USMLE USMLE Certifications USMLE-STEP-1 Questions & Answers

• Question 521:

  When dividing cells are progressing through the cell cycle, many checks are imposed to ensure that the process is occuring with fidelity. A critical checkpoint in the cell cycle occurs in response to DNA damage, such as that induced by ultraviolet light. Which of the following cellcycle proteins is involved in DNA damagemediated cell-cycle arrest?

  A. CDK2

  B. cyclin A

  C. cyclin D

  D. E2F

  E. p53

  Correct Answer: E

• Question 522:

  Numerous inherited disorders are the result of the expansion of trinucleotide (triplet) repeats either within the coding regions of genes or the untranslated regions of the resultant RNAs. Which of the following diseases has been shown to be caused by triplet expansion?

  A. cystic fibrosis (CF)

  B. Duchenne muscular dystrophy (DMD)

  C. FH

  D. Huntington disease (HD)

  E. Menkes disease

  Correct Answer: D

  Section: Biochemistry HD is an autosomal-dominant disorder leading to progressive memory loss, personality changes, and peculiar motor problems such as involuntary movements of the arms and legs. The disease results from the expansion of a CAG triplet in the amino terminus of the HD protein, referred to as huntingtin. The triplet repeat number ranges from 10 to 30 on the normal chromosome and from 36 to 121 on the HD chromosome. There is a general correlation between the length of the repeat and the age of onset of symptoms. CF (choice A) is primarily due to a common mutation (in 70% of cases) that deletes three nucleotides in exon 10 of the CF gene, which codes for the CF transmembrane conductance receptor (CFTR). Over 600 other mutations have been identified in the CF gene. DMD (choice B) results from deletions in one or more of the exons of the DMD gene, which encodes the protein referred to as dystrophin. FH (choice C) results from defects in the gene encoding the LDL receptor. These defects encompass insertions and deletions that can be found throughout the length of the LDLreceptor gene. Menkes disease (choice E) is due to defects in copper absortion leading to defective function of numerous enzymes that need copper as a cofactor.

• Question 523:

  A 17-year-old man who reports to his physician that he is incapable of obtaining an erection is also quite embarrassed by the apparent enlargement of his breasts (gynecomastia). These symptoms, when present in males, are associated with an excessive production of which of the following hormones?

  A. corticotropin-releasing hormone

  B. GnRH

  C. growth hormone

  D. melanocyte-stimulating hormone

  E. prolactin

  Correct Answer: E

  Section: Biochemistry Prolactin is necessary for initiation and maintenance of lactation. Physiologic levels act only on breast tissue primed by female sex hormones. Endocrine dysfunction leading to excessive prolactin production is associated with breast enlargement and impotence in males. Excessive production of corticotropinreleasing hormone (choice A) would result in an increase in adrenocorticotropic hormone (ACTH) production, which would lead to enhanced glucocorticoid and mineralocorticoid production. Excessive production of GnRH (choice B) would lead to increased production of luteinizing hormone (LH), follicle-stimulating hormone (FSH), and chorionic gonadotropin (hCG), with consequent effects on the female reproductive system. Excessive production of growth hormone (choice C) leads to gigantism if it occurs prior to epiphysial plate closure. If excessive release occurs following epiphysial plate closure acromegaly results, with characteristic facial changes (protruding jaw, enlarged nose) and enlarged feet, hands, and skull. Excessive production of melanocyte-stimulating hormone (choice D) would lead to hyperpigmentation of the skin.

• Question 524:

  The anticancer drug, Taxol, has been an effective chemotherapeutic agent in the fight against ovarian cancer. Which of the following represents the mechanism of action of Taxol?

  A. binds to the activated form of the RAS protein, which in turn interferes with the signaling cascades involving this protein

  B. binds to microtubules, which stabilizes them preventing their shortening and interfering with cell division

  C. interacts with the tumor supressor protein (pRB) encoded by the retinoblastoma susceptibility gene inducing its suppressive activity

  D. interacts with topoisomerase II preventing its role in DNA synthesis, which effectively terminates replication

  E. interferes with steroid hormone receptor interaction with DNA, thus preventing the growth induction by this class of hormone

  Correct Answer: B

  Section: Biochemistry The compound paclitaxel (the generic name for Taxol) interferes with the normal function of microtubule growth leading to increased stability of the structure. This effect of Taxol destroys the cell ability to use its cytoskeleton in a flexible manner. Specifically, Taxol binds to the tubulin protein of microtubules and locks them in place. The resulting microtubule/Taxol complex does not have the ability to disassemble. This adversely affects cell function because the shortening and lengthening of microtubules (termed dynamic instability) is necessary for their function as a transportation highway for the cell. Chromosomes, for example, rely on this property of microtubules during mitosis. Therefore, the inability of cells to organize and undertake the necessary movements of mitosis prevents cell division. Since cancer cells are rapidly proliferating cells, the action of Taxol inhibits their growth. None of the other choices (A, CE) reflect the actions of Taxol on cells.

• Question 525:

  A 32-year-old woman is diagnosed with hypertension, hypernatremia, hypokalemia, and alkalosis. Measurements of plasma glucocorticoid levels show them to be within the normal range; however, renin and angiotensin II levels are suppressed. Ultrasound indicates the possible existence of an adrenal cortical mass. These symptoms are likely due to excess production of which of the following hormones?

  A. aldosterone

  B. androstenedione

  C. dehydroepiandrosterone (DHEA)

  D. estradiol

  E. testosterone

  Correct Answer: A

  Section: Biochemistry Primary aldosteronism (Conn syndrome) leads to elevated production of aldosterone and the symptoms presented by the patient. This disorder is due to small adenomas of the glomerulosa cells of the kidney. The associated hyperkalemia, hypertension, and hypernatremia lead to a reduction in renin release from the juxtaglomerulosa cells. Renin is required for the conversion of angiotensinogen (released from the liver) to angiotensin I (which is in turn converted to angiotensin II by converting enzyme), and so reduced levels of renin lead to reduced levels of angiotensin II. Androstenedione (choice B) is produced from pregnenolone via the 17alpha-hydroxylase pathway of androgen synthesis in the adrenal cortex or of estrogen synthesis in the ovary. Androstenedione is derived from DHEA(choice C) and is converted to estradiol (choice D) in the ovary or to testosterone (choice E) in the testis. Estrogens are responsible for maturation of the ovaries and testosterone maturation of sperm. None of these steroids regulate sodium or potassium and would therefore not lead to the symptoms presented if elevated or reduced.

• Question 526:

  One important function of nitric oxide (NO) is the induction of vascular smooth muscle relaxation in response to acetylcholine. The production of NO requires which amino acid?

  A. arginine

  B. asparagines

  C. cysteine

  D. glutamine

  E. lysine

  Correct Answer: A

  Section: Biochemistry NO is generated from arginine in a reaction catalyzed by NOS. The other product of the reaction is citrulline. None of the other amino acids (choices B, C, D, and E) are substrates for NOS.

• Question 527:

  In renal insufficiency, calcium absorption is reduced and leads to increased bone resorption, a condition referred to as renal osteodystrophy. Treatment with which of the following can assist in the amelioration of the symptoms of this condition?

  A. antidiuretic hormone

  B. calcitonin

  C. calcitriol

  D. growth hormone

  E. parathyroid hormone

  Correct Answer: C

• Question 528:

  A 37-year-old man presents with tophaceous deposits within the articular cartilage, synovium, tendons, tendon sheaths, pinnae, and the soft tissue on the extensor surface of the forearms. These clinical observations suggest the patient is suffering from which of the following disorders?

  A. ADA deficiency

  B. gout

  C. Lesch-Nyhan syndrome

  D. purine nucleoside phosphorylase(PNP) deficiency

  E. von Gierke disease

  Correct Answer: B

  Section: Biochemistry Gout is characterized by elevated levels of uric acid in the blood and urine. Uric acid is the end-product of purine catabolism and excess production results from a variety of metabolic abnormalities that lead to overproduction of purines via the de novo pathway. Uric acid is very insoluble and when generated in large amounts will precipitate as uric acid crystals in the joints of the extremities and in renal interstitial tissue. These deposits are gritty or sandy in nature and thus are termed tophaceous deposits. ADA (choice A) and PNP (choice D) deficiencies result in various degrees of immune dysfunction and are not associated with gouty episodes. Lesch-Nyhan disease (choice C) is due to loss of HGPRT activity. Characteristic symptoms include severe mental retardation and self-mutilation. A much less severe symptom associated with Lesch-Nyhan disease is hyperuricemia, which leads to gouty episodes. However, other symptoms are of such severity that patients die very early in life. Deficiency in glucose-6-phosphatase results in von Gierke disease (choice E). Clinical symptoms of von Gierke disease include fasting hypoglycemia, lactic acidemia, and mild gouty episodes. The hyperuricemia of von Gierke disease is seldom severe enough to lead to the level of gouty deposits observed in this patient.

• Question 529:

  The initial reaction of the de novo synthesis pathway for pyrimidine nucleotides begins with glutamine and C and is complete with the formation of uridine monophosphate. Which of the following

  

  represents the ratelimiting enzyme in this pathway?

  A. aspartate transcarbamoylase

  B. orotate monophosphate decarboxylase

  C. phosphoribosylpyrophosphate (PRPP) amido transferase

  D. PRPP synthetase

  E. ribonucleotide reductase

  Correct Answer: A

  Section: Biochemistry The first reaction of de novo pyrimidine biosynthesis is catalyzed by ATC. This reaction is also the rate-limiting step in this pathway. OMP decarboxylase (choice B) catalyzes the decarboxylation of OMP, yielding UMP. PRPP amido transferase (choice C) is an enzyme of the de novo purine biosynthesis pathway. PRPP synthetase (choice D) catalyzes the production of PRPP (used in the synthesis of purines and pyrimidines) from ribose-5-phosphate and ATP. Ribonucleotide reductase (choice E) is required for the reduction of ribonucleotides to deoxyribonucleotides.

• Question 530:

  Some viruses, for example, poliovirus, contain a protease that cleaves one of the eukaryotic initiation factors allowing for cap-independent translational initiation of viral RNAs at internal ribosome entry site (IRES). Which of the following factors is the target of these viral proteases?

  A. eIF-2

  B. eIF-2B

  C. eIF-4A

  D. eIF-4E

  E. eIF-4G

  Correct Answer: E

  Section: Biochemistry Eukaryotic viral RNAs are not capped and the host 40S ribosome interacts with these viral RNAs at an IRES. Capped eukaryotic mRNAs are recognized by the complex of eIF- 4E and eIF-4G, which in turn allows for interaction of the 40S ribosome with the mRNA (see below figure).

  

  Initiation factor eIF-4E physically binds the cap structure in eukaryotic mRNAs and this function is facilitated by interaction of eIF-4E with eIF-4G. Thus, loss of the interaction of eIF-4E with the protease cleaved eIF- 4G results in loss of translational initiation from capped mRNAs. However, the binding of 40S ribosomes to viral RNAs does not require eIF-4E, only eIF-4G. The protease encoded by the poliovirus cleaves the eIF- 4E binding site on eIF-4G. None of the other translation factors (choices AC, E) are targets for poliovirus protease.