Free Online USMLE Step 1 Questions and Answers 6

4. A cell biologist is studying the role of ribonucleoproteins in normal cellular function. He prepares a cell extract using a specific cell type obtained from a 73-year-old man. Ribonucleoproteins are separated and purified from the cell extract for structural and functional analyses. These cells are found to express higher amounts of a particular protein in comparison to other cell types. This protein has reverse transcriptase activity that functions to add TTAGGG repeats to the 3′ end of chromosomes. Which of the following cell types was most likely studied in this experiment?

A. Epidermal basal cells
B. Erythrocytes
C. Myocardial cells
D. Neurons
E. Pancreatic β cells

Correct Answer: A

Answer Explanation:

Stem cells have very long telomeres due to their high telomerase activity, allowing them to proliferate indefinitely in a controlled manner. In contrast, most terminally differentiated adult somatic cells (eg, myocardial cells, neurons, pancreatic~ cells) have short telomeres as they do not express telomerase and their telomeres shorten with every cell division (Choices C, D, and E). Critical shortening in telomere length is thought to be a signal for programmed cell death. In fact, syndromes of premature aging (eg, Bloom syndrome) are associated with shortened telomeres. In contrast, cancer cells upregulate their telomerase activity, preventing cell death by maintaining their telomere length.
(Choice B) Erythrocytes have no nuclei and therefore have no potential to divide.

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5. A 9-month-old boy is brought to the clinic for a routine follow-up. His mother is concerned because the patient is not yet able to sit up unsupported. He was born at term and has had muscle weakness since birth. Vital signs are normal. The patient is alert but has diminished tone. Examination shows a prominent forehead with a depressed nasal bridge. Eye examination shows epicanthal folds. Analysis of lysosomal acid hydrolases shows an increased concentration within the serum and a decreased level within the cultured skin fibroblast cells. This patient most likely has a defect in which of the following steps of enzyme production?

A. DNA methylation
B. Posttranslational modification
C. Protein folding
D. Splicing
E. Translation

Correct Answer: B

Answer Explanation:

(Choice A) DNA methylation silences gene transcription so that the affected protein would be completely absent. A pathologic example of DNA methylation is the inhibition of tumor suppressor genes as a mechanism for cancer development.
(Choice C) A defect in protein folding, as seen with the CFTR protein in cystic fibrosis, would result in abnormal protein functioning. Protein concentration may be normal or low due to more rapid degradation of the abnormal protein.
(Choices D and E) Splicing mutations can result in inappropriate removal of exons or persistence of intrans; this creates aberrant mRNA that is presented for translation and results in production of an abnormal protein. Similarly, a defect in translation results in insufficient or abnormal protein synthesis, which is reflected both in the serum and intracellularly.

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6. An infant born to a 34-year-old woman has a flat facial profile, prominent epicanthal folds, and a holosystolic murmur heard loudest at the left sternal border. Karyotype analysis is consistent with trisomy 21. Maternal and paternal karyotypes are normal. A restriction fragment length polymorphism (R FLP) analysis is conducted to determine the parental origin of the extra chromosome. DNA samples from the child, mother, and father are obtained and the DNA is fragmented with a restriction enzyme. The fragments are then sorted by size using the Southern blot technique. Labeling is done using a probe that binds to a specific DNA sequence close to the centromere of chromosome 21 . RFLP analysis for the child, mother, and father is shown below.

Usmle5 1

In which of the following meiosis events did the nondisjunction most likely occur?

A. Maternal meiosis I
B. Maternal meiosis II
C. Paternal meiosis I
D. Paternal meiosis II

Correct Answer: A

Answer Explanation:

(Choice B) If the mother had a failure in meiosis II, the child’s RFLP analysis would reveal only two bands (see example). There would be a single band from the father and a darker, thicker band from the mother. The darker, thicker band signifies the inheritance of both sister chromatids, which wi ll produce equal-size restriction fragments (but twice the normal amount).
(Choice C) If the father had a failure in meiosis I, the child’s RFLP analysis would reveal 3 bands (see example). In this instance, there would be a single band inherited from the mother with two bands inherited from the father.
(Choice D) If the father had a failure in meiosis II, the child’s RFLP analysis would reveal two bands (see example). In this instance, there would be a single band inherited from the mother and a second band from the father. The father’s band would be darker and thicker, signifying the inheritance of both sister chromatids.

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