Bio/Biochem Flashcards
Inhibition of phosphofructokinase-1 by ATP is an example of:
allosteric regulation. feedback inhibition. competitive inhibition. A.I only B.III only C.I and II only D.II and III only
The answer to this question is C because ATP, the end product of glycolysis, downregulates through feedback inhibition the activity of phosphofructokinase-1 by binding to a regulatory site other than the active site of the enzyme (allosteric regulation). In contrast, competitive inhibition involves competition for binding to the active site.
Hydrophobic residues are _____.
nonpolar
Vasopressin regulates the insertion of aquaporins into the apical membranes of the epithelial cells of which renal structure?
A.Collecting duct
B.Proximal tubule
C.Bowman’s capsule
D.Ascending loop of Henle
The answer to this question is A because vasopressin regulates the fusion of aquaporins with the apical membranes of the collecting duct epithelial cells.
A prion is best described as an infectious:
A.prokaryote.
B.transposon.
C.protein.
D.virus.
The answer to this question is C because a prion is an abnormally folded protein that induces a normally folded version of the protein to also adopt the abnormal structure, which is often deleterious.
The precursor of EGP is translated from a transcript that has had one nontemplated nucleotide added to the open reading frame. This change does not create or eliminate a stop codon. Compared with the protein sGP, which is produced from the unedited transcript, EGP most likely has the same primary:
A.amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence.
B.carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence.
C.sequence as sGP except that EGP has one additional amino acid.
D.sequence as sGP except that EGP has one less amino acid.
The answer to this question is A because the addition of one nucleotide to the open reading frame of EGP results in a frameshift mutation and an aberrant carboxy-terminal domain.
Under anaerobic conditions, how many net molecules of ATP are produced by the consumption of 5 moles of glucose?
A.3 × 1024
B.6 × 1024
C.9 × 1024
D.1.2 × 1025
The correct answer is B.
Under anaerobic conditions, 2 moles of ATP are produced from each mole of glucose. Thus, 10 moles of ATP would be generated from 5 moles of glucose. Since there are 6 × 1023 molecules per mole, 10 moles of ATP is equal to 6 × 1024 molecules.
In humans, the characteristic tissue of which of the following organs is NOT derived from mesoderm?
A.Brain
B.Heart
C.Kidney
D.Skeletal muscle
The answer to this question is A because the brain is part of the central nervous system, which is derived from ectoderm. Heart, kidney, and skeletal muscle are derived from mesoderm.
Microtubules are cellular structures that originate from ______.
centrosomes
Pyrimidine dimers are ____________________________________.
molecular lesions formed from thymine or cytosine bases in DNA via photochemical reactions.
The kinetic properties of two enzymes are shown in the table. Both enzymes have the same substrate.
Enzyme kcat (s–1) Kd (mM) Hill coefficent Isoelectric point
X 1.5 × 106 10.0 2.25 7.9
Y 2.1 × 106 18.0 1.01 3.3
Based on the data, which statement best describes how Enzyme X differs from Enzyme Y?
The maximal velocity of the reaction catalyzed by Enzyme X is higher than that of the reaction catalyzed by Enzyme Y.
Enzyme X has a higher molecular weight than Enzyme Y.
Enzyme X exhibits cooperativity, whereas the activity of Enzyme Y does not.
Enzyme X has a lower binding affinity for the substrate than Enzyme Y.
The answer to this question is C because the data in Table 1 show that Enzyme X has a Hill coefficient greater than 1, which means it exhibits cooperativity. In contrast, Enzyme Y has a Hill coefficient that is essentially 1, which means it does not.
