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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 22 )

Updated On: 30-Jan-2026
Page 22 of 164
PDF with 811 Questions

A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0° C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0° C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet.
Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at -20° C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA).
In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

What is the molality of a stock solution that is 10% SDS by mass?

  1. 0.028 m
  2. 0.100 m
  3. 0.347 m
  4. 0.385 m

Answer(s): D

Explanation:

To answer this, you have to know the definitions of two measures of solution concentration; percentage by mass and molality. The percentage by mass of a solution is the mass of the solute divided by the total mass of the solution, multiplied by 100. Thus, a 10% SDS solution has 10% of its mass in the form of sodium dodecyl sulfate and 90% of its mass in the form of water. The molality of a solution is the number of moles of solute per kilogram of solvent. Let's assume that this solution contains one kilogram of water. Remember that the water makes up only 90% of the total weight of the solution. We can find the total weight of the solution by using the formula 1000 equals zero point nine x. This tells us that the total weight of the solution is one thousand, one hundred eleven grams, so the weight of the SDS must be one hundred eleven grams. Now we need to know how many moles of SDS there are in the 111 grams. The number of moles is equal to the mass of the SDS over its molecular weight; this comes to 0.38 moles. Since there are 0.385 moles of SDS for every thousand grams of water, the solution is 0.385 molal, choice D.



A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0° C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0° C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet.
Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at -20° C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA).
In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.


Tris (Tris(hydroxymethyl)aminomethane) is generally used as a buffer. If pH 8.0 is a good buffering region for Tris, then:
I) the pKa of Tris must be near pH 8.0
II) if Tris is titrated with acid, the titration curve will possess a steep region near pH 8.0.
III) a great deal of NaOH would have to be added to pH 8.0 Tris in order to significantly affect the pH.

  1. I only
  2. III only
  3. I and II only
  4. I and III only

Answer(s): D

Explanation:

A buffer is a mixture of either a weak base and its conjugate acid, or a weak acid and its conjugate base. Tris, as it happens, is a base. Buffer solutions resist changes in pH when acid or base is added. If pH 8 is a good buffering region for Tris, the ratio of base (Tris) to its conjugate acid (protonated Tris) will be near to 1. The pH of this solution is equal to the pKa + log [base] / [conjugate acid]. Therefore, if the ratio of base to its conjugate acid is near to 1, the pKa must be near to the pH. Thus Roman numeral I is a true statement. To evaluate Roman numeral II, we have to recall what a titration curve looks like. Titration is a procedure used for determining the normality of an acid or base. The procedure consists of adding an acid to a base, or a base to an acid, until the pH of the mixture reaches 7. The titration curve is a plot showing the pH of the solution as a function of the amount of acid or base added. Since the pH, as the dependent variable, is plotted on the y axis, a steep part of a titration curve represents a rapid change in pH. But we just said that a buffer solution in its effective pH range, such as a pH 8 Tris solution, resets pH change, so the titration curve for Tris will actually be quite flat near pH 8, and so Roman numeral II is false. Adding sodium hydroxide, which is a base, to a pH 8.0 Tris solution also wouldn't change the pH easily, so a large amount of the base would have to be added to
affect the pH very much, and Roman numeral III is true. Since I and III are true, the correct answer choice is D.



Which of the following statements identifies a chemically based sensory system?
I). Gustatory system
II). Auditory system
III). Olfactory system

  1. I only
  2. II only
  3. III only
  4. I and III only

Answer(s): D

Explanation:

Statement I is correct. The gustatory system, or our sense of taste, is mediated by the chemical interaction between the compounds in the food we eat and our taste bud receptors.
Statement II is incorrect. The auditory system is based on the mechanotransduction capabilities of inner ear cells, which take a mechanical stimulus (sound waves) and convert it into a neural stimulus.
Statement III in correct. The olfactory system, or our sense of smell, is mediated by the chemical interaction between the compounds in the air and our olfactory receptors.
Answer choice D is, therefore, the correct answer.



According to attachment theory, which of the following children is most likely to attach to a male psychologist, previously unknown to the child, in the course of a psychological study?

  1. A two-month old female infant raised in a safe, stable environment
  2. A five-month old male infant raised in a safe, stable environment
  3. An eight-month old male infant raised by a single caregiver who frequently neglect the child
  4. A thirteen-month old female infant raised by two caregivers who occasionally neglect the child

Answer(s): A

Explanation:

Attachment theory describes a series of steps that infants will progress through as they grow. During the first three months of life, an infant will indiscriminately attach to any person and will respond equally to any caregiver. Thus (A) is an apt description of attaching to a previously unknown adult. Around 4 to 6 months' babies will begin to recognize certain caregivers but will still accept care from anyone. Thus in (B) the baby will probably accept care from the psychologist, but the infant in (A) is much more likely to attach to the psychologist. From 6 to 9 months a baby will exhibit a strong attachment preference for a single caregiver, although the pattern of that attachment will vary based on the relationship that has developed between the caregiver and the child. Despite the neglect, the child in (C) will still have a preference for a single caregiver.
After 9 months, children slowly develop increasing independence and will slowly form multiple attachments. The child in (D) will, thus, begin to develop attachments to both caregivers, but not to the psychologist, who is a stranger.



Each of the following is an example of the transmission of knowledge through symbolic culture EXCEPT:
I). A young macaque monkey learning to rinse off food in the ocean from an older monkey, even when the food is not covered in dirt or sand.
II). A child learning the rules of baseball from a parent.
III). A new group of inductees in a military organization experiencing hazing rituals from older students, which they then later carry out on new recruits.

  1. I only
  2. III only
  3. I and III only
  4. II and III only

Answer(s): A

Explanation:

Symbolic culture is a form of knowledge transmission that only humans possess. It entails beliefs about the world, and behaviors elicited in response to those beliefs, which are symbolic ­ that is, non-physical. While many animal species may develop a form of culture, such as the macaques mentioned in (I), it is not symbolic culture. Things like the rules of baseball, or the traditions by which new members are brought into a group relate to non-physical things, and are thus examples of symbolic culture.



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