Resit Research Exam Semester 3 – 2020-2021 - Resit May 6, 2021, 13.00-15.30 hr
NOTE ON SCIENTIFIC INTEGRITY:
- By taking the exam, the student declares that no plagiarism is or will be committed. If the lecturer has the suspicion that fraud has been commit- ted, the student will be contacted. If needed, the case will be redirected to the Examination Board.’
- We expect you to take this exam individually, without consulting fellow students or others
INSTRUCTIONS
§ Hand in your answers before 15.30 hr. Students with special facilities can submit until 16.00 hr.
§ Write your name and student number on the first page of each question!
§ The questions must be answered in English. If you cannot remember a specific English term, you may use the Dutch term.
§ Be precise in your answers. Adding correct but irrelevant information will not increase your score. Adding incorrect information, even if it is irrelevant, will lower your score.
§ You are allowed to use a calculator of the type Casio FX-82MS.
§ During the exam, you may want to consult 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
https://libguides.ru.nl/friendly.php?s=ebooks:
§ The form with statistical formula’s is available at Brightspace.
Name:
Student Number:
Question 1 – Cancer etiology and prognosis – Prof. Dr. B. Kiemeney (20 points)
Sugary drink consumption and risk of cancer: results from NutriNet-Santé prospective cohort.
Abstract, British Medical Journal (BMJ)
Introduction: To assess the associations between the consumption of sugary drinks (such as sugar sweetened beverages and 100% fruit juices), artificially sweetened beverages, and the risk of cancer.
Design: Population based prospective cohort study.
Setting and participants: Overall, 101 257 participants aged 18 and over (mean age 42.2, SD 14.4; median follow-up time 5.1 years) from the French NutriNet-Santé cohort (2009-2017) were included. Consumptions of sugary drinks and artificially sweetened beverages were assessed by using repeated 24-hour dietary records, which were
designed to register participants' usual consumption for 3300 different food and beverage items.
Main outcome: Prospective associations between beverage consumption and the risk of overall, breast, prostate, and colorectal cancer were assessed by Cox proportional hazard models, accounting for competing risks. Hazard ratios were computed.
Results: The consumption of sugary drinks was significantly associated with the risk of overall cancer (n=2193 cases, hazard ratio for a 100mL/day increase 1.18, 95%
confidence interval 1.10 to 1.27, P<0.0001) and breast cancer (N=693, HR=1.22, 95%CI 1.07 to 1.39, P=0.004). The consumption of artificially sweetened beverages was not associated with the risk of cancer. In specific subanalyses, the consumption of 100% fruit juice was significantly associated with the risk of overall cancer (N=2193, HR=1.12, 95%CI 1.03 to 1.23, P=0.007).
Conclusion: In this large prospective study, the consumption of sugary drinks was positively associated with the risk of overall cancer and breast cancer. 100% fruit juices were also positively associated with the risk of overall cancer. These results need
replication in other large-scale prospective studies. They suggest that sugary drinks, which are widely consumed in Western countries, might represent a modifiable risk factor for cancer prevention.
A. Please take a look at Table 1.1 and explain in words the association given by the Hazard Ratio for “Sugary drinks except 100% fruit juices” and “Breast cancer”. (3 points)
- There is a 1.23 times increased risk/hazard (0.5 pt)
- … of getting Breast Cancer (0.5 pt)
- … for each 100 mL/day increase in sugary drinks except 100% fruit juices (1 pt) - This effect is statistically significant …. (0.5 pt)
- … because 1 is not in the 95% CI. (0.5 pt)
B. What is the risk of cancer of someone who drinks half a liter per day (=500 mL/day) of
“sugary drinks except 100% fruit juices” compared to someone who doesn’t consume such drinks at all? Please make use of Table 1.1. (3 points)
That person’s risk is 2.39 fold increased (1.195 = 2.39). (3 pts) Table 1.1: Hazard ratio’s for different types of cancers
All cancers Hazard ratio (95% CI)
Breast Cancer Hazard ratio (95% CI)
Colorectal Cancer Hazard ratio (95% CI) Sugary drinks except
100% fruit juices, per 100 mL/day:
1.19 (1.08 to 1.32) 1.23 (1.03 to 1.48) 1.11 (0.72 to 1.71) 100% fruit juices,
per 100 mL/day: 1.12 (1.03 to 1.23) 1.15 (0.97 to 1.35) 1.05 (0.75 to 1.46)
Figure 1.2: Cancer-free survival curve for participants using < 15 sugary drinks a week and > 15 sugary drinks a week. Dotted line is < 15 sugary drinks / week, solid line is ≥ 15 sugary drinks a week.
C. Please take a look at Figure 1.2. Is it possible to determine the median cancer-free survival time for the group of people that drink ≥ 15 sugary drinks / week? Please motivate your answer (you may draw inside the figure). (2 points)
You cannot estimate the median cancer-free survival time (1pts), because the median has not been reached yet (1 pt). or if they draw a line at the 50% of the y-axis (1pt)
D. A newspaper picked up this study and printed a new headline “Sugary drinks cause cancer!”. Provide 2 reasons why this conclusion is overstated. (4 points)
- There are still other (residual) confounding factors that might obscure the real association
- One study is not enough, replication is needed as stated in the conclusion
- Median follow-up time is 5.1 years, therefore only mid-term relationships could be examined. Some carcinogenic processes take decades.
2 pts per good reason
E. Suppose that 60% of the population drinks 100 mL of “sugary drinks except 100% fruit juices” per day and no-one drinks more than that. In that case, what is the population attributable risk for all cancers of 100 mL of “sugary drinks except 100% fruit juices” per day? Please explain your answer. (3 points)
!.#(%.%&'%)
!.#(%.%&'%))% = 0.10 thus 10 % (3 pts)
F. In this study, daily intake of sugar from sugary drinks was positively associated with overall cancer and breast cancer. What would happen to the Hazard Ratio’s (HR) for sugary drinks if the authors decided to correct in their analysis for sugar from sugary drinks? (3 points) And what would that imply? (2 points) (in total 5 points)
- The hazard ratio would probably go towards the null value of 1, or even become 1.
(3 pts) OR
- The hazard ratio will change drastically (extreme high numbers / CI’s) (3 pts).
- This would imply that there might be very strong relationship between sugar itself and getting cancer, instead of sugary drinks and getting cancer. (2 pts)
Name:
Student Number:
Question 2 – Modelling physiological systems – Dr. T. Oostendorp (15 points)
Angle meters
In movement sciences, angle meters are used to record the orientation of limbs (the same principle is used in smart phones). Figure 2.1 and 2.2 show an example.
Figure 2.1. Angle meter attached to Figure 2.2 Inside of the angle meter
the upper leg.
The piezoelectric element is compressible, and acts as a spring with spring constant .. The equilibrium length of the element is ℓ!. The element cannot bend sideways.
When being compressed, the piezoelectric element exerts a frictional force on the weight that is proportional to the velocity by which the element is being compressed: 0friction =
−2343 ℓ(5), with ℓ(5) the length of the element. The mass of the weight is 6, the mass of the piezoelectric element can be ignored. 7(5) is the angle between the axis of the angle meter and the horizontal plane (see figure 2.2), and 8 is the gravitation constant.
The angle meter is attached in such a way that displacement can be ignored; the only rele- vant motion is rotation.
A. Show that the differential equation for ℓ(5) is 6 9:
95:ℓ(5) = −2 9
95ℓ(5) − .(ℓ(5) − ℓ!) − 68 sin(7(5)) (4 points)
We start with Newton:
6 ∙ < = sum of all forces 6 9:
95:ℓ(5) = 0friction+ 0spring+ 0gravity
The spring force is proportional to the extension from the equilibrium length:
0spring = −.(ℓ(5) − ℓ!)
(there is a minus sign, since a lengthening the element will pull the weight backwards) The friction force is given: : 0friction = −2343 ℓ(5)
From fig 2.2 we see that ℓ(5) = >?(5) − >@(5)
The gravitation force along the axis of the angle meter is 0gravity = −68 sin(7(5))
Combining all of this yields 6 9:
95:ℓ(5) = −2 9
95ℓ(5) − .(ℓ(5) − ℓ!) − 68 sin(7(5))
The electric signal A(5) produced by the angle meter is proportional to the deformation of the piezoelectric element: A(5) = B(ℓ(5) − ℓ!).
B. What is A(5) when the angle meter is maintained at vertical orientation (7(5) = 90°) for a long time? (3 points)
0 = 0 − .(ℓ(5) − ℓ!) − 68 sin(90°).
This leads to
A(5) = <(ℓ(5) − ℓ!) = −<68 .
The step response of an angle meter can be obtained by instantaneously changing the ori- entation from 0° to 90°. Figure 2.3 shows the step responses of two different angle meters, A and B. Both angle meters are the same in every aspect, except that for one angle meter, the value of the friction parameter 2 is larger than for the other one.
Figure 2.3 Step responses of two different angle meters, A and B.
C. Which of the angle meters, A or B, is the one with the larger value for 2? Explain your answer. (3 points)
It takes much longer for B to reach the equilibrium, so it’s friction must be larger.
D. Fill in the missing parts, indicated by highlighted dots, of the script below. (5 points) duration <- 10
deltaT <- 0.01 nSteps <- ………
m <- 1 k <- 1000 b <- 50 L0 <- 5 g <- 9.81 c <- 100
t <- numeric(nSteps) v <- numeric(nSteps) L <- numeric(nSteps) S <- numeric(nSteps) v[1] <- 0
L[1] <- 5 S[1] <- 0
for (i in 1:nSteps) {
t[i+1] <- t[i]+deltaT if (t[i+1]<5)
phi=0 else
phi=pi/2 Ftot = ………
a=Ftot/m ………
L[i+1]=L[i]+deltaT*v[i+1]
S[i+1]=c*(L[i+1]-L0)
}
plot(t, S, type='l', xlab='Time (s)', ylab='S (V)', main='Angle meter output')
duration <- 10 deltaT <- 0.01
nSteps <- duration/deltaT m <- 1
k <- 1000 b <- 50 L0 <-5 g <- 9.81 c <- 100
t <- numeric(nSteps) v <- numeric(nSteps) L <- numeric(nSteps) S <- numeric(nSteps)
v[1] <- 0 L[1] <- 5 S[1] <- 0
for (i in 1:nSteps) {
t[i+1] <- t[i]+deltaT if (t[i+1]<5)
phi=0 else
phi=pi/2
Ftot = -m*g*sin(phi)-k*(L-L0)[i]-b*v[i]
a=Ftot/m
v[i+1]=v[i]+deltaT*a
L[i+1]=L[i]+deltaT*v[i+1]
S[i+1]=c*(L[i+1]-L0)
}
plot(t, S, type='l', xlab='Time (s)', ylab='S (V)', main='Angle meter output')
Name:
Student Number:
Question 3 – Molecular Cancer Research – Dr. F. Doubrava-Simmer & Dr. E. Ooster- wijk (15 points)
Part I. Evaluation of Western Blot setup
For visualization of the proteins ERK, AKT and GAPDH in a Western blot experiment, a re- searcher had the following antibodies at his disposal:
- Mouse-anti-human GAPDH antibody - Rabbit-anti-human ERK antibody - Goat-anti-human AKT antibody
- Goat-anti-rabbit antibody with green fluorescent label - Goat-anti-mouse antibody with green fluorescent label - Rabbit-anti-goat antibody with green fluorescent label
A. Which antibody combinations should the researcher use? In addition, explain the ra- tionale for the use of each selected antibody. (3 points)
1. Rabbit-anti-human ERK ab, goat-anti-rabbit green fluorescent labeled to demonstrate ERK
2. Goat-anti-human AKT ab, rabbit-anti-goat green fluorescent labeled to demonstrate AKT 3. Mouse- anti- human GAPDH ab, goat- anti-mouse green fluorescent labeled
Part II. Evaluation of Western Blot analysis
To investigate the effect of the MCT1 specific inhibitor AZD3965 in combination with hy- poxia, a Western blot was performed using antibodies for MCT1, MCT4 and Tubulin and protein extracts from two cell lines derived from small cell lung cancer tissue DMS114 and DMS79. The result is shown in Figure 3.1.
Figure 3.1 Western blot analysis of MCT1, MCT4 and tubulin.
B. What is your interpretation of the result for tubulin in Figure 3.1? And, what does this mean for the interpretation of the MCT1 and MCT4 protein differences observed in the cell lines DMS114 and DMS79? (2 points)
The amount of tubulin is similar for all lanes of the two respective cell lines, thus all lanes were loaded with approximately similar amounts of protein (1 point). This means that the observed expression differences in MCT1 and MCT4 proteins are due to true expression and not to differences in loading amounts (1 point)
C. What is the effect of hypoxia on MCT1 expression? And, what is the effect of hypoxia on MCT4 expression? (2 points)
Hypoxia leads to an increase in MCT1 expression in DMS114, but a decrease in DMS79.
Hypoxia leads to an increase in MCT4 expression in both cell lines.
D. What is the effect of the inhibitor on MCT1 expression under hypoxic conditions? Explain the effect for both cell lines. (2 points)
Compare lane 3 and 4 of the DMS114 cells or the DMS79 cells: under hypoxic conditions the inhibitor does not influence MCT1 expression levels substantially (2 points).
Part III. Evaluation of Western Blot analysis & Signal Transduction
The BRAF V600E mutation occurs in approximately 8% of the colorectal cancers. To inves- tigate the effect of the BRAF inhibitor Dabrafinib, a Western blot was performed using nine different antibodies, and protein extracts from the colon cancer cell line HT29. The result is shown in Figure 3.2.
Figure 3.2 Western blot analysis of HT29 cells after drug treatment.
E. What does the “p” signify in p-EGFR? And explain the difference between p-EGFR and EGFR. (1 point)
p = phosphorylation. p-EGFR indicates the analysis of phosphorylated EGFR. This indi- cates activation of the protein. With EGFR you analyse the presence of the protein inde- pendent of activation. This indicates expression of the protein.
F. Explain the effect of the BRAF-inhibitor Dabrafenib on the colon cancer cell line. (2 points) In the colon cancer cell line pERK and pMEK reduced upon dabrafenib treatment, indicating the ERK pathways is inhibited. BRAF is upstream of these kinases. (1 point)
Moreover, pAKT and pEGFR are particularly detected upon Dabrafinib treatment. This means that inhibition of BRAF in colon cancer cells leads to activation of the AKT pathway.
(1 points).
G. Cetuximab is another targeted inhibitor with a different mode of action. This drug is a monoclonal antibody that binds to EGFR. The binding is with a higher affinity than that of the endogenous ligands, and in this way the drug inhibits EGFR activity.
What would be the effect on the HT29 cells of treatment with Cetuximab instead of Dabrafenib? Describe the expected image for the nine antibodies compared to the un- treated result from Figure 3.2. Also, include an explanation in your answer. (3 points) 1) The inhibitor will not affect expression. Thus, for the Abs targeting EGFR, MEK, ERG and AKT the intensity will be similar compared to no treatment. Additionally, HSP90 should be the same intensity, because this is the loading control in this experiment.
2) pMEK and pERK will be similar, because these are downstream of activated BRAF, and are therefore independent of EGFR activation.
3) Activation of EGFR is prevented by the drug, leading to lower intensity. AKT is down- stream of EGFR and therefore activation of pAKT will also be reduced.
Name:
Student Number:
Question 4 – Statistics: Meta-analysis and more – Dr. J. in ‘t Hout (15 points) Based on the following article:
Henssler, J., Müller, M., Carreira, H., Bschor, T., Heinz, A., & Baethge, C. (2020). Con- trolled drinking—non-abstinent versus abstinent treatment goals in alcohol use disorder: a systematic review, meta-analysis and meta-regression. Addiction.
The proportion of untreated patients with alcohol use disorder exceeds that of any other mental health disorder, and treatment alternatives are needed. A widely discussed strat- egy is whether controlled drinking up to a certain amount versus total abstinence is pre- ferred.
A literature search was conducted until February 2019 aiming at controlled (randomized and non-randomized) clinical trials (RCTs and non-RCTs) among adult populations with alcohol use disorder, with an intervention group aiming at Controlled Drinking (CD) and a control group aiming for alcohol abstinence (AA). This resulted in two RCTs and 12 non- RCTs (where participants could choose their preferred approach).
Figure 4.1 Forest plot of odds ratios of number of responders in the Controlled Drinking group versus the Alcohol Abstinence group (response: abstinence or controlled/low risk drinking; OR: Odds Ratio, SE: standard error, ln: natural logarithm, CI: confidence inter- val).
A. Figure 4.1 shows the overall results of the meta-analysis. Explain the “diamond” of the random effects model at the bottom of the forest plot in Figure 4.1. (1 point)
Represents a confidence interval of the pooled effect, center: point estimate, left and right:
lower and upper limit of 95% confidence interval.
B. Interpret in words the pooled result for the CD versus AA regimen (which regimen seems most beneficial, and its statistical significance) for the fixed effect model. Ex- plain your answer. (2 points)
There are less responders in the CD group compared to the AA group (OR point esti- mate 0.59), so the AA treatment seems beneficial. This difference is statistically signifi- cant as the 95% CI excludes the value 1.
C. Explain why the weight of the study of Adamson 2010 is relatively larger in the fixed ef- fect analysis than in the random effects analysis. Use the formulas for the weights in both settings. (2 points)
The weight in a fixed effect meta-analysis is equal to 1/SE2. (0.5p)
The overall i2 is 63% which means that there is quite some heterogeneity between the studies: tau2 >0. This tau2 will be added to the denominator of the weights for all stud- ies: wgt ~1/(SE2+tau2). As a result of this, the weights of all studies in a RE meta- analysis will be more similar than in a FE meta-analysis. Consequently, the weight (%) of a large study (Adamson 2010) will be more extreme, thus larger, in a FE meta-analy- sis than in a RE meta-analysis.(1.5p)
D. The number of months of follow-up of the participants in the studies may also play a role in the results. Below you can find the results of a meta-regression for follow-up du- ration. Explain how a meta-regression is conducted in this situation (dependent varia- ble, independent variable, statistical model, experimental unit). (2 points)
estimate se zval pval ci.lb ci.ub
intrcpt -1.0035 0.3030 -3.3118 0.0009 -1.5974 -0.4096 ***
Follow_up 0.0427 0.0187 2.2907 0.0220 0.0062 0.0793 * Meta-regression is here a linear regression on the study estimates. Dependent vari- able ln(OR), Experimental unit: study. Independent variable: Follow-up duration. In this regression each study is an observation.
E. Explain whether the results of the CD or the AA regimen improve after longer follow-up.
(2 points)
Odds ratio is for CD vs. AA. After long follow-up the OR becomes larger, which means that the number of responders in the CD group increases, so the results of the CD regimen improve.
F. Give two valid reasons for the observed heterogeneity of 63% (apart from duration of follow-up). (2 points)
RCT vs observational design, wide range of calendar years, possible variation in definition of success, psychological/social support, severity of alcohol problem, cul- tural differences, etc.
G. The authors also conducted meta-regression for the relation with calendar year. The results are presented in figure 4.2. What is the name of this plot? (1 point)
Fig. 4.2 Plot of meta-regression results.
H. How is the size of the circles related to the imprecision of the study estimates? (1 point) The larger the bubble, the more precise the study estimate of the OR.
Year
Treatment effect (log odds ratio)
1980 1990 2000 2010
-2.5-2.0-1.5-1.0-0.50.00.5
Fig. 4.3 Funnel plot.
I. Figure 4.3 shows a funnel plot. What is the aim of a funnel plot, and related to that: why is the x-axis on logarithmic scale? (2 points)
Aim: evaluation of asymmetry in the plot. Such asymmetry may be caused by publica- tion bias. In order to evaluate asymmetry, the OR is plotted on logarithmic axis, be- cause the ln(OR) is symmetric and normally distributed. This helps in evaluating the asymmetry.
0.1 0.2 0.5 1.0 2.0 5.0
0.80.60.40.20.0
Odds Ratio
Standard Error
Name:
Student Number:
Question 5 – Measuring and modelling reflexes – Dr. E. Tanck & Dr. T. Oostendorp (20 points)
Statics
A young woman (60 kg) is doing static exercises to strengthen the m. triceps (see figure 5.1 in which two positions of the static exercise are visible). A physical therapist is inter- ested to know how high the ground reaction force is when the arm of the woman is (about) straight and the rope is at (about) 60 degrees from the horizontal plane (figure on the right).
Known data: The weight of the woman is 60kg. The force that the woman is applying to the rope is 100N. Use g = 10 m/s2 for the gravitational acceleration.
Figure 5.1 Exercises to strengthen the m. triceps.
A. Describe precisely what free body diagram (FBD) you would draw to calculate the ground reaction force. Describe all components of the FBD including the direction and location of these components. (5 points)
Answer
• Contour: woman with virtual cut at rope (1 point).
• Fg: gravitational force of woman, vertical direction, located at center of mass (1p).
• Fr: force of rope. Location at hand (1p). In the direction of rope; from hands
• Fnx and Fny: ground reaction force components in x and y direction. Loca- tion: below feet (1p).
• x-y coordinate system
B. The woman is doing another exercise and is standing on her toes (see FBD of her foot in figure 5.2, NB: figure is not on scale). You see a FBD with resultant ankle forces (FRX
and FRY) in the joint center, and a resultant ankle moment (MR).
Figure 5.2 Free Body Diagram of the foot of the woman.
Write down the equilibrium equations and draw the moment arms in figure 5.2 (4 points)
∑
∑
∑
Answer (1 point each)
∑ Fx = 0 Frx = 0
∑ Fy = 0 Fn-Fgf-Fry = 0
∑ Mp = 0 (p=at contactpoint Frx/Fry)
Answer drawings (1p)
f n
C. At the right of figure 5.3 you see also a FBD with joint contact forces (Fcx and Fcy) and the force in the achilles tendon (Fa). The biomechanical situation of the left and right FBD is similar. Known data: Fn=300N; Fgf= 20N; FRX=0N; FRY=280N; MR= 15Nm;
Fa=750N (direction at 80 degrees from horizontal plane).
Figure 5.3 Free Body Diagrams of the foot of the woman.
Calculate Fcx, Fcy and the moment arm of Fa to the joint center. Motivate your answers.
(6 points)
Answer (2 points each)
Frx=Fcx+Fax
0 = Fcx+750(cos80) à Fcx= -130N
Fry=Fcy-Fay
280=Fcy-750(sin80) à Fcy=280+750(sin80) = 1018 N Mr=Fa.a
15=750.a à a=15/750 = 0,02m
The ground reaction force is recorded using a force plate. Figure 5.4 shows the signal of the force plate as the woman is changing her posture.
Figure 5.4
D. What sample rate should be used in order to obtain a clear rendering of the signal? Ex- plain your answer. (2 points)
The figure shows that the highest frequency component in the signal is about 0.2/5=0.04 s (this estimate does not have to be very precise). This corresponds to a frequency of 1/0.04
= 25 Hz. You need a sample rate of about 10 samples per period of the highest frequency component. In this case that is 250 Hz.
In order to avoid aliasing, the signal must be filtered prior to sampling.
E. What kind of filter should be used, and what should be its cut-off frequency? Explain your answer. (3 points)
Aliasing will occur for frequencies higher than half the sample rate, so higher than 125 Hz (this should be consistent with your answer to question d). To make sure that the signal does not contain frequencies higher than 125 Hz, you need to use a low-pass filter with 125 Hz as the cut-off frequency.
End of the exam. Did you write your name and student number on the first page of each question?
