SCIENCE · 2026/2027 LATEST UPDATE
WGU C491 Integrated Natural Science Exam 2026
Biology, Chemistry, Earth Science & Physics -
Complete Study Guide
100+ Verified Exam Questions · Rated A+ · Guaranteed Results
✔ Biology (Cells, Genetics, Evolution) ✔ Earth Science (Geology, Meteorology, Astronomy)
✔ Chemistry (Atomic Structure, Bonding, Reactions)
✔ Physics (Motion, Forces, Energy)
Instant PDF Download · 100% Verified Answers · Score 90%+
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 1 of 53
, BEST SELLER · SCIENCE
WGU C491 Integrated Natural Science Exam 2026
Biology, Chemistry, Earth Science & Physics
2026/2027 - Q&A with Verified Answers
Every section. Every topic. 100% correct answers guaranteed. Used by
5,000+ science students to pass on their first attempt.
Biology (Cells, Genetics, Evolution)
Chemistry (Atomic Structure, Bonding, Reactions)
Earth Science (Geology, Meteorology, Astronomy)
Physics (Motion, Forces, Energy)
A+
Rated
Used by 5,000+ students · 100% Verified Answers · 2026/2027 Latest Update
Instant PDF Download
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 2 of 53
, SECTION 1 | Biology (Cells, Genetics, Evolution) | Q1-Q25 | WGU C491 Integrated Natural Science Exam 2026 Biology, Chemis
Q1 Question 1 of 100
A researcher examines a cell under a microscope and observes a large
membrane-bound organelle containing the genetic material. The cell is identified as
eukaryotic because:
A. The cell contains ribosomes for protein synthesis
B. The DNA is enclosed within a nucleus surrounded by a double membrane
C. The cell has a plasma membrane that regulates transport
D. The cell uses ATP as its primary energy currency
Correct Answer: B
Rationale:
Eukaryotic cells are distinguished by having a true nucleus enclosed by a nuclear envelope (double
membrane), which separates genetic material from the cytoplasm. Prokaryotic cells also have
ribosomes, plasma membranes, and use ATP, so options B, C, and D are features shared by all cell
types rather than unique to eukaryotes.
Q2 Question 2 of 100
During cellular respiration, a muscle cell in a sprinter's leg produces pyruvate through
glycolysis. Under anaerobic conditions, pyruvate is converted to lactate rather than
entering the Krebs cycle. This conversion allows the cell to:
A. Produce more ATP than aerobic respiration
B. Regenerate NAD+ so glycolysis can continue producing ATP
C. Store energy in lactate for later use during exercise
D. Reduce the buildup of carbon dioxide in the cell
Correct Answer: B
Rationale:
Under anaerobic conditions, the electron transport chain stops because oxygen is unavailable as the
final electron acceptor. Fermentation converts pyruvate to lactate while oxidizing NADH back to NAD+,
which is essential for glycolysis to continue. Option B is incorrect because anaerobic respiration
produces far less ATP than aerobic respiration (2 vs. 36-38 ATP per glucose).
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 3 of 53
, Q3 Question 3 of 100
A geneticist crosses two pea plants heterozygous for seed color (Yy x Yy), where yellow
(Y) is dominant over green (y). The expected phenotypic ratio in the offspring is:
A. 1 yellow : 1 green
B. 3 yellow : 1 green
C. 1 YY : 2 Yy : 1 yy
D. All yellow seeds
Correct Answer: B
Rationale:
A monohybrid cross between two heterozygotes (Yy x Yy) produces offspring with the genotypic ratio 1
YY : 2 Yy : 1 yy. Since Y is dominant, both YY and Yy appear yellow, giving a phenotypic ratio of 3
yellow : 1 green. Option C lists the genotypic ratio, not the phenotypic ratio, and option B would result
from a test cross.
Q4 Question 4 of 100
A biologist studying DNA replication observes that new nucleotides are added only in the
5' to 3' direction. This unidirectional synthesis explains why the lagging strand is
synthesized as:
A. Short Okazaki fragments that are later joined by DNA ligase
B. A continuous strand matching the leading strand
C. A single long fragment starting from the origin of replication
D. A circular template that unwinds during replication
Correct Answer: B
Rationale:
DNA polymerase can only add nucleotides in the 5' to 3' direction. On the lagging strand, the template
runs 3' to 5', so synthesis occurs in short Okazaki fragments in the opposite direction of the replication
fork movement. DNA ligase then joins these fragments. Option B describes the leading strand, not the
lagging strand.
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 4 of 53
WGU C491 Integrated Natural Science Exam 2026
Biology, Chemistry, Earth Science & Physics -
Complete Study Guide
100+ Verified Exam Questions · Rated A+ · Guaranteed Results
✔ Biology (Cells, Genetics, Evolution) ✔ Earth Science (Geology, Meteorology, Astronomy)
✔ Chemistry (Atomic Structure, Bonding, Reactions)
✔ Physics (Motion, Forces, Energy)
Instant PDF Download · 100% Verified Answers · Score 90%+
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 1 of 53
, BEST SELLER · SCIENCE
WGU C491 Integrated Natural Science Exam 2026
Biology, Chemistry, Earth Science & Physics
2026/2027 - Q&A with Verified Answers
Every section. Every topic. 100% correct answers guaranteed. Used by
5,000+ science students to pass on their first attempt.
Biology (Cells, Genetics, Evolution)
Chemistry (Atomic Structure, Bonding, Reactions)
Earth Science (Geology, Meteorology, Astronomy)
Physics (Motion, Forces, Energy)
A+
Rated
Used by 5,000+ students · 100% Verified Answers · 2026/2027 Latest Update
Instant PDF Download
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 2 of 53
, SECTION 1 | Biology (Cells, Genetics, Evolution) | Q1-Q25 | WGU C491 Integrated Natural Science Exam 2026 Biology, Chemis
Q1 Question 1 of 100
A researcher examines a cell under a microscope and observes a large
membrane-bound organelle containing the genetic material. The cell is identified as
eukaryotic because:
A. The cell contains ribosomes for protein synthesis
B. The DNA is enclosed within a nucleus surrounded by a double membrane
C. The cell has a plasma membrane that regulates transport
D. The cell uses ATP as its primary energy currency
Correct Answer: B
Rationale:
Eukaryotic cells are distinguished by having a true nucleus enclosed by a nuclear envelope (double
membrane), which separates genetic material from the cytoplasm. Prokaryotic cells also have
ribosomes, plasma membranes, and use ATP, so options B, C, and D are features shared by all cell
types rather than unique to eukaryotes.
Q2 Question 2 of 100
During cellular respiration, a muscle cell in a sprinter's leg produces pyruvate through
glycolysis. Under anaerobic conditions, pyruvate is converted to lactate rather than
entering the Krebs cycle. This conversion allows the cell to:
A. Produce more ATP than aerobic respiration
B. Regenerate NAD+ so glycolysis can continue producing ATP
C. Store energy in lactate for later use during exercise
D. Reduce the buildup of carbon dioxide in the cell
Correct Answer: B
Rationale:
Under anaerobic conditions, the electron transport chain stops because oxygen is unavailable as the
final electron acceptor. Fermentation converts pyruvate to lactate while oxidizing NADH back to NAD+,
which is essential for glycolysis to continue. Option B is incorrect because anaerobic respiration
produces far less ATP than aerobic respiration (2 vs. 36-38 ATP per glucose).
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 3 of 53
, Q3 Question 3 of 100
A geneticist crosses two pea plants heterozygous for seed color (Yy x Yy), where yellow
(Y) is dominant over green (y). The expected phenotypic ratio in the offspring is:
A. 1 yellow : 1 green
B. 3 yellow : 1 green
C. 1 YY : 2 Yy : 1 yy
D. All yellow seeds
Correct Answer: B
Rationale:
A monohybrid cross between two heterozygotes (Yy x Yy) produces offspring with the genotypic ratio 1
YY : 2 Yy : 1 yy. Since Y is dominant, both YY and Yy appear yellow, giving a phenotypic ratio of 3
yellow : 1 green. Option C lists the genotypic ratio, not the phenotypic ratio, and option B would result
from a test cross.
Q4 Question 4 of 100
A biologist studying DNA replication observes that new nucleotides are added only in the
5' to 3' direction. This unidirectional synthesis explains why the lagging strand is
synthesized as:
A. Short Okazaki fragments that are later joined by DNA ligase
B. A continuous strand matching the leading strand
C. A single long fragment starting from the origin of replication
D. A circular template that unwinds during replication
Correct Answer: B
Rationale:
DNA polymerase can only add nucleotides in the 5' to 3' direction. On the lagging strand, the template
runs 3' to 5', so synthesis occurs in short Okazaki fragments in the opposite direction of the replication
fork movement. DNA ligase then joins these fragments. Option B describes the leading strand, not the
lagging strand.
WGU C491 Integrated Natural Science Exam 2026 Biology, Chemistry, Earth Science & Physics - 2026/2027 | Passing Score: 80% | Page 4 of 53
