9700/11

Biology 9700/11October/November 2011

Cambridge AS Level · Multiple Choice (AS Level) · answer key with instant marking and worked solutions

40
questions
40
marks
60
minutes

Topics Cell Structure · Biological Molecules · Nucleic Acids and Protein Synthesis · Transport in Mammals · Gas Exchange · Cell Membranes and Transport · +6 more

Tap an option under each question to check it — your score builds as you go.

Q11MCell StructureFree sample

What is the diameter of a typical prokaryote, such as Streptococcus?

Options

A   7.5×101 nm7.5 \times 10^1\ \text{nm}
B   7.5×102 nm7.5 \times 10^2\ \text{nm}
C   7.5×100 µm7.5 \times 10^0\ \text{µm}
D   7.5×101 µm7.5 \times 10^1\ \text{µm}

DifficultyMedium-Easy
Worked solution

Working

Typical prokaryotes (e.g. Streptococcus) have diameters in the range of approximately 0.5–2 µm. Converting the options to µm:

  • A: 7.5×101 nm=75 nm=0.075 µm7.5 \times 10^1\ \text{nm} = 75\ \text{nm} = 0.075\ \text{µm} — too small
  • B: 7.5×102 nm=750 nm=0.75 µm7.5 \times 10^2\ \text{nm} = 750\ \text{nm} = 0.75\ \text{µm} — correct
  • C: 7.5×100 µm=7.5 µm7.5 \times 10^0\ \text{µm} = 7.5\ \text{µm} — too large
  • D: 7.5×101 µm=75 µm7.5 \times 10^1\ \text{µm} = 75\ \text{µm} — far too large

Answer

B

Final answer

B

Detailed explanation

Background Concept

Prokaryotic cells (bacteria and archaea) are much smaller than eukaryotic cells. A typical bacterium such as Streptococcus has a diameter of roughly 0.5–2 µm. By comparison, a typical eukaryotic animal cell is about 10–100 µm across, and a typical plant cell about 10–200 µm. The small size of prokaryotes is one reason they have a very high surface area to volume ratio, which is important for the rapid exchange of materials across the plasma membrane by diffusion.

Unit conversions in microscopy:

  • 1 µm=103 nm=1000 nm1\ \text{µm} = 10^3\ \text{nm} = 1000\ \text{nm}
  • 1 nm=103 µm=0.001 µm1\ \text{nm} = 10^{-3}\ \text{µm} = 0.001\ \text{µm}

Understanding the Question

This is a multiple-choice question (Paper 1) that tests two things at once: (1) knowledge of the typical size of a prokaryotic cell, and (2) the ability to convert between nanometres and micrometres. The four options are all expressed in different orders of magnitude or units, so a quick unit conversion exposes the correct answer.

Approach

Convert each option into a single, common unit (µm is convenient) and compare the resulting number to the known size range of a typical bacterium (~0.5–2 µm diameter). The option that falls within this range is the answer.

Step-by-Step Reasoning

  • Option A: 7.5×101 nm=75 nm7.5 \times 10^1\ \text{nm} = 75\ \text{nm}. Convert: 75÷1000=0.075 µm75 \div 1000 = 0.075\ \text{µm}. This is far too small — only large viruses approach this size.
  • Option B: 7.5×102 nm=750 nm7.5 \times 10^2\ \text{nm} = 750\ \text{nm}. Convert: 750÷1000=0.75 µm750 \div 1000 = 0.75\ \text{µm}. This lies comfortably within the 0.5–2 µm range for typical bacteria, including Streptococcus.
  • Option C: 7.5×100 µm=7.5 µm7.5 \times 10^0\ \text{µm} = 7.5\ \text{µm}. This is too large for a prokaryote; it is more in the size range of small eukaryotic cells or organelles such as the nucleus.
  • Option D: 7.5×101 µm=75 µm7.5 \times 10^1\ \text{µm} = 75\ \text{µm}. This is comparable to a typical eukaryotic animal cell and far too large for a bacterium.

Therefore, option B is the only value consistent with a typical prokaryotic diameter.

Key Takeaways

  • Typical prokaryotic cell diameter: ~0.5–2 µm (or 500–2000 nm).
  • 1 µm=103 nm1\ \text{µm} = 10^3\ \text{nm}, so always convert to a common unit before comparing values.
  • Prokaryotes are roughly 10–100× smaller than eukaryotic cells.

Common Mistakes

  • Not converting units before comparing — a number in nm looks larger than the equivalent value in µm and can mislead.
  • Confusing prokaryote size with virus size (~20–300 nm) or eukaryotic cell size (~10–100 µm), leading to the wrong order of magnitude.

Things to Be Careful About

  • Standard form in the options (7.5×1017.5 \times 10^1, 7.5×1027.5 \times 10^2, etc.) must be evaluated correctly — 7.5×100=7.57.5 \times 10^0 = 7.5, not 0.
  • The answer is the option whose converted value sits in the typical prokaryotic range, not just the option that looks smallest.
Techniques used
convert between units of length (nm and µm)recall the typical size of a prokaryotic cell

The rest of this paper

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  • Q40(outdated) Ecology1M
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