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UNIT 4 SAMPLE PROBLEMS Note: This problem set is not intended to be comprehensive, nor should you memorize these particular questions and answers. They are sample problems are designed to give you an idea of what questions will look like on the exam. As always, complete preparation requires mastering the material covered by the learning objectives. (4-‐3; Ap)1. One milliliter of a bacterial suspension containing 5 X 107 cells/mL is transferred to sterile nutrient broth. The generation time is 60 minutes. How many cells will there be after 3 hours of maximum growth? c. 4.0 X 108 cells CALCULATION: Nt = N0 X 2n = 5 X 107 cells X 23 = 4.0 X 108cells (Note: the answer could also be given in cells/mL because only one mL was used. (4-‐3; Ap)2. At 10:00 AM, you determine a cell density of 4 X 108 cells/mL for a culture grown under optimal conditions. Prior work has shown that this microbe grown under identical conditions has a 40-‐minute generation time. Using this information, what was the cell density at 8:00 AM? d. 5.0 X 107 cells/mL n
CALCULATION: N = N X 2 t
0
(4-‐5;T) 3. (4-‐7;T) 4. (4-‐8;T) 5. (4-‐8;K) 6. (4-‐13;T)7. (4-‐13;T)8. (4-‐12;T)9. "
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REARRANGE: N0 = # = = 5.0 X 107 cells/mL %
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Which IS NOT true of BOTH open and closed system growth? BOTH !sigmoid growth curve. c. produce a t!
ypical When turbidity decreases below the desired level in a chemostat, what should be done? b. No medium should be added until the culture grows and reaches the desired turbidity. Shown are three agar deep tubes, which have been inoculated with different bacteria and incubated for 48 hours. Which bacteria should grow in an aerobic environment (assuming all other metabolic requirements are satisfied)? d. The bacteria in Tubes A & B Which term best describes the organism growing in Tube B in Question #5? c. facultative anaerobe Examine the diagram of the Nutrient Agar plate at the right. Which technique was used to produce these results? a. spread plate for isolation If the technique in Question #7 is used to start a pure culture, what assumption(s) is/are being made? b. Each colony on the plate came from a single colony forming unit (CFU). What is the application of an enrichment culture? c. Specific, desired organisms are isolated. (4-‐12;T)10. What is the theory behind an enrichment culture? a. The medium is formulated so it favors a particular kind of organism. (4-‐14;T)11. You formulate a medium consisting of Nutrient Agar and a lethal concentration of the antibiotic tetracycline. How is this medium classified? c. selective medium ( Unit 4: Microbial Growth – Page 1 (4-‐15;T)12. Shown is the recipe of a commonly used medium. An organism able to grow on this medium must be a/an Glucose 10.0 g b. nitrogen-‐fixer K2HPO4 0.5 g MgSO4 0.2 g NOTE: Always look for the carbon and CaSO4 0.1 g nitrogen sources first. This medium has NaCl 0.2 g no nitrogen in it; the only source being CaCO3 5.0 g the environment (e.g., air). The organism CaCl 0.1 g must be able to fix nitrogen from the air in order to grow on this medium. FeSO4 10.0 mg Na2MoO45.0 mg H2O 1000 mL (4-‐14;T)13. How is the medium in Question #12 classified? b. enrichment (Only nitrogen fixing bacteria will be able to grow. No inhibitor is added, so it can’t be a selective medium.) THESE PRACTICE PROBLEMS ALSO APPLY TO LABORATORY OBJECTIVES. NOTE: Questions 14, 15, 16 and were included in case we had time to do a direct count in lab. We didn’t, and you aren’t responsible for these. (Lab;Ap)14. You perform a direct count on a sample that was diluted with 4 parts stain, 5 parts alcohol, to 1 part of sample. You count 37 cells in 8 small squares. What was the original cell density? Give your answer in cells/mL. 1) D2 = (V1 x D1) ÷ (V2) = (1 part x 1) ÷ (1 part + 4 parts + 5 parts) = 10-‐1 2) 37 cells/8 squares = 4.6 cells/square 3) Original Cell Density = (Average Cells per Square) ÷ (Volume per square x DF) OCD = (4.6 cells/square) ÷ (5x10-‐8mL/square x 10-‐1) OCD = 9.2 x 108 cells/mL (Lab;Ap)15. After making a 10-‐2 dilution of a broth culture, you perform a direct count and get an average of 6 cells per small square. What was the original cell density? Give your answer in cells/mL. Original Cell Density = (Average Cells per Square) ÷ (Volume per square x DF) OCD = (6 cells/square) ÷ (5x10-‐8mL/square x 10-‐2) OCD = 1.2 x 1010 cells/mL (Lab;Ap)16. You perform a direct count on a sample that was diluted by a factor of 10. You count 128 cells in one BIG square (see Exercise 6-‐3 in Leboffe and Pierce). What was the original cell density? Give your answer in cells/mL. 1) (128 cells/ big square) ÷ (16 small squares/big square) = 8 cells/small square 2) Original Cell Density = (Average Cells per Square) ÷ (Volume per square x DF) OCD = (8 cells/square) ÷ (5x10-‐8mL/square x 10-‐1) OCD = 1.6 x 109 cells/mL Unit 4: Microbial Growth – Page 2 (Lab;Ap)17. After incubation of plates made from a dilution series, you count 197 colonies on a plate with a plate dilution (final sample volume) of 10-‐6 mL. What was the original cell density? Give your answer in CFU/mL. Original Cell Density = (# Colonies) ÷ (Sample Volume) OCD = (1.97 x 102 colonies) ÷ (10-‐6) OCD = 1.97 x 108 CFU/mL (Lab;Ap)18. A dilution series is made and a 100µL of the 10-‐5 through 10-‐8 dilutions is plated on to trypticase soy agar. After 24 hours of incubation, 247 colonies are counted on the 10-‐6 plate (that is, the final sample volume = 10-‐6 mL). What was the original cell density? Give your answer in CFU/mL. Original Cell Density = (# Colonies) ÷ (Sample volume) OCD = (2.47 x 102 colonies) ÷ (10-‐6) OCD = 2.47 x 108 CFU/mL (NOTE: I converted 100 µL to 0.1 mL [because our units of density are in CFU/mL] and also changed "colonies" to "CFU" since each colony is assumed to come from a single CFU.) (Lab;Ap)19. Continuing with the experiment outlined in #18, how many colonies would you expect on the 10-‐7 plate? Original Cell Density x Sample Volume = # Colonies 2.47 x 108 CFU/mL x 10-‐7 mL = 2.47 x 101 colonies = 25 colonies This plate would not be considered countable since it is out of the 30-‐300 range. (Lab;Ap)20. You count 56 colonies on a plate inoculated with 100µL of a sample diluted by a factor of 10-‐5. What was the original cell density? Give your answer in CFU/mL. Original Cell Density = (# Colonies) ÷ (Volume Plated x DF) OCD = (5.6 x 101 colonies) ÷ (0.1 mL x 10-‐5) OCD = 5.6 x 107 CFU/mL Note that in this question, the dilution given is for the dilution tube, not the plate dilution (sample volume on the plate). (Lab;Ap)21. A dilution series was made in which samples were diluted by factors of ten between 10-‐2 and 10-‐8, inclusive. After incubation, you count 25 colonies on the plate inoculated with 100µL of a sample diluted by a factor of 10-‐4 (note: this is the dilution of the tube, not the plate dilution). Which plate should give you a statistically reliable number of colonies? Original Cell Density = (# Colonies) ÷ (Volume Plated x DF) OCD = (2.5 x 101 colonies) ÷ (0.1 mL x 10-‐4) OCD = 2.5 x 106 CFU/mL (Volume Plated x DF) = (# Colonies) ÷ OCD (10-‐1 mL x DF) = (2.5 x 102 Colonies) ÷ (2.5 x 106 CFU/mL) DF = 10-‐4 ÷ 10-‐1 = 10-‐3 That is, you should plate 0.1 mL of a 10-‐3 DF. This would also be known as a 10-‐
4 plate dilution or more correctly, 10-‐4 mL sample volume. Page 3 (Lab;Ap)22. Using a direct count, you have determined that a culture has a cell density of 1.24 x 109 cells/mL. What plate dilution (final sample volume) should yield a countable plate in a viable count? Plate Dilution = (# Colonies) ÷ OCD Plate Dilution = (1.24 x 102 Colonies) ÷ (1.24 x 109 cells/mL) Plate Dilution = 10-‐7 (Lab;Ap)23. Using the turbidometric method and a calibration curve, you determine that a culture has a cell density of 3.2 x 106 cells/mL. What plate dilution (final sample volume) should yield a countable plate using the viable count? Plate Dilution = (# Colonies) ÷ OCD Plate Dilution = (3.2 x 101 Colonies) ÷ (3.2 x 106 cells/mL) Plate Dilution = 10-‐5 UNIT 5 SAMPLE PROBLEMS
Note: This problem set is not intended to be comprehensive, nor should you memorize these particular questions and answers. They are sample problems are designed to give you an idea of what questions will look like on the exam. As always, complete preparation requires mastering the material covered by the learning objectives. (5-‐1;T) 1. Examine the metabolic pathway below. A à B à C à D Which describes Compound B? b. intermediate (5-‐2;T) 2. Examine the reaction below. A à B + heat This reaction illustrates c. the First and Second Laws of Thermodynamics. (5-‐3;T) 3. Which describes the energy-‐releasing reaction involving ATP? e. All of these are true. (5-‐3;Ap)4. Nitrogen fixation occurs by the following process: N2 + 8H+ + 8e– + 16ATP à 2NH3 + H2 + 16ADP + 16Pi Based on this equation, nitrogen fixation e. requires more than 116 Kcal of energy. (5-‐4;T) 5. Which DOES NOT describe enzymes? Enzymes c. become a product of the reaction. (5-‐1,5;T)6. Examine the reaction below. Glutamate + NH3 + ATP à Glutamine + ADP + Pi With respect to the glutamate and glutamine, this reaction is: e. anabolic and endergonic. Page 4 (5-‐1,5; T) 7. Examine the reaction below. H
H
P
O
C
C
C
O- + ADP
> H
(5-‐5;T) 8. (5-‐6;T) 9. (5-‐7;T)10. O
O
C
C
C
O- + ATP
H
Pyruvic Acid
Phosphoenolpyruvate
H
With respect to the phosphoenolpyruvate and pyruvic acid, this reaction is: d. catabolic and exergonic. Which best describes an endergonic reaction? c. The products have more free energy than the reactants and there is a +ΔG. Examine the coupled reactions below. A à B ΔG= +2.1 Kcal/mole C à D ΔG= -‐3.8 Kcal/mole What is the free energy change when these reactions occur together? a. -‐1.7 Kcal/mole Which is most likely the independent variable in the graph below? a. temperature (5-‐8;In)11. Examine the sequence of reactions shown below. These reactions are involved in ATP synthesis and each step is catalyzed by the enzyme represented by a number. 1 2 3 4 5 6 A B C D E F G NAD NADH Some enzymes may be regulated by inhibition, whereas others are regulated by activation. Enzymes 1 and 6 have allosteric sites for regulation in addition to the active site. Which is most likely a true statement? e. An increase in the concentration of ADP would increase the rate of the reactions by allosteric activation of enzyme 1. Examine the reaction below. +
Glucose 6-‐phosphate + NADP à 6-‐Phosphogluconic Acid + NADPH + H+ (5-‐9;T)12. (5-‐9;T)13. With respect to the conversion of Glucose 6-‐phosphate to 6-‐Phosphogluconic acid, which is true? a. Glucose 6-‐phosphate is being oxidized. H
Examine the reaction below.
C7H15CH 2OH + NAD+
n-Octanol
C7H15C
n-Octanal
With respect to the conversion of n-‐octanol to n-‐octanal, which is true? e. “b” and “c” are true. Page 5 O + NADH + H+