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162101 Biology of Cells 2014 Lab Theory Test Feedback. General comments. The statistics show that most people coped well with the test and have strengthened their marks before the exam. Calculation questions were done very well on the whole. If you see ECF or ecf (error carried forward) on your paper, this means that you made an error early in a calculation, but you have been given credit for correct calculations that follow, despite the wrong numbers. However, there is an additional penalty for a wrong final answer. Questions 1 and 2 had the same wording as for the 2013 paper; only the numbers were changed. It was very clear that students who could not cope with these questions did not prepare adequately for the test. Worked calculation answers and the expected graph for the 2013 paper were available on Stream and there were many practice examples of both types of question in the lab manual. Question 1 A graph is constructed to show if there is a relationship between what is changed (e.g. amount of protein) and what is measured (e.g. absorbance). A straight line relationship shows that, within the linear range, more protein results in a higher absorbance in a predictable way, shown by the line. However, the Biuret assay has a very limited range over which this linear relationship applies, and at a certain point (past the linear range of the graph), the absorbance values give you NO information about how much protein is there. For this reason, we tell you not to extend the best fit line beyond the end of the linear range. Not as a dotted line, and definitely not as a solid line, no matter what you were taught at school. Quite a few students calculated the ‘protein concentration’ of the unknown solution from the absorbances of both tubes 9 & 10 and got different answers. This shows a lack of understanding of what the test was about – you were trying to determine the protein concentration of an unknown solution. All that was different between tubes 9 and 10 was the volume of unknown solution you added to the assay. You cannot have two different answers to the concentration of protein in a single solution. One must be wrong. The absorbance of tube 10 was beyond the linear range and should NOT have been used to calculate concentration. Best fit lines: We do not expect you to calculate a statistical fit to your results in this paper. But we do expect you to be able to draw a best fit line by eye to your results. This should be a straight line that passes through or close to as many points in the linear range as possible, including zero. No protein = absorbance of zero, so it is a valid point. To do this, you use a ruler to find the line that looks right, with about as many points above the line as below it, and usually some right on the line. You were given seven points on the line (within the linear range) instead of the four you had in the lab, in order to make it easier to do this. If you didn’t know what we meant by ‘best fit line’, you should have asked for help in the weeks prior to the test. DO NOT join each point to the next with straight lines. Please note: Lab manual, p24 for an explanation of a standard curve, p 27 3(ii) ‘with a straight line of best fit drawn through the experimental points’ and p39 5(ii) ‘Using a ruler, draw a straight line of best fit through the linear portion…… If there are any non-linear absorbances at one end of the plot, do not incorporate these points in your ‘line of best fit’ otherwise your line will deviate from a straight line.’ Lines drawn freehand were unacceptable.

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Horizontal (x) axis. The lab book instructions are equally clear about this. You should have plotted amount of protein (in mg), in preference to concentration, and certainly not volume in mL. Plotting concentration is not wrong, but involves more calculations, so more chances for errors. Many students plotted amount and called it concentration, which is NOT correct. Part C of Experiment 1 explains how to plot the graph correctly, as do standard curve example questions 1 & 2 (lab manual pages 49-50). We are aware that some of the graphs in this course are not straight lines (the bacterial growth curves and the DNA gel calibration curve. Sometimes the relationship is not a straight line, but a logarithmic or some other mathematical relationship. This appears to have confused some students. However, the instructions and examples for the standard curve graphs were very detailed and explicit. If you lost any other marks for question 1, please look at the file ‘Plotting a good graph’ which is in the file with previous lab test papers. Comments about the other questions are shown with the questions and example answers. Important: Markers were required to follow a strict marking schedule. Your marks will not be altered if this would result in different marking for some students and not others. Please Note. Lab calculations and questions related to the experiments will not be tested in the final exam, but parts of Experiments 7 and 8 are relevant to lectures and the paper content as follows:

The principles of PCR: How it works, how it amplifies a specific piece of DNA, the type of primers used.

How to interpret diagrams representing the results of DNA gel electrophoresis (but not the calibration of gels).

Pedigrees and probabilities in genetics. Why and how restriction endonucleases are used in DNA Technology.

Make sure you revise these before the exam.

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QUESTION 1 (10 marks)

Suppose you wish to determine the protein concentration of an unknown solution using the Biuret assay. You have a stock protein solution with a concentration of 5 mg/mL. You prepare tubes 1 to 8 as described in the table below, and measure their absorbance at 540 nm. The experimental results are shown in the second table. Tube # Volume of

stock solution Volume of H2O

Volume of Biuret reagent

Absorbance at 540 nm

1

0

4.0 mL

4.0 mL

0

2 0.2 mL 3.8 mL 4.0 mL 0.055 3 0.5 mL 3.5 mL 4.0 mL 0.170 4 1.0 mL 3.0 mL 4.0 mL 0.350 5 1.5 mL 2.5 mL 4.0 mL 0.470 6 2.0 mL 2.0 mL 4.0 mL 0.655 7 2.5 mL 1.5 mL 4.0 mL 0.840 8 3.0 mL 1.0 mL 4.0 mL 0.850 You assay two different volumes of your unknown protein sample at the same time and obtain the results shown below. Tube # Volume of

unknown protein solution

Volume of H2O

Volume of Biuret reagent

Absorbance at 540 nm

9

0.7 mL

3.3 mL

4.0 mL

0.495

10 1.5 mL 2.5 mL 4.0 mL 0.870 Using a standard curve, calculate the concentration of protein in the unknown solution to 3 significant figures. Show all your working. Graph paper is provided on the next page.

Tube #9: 7.5 mg (7.5 ± 0.5) 0.7 ml Conc. unknown = 10.7 mg/ml

Calculate amount of protein in each tube. 1: 0 mL x 5 mg/mL = 0 mg 2: 0.2 mL x 5 mg/mL = 1.0 mg 3: 0.5 mL x 5 mg/ml = 2.5 mg 4: 1.0 mL x 5 mg/mL = 5.0 mg 5: 1.5 mL x 5 mg/ml = 7.5 mg 6: 2.0 mL x 5 mg/ml = 10 mg 7: 2.5 mL x 5 mg/mL = 12.5 mg 8: 3.0 mL x 5 mg/mL = 15 mg

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QUESTION 2 a) Suppose you start a new job working in a laboratory. You are asked to

prepare exactly 1150 mL of an 18 mg/mL solution from a stock solution of 475 mg/mL. How much stock solution will you need and how much distilled water will you need to add to dilute it? (3 marks)

C1V1 = C2V2

V1 = C2V2

C1

= (18 mg/mL) (1150 ml) 475 mg/mL = 43.6 ml (43.57 acceptable)

Volume of stock = 43.6 mL Volume of water (diluent) = 1150 mL – 43.6 mL = 1106.4 mL

b) Suppose 11 mL of a stock solution with concentration 27 mg/mL is added to

121 mL of diluent. Then 4 mL of the resulting solution is added to 20 mL of diluent. What will be the concentration of the final solution? Show all your working. (3 marks) (V2 = V1 + Vdil so 35 + 7 = 42mL) (not essential to be shown) Df1 = 132 mL

11 mL = 12

Df2 = 24 mL  4 mL = 6 TDF = 12 x 6 = 72

 

c) Convert the following to the units shown. (4 marks) (i) 9.57 nm = _______________ m 9.57 x 10-9 (ii) 78.5 L = _______________ mL 0.0785 or 7.85 x 10-2 (iii) 0.019 mg/mL = _______________ g/mL 19 (iv) 567 m = _______________ nm 5.67 x 105 Many students made errors of conversion and lost several marks.  This also showed in subsequent calculations.  You need to practice until you are confident. 

Df = C1 C2

C2 = C1 TDF

= 27 mg/mL 72

= 0.375 mg/mL

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QUESTION 3 A student designed an experiment to test the hypothesis, “washing hands with soapy water removes more microorganisms than washing with water.” An agar plate was divided into 4 sectors and the following treatments were carried out:

1. No treatment 2. Hands were rubbed together then a finger was touched to the plate 3. Hands were rubbed together using tap water then a finger was touched to the plate 4. Hands were rubbed together using soapy water then a finger was touched to the plate The plate was inverted and incubated at 250C for 24 hours.

(a) Was the hypothesis supported YES/NO? Give a reason for your answer. (2 marks)

NO the hypothesis is not supported. There is no (statistically significant) difference between the colony counts for tap

wash vs soapy wash treatments. Or Both water and soapy water appear to be equally effective at lifting microorganisms from the skin that are then readily transferred to the plate) Most students understood that the hypothesis wasn't supported due to the similarity of the colony numbers in sections 3 and 4. 

(b) What sort of control is sector 1 and what does it tell you? (2 marks)

Sector 1 is a negative control, indicating that the plate was not contaminated prior to the experiment. Students confused positive controls and negative controls. If you get the answer you want from a control that doesn't make it positive. There was no true 'positive' control in this experiment.

(c) What is a possible explanation for the low colony count in sector 2. (2 marks) A dry hand/finger does not transfer so many microorganisms to the plate as a

wet hand/finger. Microorganisms may be lifted/suspended in water, resulting in greater transfer to another surface/object than just leaving the hands dry. Many students wrote ‘mislabelled’.  This was not accepted, as we expected you to assume this was not the case due to the information legend next to the plate and to find other reasons for this result. Rubbing your hands together when dry doesn't produce enough friction heat to kill the bacteria on your hands. Many students didn't understand how water is able to mobilize bacteria on the hand to allow for better transfer in comparison to dry hands. ‘Phenomenon’ is not the correct term when describing the process in an experiment.   

(d) List two good points about the experimental design (2 marks)

Rubbing hands together was a good idea to ensure uniformity of microorganisms on hands. But rubbing your hands together will not transfer all the bacteria from one hand/finger to the other. 

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Wetting hands and testing with and without soap limited the number of variables. A negative control was used (Sector 1)

(e) List two limitations of the experimental design (2 marks)

No time duration given for washing. Wash time for sectors 3 & 4 could have been different, therefore potentially affecting the results Unclear if ‘a finger’ was from the same or different hands – needed to specify this. Lack of replicates Colonies were counted, but different types of colonies were not recorded. The numbers seen on the plate could still be due to normal flora (see below). 

Many students didn't get marks for sections (d) and (e).  Think about HOW you could have improved the experiment, and what features of the experiment you think are good (controls, hypothesis, not too many variables).  These are different and specific to each experiment.  You can’t rely on model answers to apply to all experiments. 

 Basic misunderstandings in this question/experimental session. We have many bacteria which normally live on our skin, and do no harm.  Bacteria are essential for us to live.  What we want to remove by hand washing are the extra bacteria we pick up from the environment, which may cause disease, particularly if we transfer them to our mouth, eyes, ears etc or which could enter a wound.  However, when we touch an agar plate we are transferring some of the normal skin flora and some of the extra ‘dirty hands’ environmental bacteria. You cannot sterilize your hands – certainly not by washing them. Antibacterial hand wash will remove some, but not all bacteria from the skin (hand sanitizer or ethanol would kill more of them, but not ALL).  

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QUESTION 4 (a) Which two things need to be in focus for a microscope to be correctly set up

for critical illumination? (1 mark) The object on the prepared slide and the light source.

(b) What procedure was used during slide preparation to make prokaryotic (bacterial) cells visible under the microscope? (1 mark)

Gram staining

(c) Fill in the table below to distinguish between E. coli and M. luteus. (4 marks)

Bacteria Gram + or -

Colour Shape The cell wall has a layer of peptidoglycan that is relatively thick/thin (choose one)

E. coli - Red/pink Rod thin

M. luteus + Purple/black Spherical/coccus thick

Some marks were given if the student mixed up the two bacterial species but their answers were otherwise consistent. 

(d) Write the correct term in the space to complete the sentence: Use of immersion oil increases resolution because it has a higher

_________________________ than air. (see expt 3) (1 mark) (e) A student did serial dilutions of a bacterial culture of 10-fold, 6-fold and

another 10-fold. What was the total dilution factor? (1 mark) TDF = 10 x 10 x 6 = 600 fold  

(f) What volume of a bacterial suspension of 2 x 104 cells/mL would the student

have to use to plate out 300 cells? (1 mark) Volume in mL = Cells/(cells/mL)

= 300/2 x 104 mL = 150 x 10-4 mL = 15 x 10-3 mL = 15 L or 0.015 mL (g) If a student spread 300 cells on a plate but only had 150 colonies grow, how

could they explain this result? Not all viable (alive). Assume the colonies were spread correctly and look for a deeper answer. 

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QUESTION 5 (a) A stage micrometer is etched with a 2 mm line that is divided into 100 equal

divisions. When viewed with an eyepiece micrometer at a total magnification of 400x, it is found that 14 stage micrometer divisions correspond to 87 eyepiece micrometer divisions. Viewed at 400x magnification, a spherical cell has a diameter of 21 eyepiece divisions. What is the diameter of the cell in micrometers (m)? Show all of your working. (4 marks) 1 SMD = 2mm/100 x 1000 m/mm = 20 m  

87 EPD = 14 SMD = 14 x 20 m = 280 m 1 EPD = 280 m/87 = 3.22 m Cell diameter = 21 EPD = 21 x 3.22 m = 67.6 m  

(b) A cytometer has the dimensions shown below. A student counted 486

yeast cells in the shaded portion of the cytometer grid. Calculate the cell density of the yeast cell suspension in cells/mL. Show all of your working. Note that 1 mL = 103 mm3. (3 marks)

Depth = 0.2 mm 

Volume of grid used : 2.5 mm x 0.5 mm x 0.2 mm = 0.25 mm3 or 0.5 x 0.5 x 0.2 x 5 = 0.25 mm3 or (2.5 mm x 2.5 mm x 0.2 mm)*5 = 0.25 mm3 0.25 mm3/103 mm3 x 1 mL = 2.5 x 10-4 mL        

       Cell density = # cells/volume       = 486/2.5 x 10‐4 cells/mL       = 1.94 x 106 cells/mL  

(c) If the cell density of a yeast cell suspension was 9.97 x 105 cells/mL, and the cells were counted in the same shaded area of the grid shown above in (b), how many cells would be counted? ( 2 marks) The volume would be the same: 2.5 x 10-4 mm3. # cells = cell density x volume = 9.97 x 105 cells/ml x 2.5 x 10-4 ml = 249 cells (no part cells! ) An answer of 10‐6 cells (= 1 millionth of a cell) or similar should have rung alarm bells! or Students could have done this as a ratio of cell densities using the cell number from part (b):   

486 x 9.97 x 105/(1.94 x 106) = 249 cells  (d) Would it be possible to count E. coli cells using a cytometer? Explain. (1 mark)

No, they are too small, can’t use immersion oil with cover slip, lens too long, can’t treat cytometer with gram stain, can’t usually see bacteria without the stain. So many reasons, and you only needed one.  This type of question was why we recommended you try to visualise being back in the lab looking down the microscope and not try to look for one model answer. 

2.5 mm 

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QUESTION 6 (a) Vicia faba root tips were microscopically examined to determine the number

of cells in each stage of the cell cycle. The treatment consisted of root tips incubated in 0.1% colchicine for 5 hours at 20C. 400 cells were examined at random for the untreated control and the colchicine treatment. Calculate the length of the cell cycle for this organism. Show all of your working. (3 marks)

Number of cells in …. Interphase Prophase Metaphase Anaphase Telophase

Control (Slide 1)

299 28 22 38 13

Treatment (Slide 2)

319 35 46 0 0

length of cell cycle = 100% x time (hr) % cells in met after - % cells in met before

  

% met after = 46/400 = 11.5% % met before = 22/400 = 5.5%   Treatment time = 5 h  

Cell cycle length = 100/(11.5 – 5.5) x 5 = 83.3 h         The most common mistake was to use the numbers in the table and not to calculate percentages 

(b) Using your result from part (a) above, calculate how long untreated cells of

these beans will spend in: (3 marks) (i) Prophase. % untreated cells in interphase = 28/400 = 7%   

Time in interphase = 7% (0.07) x 83.3 h = 5.83 h   (ii) Anaphase 38/400 x 83.3 h = 7.91 h

  Full marks were given for correct calculations even if the cell cycle length from (a) was incorrect.

(c) Why is it necessary to select only the very end of a root tip when preparing a

slide to study mitosis? (1 mark) Because this is the only part of the root containing rapidly dividing cells (meristem).

This is where the root is growing, so mitosis is occurring in many cells. ‘Mitosis’ is in the question, so an explanation of why this was occurring in the root tip was required. Avoid sweeping statements:  the root tip is not the ‘only place in the plant where cell division (or mitosis) is occurring’. 

(d) The frequency of penicillin-resistant mutants in a particular bacterial culture is

7.45 x 10-6. How many penicillin-resistant cells will there be in 500 µl of the culture with a total cell density of 8.95 x 108 cells/ml? Show all of your working. (3 marks) 500 µl = 0.5 ml or total cells = cells/mL x cells 0.5 ml x 8.95 x 108 cells/ml = 4.475 x 108 cells in total

(Best not to round to 3 sig figs at this stage – it affects the answer quite a lot) If mutant freq = 7.45 x 10-6 then no. of cells in the culture is 4.475 x 108 cells x 7.45 x 10-6    = 3.33 x 103 cells (3333 or 3334 cells)

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QUESTION 7 (a) Complete the table to show how you detected the following cellular

components after differential centrifugation of pea homogenate. Include the colours you observed. Include microscopic evidence, biochemical assays or stains as appropriate and mention any diagnostic colours observed in each case. (3 marks)

Components detected

Methods used to detect component

Starch granules Blue/black (not just ‘dark’) with iodine stain, observed using microscope. Not gram stain procedure.

Chloroplasts Green colour observed in pellet and in microscope Not a specific test for ‘green’ 

Mitochondria Red colour in tetrazolium assay. Mitochondria are not red coloured and are too small to see by eye. 

 Had to state methods, assays and colours. 

  (b) Explain what differential centrifugation is and why it was used. (2 marks)

Samples are centrifuged, and the supernatant and precipitate separated. Further steps of centrifugation for longer and at faster speeds are used to separate fractions of smaller and smaller particles. Used as a method of separation by density (weight, size) (and often as a start for purification of some cell components). Many people described a single centrifugation step, but not multi‐step differential centrifugation or why it was used. 

(c) Explain whether pure preparations of the components listed in the table

above were obtained. (2 marks) Pure components are not obtained - some components (eg starch granules) vary in size/mass density so are not totally

separated - some different components may have the same size/mass density as each other so not

be separated.   Many students asked about this question before the test.  Pure means pure.  Consider what 

is being separated and the basis of separation of the method and think about what is happening during the procedure to answer a question of this type.  

 (d) Name and briefly describe what happens in each of the three steps in each

PCR cycle. (3 marks) (i) Denaturing/denaturation - to separate double-stranded DNA molecules into

single strands (ii) Annealing/primer attachment - of primers to the single-stranded

template DNA (iii) Extension/DNA synthesis - DNA synthesis extending from the end of each

primer, making a double-stranded DNA molecule (the PCR product). Primers do not move along the DNA strand.  The DNA primers hydrogen‐bond to the complementary antiparallel sequence of the DNA and then the Taq DNA polymerase catalyses formation of the rest of the DNA strand until there are no more bases to copy.  RNA Is not involved. 

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QUESTION 8 A 15.3 kb fragment of linear DNA was cut with the restriction enzymes EcoRI and BamHI to yield the DNA fragment sizes shown in the table.

Restriction enzyme(s)

EcoRI BamHI EcoRI + BamHI

Sizes of fragments (kb)

6.0, 5.3, 4.0 7.5, 4.8, 3.0 6.0, 4.0, 3.0, 1.5, 0.8

a) Draw the pattern of DNA fragments you would expect to see after gel

electrophoresis of vector cut with EcoRI or BamHI or with both enzymes together. Clearly indicate their sizes by reference to the size marker ladder shown. (5 marks)

Plasmid vector digested with

Size marker EcoRI BamHI EcoRI+BamHI Ladder

Easy marks – most students did this OK 

b) Draw lines on the diagram below to indicate where these enzymes cut the vector. Label the distances (kb) between the cutting positions. (5 marks)

Units were required and each cut had to be labelled with the correct enzyme. 

EcoR1 EcoR1

BamH1BamH1

5 kb   3 kb 4 kb 1.5kb 0.8 kb