All MCAT Biology Resources
Example Questions
Example Question #31 : Enzymes And Enzyme Inhibition
Which of the following is not an example of positive feedback?
As blood calcium levels increase, parathyroid hormone (PTH) is reduced.
A forest fire slowly expands outward, which provides it with even more fuel to burn.
As more buffalo begin to run in a herd, the overall level of panic increases. This results in even more buffalo running.
During childbirth, oxytocin creates a stimulus which causes the hypothalamus to release more oxytocin.
As blood calcium levels increase, parathyroid hormone (PTH) is reduced.
Negative feedback provides the body with a method for shutting down a reaction once sufficient product has been created. Parathyroid hormone (PTH) is responsible for increasing blood calcium levels, but once the level is sufficient, the parathyroid glands detect the sufficient calcium level and no longer produce PTH. PTH works in coordination with calcitonin to maintain this balance via its negative feedback loop.
Positive feedback, in contrast, involves the exponential increase of a reaction upon detection. Very few examples of positive feedback exist in the body, though oxytocin follows this model during childbirth.
Example Question #32 : Enzymes And Enzyme Inhibition
Sulfanilamide is an antibiotic that resembles the intermediate, 4-aminobenzoic acid (PABA), in the metabolic pathway to create folic acid. It binds to the active site of the enzyme that normally binds to PABA, and inhibits the binding of PABA temporarily. Since folic acid is necessary for bacterial growth, this antibiotic helps inhibit the spread of infection in humans.
Based on this information, what type of inhibitor is sulfanilamide?
Competitive inhibitor
Irreversible inhibitor
Noncompetitive inhibitor
Uncompetitive inhibitor
Competitive inhibitor
Competitive inhibitors block substrates by binding noncovalently to the active site on enzymes. This prevents the substrate from entering the active site.
Noncompetitive inhibitors, in contrast, will act on regions outside of the active site to prevent binding. Uncompetitive inhibition is a specialized form of noncomptitive inhibition in which the inhibitor binds to the enzyme complex after the substrate has entered the active site. Irreversible inhibitors form covalent bonds, and fall outside the common inhibitor classifications.
Example Question #31 : Enzymes And Enzyme Inhibition
Which of the following strategies of enzymatic inhibition is used by noncompetitive inhibitors?
Bind to the active site and prevent substrate from binding
Induce another molecule to bind to the active site of the enzyme
Bind to an allosteric site to cause a conformational shift in the enzyme
Bind to substrate so that it cannot bind to the active site
Target the enzyme for destruction using a protease
Bind to an allosteric site to cause a conformational shift in the enzyme
Noncompetitive inhibitors of enzymes function by binding to a site other than the active site on the enzyme, known as an allosteric site. This causes the enzyme to go through a conformational shift and inhibits binding of the substrate.
Competitive inhibitors bind to the enzyme active site to prevent substrate action. Uncompetitive inhibitors are a sub-category of noncompetitive inhibitors, and bind to an allosteric site only after the substrate has entered the active site, thus preventing the substrate from leaving.
Example Question #34 : Enzymes And Enzyme Inhibition
An inhibitor is added, disrupting the function of a particular enzyme. The experimenter adds more substrate, and enzyme function increases again. These results indicate the involvement of what type of inhibitor?
Competitive
Uncompetitive
There is not enough information to determine
Noncompetitive
Allosteric
Competitive
Competitive inhibitors bind to the active site of the enzyme, blocking substrate entry. Increasing the concentration of the substrate helps overcome this type of inhibition by increasing the proportion of substrate to inhibitor in solution. By increasing substrate, the active site is more likely to bind substrate than inhibitor.
Noncompetitive inhibitors and uncompetitive inhibitors are types of allosteric inhibitors, meaning that they bind the enzyme in a site other than the active site. Increasing substrate concentration cannot overcome these types of inhibition because the inhibitor and enzyme are not in direct competition for binding sites.
Example Question #81 : Macromolecules
What type of inhibition affects both the Michaelis constant the maximum reaction rate of an enzyme?
Competitive inhibition
Uncompetitive inhibition
All types of inhibition will affect the Michaelis constant or the maximum reaction rate, but never both
All types of inhibition will affect both values
Non-competitive inhibition
Uncompetitive inhibition
Enzymes are catalysts that speed up the rate at which a reaction occurs. Competitive inhibitors bind to the active site, affecting the Michaelis constant, but not the maximum reaction rate. Competitive inhibitors can be overcome by adding more substrate to displace the inhibitor. Non-competitive inhibitors bind to an allosteric site, affecting the maximum reaction rate, but not the Michaelis constant. No amount of additional substrate can overcome the inhibitor's effect.
In uncompetitive inhibition, the inhibitor binds to the enzyme-substrate complex. The substrate is required to bind the active site, affecting the Michaelis constant, and is then held in place by the inhibitor, affecting the maximum reaction rate. Uncompetitive inhibitors are, thus, as special class of inhibitor that will affect both values.
Example Question #81 : Macromolecules
An unknown molecule is added to an enzyme-catalyzed reaction, immediately decreasing its rate. If the addition of more substrate has no effect, but the addition of an antibody for the unknown molecule restores the initial reaction rate, what form of inhibition is most likely occurring?
Irreversible inhibition
Competitive inhibition
Uncompetitive inhibition
Feedback inhibition
Noncompetitive inhibition
Noncompetitive inhibition
This question is referring specifically to the different modes of enzyme inhibition. The given fact that increasing substrate concentration does not restore enzyme function indicates that the inhibitor is binding to the enzyme at an allosteric site (eliminating competitive inhibition). The given fact that inhibitor-specific antibodies restored enzyme function indicates that the inhibition is reversible.
Uncompetitive inhibition is a specific type of noncompetitive inhibition in which the inhibitior binds to the enzyme-substrate complex. We are unable to conclude that this is the case based on the given information alone.
Example Question #42 : Enzymes And Enzyme Inhibition
A graduate student needs to cut a DNA plasmid using two different restriction enzymes in a buffer. He can use any two of the following enzymes in any of the four given buffers.
Enzyme |
Efficiency in Buffer (%) |
|||
|
1 |
2 |
3 |
4 |
AgeI |
100 |
50 |
10 |
75 |
ClaI |
10 |
50 |
50 |
100 |
PsiI |
10 |
100 |
10 |
100 |
Which two enzymes and buffer should the student choose for the digestion?
ClaI and AgeI in buffer 2
ClaI and PsiI in buffer 4
ClaI and AgeI in a mixture of buffers 1 and 4
PsiI and AgeI in buffer 2
ClaI and PsiI in buffer 4
Because ClaI and PsiI both have 100% efficiency in buffer 4, that is the most ideal choice. A combination of buffers is never a valid option, as diluting the buffers fundamentally changes the composition of the reaction and the efficiency of the enzyme. In the absence of two enzymes that can function with 100% efficiency, it is best to find a combination with the greatest possible efficiency or to complete the digest sequentially in the appropriate buffers.
Example Question #82 : Proteins
Sildenafil (commonly called Viagra) is a common drug used to treat erectile dysfunction and pulmonary arterial hypertension. Sildenafil's effect comes from its ability to cause vasodilation in smooth muscle cells. For this problem, we're only going to consider its effects on erections in males.
Erectile dysfunction is a common medical problem in older men. Its most significant effect is the prevention of erections. Erections occur when there is an increase in blood flow via enlargement of an artery (vasodilation). Understanding the mechanism by which vasodilations occur is important in order to treat erectile dysfunction.
Erections occur when nitric oxide is released from an area in the penis and binds to guanylate cyclase in other cells of the penis, which creates cyclic guanosine monophosphate (cGMP) from GTP. cGMP causes a relaxation of the arterial wall in order to increase blood flow to the region, thereby causing an erection. cGMP is broken down over time by cGMP-specific phosphodiesterase type 5 (PDE5) into GTP, which reverses the effect and causes vasoconstriction on the arterial wall. Combatting this effect is the major method by which Viagra functions.
Which of the following reactions would you expect PDE5 to catalyze?
For this problem, we need to know what phosphodiesterases do. Phosphodiesterases catalyze the breakdown of phosphodiester bonds via hydrolysis. Cyclic GMP contains an internal phosphodiester bond. Therefore its breakdown would result in the formation of GMP. The hydrolysis of cGMP should not yield GMP and a phosphate, since cyclic GMP only has one phosphate group.
Example Question #1 : Structure Of Dna And Rna
Which of the following statements regarding the double-helix of DNA is true?
Each strand is parallel with 5' to 3' direction
All hydroxyl groups of pentoses are involved in linkages
Both strands are identical to one another
The nitrogenous bases are perpendicular to the axis
The nitrogenous bases are perpendicular to the axis
The DNA double helix is a structure of double stranded nucleic acids, and is held together by base pairing of these nucleic acids perpendicular to the helical axis. The 5’ carbons and 3’ carbons between the pentoses are involved in phosphodiester bonds, while the 1’ carbons are involved in N-glycosidic bonds with the bases. Not all hydroxyl groups are involved in bonding, however, as there is a free 3’ carbon hydroxyl group on the extreme 3’ end of linked nucleoside monophosphate monomers. Both strands are sequentially unique as base pairing occurs between a purine and pyrimidine only. Finally, the strands are antiparallel, meaning they run in opposition to one another. Thus, the correct answer is that the nitrogenous bases are perpendicular to the axis.
Example Question #2 : Structure Of Dna And Rna
Human chromosomes are divided into two arms, a long q arm and a short p arm. A karyotype is the organization of a human cell’s total genetic complement. A typical karyotype is generated by ordering chromosome 1 to chromosome 23 in order of decreasing size.
When viewing a karyotype, it can often become apparent that changes in chromosome number, arrangement, or structure are present. Among the most common genetic changes are Robertsonian translocations, involving transposition of chromosomal material between long arms of certain chromosomes to form one derivative chromosome. Chromosomes 14 and 21, for example, often undergo a Robertsonian translocation, as below.
A karyotype of this individual for chromosomes 14 and 21 would thus appear as follows:
Though an individual with aberrations such as a Robertsonian translocation may be phenotypically normal, they can generate gametes through meiosis that have atypical organizations of chromosomes, resulting in recurrent fetal abnormalities or miscarriages.
Which of the following is true of the DNA component of a standard chromosome 14?
I. The intra-strand bonds are covalent
II. The inter-strand bonds are covalent
III. During replication, Okazaki fragments are found on the leading strand
II and III
I, II, and III
I and II
I and III
I only
I only
Intra-strand bonds are the covalent bonds between sugars and phosphates in the sugar-phosphate backbone. The other choices are incorrect, as inter-strand bonds in nucleic acids are hydrogen bonds between nitrogenous bases. Additionally, Okazaki fragments are found on the lagging strand of DNA during replication, because DNA on both the leading and lagging strands must be synthesized by DNA polymerase in the 5' to 3' direction.