BIOD 171
OBJECTIVE ASSESSMENT - EXAM
Microbiology BIOD 171
Module 2 Exam - Portage Learning
2026/2027 Actual Exam
Portage Learning - Online College Credit Course
50 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
* Microbial Metabolism & Energy * Microbial Classification & Diversity
* Microbial Genetics & Gene Transfer * Bacterial Pathogenesis & Virulence
* Microbial Growth & Control Methods * Applied Clinical Microbiology
COVER PAGE - 1
, SECTION 1 | Microbial Metabolism | Q1-Q10 | BIOD 171 2026/2027
Q1 Question 1 of 50
A 34-year-old laboratory technician is reviewing bacterial metabolism pathways. She notes that
during glycolysis, a specific phosphorylated 6-carbon sugar is split into two 3-carbon
molecules. Which intermediate is directly formed when fructose-1,6-bisphosphate is cleaved by
aldolase?
A. Glyceraldehyde-3-phosphate and dihydroxyacetone phosphate
B. Two molecules of pyruvate
C. Two molecules of glucose-6-phosphate
D. Phosphoenolpyruvate and oxaloacetate
Correct Answer: A
Rationale:
Aldolase cleaves fructose-1,6-bisphosphate into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. These two
3-carbon intermediates are isomers and both continue through glycolysis. Pyruvate is the end product of glycolysis, not the
immediate cleavage product of aldolase.
Q2 Question 2 of 50
A microbiology student observes that certain bacteria can grow in anaerobic environments by
regenerating NAD+ from NADH without using oxygen as the final electron acceptor. Which
metabolic process allows this regeneration while also producing a reduced organic compound
as a byproduct?
A. Aerobic respiration
B. Photosynthesis
C. The Calvin cycle
D. Fermentation
Correct Answer: D
Rationale:
Fermentation regenerates NAD+ from NADH by transferring electrons to an organic molecule, producing reduced organic
byproducts like ethanol or lactic acid. Aerobic respiration uses oxygen as the final electron acceptor, not organic molecules.
Photosynthesis and the Calvin cycle are anabolic processes unrelated to NAD+ regeneration.
BIOD 171 - 2026/2027 | Passing Score: 80% | Page 2 of 27
, SECTION 1 | Microbial Metabolism | Q1-Q10 | BIOD 171 2026/2027
Q3 Question 3 of 50
During a clinical rotation, a student learns that Mycobacterium tuberculosis survives within
macrophages by using the glyoxylate cycle. What is the primary advantage of this modified
citric acid cycle for bacteria living on two-carbon compounds?
A. It produces more ATP per glucose than the standard citric acid cycle
B. It eliminates the need for oxygen as a terminal electron acceptor
C. It generates NADPH for biosynthetic reactions exclusively
D. It allows net synthesis of carbohydrates from acetyl-CoA
Correct Answer: D
Rationale:
The glyoxylate cycle bypasses the two decarboxylation steps of the citric acid cycle, allowing net conversion of acetyl-CoA
into oxaloacetate and subsequently into carbohydrates. It does not produce more ATP; rather, it conserves carbon for
biosynthesis. Oxygen requirements are unaffected by this pathway.
Q4 Question 4 of 50
A researcher is studying the electron transport chain of Escherichia coli grown under aerobic
conditions. She identifies a mobile electron carrier that transfers electrons from Complex I or
Complex II to Complex III. Which molecule serves this function in bacterial electron transport?
A. NADH dehydrogenase
B. Cytochrome c oxidase
C. Ubiquinone (coenzyme Q)
D. Ferredoxin
Correct Answer: C
Rationale:
Ubiquinone (coenzyme Q) is a lipid-soluble mobile electron carrier that shuttles electrons between Complex I/II and
Complex III in the electron transport chain. NADH dehydrogenase is Complex I itself, not a mobile carrier. Cytochrome c
oxidase is Complex IV. Ferredoxin participates in other redox reactions but not typically as the mobile carrier between these
complexes.
BIOD 171 - 2026/2027 | Passing Score: 80% | Page 3 of 27
OBJECTIVE ASSESSMENT - EXAM
Microbiology BIOD 171
Module 2 Exam - Portage Learning
2026/2027 Actual Exam
Portage Learning - Online College Credit Course
50 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
* Microbial Metabolism & Energy * Microbial Classification & Diversity
* Microbial Genetics & Gene Transfer * Bacterial Pathogenesis & Virulence
* Microbial Growth & Control Methods * Applied Clinical Microbiology
COVER PAGE - 1
, SECTION 1 | Microbial Metabolism | Q1-Q10 | BIOD 171 2026/2027
Q1 Question 1 of 50
A 34-year-old laboratory technician is reviewing bacterial metabolism pathways. She notes that
during glycolysis, a specific phosphorylated 6-carbon sugar is split into two 3-carbon
molecules. Which intermediate is directly formed when fructose-1,6-bisphosphate is cleaved by
aldolase?
A. Glyceraldehyde-3-phosphate and dihydroxyacetone phosphate
B. Two molecules of pyruvate
C. Two molecules of glucose-6-phosphate
D. Phosphoenolpyruvate and oxaloacetate
Correct Answer: A
Rationale:
Aldolase cleaves fructose-1,6-bisphosphate into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. These two
3-carbon intermediates are isomers and both continue through glycolysis. Pyruvate is the end product of glycolysis, not the
immediate cleavage product of aldolase.
Q2 Question 2 of 50
A microbiology student observes that certain bacteria can grow in anaerobic environments by
regenerating NAD+ from NADH without using oxygen as the final electron acceptor. Which
metabolic process allows this regeneration while also producing a reduced organic compound
as a byproduct?
A. Aerobic respiration
B. Photosynthesis
C. The Calvin cycle
D. Fermentation
Correct Answer: D
Rationale:
Fermentation regenerates NAD+ from NADH by transferring electrons to an organic molecule, producing reduced organic
byproducts like ethanol or lactic acid. Aerobic respiration uses oxygen as the final electron acceptor, not organic molecules.
Photosynthesis and the Calvin cycle are anabolic processes unrelated to NAD+ regeneration.
BIOD 171 - 2026/2027 | Passing Score: 80% | Page 2 of 27
, SECTION 1 | Microbial Metabolism | Q1-Q10 | BIOD 171 2026/2027
Q3 Question 3 of 50
During a clinical rotation, a student learns that Mycobacterium tuberculosis survives within
macrophages by using the glyoxylate cycle. What is the primary advantage of this modified
citric acid cycle for bacteria living on two-carbon compounds?
A. It produces more ATP per glucose than the standard citric acid cycle
B. It eliminates the need for oxygen as a terminal electron acceptor
C. It generates NADPH for biosynthetic reactions exclusively
D. It allows net synthesis of carbohydrates from acetyl-CoA
Correct Answer: D
Rationale:
The glyoxylate cycle bypasses the two decarboxylation steps of the citric acid cycle, allowing net conversion of acetyl-CoA
into oxaloacetate and subsequently into carbohydrates. It does not produce more ATP; rather, it conserves carbon for
biosynthesis. Oxygen requirements are unaffected by this pathway.
Q4 Question 4 of 50
A researcher is studying the electron transport chain of Escherichia coli grown under aerobic
conditions. She identifies a mobile electron carrier that transfers electrons from Complex I or
Complex II to Complex III. Which molecule serves this function in bacterial electron transport?
A. NADH dehydrogenase
B. Cytochrome c oxidase
C. Ubiquinone (coenzyme Q)
D. Ferredoxin
Correct Answer: C
Rationale:
Ubiquinone (coenzyme Q) is a lipid-soluble mobile electron carrier that shuttles electrons between Complex I/II and
Complex III in the electron transport chain. NADH dehydrogenase is Complex I itself, not a mobile carrier. Cytochrome c
oxidase is Complex IV. Ferredoxin participates in other redox reactions but not typically as the mobile carrier between these
complexes.
BIOD 171 - 2026/2027 | Passing Score: 80% | Page 3 of 27
