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(Bio)chemistry: TestAS practice testlets

Biomolecules, enzymes, metabolism, acids and bases and concentrations, explained and then calculated.

testas.de lists this topic for the Life Sciences module. In the test it comes as testlets: a short text, table or formula, then 4 to 8 questions you answer from it.

Topic list: testas.de

Two practice testlets

Full exam-length testlets from the Kollego course, generated the same way as our mocks. Answer from the material, then open each answer to check.

Testlet 1

Enzyme kinetics and inhibition

Enzymes are biological catalysts, mostly proteins, that speed up chemical reactions by lowering their activation energy without being used up. The molecule an enzyme acts on, the substrate, binds to a specific region, the active site, whose shape and chemical properties fit the substrate. After the reaction, the product is released and the enzyme can bind the next substrate molecule.

The rate of an enzyme reaction depends on the substrate concentration. At low concentrations, the rate rises almost in proportion to the concentration, because most active sites are free. At high concentrations, nearly all active sites are occupied, and the rate approaches a maximum, Vmax, which cannot be exceeded however much substrate is added; the enzyme is saturated. This behaviour is described by the Michaelis-Menten equation. The Michaelis constant Km is the substrate concentration at which the rate is half of Vmax. A low Km means that the enzyme reaches half its maximum rate at a low substrate concentration, which indicates a high affinity for the substrate.

v = Vmax × [S] ÷ (Km + [S])

Inhibitors reduce enzyme activity. A competitive inhibitor resembles the substrate and competes for the active site. Its effect can be overcome by adding more substrate, so Vmax is unchanged, but more substrate is needed to reach half of Vmax: the apparent Km increases. A non-competitive inhibitor binds elsewhere on the enzyme and changes its shape, so that some enzyme molecules cannot work at all. Adding more substrate does not help; Vmax falls, while Km stays the same. Many drugs are enzyme inhibitors, and cells regulate metabolic pathways by inhibiting key enzymes.

Enzymes also depend on their environment. Activity rises with temperature up to an optimum, which is about 37 °C for most human enzymes, and falls steeply above it as the protein denatures. Each enzyme also has a pH optimum: pepsin works best at about pH 2 in the stomach, trypsin at about pH 8 in the small intestine.

Knowing Km and Vmax allows the rate at any substrate concentration to be calculated with the equation. Biochemists also use these values to compare enzymes from different organisms, to identify how a drug inhibits its target and to design industrial processes in which enzymes are used, for example in detergents or food production.

An enzyme was studied in the laboratory at 37 °C and pH 7. Its Vmax is 100 µmol of product per minute, and its Km is 2 mmol/L.

Reaction rate at different substrate concentrations
[S] (mmol/L)12618
Rate (µmol/min)33507590
  1. 1. What is the reaction rate when the substrate concentration equals Km?

    • a33 µmol/min
    • b50 µmol/min
    • c25 µmol/min
    • d100 µmol/min
    Show answer

    B: 50 µmol/min

    The text defines Km as the substrate concentration at which the rate is half of Vmax. Vmax = 100, so at [S] = Km the rate is 100 ÷ 2 = 50 µmol/min (the table also shows 50 at 2 mmol/L).

  2. 2. What rate does the equation predict at [S] = 8 mmol/L?

    • a90 µmol/min
    • b80 µmol/min
    • c75 µmol/min
    • d40 µmol/min
    Show answer

    B: 80 µmol/min

    v = Vmax × [S] ÷ (Km + [S]) = 100 × 8 ÷ (2 + 8) = 800 ÷ 10 = 80 µmol/min.

  3. 3. Why does tripling [S] from 6 to 18 mmol/L raise the rate only from 75 to 90?

    • aThe product inhibits the enzyme competitively
    • bKm increases when more substrate is added
    • cMost active sites are already occupied
    • dThe enzyme is denatured at high substrate levels
    Show answer

    C: Most active sites are already occupied

    The text says at high concentrations 'nearly all active sites are occupied' and the rate only approaches Vmax. At 6 mmol/L the enzyme already runs at 75 of 100, so little room is left.

  4. 4. A competitive inhibitor is added. Which change is expected?

    • aVmax falls, Km is unchanged
    • bBoth Km and Vmax fall
    • cKm rises, Vmax is unchanged
    • dNeither Km nor Vmax changes
    Show answer

    C: Km rises, Vmax is unchanged

    The text says a competitive inhibitor can be overcome by more substrate, 'so Vmax is unchanged', but more substrate is needed to reach half of Vmax: 'the apparent Km increases'.

  5. 5. With a non-competitive inhibitor, the rate at very high [S] is 60 µmol/min. What is the new Vmax?

    • aAbout 30 µmol/min
    • bAbout 60 µmol/min
    • cAbout 120 µmol/min
    • d100 µmol/min
    Show answer

    B: About 60 µmol/min

    At very high [S] the rate equals Vmax. The text says a non-competitive inhibitor lowers Vmax and extra substrate does not help, so the new Vmax is the measured 60 µmol/min.

  6. 6. A second enzyme has a Km of 0.2 mmol/L for the same substrate. Which statement is correct?

    • aIt reaches half its maximum rate only at higher [S]
    • bIt has a lower affinity for the substrate
    • cIt has a higher affinity for the substrate
    • dIt must have a lower Vmax than the first enzyme
    Show answer

    C: It has a higher affinity for the substrate

    The text says a low Km means half the maximum rate is reached at a low substrate concentration, which 'indicates a high affinity'. 0.2 mmol/L is lower than the first enzyme's 2 mmol/L.

Testlet 2

Preparing solutions in the laboratory

Chemists count particles in moles. One mole contains about 6.022 × 10²³ particles, a number chosen so that the mass of one mole of a substance in grams equals its relative molecular mass. The molar mass of glucose (C6H12O6) is therefore about 180 g/mol, and that of sodium chloride (NaCl) about 58.4 g/mol. The amount of a substance in moles is its mass divided by its molar mass.

The concentration of a solution is most often given as molarity, the number of moles of solute per litre of solution, with the unit mol/L, often written M. A 1 M glucose solution contains 180 g of glucose per litre. To prepare a solution, the required amount of substance is calculated as concentration times volume, converted into grams, dissolved in part of the solvent and then made up to the final volume.

n = m ÷ M c = n ÷ V m = c × V × M

Concentrations are also given as mass per volume (for example g/L or mg/mL) or as percentages. In biology and medicine, "0.9% sodium chloride", physiological saline, means 0.9 g of NaCl per 100 mL, or 9 g per litre. Its molarity is 9 ÷ 58.4, about 0.154 mol/L. Because each NaCl unit releases two ions in water, the solution contains about 0.308 mol/L of dissolved particles, which is close to the osmotic concentration of blood plasma.

Dilution is one of the most common laboratory tasks. When a solution is diluted, the amount of solute stays the same while the volume increases, so the product of concentration and volume is constant. For large dilution factors, laboratories use serial dilutions: a series of steps, each diluting the previous solution by the same factor. Three successive 1 : 10 dilutions give a total dilution of 1 : 1,000. Serial dilutions are used, for example, to count bacteria, since a sample with millions of cells per millilitre must be diluted before colonies can be counted on a plate.

c1 × V1 = c2 × V2

Careful work also requires attention to units. Volumes in millilitres must be converted into litres before they are multiplied by a concentration in mol/L, and masses in milligrams into grams. A quick plausibility check, such as asking whether a few grams of substance for a few hundred millilitres of solution is reasonable, catches many errors before a solution is prepared.

Molar masses
SubstanceMolar mass (g/mol)
Glucose, C6H12O6180
Sodium chloride, NaCl58.4
Water, H2O18
  1. 1. How many grams of glucose are needed for 250 mL of a 0.2 mol/L solution?

    • a36.0 g
    • b45.0 g
    • c9.0 g
    • d0.9 g
    Show answer

    C: 9.0 g

    250 mL = 0.25 L. Moles: n = c × V = 0.2 × 0.25 = 0.05 mol. Mass: m = n × M = 0.05 × 180 = 9 g.

  2. 2. How many moles are contained in 18 g of glucose?

    • a10.00 mol
    • b0.10 mol
    • c0.01 mol
    • d1.00 mol
    Show answer

    B: 0.10 mol

    n = m ÷ M = 18 g ÷ 180 g/mol = 0.1 mol.

  3. 3. What volume of a 2 mol/L stock solution is needed to prepare 500 mL of a 0.1 mol/L solution?

    • a250 mL
    • b10 mL
    • c50 mL
    • d25 mL
    Show answer

    D: 25 mL

    c1 × V1 = c2 × V2: 2 × V1 = 0.1 × 500 = 50, so V1 = 50 ÷ 2 = 25 mL.

  4. 4. A bacterial sample is diluted 1 : 10 four times in series. What is the total dilution?

    • a1 : 400
    • b1 : 10,000
    • c1 : 1,000
    • d1 : 40
    Show answer

    B: 1 : 10,000

    The text says the factors of serial dilutions multiply: three 1 : 10 steps give 1 : 1,000. Four steps: 10 × 10 × 10 × 10 = 10,000, so 1 : 10,000.

  5. 5. After a total dilution of 1 : 10,000, 0.1 mL is spread on a plate and 45 colonies grow. How many bacteria per mL did the original sample contain?

    • a450
    • b4,500,000
    • c45,000
    • d450,000
    Show answer

    B: 4,500,000

    45 colonies grew from 0.1 mL of the diluted sample: 45 ÷ 0.1 = 450 per mL. The original was 10,000 times more concentrated: 450 × 10,000 = 4,500,000 per mL.

  6. 6. About how many moles of dissolved particles does 1 L of 0.9% saline contain?

    • aAbout 0.31 mol
    • bAbout 9 mol
    • cAbout 0.9 mol
    • dAbout 0.15 mol
    Show answer

    A: About 0.31 mol

    The text says 0.9% saline is 9 g NaCl per litre, 9 ÷ 58.4 ≈ 0.154 mol/L. Each NaCl gives two ions: 0.154 × 2 ≈ 0.31 mol of particles in 1 L.

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