REPLICATION
A 7-year-old boy is being evaluated for a genetic disorder causing developmental delay. During laboratory analysis, researchers identify a defect in an enzyme required during the initiation phase of DNA replication. Without this enzyme, DNA polymerase is unable to begin synthesis of a new DNA strand despite the presence of an intact DNA template and abundant nucleotides.
Which of the following enzymes is most directly responsible for allowing DNA polymerase to initiate DNA synthesis?
A. DNA ligase
B. DNA primase
C. Topoisomerase II
D. DNA polymerase δ
E. Helicase
Answer:
B. DNA primase
Explanation:
DNA polymerases cannot initiate DNA synthesis de novo and require a free 3′-OH group. DNA primase synthesizes a short RNA primer that provides this 3′-OH end, allowing DNA polymerase to begin replication. This occurs during initiation of DNA replication
A researcher is studying gene regulation in human hepatocytes. She introduces a mutation into a DNA sequence located approximately 25 base pairs upstream of the transcription start site. Following the mutation, RNA polymerase II is unable to efficiently initiate transcription despite the presence of all necessary transcription factors.
Which of the following DNA elements was most likely mutated?
A. Polyadenylation signal
B. TATA box
C. Exon
D. Enhancer
E. Intron
Answer:
B. TATA box
Explanation:
The TATA box is a promoter consensus sequence located approximately 25 base pairs upstream of the transcription start site. TFIID, through its TATA-binding protein (TBP), recognizes this sequence and helps assemble the transcription initiation complex, allowing RNA polymerase II to begin transcription. A mutation in the TATA box can significantly impair transcription initiation.
A 6-year-old boy is enrolled in a research study evaluating protein synthesis in human cells. Investigators isolate a newly synthesized mRNA transcript and observe that translation begins only after a specific codon is recognized by an initiator tRNA carrying methionine.
Which of the following codons most likely initiates translation in this patient’s cells?
A. UAA
B. UAG
C. UGA
D. AUG
E. GUG
Answer:
D. AUG
Explanation:
Translation initiation begins when the 40S ribosomal subunit scans the mRNA and identifies the start codon AUG. The initiator methionine tRNA pairs with this codon, allowing assembly of the complete 80S ribosome and initiation of protein synthesis
A 24-year-old woman undergoing infertility evaluation asks how mature gametes are initially formed during embryonic development. Her physician explains that germ cells arise early in embryogenesis and migrate to the developing gonads before differentiating into male or female gametes.
From which of the following structures do primordial germ cells originate?
A. Neural crest
B. Intermediate mesoderm
C. Epiblast
D. Primitive streak
E. Trophoblast
Answer:
C. Epiblast
Explanation:
Primordial germ cells (PGCs) are formed in the epiblast during the second week of development. They subsequently migrate through the yolk sac region and travel to the developing gonads, where they proliferate and differentiate into oogonia or spermatogonia
A 4-year-old boy is brought to the pediatrician because of difficulty sweating and sparse hair growth. Physical examination demonstrates hypoplastic teeth and reduced sweat gland function. These structures all share a common embryologic origin.
Which of the following germ layers is the most likely source of the affected tissues?
A. Endoderm
B. Intermediate mesoderm
C. Lateral plate mesoderm
D. Paraxial mesoderm
E. Ectoderm
Answer:
E. Ectoderm
Explanation:
The ectoderm gives rise to the epidermis and its derivatives, including hair follicles, sweat glands, and tooth enamel, as well as the nervous system. Defects in ectodermal derivatives commonly present with abnormalities involving the skin, hair, teeth, and glands
A 24-year-old graduate student is participating in a research study involving DNA replication. Investigators expose dividing cells to a drug that selectively inhibits the enzyme responsible for relieving torsional strain generated ahead of the replication fork. Following treatment, DNA replication stalls despite successful unwinding of the DNA double helix.
Inhibition of which of the following enzymes most likely produced these findings?
A. DNA ligase
B. DNA polymerase ε
C. Topoisomerase
D. Telomerase
E. RNase H
Answer:
C. Topoisomerase
Explanation:
As helicase unwinds DNA, positive supercoils accumulate ahead of the replication fork. Topoisomerases relieve this torsional stress through reversible DNA strand breaks, permitting continued replication. Inhibition causes excessive supercoiling and replication fork arrest.
A molecular biologist is analyzing a newly discovered human gene. During transcription, she notices that the synthesized RNA sequence is nearly identical to one DNA strand except that uracil replaces thymine. She concludes that this DNA strand is not being directly used as the template for RNA synthesis.
Which of the following strands is identical in sequence to the newly synthesized RNA molecule?
A. Antisense strand
B. Template strand
C. Coding strand
D. Lagging strand
E. Leading strand
Answer:
C. Coding strand
Explanation:
RNA polymerase reads the template (antisense) strand in the 3′→5′ direction while synthesizing RNA in the 5′→3′ direction. The resulting RNA sequence is complementary to the template strand and nearly identical to the coding (sense) strand, except that uracil replaces thymine
A scientist creates a mutation in a eukaryotic ribosome that prevents incoming charged tRNAs from entering their normal binding site. As a result, elongation cannot proceed because new amino acids are unable to be added to the growing peptide chain.
Which of the following ribosomal sites is most directly affected?
A. E site
B. P site
C. A site
D. Exit channel
E. Peptidyl transferase center
Answer:
C. A site
Explanation:
During elongation, aminoacyl-tRNAs carrying new amino acids enter the A (aminoacyl) site of the ribosome. The growing peptide chain is held in the P (peptidyl) site, and empty tRNAs leave through the E (exit) site. Blocking the A site prevents addition of new amino acids and arrests elongation
A newborn girl is found to have streak gonads due to abnormal embryologic development. Histologic examination reveals complete absence of primordial germ cells within the gonadal ridges. Investigators suspect a defect in migration of these cells during embryogenesis.
Normally, primordial germ cells migrate from the yolk sac to the developing gonads during which of the following time periods?
A. Second week to third week
B. Third week to fourth week
C. Fourth week to end of fifth week
D. Seventh week to eighth week
E. Tenth week to twelfth week
Answer:
C. Fourth week to end of fifth week
Explanation:
Primordial germ cells are formed in the epiblast during the second week. During the fourth week they begin migrating from the yolk sac toward the developing gonads and arrive by the end of the fifth week. Failure of migration may result in gonadal defects and can contribute to germ-cell tumors.
A pregnant woman who did not take prenatal vitamins undergoes a fetal ultrasound at 13 weeks gestation. The fetus is found to lack a major portion of the cranial vault and cerebral hemispheres. The anomaly is traced to failure of closure of a structure that normally closes around day 25 of embryonic development.
Failure of closure of which of the following structures most likely caused this defect?
A. Posterior neuropore
B. Primitive streak
C. Neural crest
D. Cranial neuropore
E. Notochord
Answer:
D. Cranial neuropore
Explanation:
Anencephaly results from failure of closure of the cranial (anterior) neuropore, which normally closes around day 25. During neurulation, neural folds fuse to form the neural tube. Failure of cranial neuropore closure leads to severe defects of the brain and skull
A 6-year-old boy is brought to the clinic because of severe photosensitivity. His parents report that he develops extensive freckling after minimal sun exposure. Physical examination reveals multiple hyperpigmented lesions on sun-exposed areas of the skin. Biopsy demonstrates early malignant transformation. Genetic testing shows a mutation affecting a protein involved in repairing UV-induced thymine dimers.
Which of the following DNA repair pathways is most likely defective?
A. Base excision repair
B. Mismatch repair
C. Homologous recombination
D. Nucleotide excision repair
E. Nonhomologous end joining
Answer:
D. Nucleotide excision repair
Explanation:
This patient has Xeroderma Pigmentosum, an autosomal recessive disorder caused by defective nucleotide excision repair (NER). NER removes bulky, helix-distorting DNA lesions such as UV-induced thymine dimers. Failure of this mechanism leads to extreme UV sensitivity and markedly increased risk of skin cancer
A 42-year-old man ingests a toxic mushroom during a camping trip. Several days later, he develops severe hepatic failure. Laboratory analysis demonstrates inhibition of RNA polymerase II, causing marked reductions in mRNA synthesis and impaired protein production.
Which of the following toxins is most likely responsible for this patient's condition?
A. Rifampin
B. Actinomycin D
C. Cyclophosphamide
D. α-Amanitin
E. Methotrexate
Answer:
D. α-Amanitin
Explanation:
α-Amanitin, found in Amanita phalloides mushrooms, inhibits RNA polymerase II, preventing elongation of mRNA transcripts. Because mRNA production is essential for protein synthesis, highly metabolic tissues such as the liver are particularly susceptible to injury
A 19-year-old man undergoes genetic testing after several relatives are diagnosed with β-thalassemia. Sequencing reveals a mutation affecting the translation initiation sequence of the β-globin gene. Although the remainder of the coding sequence is intact, very little β-globin protein is produced.
Which of the following genetic abnormalities most likely explains these findings?
A. Missense mutation in the middle of the coding region
B. Silent mutation in the third codon position
C. Mutation of the AUG start codon
D. Mutation of a stop codon to a sense codon
E. Degenerate codon substitution
Answer:
C. Mutation of the AUG start codon
Explanation:
The start codon is essential for establishing the correct reading frame and initiating translation. A mutation involving AUG may prevent ribosome assembly and initiation of protein synthesis, resulting in markedly reduced protein production despite an otherwise intact gene. The lecture specifically identifies start-codon mutations as a cause of β-thalassemia.
A 13-year-old girl presents shortly after menarche. An LH surge induces ovulation, and a primary oocyte completes meiosis I. The resulting daughter cells contain unequal amounts of cytoplasm.
Which of the following cells is directly produced by completion of meiosis I in the female?
A. Oogonium and first polar body
B. Secondary oocyte and first polar body
C. Mature ovum and second polar body
D. Secondary oocyte and second polar body
E. Primary oocyte and secondary oocyte
Answer:
B. Secondary oocyte and first polar body
Explanation:
Primary oocytes remain arrested in prophase I from fetal life until puberty. Just before ovulation, an LH surge triggers completion of meiosis I, producing a secondary oocyte and the first polar body. The secondary oocyte then enters meiosis II and becomes arrested in metaphase II until fertilization occurs
A newborn is noted to have congenital absence of several dorsal root ganglia. Further evaluation reveals a defect in migration of a population of embryonic cells that normally undergoes epithelial-to-mesenchymal transition and migrates away from the developing neural tube.
Which of the following additional structures is most likely absent in this patient?
A. Vertebral bodies
B. Adrenal cortex
C. Hepatocytes
D. Adrenal medulla
E. Renal tubules
Answer:
D. Adrenal medulla
Explanation:
Neural crest cells migrate from the neural tube and give rise to multiple structures including dorsal root ganglia, sympathetic ganglia, Schwann cells, enteric neurons, melanocytes, and the adrenal medulla. Loss of neural crest migration would affect all of these derivatives.
An 11-year-old girl is evaluated for fatigue and recurrent infections. Physical examination reveals short stature, café-au-lait spots, thumb hypoplasia, and a triangular face. Laboratory studies show pancytopenia. Chromosomal breakage testing using diepoxybutane demonstrates marked chromosomal fragility. Further analysis reveals defective interaction between proteins involved in DNA repair and BRCA-mediated pathways.
Which of the following DNA repair mechanisms is most likely impaired in this patient?
A. Base excision repair
B. Mismatch repair
C. Homologous recombination
D. Nucleotide excision repair
E. Direct repair
Answer:
C. Homologous recombination
Explanation:
This patient has Fanconi anemia. Hallmark findings include pancytopenia, short stature, radial ray abnormalities, café-au-lait spots, and chromosomal instability. Fanconi anemia results from defective repair of DNA interstrand crosslinks and impaired homologous recombination pathways that function closely with BRCA1 and BRCA2 proteins. Homologous recombination is a high-fidelity mechanism for double-stranded DNA break repair that requires a homologous template
A geneticist identifies a mutation affecting a transcription factor required for opening double-stranded DNA during the initiation phase of transcription. Cells containing the mutation show impaired formation of the transcription bubble despite normal promoter recognition and TATA-binding protein function.
Which of the following transcription factors is most likely defective?
A. TAF
B. TFIID
C. TBP
D. TFIIH
E. RNA Polymerase I
Answer:
D. TFIIH
Explanation:
TFIIH possesses helicase activity that unwinds DNA near the transcription start site, creating the transcription bubble required for RNA synthesis. A defect in TFIIH would prevent strand separation even if promoter recognition by TFIID and TBP remains intact
A 7-year-old child develops severe pharyngitis followed by myocarditis. Laboratory testing demonstrates exposure to a bacterial toxin that irreversibly inhibits elongation during protein synthesis. Cells show normal initiation of translation but are unable to continue peptide chain extension.
Which of the following toxins is the most likely cause of this patient's illness?
A. Cholera toxin
B. Tetanus toxin
C. Botulinum toxin
D. Diphtheria toxin
E. Shiga toxin
Answer:
D. Diphtheria toxin
Explanation:
Diphtheria toxin inhibits protein synthesis by blocking elongation, ultimately leading to cell death. This mechanism produces the characteristic systemic manifestations of Corynebacterium diphtheriae infection, including myocarditis and neurologic complications. The lecture specifically lists diphtheria toxin as a translation inhibitor that blocks elongation.
A 17-year-old boy is evaluated for delayed puberty and infertility. Laboratory studies show low testosterone levels despite elevated LH concentrations. Testicular biopsy demonstrates impaired spermatogenesis. The affected cells normally respond to LH stimulation and synthesize testosterone, which subsequently acts on Sertoli cells to support sperm production.
Which of the following cells is most likely dysfunctional in this patient?
A. Sertoli cells
B. Spermatogonia
C. Leydig cells
D. Primary spermatocytes
E. Spermatids
Answer:
C. Leydig cells
Explanation:
LH acts on Leydig cells within the testes to stimulate testosterone production. Testosterone then acts on Sertoli cells to support spermatogenesis. Defective Leydig cell function results in decreased testosterone production and impaired sperm development despite elevated LH levels
A research laboratory selectively ablates paraxial mesoderm in a developing embryo. Several weeks later, the embryo demonstrates severe abnormalities of the axial skeleton and trunk musculature.
Which of the following embryologic structures normally arises from paraxial mesoderm?
A. Gonads
B. Vertebrae and ribs
C. Gut tube
D. Heart tube
E. Kidney collecting system
Answer:
B. Vertebrae and ribs
Explanation:
Paraxial mesoderm forms somites, which subsequently differentiate into sclerotome, dermatome, and myotome. The sclerotome forms the vertebrae and ribs, whereas the myotome forms skeletal muscle and the dermatome contributes to dermis. [TRILAMINAR...ERIVATIVES | PDF]
A pharmaceutical company is developing a novel antiviral medication structurally similar to guanosine. During preclinical testing, infected cells treated with the drug demonstrate premature termination of viral DNA synthesis shortly after incorporation into the growing DNA strand. Biochemical analysis shows that once the drug is incorporated, no additional nucleotides can be added to the newly synthesized DNA molecule.
Which of the following molecular features of this drug most directly explains its mechanism of action?
A. Inhibition of topoisomerase-mediated relaxation of supercoils
B. Lack of a 3′-hydroxyl group required for phosphodiester bond formation
C. Irreversible inhibition of DNA ligase activity
D. Inactivation of helicase at the replication fork
E. Prevention of RNA primer synthesis by primase
Answer:
B. Lack of a 3′-hydroxyl group required for phosphodiester bond formation
Explanation:
DNA polymerase elongates DNA strands by adding nucleotides to the free 3′-OH group of the growing DNA chain. Certain antiviral nucleoside analogs, such as acyclovir, lack this critical 3′-OH group. Once incorporated into DNA, no additional phosphodiester bonds can be formed, resulting in immediate chain termination and cessation of DNA replication. This mechanism selectively inhibits viral DNA synthesis
A 5-year-old child is found to have a mutation in a small nuclear ribonucleoprotein (snRNP) complex involved in pre-mRNA processing. Laboratory studies reveal normal transcription initiation, elongation, and termination. However, mature mRNA production is severely impaired, resulting in multiple abnormal proteins with retained noncoding sequences.
Which of the following cellular processes is most directly affected by this mutation?
A. Addition of the 7-methylguanosine cap
B. Polyadenylation of the 3′ end
C. Recognition of the TATA box
D. Removal of introns from pre-mRNA
E. Synthesis of rRNA by RNA polymerase I
Answer:
D. Removal of introns from pre-mRNA
Explanation:
snRNPs are key components of the spliceosome, which removes introns and joins exons during pre-mRNA processing. Failure of spliceosome function results in retention of introns within mRNA transcripts, producing abnormal proteins or preventing proper translation. Alternative splicing is a major source of protein diversity in eukaryotes
A 23-year-old woman participates in a genetics experiment evaluating the consequences of nucleotide insertions within coding regions. Investigators insert a single nucleotide immediately after the start codon of a gene. Translation still initiates normally, but the resulting protein is nonfunctional because nearly every amino acid downstream differs from the original sequence.
Which of the following properties of the genetic code best explains this result?
A. Degeneracy
B. Universality
C. Specificity
D. Reading frame dependence
E. Wobble pairing
Answer:
D. Reading frame dependence
Explanation:
The genetic code is read in nonoverlapping groups of three nucleotides (codons) beginning at the start codon. Insertion of a single nucleotide shifts the reading frame, causing all downstream codons to be interpreted incorrectly. This frameshift mutation often produces severely abnormal proteins and may introduce premature stop codons.
A reproductive endocrinologist is studying two patients undergoing gametogenesis. One cell sample is obtained from a fetal ovary, while the other is obtained from a seminiferous tubule of an adult male. The ovarian cell has remained arrested for years, whereas the male germ cell is actively completing meiosis.
Which of the following best describes the normal meiotic arrest that occurs in human oogenesis?
A. Oogonia arrest in metaphase I until birth
B. Primary oocytes arrest in prophase I until puberty
C. Secondary oocytes arrest in prophase II until fertilization
D. Oogonia arrest in metaphase II until ovulation
E. Primary oocytes arrest in anaphase I until menarche
Answer:
B. Primary oocytes arrest in prophase I until puberty
Explanation:
During fetal development, oogonia differentiate into primary oocytes and enter meiosis I. Near birth, all surviving primary oocytes become arrested in the diplotene stage of prophase I due to the action of oocyte maturation inhibitor (OMI). They remain arrested until puberty, when selected follicles resume development. Completion of meiosis I occurs just before ovulation.
A newborn presents with unilateral renal agenesis and abnormalities of the reproductive tract. Genetic testing identifies a mutation affecting development of the embryonic tissue that temporarily connects paraxial mesoderm with the lateral plate mesoderm. Failure of differentiation of this structure disrupted formation of nephrotomes and the nephrogenic cord.
Which of the following embryologic tissues was most likely affected?
A. Neural crest
B. Endoderm
C. Intermediate mesoderm
D. Somatic mesoderm
E. Splanchnic mesoderm
Answer:
C. Intermediate mesoderm
Explanation:
Intermediate mesoderm gives rise to the urogenital system, including nephrotomes, the nephrogenic cord, kidneys, and gonads. Defects in intermediate mesoderm commonly result in concurrent urinary and reproductive abnormalities because both systems share this embryologic origin.