標題:
問BIO 1999 CE MC Q.8 and Q.9
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發問:
問BIO 1999 CE MC Q.8 and Q.9 可唔可以解釋下個答案問黎?
最佳解答:
Q.8=C Starch is insoluble in water , the presence of starch does not affect the water potential of the solution . In solution 2 , starch is broken down by amylase into maltose which is soluble in water . The maltose produced reduces the water potential of solution 2 . Thus water potential of solution 2 is much lower than that of solution 1 . Net movement of solution 2 is faster than that solution 1 . Hence , liquid level in he capillay tube of set-up 2 is much higer than that of set - up 1 . Dialysis tubings prevent movements of larger molecules such as starch and sugar molecules , but water molecules are small enough to pass through . Q.9=C (1)is incorrect because a narrower dialysis tubing reduces the surface area for osmosis as well as the sugar content , which the latter implies that less water is required for the solution to reach the same sugar concentration . The rate of increase in water level will be lower . (2)is correct because an increase in unit volume of solution in set - up 2 produced a larger increase in liquid level since the cross setion of the tube is smaller . The rate , therefore , increasees . (3)is correct because raising the temperature can increase the enzymactic digestion of starch by amylase , as well as the rate of osmosis process .
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