biomedical sub test[kalplan 2001] Discussion is welcome!

zhangheng1020UID：2130513

Explanation:
         The correct answer is A. Glucose, galactose, and fructose are all reducing sugars, and elevations of all of  these sugars can be detected with Clinitest tablets. Neither lactose nor pyruvate can be detected, thus  eliminating lactase and pyruvate dehydrogenase as plausible choices. This leaves three possibilities: fructokinase, fructose 1-phosphate aldolase, and galactose 1-P-uridyl transferases. Of these, only fructokinase deficiency produces a mild (usually completely asymptomatic) condition known as fructosuria. Fructose 1-phosphate aldolase deficiency (choice B) produces severe hereditary fructose intolerance.
         Galactose 1-P-uridyl transferase deficiency (choice C) produces classic galactosemia.
         Lactase deficiency (choice D) produces lactose intolerance.
         Pyruvate dehydrogenase deficiency (choice E) produces severe disease (e.g., a subset of Leigh's disease).

zhangheng1020UID：2130513

Addition of which of the following exhaustively 14C labeled substrates would lead to evolution of 14CO2 from a cell-free suspension containing all the enzymes and substrates required for the synthesis of uridylic acid?
                A. Aspartate
                B. Carbamoyl phosphate
                C. Glutamine
                D. Glycine
                E. N10-Formyltetrahydrofolate

zhangheng1020UID：2130513

Explanation:
        The correct answer is A. In the first step of pyrimidine synthesis, carbamoyl phosphate condenses with aspartate to form carbamoyl aspartate, in a reaction catalyzed by aspartate transcarbamoylase. In subsequent steps, ring closure occurs with the loss of water, followed by oxidation to yield orotic acid. Addition of ribose-5-phosphate produces orotidylic acid, which is decarboxylated by orotidylate decarboxylase to yield uridylic acid. The carbon dioxide that is evolved is derived from the alpha carboxyl group of aspartate.Carbamoyl phosphate (choice B) condenses with aspartate with the loss of inorganic phosphate to produce carbamoyl aspartate. The carbamoyl moiety of carbamoyl phosphate is retained.Glutamine (choice C), glycine (choice D) and N10-formyltetrahydrofolate (choice E) are all used in purine synthesis. Glutamine also donates an amino group to UTP to form CTP, but this step occurs after the synthesis of uridylic acid is complete.

[ Last edited by zhangheng1020 on 2005-12-1 at 11:23 ]

zhangheng1020UID：2130513

A histological section of the left ventricle of a deceased 28-year-old white male shows classic contraction band necrosis of the myocardium. Biological specimens confirm the presence of cocaine and metabolites. Activity of which of the following enzymes was most likely increased in the patient's myocardial cells shortly prior to his  death?
                A. Phosphoenolpyruvate carboxykinase
                B. Phosphofructokinase-1
                C. Pyruvate dehydrogenase
                D. Succinate dehydrogenase
                E. Transketolase

zhangheng1020UID：2130513

Explanation:
        The correct answer is B. Cocaine causes contraction band necrosis by blocking the reuptake of norepinephrine, resulting in excessive vasoconstriction of coronary vessels, leading to ischemia and infarction of heart tissue. Under these pathological conditions, myocardial cells switch to anaerobic metabolism and therefore glycolysis  becomes the sole source of ATP via substrate-level phosphorylations by phosphoglycerate kinase and pyruvate kinase. Phosphofructokinase-1 (PFK-1) is the rate-limiting enzyme of glycolysis, and its activity would therefore be increased.
        Phosphoenolpyruvate carboxykinase (choice A) is a regulatory enzyme in gluconeogenesis, which is induced by cortisol, epinephrine, and glucagon. It functions in the hepatic synthesis of glucose when energy levels from beta-oxidation of fatty acids are adequate.
        Pyruvate dehydrogenase (choice C) produces acetyl-CoA from pyruvate and coenzyme A, bridging glycolysis and the Krebs cycle. It requires 5 cofactors, including NAD and FAD, which would no longer be produced by the electron transport under hypoxic conditions, decreasing its activity.
        Succinate dehydrogenase (choice D) is a key enzyme of the Krebs cycle, producing a reduced equivalent of FAD to feed into the electron transport chain. It is also known as Complex II. The Krebs cycle only functions if oxygen is in appropriate concentrations since it is regulated by the levels of NADH, which is only consumed by the electron transport chain if there is enough oxygen. The absence of oxygen leads to an accumulation of NADH and a subsequent decrease in the enzyme activities of the Krebs cycle.
        Transketolase (choice E) is a thiamine requiring enzyme of the non-oxidative half of the hexose monophosphate shunt. The shuffling of sugars in the second half of this pathway results in the reentry of glyceraldehyde-3-phosphate and fructose-6-phosphate into the glycolytic pathway. Transketolase activity in red  blood cells is used as a clinical marker of thiamine deficiency, markedly decreasing in disorders such as Wernicke-Korsakoff syndrome.

[ Last edited by zhangheng1020 on 2005-12-1 at 11:28 ]

zhangheng1020UID：2130513

An 8-month-old child is brought to a pediatrician because of the mother's concern about the boy's tendency to compulsively bite his fingers. On questioning, the mother reported that she has noticed yellow-orange crystals in his diapers, but has not mentioned them to anyone. A genetic defect in which of the following pathways should be suspected?
                A. Aromatic amino acid metabolism
                B. Branched chain amino acid metabolism
                C. Purine metabolism
                D. Pyrimidine metabolism
                E. Sulfur-containing amino acid metabolism

zhangheng1020UID：2130513

Explanation:
         The correct answer is C. The disease is Lesch-Nyhan syndrome, and the yellow-orange crystals of uric acid in the diaper are an important, but often neglected, clue to early diagnosis. Lesch-Nyhan syndrome is characterized by a tremendous overproduction of purines, because the reutilization of purines via the purine salvage pathway is blocked by a near total absence of hypoxanthine-guanine phosphoribosyl-transferase  (HGPRT) activity. Patients with this severe X-linked disease, for reasons that are unknown, show aggressive behavior that leads to self-mutilation. They may also develop gouty arthritis or gouty nephropathy.
         Phenylketonuria is an example of a disorder of aromatic amino acid metabolism (choice A) characterized by mental retardation. Maple syrup urine disease is an example of a disorder of branched chain amino acids (choice B) causing motor  abnormalities and seizures.
         Orotic aciduria is an example of a disorder of pyrimidine metabolism (choice D), characterized by retarded growth and development as well as megaloblastic anemia.
         Homocystinuria is an example of a disorder of sulfur-containing amino acids (choice E), characterized by mental retardation, dislocation of the lenses, osteoporosis, and thromboses.

樱桃小丸子/丸丸UID：2160265

呵呵呵

zhangheng1020UID：2130513

Which of the following will be unchanged in a Lineweaver-Burk plot of an enzyme with and without a competitive inhibitor?
                 A. Km
                 B. Slope
                 C. x-intercept
                 D. y-intercept

樱桃小丸子/丸丸UID：2160265

丙酮酸尿症  我考试就考过着个题

樱桃小丸子/丸丸UID：2160265

晕了

zhangheng1020UID：2130513

Explanation:
         The correct answer is D. It is worth taking the time to learn how to read a Lineweaver-Burk plot. Lineweaver-Burk plots are used to determine the Vmax and Km of an enzyme; they are also used to differentiate between competitive and noncompetitive inhibition.
         Note that in a Lineweaver-Burk plot, the slope is Km/Vmax, the x-intercept is -1/Km, and the y-intercept is 1/Vmax. In the presence of a competitive inhibitor, the Km(choice A) and therefore the slope (choice B) are both increased. Similarly, if Km is increased, -1/Km will become less negative and the x-intercept will shift to the right. Intuitively, this makes sense since a competitive inhibitor will increase the amount of substrate needed to reach half-maximal velocity (definition of Km). In contrast, the Vmax, and hence the y-intercept, is unchanged (choice D).

[ Last edited by zhangheng1020 on 2005-12-1 at 11:57 ]

樱桃小丸子/丸丸UID：2160265

疯了

zhangheng1020UID：2130513

Which of the following metabolic alterations would most likely be present in a chronic alcoholic compared to a  non-drinker?
                A. Fatty acid oxidation is stimulated
                B. Gluconeogenesis is stimulated
                C. Glycerophosphate dehydrogenase is stimulated
                D. The ratio of lactate to pyruvate is decreased
                E. The ratio of NADH to NAD+ is increased

zhangheng1020UID：2130513

Explanation:
         The correct answer is E. The principal route of metabolism of ethanol is via alcohol dehydrogenase, which uses hydrogen from ethanol to form NADH from NAD+, markedly increasing the ratio of NADH to NAD+. The relative excess of NADH has a number of effects, including inhibiting, rather than stimulating fatty acid oxidation (choice A); inhibiting gluconeogenesis rather than stimulating it (choice B); inhibiting, rather than stimulating (choice C) glycerophosphate dehydrogenase; and favoring the formation of lactate rather than pyruvate from glycolysis (thereby increasing, rather than decreasing the lactate/pyruvate ratio; choice D).