2019 Bio H2 MCQ Explanations

1. N19Q1 D Not only should any correct option be biologically correct, but it must also answer the question. Statement 4 does not explain why cells are considered to be the smallest unit of all life since it only relates to multicellular organisms. 2. N19Q2 B Options A and D were the most frequently selected incorrect answers. Information indicates that the stain is specific for phospholipids rather than proteins. An explanation for the presence of the red -coloured structures must account for the removal of the red colour from most of the cell. 3. N19Q3 A Options B and D are incorrect because nuclear envelope and 80S ribosomes are found in eukaryotic cells not in E. coli (prokaryotic cell). Option C is incorrect because RNA polymerase is used to transcribe DNA to mRNA not translate mRNA to protein. 4. N19Q4 B Statement 1 is true. At optimum temp of 86°C for thermophilic bacteria, the percentage unfolding of the enzyme is more than half at 58%. Statement 2 is true. The psychrophilic bacteria begins unfolding at 38°C and is 100% unfolded at 48°C which is within the 15°C range. Statement 3 is false. At 62°C, for mesophilic bacteria, the enzyme is 100% unfolded but the activity of the enzyme remains high at about 97%. 5. N19Q5 C Graph X reaches a higher concentration of product within the same time limit. Hence, it has a higher rate of enzyme reaction which could be due to higher enzyme concentration or higher substrate concentration. (Hence, options narrowed to B and C). Graph Y has a steeper initial gradient but reaches the same concentration of product as the original reaction. Hence, it has the same concentration of substrate but higher rate of reaction due to higher temperature (greater number of effective collisions between E and S). This excludes option B and leaves answer as option C. To confirm, graph Z has a lower initial rate of reaction and reaches a lower maximum product concentration in the same time frame. Hence, graph Z has a lower substrate concentration. 6. N19Q6 C From the figure, it can be seen that the structure of arabinose and glucose are similar except for slight differences. The graph shows that as concentration of glucose (substrate) increases, the effects of arabinose inhibition can be overcome and the graph will reach the same V max as the uninhibited enzyme graph. Hence, this is an example of competitive inhibition. Note: option B is not correct because arabinose does not have the same structure as glucose but a similar structure. 7. N19Q7 D Only totipotent stem cells can give rise to genetically modified individuals 8. N19Q8 B H: refers to the entire DNA strand including the nitrogenous bases, hence it is the polynucleotide strands. J: enzyme adding free dNTPs to the 3’ end of the leading strand → DNA polymerase K: enzyme joining the ends of the Okazaki fragments together → DNA ligase 9. N19Q9 A H7N10 does not exist, hence options B & C cannot be the answer. For o