The Biologic Basis for Disease in Adults and Children
9th Edition
• Author(s)Julia Rogers
TEST BANK
McCance & Huether — Pathophysiology, 9th Ed. — Chapter 1:
Cellular Biology.
Chapter 2, Section: Prokaryotes and Eukaryotes — Cellular
Biology
Stem
A hospitalized patient’s wound culture shows a bacterial species
lacking a true nucleus and membrane-bound organelles. Which
cellular feature best distinguishes this organism from the
patient’s eukaryotic host cells?
Options
A. Presence of histone-associated DNA packaging
B. DNA contained within a membrane-bound nucleus
C. Circular DNA not enclosed by a nuclear membrane
D. Presence of membrane-bound mitochondria
,Correct Answer
C
Rationales
Correct: Bacterial (prokaryotic) genomes are typically circular
and not enclosed by a nuclear membrane, distinguishing them
from eukaryotic cells.
A: Histone-associated DNA packaging is characteristic of
eukaryotes; prokaryotes generally lack histones.
B: DNA in eukaryotes is contained within a membrane-bound
nucleus — this describes the host, not the bacterium.
D: Membrane-bound mitochondria are eukaryotic organelles
absent in prokaryotes.
Teaching Point
Prokaryotes have circular DNA that is not enclosed by a nucleus.
2.
Chapter 2, Section: Cellular Functions — Homeostasis and
Adaptation
Stem
A patient with chronic hypertension shows left ventricular
hypertrophy. Which cellular adaptation best explains the
increase in cardiac myocyte size?
Options
A. Metaplasia due to pressure overload
B. Hyperplasia through increased cell division
,C. Hypertrophy via increased synthesis of contractile proteins
D. Atrophy from decreased workload
Correct Answer
C
Rationales
Correct: Cardiac myocytes respond to increased workload by
hypertrophy—enlargement through increased synthesis of
contractile proteins (actin/myosin).
A: Metaplasia is replacement of one cell type by another and is
not the mechanism for increased myocyte size.
B: Cardiac myocytes are terminally differentiated with limited
proliferation, so hyperplasia is not the main response.
D: Atrophy is decreased cell size due to reduced workload, the
opposite of observed change.
Teaching Point
Hypertrophy increases cell size by synthesizing more contractile
proteins.
3.
Chapter 2, Section: Structure/Function of Cellular Components
— Mitochondria
Stem
A patient’s muscle biopsy shows reduced ATP synthesis and
accumulation of lactic acid during minimal exertion. Which
mitochondrial defect most likely explains these findings?
, Options
A. Increased activity of pyruvate dehydrogenase complex
B. Impaired electron transport chain (oxidative
phosphorylation)
C. Enhanced β-oxidation of fatty acids
D. Increased mitochondrial fusion leading to larger organelles
Correct Answer
B
Rationales
Correct: Impaired electron transport/oxidative phosphorylation
reduces ATP production, forcing anaerobic glycolysis and lactic
acid accumulation.
A: Increased pyruvate dehydrogenase activity would promote
aerobic ATP production, not lactic acidosis.
C: Enhanced β-oxidation would increase acetyl-CoA for the TCA
cycle and likely improve ATP generation.
D: Mitochondrial fusion alone doesn’t explain reduced ATP and
lactic acid buildup; functional ETC impairment does.
Teaching Point
ETC dysfunction reduces ATP and increases anaerobic glycolysis
and lactate.
4.
Chapter 2, Section: Structure/Function of Cellular Components
— Endoplasmic Reticulum & Golgi