Assignment 13. In the Griffith experiment, why did mice die when injected with live R bacteria plus heatkilled

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1 Assignment Multiple-choice (1 point) In the Griffith experiment, why did mice die when injected with live R bacteria plus heatkilled S bacteria? Some of the S bacteria were still alive. The R bacteria had mutated to become virulent. The R bacteria had taken up the virulency factor from the dead S bacteria. The virulency factor in dead S bacteria was sufficient to kill the mice. The mice died from a cause unrelated to pneumonia. 2. Multiple-choice (1 point) Before beginning an experiment involving bacterial cultures, a recommended method for sterilizing the work area is flaming the work surface with a wire loop cleansing the work surface with ethanol alcohol washing the work surface with soap and water pouring alcohol on the work surface, then setting it on fire wiping the work surface with a dry paper towel 3. Multiple-choice (1 point) In an experiment involving bacterial cultures, what is the purpose of streaking? It is a way to suspend bacteria in liquid media of a tube. It is a way to sterilize the wire loop used for bacterial transfer. It is a way to spread bacteria over the entire surface of the agar in a Petri dish. It is a way to spread bacteria in a small region of a Petri dish so that individual colonies are more easily seen. It is a way to grow large numbers of bacteria in liquid media. 4. Essay (3 points) In which section of the dish shown in the simple transformation experiment would you expect transformation of bacteria to occur? Write a short essay that explains your answer.

2 5. Multiple-choice (1 point) In the above experiment, why was there no growth in dish sector RL? The bacteria in this sector were killed before placing them on the agar. The bacteria in this sector have been transformed. The bacteria in this sector have not been transformed. The bacteria in this sector are not resistant to streptomycin. The bacteria in this sector are slower growing, so cannot be seen yet. 6. Essay (2 points) In step 2 of the experiment, bacteria are transferred from the S, R, and S + RL sections of the dish to a new dish containing nutrients plus streptomycin. Students performing this experiment are told that if growth occurred in the sector RL of step 1, they were not to use the S+RL bacteria from their own plates but to borrow a sample of this culture from students obtaining the successful results in the photographs. Write a short essay that explains why students should not use their S + RL culture if the RL culture in their dish grew. 7. Multiple-choice (1 point) Which of the following best explains the final results (step 3) of the experiment? Bacteria grew in R because this strain was streptomycin-resistant and in S + RL because the S bacteria were transformed. Bacteria grew in S because this strain was streptomycin-resistant and in S + RL because the R bacteria were transformed. Bacteria grew in R and in S + RL because the bacteria in both sectors were transformed. Bacteria grew only in S + RL because only these bacteria were transformed. Bacteria grew in all sectors because the S bacteria were transformed. 8. Essay (2 points) Mistakes are sometimes made in experiments. Write a short essay that indicates how the dish in step 3 would look if the following mistakes had been made: a) S bacteria (instead of R bacteria) were used during the detergent/heating step. b) The wire loop was not sterilized between bacterial transfers during the experiment.

3 9. True/false (2 points) Indicate whether each statement is true or false by writing true or false in each blank. If a bacterium contained the pglo plasmid, which of the following could it do? Survive in an environment containing streptomycin. Survive in an environment containing ampicillin. Emit a green glow if placed under an ultraviolet light. Pass the GFP gene to other bacteria if killed in their presence. 10. Multiple-choice (1 point) In the GFP experiment, bacteria took up the pglo plasmids when they were grown on a medium containing ampicillin put under an ultraviolet light lysed with a detergent killed by boiling frozen, then briefly heated 11. Multiple-choice (1 point) Looking at set up for the GFP experiment (in which the dishes are numbered), which dish(es) will contain transformed bacteria? Dish 1 only Dish 2 only Dish 3 only Dishes 1 and 2 Dishes 3 and Multiple-choice (1 point) The function of arabinose in the GFP experiment is to Provide a way to control the expression of the GFP gene. Allow the bacteria to grow in a nutrient-free environment. Allow the bacteria to grow in the presence of ampicillin. Provide a way to insert the GFP gene into the bacteria. None of these is the function of arabinose.

4 13. Essay (2 points) Study the results of the GFP experiment, then write a short essay that answers the following questions: a) Why is there no growth in the LB/AMP dish on the top row? b) Why is there more growth in the LB dish than in the LB/AMP dish beneath it (on the bottom row)? 14. Multiple-choice (1 point) At the end of the GFP experiment, what two conditions are required for bacteria to fluoresce under ultraviolet light? They must contain the pglo plasmid and grow in the presence of ampicillin. They must contain the pglo plasmid and grow in the presence of arabinose. They must grow in the absence of both ampicillin and arabinose. They must have lost the pglo plasmid and grow in the presence of arabinose. They must have lost the pglo plasmid and grow in the presence of ampicillin. 15. True/false (2 points) At the end of the GFP experiment, (1) one of the fluorescent colonies is suspended in a large flask of growth medium and incubated. Then (2) the bacteria are lysed, and (3) the bacterial suspension is poured through a chromatography column. Indicate whether each statement is true or false by writing true or false in the blanks. Fluorescent colonies were suspended in growth media and incubated to increase the number of transformed bacteria. Bacteria were lysed to remove bacteria that had not been transformed. The purpose of column chromatography was to obtain a pure sample of GFP from the bacterial suspension. The matrix of the column bound everything in the bacterial suspension except the GFP. For questions 16 and 17: The next two questions require you to locate a point mutation in the DNA that codes for miacalcin, a small protein (32 amino acids) that is used to stimulate bone formation in menopausal women. The amino acid sequence of the protein is listed below.

5 5 Cys- Ser- Asn- Leu Ser- Thr Cys- Val- Leu- Gly- Lys- Leu- Ser Gln- Glu- Leu- His- Lys- Leu Gln- Thr- Tyr- Pro- Arg- Thr Asn- Thr- Gly- Ser- Gly- Thr- Pro Before you begin work on the problems, use your codon table to fill out this table to show the mrna codons for each amino acid found in miacalcin. (Remember that some amino acids have more than one codon.) Position in Amino the protein Acid 1 Cys 2 Ser 3 Asn 4 Leu 6 Thr 8 Val 10 Gly 11 Lys 14 Gln 15 Glu 17 His 22 Tyr 23 Pro 24 Arg codons Below are 3 different mutated DNA segments that originally coded for the indicated 3 regions of Miacalcin. For each segment, you must determine the mutation that occurred and the amino acids that would be coded for by the new mrna. There is only one small change, point mutation, in each of the mutated DNA fragments. The change may be an addition, deletion, or simply a change from one nucleotide to another. The mutation in segment 1 serves an example to help you solve the problem. You must do the work yourself to find the mutations in segments 2 and 3. (Segment 1) Mutated DNA, positions 1 6 in the miacalcin protein ACG, AGT, TTC, GGA, GAG, CTG, A

6 How to tackle this problem: First determine the mrna that would result from the mutated DNA. Sequence of mrna: UGC, UCA, AAG, CCU, CUC, GAC, U Next determine the new amino acid sequence that would be translated from this mrna. New amino acid sequence for segment 1: cys, ser, lys, pro, leu, asp Notice that the amino acid sequence differs from the normal protein at the third amino acid which is lys (lysine) instead of asn (asparagine). The codons for lysine are AAA and AAG. The codons for asparagine are AAU or AAC. So it looks like there was a change in the DNA that resulted in an A or G to be substituted for a U or C in the 3 rd codon. Also note that the mutated DNA is one nucleotide longer than the original which suggests that an extra nucleotide has been inserted somewhere in this segment. Since the last letter in the 3 rd DNA triplet is C, a C was probably inserted into the original DNA fragment at this point. This would result in an AAG codon and shifting of the triplet code in codons 4-6. To verify that this is the correct answer, take the C out of the mutated fragment and rewrite the DNA, mrna and amino acid sequence. DNA: ACG, AGT, TTG, GAG, AGC, TGA mrna: UGC, UCA, AAC, CUC, UCG, ACU amino acid sequence: Cys-Ser-Asn-Leu-Ser-Thr This is the correct amino acid sequence (see first segment of the Miacalcin peptide), so the answer must be correct: a C nucleotide was inserted at the end of the 3 rd triplet in the DNA. 16. Essay (3 points) (Segment 2) Mutated DNA, positions 7-13 ACA, CAC, GAC, CCT, ATC, GAT, AGT Write the new m RNA: Write the new amino acid sequence: Indicate whether the mutation was a nucleotide insertion, deletion, or substitution and where in the DNA fragment it occurred. Then explain how the mutation affected the amino acid sequence in segment Essay (3 points) (Segment 3) Mutated DNA, positions GTC, TCA, ATG, TGT, TCG, AG

7 Write the new m RNA: Write the new amino acid sequence: Indicate whether the mutation was a nucleotide insertion, deletion, or substitution and where in the DNA fragment it occurred. Then explain how the mutation affected the amino acid sequence in segment 3.