGTGAGC ACAAGAAGGA AGAGGAACAC AAGAAACACG TTGACGTCCA CAATCAAAGA CAAGATCCAC GGTGGTGAAG GTAAAAGCCA CGACGGAGAA GGCAAAAGCC ACGACGGTGA GAAGAAAAAG AAGAAGGACA AGAAGGAGAA GAAACATCAT GATGATGGTC ACCACAGCAG CAGCAGTGAC AGCGACAGCG