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What are the results of the Benedict's test and iodine test for starch? Explain the functions of vitamins, proteins, and mineral salts. Calculate the volume of air escaped from soil samples A and B.

Biology

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Here are the solutions to the questions:

1. (a)

  • i) The steps describe the Benedict's test, which is used to detect the presence of reducing sugars.
  • ii) An alternative reagent that could be used in place of Benedict's solution is Fehling's solution.

1. (b)

  • i) The student is likely to observe a blue-black color on the piece of bread.
  • ii) The conclusion drawn is that starch is present in the bread.

1. (c)

  • I. Vitamins: Regulate body processes, protect against diseases, and maintain overall health.
  • II. Proteins: Essential for the growth and repair of body tissues, and for the production of enzymes and hormones.
  • III. Mineral salts: Important for the formation of bones and teeth, nerve and muscle function, and maintaining fluid balance.

2. (i)
To complete the table, we read the final volume from the diagrams and calculate the volume of air escaped. The initial volume of each soil sample is 100 cm3100 \text{ cm}^3. The scale on the diagrams has major markings at 10 cm310 \text{ cm}^3 and 50 cm350 \text{ cm}^3, with 4 divisions between them. This means each division represents (50−10)/5=8 cm3(50-10)/5 = 8 \text{ cm}^3.

  • Soil A: Final volume = 10 cm310 \text{ cm}^3. Volume of air escaped = 100−10=90 cm3100 - 10 = 90 \text{ cm}^3.
  • Soil B: Final volume = 10+(3×8)=10+24=34 cm310 + (3 \times 8) = 10 + 24 = 34 \text{ cm}^3. Volume of air escaped = 100−34=66 cm3100 - 34 = 66 \text{ cm}^3.
  • Soil C: Final volume = 10+(2×8)=10+16=26 cm310 + (2 \times 8) = 10 + 16 = 26 \text{ cm}^3. Volume of air escaped = 100−26=74 cm3100 - 26 = 74 \text{ cm}^3.
  • Soil D: Final volume = 10+(4×8)=10+32=42 cm310 + (4 \times 8) = 10 + 32 = 42 \text{ cm}^3. Volume of air escaped = 100−42=58 cm3100 - 42 = 58 \text{ cm}^3.
  • Soil E: Final volume = 10+(1×8)=10+8=18 cm310 + (1 \times 8) = 10 + 8 = 18 \text{ cm}^3. Volume of air escaped = 100−18=82 cm3100 - 18 = 82 \text{ cm}^3.
Soil sampleFinal volume/cm3^3Volume of air escaped/cm3^3
Soil A1090
Soil B3466
Soil C2674
Soil D4258
Soil E1882

2. (ii)
To calculate the percentage of air content of A:

  1. Identify the volume of air escaped for Soil A.

    Volume of air escaped for A=90 cm3\text{Volume of air escaped for A} = 90 \text{ cm}^3
  2. Use the formula for percentage of air content.

    Percentage of air content=Volume of air escapedInitial volume of soil×100%\text{Percentage of air content} = \frac{\text{Volume of air escaped}}{\text{Initial volume of soil}} \times 100\%
  3. Substitute the values and calculate.

    Percentage of air content for A=90 cm3100 cm3×100%=90%\text{Percentage of air content for A} = \frac{90 \text{ cm}^3}{100 \text{ cm}^3} \times 100\% = 90\%

    The percentage of air content of A is \boxed{\text{90%}}.

    2. (iii)
    The soil sample that retains the largest amount of water will be the one with the largest pore space, which corresponds to the largest volume of air escaped.
    From the table in 2(i), Soil A has the largest volume of air escaped (90 cm390 \text{ cm}^3). Therefore, Soil A can retain the largest amount of water.

    The soil sample that would retain the largest amount of water is Soil A\boxed{\text{Soil A}}.

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