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MCAT Sample Questions : Biological Sciences

Passage II

Milk, one of the most nutritionally complete food sources in nature, contains water, vitamins, minerals, proteins, carbohydrates, and lipids.

An important milk protein is casein, which typically exists in milk as its calcium salt, calcium caseinate. Calcium caseinate is soluble at the normal pH of milk (6.6), but it becomes an insoluble form of casein in solutions that have a pH of 4.6 (the isoelectric point of calcium caseinate).

Milk contains two other important types of proteins, the lactalbumins and the lactoglobulins, both of which can be denatured and precipitated by heat.

The main carbohydrate in milk is the disaccharide lactose (shown below), which can be hydrolyzed with acid to two monosaccharides, galactose and glucose. In the body, lactose is digested to galactose and glucose by the enzyme lactase.

The first step in isolating several of the components of milk is to add a dilute solution of acetic acid to nonfat milk. The solid material that precipitates (Solid 1) is removed. Care must be taken to avoid adding an excess of acetic acid.

The remaining acidic solution is treated with powdered calcium carbonate and heated to boiling, after which time another solid (Solid 2) forms. Solid 2 is removed by filtration, and hot ethanol is added to the aqueous filtrate. The solution is allowed to cool overnight, during which time lactose crystallizes in the flask.

Following are some sample questions on this passage:

  1. The formation of curds in sour milk is most likely a result of the precipitation of:

    1. casein as the pH decreases.
    2. casein as the pH increases.
    3. lactose as the pH decreases.
    4. lactose as the pH increases.

    Answer: A

    Explanation: The passage states that an insoluble form of casein forms as the pH decreases from 6.6 to 4.6. Sour milk is acidic. Thus, answer choice A is the best answer.

  2. In the procedure in the passage, the powdered calcium carbonate is most likely used to:

    1. neutralize any acetic acid that may be present.
    2. dissolve any Solid 1 that remains.
    3. react with calcium caseinate to form casein.
    4. increase the solubility of Solid 2 in the aqueous solution.

    Answer: A

    Explanation: The passage mentions three specific classes of compounds: (1) Calcium caseinate, which precipitates at low pH as casein; (2) Proteins (lactalbumins and lactoglobulins), which precipitate when denatured by heat; and (3) Lactose, which hydrolyzes to galactose and glucose. In the procedure mentioned in the passage, the powdered calcium carbonate is probably added to neutralize any remaining acetic acid. Solid 1 (casein) forms on the addition of acetic acid and is removed by filtration, so it is not present when the CaCO3 is added. Solid 2 (denatured and precipitated proteinaceous material) is insoluble in the CaCO3. Thus, answer choice A is the best answer.

  3. The conversion of lactose to glucose and galactose involves the net addition to the lactose molecule of which of the following species?

    1. H2
    2. H2O
    3. O2
    4. H3O+

    Answer: B

    Explanation: As seen in the equation, the hydrolysis of lactose results in the net addition of H2O during the cleavage of the disaccharide. Thus, answer choice B is the best answer.

  4. The major components of Solid 2 are most likely:

    1. lactose and casein only.
    2. casein, the lactalbumins, and the lactoglobulins only.
    3. the lactalbumins and the lactoglobulins only.
    4. lactose, the lactalbumins, and the lactoglobulins only

    Answer: C

    Explanation: Solid 2 forms after the solution is heated to boiling, a condition that denatures and precipitates the lactalbumins and lactoglobulins. Thus, answer choice C is the best answer

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