Biochemistry Exam 2 Chicago State University School of Pharmacy.
Access The Exact Questions for Biochemistry Exam 2 Chicago State University School of Pharmacy.
💯 100% Pass Rate guaranteed
🗓️ Unlock for 1 Month
Rated 4.8/5 from over 1000+ reviews
- Unlimited Exact Practice Test Questions
- Trusted By 200 Million Students and Professors
What’s Included:
- Unlock Actual Exam Questions and Answers for Biochemistry Exam 2 Chicago State University School of Pharmacy. on monthly basis
- Well-structured questions covering all topics, accompanied by organized images.
- Learn from mistakes with detailed answer explanations.
- Easy To understand explanations for all students.
Free Biochemistry Exam 2 Chicago State University School of Pharmacy. Questions
Match each Phospholipid to its correct primary function.
A) Phosphatidylinositol
B) Phosphatidylethanolamine
C) Phosphatidylcholine
D) Phosphatidylserine
-
Acts as an "eat me" signal during apoptosis by flipping to the outer leaflet of the plasma membrane.
-
Serves as a precursor for second messengers IP3 and DAG in intracellular signaling.
-
Major component of pulmonary surfactant and the outer leaflet of the plasma membrane.
-
Supports membrane curvature and serves as a biosynthetic precursor to phosphatidylcholine.
Explanation
Correct Answers: A-2, B-4, C-3, D-1
A) Phosphatidylinositol → 2 — acts as the precursor for IP3 and DAG, two critical second messengers that regulate intracellular calcium release and protein kinase C activation respectively.
B) Phosphatidylethanolamine → 4 — supports membrane curvature essential for vesicle formation and autophagy, and is converted to phosphatidylcholine through three sequential methylation reactions catalyzed by phosphatidylethanolamine N-methyltransferase (PEMT).
C) Phosphatidylcholine → 3 — the most abundant phospholipid in mammalian membranes, concentrated in the outer leaflet, and a critical component of lung surfactant that reduces surface tension and prevents alveolar collapse.
D) Phosphatidylserine → 1 — normally restricted to the inner membrane leaflet, but translocates to the outer leaflet during programmed cell death, serving as a recognition signal that triggers phagocytic clearance by macrophages.
Among two different types of bonds shown in the figure below, what type of bond is present in sucrose (Table sugar)?
-
α-1,2-bond
-
β-1,4-bond
Explanation
Correct Answer: A) α-1,2-bond
Sucrose (table sugar) is a disaccharide composed of glucose and fructose linked together by an α-1,2-glycosidic bond. This bond forms between the anomeric carbon (C1) of α-glucose and the anomeric carbon (C2) of β-fructose. This linkage is unique because it involves both anomeric carbons of the two monosaccharides, making sucrose a non-reducing sugar — it has no free anomeric carbon available to act as a reducing agent.
Which of the following fatty acids is antiinflammatory in nature?
-
Structure 1
-
Structure 2
-
Structure 3
-
Structure 4
Explanation
Correct Answers: A and D
A) Structure 1 — Structure 1 depicts a monounsaturated omega-9 fatty acid (oleic acid), which has demonstrated anti-inflammatory properties by reducing pro-inflammatory cytokine production and is a major component of olive oil, well known for its anti-inflammatory benefits.
D) Structure 4 — Structure 4 depicts a polyunsaturated omega-3 fatty acid (such as alpha-linolenic acid or EPA/DHA), which are the most well-established antiinflammatory fatty acids. Omega-3 fatty acids inhibit the arachidonic acid pathway, reducing the production of pro-inflammatory eicosanoids such as prostaglandins and leukotrienes, and promote the synthesis of anti-inflammatory resolvins and protectins.
The kinase enzyme which is involved in the conversion of glucose to glucose-6-phosphate also requires:
-
Mg2+
-
Zn2+
-
Ca2+
-
Mn2+
Explanation
Correct Answer: A) Mg2+
The enzyme hexokinase catalyzes the first step of glycolysis — the phosphorylation of glucose to glucose-6-phosphate using ATP. This reaction requires Mg2+ (magnesium ions) as an essential cofactor. Mg2+ forms a complex with ATP (Mg-ATP), stabilizing the negative charges on the phosphate groups and making the terminal phosphate more electrophilic and accessible for transfer to glucose. Mg2+ is a critical cofactor for virtually all kinase enzymes that utilize ATP as a phosphate donor.
Which of the following is very stable?
-
Cis Fatty Acids
-
Trans Fatty Acids
Explanation
Correct Answer: B) Trans Fatty Acids.
Trans fatty acids are significantly more stable than cis fatty acids due to their molecular geometry. In trans fatty acids, the hydrogen atoms are positioned on opposite sides of the double bond, creating a straighter, more linear chain that allows for tighter packing and stronger van der Waals interactions between molecules. This results in higher melting points and greater thermodynamic stability. Cis fatty acids have hydrogen atoms on the same side of the double bond, creating a kinked structure that prevents tight packing, reduces stability, and lowers the melting point.
Nucleotides in DNA are linked together by:
-
Phosphodiester bonds
-
Peptide bonds
-
Glycosidic bonds
-
Hydrogen bonds
Explanation
Correct Answer: A) Phosphodiester bonds.
Nucleotides in DNA are covalently linked by phosphodiester bonds, which form between the 3'-hydroxyl group of one nucleotide's sugar and the 5'-phosphate group of the next nucleotide, creating the sugar-phosphate backbone of the DNA strand. Peptide bonds link amino acids in proteins, glycosidic bonds link monosaccharides in carbohydrates, and hydrogen bonds hold the two complementary DNA strands together between base pairs — but they do not link nucleotides within the same strand.
Vitamin responsible for decreased vision in darkness.
-
Vitamin D
-
Vitamin A
-
Vitamin K
-
Vitamin B12
Explanation
Correct Answer: B) Vitamin A.
Vitamin A (retinol) is essential for vision, particularly in low-light and dark conditions. It is converted to retinal, which combines with the protein opsin to form rhodopsin — the light-sensitive pigment found in rod cells of the retina. Rod cells are responsible for scotopic (dim light/night) vision. Vitamin A deficiency leads to night blindness (nyctalopia) — the inability to adapt to darkness — and if severe, can progress to complete blindness. Vitamins D, K, and B12 have no direct role in visual phototransduction.
Cellulose differs from starch primarily because cellulose contains:
-
α-1,4 glycosidic bonds
-
α-1,6 glycosidic bonds
-
Peptide bonds
-
β-1,4 glycosidic bonds
Explanation
Correct Answer: D) β-1,4 glycosidic bonds.
Cellulose is a structural polysaccharide composed of glucose units linked by β-1,4 glycosidic bonds. This β-linkage causes the glucose units to alternate orientation, creating long, straight, rigid chains that form strong hydrogen bonds with neighboring chains, resulting in the tough fibrous structure of plant cell walls. Starch, in contrast, uses α-1,4 glycosidic bonds (with α-1,6 at branch points in amylopectin), which create a helical, coiled structure that is digestible by human amylase enzymes. Human enzymes cannot break β-1,4 bonds, which is why cellulose is indigestible dietary fiber.
Which of the following statements regarding pyruvate dehydrogenase phosphatase (PDP) is true?
-
PDP activates Pyruvate Kinase which activates TCA cycles producing more energy.
-
PDP activates α-ketoglutarate dehydrogenase complex which activates TCA cycles producing more energy.
-
PDP inactivates Pyruvate Kinase which activates TCA cycles producing more energy.
-
PDP activates Pyruvate dehydrogenase complex which activates TCA cycles producing more energy.
Explanation
Correct Answer: D) PDP activates Pyruvate dehydrogenase complex which activates TCA cycles producing more energy.
PDP dephosphorylates (removes a phosphate group from) the Pyruvate Dehydrogenase Complex (PDC), which activates it. The active PDC then converts pyruvate to Acetyl-CoA, which enters the TCA cycle, ultimately producing more energy in the form of ATP. PDP does not act on Pyruvate Kinase or α-ketoglutarate dehydrogenase complex — those are separate enzymes with distinct regulatory mechanisms.
Which of the following is responsible for donating electrons and protons to oxygen to form water in the ETC complex?
-
Complex I
-
Complex II
-
Complex III
-
Complex IV
Explanation
Correct Answer: D) Complex IV.
Complex IV (cytochrome c oxidase) is the final electron acceptor in the electron transport chain and is solely responsible for the reduction of molecular oxygen (O₂) to water (H₂O). It accepts electrons from cytochrome c and combines them with protons (H⁺) from the matrix and molecular oxygen to form water. This is the terminal reaction of the ETC: 4e⁻ + 4H⁺ + O₂ → 2H₂O. Complexes I, II, and III transfer electrons but do not directly reduce oxygen to water.
How to Order
Select Your Exam
Click on your desired exam to open its dedicated page with resources like practice questions, flashcards, and study guides.Choose what to focus on, Your selected exam is saved for quick access Once you log in.
Subscribe
Hit the Subscribe button on the platform. With your subscription, you will enjoy unlimited access to all practice questions and resources for a full 1-month period. After the month has elapsed, you can choose to resubscribe to continue benefiting from our comprehensive exam preparation tools and resources.
Pay and unlock the practice Questions
Once your payment is processed, you’ll immediately unlock access to all practice questions tailored to your selected exam for 1 month .