. When holding an object, does the number of motor units remain the same? Are the same motor units used for the duration of holding an object? Why or why not

Electromyography Lab
Personnel:
1 Subject and 1 Tester
Equipment:
Biopac MP36 system
Hand dynamometer
EMG electrodes
Procedures:
Use Biopac Electromyography II Lab (BSL-02)
Test #1:
The subject will perform 5 successive squeezes of the hand dynamometer (isometric contractions). The contractions will last 2-3 seconds with 5 seconds in between contractions. Each contraction will be of increasing effort (i.e., 20% effort, 40% effort, etc…) Press RECORD to begin recording data and press STOP to end recording.
Reflex time will be measured as the time from rest to maximum EMG signal during trial #5 (maximum contraction).
Test #2
The Subject will perform a MAXIMAL squeeze (contraction) of the hand dynamometer for 60 seconds. The contraction MUST BE MAXIMAL (DO NOT “PACE” YOURSELF). Press RECORD to begin recording data and press STOP to end recording.
Data – SUBJECT #1:
Test #1 (Motor Unit Recruitment)
Dominant arm Non-dominant arm
Trial
# Approximate Force Increment Force at peak
(kg) Avg Integrated EMG (mV) Force at peak
(kg) Avg Integrated EMG (mV)
1 20% 5.14kg  0.045 5.24 0.052
2 40% 9.67 0.85 9.41 0.081
3 60% 17.52 .0.147 14.77 0.13
4 80% 25.38 0.232 23.26 0.199
5 Maximum 29.46 0.275 26.89 0 .256
Test #1 (reflex time) : ____0.83__________________________
(time to maximal EMG during trial #5)
Test #2 (fatigue)
Dominant arm
Max clench Force (kg) 50% of Max clench Force (kg) Time to Fatigue EMG at Max Clench Force (mV) EMG at 50% of Max Clench Force (mV) % Change in EMG (%)
31.35 15.68  31.76 .34 19.147 50%
Max Clench Force = Maximum force recorded during the entire test
50% of Max Clench Force = (Max Clench Force / 2)
Time to fatigue = Amount of time it took drop to the 50% of Max Clench Force value during the test
EMG at Max Clench Force = EMG (mV) when Max Clench Force occurred
EMG at 50% of Max Clench Force = EMG (mV) when 50% of Max Clench Force occurred
% Change in EMG (%) = ((100% EMG – 50% EMG / 100% EMG) * 100)
Data – SUBJECT #2:
Test #1 (Motor Unit Recruitment)
Dominant arm Non-dominant arm
Trial
# Approximate Force Increment Force at peak
(kg) Avg Integrated EMG (mV) Force at peak
(kg) Avg Integrated EMG (mV)
1 20% 2.76 0.055 3.96 0.074
2 40% 5.18 0 .109 6.23 0.119
3 60% 8.56 0.167 8.86 0.171
4 80% 10.26 0.194 11.75 0.204
5 Maximum 13.6 0.206 12.56 0.224
Test #1 (reflex time) : 0.502_
(time to maximal EMG during trial #5)
Test #2 (fatigue)
Dominant arm
Max clench Force (kg) 50% of Max clench Force (kg) Time to Fatigue EMG at Max Clench Force (mV) EMG at 50% of Max Clench Force (mV) % Change in EMG (%)
20.66 10.33 30.76  0.39 0.18 50%
Max Clench Force = Maximum force recorded during the test
50% of Max Clench Force = (Max Clench Force / 2)
Time to fatigue = Amount of time it took drop to 50% of Max Clench Force
EMG at Max Clench Force = EMG (mV) when Max Clench Force occurred
EMG at 50% of Max Clench Force = EMG (mV) when 50% of Max Clench Force occurred
% Change in EMG (%) = ((100% EMG – 50% EMG / 100% EMG) * 100)
Questions:
1. When holding an object, does the number of motor units remain the same? Are the same motor units used for the duration of holding an object? Why or why not?

2. During the 60 second Fatigue test, do muscle forces and EMG (mV) change at the same rate during the test? If not, what accounts for this difference?

3. As you fatigue, muscle force decreases. What physiological processes explain this decline?

4. During EMG measurement, we generally use “integrated EMG” v. the actual raw EMG data. Why? Explain.

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