All MCAT Biology Resources
Example Questions
Example Question #91 : Organic Chemistry, Biochemistry, And Metabolism
Approximately where would a carbonyl peak be found on an IR spectrum?
1000cm-1
1700cm-1
2800cm-1
The peak location will vary depending on the compound being analyzed.
1700cm-1
It is important to memorize a couple key functional groups, and where they are located on an IR spectrum. If you see a sharp peak near 1700cm-1, you can assume it is made by a carbonyl group.
Similarly, a wide peak around 3000cm-1 will be made by a hydroxyl group.
Example Question #92 : Organic Chemistry, Biochemistry, And Metabolism
The visible spectrum is typically between 390-700nm. Shorter wavelengths are responsible for the purple end of the color spectrum, while red is perceived in the eye from the longer wavelengths. -carotene is able to absorb light at a maximum value of 497nm.
-carotene is found in carrots. What would need to happen to -carotene in order for the color of carrots to become more yellow?
The number of amount of double bonds must be increased
The number of amount of conjugated double bonds must be reduced
The -carotene must become less saturated
The amount of -carotene in carrots must be reduced
The number of amount of conjugated double bonds must be reduced
The wavelength of light absorbed by a pigment is related to the number of conjugated double bonds in the molecule. The eye will begin to perceive color when the number of conjugated double bonds in a molecule becomes around eight. At this point, the color yellow can be perceived by the eye. In order for -carotene to give carrots a more yellowish color, it must have a change in the number of conjugated double bonds so that the pigment absorbs a shorter wavelength of light. This way, it will emit complementary light with a wavelength closer to yellow on the visible spectrum.
Example Question #1 : Compound Purification
To isolate DNA from a cell pellet, a student is performing an extraction using phenol and chloroform. The cells were lysed with 0.5mL of lysis buffer. 0.5mL extraction solution, consisting of phenol:chloroform:isoamyl alcohol (25:24:1), was added to the lysate. The sample was centrifuged at high speed for five minutes. The liquid in the tube was separated into three phases after centrifugation (see diagram).
In which layer will nucleic acids (RNA and DNA) be found?
Cannot be determined
3
1
2
1
RNA and DNA are hydrophilic substances, and will be found in the aqueous layer after the cells are lysed. Based on molecular weights of water (18), phenol (94), and chloroform (119), it is evident that the aqueous layer will be on the top, denoted as layer 1.
Example Question #2 : Compound Purification
To isolate DNA from a cell pellet, a student is performing an extraction using phenol and chloroform. The cells were lysed with 0.5mL of lysis buffer. 0.5mL extraction solution, consisting of phenol:chloroform:isoamyl alcohol (25:24:1), was added to the lysate. The sample was centrifuged at high speed for five minutes. The liquid in the tube was separated into three phases after centrifugation (see diagram).
In which layer will lipids and other cellular debris be found?
1
Cannot be determined
2
3
3
Lipids and other cellular debris are hydrophobic, and can be dissolved in the organic (bottom) layer. We know that the organic solvents will settle in layer 3 due to their relative densities compared to water. Phenol and chloroform are heavier molecules than water, with molecular weights of 94 and 119 respectively, and will thus be found in the bottom layer.
Example Question #3 : Compound Purification
To isolate DNA from a cell pellet, a student is performing an extraction using phenol and chloroform. The cells were lysed with 0.5mL of lysis buffer. 0.5mL extraction solution, consisting of phenol:chloroform:isoamyl alcohol (25:24:1), was added to the lysate. The sample was centrifuged at high speed for five minutes. The liquid in the tube was separated into three phases after centrifugation (see diagram).
What is the nature of each phase?
1: Phenol
2: Chloroform
3: Isoamyl alcohol
1: Organic
2: Interphase
3: Aqueous
1: Aqueous
2: Interphase
3: Organic
1: Phenol
2: Isoamyl alcohol
3: Chloroform
1: Chloroform
2: Isoamyl alcohol
3: Phenol
1: Aqueous
2: Interphase
3: Organic
The purpose of a phenol:chloroform extraction is to separate organic and aqueous cellular components. The lysis buffer is an aqueous solution, while the phenol:chloroform extraction solution is organic. Generally, organic solutions are more dense that aqueous solutions, resulting in the aqueous solution being on the top and organic being on the bottom. The interphase consists of proteins containing both hydrophilic and hydrophobic domains.
You can also determine relative density by calculating the molecular weights of the solvents. Doing so will show that water (MW = 18) is much lighter than phenol (MW = 94) or chloroform (MW = 119).
Example Question #3 : Compound Purification
To isolate DNA from a cell pellet, a student is performing an extraction using phenol and chloroform. The cells were lysed with 0.5mL of lysis buffer. 0.5mL extraction solution, consisting of phenol:chloroform:isoamyl alcohol (25:24:1), was added to the lysate. The sample was centrifuged at high speed for five minutes. The liquid in the tube was separated into three phases after centrifugation (see diagram).
In which layer will proteins be found?
1
3
Cannot be determined
Both 1 and 3
2
2
Proteins generally contain both hydrophobic and hydrophilic domains, and therefore, do not completely associate with either the organic or aqueous layers. They can be found in the interphase layer, denoted as layer 2. This allows different domains of the protein to interact with each phase. Hydrophilic regions will interact with the aqueous phase (layer 1), while hydrophobic regions will orient toward the organic phase (layer 3).
Example Question #5 : Compound Purification
To isolate DNA from a cell pellet, a student is performing an extraction using phenol and chloroform. The cells were lysed with 0.5mL of lysis buffer. 0.5mL extraction solution, consisting of phenol:chloroform:isoamyl alcohol (25:24:1), was added to the lysate. The sample was centrifuged at high speed for five minutes. The liquid in the tube was separated into three phases after centrifugation (see diagram).
The student saves the layer containing nucleic acids and transfers it to a new tube. The student then adds ethanol and salt, and centrifuges the solution at high speed. Where is the DNA in this new solution after centrifugation?
Interphase
DNA cannot be isolated by adding ethanol
Pellet
Supernatant
Organic layer
Pellet
After the first extraction, the DNA will be found in the aqueous phase due to its negative charge and hydrophilic nature. Transferring the aqueous phase (with DNA) and adding ethanol and salt will not result in an extraction; there will only be one phase after centrifugation: the aqueous phase. Following centrifugation, the DNA and salt precipitate from the aqueous ethanol solution and form a DNA pellet.
Example Question #1481 : Mcat Biological Sciences
Which of the following properties can be used to differentiate between enantiomers?
Melting temperature
Optical activity
Density
Heat of vaporization
Optical activity
Chiral molecules are a unique type of isomer, defined as two molecules with inverted stereochemistry, which makes them mirror images of one another. These isomers will ahve the same molecular weight, as well as most physical properties. Enatiomers cannot be distinguished based on melting point, heat of vaporization, or density. The primary, defining characteristic of a chiral molecule is its ability to effect plane polarized light. One isomer will rotate the light to the right, while the other will rotate the light to the left.
Example Question #1 : Other Techniques
Which of the following properties cannot be used to differentiate between myoglobin and hemoglobin?
Location
Affinity for oxygen
Number of subunits
Bound reduced iron
Bound reduced iron
Myoglobin and hemoglobin are the proteins primarily responsible for binding, transporting, and storing oxygen. Myoglobin is primarily located in muscle tissue, and has a very high affinity for oxygen. Hemoglobin is located in the blood, and has a slightly lower oxygen binding affinity. Thus, as hemoglobin carries oxygen past muscle tissue, oxygen can be released and will preferentially bind to myoglobin. Hemoglobin contains four subunits, each capable of binding one oxygen molecule, while myoglobin consists of only one subunit.
Both hemoglobin and myoglobin contain heme units, which structurally and functionally rely on the incorporation of reduced iron. As such, this property cannot be used to differentiate these two proteins.
Example Question #3 : Other Techniques
Which macromolecule is classified by physical properties rather than by monomer?
Carbohydrates
Nucleic acids
Lipids
Proteins
Lipids
Unlike carbohydrates, proteins, and nucleic acids, lipids do not have a uniform monomer structure. Rather, they are classified by their hydrophobic properties and a variety of beasic structural categorizations. Some examples of lipid classifications are glycerophospholipids, sphingolipids, and steroids.
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