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Wilkes University NSG 530 Exam 1 Advanced Pathophysiology 2026/2027 Intensive Study Guide for Nursing Graduate Students with Practice Exams and High-Yield Review Topics

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Comprehensive NSG 530 Exam 1 Advanced Pathophysiology Study Guide 2026/2027 designed to help graduate nursing students prepare for quizzes, tests, and major course examinations. Covers essential pathophysiology concepts commonly assessed on Exam 1, including cellular adaptations, genetic influences on disease, inflammatory and immune responses, fluid and electrolyte balance, acid-base regulation, tissue injury and repair, mechanisms of disease development, and foundational principles of human pathophysiology. Includes practice exams, high-yield review topics, study exercises, detailed notes, concept summaries, and exam-focused preparation content to strengthen pathophysiological knowledge and improve academic performance. Ideal for students seeking structured revision support and comprehensive preparation for Wilkes University NSG 530 Exam 1.

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Institución
CNA - Certified Nursing Assistant
Grado
CNA - Certified Nursing Assistant

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2026/2027



Wilkes University NSG 530 Exam 1
Advanced Pathophysiology
2026/2027 Intensive Study Guide for
Nursing Graduate Students with
Practice Exams and High-Yield
Review Topics
Question 1:

An ordered photographic display of a set of chromosomes from a single cell is best
described as:

A. Metaphase spread
B. Autosomal spread
C. Karyotype
D. Anaphase spread

Correct Answer: C. Karyotype

Rationale:
A karyotype is the organized visual representation of all chromosomes in a cell,
arranged in homologous pairs according to size and banding pattern. It is used
clinically to detect chromosomal abnormalities such as trisomies or deletions. A
metaphase spread refers only to the stage in which chromosomes are most visible
during cell division, but it is not an organized display. An autosomal spread is not a
standard genetic term, and an anaphase spread refers to chromosomes separating
during anaphase, which is not used for diagnostic imaging.


Question 2:

Failure of homologous chromosomes to separate during meiosis is termed:

A. Aneuploidy
B. Nondisjunction
C. Polyploidy
D. Anaplasia

Correct Answer: B. Nondisjunction

,2026/2027

Rationale:
Nondisjunction is the failure of homologous chromosomes or sister chromatids to
separate properly during meiosis, leading to abnormal chromosome numbers in
daughter cells. This is the primary mechanism behind many chromosomal disorders.
Aneuploidy refers to the abnormal number of chromosomes resulting from
nondisjunction, but not the mechanism itself. Polyploidy refers to complete extra sets
of chromosomes, and anaplasia refers to loss of cellular differentiation, commonly
seen in cancer.


Question 3:

A somatic cell that does not contain a multiple of 23 chromosomes is called:

A. Euploid cell
B. Aneuploid cell
C. Polyploid cell
D. Haploid cell

Correct Answer: B. Aneuploid cell

Rationale:
An aneuploid cell has an abnormal number of chromosomes that is not a complete
multiple of 23, often due to nondisjunction. This can result in genetic disorders such
as trisomy 21. Euploid cells contain complete sets of chromosomes (e.g., 46 in
humans). Polyploid cells contain more than two full sets of chromosomes, and haploid
cells contain only one set (23 chromosomes), such as gametes.


Question 4:

A fetus is born stillborn with 92 chromosomes. This condition is best described as:

A. Euploidy
B. Triploidy
C. Tetraploidy
D. Aneuploidy

Correct Answer: C. Tetraploidy

Rationale:
Tetraploidy occurs when a cell contains four complete sets of chromosomes (92 in
humans), which is typically incompatible with life. Triploidy involves three sets (69
chromosomes). Euploidy refers to a normal complete set of chromosomes, and
aneuploidy refers to an abnormal number that is not a full set, such as trisomies.


Question 5:

A chromosomal mosaic individual may:

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A. Be a carrier of a genetic disease
B. Have a mild form of the genetic disease
C. Have two genetic diseases
D. Be sterile due to genetic disease

Correct Answer: B. Have a mild form of the genetic disease

Rationale:
Mosaicism occurs when an individual has two or more genetically different cell lines,
which often leads to a milder phenotype because not all cells are affected. This can
reduce disease severity compared to full chromosomal abnormalities. Being a carrier
refers to single-gene inheritance patterns, not mosaicism. Having two diseases or
sterility is not a defining feature.


Question 6:

The most common cause of Down syndrome is:

A. Paternal nondisjunction
B. Maternal translocations
C. Maternal nondisjunction
D. Paternal translocations

Correct Answer: C. Maternal nondisjunction

Rationale:
Down syndrome is most commonly caused by nondisjunction during maternal meiosis,
leading to trisomy 21. This accounts for the majority of cases. Translocations are less
common causes. Paternal nondisjunction is rare in comparison to maternal origin.


Question 7:

A major risk factor for Down syndrome is:

A. Fetal exposure to mutagens
B. Increased paternal age
C. Family history
D. Maternal age over 35

Correct Answer: D. Maternal age over 35

Rationale:
Advanced maternal age significantly increases the risk of chromosomal
nondisjunction during meiosis, leading to conditions such as Down syndrome.
Although paternal age and environmental mutagens may contribute to genetic risk,
they are less strongly associated. Family history is not a major risk factor in most
cases because most occurrences are sporadic.

, 2026/2027


Question 8:

A child with a single X chromosome and no second sex chromosome has:

A. Down syndrome
B. Cri du chat syndrome
C. Turner syndrome
D. Edward syndrome

Correct Answer: C. Turner syndrome

Rationale:
Turner syndrome is characterized by monosomy X (45,X), leading to short stature,
gonadal dysgenesis, and infertility. Down syndrome involves trisomy 21, Cri du chat
involves deletion of chromosome 5, and Edward syndrome is trisomy 18.


Question 9:

Cystic fibrosis occurring in a child whose parents are first cousins is most likely due
to:

A. X inactivation
B. Genomic imprinting
C. Consanguinity
D. Obligate carriers

Correct Answer: C. Consanguinity

Rationale:
Cystic fibrosis is an autosomal recessive disorder, and consanguinity increases the
probability that both parents carry the same recessive gene mutation. X inactivation
and imprinting involve gene expression regulation, not inheritance of recessive
disorders. Obligate carriers are individuals who must carry a mutation based on
family history, not a cause.


Question 10:

Duchenne muscular dystrophy is inherited as a:

A. Sex-linked dominant trait
B. Sex-influenced trait
C. Sex-limited trait
D. Sex-linked recessive trait

Correct Answer: D. Sex-linked recessive trait

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Institución
CNA - Certified Nursing Assistant
Grado
CNA - Certified Nursing Assistant

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Subido en
4 de junio de 2026
Número de páginas
35
Escrito en
2025/2026
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