Research Exam Semester 3 – 2018-2019 January 18, 2019

Research Exam Semester 3 – 2018-2019 January 18, 2019

During the exam, you have access on a computer to these books:

• Baynes & Dominiczak: Medical Biochemistry

• Campbell: Statistics at square one

• Donders: Literature Measurement errors

• Fletcher: Clinical Epidemiology

• van Oosterom en Oostendorp: Medische Fysica

• Petrie and Sabin: Medical Statistics at a Glance

• Turnpenny: Emery's Elements of Medical Genetics

• Form with statistical formula’s

You are allowed to use a calculator of the type Casio FX-82MS.

The questions must be answered in English. If you cannot remember a specific English term, you may use the Dutch term.

Write your name and student number on the first page of each question!

Name:

Student Number:

Question 1

Q5: Cancer etiology and prognosis – prof. dr. B. Kiemeney (20 points)

Molecular Markers Increase Precision of the European Association of Urology Non–

Muscle-Invasive Bladder Cancer Progression Risk Groups Abstract, Journal of clinical research 2018

Purpose: The European Association of Urology (EAU) guidelines for non-muscle-invasive bladder cancer (NMIBC) recommend risk stratification based on clinicopathologic

parameters. Our aim was to investigate the added value of biomarkers to improve risk stratification of NMIBC.

Experimental Design: We prospectively included 1,239 patients in follow-up for NMIBC in six European countries between the years 2008 and 2013. Fresh-frozen tumor samples were analyzed for GATA2, TBX2, TBX3, and ZIC4 methylation (Note: methylation of promoters of genes usually repress gene transcription) and FGFR3, TERT, PIK3CA, and RAS mutation status (Note: gene mutations may alter gene activity). A regression analysis identified markers that were significantly associated with progression to muscle-invasive disease.

Results: In our cohort, 276 patients had a low, 273 an intermediate, and 555 a high risk of tumor progression based on the EAU NMIBC guideline. Fifty-seven patients (4.6%)

progressed to muscle-invasive disease. Overall, wild-type FGFR3 and methylation of GATA2 and TBX3 were significantly associated with progression (Hazard Ratio (HR) = 0.34, 2.53, and 2.64, respectively).

Conclusions: We conclude that the addition of selected biomarkers to the EAU risk stratification increases its accuracy and identifies a subset of NMIBC patients with a very high risk of progression.

Table 1.1 - Results of the regression analysis that was performed on the progression to non-muscle-invasive bladder cancer

Factor Univariable Multivariable

HR (CI 95%) p-Value HR (CI 95%) p-Value Age (years) 1.04 (1.01–1.07) 0.004

Biomarkers

FGFR3 mutation GATA2 methylation TBX3 methylation

0.34 (0.71 - 0.68) 2.53 (1.36 - 4.71) 2.64 (1.41 - 4.92)

0.002 0.003 0.002

0.43 (0.21 - 0.87) 2.23 (1.16 - 4.31) 1.85 (0.94 - 3.65)

0.019 0.016 0.076

a. Which regression analysis was performed in this study (1pt)?

Provide 1 reason why this specific regression analysis is used (2pt) Cox regression / Cox proportional hazard regression (1pt)

Because we deal with time to event data (2pt)

b. One of the reported limitations in this study is that they found a very low number of people who had progression to muscle-invasive disease when compared to (much) older studies. Provide an explanation for the lower number found in this study (3pts).

- This is likely due to improved treatment over time (3 pts)

- This is likely due to improved treatment in the past two decades, possibly partly based on the use of a prognostic model. (3pts)

- Too short follow-up in part of the patients (3 pts)

c. Please take a look at Table 1.1 and explain in words the results for the factor ‘Age (Years)’.

(4 pts)

- There is a 1.04 times increased risk (0.5 pt) of getting progression (0.5 pt) ….

- … for each year increase in age (1 pt).

- This effect is statistically significant ….(1 pt)

- … because the p-value is <0.001 / because 1 is not in the 95% CI. (1 pt)

The authors of this article also looked at the classification groups and progression-free survival in a Kaplan-Meier Curve using the EAU-classification model combined with the FGFR3 and GATA2 status. They obtained the following figure 1.1.

Figure 1.1: Progression Free Survival curve for patients with Non-Muscle-Invasive Bladder Cancer. Dotted line, good health status; dashed line, moderate health status; solid line, poor health status. P-value is based on log-rank test

d. In figure 1.1, what is the 3-year progression free survival risk for the poor health risk group?

(2 pts)

The 3-year progression-free survival risk is around 0.8 or 80%. (2pts)

e. The log-rank test tells you that p < 0.01. Please provide one reason why only a log- rank test is not very informative. (3 pts)

- It only tells you that (at least) one of the curves is statistically different from another curve, but not which curves that might be. (3 pts)

- It only tells you that at least 1 of the curves are statistically significant different from each other, but it does not quantify this difference. (3 pts)

f. Take a look at the figure 1. Is this risk classification useful in the clinic? Please explain your answer. (5 pts)

- Yes, it might be because you now know that patients with a low risk should not be treated aggressively. But you don’t know what to do with the two other groups (5 pts)

- No, not really. The intermediate and poor risk groups have a risk of progression but if you treat everyone in these groups aggressively you are over treating many patients. (5 pts)

- No, the discrimination between the groups is too small (3 pts) - Yes, because the difference is statistically significant (1 pts)

No points will be given for just a yes or no answer.

Name:

Student Number:

Question 2

Q5, Q6: Modelling physiological systems – dr. T. Oostendorp (15 points)

Eye rotation

When studying the motion of the eye lid, the eye lid can be modelled as a mass-spring system. The agonist and antagonist muscle are modelled together as a single spring with spring constant 𝑘.

and resting position 𝑦₀. When the muscle moved the eye lid from open position 𝑦_b to closed position 𝑦_a (see figure 2.1), the resting position of the muscle pair changes value from 𝑦_b to 𝑦_a.

Two forces are involved:

1. The muscle force: 𝐹_muscle = 𝑘 𝑦₀− 𝑦 ,

2. The friction force 𝐹_friction, which is proportional to the eye lid velocity.

a. Explain, using Newton’s law (𝐹 = 𝑚 ∙ 𝑎) that the differential equation for the eye lid position 𝑦(𝑡) is (4 pt):

𝑚 𝑑^-

𝑑𝑡^-𝑦 𝑡 = 𝑘 𝑦₀− 𝑦(𝑡) − 𝛽 𝑑 𝑑𝑡𝑦(𝑡)

We start with Newton’s law: 𝐹 = 𝑚 ∙ 𝑎

The acceleration 𝑎 is the second order derivative of the position: 𝑎 =_/1^/0₀𝑦 𝑡 This yields

𝑚 𝑑^-

𝑑𝑡^-𝑦 𝑡 = 𝐹_muscle+ 𝐹_friction

𝐹_muscle is given above. 𝐹_friction is proportional to the velocity which is _/1^/ 𝑦(𝑡)

Figure 2.1 Eye lid geometry

b. Why is there a minus sign in front of 𝛽_/1^/ 𝑦(𝑡) in the differential equation? (2 pt)

The friction force is always in the opposite direction of the velocity, hence the minus sign.

We will use this model to study the motion the eye lid makes when it changes position from ‘closed’ to ‘open’.

c. Is 𝑦₀ a parameter or a variable in this study? Explain your answer. (2 pt)

It is a variable. It represents the resting position of the combined muscles. Hence, it does not represent a constant property of the body, but a state the body is in. The values of 𝑦₀ changes in order to open and close the eyes.

d. The figure below is the incomplete Simulink diagram for this study. Complete the diagram. (7 pts)

NB: Het k/m-blokje had ná het sum-blokje moeten zitten. Het wordt studenten niet kwalijk genomen als ze daardoor in de war raakten.

Name:

Student Number:

Question 3

Q5: Molecular Cancer Research – dr. P. Groenen (15 points)

I. Western blot: procedures and interpretation

Small cell lung cancer (SCLC), the most aggressive type of lung cancer, accounts for approximately 15% of lung cancer cases and is responsible for 25% of lung cancer-related deaths. Here, investigators studied AZD3965, an MCT1 specific inhibitor. They also

investigated the effects of hypoxia. Two cell lines were studied (COR-L103 and NCI- H1048), tubulin was used as reference protein.

Figure 3.1. Protein expression analyses of MCT1, MCT4 and tubulin in COR-L103 and NCI-H1048 cell lines treated with the MCT1 inhibitor AZD3965 in either a normoxic or a hypoxic environment.

a. Based on the results in Figure 3.1, what is the correct interpretation about the protein loading and what does this mean for the interpretation of the MCT1 and MCT4 protein differences observed in the cell lines COR-L103 and NCI-H1048? (4pts)

The amount of tubulin is similar in all lanes of the two respective cell lines, thus all lanes were loaded with approximately similar amounts of protein. (2 pnts) This means that the observed expression differences in MCT1 and MCT4 proteins are due to true expression level differences and not to differences in loading amounts. (2pnts)

b. Explain the effects of hypoxia on the AZD3965 inhibitor sensitivity for the two cell lines as shown in Figure 3.1. (4 pts)

Under hypoxic conditions MCT1 expression is downregulated in COR-L103 cells. The additional effect of the inhibitor under hypoxic conditions cannot be evaluated: the signal of lane 3 and 4 is equal. Also for NCI-H1048 hypoxia leads to lower MCT1 expression.

Addition of the inhibitor shows a clear decrease of MCT1 under hypoxic conditions.

II. Tumor heterogeneity

Targeted therapies have been developed to counteract processes of cancer proliferation by inhibition of phosphorylated kinases. However, in tumors several different cell

populations may arise, leading to tumor heterogeneity, as schematically illustrated in Figure 3.2.

In order to determine the diagnosis and treatment options for a patient with this tumor, tumor cells are obtained by collecting a tumor biopsy. In this biopsy, all five subclones depicted in Figure 3.2 are included (red, green, yellow, light blue and dark blue).

By DNA extraction and genetic analysis, five mutations are identified in this biopsy:

- an activating mutation in the PI3K oncogene - an activating mutation in the BRAF oncogene - an activating mutation in GNAQ/11 oncogene

- an activating mutation in the receptor tyrosine kinase (RTK) CKIT - an inactivating mutation in the tumor suppressor PTEN

Figure 3.2 Schematic illustration of the development of tumor heterogeneity. The different tumor populations are indicated with different colors.

Figure 3.3 Schematic overview of oncogenic intracellular pathways.

c. Based on the presence of these mutations and on the pathway scheme (Figure 3.3), what is the best combination of two targeted therapies from the list below that will effectively inhibit oncogenic signaling in all five subclones. Explain why this is the best combination therapy. (4 pts)

- MEK inhibition - BRAF inhibition - CKIT inhibition - PI3K inhibition - PTEN inhibition - GNAQ/11 inhibition

The right combination of targeted therapies is MEK inhibition + PI3K inhibition, since both these pathways are overactivated (2 points). Furthermore, activating signals from

GNAQ/11, CKIT and BRAF converge at MEK. Hence, BRAF inhibition and CKIT inhibition would only target a few subclones but not all. (2 points)

d. From a patient with melanoma, the melanoma tissue is obtained to identify the oncogenic pathway that is activated. By genetic analysis, an activating BRAF V600E mutation is identified in the melanoma. Next, a student in the lab performs a Western blot to determine the amount of phosphorylated ERK (pERK). The following antibodies are available: an antibody directed against phosporylated ERK and an antibody against tubulin. In addition, a tissue that has no activating BRAF mutation is available.

Please draw the Western blot that shows the activation status of the pathway in the melanoma. Please mind the incorporation of the necessary controls in the experiment (3 pts).

Model answer Blot:

Tissue 1 pERK phosphorylation band 1 point Control tissue Showing no phosphorylated ERK 1 point Tubulin antibody

used as a control

Showing similar loading of the two protein lysates

1 point

Name:

Student Number:

Question 4

Q6: Statistics – dr. J. in’t Hout (15 points)

Background The benefits of blood pressure lowering treatment for prevention of

cardiovascular disease are well established. However, the extent to which these effects differ by drug class or other factors is less clear. Ettehad et al. (Lancet, 2016) performed a systematic review and meta-analysis to clarify these differences.

Medline was searched for large-scale blood pressure lowering trials. Randomised controlled trials of blood pressure lowering treatment were eligible for inclusion if they included a minimum of 1000 patient-years of follow-up in each study arm.

In total, the authors included 123 studies with 613,815 participants.

a. How realistic is the danger of publication bias in this meta-analysis? (1p) Explain. (2p)

Not much danger of publication bias. (1p)

Trials were eligible for inclusion if they included a minimum of 1000 patient-years of follow- up in each study arm. These are large-scale trials, and consequently the chance that the results were published is also large, irrespective of the findings in the trial / the chance that the results of such big trials were not published is small.(2p)

b. The authors performed a fixed-effect meta-analysis. How are the weights calculated that are assigned to the studies, using a fixed-effect model? (2p)

The weight assigned to a study in a fixed-effect model is equal to the inverse (1p) of the variance of that study (i.e. the variance of the treatment effect estimate = SE²) (1p).

So 1/ SE². (total 2p)

c. Consider a small study, taken from a meta-analysis with quite some between-study heterogeneity. Is this study relatively more important (for the pooled result) in a fixed- effect or in a random-effects meta-analysis? (1p)

Explain your answer (2p).

The study is relatively more important in a random-effect MA (1p).

If there is between-study heterogeneity, the weights assigned to the studies in a random- effects MA will be more similar to each other than in a fixed-effect MA, because the weights are 1/(var+tau2) (1p). (If tau2 is very large, all weights are similar.)

Consequently small studies get relatively larger weights in a random-effects MA than in a fixed-effect MA. (1p)

d. To evaluate the relation between reduction in major cardiovascular disease events in relation to the achieved blood pressure reduction, the authors performed meta-

regression. This resulted in an estimated risk ratio (RR) of 0.80 per 10 mmHg of reduction.

Assume a cardiovascular disease event rate of 10% in the control group. Explain which event rate you expect in the experimental group if the difference in SBP between the control and experimental group is 15 mmHg? (4p)

An expected event rate of 7% in the experimental group (2p).

Per 10 mm Hg reduction in SBP the risk of major cardiovascular disease events is

estimated to be reduced with an RR of 80% à If the difference in SBP between the control and experimental group is 15 mmHg we expect 15/10 * 20% = 30% less events in the experimental group than in the control group (2p)

Figure 4.1. Meta-regression plot of the percentage risk reduction in all-cause mortality (y-axis) regressed against the difference in achieved systolic blood pressure (SBP) between treatment arms (x-axis).

e. Figure 4.1 shows the fixed-effect meta-regression results for all-cause mortality in relation to systolic blood pressure reduction. Studies are indicated with small and large circles (bubbles). Relate the bubble size to the importance of the studies (1p) and the weights of the studies (2p).

The circle sizes reflect the importance of the study in the meta-regression: the larger the more important the study is. (1p).

The bubble sizes (areas) reflect the weights of the studies, i.e. 1/variance. Studies with small variances (standard errors^2) have large weights and will be indicated with large circles. (2p)

Name:

Student Number:

Question 5

Q6: Measuring and modelling reflexes – dr E. Tanck & dr. T. Oostendorp (20 points)

Static exercise

A young man (80 kg) lifts his right leg from a vertical position to horizontal to train the quadriceps muscle and holds his leg in a position of 45° (see figure below). A physical therapist is interested to know how high the forces are in the hip joint and the m.

quadriceps when the leg is at a position of 45° during this static exercise. You are assisting him with the calculations.

Known data: the young man weights 80 kg, his leg weights 8 kg, use g = 10 m/s² for the gravitational acceleration. You can assume that all distances, masses, angles and attachment points for ligaments and muscles are known.

Figure 5.1

a. Create a free body diagram (FBD) to calculate the forces in the hip joint and the m.

quadriceps. Make your diagram large enough so everything can be labeled clearly.

Create a legend to specify all components of the FBD. (7 points)

Full page needed for question a

b. Give the Equilibrium equations that belong to your FBD (4 points). You do not have to calculate the forces.

Halve a page needed for question b

c. What happens to the force in the m. quadriceps when the man lifts his leg to 90°

degrees, so parallel to the floor? Choose from: higher than, lower than or equal to the force when lifting to 45°, and explain your answer using an equilibrium equation. (4 points)

Quarter page needed for question c

The quadriceps is innervated by the femoral nerve. A researcher wants to record the activity of the femoral nerve. He finds that the amplitude of the ElectroNeuroGram (ENG) from the femoral nerve is about 40 𝜇V, while the amplitude of ElectroMyoGram (EMG) from the quadriceps muscle is about 700 𝜇V.

d. Explain why the amplitude from the quadriceps is so much larger than that from the femoral nerve. (2 points)

The cell membranes produce the currents the generate the EMG and ENG. Muscles contain a lot more membrane than a nerve, and consequently produce a lot more current.

Because the ENG from the femoral nerve is so small, the researcher plans to use signal averaging in order in improve the signal-to-noise ratio.

e. Explain why it does not make sense to use signal averaging when recording voluntary activity of the femoral nerve. (3 points)

In order to average multiple recordings of a signal, the recordings have to be aligned. In responses to electric stimulation, you can align the recordings because there is a well- defined, simultaneous starting point of all fibers in the nerve. In voluntary activity, the activity in the fibers in not synchronized, and it is not possible to determine a reproducible starting point.

End of the exam. Did you write your name and student number on the first page of each question?