Most cytochrome P450 enzymes alter the activity of drugs by:
A.
phosphorylating them.
B.
dephosphorylating them.
C.
oxidizing them.
D.
reducing them. - ANSWER C.
oxidizing them.
cytochrome p450 acts as monoox
...
Most cytochrome P450 enzymes alter the activity of drugs by:
A.
phosphorylating them.
B.
dephosphorylating them.
C.
oxidizing them.
D.
reducing them. - ANSWER C.
oxidizing them.
cytochrome p450 acts as monooxygenase, where an oxygen atom is inseted into substrate, thereby resulting in oxidation of substrate.
Given that antibiotics like erythromycin are metabolized by estrogen-sensitive P450 enzymes within the liver, which graph best predicts the expected relative half-life of erythromycin in adult males versus adult females? - ANSWER A
This is the only graph that correctly illustrates a quicker metabolism—or shorter half-life—of erythromycin in females. Given the continuous presence of GH in females, the cytochrome P450 enzymes are up-regulated, resulting in more rapid metabolism of certain drugs (like erythromycin). As a result, the half-life of erythromycin is expected to be lower in females (in other words, it would take less time to metabolize the drug due to the greater presence of P450 enzymes).
The researchers want to generate a mutant mouse strain that expresses a constantly active variant of ACC2 by replacing the amino acid at position 212 with a different residue. Which residue is the best choice as a replacement?
A.
Tyrosine
B.
Serine
C.
Alanine
D.
Threonine - ANSWER C.
Alanine
According to the passage, AMPK inactivates ACC2 through phosphorylation of the residue at position 212. To create a constitutively active variant, the residue at position 212 should be substituted with a residue that cannot be phosphorylated. Of all the amino acids listed, only alanine cannot be phosphorylated.
Phosphorylation= tyrosine, serine, and threonine
Bacteriophages were labeled with radioactive phosphorous (32P) and sulfur (35S). The labeled bacteriophages were then allowed to infect their host cells. At the end of the experiment, the 32P label was found only inside the host cells and the 35S label was found only outside the host cells. This experiment shows that:
A.
bacteriophages consist only of DNA and protein.
B.
only DNA, not protein, can enter the host cell.
C.
only protein, not DNA, can enter the host cell.
D.
both DNA and protein can enter the host cell. - ANSWER B.
only DNA, not protein, can enter the host cell.
32P incorporates into DNA
35S incorporates into proteins
if only 32 P is found inside that means only DNA can enter the host cell
Which statement describes ATP consumption and production during the preparatory and payoff phases of glycolysis, respectively?
A.
Two ATP molecules consumed, four ATP molecules produced
B.
Four ATP molecules consumed, eight ATP molecules produced
C.
Four ATP molecules consumed, two ATP molecules produced
D.
Eight ATP molecules consumed, four ATP molecules produced - ANSWER A.
Two ATP molecules consumed, four ATP molecules produced
2 ATP in and 4 ATP out in glycolysis
As blood passes through actively contracting skeletal muscle tissue, the affinity of hemoglobin for oxygen in the muscle tissue:
A.
increases as a result of an increase in plasma temperature.
B.
increases as a result of an increase in plasma PO2.
C.
decreases as a result of a decrease in plasma pH.
D.
decreases as a result of a decrease in plasma PCO2. - ANSWER C.
decreases as a result of a decrease in plasma pH.
affinity would decrease with decrease in plasma pH (during prolonged exercise, anaerobic respiration decreases plasma pH)
affinity decrease with increase in plasma temperature
affinity increase with pressure increase however pressure in muscle cells decreases with exercises
affinity increase with decrease in plasma CO2
Based on the passage, what can be determined about the composition of the mRNA that encodes the protein from which ghrelin is cleaved?
It is composed of:
A.
more than 28 amino acids.
B.
exactly 84 nucleotides.
C.
exactly 87 nucleotides.
D.
more than 87 nucleotides. - ANSWER D.
more than 87 nucleotides.
Each amino acid is coded by 3 nucleotides; thus, 84 nucleotides are needed to code for the 28-amino-acid peptide ghrelin. To this number, the triplet that codes for the STOP codon needs to be added. Therefore, 87 is the minimum number of nucleotides for the 28-amino-acid peptide ghrelin. However, the passage indicates that ghrelin is cleaved from a longer peptide chain. For this reason, the total number of nucleotides is higher than 87.
Hormone replacement therapy is often given to children who have Prader-Willi syndrome once they have reached the hyperphagic stage. Based on the passage, the most likely reason for this therapy would be to prevent:
A.
giantism.
B.
weight loss.
C.
short stature.
D.
muscle rigidity. - ANSWER C.
short stature.
Erythromycin interferes with protein synthesis by binding to which ribosomal subunit?
A.
50S
B.
60S
C.
70S
D.
80S - ANSWER A.
50S
Based on the passage, erythromycin inhibits bacterial protein synthesis by binding to the 23S rRNA component of the large subunit of the bacterial ribosome. The large subunit of the bacterial ribosome has a sedimentation coefficient of 50 Svedberg (S).
Substituting the residue at position 54 of BvgA with which amino acid will LEAST likely prevent the expression of the PTx operon?
A.
Glycine
[Show More]
