Exam style questions Flashcards
The null hypothesis is:
the assumption there is no relationship or difference between the variables you are testing.
A final year student has carried out their Applied Sciences research project which measured the concentration of the inflammatory marker MMP-8 in samples after treatment with either a statin or an inactivated version of the drug, using an explant model of carotid artery disease. The student tested 10 samples in each treatment group. After doing the test, the student calculates the following p value: p<0.0001
What would be an appropriate statistical test to use (1 mark)?
t test
A final year student has carried out their Applied Sciences research project which measured the concentration of the inflammatory marker MMP-8 in samples after treatment with either a statin or an inactivated version of the drug, using an explant model of carotid artery disease. The student tested 10 samples in each treatment group. After doing the test, the student calculates the following p value: p<0.0001
What would be the null and alternative hypotheses (2 marks in total)
H0: There is no association between treatment with the statin and concentration of MMP-8
H1: There is an association between treatment with the statin and concentration of MMP-8
A final year student has carried out their Applied Sciences research project which measured the concentration of the inflammatory marker MMP-8 in samples after treatment with either a statin or an inactivated version of the drug, using an explant model of carotid artery disease. The student tested 10 samples in each treatment group. After doing the test, the student calculates the following p value: p<0.0001
What is your interpretation of this p value (1 marks) ?
There is a highly significant relationship between treatment of the statin and levels of the inflammatory marker MMP-8.
A final year student has carried out their Applied Sciences research project which measured the concentration of the inflammatory marker MMP-8 in samples after treatment with either a statin or an inactivated version of the drug, using an explant model of carotid artery disease. The student tested 10 samples in each treatment group. After doing the test, the student calculates the following p value: p<0.0001
Suggest an appropriate follow-up experiment that builds on this work? (1 mark)
This could be variable, many correct answers here – examples include identifying the most effective concentration, testing the drug in animal models of the disease, assessing whether MMP-8 concentration is associated with risk of stroke in humans, whether treating with statins can reduce the risk of stroke…..
The Bernoulli Effect allows particles to align in the centre of the fluid stream by:
creating a low pressure area in the centre of the fluid flow.
Fluorescence can be used to collect information about:
internal cellular structures.
Give the principles behind the Three Colour Lymphocyte immunofluoresence (LIFT) test for HLA Antibody detection. (5 Marks)
1) HLA typed lymphocytes required
2) 3 Ab used
a. Anti CD3/PE (T-Cells)
b. Anti CD19/PerCP (B-Cells)
c. Anti IgG(IgM)/FITC
3) Lymphocytes detected by FSC/SSC and sub divided into T and B-Cells by PE/PerCP
4) Positive results are determined by the binding of Anti IgG(IgM)/FITC
5) Multiple individual typed cells required so slow
Which chemical can be used to cross link DNA to protein?
Formaldehyde
From the ChIP data below, calculate the amount of DNA binding in tube A and B as a % of the input. Which of the options below is true?
- % input of A = 6%
- A and B binds DNA almost equally
- B binds DNA at more sites than A
- A binds DNA at more sites than B
- None of the above
A and B binds DNA almost equally
The value of the final enrichment factor obtained at the end of a strategy is determined by which of the following? (2 marks)
- The protein being purified.
- The efficiency of the different steps.
- The type of purification steps used.
- The origin of the starting source
The origin of the starting source
The efficiency of the different steps
Explanation: The purification factor is calculated by dividing the specific activity of the pooled fraction collected after an individual step by that of the initial starting material therefore the final factor obtained will depend on the origin of the starting source – the fraction of this that is our protein of interest. It will also depend on the efficiency of the different steps as these will determine the amount of contaminating protein co-purified.
Using ammonium sulphate to carry out a differential solubility step and obtaining the protein of interest in the supernatant, which of the following CANNOT be used as the next step immediately? (2 marks)
- Ion exchange chromotography.
- Gel filtration chromatography.
- Affinity chromatography
- Hydrophobicity chromatography.
Ion exchange chromatography
Affinity chromatography
Explanation: The supernatant will contain a high salt (ammonium sulphate) concentration and thus would require a desalting step before being applied to an ion exchange or affinity column. The salt concentration is irrelevant for gel filtration, and binding protein to a hydrophobic column is carried out in the presence of a high salt concentration.
If the probability of making a type II error is 0.2, what is the power of the experiment? (2 marks)
0.8
If the power of a research study is low then: (2 marks)
The experiment will likely be inconclusive
Define statistical power and effect size in a power calculation. Briefly discuss the consequences of testing more patients than is necessary in a clinical trial (5 marks).
Statistical power – The probability of correctly rejecting a false null hypothesis (or the chance of finding a sig. diff. when there is one to be found) (2 marks). Effect size is the quantitative measurement of the magnitude of difference that is expected to be observed (1 mark). Many issues - Financial – waste of money on an overpowered experiment (1 mark). Ethical - potentially exposes an excess number of patients to harm (1mark).
What conclusion can you draw from the figure below? (2 marks)
a. pRb is interacting specifically with E2F1
b. E2F1 is binding non-specifically to the beads
c. Anti-pRb does not interact with pRb
d. pRb is not present in the assay
e. E2F1 is binding non-specifically to the primary antibodies
E2F1 is binding non-specifically to the beads
In order to biochemically characterise mutated p53 in U937 cancer cells, researchers wanted to isolate this protein. In order to do this whole cell lysates from U937 cells were mixed with rabbit anti-p53 antibody (Panel A, below) or with a control rabbit IgG immunoglobulin (Panel B, below).This matrix consists of protein-A (which specifically binds antibodies) attached to agarose beads.
Fractions 1– 4 were collected during washing the columns with a buffer that removes proteins that did not bind to the agarose beads.
Fractions 5–7 are the results of elution with a buffer that disrupts bonds between protein A and the antibodies.
Samples of chromatography fractions were then separated by SDS-PAGE and probed using a mixture of anti-p53 and anti-IgG-heavy chain antibodies.
What is the identity of band a in both blots?
Band a= p53
In order to biochemically characterise mutated p53 in U937 cancer cells, researchers wanted to isolate this protein. In order to do this whole cell lysates from U937 cells were mixed with rabbit anti-p53 antibody (Panel A, below) or with a control rabbit IgG immunoglobulin (Panel B, below).This matrix consists of protein-A (which specifically binds antibodies) attached to agarose beads.
Fractions 1– 4 were collected during washing the columns with a buffer that removes proteins that did not bind to the agarose beads.
Fractions 5–7 are the results of elution with a buffer that disrupts bonds between protein A and the antibodies.
Samples of chromatography fractions were then separated by SDS-PAGE and probed using a mixture of anti-p53 and anti-IgG-heavy chain antibodies.
What is the class of molecule represented by band b in both blots?
ii) Band b = Antibodies (will accept immunoglobulin or IgG)
In order to biochemically characterise mutated p53 in U937 cancer cells, researchers wanted to isolate this protein. In order to do this whole cell lysates from U937 cells were mixed with rabbit anti-p53 antibody (Panel A, below) or with a control rabbit IgG immunoglobulin (Panel B, below).This matrix consists of protein-A (which specifically binds antibodies) attached to agarose beads.
Fractions 1– 4 were collected during washing the columns with a buffer that removes proteins that did not bind to the agarose beads.
Fractions 5–7 are the results of elution with a buffer that disrupts bonds between protein A and the antibodies.
Samples of chromatography fractions were then separated by SDS-PAGE and probed using a mixture of anti-p53 and anti-IgG-heavy chain antibodies.
** What conclusions can be drawn from the western blot image (2 marks)**
A. Proteins bound non-specifically to the agarose beads contaminated the p53 immunoprecipitate
B. Most, but not all p53 can be isolated from U937 cells with this procedure
C. The yield of p53 could be further increased by adding more anti-p53 to the U937 extract
D. A, B, and C are all correct
E. B & C are correct
B. Most, but not all p53 can be isolated from U937 cells with this procedure
C. The yield of p53 could be further increased by adding more anti-p53 to the U937 extract
Coomassie staining of the gels would indicate whether the p53 sample is contaminated with other proteins but since no such results are shown, we cannot make this conclusion, so we can discount A. As some of p53 (but not the antibody) appears during washing the column (fraction 1), this indicates that the anti-p53 antibody was present in quantities not sufficient to bind all p53 molecules so B and C are correct.
A researcher obtained the shown profile on a 1D-SDS PAGE gel and associated result from immunoblotting. They decided to use the Sephadex G-100 (fractionation range 4000 – 150000Da) as opposed to Ultrogel AcA 54 (fractionation range 6000 – 70000Da) to perform gel filtration as a step in their purification strategy. The rationale behind this decision was: (3 marks).
a) Its fractionation range matches the sample profile better
b) The protein of interest forms a dimer of 70kDa
c) Sephadex G-100 spans a wider range of molecular weights d) It is more stable at the pH of the mobile phase used
a) Its fractionation range matches the sample profile better
b) The protein of interest forms a dimer of 70kDa
c) Sephadex G-100 spans a wider range of molecular weights
In this case the protein of interest is a dimer with a molecular weight 70kDa. Since 70kDa lies at the upper limit of the Ultrogel AcA 54 matrix, Sephadex G-100 is a better choice to encompass the entire profile of the protein of interest.
Given the following data, what enrichment factor would be obtained after carrying out this purification
step? (Give your answer to one decimal place)
Correct answer is 1.6.
Explanation: Correct – the enzyme activity itself is largely unchanged (99.9% yield), but the total protein is reduced, indicating an enrichment. The amount that the enzyme has been enriched by (the enrichment factor) is therefore 1234/763 = 1.6.
What is the mean value of this experimental set of readings? (give your answer to 2 decimal places) (1 mark)
95.62 +/-1
What is the standard deviation? (give your answer to 2 decimal places) (1 mark)
16.29+/-1
Using an alternative technique B the mean value of Creatine Kinase enzymatic activity was determined to be 68.72 with a standard deviation of 8.82.
Determine the % Coefficient of variation for technique 1 and technique 2 (give your answer to 2 decimal places) (2 marks)
CV= SD/MEAN X100
A) CV = 16.29/95.62*100 = 17.04% +-0.01%
B) CV = 8.82/68.72*100 = 12.83% +- 0.01%
Technique B the mean value of Creatine Kinase enzymatic activity was determined to be 68.72 with a standard deviation of 8.82.
Identify which of the above technique is more accurate and reliable using the values obtained above (1 mark)
B since it has the lowest variance therefore it means it has higher precision than A.
