Research Exam Semester 3 – 2020-2021 January 15, 2021:
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In case of technical problems, mail Merel.Adjobo-Hermans@radboudumc.nl or call 06-19401341.
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.
During the exam, you may want to consult these books (https://lib- guides.ru.nl/friendly.php?s=ebooks):
• 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
Question 1 – Cancer etiology and prognosis – Prof. Dr. B. Kiemeney (20 points)
Note: the following abstract is fictitious and does not relate to a real study.
Association between playing on synthetic turf pitches with rubber granulate and leu- kemia in children, a case-control study in the USA
Abstract
Introduction: There is an increased use of synthetic grass fields in several sports, like hockey and soccer. Usually, such fields contain rubber granulate recycled from used car tires. This rubber granulate contains dozens of harmful substances like the genotoxic poly- cyclic aromatic hydrocarbons (PAHs) and Bisphenol A (BPA). Internal exposure to these substances is possible through oral or dermal uptake. An increased risk of leukemia has been suggested for children playing on synthetic grass fields.
Methods: We conducted a population-based case control study involving 246 cases and 2587 controls, aged 8 until 18, from January 2016 to December 2019 in the USA. Odds Ratios (OR) were calculated while controlling for several confounders simultaneously, for playing at synthetic grass fields as well as per unit of time and duration of sporting at such fields.
Results: 120 cases and 1593 controls had no or only rare exposure to rubber granulate on sports fields. 126 cases and 994 controls reported exposure (≥1 hours per week for at least one year): OR 1.15 (95% CI: 0.85 – 1.54). Children who had exposure for more than 5 years had an OR=1.32 (95% CI: 0.92-1.87) while for children with exposure £5 years the OR=0.99 (95% CI: 0.69-1.41). There was no difference in risk between children who played more intensively on synthetic fields (> 2 hours per week) versus children who played less intensively (£2 hours per week): both OR=1.11.
Conclusion: The present study supports an association between exposure to rubber granulate and leukemia risk, especially for longer durations of exposure. Studies with more power are needed to confirm this finding.
A. The authors obtained Odds Ratios (ORs) for this study while controlling for several confounders simultaneously. Which statistical method should you use to obtain such adjusted ORs? (1 point)
Logistic regression analysis (1 pt)
B. Provide two potential confounders for which you would adjust in these analyses and explain for each factor how this factor may work as a confounder. (3 points / con- founder; max 6 points)
1 point for confounder; 2 points for correct explanation
Age (older children have another risk of leukemia and can have a longer history of expo- sure).
Gender (boys and girls may have a different risk of leukemia and boys may be exposed more to synthetic sports fields)
Lifestyle factors such as smoking (older children may have smoked since their early pu- berty and this may be related to sports activities).
Environmental factors / SES (higher SES children may sport more and live in healthier envi- ronments).
Any other sensible answer.
C. The authors analysed intensity and duration of exposure as determinants. Explain how they could have done these analyses more effectively. (5 points)
It would have been more effective to make a combined variable (1 point) by the product (2 points) of the number of hours per week and the number of years exposed (comparable with the number of pack-years with smoking) and to analyse this as a continuous variable (2 points).
D. The authors used a case-control study to examine this association. They did not con- sider conducting a cohort study. Name, for this specific research question, two ad- vantages and two disadvantages of such a cohort study. Explain your answers. (1 point / each correct (dis)advantage, max 4 points)
Advantage:
There will likely be less misclassification of exposure because children / parents don’t have to recall exposure.
You don’t have to focus on leukemia anymore; there may be more risks of the exposure. In a cohort study you can collect all.
You can also collect data on internal exposure which would give much stronger evidence.
Any other sensible answer.
Disadvantage:
For good exposure measurements, you would have to go back to the study participants in a periodic way. This makes it less feasible.
You would need to do an enormously large study to collect a sufficient number of leukemia patients for a well powered study.
It may be very difficult to keep tracking your study participants because at these agents par- ents tend to move a lot.
Any other sensible answer.
Note: advantages that are just the opposite of disadvantages or the other way around will not be credited.
E. Calculate the Population attributable fraction (PAF) for exposure to rubber granulate through sports on synthetic fields (≥1 hour per week for at least 1 year) and provide the interpretation of its result (4 points).
q = 994 / 2587 = 0.38
!(#$%&)
!(#$%&)(&
=
).+,(&.&-%&)).+,(&.&-%&)(& = 0.054 à 5.4% (1 point)
Thus if we would remove exposure to rubber granulate on synthetic sports fields, in this population 5.4% of leukemia would be prevented, provided that the results are valid and rubber granulate is indeed the causative factor. (3 points)
Question 2 – Modelling the Acoustic Reflex – Dr. T. Oostendorp (15 points)
The acoustic reflex protects the fragile hair cells in the inner ear against loud noise. When a loud noise is perceived, the Tensor tympani muscle (see figure 2.1) contracts and pulls the malleus (Hamer in Dutch), and with it the tympanic membrane (ear drum), inward. As a consequence, the tympanic membrane becomes less sensitive to sound. In this question we will model the acoustic reflex.
Figure 2.1 Middle ear anatomy.
Figure 2.2 shows the diagram of the model we will use.
Figure 2.2 Diagram of model for acoustic reflex. In this diagram, the Tensor tympani muscle has contracted and consequently the tympanic membrane has stretched. 𝑥 = 0 is the position of the center of the tympanic membrane when it is not stretched (i.e. when the Tensor tympani muscle is relaxed).
When the tympanic membrane is pulled away from its equilibrium position at 𝑥 = 0, it ex- erts a force 𝐹2 = −𝑘 𝑥(𝑡), with 𝑥(𝑡) the position of the center of the membrane.
The malleus, with mass 𝑚, is fixed tightly to the membrane. The mass of the membrane and muscle is neglected in this model.
The Tensor tympani muscle exerts a force 𝐹8, that is controlled by the acoustic reflex.
Finally, when the malleus and membrane move, a friction force occurs that is proportional to their velocity: 𝐹9 = −𝑏;<; 𝑥(𝑡).
A. Explain that the differential equation for this model is (3 points):
𝑚 𝑑>
𝑑𝑡>𝑥(𝑡) = 𝐹8− 𝑘 𝑥(𝑡) − 𝑏 𝑑 𝑑𝑡 𝑥(𝑡)
Newton’s law: 𝐹 = 𝑚 ∙ 𝑎, with F the sum of all forces, and a the acceleration ;A
;<A𝑥(𝑡).
There are 3 forces: the muscle force 𝐹8, the force by the membrane 𝐹8 and the friction force 𝐹9. Substituting the strength of these forces leads to the above equation.
Naively, you might expect that evolution would try to minimize the friction force, in order to make the reflex fast. Actually, it would not be beneficial if the friction force would be zero or very small.
B. What undesired effect would occur if the friction force would be zero? (3 points)
The membrane would start to oscillate at the slightest action of the tensor tympani muscle.
C. What will be the equilibrium position 𝑥& of the center of the membrane when the muscle exerts a constant force 𝐹&? (3 points)
In equilibrium, all derivatives are zero. The differential equation then reduces to 0 = 𝐹8− 𝑘 𝑥(𝑡)
So, in equilibrium 𝐹8 = 𝑘 𝑥(𝑡), which means that 𝑥&= 𝐹&/𝑘.
As long as there is no sound, the force exerted by the Tensor tympani muscle is 0. We will use the model to simulate what happens when there is no sound from t=0 to t=10 s, and then a loud, constant sound after 10 s, which makes the Tensor tympani muscle to exert a constant force 𝐹&.
For 20/21 students who learned to use R-scripting:
D. Fill in the missing parts, indicated by highlighted dots, of the script below. (6 points)
duration <- 20 deltaT <- 0.01
nSteps <- duration/deltaT m <- 1
k <- 100 b <- 20 F1 <-10
t <- numeric(nSteps) v <- numeric(nSteps) x <- numeric(nSteps) v[1] <- 0
x[1] <- 0
for (i in 1:nSteps) {
t[i+1] <- t[i]+deltaT if (………)
F = ………
else F = F1
Ftot = F-k*x[i]-b*v[i]
a=Ftot/m
v[i+1]=v[i]+deltaT*a ………
}
plot(t, x, type='l', xlab='Time (s)', ylab='x (mm)', main='tympanic membrane stretch')
duration <- 20 deltaT <- 0.01
nSteps <- duration/deltaT m <- 1
k <- 100 b <- 20 F1 <-10
t <- numeric(nSteps) v <- numeric(nSteps) x <- numeric(nSteps) v[1] <- 0
x[1] <- 0
for (i in 1:nSteps) {
t[i+1] <- t[i]+deltaT if (t[i+1]<10)
F=0 else F=F1
Ftot = F-k*x[i]-b*v[i]
a=Ftot/m
v[i+1]=v[i]+deltaT*a
x[i+1]=x[i]+deltaT*v[i+1]
}
plot(t, x, type='l', xlab='Time (s)', ylab='x (mm)', main='tympanic membrane stretch')
For 19/20 (and earlier) students who learned to use Simulink:
D1 What type of block is missing at the red cross? (2 points) An integrator
D2 Is k or b the correct parameter for the triangular block marked y? (2 points) b
D3 Give the correct sequence of plusses and minuses (from top to bottom) of the addi- tion block (left of the 1/m block). (2 points)
+--
Question 3 – Molecular Cancer Research – Dr. F. Doubrava-Simmer &
Dr. E. Oosterwijk (15 points)
Part I – Signal transduction pathways in cancer – F. Doubrava-Simmer (6 points) The figure below shows a schematic overview of intracellular signaling pathways that are often altered in cancer due to specific mutations. These intracellular signaling pathways drive cell proliferation and survival. They become overactive.
Figure 3.1 Intracellular signal pathways.
Targeted therapies have been developed to counteract these processes. Examples of tar- geted inhibition are:
- MEK inhibition - BRAF inhibition - KIT inhibition - PI3K inhibition - PTEN inhibition - GNAQ/11 inhibition
In order to determine the diagnosis and treatment options for a patient with a lung tumor, a bronchial brush sample is taken. Cytological and molecular analyses are performed. By DNA extraction and sequencing, five mutations are identified in the tumor cells:
- an activating mutation in the PI3K oncogene
- an activating mutation (V600) in the BRAF oncogene - an activating mutation in GNAQ/11 oncogene
- an activating mutation in the receptor tyrosine kinase (RTK) KIT (CKIT) - an inactivating mutation in the tumor suppressor PTEN
A. Based on the presence of these mutations, figure 3.1 and the example targeted therapies listed above, choose two types of targeted therapy from the list that in combination could effectively inhibit oncogenic signaling in this tumor. Explain your answer. (3 points)
Answer: The right combination of targeted therapies is MEK inhibition + PI3K inhibi- tion, since both these pathways are overactivated. Furthermore, activating signals from GNAQ/11, EGFR and BRAF converge at MEK. Hence, BRAF inhibition and EGFR inhibition would only target a few subclones but not all.
B. Targeted therapies are generally not completely without side-effects and resistance is also almost always developed. These are disadvantages that should be taken along in making a treatment decision.
Explain how tumors can acquire therapy resistance. (3 points)
Answer: Therapeutic selective pressure leads to expansion of a pre-existing sub-clonal population or evolution of drug-tolerant cells (a few cells may barely survive but manage to
acquire a new driving mutation). The mutations that are acquired lead to: 1) activation of other component of the pathway, making the pathway independent of the component that is inhibited; 2) change in the conformation of the target that prevents the effect of the inhib- itor; 3) alternative pathway activation.
Part II – Experimental steps of Western Blotting – F. Doubrava-Simmer (3 points) Simply put, Western blots can be used to look at proteins from cell extracts. However, set- ting up a good experiment requires many considerations, such as: which antigen should I target; which controls should I use, what buffer composition will work best etc.
Here we give you the details of two antibodies targeting the AKT protein.
Property Anti-Akt1 antibody Anti-phospho-Akt antibody
antibody product type primary antibodies primary antibodies
immunogen
Akt1 antibody was raised against a 16 amino acid pep- tide from near the N-terminus of human Akt1
Peptide encompassing and in- cluding phosphorylated
Ser473 of human Akt1.
conjugate unconjugated unconjugated
clone polyclonal monoclonal
biological source rabbit mouse
species reactivity human, rat, mouse human, rat, mouse
application - IHC suitable suitable
application - ELISA suitable suitable
application - western blot suitable suitable
C. Explain the three differences between these two antibodies, and why these differences are important. (3 points)
Answer:
1) The immunogen/antigen is different: the left AB targets the start of the protein (that pro- vides a lipid-binding module to direct Akt to PIP3 at the cell membrane that is necessary for its activation), whereas the AB on the right targets the region containing the Ser473 (C- terminal hydrophobic motif) and ONLY binds when this amino acid is phosphorylated.
Therefor with the left AB you can analyse the presence of the AKT protein, and with the right AB you can analyse activation of AKT.
2) The left AB is polyclonal; the right is monoclonal. A polyclonal antibody is actually a mix of antibodies (each antibody recognizes a different epitope on the same antigen), whereas a monoclonal antibody is just one antibody that specifically detect only one particular epitope on the antigen. Effectively, polyclonal antibodies may give a stronger signal or more background compared to monoclonal antibodies.
3) The host species of the two primary antibody is different. The AB on the left was gener- ated in a rabbit the one on the right in a mouse. This is important to take into account when choosing the secondary antibodies. The secondary antibodies should be against the host species of the primary antibody.
Part III – Hypoxia in cancer AND analysis of experiments – E. Oosterwijk (6 points) During hypoxia many genes are upregulated, amongst others MCT1. A large drug com- pany developed an MCT1 inhibitor to investigate whether it might be useful as therapeutic
intervention (kill hypoxic cells). In the experiment below, four cell lines were exposed to normoxia (N) or hypoxia (H) combined with the MCT1 inhibitor AZD3965.
Figure 3.2 Expression of MCT1 and MCT4 protein assayed by Western blotting.
D.Based on Figure 3.2, what conclusion can be drawn about the MCT1 expression levels of the 4 cell lines in the untreated condition? (1 point)
Answer: The MCT1 expression levels differ with NCI-H1048> COR-L103/DMS79>
DSM114
E. Based on Figure 3.2, explain the observed MCT1 expression patterns under hypoxic conditions, and whether the inhibitor is specific for MCT1 or also works on MCT4. What overall conclusion can be drawn about this inhibitor? (2 points)
Answer: Under hypoxic conditions, MCT1 levels in COR-L103 and DMS79 are already greatly reduced and the addition of the inhibitor does not add anything substantial. The drug inhibits MCT-1 levels in NCI-H1048 but not in DSM114. MCT4 levels are not inhibited by the drug. In conclusion, the inhibitor is specific, but the effect is cell line dependent.
Figure 3.3 Analysis of MCT4 overexpression.
F. MCT4 expression is often implicated as resistance factor for MCT1 inhibition. In Figure 3.3 the effect of MCT4 overexpression in the AZD3965-sensitive cell line NCI-H1048 was determined. NCI-H1048 cells were engineered to inducibly overexpress GFP or MCT4 in response to doxycycline treatment. Is the induction by doxorubicin successful? Explain what you see in the figure. (1 point) Answers: Yes, the induction worked. After doxorubicin treatment MCT4 is induced under normoxic and hypoxic conditions as visible in the doxo- rubicin treated versus non-treated cells.
G. Is MCT4 expression hypoxia dependent in the NCI-H1048 cell line? Explain your an- swer. (1 point) Answer: MCT4 induction is NOT hypoxia dependent, as can be seen in fig- ure A and B. Levels do not differ.
H. Why was CAIX expression included in Figure 3.3? (1 point) Answer: CAIX was included to control for the induction of hypoxia-regulated genes
Question 4 – Statistics: Meta-analysis and more – Dr. J. in ‘t Hout (15 points)
Based on the following article: S. Sze et al., Ethnicity and clinical outcomes in COVID-19:
A systematic review and meta-analysis, EClinicalMedicine (2020), https://doi.org/10.1016/j.eclinm.2020.100630 (from Lancet)
Background Understanding the relationship between ethnicity and COVID-19 is an urgent research priority, in order to reduce the disproportionate burden of disease in Black, Asian and other minority ethnic groups in the USA and UK. In particular, it is important to disen- tangle whether worse reported outcomes in ethnic minority groups are attributable to an increased risk of becoming infected, developing severe COVID-19 pneumonia or death.
The authors therefore conducted a systematic review and meta-analysis of both published and preprint research to study the association of ethnicity with COVID-19 infection, ITU (in- tensive therapy unit) admission, and death.
Fig. 4.1. Pooled adjusted risk of SARS-CoV-2 infection by ethnicity (Reference group:
white, RR = Risk Ratio).
A. What is the name of this plot in Figure 4.1? (1 point) Forest plot
B. Why are the I-squared value (1 point) and p-value (1 point) for the native Americans missing? (in total 2 points)
They are missing as there is only 1 study; at least 2 studies are needed to calculate the I- squared. If there is no I-squared, one cannot calculate whether the corresponding p-value is larger than 0 or not à also the p-value is missing.
C. Interpret in words the pooled result (RR and its significance) for the Hispanics. (2 points)
Adjusted RR 1.77, 95% CI 1.39-2.25. i2=0, p>0.8
- The pooled adjusted risk for COVID-19 infection in the Hispanics is 1.77 higher than for white people.
- This difference is statistically significant: the 95% CI excludes 1.
The figure shows the adjusted RR: the authors included the models which most closely matched their a priori chosen confounders of age, sex, deprivation, obesity, and comorbid- ities. Despite this adjustment, the researchers used a random-effects model.
D. (2 points) Explain the major assumption in a random-effects model with regard to the ethnicity effects (1 point) and explain the difference with a fixed-effect model (1 point).
i. The RE model assumes that the underlying true effects in the studies have been sampled from a distribution of true effects (not one common true effect). (1p) ii. In a fixed-effect model it is assumed that there is one common true effect underlying to all studies. (1p)
E. Indicate exactly two valid reasons why the authors have chosen for a random effects model. (2 points)
The authors may have chosen for a RE model because:
- probably not all studies have used the a priori defined confounders to adjust their treatment effect estimates.
-the effect of ethnicity may depend on more factors than the ones for which was corrected.
- the definition of ethnicity may vary across studies, also dependent on number and type of di- agnostic tests available and testing policy.
- definition of the outcomes (e.g. infection) may vary across studies (two valid reasons needed, 1p per valid reason)
There was a large variation in study sizes: some studies were based on at most 100 per- sons whereas others were based on over 10 million.
F. Show how the study weights in a random-effects model are calculated. (1 point) The weight for a study in a random effects meta-analysis is equal to
weight = 1 / (SE2 + tau2),
G. Is the weight of a small study in this random-effects meta-analysis smaller, equal or larger than it would have been in a fixed-effect meta-analysis of the same data? Men- tion also the role of the heterogeneity observed in this meta-analysis in your answer. (2 points)
The weight in a fixed effect meta-analysis is equal to 1/SE2. (0.5p)
The overall i2 is 84% 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 studies. This means that 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 small study will be larger in a RE meta-analysis than in a FE meta-analysis.(1.5p)
Fig. 4.2 Asymmetry plot for studies on infection rates of black vs. white people.
H. The plot in Figure 4.2 is created to evaluate the effect of a certain phenomenon. Which phenomenon, and what is the name of such a plot? (1p)
Funnel plot to evaluate possible publication bias (1p)
I. What does Figure 4.2 suggest with respect to between-study heterogeneity? Explain.
(1p)
It shows a substantial between-study heterogeneity as most studies are outside the tri- angle.
J. What does Figure 4.2 suggest with respect to publication bias? Explain. (1p)
Suggests no publication bias, there is no clear imbalance between left and right part (see also Eggers test).
Question 5 – Measuring and modelling reflexes – Dr. E. Tanck & Dr. T.
Oostendorp (20 points)
Statics
A professional baseball player has an injury of the deltoid anterior muscle due to overuse and repetitive motion of the arm. The deltoid anterior muscle has its insertion on the humerus and the origin on the clavicle (see fig 5.1). Other muscles involved in the shoulder flexion movement may be ignored.
Fig 5.1 left: The deltoid anterior muscle has its insertion on the humerus and the origin on the clavicle; right: the man is moving his arm with the weight from vertical to horizontal po- sition.
After some rest, the physiotherapist recommends to do some exercises to strengthen the shoulder muscle. One of the exercises is a single-arm shoulder flexion with a low-weight dumbbell of 1 kg. The physiotherapist likes to know the forces in the shoulder joint (contact force) and the deltoid anterior muscle when the arm is in static equilibrium in the horizontal position.
Known data: The weight of the man is 80kg. Use g = 10 m/s2 for the gravitational accelera- tion.
A. Describe precisely what free body diagram (FBD) you would draw to calculate the forces in the shoulder joint (contact force) and the deltoid anterior muscle. Describe all components of the FBD including the direction and location of these compo- nents. (5 points)
Answer
• Contour: upper arm, virtual cut at shoulder joint (1 point).
• Fg: gravitational force of arm, vertical direction, located at center of mass arm (1p).
• Fw: force of weight, vertical direction, located at hand (1p).
• Fcx and Fcy: contact force components in x and y direction. Location: con- tact point at humurus (1p).
• Fd: force of m. deltoid ant. Direction: from humerus to clavicle. Location: at attachment point of humerus (see fig 5.1) (1p.)
• x-y coordinate system
B. Another man is doing another exercise and is standing on his toes (see FBD of the foot in figure 5.2, NB figure is not on scale). You see a FBD with joint contact forces (Fcx and Fcy) and the force in the achilles tendon (Fa). The direction of Fa is 80 de- grees from the horizontal plane.
Figure 5.2 Free Body Diagram of the foot of the man
Write down the equilibrium equations and use [invoegen vormen / tekstvak] to draw the moment arms in figure 5.2 (4 points)
∑
∑
∑
Answer (1 point each)
∑ Fx = 0
Fcx + Fa (cos80) = 0
∑ Fy = 0
Fn-Fgf-Fcy+Fa (sin80) = 0
∑ Mp = 0 (p=at contactpoint Fcx/Fcy) Fn.n-Fgf.f-Fa.a=0
Drawings (1 point)
C. At the left of figure 5.3 you see also a FBD with resultant ankle forces (FRX and FRY) in the joint center, and a resultant ankle moment (MR). The biomechanical situation of the left and right FBD is similar. Known data: Fcx= -173,6N; Fcy= 1364,8N;
Fn=400N; Fgf= 20N; Fa=1000N; moment arm of Fa to the joint center = 0,04 m
Figure 5.3 Free Body Diagrams of the foot of the man.
Calculate Frx, Fry and Mr using the known data and the FBDs. Motivate your answers. (6 points)
Answer (2 points each) a f
n
Frx=Fcx+Fax
Frx= - 173,6+1000(cos80)= 0 N
Fry=Fcy-Fay
Fry=1364,8-1000(sin80) = 380 N
Mr=Fa.a
Mr=1000.0,04= 40 Nm
Figure 5.4 Recorded EMG.
D. In order to assess the activity of the deltoid anterior muscle, its EMG is recorded us- ing surface electrodes. Figure 5.4 shows an example of the recorded EMG. Explain why you can conclude that during this particular recording the muscle was exerting only minimal force. (3 points)
During this recording, the activity of individual motor units can be seen. If the amount of force increases, more motor units are recruited. The electric activity of the motor units will merge, and the individual motor units will no longer be visible.
E. A colleague thinks that the nerve innervating the deltoid anterior muscle might be damaged. He suggests to record the activity of that nerve during voluntary move- ment.
Explain why, in contrast to muscles’ activity, it is not possible to record nerve activity during voluntary movement. (2 points)
Compared to muscles, nerves contain very little tissue, and consequently a small amount of ion channels. During voluntary movement, only a small fraction of the ion channels will be active simultaneously. Hence, the surface potential that is generated by all active ion chan- nels is below the level of noise that is always present in skin electrode recordings. Conse- quently, the voluntary activity of the nerve cannot be recorded.
End of the exam.
