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State three structural similarities between oviduct and sperm duct AND 11. A group of students observed five cells along the diameter of field of view shown below, using eyepiece lens magnification of

Biology

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a

480 µm\text{480 µm}

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10. State three structural similarities between oviduct and sperm duct

  1. Both are tubular structures.
  2. Both have muscular walls (smooth muscle) for peristaltic movement.
  3. Both are lined by epithelial tissue.

11. A group of students observed five cells along the diameter of field of view shown below, using eyepiece lens magnification of ×5\times 5 and objective lens magnification of ×15\times 15.

Part a· 3 steps

Estimate the actual length of one cell in micrometres

  1. Calculate the total magnification.

    Total Magnification=Eyepiece Magnification×Objective Magnification\text{Total Magnification} = \text{Eyepiece Magnification} \times \text{Objective Magnification} Total Magnification=5×15=75×\text{Total Magnification} = 5 \times 15 = 75 \times

  2. Determine the diameter of the field of view at the given magnification.

    Assumption: The diameter of the field of view at 100×100 \times magnification is 1800 µm1800 \text{ µm}.

    Diameter at 75×=Diameter at 100××10075\text{Diameter at } 75 \times = \text{Diameter at } 100 \times \times \frac{100}{75} Diameter at 75×=1800 µm×43\text{Diameter at } 75 \times = 1800 \text{ µm} \times \frac{4}{3} Diameter at 75×=2400 µm\text{Diameter at } 75 \times = 2400 \text{ µm}

  3. Calculate the actual length of one cell.

    Since five cells span the diameter:

    Actual length of one cell=Diameter of field of viewNumber of cells\text{Actual length of one cell} = \frac{\text{Diameter of field of view}}{\text{Number of cells}} Actual length of one cell=2400 µm5\text{Actual length of one cell} = \frac{2400 \text{ µm}}{5} Actual length of one cell=480 µm\text{Actual length of one cell} = \boxed{\text{480 µm}}

Part b

State any two limitations of using the above method in estimating cell size

  1. Cells may not be uniform in size or shape, leading to an average estimate that doesn't reflect individual cell variations.
  2. Cells may overlap or have spaces between them, making it difficult to accurately count them or measure the total length they occupy.

12. State two reasons why animals have a complex excretory system

  1. To efficiently remove toxic metabolic waste products such as urea, uric acid, and ammonia, which are generated from metabolic processes.
  2. To maintain homeostasis by regulating water balance, electrolyte concentrations, and blood pH within narrow limits.
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