J. A. CAMPBELL H a n s y Mudd Collage Claramont, California 9171 1
Questions Q112. T h e human heart consumes a minimum of about 9 ml 02/100 (g-min), of which about 2 ml is required t o maintain basic cellular processes, 1m l t o make the heart beat (without doing any pumping), and 4 ml to overcome friction in t h e muscle fibers and heart fluids. One ml of 0 2 generates 2 kg-m of energy = 4.8 cal. If the arterial blood pressure is 120 torr, what volume of blood does the average heart p u m p per minute? [The experimental figure is about 40 m l of blood/heartheat.] Q113. An experimental project a t Puerto Penasco, Sonora, Mexico grows vegetables in a colorless plastic-covered greenhouse whose atmosphere consists partly of the exhaust of the local diesel-electric generator. The atmosphere is continually recirculated and blown counter-current through sprayed sea water from the radiator of the diesel engine. Two effects are noted: (1) only about 5% of the normal amount of additional, fresh irrigation water is needed for t h e plants, (2) plants grow up t o twice a s fast. Interpret these two results in terms of molecular behavior. Q114. Most metabolic reaction steps fall into one of six types: photon gain or loss, electron transfer, H atom transfer, insertion or splitting out of H,+ OH,- or H z 0 (hydrolysis), insertion or splitting out of COz (carboxyla-
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tion and decarboxylation), group transfer (-NH2, -OP03, -CH3, -CHO, -COCH3, peptide and glycosides are known). Do you see anything in common among these reactions? If not, try only the first four. Q115. Osmotic pressure may be expressed, especially by biologists, in several units. T h e following are all for the same aqueous sodium chloride solution: 0.171 M , 0.32 osmoles, A 0.595"C, 7.17 atm. Define clearly what each of these units means and how its value would be determined experimentally for a solution of unknown concentration. Q l l 6 . A. Champagnot, et al. [Nature, 197 (4862), 13 (1963)l is quoted a s obtaining protein-rich products with more water-soluble vitamins than found in cereal when he cultured yeast on petroleum. What questions does this statement raise in your mind? This column consists of questions (plus possible, but certainly not uniquely satisfactory, answers) requiring no more than a cancurrent first-year, college level course, a data handbook, and a willinmess to a.~. ~ fundamental l v chemical ideas to the svstems whwh surround u * , o r r v r n are lnsidr u i l Cmtrihuiow t o r plss~hlcinrlu-wn .rrr .oltr~lcd lnitlat4 in rhc .lnt~nnrv IqT2 i~sucol This Journal.
Answers A112. 9-2-1-4 = 2 ml 02/(100 g-min) available for PAV work. Note one significant figure.
A114. All involve the gain or loss of one particle by one molecule to one other molecule. Thus, each step is probably a bimolecular reaction in which no more than one bond breaks 2[ml 02/(100 g-min)]x 4.8 [eal/ml Oz] = and/or one bond forms. The insertion, or splitting out, of HzO lO[cal/(lOOg-min)]= PAV may he an exception but H molecules are exceptionally small and mobile so may more readily participate in slightly more 10[ca1/(100g-mini] x 0.04l)l atm/cal] = complicated mechanistic steps. The rest of the entities men0.411 atm/(100 g-min)]= PAV tioned usually bond directly with their single "free" bond. A115 0.171 M = moles/l of aqueous NaCI. 0.32 osmoles = AV = 0.4 [I atm/(100 g-min)](120/760) atm X [I20 beats/min]; molarity of a nonelectrolyte (such as sugar) giving the same count your pulse to get this last figure. osmotic pressure as 0.171 M NaCI. I 0.59S"C = lowering in = 0.02[1/(100g x heat)] freezing point in 0.171 M NaCI,,,, compared to pure HzO. 7.17 atm = pressure which must be applied to a 0.171 M NaThe density of heart tissue is about 1 g/cm3, volume about CI,,,, solution to increase the escaping tendency of the water 300-400 em3 (about half of which is chambers). So tissue volto that in pure HzO. Each can be determined fbr an unume should be about 200 em3 or 200 g and I V = 2 X 0.02 = knownsolutionsssuggestedineachdefinition. 0.04 1 or 40 ml/beat. A116. Yeast growing on petroleum would he expected to AIIJ. 1) The closed greenhouse and the high humidity readapt to their food and synthesize mostly non-polar, water insulting from blowing the atmosphere through the heated water soluble substances partly because of the availability of such from the radiator minimize net evaporative lasses from plant starting materials favoring "on-polar products and partly beand sail so that the water actually needed for incorporation cause of a scarcity of the atoms necessary to make polar, into the growing plant is a larger fraction of the total (smaller) water-soluble ones. On the other hand, vitamins are seldom water requirement. 2) The faster growth is probably due to a ) prepared in large yields. In an aqueous medium they might be the higher average temperature in the house .(day and night), extracted by the water and escape from the yeast. In a hydrob) the more favorable water supply, c ) absence of wind and carbon medium there would be little tendency for any waterwind-blown contaminants. These facton favor a high rate by soluble vitamins to enter the petroleum phase, and they might, supplying the necessary activation energy, a higher eoncentratherefore, be prepared in higher concentration under such cantion of water,. and lower concentrations of possible inhibitors ditions. The question is which effect predominates and is eito the rate determining steps. ther at'the above mechanisms a possible interpretation?
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