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Bio Mc hkcee level (3,35題)

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圖片參考:http://imgcld.yimg.com/8/n/HA00861954/o/701004110066613873382451.jpg 3,35題 ans係A,C,, why?????????? 更新: Green is dominant character if E with C produce yellow offspring(homozygous recessive) ,, The offspring must received a yellow allele from E,, so E is heterozygous. what do you mean ''cannot determine the yellow seed allel is come from E or C.'' ~?? 更新 2: >>We cannot determine C or E's genotype since they have the same phenotype. C must received a yellow allele from B, where B is homozygous recessive. And phenotype of C is green C must be heterozygous 更新 3: Btw,, what do you mean''by comparison''? thx a lot 更新 4: It is crosses(to produce offspring), not compare their appearance(it is to compare appearance of their offspring) . why option (1) is wrong~~??

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2. A 頭30秒的amount of gas 來compare. 3. B Actually, I think the answer is B, Since amount of enzyme doubled, rate of breakdown increases. Then since the amount of hydrogen peroxide is the same, overall volume of gas produce is the same. 35. C , (2) and (3) are correct. Since if E with C, it cannot determine the yellow seed allel is come from E or C. If E with E, it can determine the genotype of E if the offspring has yellow seed allele If E and D, it can determine the genotype of E if the off spring has yellow seed allele, and if there is no yellow seed in the next generation, E has genotype of GG, so that the offspring is Gg 2010-04-11 23:58:11 補充: Sorry that I think wrongly, The option (1)C with E means comparasion, We cannot determine C or E's genotype since they have the same phenotype. However, (2) is true, (3) E with D, by comparasion, we can determine E havs a different genotype with D 2010-04-12 12:06:18 補充: Comparasion means compare their phenotype, in order to guess their genotype. Make it easier, to compare their appearance, so that to guess the pair of alleles the seeds have. 2010-04-12 17:38:56 補充: Since C,E have the same phenotype, you cannot distinguish what's their genotype difference by just observing the appearance of C and E

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