USABO Semifinal Examination 2011 - Part A - CELL AND MOLECULAR QUESTIONS 1 to 12

Sep 29, 2026

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USABO Semifinal Examination 2011 Part A includes CELL AND MOLECULAR QUESTIONS 1–12 spanning core topics in cell and molecular biology. Questions address nucleotide composition and DNA structure interpretation, DNA amount versus time during cell division, identification of protein structures contacting chromosomes, troubleshooting PCR amplification using polymerase and primer/template inputs, and inhibitor effects on enzyme kinetics. Additional items test stop codon implications for genetic principles, comparative existence of cellular structures across cell types, oxidative phosphorylation uncoupling versus inhibition, recombination frequency inference from probe hybridization data, and degenerate probe design from protein sequence. Numerical computation includes maximum ethanol production under anaerobic metabolism.

LanguageEnglish

CategoryExam

FormatPDF

Pages43

Reading Time1 h 48 min

AccessFree to download

CELL AND MOLECULAR QUESTIONS 1 to 12

Cell division and DNA variation

Chromosomes, proteins, and genetic mechanisms

PCR, enzyme kinetics, and respiration

Probes, recombination, and metabolic calculations

USABO SEMIFINAL EXAMINATION March 16 to 25, 2011 2011 USABO Semifinal Exam DRAFT Part A | Cell and Molecular Biology (20%) | 12 questions 1-12 | | --- | --- | | Plant Anatomy and Physiology (15%) | 9 questions 13-21 | | Animal Anatomy and Physiology (25%) | 15 questions 22-36 | | Ethology (5%) | 3 questions 37-39 | | Genetics and Evolution (20%) | 12 questions 40-51 | | Ecology (10%) | 6 questions 52-57 | | Biosystematics (5%) | 3 questions 58-60 | Part A CELL AND MOLECULAR QUESTIONS 1 to 12 1. If you extract the genetic material of the Bacteriophage (􀀩X174), you will find that its composition is 25% A, 33% T, 24% G, and 18% C. How would you interpret these results? A. The experiment's results must be erroneous; something went wrong B. We could admit that the %A approximately equals that ofT, and the same is true for C and G. Consequently, Chargaff's rules are followed and the DNA is double stranded C. Because the %A does not equal the %T, nor does the %G equal the %C, the DNA is single-stranded; it is replicated by special enzymes, following a particular replication pattern, with the single-stranded chain as a template D. Because the %A does not equal the %T, nor does the %G equal the %C, the DNA must be single-stranded; It replicates by synthesizing a complementary strand and uses the double stranded form as a template E. Bacteriophage (􀀩X174) is an RNA virus Answer D 2. The diagrams below refer to variation in the amount of DNA (y-axis) as a function of time (x-axis) during cell division. (The units are arbitrary.) Which of these diagrams depict what happens during meiosis and mitosis, respectively? A. 1 and 2 B. 1 and 3 C. 2 and 3 D. 3 and 4 E. 4 and 1 Answer E 3. Which of the following cellular proteins make direct contact with chromosomes? A. Centrosomes and microtubule-associated proteins B. Kinesins, myosins, and actin microfilaments C. Histones, condensins, and synaptonemal complexes D. All of the above E. None of the above Answer C 4. Your colleague’s PCR reaction failed to amplify the human rhodopsin gene. Her lab notebook indicates that she added dNTPs, the Rhodopsin template gene, complementary DNA primers, buffer, and an E. Coli DNA polymerase and ran the reaction under the proper conditions. How would you advise her to fix thereaction? A. Add a bacterial promoter to the Rhodopsin template B. Add extra ATP so there is enough energy to catalyze the reaction C. Add dNTPs to facilitate elongation D. Use a different polymerase E. Use RNA primers instead Answer: D 5. For a constant concentration of an enzyme, which of the following is true of anoncompetitive inhibitor? A. km is lowered; vmax remains constant B. km is increased; vmax remains constant C. km remains constant; vmaxis lowered D. km remains constant; vmax is increased E. km remains constant; vmax remains constant Answer C 6. The mitochondrion has more than three stop codons. Which biological principle does this violate? A. Central dogma B. Endosymbiotic theory C. Conservation DNA replication D. Complementary base-pairing E. Universal genetic code Answer: E 7. Which of the following structures and processes can exist in eukaryotic cells and in all prokaryotic cells? I. Nuclear envelope II. Ribosomes III. Introns IV. ATP synthesis V. rRNA 18S VI. Cell membrane VII. DNA polymerase VIII. Cytoskeletal elements A. I, II, III and VIII B. II, IV, VI and VIII C. I, III, V and VII D. II, IV, V and VI E. II, III, VI and VIII Answer: D 8. Comparing the effect of an inhibitor with an uncoupler of oxidative phosphorylation, the: A. inhibitor would allow electrons to pass through the electron transport chain B. inhibitor would increase the pumping of protons by the electron transport chain C. uncoupler would increase heat production by the mitochondria D. uncoupler would inhibit the reduction of oxygen by the electron transport chain E. unc

FAQs

How should the genetic material composition of bacteriophage X174 be interpreted?

The provided percentages indicate that Chargaff’s rules are followed, consistent with double-stranded DNA.

Which diagram pair corresponds to meiosis and mitosis based on DNA amount versus time?

The correct choice is the pair that matches the expected DNA content changes during meiosis and mitosis, given in the answer as E.

What is the best advice to fix a failed PCR reaction for amplifying the human rhodopsin gene?

The reaction should use a different polymerase, since the listed components and conditions are otherwise indicated as proper, with the provided answer D.

Sep 29, 2026

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