heartbeat at the room temperature;
2. Measure heartbeat at the increase in temperature by 10 °C;
3. Measure heartbeat at the decrease in temperature by 10 °C;
4. Calculate the temperature coefficient and activation energy using the formulas;
5. Make conclusions of the observed phenomena and prepare a report.
Equipment and Materials:
Thermometer (0 – 30 °C), vessel for the heart (50 mL), thermostat, ice, hot water, Ringer-Lock solution for the cold-blooded, stopwatch, dissecting tools (scissors, scalpel, dissecting needle, tweezers, gauze.
Procedure:
In this lab work students should register the heartbeat of frog’s heart at 2 or 3 different temperatures, with differences between each other by 10 °C).
After the immobilization of the frog, the heart should be taken from the chest and placed it in a vessel full by 2/3 with Ringer-Lock solution for the cold-blooded.
1st measurement is after 5-7 min. Heartbeat during 1 min is measured 3-4 times and the average is calculated.
Picture 1.1. 1 – thermometer; 2 – vessel; 3 – Ringer-Lock solution; 4 – frog’s heart
Measure room temperature. Place the vessel with the heart in the thermostat with the temperature 10 °C higher than room T. After 3-4 min perform the 2nd measurement with the calculation of the average.
Return the vessel to the room conditions and observe restoring of the initial heartbeat. After 3-5 min the heartbeat during 1 min is measured 3-4 times and the average is calculated.
Put the heart into the camera being cooled by the mixture of ice and water. When the temperature would be 10 °C lower than room’s temperature, heart rate should be measured again, the measurements should be done 3-4 times and calculated mean of heart rate.
Return the vessel to the room conditions and observe restoring of the initial heartbeat again. After 3-5 min the heartbeat during 1 min is measured 3-4 times and the average is calculated.
All data should be written in table 1.1.1
Table 1.1.1
Example:
During the experiment we have been seeing that the frog’s heart at 18 °C (Т1 = 273 + 18 = 291) has HR (heart rate) = 31 beat/minute, and at 28 °C (Т2 = 273 + 28 = 301) – 60 beat/minute. So, Q10=60/31=1,9.
In the table Bradis value of lgQ10 is 0,27875. By inserting obtained data into the formula (1.2) calculate the energy of activation:
Report design. The results are inserted into the table and the calculations are made. Define measurement deviations and make conclusions.
Laboratory work № 2
Determination of the temperature coefficient and calculation of the action energy of respiration of elodea plant branch
Objective: to get sure with the experiment that physical chemical laws are maintained in the living systems.
Tasks:
1. Count the amount of gas bubbles at room temperature;
2. Count the amount of gas bubbles during the rise of the temperature by 10 °C;
3. Count the amount of gas bubbles during the declination of the temperature by 10 °C;
4. To calculate the temperature coefficient and the energy of activation using formulas.
5. To make a conclusion and process the results.
Equipment and materials: Elodea plant in the small vessel, two big vessels with water more and less than room temperature by 10 °C; thermometers, timer, Bradice table.
Procedure:
To complete this work you need observe the gas bubbles emission by the elodea plant in the 3 different vessels with waters with temperatures which differ from each other by 10 °C.
Elodea is a typical alga of most of the aquariums, which with sufficient amount of light emits the oxygen bubbles from the tips of its leaves. You can count the quantity of emitted bubbles by watching the plant.
The first count of gas bubbles is done after 5-7 minutes after termostation of the vessel with plant. To make it, you should place the vessel with elodea into the bigger vessel with fixed temperature of the water. Then, you should count the emitted quantity of gas bubbles during 1 minute. You should repeat it for 3-4 times and the find an average quantity. Also you should define the temperature under the normal (room) conditions.
Then you should transport the vessel with plant into a thermostat with water hotter by 10 °C than in the first one. After 3-5 minutes count again the quantity of bubbles, repeat it 3-4 times and the find the average.
After that, you should transport the vessel with elodea into the first one with room temperature to retain the initial gas emission state. Then after 3-5 minutes repeat the counting like previous times.
Repeat all the same but with the vessel with colder water.
Fill the table 1.2.1.
EXAMPLE.
During the experiment following was obtained: at the temperature 18 °C (Т1 =273 + 18=291) 31 gas bubble was emitted but at the 28 °C (Т2 =273 + 28=301) 60 bubbles per minute. Thus, Q10 = 60/31 = 1,9. Using the table Bradis find lgQ10= 0,27875. By inserting obtained data into the formula (1.2) calculate the energy of activation:
Е=0,46 · 291·301·0,27875 = 11,239 kkal/mol.
Тable 1.2.1.
Report design: Fill the table the obtained results and calculate the energy of activation using formulas. Find the error of calculations and make conclusion.
Chapter 1. «Thermodynamics» questions:
How the dependence, which is expressed via this formula, is called?
1. To which type of thermodynamic systems living organisms belong?
2. Why our planet is considered an open thermodynamic system?
3. Which animals regulate their body temperature by homeostatic regulation (change of the metabolic processes) and change in the behavior.
4. What is an energy of activation – Е?
5. Give an examples of warm-blooded and cold-blooded animals.
Chapter 2
WORKING PRINCIPLE OF PH-SENSOR
Chemically clear water partially dissociates on ions:
Thereby forming a H + ion rapidly hydrated (attaches to itself a water molecule) to form hydronium ion H3O+. However, for simplicity instead of hydronium talk about hydrogen ion. The law of mass action applied to the dissociation of water of any aqueous solution expressed in this equation:
where, K – a constant of dissociation of water or its solution.
The constant of dissociation of water at 22 °C is equal 1,8 °C10-16, that is degree of dissociation of water is very small. Therefore concentration of not dissociated molecules [H2O] can be considered a constant. At its expression in moths on 1 l it turns out:
From