Biochemistry/Biology Flashcards
D. SDS-Page under reducing conditions
The answer to the question is D because the two proteins have the same molecular mass and pI, which means they will not be well-resolved on either native PAGE or isoelectric focusing.
Only SDS-PAGE under reducing conditions would show two bands for Protein 1 at distinctly different molecular weights than the single band for Protein 2.
The answer to the question is B because the data in the table indicate that the inhibitor is a competitive inhibitor (apparent KM is increased while the apparent Vmax is unchanged).
Competitive inhibitors bind in the enzyme active site to the exclusion of the substrate.
The answer is C because each cycle of glycolysis produces two NADH molecules and requires an input of one glucose molecule.
The key is D because lipids with unsaturated acyl chains do not pack well against one another, increasing the level of disorder. Increased temperature also increases the level of disorder in the membrane because there is more movement at higher temperatures.
The answer is A because the molecule with the most negative standard free energy of hydrolysis is also the molecule that is able to spontaneously transfer a phosphate group to create phosphorylated compounds of lower free energy. Therefore, if ATP can spontaneously transfer phosphate to glucose and PEP can spontaneously transfer phosphate to ADP, then PEP has the most negative standard free energy of hydrolysis followed by ATP. The absolute value of the most negative number is the largest value.
The answer is A because the isoelectric point of an amino acid is the highest for basic amino acids, and arginine and lysine are both basic amino acids. However, arginine has a higher isoelectric point because it is more basic.
The answer is C because in order to hydrogen bond with glutamate, an amino acid side chain must contain a hydrogen bond donor group to match with the glutamate acceptor. Glutamine has an NH2 group that can act in this way.
Glutamate has a negative C=O group. Need a group with Hs for bonding.
The answer is D because uncompetitive inhibitors affect both the kcat and KM of an enzymatically catalyzed reaction in a way that does not change the slope of the Lineweaver-Burk plot, which means the KM/Vmax is not changed.
The answer is B because the slope of a Lineweaver-Burk plot is equivalent to KM/Vmax.
From the data it is apparent that the Vmax is 200 μM/min, and the KM is therefore the substrate concentration at a V0 of 100 μM/min, or 10 μM. The numerical value of the slope is then 10/200 or 0.05.
The answer is B because the protein with the lowest binding dissociation constant has the highest affinity for its ligand. The dissociation constant represents the ligand concentration that is necessary to achieve 50% binding
The answer is C because the activity is a measure of the amount of enzyme per milligram of total protein. This provides a measure of the purity of an enzymatic mixture.
The answer is A because in the deamination of cytosine, the amine group of cytosine is replaced with a carbonyl group, resulting in the structure of uracil.
As sperm cells leave the testis they travel through the epididymis to the vas deferens and into the urethra. The sperm then enter the female’s vagina, travel through the cervix and uterus, and enter the fallopian tube, where fertilization most commonly takes place. Of the options listed, only A correctly identifies both the structures and the order. Thus, A is the best answer.
A sharp rise in osmotically active albumin in the serum would increase the flow of interstitial fluid into the bloodstream and result in an increase in blood pressure, not a decrease.
Therefore, A is incorrect, and D is correct. B is incorrect because the immune response does not depend on the level of serum albumen. C is incorrect because albumen would not normally pass through capillary walls.
If all other steps occur normally, then four sperm would be produced. Because replication occurred twice instead of once prior to tetrad formation, each sperm would have twice the normal amount of DNA.
As a result, four diploid sperm would be produced instead of four haploid sperm. A, C, and D are incorrect because they describe results that contain an incorrect number of sperm or ploidy number for the sperm produced. Thus, B is the best answer.
The number of different possible gametes that can be formed by diploid organisms as a result of independent assortment of chromosomes during meiosis can be calculated using the formula 2n where n is the haploid number of chromosomes. In this case, the haploid number is 3, making the number of different haploid cells 23, or 8. Thus, B is the best answer.
The knee-jerk reflex is a simple monosynaptic stretch reflex. A tap to the tendon that connects the quadriceps to the patella activates a sensory neuron that directly synapses with a motor neuron in the spinal cord, causing the quadriceps to contract. Thus, A is the best answer.
The number of individuals carrying the described trait is 480. The gene frequency of the recessive trait, q, can be ascertained by taking the square root of 16/100 or 0.4. Given this frequency, the frequency of p must be 1.0 - 0.4 = 0.6. According to the Hardy-Weinberg law, the frequency of carriers is thus given by 2 pq or 2 X .4 X .6 = .48 or 48%. 48% of 1000 = 480. Thus, answer choice C is the correct answer.
In gastrulation, the three cell regions, or germ layers, are formed in the embryo.
From the outer layer (ectoderm) come cells of the nervous system and epidermis,
from the inner layer (endoderm) come cells of the lining of the digestive tube and associated organs,
and from the middle layer (mesoderm) come the blood cells, connective tissues (bones, muscles, and tendons), and several organs (kidney, heart, gonads). Thus, the eye and spinal cord are formed from ectoderm, the liver is formed from endoderm, and the heart and bone are formed from mesoderm. Answer choice B is the best answer.
The stem asks students to identify the procedure that would be LEAST likely to prevent synthesis of the superantigen. Foils B (binding of the operator), C (binding of the mRNA), and D (insertion of a stop codon) would all be effective ways to prevent synthesis of the superantigen. Adding tRNA nucleotides would be unlikely to prevent synthesis of the antigen. Therefore, the key is A.
The correct answer is option B, epinephrine.
Epinephrine is secreted by the adrenal medulla and its secretion is controlled by the autonomic nervous system. Secretion of the other three hormones is regulated by secretion of anterior pituitary hormones. Cortisone secretion is regulated by adrenocorticotropic hormone (ACTH), progesterone by luteinizing hormone (LH), and thyroxin by thyroid stimulating hormone (TSH).
The question shows a short polynucleotide containing A, U, C, and G. Given that it is a polynucleotide containing uracil (U), one can conclude that it is RNA. Therefore, the key is C.
