标题: biomedical sub test[kalplan 2001] Discussion is welcome! [打印本页] 作者: zhangheng1020 时间: 2005-12-1 08:21:32 标题: biomedical sub test[kalplan 2001] Discussion is welcome!
kaplan 2001 的 sub biochemistry资料是我的朋友给我的，当时下载的时候，转到.DOC格式上很多的错行和乱码，所以我一个个的贴到这里，一边看，同时整理。
前60道题是独立的，
后面题是前面题目的乱序。时间不够可以只看前面的。
1
A chronic alcoholic develops severe memory loss with marked confabulation. Deficiency of which of the following vitamins would be most likely to contribute to the neurologic damage underlying these symptoms?
A. Folic acid
B. Niacin
C. Riboflavin
D. Thiamine
E. Vitamin B12
[ Last edited by zhangheng1020 on 2005-12-2 at 15:22 ]作者: zhangheng1020 时间: 2005-12-1 08:22:07 标题: 1
Explanation:
The correct answer is D. Wernicke-Korsakoff syndrome refers to the constellation of neurologic symptoms caused by thiamine deficiency. Among these, a severe memory deficit, which the patient may attempt to cover by making up bizarre explanations (confabulation), is prominent. Anatomical damage to the mamillary bodies and periventricular structures has been postulated as the cause. In the U.S., severe thiamine deficiency is seen most commonly in chronic alcoholics. Thiamine deficiency can also damage peripheral nerves ("dry" beriberi) and the heart ("wet" beriberi). Folic acid deficiency (choice A) produces megaloblastic anemia without neurologic symptoms. Niacin deficiency (choice B) produces pellagra, characterized by depigmenting dermatitis, chronic diarrhea, and anemia. Riboflavin deficiency (choice C) produces ariboflavinosis, characterized by glossitis, corneal opacities, dermatitis, and erythroid hyperplasia. Vitamin B12 deficiency (choice E) produces megaloblastic anemia accompanied by degeneration of the posterolateral spinal cord.
[ Last edited by zhangheng1020 on 2005-12-1 at 23:06 ]作者: zhangheng1020 时间: 2005-12-1 08:23:22 标题: 2
A 25-year-old woman with sickle cell anemia complains of steady pain in her right upper quadrant with radiation to the right shoulder, especially after large or fatty meals. Her physician diagnoses gallstones. Of which of the following compounds are these stones most likely composed?
A. Calcium bilirubinate
B. Calcium oxalate
C. Cholesterol
D. Cholesterol and calcium bilirubinate
E. Cystine
[ Last edited by zhangheng1020 on 2005-12-1 at 23:06 ]作者: zhangheng1020 时间: 2005-12-1 08:25:37 标题: 2
Explanation:
The correct answer is A. Bilirubin is a degradative product of hemoglobin metabolism. Bilirubin (pigment) stones are specifically associated with excessive bilirubin production in hemolytic anemias, including sickle cell anemia. Bilirubin stones can also be seen in hepatic cirrhosis and liver fluke infestation. Calcium oxalate stones (choice B) and cystine stones (choice E) are found in the kidney, rather than the gallbladder. Pure cholesterol stones (choice C) are less common than mixed gallstones, but have the same risk factors, including obesity and multiple pregnancies
Mixed stones (choice D) are the common "garden variety" gallstones, found especially in obese, middle aged patients, with a female predominance.
[ Last edited by zhangheng1020 on 2005-12-1 at 23:07 ]作者: zhangheng1020 时间: 2005-12-1 08:26:23 标题: 3
Two sisters are diagnosed with hemolytic anemia. Their older brother was previously diagnosed with the same disorder. Two other brothers are asymptomatic. The mother and father are second cousins. Deficiency of which of the following enzymes would be most likely to cause this disorder?
A. Debranching enzyme
B. Glucose-6-phosphatase
C. Glucose-6-phosphate dehydrogenase
D. Muscle phosphorylase
E. Pyruvate kinase
[ Last edited by zhangheng1020 on 2005-12-1 at 23:08 ]作者: zhangheng1020 时间: 2005-12-1 08:30:42 标题: 3
Explanation:
The correct answer is E. In general, you should associate hemolytic anemia with defects in glycolysis or the hexose monophosphate shunt (pentose phosphate pathway). Only two enzymes of those listed in the answer choices specifically involve these pathways and cause hemolytic anemia: pyruvate kinase and glucose-6-phosphate dehydrogenase. Glucose-6-phosphate dehydrogenase (G6PD) deficiency is inherited as an X-linked recessive trait, so females would not be affected. Pyruvate kinase is a glycolytic enzyme; pyruvate kinase deficiency is an autosomal recessive disorder, affecting males and females approximately equally. If this enzyme is deficient, red cells have trouble producing enough ATP to maintain the Na+/K+ pump on the plasma membrane, secondarily causing swelling and lysis.
Debranching enzyme (choice A) defects produce Cori's disease, one of the glycogen storage diseases.
Defects in glucose-6-phosphatase (choice B) produce Von Gierke's disease, one of the glycogen storage diseases.
Glucose-6-phosphatase dehydrogenase (choice C) deficiency produces an X-linked hemolytic anemia.
Defects in muscle phosphorylase (choice D) produce McArdle's disease, one of the glycogen storage diseases.
[ Last edited by zhangheng1020 on 2005-12-1 at 23:08 ]作者: zhangheng1020 时间: 2005-12-1 08:31:19
4
Which of the following amino acids would most likely be found on the surface of a protein molecule?
A. Alanine
B. Arginine
C. Isoleucine
D. Leucine
E. Phenylalanine
F. Tryptophan作者: zhangheng1020 时间: 2005-12-1 08:33:50 标题: 4
Explanation:
The correct answer is B. This question requires two logical steps: first, you need to appreciate that the hydrophilic amino acids are more likely to appear on the surface of a protein molecule, while hydrophobic amino acids are most likely be found in its interior. Next, you need to figure out which of the amino acids listed is hydrophilic. If you recall that arginine is a basic amino acid that is positively charged at physiologic pH, you should be able to answer this question right away.
All of the other choices have neutral side chains and are uncharged at physiologic pH. They would most likely be found in the hydrophobic core of the protein structure. Alanine (choice A), isoleucine (choice C), and leucine (choice D) all have aliphatic side chains; phenylalanine (choice E) and tryptophan (choice F) have aromatic side chains.作者: zhangheng1020 时间: 2005-12-1 08:35:12 标题: 5
5
A Southeast Asian immigrant child is noted to be severely retarded. Physical examination reveals a pot-bellied, pale child with a puffy face. The child's tongue is enlarged. Dietary deficiency of which of the following substances can produce this pattern?
A. Calcium
B. Iodine
C. Iron
D. Magnesium
E. Selenium作者: zhangheng1020 时间: 2005-12-1 08:36:30 标题: 5
Explanation:
The correct answer is B. The disease is cretinism, characterized by a profound lack of thyroid hormone in a developing child, leading to mental retardation and the physical findings described in the question stem. Cretinism can be due to dietary deficiency of iodine (now rare in this country because of iodized salt), to developmental failure of thyroid formation, or to a defect in thyroxine synthesis.
Calcium deficiency (choice A) in children can cause osteoporosis or osteopenia.
Iron deficiency (choice C) can cause a hypochromic, microcytic anemia.
Magnesium deficiency (choice D) is uncommon, but can cause decreased reflexes, and blunts the parathyroid response to hypocalcemia.
Selenium deficiency (choice E) is rare, but may cause a reversible form of cardiomyopathy.作者: zhangheng1020 时间: 2005-12-1 08:37:36 标题: 6
To which of the following diseases is pyruvate kinase deficiency most similar clinically?
A. α-thalassemia
B. β-thalassemia
C. Glucose-6-phosphate dehydrogenase deficiency
D. Hereditary spherocytosis
E. Iron deficiency anemia作者: zhangheng1020 时间: 2005-12-1 08:39:29 标题: 6
Explanation:
The correct answer is C. Both pyruvate kinase deficiency and glucose-6-phosphate dehydrogenase deficiency are red cell enzyme deficiencies characterized clinically by long "normal" periods interspersed with episodes of hemolytic anemia triggered by infections and oxidant drug injury (antimalarial drugs, sulfonamides, nitrofurans). In both of these conditions, the cell morphology between hemolytic episodes is usually normal or close to normal.
The α(choice A) and β(choice B) thalassemias, in their major forms, are characterized by persistent severe anemia. In the trait forms, they are charactertized by mild anemia.
Hereditary spherocytosis (choice D) is characterized by intermittent hemolysis, but, unlike pyruvate kinase deficiency and glucose-6-phosphate dehydrogenase deficiency, oxidant drugs are not a specific trigger for hemolysis.
Iron deficiency anemia (choice E) is characterized by chronic anemia with hypochromic, microcytic erythrocytes.作者: zhangheng1020 时间: 2005-12-1 08:41:03 标题: 7
A baby that was apparently normal at birth begins to show a delay in motor development by 3 months of age. At one year of age, the child begins to develop spasticity and writhing movements. At age three, compulsive biting of fingers and lips and head-banging appear. At puberty, the child develops arthritis, and death from renal failure occurs at age 25. This patient's condition is due to an enzyme deficiency in which of the following biochemical pathways?
A. Ganglioside metabolism
B. Monosaccharide metabolism
C. Purine metabolism
D. Pyrimidine metabolism
E. Tyrosine metabolism作者: zhangheng1020 时间: 2005-12-1 08:44:39 标题: 7
Explanation:
The correct answer is C. The patient has a classical case of Lesch-Nyhan syndrome, an X-linked disorder due to severe deficiency of the purine salvage enzyme hypoxanthine-guanine phosphoribosyl transferase (HPRT). This defect is associated with excessive de novo purine synthesis, hyperuricemia, and the clinical signs and symptoms described in the question stem. The biochemical basis of the often-striking self-mutilatory behavior (which may require restraints and even tooth extraction) has never been established. Treatment with allopurinon inhibits xanthine oxidase and reduces gouty arthritis, urate stone formation, and urate nephropathy. It does not, however, modify the neurologic/psychiatric presentation.作者: zhangheng1020 时间: 2005-12-1 08:47:03 标题: 8
An obese individual is brought to the emergency room by a concerned friend. The patient has been on a self-imposed "starvation diet" for four months, and has lost 60 pounds while consuming only water and vitamin pills. If extensive blood studies were performed, which of the following would be expected to be elevated?
A. Acetoacetic acid
B. Alanine
C. Bicarbonate
D. Chylomicrons
E. Glucose作者: zhangheng1020 时间: 2005-12-1 08:48:30 标题: 8这个大哥好强，4个月
Explanation:
The correct answer is A. Long-term starvation induces many biochemical changes. Much of the body's energy requirements are normally supplied by serum glucose, but in starvation are supplied by both glucose and lipid-derived ketone bodies, including acetoacetic acid and beta-hydroxybutyric acid. Glucose cannot be synthesized from lipids, and is instead made from amino acids such as alanine in the process of gluconeogenesis.
Serum alanine (choice B) drops dramatically in starvation, due to its conversion to glucose.
Bicarbonate (choice C) levels drop as the bicarbonate buffers the hydrogen ions produced by the ketone bodies.
Chylomicrons (choice D) are the lipid form seen after absorption of dietary fat, and would drop because the person is not feeding.
Glucose (choice E) is maintained in the blood at a much lower than normal level during starvation.作者: zhangheng1020 时间: 2005-12-1 08:49:28 标题: 9
A 15-year-old girl is seen by a dermatologist for removal of multiple squamous cell carcinomas of the skin. The patient has nearly white hair, pink irises, very pale skin, and a history of burning easily when exposed to the sun.This patient's condition is caused by a disorder involving which of the following substances?
A. Aromatic amino acids
B. Branched chain amino acids
C. Glycolipids
D. Glycoproteins
E. Sulfur-containing amino acids作者: zhangheng1020 时间: 2005-12-1 08:51:46 标题: 9 这个简单，一看就是白化病
Explanation:
The correct answer is A. The disease is albinism. The most common form of albinism is caused by a deficiency of copper-dependent tyrosinase (tyrosine hydroxylase), blocking the production of melanin from the aromatic amino acid tyrosine. Affected individuals lack melanin pigment in skin, hair, and eyes, and are prone to develop sun-induced skin cancers, including both squamous cell carcinomas and melanomas. Maple syrup urine disease is an example of a disorder of branched chain amino acids (choice B) causing motor abnormalities and seizures.
Tay-Sachs disease is an example of a disorder of glycolipids (choice C). In this disorder, a deficiency of hexosaminidase A leads to accumulation of ganglioside GM2.
Hunter's disease is an example of a disorder of glycoproteins (choice D). This mucopolysaccharidosis is inherited as an autosomal recessive trait.
Homocystinuria disease is an example of a disorder of sulfur-containing amino acids (choice E).作者: zhangheng1020 时间: 2005-12-1 08:54:41 标题: 10
A 7-year-old boy is referred to a specialty clinic because of digestive problems. He often experiences severe abdominal cramps after eating a high fat meal. He is worked up and diagnosed with a genetic defect resulting in a deficiency of lipoprotein lipase. Which of the following substances would most likely be elevated in this patient's plasma following a fatty meal?
A. Albumin-bound free fatty acids
B. Chylomicrons
C. HDL
D. LDL
E. Unesterified fatty acids作者: zhangheng1020 时间: 2005-12-1 08:58:35 标题: 10 脂类代谢机制
Explanation:
The correct answer is B. After eating a high fat meal, triglycerides are processed by the intestinal mucosal cells. They are assembled in chylomicrons and eventually sent into the circulation for delivery to adipocytes and other cells. Chylomicrons are too large to enter cells, but are degraded while in the circulation by lipoprotein lipase. A defect in this enzyme would result in the accumulation of chylomicrons in the plasma.
Albumin-bound free fatty acids (choice A) is incorrect because fatty acids leave the intestine esterified as triglycerides in chylomicrons.
HDL (choice C) is not a carrier of dietary fat from the intestine.
LDL (choice D) would be not be elevated in this patient after a high fat meal. However, VLDL would be elevated if the patient ate a high carbohydrate meal. In this situation, the carbohydrate would be converted into fat in the liver and sent out into circulation as VLDL. VLDL would be unable to be degraded to LDL and, therefore, would accumulate.
A defect in lipoprotein lipase would cause a decrease, not an elevation of unesterified fatty acids (choice E), since the chylomicrons contain esterified fatty acids.作者: zhangheng1020 时间: 2005-12-1 08:59:21 标题: 11
A 38-year-old man in a rural area presents to a physician for an employment physical. Ocular examination reveals small opaque rings on the lower edge of the iris in the anterior chamber of the eye. Nodular lesions are found on his Achilles tendon. Successful therapy should be aimed at increasing which of the following gene products in hepatocyte cell membranes?
A. Apo B-100
B. Apo B-100 receptor
C. Apo E
D. Apo E receptor
E. Lecithin cholesterol acyltransferase作者: zhangheng1020 时间: 2005-12-1 09:02:19 标题: 11 脂类的转化代谢
Explanation:
The correct answer is B. This man has characteristic signs of familial hypercholesterolemia, an autosomal dominant disorder affecting about 1 in 500 persons. The xanthomas on the Achilles tendon and the arcus lipoides (the opaque rings in the eye) are pathognomonic. Affected individuals have very high LDL cholesterol because of deficient endocytosis of LDL particles by LDL receptors. These receptors recognize the apo B-100 protein cotransported with cholesterol esters in LDL. Treatments aim at increasing genetic expression of LDL receptors (i.e., apo B-100 receptors) to enhance clearance of LDL particles. Dietary changes, a resin drug, niacin, or an HMG-CoA reductase inhibitor could be tried.
Apo B-100 (choice A) is the apoprotein of liver-produced lipoproteins such as VLDL, IDL and LDL. It is therefore not in the hepatic cell membranes, and it might be expected to decrease with decreasing concentrations of circulating LDL.
Apo E (choice C) is an apoprotein found on VLDL, IDL, and chylomicrons, allowing "scavenging" by the liver of remnants or of the lipoprotein itself. It is not found in the hepatocyte membrane.
Apo E receptor (choice D) would actually be increased by the treatment with hypocholesterolemic agents. However, the apo E receptor is not involved in the scavenging of LDL particles.
Lecithin cholesterol acyl transferase (choice E) or LCAT, is activated by apo AI, and esterifies free cholesterol in plasma. Plasma levels of HDL cholesterol and apo AI are inversely related to the risk of coronary heart disease.作者: zhangheng1020 时间: 2005-12-1 09:03:21 标题: 12
Which of the following structures is common to all sphingolipids?
A. Carnitine
B. Ceramide
C. Diacylglycerol
D. Sphingomyelin
E. Squalene作者: zhangheng1020 时间: 2005-12-1 09:08:29 标题: 12 A or D or C?
Explanation:
The correct answer is B. Sphingolipids are a class of lipids that are structural components of membranes. Ceramide is a component of sphingolipids. Ceramide is composed of sphingosine, a long-chain amino alcohol with a saturated fatty acid linked to the amino group. Sphingolipids can be differentiated on the basis of the "X" group that is esterified to the terminal hydroxyl group of ceramide.
Carnitine (choice A) is involved in the oxidation of fatty acids. Carnitine is important in transferring fatty acids from the cytoplasm into the mitochondria (the carnitine shuttle).
Diacylglycerol (choice C) is the alcohol common to all phospholipids. The second alcohol (e.g., choline, ethanolamine, serine) contributes the polar head that distinguishes the different classes of phospholipids. Like sphingolipids, phospholipids are found in membranes.
Sphingomyelin (choice D) is a sphingolipid with phosphocholine as its "X" group. It is a component of the myelin sheath.
Squalene (choice E) is a 30-carbon intermediate in the synthesis of cholesterol.作者: zhangheng1020 时间: 2005-12-1 09:10:09 标题: 13
A 47-year-old male patient presents with painful arthritis in the right big toe and uric acid renal stones. He has been taking allopurinol for his condition. What biochemical defect would likely be found in this patient?
A. A defect in urea synthesis
B. An abnormality of the purine degradation pathway
C. An inability to synthesize non-essential amino acids
D. Defective topoisomerases
E. Increased levels of leukotrienes作者: zhangheng1020 时间: 2005-12-1 09:11:25 标题: 13 A刚好相反,这个也比较简单
Explanation:
The correct answer is B. This patient has gout, characterized by painful joints due to the precipitation of uric acid crystals caused by excessive production of uric acid (a minority of cases are associated with underexcretion of uric acid). Kidney disease is also seen due to accumulation of uric acid in the tubules. The disease mostly affects males, and is frequently treated with allopurinol, an inhibitor of xanthine oxidase. Xanthine oxidase catalyzes the sequential oxidation of hypoxanthine to xanthine to uric acid.
A defect in urea synthesis (choice A) would result in the accumulation of ammonia.
Phenylketonuria is a disease in which tyrosine cannot be produced from phenylalanine (choice C). It is characterized by a musty body odor and mental retardation.
Defective topoisomerases (choice D) would affect DNA unwinding, and therefore replication.
Leukotrienes (choice E) are potent constrictors of smooth muscle and would more likely lead to bronchoconstriction.作者: zhangheng1020 时间: 2005-12-1 09:12:08 标题: 14
A newborn vomits after each feeding of milk-based formula, and does not gain weight. Biochemical testing reveals a severe deficiency of galactose-1-phosphate uridyltransferase, consistent with homozygosity. If this condition goes untreated, what is the likely outcome for this patient?
A. Benign disease except for cataract formation
B. Chronic emphysema appearing in early adulthood
C. Chronic renal failure appearing in adolescence
D. Death in infancy
E. Gastrointestinal symptoms that remit with puberty作者: zhangheng1020 时间: 2005-12-1 09:16:15 标题: 14
Explanation:
The correct answer is D. Galactosemia occurs in two very different clinical forms. Deficiency of galactokinase produces very mild disease with the only significant complication being cataract formation. In contrast, homozygous deficiency of galactose-1-phosphate uridyltransferase produces severe disease culminating in death in infancy. In addition to galactosemia and galactosuria, these patients have impaired renal tubular resorption leading to aminoaciduria, gastrointestinal symptoms, hepatosplenomegaly, cataracts, bleeding diathesis, hypoglycemia, and mental retardation. Pathologically, the CNS shows neuronal loss and gliosis and the liver shows fatty change progressing to cirrhosis.
Benign disease with cataract formation (choice A) is characteristic of galactokinase deficiency.
Chronic emphysema (choice B) is not associated with homozygous galactose-1-phosphate uridyltransferase deficiency, but rather with alpha 1-antitrypsin deficiency.
Impaired tubular reabsorption (producing aminoaciduria) is seen within a few days or weeks of feeding milk to an infant with severe galactosemia, as opposed to chronic renal failure appearing in adolescence (choice C).
Gastrointestinal symptoms (choice E) certainly occur in homozygous galactose-1-phosphate uridyltransferase deficiency, but they would not be expected to remit with puberty. Instead, most untreated infants with this disorder show failure to thrive and die in infancy from wasting and inanition.作者: zhangheng1020 时间: 2005-12-1 09:17:21 标题: 15
A 20-year old female who is 2 months pregnant remembers that she had phenylketonuria (PKU) as a child and required a special diet. Tests confirm markedly elevated maternal serum levels of phenylalanine and phenylacetic acid. Genetic studies have not been performed on the father. What should the physician tell the parents regarding the welfare of the child?
A. Childhood phenylalanine restriction is sufficient to protect the health of her child.
B. Further information is required to ascertain if the fetus is at risk.
C. The fetus is at no health risk if it is heterozygous for the PKU gene.
D. The fetus is at no health risk if phenylalanine levels are normalized by the third trimester.
E. The mother's hyperphenylalaninemia may have already harmed the fetus.作者: zhangheng1020 时间: 2005-12-1 09:20:41 标题: 15 2 MONTHS !!!
Explanation:
The correct answer is E. Phenylalanine crosses the placenta and, if maternal serum levels are elevated, acts as a teratogen to the developing fetus. This condition is known as maternal PKU. Although the mother can fare well with substantial elevations in serum phenylalanine concentration, the children born to such women are usually profoundly retarded and may have multiple birth defects.
Although dietary modifications (choice A) can prevent the neurological and dermatologic manifestations of PKU in a child, the fetus is still at risk from maternal PKU.
Further information regarding the cause of this woman's hyperphenylalaninemia (choice B) is not needed, since the fetus is exposed to teratogenic levels of phenylalanine.
Children born to mothers with untreated PKU develop maternal PKU even if they are heterozygous for the PKU gene (choice C). Fetal phenylalanine hydroxylase cannot compensate for the high maternal levels of phenylalanine.
The critical period in development during which teratogenic materials affect the growing organs is between the 3rd and 8th weeks of gestation. By the end of the 2nd month (compare with choice D), the damage caused by the maternal PKU has already occurred.作者: zhangheng1020 时间: 2005-12-1 09:22:33 标题: 16
A 24-year-old graduate student presents to a physician with complaints of severe muscle cramps and weakness with even mild exercise. Muscle biopsy demonstrates glycogen accumulation, but hepatic biopsy is unremarkable. Which of the following is the most likely diagnosis?
A. Hartnup's disease
B. Krabbe's disease
C. McArdle's disease
D. Niemann-Pick disease
E. Von Gierke's disease作者: zhangheng1020 时间: 2005-12-1 09:25:02 标题: 16
Explanation:
The correct answer is C. A variety of glycogen storage diseases exist, corresponding to defects in different enzymes in glycogen metabolism; most of these involve the liver. McArdle's disease (Type V glycogen storage disease), due to a defect in muscle phosphorylase, is restricted to skeletal muscle. The presentation described in the question stem is typical. Many affected individuals also experience myoglobinuria. Definitive diagnosis is based on demonstration of myophosphorylase deficiency.
Hartnup's disease (choice A) is a disorder of amino acid transport.
Krabbe's disease (choice B) is a lysosomal storage disease.
Niemann-Pick disease (choice D) is a lysosomal storage disease.
Von Gierke's disease (choice E) is a glycogen storage disease with prominent involvement of liver, intestine, and kidney.作者: zhangheng1020 时间: 2005-12-1 09:25:44 标题: 17
Which of the following metabolic processes occurs exclusively in the mitochondria?
A. Cholesterol synthesis
B. Fatty acid synthesis
C. Gluconeogenesis
D. Glycolysis
E. Hexose monophosphate shunt
F. Ketone body synthesis
G. Urea cycle作者: zhangheng1020 时间: 2005-12-1 09:26:16 标题: 17
Explanation:
The correct answer is F. Of the processes listed, only ketone body synthesis occurs exclusively in the mitochondria. Other mitochondrial processes include the production of acetyl-CoA, the TCA cycle, the electron transport chain, and fatty acid oxidation.
Processes that occur exclusively in the cytoplasm include cholesterol synthesis (choice A; in cytosol or in ER), fatty acid synthesis (choice B), glycolysis (choice D), and the hexose monophosphate shunt (choice E).
Note that gluconeogenesis (choice C) and the urea cycle (choice G) occur in both the mitochondria and the cytoplasm.作者: zhangheng1020 时间: 2005-12-1 09:29:10 标题: 18
A physician from the United States decides to take a sabbatical from his responsibilities at a teaching hospital to work in a clinic in a remote part of Africa. During his first week at the clinic, he is told that he will be seeing a patient with glucose-6-phosphate dehydrogenase deficiency. Which of the following will be the most likely clinical presentation of this patient?
A. A 6-month-old child who develops severe anemia following a respiratory tract infection
B. A child who develops hemoglobinuria following a meal of beans
C. A neonate with an enlarged spleen and severe anemia
D. An adult who develops anemia following use of antimalarial drugs
E. An adult who develops severe shortness of breath during an airplane ride作者: zhangheng1020 时间: 2005-12-1 09:31:20 标题: 18 glucose-6-phosphate dehydrogenase deficiency==蚕豆病
Explanation:
The correct answer is B. In Africa, the classic presentation of glucose-6-phosphate dehydrogenase deficiency is a child who eats a meal of beans (Vicia fava) and several hours later develops hemoglobinuria and peripheral vascular collapse secondary to intravascular hemolysis as a result of the oxidant injury initiated by the fava beans. Blood studies in this setting show a rapid fall in total hemoglobin and a rise in free plasma hemoglobin, accompanied by a rise in unconjugated bilirubin and a fall in haptoglobin. The episode usually resolves spontaneously several days later. Today, the classic presentation is less common in developed countries than is a slower onset syndrome beginning 1-3 days after starting an antimalarial drug, sulfonamide, or other antioxidant drug. Rarely, glucose-6-phosphate dehydrogenase deficiency presents as neonatal jaundice or with chronic hemolysis.作者: zhangheng1020 时间: 2005-12-1 09:32:29 标题: 19
A patient with familial hypercholesterolemia undergoes a detailed serum lipid and lipoprotein analysis. Studies demonstrate elevated cholesterol in the form of increased LDL without elevation of other lipids. This patient's hyperlipidemia is best classified as which of the following types?
A. Type 1
B. Type 2a
C. Type 2b
D. Type 3
E. Type 5作者: zhangheng1020 时间: 2005-12-1 09:39:07 标题: 19 secondary, acquired forms related to nephritic and hyperthyroidism
Explanation:
The correct answer is B. Hyperlipidemia has been subclassified based on the lipid and lipoprotein profiles. Type 2a, which this patient has, can be seen in a hereditary form, known as familial hypercholesterolemia, and also in secondary, acquired forms related to nephritic syndrome and hyperthyroidism. The root problem appears to be a deficiency of LDL receptors, which leads to a specific elevation of cholesterol in the form of increased LDL Heterozygotes for the hereditary form generally develop cardiovascular disease from 30 to 50 years of age. Homozygotes may have cardiovascular disease in childhood.
Type 1 (choice A) is characterized by isolated elevation of chylomicrons.
Type 2b (choice C) is characterized by elevations of both cholesterol and triglycerides in the form of LDL and VLDL.
Type 3 (choice D) is characterized by elevations of triglycerides and cholesterol in the form of chylomicron remnants and IDL.
Type 5 (choice E) is characterized by elevations of triglycerides and cholesterol in the form of VLDL and chylomicrons.
[ Last edited by zhangheng1020 on 2005-12-1 at 09:40 ]作者: zhangheng1020 时间: 2005-12-1 09:39:48 标题: 20
During the isolation of Met-enkephalin (Tyr-Gly-Gly-Phe-Met) from post-mortem human brain tissue, researchers find that the peptide is rapidly degraded by peptidases in 1 minute at 37 C. Detailed analysis of the peptide cleavage pattern of Met-enkephalin is investigated with two candidate enzymes. Using the drug bestatin, the investigators found no detectable Tyr-Gly-Gly-Phe-Met but did find significant concentrations of Tyr-Gly-Gly. Using thiorphan, there was no detectable Tyr-Gly-Gly-Phe-Met, but there was a high concentration of Tyr. Which of the following is the best conclusion about Met-enkephalin metabolism that can be drawn from these data?
A. Bestatin inhibits an aminopeptidase, and thiorphan inhibits an endopeptidase in the degradative pathway
B. Bestatin inhibits a carboxypeptidase in the degradative pathway
C. Bestatin inhibits an endopeptidase in the degradative pathway
D. Thiorphan inhibits an aminopeptidase, and bestatin inhibits an endopeptidase in the degradative pathway
E. Thiorphan inhibits an aminopeptidase in the degradative pathway作者: zhangheng1020 时间: 2005-12-1 10:03:41 标题: 20 GEEE........ I've lost....
Explanation:
The correct answer is A. Met-enkephalin, the most abundant opioid peptide in the human brain, undergoes two routes of metabolism. One route releases a tripeptide and therefore is the result of a peptidase that cuts an amino acid bond within the molecule: an endopeptidase. The other route releases free tyrosine and therefore is an exopeptidase. Exopeptidases can remove amino acid residues from the amino- or carboxyl-terminus of the protein. By convention, all peptide sequences are given from the N to the C terminus, the direction of translation. Tyrosine is therefore at the amino-terminus of Met-enkephalin, and its release is the result of digestion by an aminopeptidase.
The scientists have used two drugs to highlight the two enzymatic pathways. With bestatin, Met-enkephalin is metabolized only to the tripeptide; therefore bestatin inhibits the aminopeptidase enzyme, preventing release of free tyrosine residues. With thiorphan, Met-enkephalin is metabolized to free tyrosine; the tripeptide is no longer formed. Thiorphan is an inhibitor of the endopeptidase. The lack of persistence of Met-enkephalin in the presence of an enzyme inhibitor is evidence that the peptide's metabolism is shifted in the direction of the noninhibited enzyme. A schematic of the metabolism would be:
Tyrosine cannot be the result of carboxypeptidase activity (choice B), since the carboxyl-terminus of Met-enkephalin is a methionine.
Bestatin inhibits an aminopeptidase, not an endopeptidase (choice C). An endopeptidase would not release a free amino acid residue.
Met-enkephalin is indeed metabolized by an aminopeptidase and an endopeptidase, but bestatin inhibits the aminopeptidase and thiorphan inhibits the endopeptidase (compare with choice D).
Thiorphan does not inhibit an aminopeptidase (choice E); furthermore, such an enzyme would release a free Tyr and a tetrapeptide.作者: zhangheng1020 时间: 2005-12-1 10:22:03 标题: 21
The parents of a 6-month-old child who was normal at birth bring her into the clinic. Since their emigration to the U.S. from Eastern Europe soon after her birth, the child has developed diminished responsiveness, progressive blindness and deafness, and recently, seizures. Serum levels of which of the following compounds would be expected to be decreased in both of the parents?
A. Dystrophin
B. Hexosaminidase A
C. Hypoxanthine-guanine phosphoribosyltransferase (HGPRT)
D. Phenylalanine hydroxylase
E. Vitamin D3作者: zhangheng1020 时间: 2005-12-1 10:24:18 标题: 21 Tay-Sachs disease
Explanation:
The correct answer is B. This patient has Tay-Sachs disease, an autosomal recessive disorder caused by the deficiency of hexosaminidase A, which leads to the accumulation of ganglioside GM2 in neurons, producing a degenerative neurologic disease. Children appear normal at birth, but then begin to suffer from diminished responsiveness, deafness, blindness, loss of neurologic function, and seizures. A cherry-red spot on the macula may be seen by ophthalmoscopic examination. Death usually occurs by 4 to 5 years of age. There is no therapy. The incidence is higher among Jews of Eastern European descent. Since the parents must be heterozygotes for the mutant hexosaminidase A allele, they would be expected to have diminished levels of the enzyme.
A defect in the dystrophin (choice A) gene produces Duchenne muscular dystrophy, characterized by onset of weakness in early childhood.
A severe deficiency in HGPRT (choice C) will lead to Lesch-Nyhan syndrome, characterized by excessive uric acid production, mental retardation, spasticity, self-mutilation, and aggressive, destructive behavior.
Deficiency of phenylalanine hydroxylase (choice D) results in classic phenylketonuria, a disease in which phenylalanine, phenylpyruvate, phenylacetate, and phenyllactate accumulate in plasma and urine. Clinically, there is a musty body odor and mental retardation.
Hypophosphatemic rickets is an X-linked dominant condition causing abnormal regulation of vitamin D3 (choice E) metabolism and defects in renal tubular phosphate transport. Symptoms include growth retardation, osteomalacia, and rickets.作者: zhangheng1020 时间: 2005-12-1 10:25:04 标题: 22
Poor oxygenation of tissues decreases the production of ATP necessary for many cellular functions. Which of the following processes is most immediately compromised in a typical cell when ATP production is inadequate?
A. Complex carbohydrate synthesis
B. Lipid synthesis
C. Na+/K+ ATPase function
D. Nucleic acid synthesis
E. Protein synthesis作者: zhangheng1020 时间: 2005-12-1 10:49:46 标题: 22
Explanation:
The correct answer is C. While ATP is important in cellular synthetic functions, its role in maintaining the Na+/ K+ exchange across the plasmalemma is actually the most immediately important function for most cells. The direct effect of this is the energy (ATP) driven exchange of 3 Na+ ions (which go from inside the cell to outside) for 2 K+ ions (which go from outside to inside). This process requires considerable energy (1 ATP per 3Na+/2K+ exchange), since both the Na+ and K+ are traveling against a concentration gradient. This direct effect of the Na+/ K+ ATPase may seem trivial, but the secondary consequences are dramatic. The Na+/ K+ ATPase helps establish the transmembrane potential of the cell (because the quantitatively uneven exchange of Na+/ K+ drives more positive ions out of the cell than in) and also both the Na+ and K+ gradients. All of these facilitate a wide variety of exchanges and transmembrane transport systems that allow entry into the cell of the large variety of small molecules and ions that it needs. The first microscopically visibl effect of significant hypoxia is cellular edema, which is a consequence of distorted water balance, also an indirect function of the Na+/ K+ ATPase.作者: zhangheng1020 时间: 2005-12-1 11:13:26 标题: 23
Urine screening of an apparently healthy pregnant woman demonstrates a positive Clinitest reaction. However, blood glucose levels were within normal limits, and more specific testing for urine glucose is negative. The woman has been unaware of any metabolic problems and has been living a normal life. Deficiency of which of thefollowing enzymes would most likely produce this presentation?
A. Fructokinase
B. Fructose 1-phosphate aldolase
C. Galactose 1-P-uridyl transferase
D. Lactase
E. Pyruvate dehydrogenase作者: zhangheng1020 时间: 2005-12-1 11:17:10 标题: 23 孕妇的生理指标要注意！！
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).作者: zhangheng1020 时间: 2005-12-1 11:18:40 标题: 24
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作者: zhangheng1020 时间: 2005-12-1 11:19:01 标题: 24
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 ]作者: zhangheng1020 时间: 2005-12-1 11:24:15 标题: 25
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作者: zhangheng1020 时间: 2005-12-1 11:26:14 标题: 25 这也是个变态的题
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 ]作者: zhangheng1020 时间: 2005-12-1 11:33:30 标题: 26
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作者: zhangheng1020 时间: 2005-12-1 11:41:31 标题: 26 Lesch-Nyhan syndrome
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.作者: 樱桃小丸子/丸丸 时间: 2005-12-1 11:44:27
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作者: 樱桃小丸子/丸丸 时间: 2005-12-1 11:47:28
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 ]作者: 樱桃小丸子/丸丸 时间: 2005-12-1 11:53:00
疯了作者: zhangheng1020 时间: 2005-12-1 11:53:05 标题: 28
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作者: zhangheng1020 时间: 2005-12-1 11:54:55 标题: 28 第二次出现了
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).作者: 樱桃小丸子/丸丸 时间: 2005-12-1 11:55:08
d作者: zhangheng1020 时间: 2005-12-1 11:55:31 标题: 29
A couple brings in their 6-month-old child because they are concerned about the child's inability to sit without support. The physician interviews the parents and ascertains that they are both Ashkenazic Jews. The doctor should inform them that, because of their heritage, their child may have an increased risk of which of the following disorders?
A. Albinism and galactosemia
B. Cystic fibrosis and Lesch-Nyhan disease
C. Gaucher's disease and Tay-Sachs disease
D. Krabbe's disease and Niemann-Pick disease
E. Metachromatic leukodystrophy and phenylketonuria作者: 樱桃小丸子/丸丸 时间: 2005-12-1 11:55:34
好强阿作者: zhangheng1020 时间: 2005-12-1 12:02:17 标题: 29 Tay-Sachs disease and Type I Gaucher's disease
Explanation:
The correct answer is C. You should associate Ashkenazic (Eastern European) Jews with two diseases: Tay-Sachs disease and Type I Gaucher's disease. Both of these diseases are sphingolipidoses. Tay-Sachs disease is the more devastating of the two, and is characterized by progressive neurologic (including visual) eterioration beginning at about 6 months of age and leading to death by age 3. In contrast, Type I Gaucher’s disease is compatible with a normal life span and causes hepatosplenomegaly with CNS involvement. (The infantile Type II and the juvenile Type III forms cause more serious disease but are not seen with increased incidence in Ashkenazic Jews.) None of the other conditions listed occur with greater frequency in Ashkenazic Jews. In this case, also note that many perfectly normal children cannot sit without support at 6 months of age, so the child may well be healthy.作者: 樱桃小丸子/丸丸 时间: 2005-12-1 12:02:57
A 2-year-old retarded child is evaluated by a metabolic specialist. The child's history is significant for failure to thrive and progressive neurologic deterioration, including deafness and blindness. Physical examination is remarkable for hepatosplenomegaly, as well as a cherry-red spot on funduscopic examination. These symptoms are consistent with a diagnosis of
A. Hunter syndrome
B. Niemann-Pick disease
C. Pompe's disease
D. tyrosinosis
E. von Gierke's disease
[ Last edited by zhangheng1020 on 2005-12-1 at 12:09 ]作者: zhangheng1020 时间: 2005-12-1 12:09:33 标题: 30
Explanation:
The correct answer is B. Hepatosplenomegaly accompanied by progressive neurologic deterioration should make you think of lipid storage diseases; Niemann-Pick disease is the only lipid storage disease in the answer choices. Niemann-Pick disease is due to a deficiency of sphingomyelinase, leading to an accumulation of sphingomyelin. It is most common among Ashkenazic Jews and generally results in death by age 2. The cherry-red spot is also a characteristic of Tay-Sachs disease, but hepatosplenomegaly suggests Niemann-Pick disease rather than Tay-Sachs. Hunter syndrome (choice A) is a mucopolysaccharidosis, inherited in an X-linked recessive fashion. Pompe's disease (choice C) is a glycogen storage disease characterized by hypotonia and cardiorespiratory failure. Tyrosinosis (choice D) is a rare abnormality of tyrosine metabolism that would not produce the listed symptoms. von Gierke's disease (choice E) is a severe form of glycogen storage disease characterized by hypoglycemia, hepatomegaly, and renomegaly.作者: zhangheng1020 时间: 2005-12-1 12:12:19 标题: 31
Which of the following enzymes is located at arrow 1 in the electron micrograph above?
A. Carnitine acyltransferase II
B. Fatty acyl CoA synthetase
C. Glucose-6-phosphate dehydrogenase
D. Hexokinase
E. Pyruvate kinase作者: zhangheng1020 时间: 2005-12-1 12:15:41 标题: 31
Explanation:
The correct answer is A. Arrow 1 indicates the inner mitochondrial membrane. Carnitine acyltransferase II is located on the inner face of the inner mitochondrial membrane. It reforms fatty acyl CoA in the mitochondrial matrix (arrow 5) from acyl carnitine, thus preparing it for mitochondrial oxidation. The acyl groups on carnitine are derived from acyl CoA esters synthesized in the outer mitochondrial membrane, which are made from free fatty acids circulating in the blood.
Fatty acyl CoA synthetases (choice B) are located in the outer mitochondrial membrane, indicated by arrow 2. Glucose-6-phosphate dehydrogenase, the first enzyme in the pentose phosphate pathway (choice C), hexokinase, the first enzyme in the glycolytic pathway (choice D), and pyruvate kinase (choice E), which produces pyruvate from phosphoenolpyruvate in glycolysis, are all located in the cytosol, indicated by arrow 3. Arrow 4 indicates smooth endoplasmic reticulum.作者: zhangheng1020 时间: 2005-12-1 12:16:26 标题: 32
32
A 72-year-old woman, in otherwise good health, presents with megaloblastic anemia. Careful evaluation reveals a folate deficiency as the cause of the anemia. Assuming the folate deficiency is due to dietary causes, which of the following is the most likely problem?
A. Lack of leafy green vegetables
B. Lack of milk products
C. Lack of red meat
D. Lack of yellow vegetables
E. Overcooked food作者: zhangheng1020 时间: 2005-12-1 12:17:45 标题: 32
Explanation:
The correct answer is E. Folates (pteroylglutamic acid and related compounds) are widely distributed in foodstuffs. Dietary deficiency is usually due to overcooked (folates are very labile) and old (folates rapidly decay with time) food.作者: zhangheng1020 时间: 2005-12-1 12:18:32 标题: 33
A 2-month-old boy is evaluated for failure to thrive. As the pediatrician is examining the patient, she witnesses a seizure. Physical examination is remarkable for hepatomegaly, a finding later confirmed by CT scan, which also reveals renomegaly. Serum chemistries demonstrate severe hypoglycemia, hyperlipidemia, lactic acidosis, and ketosis. Which of the following diseases best accounts for this presentation?
A. Gaucher's disease
B. McArdle's disease
C. Niemann-Pick disease
D. Pompe's disease
E. von Gierke's disease作者: zhangheng1020 时间: 2005-12-1 12:19:43 标题: 33 glycogen storage disease
Explanation:
The correct answer is E. von Gierke's disease is a glycogen storage disease caused by a deficiency of glucose-6-phosphatase. It typically presents with neonatal hypoglycemia, hyperlipidemia, lactic acidosis, and ketosis. Failure to thrive is common in early life; convulsions may occur due to profound hypoglycemia. The glycogen accumulation in von Gierke's disease occurs primarily in the liver and kidneys, accounting for the enlargement of these organs. Gout may develop later because of the derangement of glucose metabolism. Even if you do not remember all of the details of the presentation of these genetic diseases, you should be able
to narrow the choices: Gaucher's disease (choice A) and Niemann-Pick disease (choice C) are lipid storage diseases, and would not be expected to produce hypoglycemia. The other diseases are glycogen storage diseases, but McArdle's (choice B) and Pompe's (choice D) disease affect muscle rather than liver and would not be expected to produce profound hypoglycemia, since the liver is the major source for blood glucose.作者: zhangheng1020 时间: 2005-12-1 12:21:01 标题: 34
A newborn appears normal at birth, but develops vomiting and diarrhea accompanied by jaundice and hepatomegaly within the first few weeks of life. Within months, the baby has obvious cataracts and ascites. The infant is switched to a milk-free diet, which stabilizes but does not completely reverse his condition. By one year of age, he has developed mental retardation. Which of the following is the most likely diagnosis?
A. Cystic fibrosis
B. Galactosemia
C. McArdle's disease
D. Von Gierke's disease
E. Wilson's disease作者: zhangheng1020 时间: 2005-12-1 12:27:53 标题: 34 也是第二次出现了
Explanation:
The correct answer is B. Galactosemia is an autosomal recessive disease caused by a deficiency of galactose-1-phosphate uridyltransferase, which is necessary for the metabolism of the galactose derived from milk lactose. The condition should be suspected in infants with growth failure, cataracts, liver disease, aminoaciduria, and mental retardation. A reducing sugar (galactose) is usually present in the urine. Most of the pathology is related to the toxic effects of galactose-1-phosphate. Treatment involves strict dietary lactose restriction, which consists of more than simply withdrawal of milk products, because lactose is also present in many non-diary foods. Strict adherence to the diet can strikingly alter the course of this disease. Cystic fibrosis (choice A) is associated with maldigestion pancreatic disease, and pulmonary disease. McArdle's disease (choice C) is a glycogen storage disease that selectively affects muscle. Von Gierke's disease (choice D) is a glycogen storage disease affecting the liver and kidneys. Wilson's disease (choice E) is a caused by a metabolic abnormality in the handling of copper that can cause cirrhosis and brain damage, and usually presents in adolescence.作者: zhangheng1020 时间: 2005-12-1 12:32:40 标题: 35
35
Which of the following pairs of enzymes is required for the process of gluconeogenesis?
A. Fructose-1,6-bisphosphatase and pyruvate carboxylase
B. Glucose-6-phosphatase and phosphofructokinase-1
C. Glucose-6-phosphatase and pyruvate dehydrogenase
D. Phosphoenolpyruvate carboxykinase and glucokinase
E. Pyruvate kinase and pyruvate carboxylase作者: zhangheng1020 时间: 2005-12-1 12:34:15 标题: 35 gluconeogenesis，这道题是人都应该知道，不应该做错的
Explanation:
The correct answer is A.The three irreversible steps of glycolysis are catalyzed by hexokinase, phosphofructokinase-1 (choice B), and pyruvate kinase. In gluconeogenesis, other enzymes are needed to bypass these key steps. Pyruvate cannot be directly converted to phosphoenolpyruvate in gluconeogenesis. Therefore, pyruvate carboxylase (a mitochondrial enzyme; choice A) converts pyruvate to oxaloacetate, which can be converted to phosphoenolpyruvate by phosphoenolpyruvate carboxykinase (choice D), using two ATP equivalents per molecule of phosphoenolpyruvate. Fructose-1,6-bisphosphatase (choice A) is the enzyme that splits fructose-1,6-bisphosphate into fructose-6-phosphate and inorganic phosphate. It is also required for gluconeogenesis.
Glucose-6-phosphatase (choices B and C) is a liver enzyme that hydrolyzes glucose-6-phosphate to glucose. A deficiency of this enzyme leads to von Gierke disease, also known as glycogen storage disease type I. Pyruvate dehydrogenase (choice C) is a mitochondrial enzyme that converts pyruvate to acetyl CoA. This enzyme requires thiamine pyrophosphate, lipoamide, and FAD as cofactors. Glucokinase (choice D) is a liver enzyme that converts glucose to glucose-6-phosphate. Unlike hexokinase, it is specific for glucose and is unresponsive to the level of glucose-6-phosphate. Its function is to store excess glucose, so it has a very high Km (ie, a low affinity) for glucose, becoming active only when the concentration of glucose is very high.
Pyruvate kinase (choice E) catalyzes the conversion of phosphoenolpyruvate to pyruvate in the glycolytic pathway. It is activated by fructose-1,6-bisphosphate, the product of the committed step of glycolysis, and is allosterically inhibited by ATP, alanine, and acetyl CoA.作者: zhangheng1020 时间: 2005-12-1 12:34:43 标题: 36
A 40-year-old, formerly obese woman presents to her physician. She was very proud of having lost 80 lbs. during the previous 2 years, but now noticed that her "hair is falling out." On questioning, she reports having followed a strict fat-free diet. Her alopecia is probably related to a deficiency of which of the following vitamins?
A. Vitamin A
B. Vitamin C
C. Vitamin D
D. Vitamin E
E. Vitamin K作者: zhangheng1020 时间: 2005-12-1 12:36:16 标题: 36 oil-soluble vitamins 这个也很简单的
Explanation:
The correct answer is A. While it is hard to develop a deficiency in oil-soluble vitamins (A, D, E, K) because the liver stores these substances, deficiency states can be seen in chronic malnutrition (specifically chronic fat deprivation) and chronic malabsorption. Vitamin A is necessary for formation of retinal pigments (deficiency can cause night blindness) and for appropriate differentiation of epithelial tissues (including hair follicles, mucous membranes, skin, bone, and adrenal cortex).
Vitamin C (choice B), which is water soluble rather than oil soluble, is necessary for collagen synthesis.
Vitamin D (choice C) is important in calcium absorption and metabolism.
Vitamin E (choice D) is a lipid antioxidant that is important in the stabilization of cell membranes.
Vitamin K (choice E) is necessary for normal blood coagulation.作者: zhangheng1020 时间: 2005-12-1 12:37:04 标题: 37
5 mL of synovial fluid is aspirated from an inflamed knee joint. The fluid is yellow-white and cloudy and contains 200,000 WBC/mm3 (85% neutrophils). Needle-shaped, strongly negatively birefringent crystals are seen both within and outside neutrophils. These crystals most likely have which of the following compositions?
A. Basic calcium phosphate
B. Calcium oxalate
C. Calcium pyrophosphate dihydrate
D. Cholesterol
E. Monosodium urate作者: zhangheng1020 时间: 2005-12-1 12:37:50 标题: 37
Explanation:
The correct answer is E. All the compounds listed can produce crystals in joint fluid, but only monosodium urate (associated with gout) and calcium pyrophosphate dihydrate (associated with CPPD crystal deposition disease, also called pseudogout), and to lesser degree basic calcium phosphate (apatite-associated arthropathy), have a high likelihood of being encountered on a step 1 USMLE exam. The crystals described are those of monosodium urate. Be careful not to answer "uric acid" if that is listed as an alternative choice on an exam, since the sodium salt is the predominant species in vivo.
Basic calcium phosphate (choice A) is seen in apatite-associated arthropathy and produces spherical clumps of nonbirefringent submicroscopic crystals.
Calcium oxalate crystals (choice B) are seen in primary oxalosis and are bipyramidal, positively birefringent crystals.
Calcium pyrophosphate dihydrate crystals (choice C) are a feature of pseudogout and are rod-to-rhomboidal-shaped, weakly positively birefringent crystals.
Cholesterol crystals (choice D) are seen in chronic and chylous effusions in inflammatory and degenerative arthritis, where they form large, flat, rhomboidal plates with notched corners.作者: zhangheng1020 时间: 2005-12-1 12:44:49 标题: 38
38
Which of the following cofactors is required for decarboxylation of alpha-ketoacids?
A. Vitamin B1
B. Vitamin B2
C. Vitamin B3
D. Vitamin B5
E. Vitamin B6作者: zhangheng1020 时间: 2005-12-1 12:47:07 标题: 38 vit B
Explanation:
The correct answer is A. Vitamin B1, or thiamine, is the coenzyme required (as the pyrophosphate) for the decarboxylation of alpha-ketoacids. An example of this reaction is pyruvate decarboxylase reaction in alcoholic fermentation. Other reactions such as that catalyzed by pyruvate dehydrogenase also rely on thiamine pyrophosphate for decarboxylation, but require other cofactors as well. Thiamine is also required for the generation of pentose phosphates for nucleotide synthesis in the pentose phosphate pathway (hexose monophosphate shunt), serving as a cofactor for transketolase.
Vitamin B2(choice B), or riboflavin, is a constituent of FMN (flavin mononucleotide) and FAD (flavin adenine dinucleotide). It functions in hydrogen and electron transport.
Vitamin B3(choice C), or niacin (nicotinic acid), is a coenzyme that is also involved in hydrogen and electron transport. Nicotinic acid functions in the form of NAD and NADP.
Vitamin B5(choice D), or pantothenic acid, is conjugated with coenzyme A to act as a carboxylic acid carrier.
Vitamin B6(choice E), or pyridoxine, is required as a cofactor for pyridoxal phosphate and pyridoxamine phosphate. Both of these cofactors are essential to protein metabolism and energy production.作者: zhangheng1020 时间: 2005-12-1 12:47:34 标题: 39
A newborn presents with severe acidosis, vomiting, hypotonia, and neurologic deficits. Serum analysis reveals elevated levels of lactate and alanine. These observations suggest a deficiency in which of the following enzymes?
A. Alanine aminotransferase
B. Glutamate dehydrogenase
C. Lactate dehydrogenase
D. Pyruvate carboxylase
E. Pyruvate dehydrogenase作者: 樱桃小丸子/丸丸 时间: 2005-12-1 12:49:08
Originally posted by zhangheng1020 at 2005-12-1 12:11
Explanation:
The correct answer is E. Pyruvate dehydrogenase (PDH) catalyzes the irreversible conversion of pyruvate to acetyl-CoA. If PDH is absent, pyruvate will be used in other pathways instead. Pyruvate will be converted to alanine via alanine aminotransferase (choice A) and to lactate via lactate dehydrogenase (choice C).
Glutamate dehydrogenase (choice B) is involved in oxidative deamination, releasing ammonium ion for urea synthesis. Deficiency of this enzyme would not cause the symptoms described.
Pyruvate carboxylase (choice D) is a gluconeogenic enzyme that catalyzes the conversion of pyruvate to oxaloacetate. Deficiency of this enzyme would not cause the symptoms described.作者: zhangheng1020 时间: 2005-12-1 12:51:46 标题: 40
The activity of which of the following enzymes is directly affected by citrate?
A. Fructose-2,6-bisphosphatase
B. Isocitrate dehydrogenase
C. Phosphofructokinase I
D. Pyruvate carboxylase
E. 6-phosphogluconate dehydrogenase作者: zhangheng1020 时间: 2005-12-1 12:56:53 标题: 40 in the mitochondria
Explanation:
The correct answer is C. Citrate is produced by citrate synthase from acetyl CoA and oxaloacetate. This reaction takes place in the mitochondria, but citrate can move freely from the mitochondria into the cytosol. When the citric acid cycle slows down, citrate accumulates. In the cytosol, it acts as a negative allosteric regulator of phosphofructokinase I, the enzyme that catalyzes the committed step of glycolysis.
Fructose-2,6-bisphosphatase (choice A) breaks down fructose-2,6-bisphosphate, a potent allosteric activator of phosphofructokinase I. Fructose-2,6-bisphosphatase is activated by cyclic AMP-dependent protein kinase.
Isocitrate dehydrogenase (choice B) converts isocitrate to alpha-ketoglutarate in the citric acid cycle. It is allosterically stimulated by ADP and inhibited by ATP and NADH. This reaction produces NADH and CO2.
Pyruvate carboxylase (choice D) is a mitochondrial enzyme that converts pyruvate to oxaloacetate. It is important in gluconeogenesis and replenishes the oxaloacetate in the citric acid cycle.
6-phosphogluconate dehydrogenase (choice E) converts 6-phosphogluconate to ribulose 5-phosphate in the pentose phosphate shunt pathway.作者: zhangheng1020 时间: 2005-12-1 12:58:39 标题: 41
Which of the following enzymes is stimulated by glucagon?
A. Acetyl-CoA carboxylase
B. Glycogen phosphorylase
C. Glycogen synthase
D. HMG-CoA reductase
E. Pyruvate kinase作者: zhangheng1020 时间: 2005-12-1 12:59:49 标题: 41 依旧是死记硬背的题一个
Explanation:
The correct answer is B. Before you started analyzing all of the answer choices you should have reminded yourself that glucagon increases serum glucose. So an enzyme stimulated by glucagon might be involved in either the breakdown of glycogen to glucose (glycogenolysis) or in the creation of glucose from noncarbohydrate precursors (gluconeogenesis). Glycogen phosphorylase catalyzes the first step in glycogenolysis; it makes sense that it would be stimulated by glucagon.
Acetyl-CoA carboxylase (choice A) catalyzes the first step in fatty acid synthesis, an anabolic process that would be stimulated by insulin, not glucagon.
As its name implies, glycogen synthase (choice C) is involved in the synthesis of glycogen. Glucagon (and epinephrine) stimulate the phosphorylation and inactivation of glycogen synthase.
HMG-CoA reductase (choice D) is the key enzyme involved in the synthesis of cholesterol. Since this is an anabolic process that occurs in the well-fed state, you would expect it to be stimulated by insulin and inhibited by glucagon (which it is).
Pyruvate kinase (choice E) catalyzes the last reaction of glycolysis. You would expect it to be inhibited by glucagon (thus decreasing the amount of glucose consumption). Glucagon promotes the phosphorylation of pyruvate kinase, which renders it inactive.作者: zhangheng1020 时间: 2005-12-1 13:00:27 标题: 42
Which of the following inhibits the activity of acetyl-CoA carboxylase?
A. Citrate
B. Glucagon
C. High-carbohydrate, low-fat diet
D. Insulin作者: zhangheng1020 时间: 2005-12-1 13:13:27 标题: 42 依旧死记硬背的题
Explanation:
The correct answer is B. The key thing to remember here is that acetyl-CoA carboxylase catalyzes the first and rate-limiting step of fatty acid synthesis. If you got that far, you could have figured out which of the choices would inhibit the synthesis of fatty acids. Certainly glucagon, a catabolic hormone released in response to low blood glucose, would be a likely candidate to inhibit the synthesis of fatty acids. In fact, glucagon inhibits fatty acid synthesis by a cAMP-dependent phosphorylation of acetyl-CoA carboxylase. Conversely, glucagons stimulates fatty acid oxidation.
Citrate (choice A) is a key player in fatty acid synthesis (citrate shuttle). Therefore, the presence of citrate would stimulate, not inhibit, acetyl-CoA carboxylase.
A high-carbohydrate, low-fat diet (choice C) would stimulate, not inhibit, the synthesis of fatty acids.
In contrast to glucagon, insulin (choice D) is an anabolic hormone that promotes fatty acid synthesis and therefore would stimulate acetyl-CoA carboxylase. It does so by dephosphorylating the enzyme.作者: zhangheng1020 时间: 2005-12-1 13:33:37 标题: 43
An individual lacking the enzyme tyrosinase would be particularly predisposed to develop which of the following?
A. Glioblastoma multiforme
B. Hemangioblastoma
C. Hepatoma
D. Melanoma
E. Renal cell carcinoma作者: zhangheng1020 时间: 2005-12-1 14:02:43 标题: 43 melanin formation
Explanation:
The correct answer is D. This question is simple if you know that tyrosinase is an enzyme in the biosynthetic pathway for melanin formation from tyrosine. A lack of tyrosinase causes one form of albinism; a second form is caused by defective tyrosine uptake. Patients with albinism are vulnerable to developing cancers of the skin of all types, including basal cell carcinoma, squamous cell carcinoma, and melanoma. The melanomas are unusual in that they are non-pigmented (amelanotic) rather than black, since the patients cannot form melanin.作者: 樱桃小丸子/丸丸 时间: 2005-12-1 17:48:10
up作者: zhangheng1020 时间: 2005-12-1 22:11:49 标题: 44
44
A newborn baby has multiple hemorrhages. Clotting studies demonstrate an elevated prothrombin time. An abnormality of which of the following biochemical processes is likely present in this patient?
A. Conversion of homocysteine to methionine
B. Conversion of methylmalonyl CoA to succinyl CoA
C. Degradation of cystathionine
D. Formation of gamma-carboxyglutamate residues
E. Hydroxylation of proline