Test 4 – Data Analysis & Practical PDF

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Question 1CorrectMark 2 out of 2.Started on State Completed on Time taken Marks 38/57. Grade 67 out of 100.You are using the light microscope to measure the size of cells at x magni!cation. You have an arbitrary scale, divided into 100 divisions, that you must calibrate at x10 magni!cation, against ...

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Started on State Completed on Time taken Marks 38.67/57.00 Grade 67.84 out of 100.00 Question 1 Correct Mark 2.00 out of 2.00

You are using the light microscope to measure the size of cells at x10 magni!cation. You have an arbitrary scale, divided into 100 divisions, that you must calibrate at x10 magni!cation, against a 1mm scale that is also divided into 100 divisions. You measure that 93.8 divisions of the arbitrary scale are equal to 100 divisions on the 1mm scale. What is the factor by which you must multiply all measurements made at x10 magni!cation in order to convert them into µm? Give your answer to two decimal places.

Answer: !

100 divisions on the 1mm scale =1000µm 93.8 divisions on the arbitrary scale =1000µm 1 division on the arbitrary scale=1000/93.8; this is the conversion factor that all arbitrary measurements should be multiplied by to convert them to µm. The correct answer is: 10.66

Question 2 Partially correct Mark 8.00 out of 12.00

The table below shows the average values recorded for two organisms, observed moving on a haemocytometer. Use the data to calculate: a)

Average speed in µm/sec for organism 1

b)

Average speed in µm/sec for organism 2

c)

Average speed in body lengths/sec for organism 1

d)

Average speed in body lengths/sec for organism 2.

e)

How long in seconds it would take organism 1 to move one body length?

f) How long in seconds it would take organism 2 to move one body length? Average Body length (µm) Organism 1

Organism 2

243.40

33.70

Average Time to travel 50µm (seconds) Organism 1

Organism 2

0.57

2.56

Calculate the average speed in µm per second for organism one. Give your answer to two decimal places. =! µm/sec Calculate the average speed in µm per second for organism two. Give your answer to two decimal places. =! µm/sec Calculate the average speed in body length per second for organism one. Give your answer to two decimal places. = " body lengths per second Calculate the average speed in body length per second for organism two. Give your answer to two decimal places. = " body lengths per second How long, in seconds, would it take organism one to travel one body length? =! seconds How long, in seconds, would it take organism two to travel two body lengths? =! seconds

(a) Speed in µm/sec can be calculated by considering the distance travelled in the time taken. In this case, 50µm/0.57 = 87.72 µm/sec (b) As for (a), 50/2.56 = 19.53 µm/sec (c) The body length is also measured in µm and the value recorded for this equals one body length. Therefore to calculate the speed in body lengths per second, divide the speed in µm/sec by the body length. 87.72/ 243.40 = 0.36039441 rounded to 0.36 body lengths/sec (d) For organism 2: 19.53/33.7 = 0.58 body lengths/sec (e) time in seconds to travel one body length for org. 1: 1/0.36 = 2.77 seconds Alternatively you could calculate it this way: Body length=243.40 µm Speed in µm/sec =102.04 Time taken to travel one body length = 243.40/87.72 = 2.77 seconds (f) 1/0.58 = 1.72 seconds or 33.7/19.53 = 1.72 seconds.

Question 3 Correct Mark 3.00 out of 3.00

Use the image of the 1Kb marker at the right with the indicated size bands, to identify the size of the bands on the left, indicated as a, b and c with an arrow, in Kb.

a=! b =! c=!

The 1Kb marker size bands are given in base pairs, however the question is asking you to write your answer in Kilobases - Kb. The correct answers are: a=6 b = 2.5 c = 0.75

Question 4 Correct Mark 3.00 out of 3.00

The following gel image shows the results of three di"erent restriction enzyme digests of samples of genomic DNA and cloned plasmid DNA. Next to it, you can see the 1Kb marker used in the experiment, with the band sizes labelled.

Using as a landmark the 1Kb marker seen on the right hand side, what size (in Kb) are the fragments in lanes 2, 3 and 4? Lane 2

8 to 1

!

Lane 3

5 and 0.8

!

Lane 4

5 and 4.8

!

Your answer is correct. The 1Kb marker size bands are given in base pairs, however the question is asking you to write your answer in Kilobases - Kb. The correct answers are: Lane 2: 8 to 1 Lane 3: 5 and 0.8 Lane 4: 5 and 4.8 The correct answer is: Lane 2 → 8 to 1, Lane 3 → 5 and 0.8, Lane 4 → 5 and 4.8

Question 5 Partially correct Mark 1.50 out of 3.00

Restriction digests of the recombinant plasmid pUC19 with EcoRI, HinfI, BglII, or their combination, generate the following fragments:

Choose the correct restriction map,corresponding to the indicated genomic fragments. a.

b.

c.

# Restriction digest of this plasmid would give fragments of 1800, 1000, 2x800 and 600 with Eco and Bgl digest, matching the band sizes in the table. But EcoRI + HinfI would give 1700, 1400 and 900 that doesn’t correspond to the band sizes in the table. We had a typo in the table which meant there isn't a 100% correct answer so we have allowed part marks if you chose this option.

Your answer is partially correct. The correct answer is:

Question 6 Correct Mark 1.00 out of 1.00

In the Molecular Biology practical, we used 0.8% agarose gels to separate our DNA digests. Each week we prepare 24 agarose gels, and each requires 55mL of agarose solution. How much agarose, in grams, do the technicians need to weigh out for the whole class? Round your answer to two decimal points.

Answer: !

24 gels x 55ml per gel = 1320ml in total. Each gel has 0.8g of agarose per 100ml solution (0.8%) 1250 x 0.9 / 100 = 10.56gr

The correct answer is: 10.56 gr

Question 7 Incorrect Mark 0.00 out of 3.00

Our agarose gels use a running bu"er called TBE (contains 90mM Tris, 90mM Boric acid and 2.5mM EDTA). The technicians make up a stock solution that is ten times more concentrated than required (known as 10X TBE) and then dilute it as needed. The molecular weights of each component are as follows: Tris - 121.136 g mol-1 Boric acid - 55.65 g mol-1 EDTA (disodium salt) - 372.24 g mol-1 If the technicians are making up 6 litres of the 10X stock solution how much of each component do they weigh out? Round your answer to two decimal points. Tris = " g Boric acid = " g EDTA = " g

Tris 121.136 x 0.9 x 6 = 654.13g Boric acid 55.65 x 0.9 x 6 = 300.51g EDTA 372.24 x 0.025 x 6 = 55.84g

Question 8 Correct Mark 2.00 out of 2.00

Read the following statements relating to the transamination and deamination experiment and select the most suitable answer. 1. Glutamate dehydrogenase removes the amino group from glutamate to produce 2-oxoglutarate. 2. 2,4 dinitrophenyl-hydrazones of oxo-acids appear as purple spots on the TLC plates. 3. Glutamate dehydrogenase can use either NAD+ or NADP+ as its coenzyme. 4. Ethyl acetate is added to the tubes for visualisation of the oxo-acids to ensure that the 2,4-dinitrophenylhydrozone derivatives are concentrated in a small volume. 5. Unreacted DNPH appears as a band on the oxo-acid plate in the control sample. a. Statements 1, 3, 4 and 5 are true. b. All statements are true. c. All statements are false. d. Statements 2 and 3 are false. e. Statements 2 and 5 are true.

Your answer is correct. Statement 1 is true Statement 2 is false - the spots would be yellow not purple Statement 3 is true Statement 4 is true Statement 5 is true. The correct answer is: Statements 1, 3, 4 and 5 are true.

!

Question 9 Correct Mark 1.00 out of 1.00

The following image shows two TLC plates, similar to we ran in the virtual practical. The plate on the left (purple spots) shows three di"erent amino acids in lanes 1, 2 and 3 and our reaction mixture (after incubation in lane R). The plate on the right (yellow spots) shows three di"erent oxo-acids in lanes 4, 5 and 6 and the same reaction mixture in R.

If a single amino acid was added to the incubation mix what type of reaction most likely occurred in tube R? a. Transamination

!

b. No reaction c. Deamination

Your answer is correct. Here you are told that a single amino acid was added to the reaction mixture but after incubation and separation we can see two amino acid spots. This means there had to be a reaction of some sort. A deamination reaction would leave a single amino acid and the appearance of one oxo acid. When we saw transamination in the practical we could see a pattern as above - two amino acids and two oxo acids, after incubation. The correct answer is: Transamination

Question 10 Correct Mark 3.00 out of 3.00

When running TLC plates in the transamination practicals we used two di"erent solvent mixtures. Solvent mix 1 contained 20% ammonia and 80% ethanol while mixture 2 had 20% ammonia, 10% ethanol and 70% n-butanol. We need 30 tanks of each solvent and each tank needs to have 15mL of liquid in total. How much of each solvent do we need for one afternoons practical class? ammonia =!mL ethanol =!mL n-butanol =!mL

First you should work out the !nal volume of solvent needed in for each mixture. You are told we have 30 tanks and 15mL in each - therefore we need 450mL of each solvent mixture. Solvent 1 is 20% ammonia and 80% ethanol. 20% of 450 is 90mL so we need 90mL of ammonia. 80% of 450mL is 360mL so thats how much ethanol is needed. Then consider solvent 2. Volume is still 450mL and we need another 90mL of ammonia, 10% ethanol (so 45mL) and 70% n butanol (70% of 450 is 315mL)

Question 11 Correct Mark 2.00 out of 2.00

If 150mls of 200mM amino acid solution was placed inside a similar semipermeable membrane and equilibrated in 5L of bu"er what would be the concentration of amino acid at the end of the incubation? Give you answer in mM to 2 d.p. (do not type the units). Answer:

!

Here you needed to consider the concentration and volume of the initial solution (information given), calculate the !nal volume and then use the equation C1V1 = C2V2. Here C1 = 200mM V1 = 150mL V2 = 5L + 150mL Therefore C2 = (200*150)/5150 = 5.83mM If you used 5000mL for C2 then your answer would have been 6mM and we gave you half marks. The correct answer is: 5.83

Question 12 Partially correct Mark 0.17 out of 2.00

Our heart extract was dialysed against several litres of bu"er before use in the practicals. Which of the following components of the heart extract would be removed via the dialysis step? Select all possible correct answers. Incorrect choices will have marks deducted. a. NAD+ b. amino acids

!

c. oxo-acids d. transaminase enzymes e. mitochondria f.

nucleic acid

"

g. membrane contamination

Your answer is partially correct. You have correctly selected 1. When making the heart extract the tissue is minced and homogenised to break open the cells. The resultant mixture is then centrifuged. The centrifugation will remove unbroken cells and large organelles/membrane fragments from the mixture. The soluble components of the broken cells will be in the supernatant. Its this supernatant that is then dialysed. We place the liquid inside a bag made of a membrane with very small pores. Only the smallest molecules eg amino acids, oxo acids and NAD would be able to migrate through the pores. All other soluble components would be retained. The correct answers are: amino acids, NAD+ , oxo-acids

Question 13 Correct Mark 1.00 out of 1.00

Trypsin is a pancreatic serine protease that hydrolyzes peptide bonds. Trypsin activity is often measured by determining the rate of appearance of p-nitroaniline from a synthetic substrate, Benzoyl - L-Arginine p-Nitroanilide Hydrochloride (BAPA). p-nitroaniline strongly absorbs at 405 nm (ε405 nm = 9620 M-1 cm-1) and can be measured colorimetrically. First, you want to !nd the pH optimum by varying the pH of the bu"er used to measure trypsin activity but keeping the trypsin and BAPA concentrations constant. Here is the data you collected:

What is the pH optimum of trypsin based on your data? Select one: 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10

Your answer is correct. The correct answer is: 6.5

!

Question 14 Partially correct Mark 2.00 out of 4.00

Then, you perform an enzyme kinetics experiment using di"erent concentrations of BAPA. Here is the Michaelis-Menten plot you collected:

(1) What is the Vmax for trypsin with BAPA as substrate based on your data? 1000

" µmol / min

(2) What is the Km for trypsin with BAPA as substrate based on your data? 100

! mM

Your answer is partially correct. You have correctly selected 1. The correct answer is: Then, you perform an enzyme kinetics experiment using di"erent concentrations of BAPA. Here is the Michaelis-Menten plot you collected:

(1) What is the Vmax for trypsin with BAPA as substrate based on your data? [40] µmol / min (2) What is the Km for trypsin with BAPA as substrate based on your data? [100] mM

Question 15 Correct

Potassium thiocyanate (KSCN) is an inhibitor of trypsin. Which LineweaverBurk plot will you generate when KSCN is an uncompetitive inhibitor?

Mark 2.00 out of 2.00

Select one:

a.

b.

c.

d.

Your answer is correct. The correct answer is:

!

Question 16 Incorrect Mark 0.00 out of 3.00

Given a Vmax of 160 µmol per 10 minutes, a reaction volume of 1 ml, and a trypsin concentration of 2.4 µM, what is the kcat for trypsin in min-1 ? a. 66667 b. 15.00 c. 1.500 d. 667 e. 0.1500

"

f. 6667

Your answer is incorrect. The correct answer is: 6667

Question 17 Correct Mark 4.00 out of 4.00

You are designing an experiment to evaluate the function of mitochondrial samples, using oxygen electrode traces. In your experimental protocol, the mitochondria are added !rst, followed by ADP with oxidizable substrate (malate or succinate) and any inhibitors. In order to accurately measure the P:O ratio of succinate in the mitochondrial samples, you have determined that succinate and phosphate need to be in excess, but ADP does not. Select the statements that are correct regarding this decision. a. If ADP is in excess, the electron transport chain could be uncoupled from ATP synthesis. b. If phosphate is limiting, then the P:O ratio calculated from the experiment will be higher than the correct value.

!

c. If ADP is in excess, then the oxygen usage will be dependent on [succinate] or [phosphate].

!

d. If succinate is limiting, then complex I activity will increase, making the readings inaccurate. e. If phosphate is limiting, oxygen consumption will not be dependent on [ADP]. f.

!

If succinate is limiting, then oxygen consumption will continue after the ADP has all been converted to ATP.

Your answer is correct. The correct answers are: If ADP is in excess, then the oxygen usage will be dependent on [succinate] or [phosphate]., If phosphate is limiting, then the P:O ratio calculated from the experiment will be higher than the correct value., If phosphate is limiting, oxygen consumption will not be dependent on [ADP].

Question 18 Incorrect Mark 0.00 out of 3.00

Background information on a patient suggests a genetic disorder inherited from the father. You carry out a number of tests on the patient's sample, obtaining oxygen electrode traces for the sample in various conditions. Select the oxygen electrode traces that are consistent with a deleterious mutation in complex II. a.

b.

c.

"

d.

Your answer is incorrect. The correct answers are:

,

Question 19 Correct Mark 3.00 out of 3.00

You are conducting research on a new species of yeast, and have isolated the mitochondria in order to calculate the P:O ratio for succinate. You generate the following oxygen electrode trace.

This drawing is not to scale. a: 111 mm b: 99 mm c: 85 mm d: 65 mm e: 150 mm Given the measurements above, calculate the P:O ratio for succinate. Please give your answer to 2 decimal places, and include no units.

Answer: !

The correct answer is: 1.32...