Question

The amount of oxygen used by a house fly, at a steady temperature of 15°C, can be compared to its oxygen use at a steady temperature to 25°C. Knowing that increased temperature leads to increased biochemical reaction rates, and that flies are an incredible collection of biochemical reactions, you might reasonably predict that the rate of oxygen consumption by the flies is going to be higher at 25°C than it was at 15°C. In fact, the rate of increase for many insects is about 2.5 times more oxygen use at 25°C versus 15°C. The calculated rate of increase, called Q.10 (see Unit 6), here is equal to 2.5, the slope of a line connecting the two data points (oxygen-consumption rates on the y-axis and temperature on the x-axis). The "10" in Quo refers to the 10°C increase from 15°C to 25°C. The circadian timing mechanism of plants and animals is overall expression of a series of chemical reactions in these organisms. Does heating up plants and animals speed up these reactions? Imagine that there are 24 sequential chemical reactions, each lasting about an hour, that determine tau. If you heat these reactions up, do you expect the reactions to go faster? If so, wouldn't that shorten tau? Researchers in this field measure the tau (period) of circadian rhythms at two temperatures, calculate the ratio of the two taus, yielding Q10 to be considered and explained. 7A. Sunflowers exude a type of sap on a regular schedule each day, suggesting the operation of a biological clock. The following parameters were measured. GROWTH TEMPERATURE (°C) TAU OF EXUDATION CYCLE (HRS) VOLUME OF EXUDATION IN 24 HRS (MICRO-LITERS/PLANT) 20°C plants 25.3 89 30°C plants 24.1 223 All plants were grown in 12:12D before the experiment, and they were kept under constant dim light during data collection. Your assignment: calculate the Q10 for each of the two different variables. For the period data, calculate Q10 as a simple ratio with the data from the lower temperature as the numerator and the data from the higher temperature as the denominator. For the volume data, calculate Q10 as a simple ratio with the data from the higher temperature as the numerator and the data from the lower temperature as the denominator. For each variable, write two sentences explaining whether or not you find evidence of temperature dependence and of temperature compensation. You can score a 5% bonus challenge in this assignment by explaining why the numerators and denominators switch places in the two different types of sunflower measurements ... you can figure this out if you think of the mathematical relationship between frequency and period. Label this part of your assignment as "Attempt to score 5% bonus." 7B. Here are more "rhythms" data, from three other species. Calculate the three different Q.10 values. Calculate one for each species and for each write a paragraph discussing your results as evidence of temperature compensation. organism value of tau (hours) Euglena gracilis (alga) Euglena gracilis (alga) Phaseolus multiflorus (bean) Phaseolus multiflorus (bean) Lacerta sicula (lizard) Lacerta sicula (lizard) temperature (°C) range tested 15 25 15 25 16 26 26.2 23.2 28.3 27.2 26.1 25.0 I

Answer 1

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